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MP Board Class 11th Maths Important Questions Chapter 9 Sequences and Series Important Questions

Sequences and Series Important Questions

Sequences and Series Objective Type Questions

(A) Choose the correct answer of the following:

Question 1.
The sum of the cube of first n positive integer is :
(a) \(\frac {n(n + 1)}{2}\)
(b) \(\frac {n(n + 1)(2n + 1)}{6}\)
(c) \(\frac {n(n + 1)(n + 2)}{6}\)
(d) {\(\frac {n(n + 1)}{2}\)}²
Answer:
(d) {\(\frac {n(n + 1)}{2}\)}²

Make use of this free Radius of Convergence Calculator to get the radius of convergence of a power series. Online calculator tool gives output in no time.

Question 2.
The sum of n term of arithmetic progression is 2n + 3n², its second term will be :
(a) 10
(b) 12
(c) 16
(d) 11
Answer:
(c) 16

Question 3.
If arithmetic mean of a and b is \(\frac { { a }^{ n }+{ n }^{ n } }{ { a }^{ n – 1 }+{ b }^{ n – 1 } }\), then the value of n will be :
(a) 1
(b) 0
(c) – 1
(d) \(\frac {1}{2}\)
Answer:
(a) 1

Question 4.
Which term of the series 8, 4,0, ………….. is – 24 :
(a) 7^th
(b) 28^th
(c) 8^th
(d) 9^th
Answer:
(d) 9^th

Question 5.
If the first term of arithmetic progression be a and last term is l, then the sum of n terms will be :
(a) \(\frac {n}{2}\)[2a – (n – 1)d]
(b) \(\frac {n}{2}\)[2a + (n – 1)d]
(c) \(\frac {n}{2}\)(a + l)
(d ) \(\frac {n}{2}\)(a – l).
Answer:
(c) \(\frac {n}{2}\)(a + l)

Question 6.
The next term of the sequence 2\(\sqrt { 2 }\), \(\sqrt { 2 }\), 0, …………… is :
(a) – \(\sqrt { 3 }\)
(b) \(\frac { 1 }{ \sqrt { 2 } }\)
(c) – \(\sqrt { 2 }\)
(d) \(\sqrt { 2 }\)
Answer:
(c) – \(\sqrt { 2 }\)

Question 7.
If 2x, x + 8, 3x + 1 are in A.P, then value of x will be :
(a) 3
(b) 7
(c) 5
(d) 2
Answer:
(c) 5

Question 8.
The 15^th term from the end of the A.P. 2, 6,10, …………., 86 is :
(a) 30
(b) 32
(c) 46
(d) 48
Answer:
(a) 30

Question 9.
If first 15^th term of an A.P. is a, second term is b and n^th term is 2 a, then sum of the n terms is :
(a) \(\frac {ab}{2(b – a)}\)
(b) \(\frac {2ab}{3(b – a)}\)
(c) \(\frac {3ab}{2(b – a)}\)
(d) \(\frac {3ab}{(b – a)}\)
Answer:
(c) \(\frac {3ab}{2(b – a)}\)

Question 10.
In an A.P. S_n = 3n² + 5n and T_m = 164, then m equals to :
(a) 26
(b) 27
(c) 28
(d) None of these.
Answer:
(b) 27

Question 11.
A.M. of two number is 10 and GM. is 8, the numbers are
(a) a = 4, b = 16
(b) a = 2, b = 8
(c) a = 4, b = 9
(d) a = 2, b = 18.
Answer:
(a) a = 4, b = 16

Question 12.
The A.M. of two numbers is A and G M is G, then relation between them is :
(a) A < G (b) A = G (c) A > G
(d) None of these.
Answer:
(c) A > G

Question 13.
2^1/4.4^1/8.8^1/16 ……………. ∞ =
(a) 1
(b) 2
(c) 3/2
(d) 4
Answer:
(b) 2

Question 14.
If y = x – x² + x³ – x⁴ + ………. ∞, then value of x be (- 1 < x < 1) :
Answer:
(a) y + \(\frac {1}{y}\)
(b) \(\frac {y}{1 + y}\)
(c) y – \(\frac {1}{y}\)
(d) \(\frac {y}{1 – y}\)
Answer:
(d) \(\frac {y}{1 – y}\)

Question 15.
If the second, third and sixth terms of an A.P. are in GP, then the common ratio of the GP. is :
(a) 2
(b) 5
(c) 4
(d) 3
Answer:
(d) 3

Question 16.
Sum up to infinity of 1 + \(\frac {4}{5}\) + \(\frac { 7 }{ { 5 }^{ 2 } }\) + \(\frac { 10 }{ { 5 }^{ 2 } }\) + …………. :
(a) \(\frac {35}{16}\)
(b) \(\frac {37}{16}\)
(c) \(\frac {39}{16}\)
(d) 3
Answer:
(a) \(\frac {35}{16}\)

(B) Match the following :

Answer:

1. (d)
2. (f)
3. (a)
4. (g)
5. (b)
6. (j)
7. (c)
8. (i)
9. (e)
10. (h)

(C) Fill in the blanks :

1. – 7\(\sqrt { 3 }\) will be term of the series 5\(\sqrt { 3 }\), 3\(\sqrt { 3 }\), \(\sqrt { 3 }\), …………….
2. 116 is sum of the term of the series 25,22, 19 …………….
3. If sum of n terms of series is n² + 4n, then 15^th term of the series is …………….
4. 0 will be the term of the series 27, 24,21, 18 …………….
5. 729 will be the term of the series – \(\frac {1}{27}\), \(\frac {1}{9}\), – \(\frac {1}{3}\) …………….
6. The first term of GP. is a, common ratio r < 1 and its last term l, then its sum will be …………….
7. The ratio of the sum of first three term to the sum of first six term is 125 : 152, then common ratio will be …………….
8. First term 16 and fifth term \(\frac {1}{16}\), then its 4^th term will be …………….
9. Sum of n terms of the series x + 2x² + 4x³ + 8x⁴ + ……………. will be …………….
10. If a = 2, d = 2 and n = 50, then the last term of the series will be …………….

Answer:

1. 7
2. 12
3. 33
4. 10
5. 10
6. \(\frac {a – rl}{1 – r}\)
7. \(\frac {3}{5}\)
8. \(\frac {1}{4}\)
9. \(\frac { 1-({ 2x }^{ n })x }{ 1 – 2x }\)
10. 100

(D) Write true / false :

1. 1, 3, 5, 8, ………….. are inA.P.
2. n^th term of a G.P. is a + (n – 1 )d.
3. If 2x, x + 5 and x + 11 are in A.P., then value of x will be – 1.
4. Arithmetic mean of a and b is \(\sqrt { ab }\)
5. Sum of 9 terms of a sequence 24 + 20 + 16 + will be.
6. Four consicutive term of G.P. are \(\frac { a }{ { r }^{ 3 } }\) \(\frac {a}{r}\), ar, ar³.

Answer:

1. False
2. False
3. True
4. False
5. True
6. True.

(E) Write answer in one word / sentence :

1. If a, b, care inA.P., then find the value of ab + ac?
2. Arithmetic mean of (a + b)²and (a – b)² will be?
3. Sum of n arithmetic mean between x and 3x will be.
4. Find the sum of infinite terms of series : 9^1/3.9^1/3.9^1/27 ………….. up to ∞.
5. If a, b, c are in GP., then find the value \(\frac {1}{b}\) + \(\frac {1}{a – b}\) –\(\frac {1}{b – c}\)

Answer:

1. 2b²
2. a² + b²
3. 3nx
4. 3
5. 0

Sequences and Series Short Answer Type Questions

Question 1.
Which term of the sequence 27,24,21,18, …………. is zero? (NCERT)
Solution:
Here a = 27 and d= T₂ – T₁ = 24 – 27 = – 3.
Let the n^th term of series be 0.
∴ T_n = a + (n – 1)d
⇒ 0 = 27 + (n – 1)(- 3)
⇒ 3n – 3 = 27
⇒ 3n = 30
n = 10 or 10^th term.

Question 2.
The last term of the series 8,4,0, ……………. is – 24. Find the total number of terms. (NCERT)
Solution:
Given series 8,4, 0, … (1)
First term = a = 8 Common difference d = 4 – 8 = -4
d = 0 – 4 = – 4
∵ The common difference is same the series is in A.P.
last term l = – 24,
l = a + (n – 1)d,
= – 24 = 8 + (n – 1)(-4)
⇒ – 24 – 8 = (n – 1)(-4)
⇒ – 32 = (n – 1)(- 4)
⇒ (n – 1) = \(\frac {32}{4}\)
⇒ n – 1 = 8
⇒ n = 8 + 1
⇒ n = 9
∴ Number of terms n = 9

Question 3.
Seven times the 7^th term of a series is equal to eleven times of its 11th term. Find the 18^th term of the series. (NCERT)
Solution:
Let first term = a and common difference = d of A.P.
∴ 7^th term = a + 6d and 11^th term = a + 10 d.
∴ According to question,
7(a + 6d) = 11 + (a + 10d)
⇒ 7a + 42d = 11a + 110d
⇒ 7a – 11a = 110d – 42d
⇒ – 4a = 68 d
⇒ a = – 17d
Hence 18^th term = a + 17d
= – 17d + 17d [∵ a = – 17d]
= 0.

Question 4.
Prove that the sum of (m + n)^th term and (m – n)^th terms of an A.P. is twice of its m^th term.
Solution:
Let the first term = a and common difference = d.
T_n = a+(n – 1)d
T_{m + n}= a + (m + n – 1)d … (1)
T_{m – n} = a + (m – n – 1)d … (2)
T_m = a + (m – 1)d … (3)
T_m+n + T_m-n = a + (m + n – 1)d + a(m – n – 1)d
= 2a + (m + n – 1 + m – n – 1)d
= 2a + (2m – 2)d
= 2a + 2(m – 1)d
= 2[a+(m – 1)d]
T_m+n + T_m-n = 2T_m.

Question 5.
Insert three Arithmetic means between 3 and 19. (NCERT)
Solution:
Let three Arithmetic mean be A₁, A₂, A₃,
then 3, A₁, A₂, A₃, 19 are in A.P.
∴ 3 = a, 19 = T₅, let common difference = d
T₅ = a + 4d
T₅ = a + 4d
⇒ 19 = 3 + 4d
⇒ 16 = 4d
⇒ d = 4
Hence A₁ = 3 + 4 = 7, A₂ = 7 + 4 = 11, A₃ = 11 + 4 = 15.

Question 6.
If – 8, A₁, A₂ are in Arithmetic progression (A.P.), then find the value of A₁, A₂. (NCERT)
Solution:
– 8, A₁, A₂, 9 are in A.P.
∴ a = – 8, T₄ = 9, let common difference = d
T₄ = a + 3d
⇒ 9 = – 8 + 3 d
⇒ 3d = 11
⇒ d = \(\frac {17}{3}\)

Instruction :
Write the first five terms of each of the sequence and obtain the corresponding series.

Question 7.
(a) a₁ = 3, a_n = 3a_{n – 1} + 2, where n > 1. (NCERT)
Solution:
Given : a₁ = 3,
a₂ = 3a_{n – 1} + 2, where n > 1
a₂ = 3a_{2 – 1} + 2
⇒ a₂ = 3a₁ + 2
⇒ a₂ = 3 x 3 + 2 = 9 + 2 = 11
a₃ = 3a₁ + 2
= 3a₂ + 2
⇒ a₃ = 3(11)+ 2 = 33+ 2 = 35
a₄ = 3a_{4 – 1} + 2
= 3a₃ + 2
⇒ a₄ = 3(35) + 2 = 105 + 2 = 107
a₅ = 3a_{5 – 1} + 2
= 3a₄ + 2
= 3(107) + 2 = 321 + 2 = 323
Hence series is 3, 11, 35, 107, 323.

Question 7.
(b) a₁ = -1, a_n = \(\frac { { a }_{ n-1 } }{ n }\) where n ≥ 2.
Solution:

Question 8.
A person pays first instalment of Rs. 100 towards his loan. If he increases his instalment every month by Rs. 5, then what will be his 30^th instalment.
Solution:
Given : a = Rs. 100, d= Rs. 5, n = 30,
T_n = a + (n – 1)d
Amount of 30^th instalment
J₃₀ = 100 + (30 – 1) x 5
= 100 + 29 x 5 = 100 + 145
= Rs. 245.
Hence the 30^th instalment = Rs. 245.

Question 9.
Which term of the sequence \(\sqrt { 3 }\), 3, 3\(\sqrt { 3 }\) ……. is 729?
Solution:

Question 10.
How many terms are required in GP. 3,3²,3³ …………, so that their sum would be 120? (NCERT)
Solution:
Given : a = 3, r = \(\frac {9}{3}\) = 3, S_n = 120,
S_n = \(\frac { a({ r }^{ n }-1) }{ r-1 }\)
⇒ \(\frac { 3({ 3 }^{ n }-1) }{ 3-1 }\) = 120
⇒ 3ⁿ – 1 = \(\frac {120 × 2}{3}\)
⇒ 3ⁿ – 1 = 80
⇒ 3ⁿ = 81
⇒ 3ⁿ = 3⁴
⇒ n = 4

Question 11.
Show that ratio between sum of n terms and sum of (n +1)^th term to (2n)^th terms in a G.P. is \(\frac { 1 }{ { r }^{ n } }\)
Solution:
Let 1^st term of G.P. = a and common ratio = r
∴ n terms of GP. is a, ar, ar², ………….. , ar^{n – 1}
Let the sum to n terms be S₁
S₁ = \(\frac { a({ r }^{ n }-1) }{ r-1 }\)
GP. from (n + 1)^th term to (2n)^th term
arⁿ, ar^{n + 1}, …………. , ar^{2n – 1}
Let the sum of this G.P. up to n terms be S₂

Question 12.
If A.M. and GM. of roots of a quadratic equation are 8 and 5 respectively, then find the quadratic equation. (NCERT)
Solution:
Let the roots of equation be α and β
A.M. of roots = \(\frac {α + β}{2}\) = 8
⇒ α +β = 16
G.M. of roots = \(\sqrt {αβ }\) = 5
⇒ αβ = 25
If α, β are the roots of equation
then, x² – (α + β)x + αβ = 0
⇒ x² -16x + 25 = 0.

Question 13.
If the first and n^th term of GP. are a and b respectively and if p is the product of n terms, then prove that
p² = (ab)ⁿ. (NCERT)
Solution:
Let the common ratio of G.P. = r
Given: ar^{n – 1} = b, …(1)
then a.ar.ar²……. ar^{n – 1} = p
⇒ aⁿr^{1+2+3+……+(n – 1)} = p
⇒ aⁿr^{\(\frac {1}{2}\)(n – 1)n} = p
⇒ a²ⁿ.r^{(n – 1)n} = p²
⇒ p² = (a²r^{n – 1n}
⇒ p² = (a.ar^{n – 1})ⁿ
⇒ p² = (ab)ⁿ, [fromeqn. (1)]

Question 14.
If the 4^th term of a GP. is square of its second term and the 1^st term is -3, then find the 7^th term. (NCERT)
Solution:
Let a and r be the 1st term and common ratio of GP.
∴ T_n = ar^n-1
T₄ = ar^4-1 = ar³ … (1)
T₂ = ar^2-1 = ar … (2)
Given : T₄ = (T₂)²
⇒ ar³ = (ar)²
⇒ ar³ = a²r²
⇒ a = r = – 3 (given, a = -3)
T₇ = ar^7-1 = ar⁶
= (-3)(-3)⁶ = (-3)⁷
∴ T₇ = – 2187
Hence 7_th term of G.P. = – 2187.

Question 15.
Find the value of \(\sum _{ k=1 }^{ 11 }{ (2+{ 3 }^{ k }) }\).
Solution:

Sequences and Series Long Answer Type Questions

Question 1.
If m times of the m^th term of an A.P. is equal to n times the n^th term, then prove that (m + n)^th term of the series is zero.
Solution:
Let the 1^st term = a and common difference = d.
then t_m = a + (m – 1)d
t_n = a + (n – 1)d
Given: m[a + (n – 1)d] = n[a + (n – 1)d]
⇒ ma + m(m – 1 )d = na + n(n – 1 )d
⇒ (m² – m)d + (m – n)a = (n² – n)d
⇒ (m² – m – n² + n) d + (m – n)a = 0
⇒ [m² – n² – (m – n)] d + (m – n)a = 0
⇒ [(m – n)(m + n) – (m – n)] d+(m – n)a = 0
⇒ (m – n)[(m + n) – 1] d + (m – n)a = 0
⇒ a + (m + n – 1 )d = 0
⇒ t_m+n=0.

Question 2.
If the 6^th term of A.P. is 12 and 9^th term is 27, then find its r^th term.
Solution:
Let the 1^st term = a and common difference = d.
Given : t₆ = 12
⇒ a + 5d = 12 … (1)
t₉ = 27
⇒ a + 8d = 27 … (2)

Question 3.
If p^th term of an A.P. is \(\frac {1}{q}\) and q^th term is \(\frac {1}{p}\), then prove that (pq)^th
Solution:

Question 4.
If p^th term of an A.P. is \(\frac {1}{q}\) and q^th term is \(\frac {1}{p}\), then prove that (pq)^th terms is \(\frac {1}{2}\) (pq + 1), where p ≠ q.
Solution:

Question 5.
The sum of the series 25, 22, 19, ………….. of A.P. is 116, then find its last term. (NCERT)
Solution:
Given : a = 25, d = 22 – 25 = – 3, S_n = 116
S_n = \(\frac {n}{2}\) [2a + (n – 1)d]
⇒ 116 = \(\frac {n}{2}\)[2 x 25 + (n -1) x (-3)]
⇒ 232 = n[50 – 3n + 3]
⇒ 232 = n[53 – 3n]
⇒ 232 = 53n – 3n²
⇒ 3n² – 53n +232 = 0
⇒ 3n² – 24n – 29n + 232 = 0
⇒ 3n(n – 8) – 29(n – 8) = 0
⇒ (n – 8)(3n – 29) = 0
n = 8, or n = \(\frac {29}{3}\), (which is impossible)
∴ Last term l = a + (n – 1 )d
= 25 + (8 – 1) x (- 3)
⇒ l = 25 – 21 = 4.

Question 6.
If the A.M. of a and b is \(\frac { { a }^{ n }+{ b }^{ n } }{ { a }^{ n – 1 }+{ b }^{ n – 1 } }\), then find the value of it. (NCERT)
Solution:
A.M. of a and b = \(\frac {a+b}{2}\)
According to question,
\(\frac { { a }^{ n }+{ b }^{ n } }{ { a }^{ n – 1 }+{ b }^{ n – 1 } }\) = \(\frac {a+b}{2}\)
⇒ 2aⁿ + 2bⁿ = (a + b) (a^{n – 1} + b^{n – 1})
⇒ 2aⁿ +2bⁿ = aⁿ + bⁿ + ab^{n – 1} + ba^{n – 1}
⇒ aⁿ + bⁿ = ab^{n – 1} + ba^{n – 1}
⇒ aⁿ – ba^{n – 1} = ab^{n – 1} – bⁿ
⇒ a^{n – 1}(a – b) = b^{n – 1}(a – b)
⇒ a^{n – 1} = b^{n – 1}
⇒ \(\frac {a}{b}\)^{n – 1} = \(\frac {a}{b}\)⁰
⇒ n – 1 = 0
⇒ n = 1.

Question 7.
Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7^th and (m – 1)^th mean is 5 : 9 then, find the value of m.
Solution:
Let A₁, A₂, A₃, ……… ,A_m are m A.M. respectively inserted between 1 and 31.
Then 1, A₁, A₂, A₃, ……… A_m 31 are in A.P. Here, 1^st term = 1 and (m + 2)^th term = 31 and let the common difference of sequence be d.

Question 8.
Show that the sum of (m + n)^th and (m – n)^th term of an A.P. is equal to twice the m^th term. (NCERT)
Solution:
Let the first term = a and common difference = d.
T_n = a + (n – 1)d
T_{m + n} =a +(m + n – 1)d … (1)
T_{m – n} = a + (m – n – 1)d … (2)
T_m = a + (m – 1)d … (3)
Adding equation (1) and (2),
T_{m + n} + T_{m – n} = a + (m + n – 1)d + a + (m – n – 1)d
= 2a + (m + n – 1 + m – n – 1)d
= 2a + (2m – 2 )d
= 2a + 2 (m – 1)d
= 2[a + (m – 1)d]
∴ T_{m + n} + T_{m – n} = 2T_m [From equation (3)]

Question 9.
If the sum of three numbers in A.P. is 24 and their product is 440, then find the numbers? (NCERT)
Solution:
Let the three numbers of A.P. are a – d, a, a + d.
Given : a – d + a + a + d = 24
3a = 24
⇒ a = 8
and (a – d) × a × (a + d) = 440
a(a² – d²) = 440
⇒ 8(64 – d²) = 440
⇒ 64 – d² = \(\frac {440}{8}\)
⇒ 64 – d² = 55
⇒ d² = 64 – 55
⇒ d² = 9
⇒ d = ± 3
When a = 8 and d = 3
Then, a – d = 8 – 3 = 5, a = 8, a + d= 8 + 3 = 11
When a = 8 and d = – 3
Then, a – d = 8 + 3 = 11, a = 8, a + d = 8 – 3 = 5
Hence required numbers are 5, 8, 11 or 11, 8, 5.

Question 10.
If the sum of n, 2n and 3n terms in A.P. are S₁ S₂ and S₃, then show that S₃ = 3(S₂ – S₁).
Solution:

Question 11.
Find the sum of all natural numbers lying between 200 and 400, which are divisible by 7. (NCERT)
Solution:
Numbers divisible by 7 are 203, 210, 217, …………. , 399
Here a = 203, d = 210 – 203 = 7, l = 399
l = a + (n – 1)d
399 = 203 + (n – 1) x 7
⇒ 399 – 203 = (n – 1) x 7
⇒ (n – 1)7 = 196
⇒ (n – 1) = \(\frac {196}{7}\)
⇒ (n – 1) = 28
∴ n = 29.
Hence required sum, S₂₉ = \(\frac {n}{2}\)[a+l]
⇒ S₂₉ = \(\frac {29}{2}\)[203 + 399]
⇒ S₂₉ = \(\frac {29}{2}\)[602] = \(\frac {17458}{2}\) = 8729.

Question 12.
If the 5^th, 8^th and 11^th terms of a GP. are p, q and s respectively, then show
that q² = ps. (NCERT)
Solution:
Let the 1st term of GP. = a and common ratio = r.
Then, T_n = ar^{n – 1}
T₅ = ar^{5 – 1} =p
⇒ ar⁴ = P … (1)
T₈ = ar^8-1 = q
⇒ ar⁷ = q … (2)
T₁₁ = ar^{11 – 1} = s
⇒ ar¹⁰ = s
ps = ar⁴.ar¹⁰, [from equation (1) And (3)]
⇒ ps = a²r¹⁴
⇒ ps = (ar⁷)²
⇒ ps = q², [from equation (2)]
∴ q² = ps.

Question 13.
If the first term of G.P. a = 729 and 7^th term is 64, then find S₇. (NCERT)
Solution:

Question 14.
If the 4^th terms of a GP is square of its 2^nd term and first term is – 3, then find its 7^th (NCERT)
Solution:
Let the 1st term of GP. = a and common ratio = r.
Then, T_n = ar^{n – 1}
T₄ = ar^{4 – 1} = ar³ … (1)
T₂ = ar^2-1 = ar … (2)
Given : T₄ = T₂²
ar³ = (ar)₂
ar³ = a²r²
⇒ ar³ = a²r²
⇒ a = r = – 3 (given a = – 3)
T₇ = ar^{7 – 1} = ar⁶
= (-3)(-3)⁶ = (-3)⁷
∴ T₇ = – 2187
Hence, 7^thterm of G. P. = – 2187.

Question 15.
Find the sum of the sequence 8,88,888,8888 ……….. up to n terms. (NCERT)
Solution:
Let the sum of the n terms be S

Question 16.
Find the sunt of the numbers 7,77,777, 7777 up to n terms. (NCERT)
Solution:
Let the sum of n terms is S.

Question 17.
If the A.M. and GM. between two positive numbers a and b are 10 and 8 respectively, then find the numbers. (NCERT)
Solution:
A.M. A = \(\frac {a+b}{2}\)
a + b = 20 … (1)
G.M. G = \(\sqrt {ab}\)
ab = 64 … (2)
(a – b)² = (a + b)² – 4ab
= (20)² – 4 × 64, [From equation (2)]
= 400 – 256
(a – b)² = 144 = (12)² … (3)

Put a = 16 in equation (1), we get,
16 + b = 20
∴ b = 4
Numbers are 16 and 4.
When a – b = – 12, then
a + b = 20
a – b = – 12
On adding 2a = 8
⇒ a = 4
Put a = 4 in equation (1), we get,
4 + 6 = 20
∴ b = 16
Numbers are 4 and 16.
Hence numbers a and b are 4, 16 or 16,4.

Question 18.
The sum of two numbers is 6 times their geometric mean, show that the . numbers are in the ratio (3 + 2\(\sqrt {2}\)) ; (3 – 2\(\sqrt {2}\)). (NCERT)
Solution:
Let the numbers are a and 6.
Given: a + b = 6 \(\sqrt {ab}\)
\(\frac { a+b }{ 2\sqrt { ab } }\) = \(\frac {3}{1}\)
⇒ a + b = 3K …. (1)
and 2 \(\sqrt {ab}\) = K ⇒ 4ab = K²
(a – b)² = (a + b)² – 4ab
= (3K)² – (K)²
= 9K² – K² = 8K²
⇒ a – b = 2\(\sqrt {2}\)K …. (2)

Sequences and Series Very Long Answer Type Questions

Question 1.
If the ratio of the sum of n terms of two A.P. is 5n + 4 : 9n + 6, then find the ratio of their 18^th term.
Solution:
Let the two A.P. are :
a, a + d, a + 2d, ………….
and A, A + D, A +2D …………..

Question 2.
The ratio of the sum of m and n terms of an A.P. is m² : n². Show that the ratio of and u* term is (2m – 1) : (2n – 1). (NCERT)
Solution:
Let the A.P. are a, a + d, a + 2d, ……………
∴ m^th term of A.P. Tm = a + (m – 1)d
n^th term of A.P. Tn= a + (n – 1)d

Question 3.
If the sum of first three terms of a G.P. is \(\frac {39}{10}\) and their product is 1, then find the common ratio and the terms. (NCERT)
Solution:
Let the three terms of G.P. are \(\frac {a}{r}\), a, ar.
Given: \(\frac {a}{r}\) × a × ar = 1
⇒ a³ = 1 ⇒ a = 1
and \(\frac {a}{r}\) + a + ar = \(\frac {39}{10}\)
⇒ a(\(\frac {a}{r}\) + r + 1) = \(\frac {39}{10}\)
⇒ 1 x \(\frac { 1+{ r }^{ 2 }+r }{ r }\) = \(\frac {39}{10}\)
⇒ 10r² + 10r + 10 = 39r
⇒ 10r² – 29r + 10 = 0
⇒ 10r² – 25r – 4r + 10 = 0
⇒ 5r(2r – 5) – 2(2r – 5) = 0
⇒ (5r – 2)(2r – 5) = 0
∴ r = \(\frac {2}{5}\) and r = \(\frac {5}{2}\)
When a = 1 and r = \(\frac {2}{5}\)

Question 4.
Find four numbers forming a G.P. in which the third term is greater than the lint term by 9 and the second term is greater than 4^th term by 18. (NCERT)
Solution:
Let the four terms of G.P. be a, ar, ar²,ar³.
Given: T₃ = T₁ + 9
⇒ T₃ = T₁ + 9
⇒ T₃ = a + 9
⇒ ar² = a + 9 …. (1)
According to question,
T₂ = T₄ + 18
⇒ ar = a³ + 18
⇒ ar – ar³ = 18 …. (2)

Put r = – 2 in equation (1), we get
a(- 2)² – a = 9
⇒ 4a – a = 9
⇒ 3a = 9
⇒ a = 3
∴ Numbers are 3, 3(- 2), 3(- 2)², 3(-2)³, ……………..
3, – 6, 12, – 24, ………………

Question 5.
If S be the sum of it terms of a GP., P be the product and R be the sum of reciprocal of n terms, then prove that
P²Rⁿ = Sⁿ. (NCERT)
Solution:
Let n terms of GP. be a, ar, ar², …………… ar^{n – 1}.
According to the question,

Question 6.
If x = 1 + a + a² + ………….∞ (\(\left| a \right|\)<1)
y = 1 + b + b² + …………….∞ (\(\left| b \right|\)<1)
then prove that
1 + ab + a²b² + …………….∞ = \(\frac {xy}{x + y – 1}\)
Solution:

Question 7.
The sum of infinite terms of a Geometric Progression is 15 and sum of the square of its terms is 45. Find the Geometric Progression. (NCERT)
Solution:
Let a be the first term and r be the common ratio.
∵ \(\left| r \right|\)<1,
Then, \(\frac {a}{1 – r}\) = 1.5 …. (1)
Squaring the terms of GP. the new G.P. is
a², a² r², a² r⁴, a² r⁶, ……….
Sum of infinity of G.P. = \(\frac { { a }^{ 2 } }{ { 1 – r }^{ 2 } }\).
\(\frac { { a }^{ 2 } }{ { 1 – r }^{ 2 } }\) = 45, (given) … (2)
Squaring both sides of equation (1) and the result is divided by equation (2),
\(\frac { { a }^{ 2 } }{ { 1 – r }^{ 2 } }\)² × \(\frac { 1-{ r }^{ 2 } }{ { a }^{ 2 } }\) = \(\frac {15×15}{45}\)
⇒ \(\frac {1 + r}{1 – r}\) = 5
⇒ 1 + r = 5 – 5r
⇒ 6r = 4
⇒ r = \(\frac {2}{3}\)
Put r =\(\frac {2}{3}\) in equation (1), we get

Hence required progression is 5, 5 × \(\frac {2}{3}\), 5 × (\(\frac {2}{3}\))², …………….
Hence 5, \(\frac {10}{3}\), \(\frac {20}{9}\), ………………

Question 8.
A farmer buys a used tractor of Rs. 12,000. He pays Rs. 6,000 cash and agrees to pay the balance in annual instalment of Rs. 500 plus 12% interest on theunpaid amount How much the tractor cost him?
Solution:
Cost of tractor = Rs. 12, 000, down payment = Rs. 6,000
Balance amount = 12,000 – 6,000 = Rs. 6,000

Actual cost of tractor = 12,000 + 4,680 = Rs. 16,680. Ans.

Question 9.
Shamshad Ali buys a scooter for Rs. 22,000. He pays Rs. 4,000 cash and agrees to pay the balance in annual instalment of Rs. 1,000 plus 10% interest on the unpaid amount How much will scooter cost him? (NCERT)
Solution:
Cost of scooter = Rs. 22,000, Cash down payment = Rs. 4,000.
Remaining amount = 22,000 – 4,000 = Rs. 18,000

Actual cost = 22,000 + 17,100 = Rs. 39,100.
Total amount paid for scooter = Rs. 39,100.

Question 10.
A person writes a letter to four of his friends. He asks each one of them, to copy the letter and mail to four different persons with instructions that they move the chain similarly. Assuming that the chain is not broken and that is costs 50 paisa to mail one letter. Find the amount on postage when 8^th set of letter is mailed. (NCERT)
Solution:
First person sends 4 letters.
2^nd step, he sends 4 x 4= 16 letters
3^rd step, he sends 4 x 4 x 4 = 64 letters
Hence 4, 16, 64, 256, …………… is a Geometric series.
Here a = 4, r = \(\frac {16}{4}\) = 4
S_n = \(\frac { { a(r }^{ n } – 1) }{ r – 1 }\) [∵r>1]
∴ Total number of letters till 8^th set = S₈ = \(\frac { { 4(r }^{ 4 } – 1) }{ 4 – 1 }\)
= \(\frac {4}{3}\) (65536 – 1) = \(\frac {4}{3}\) x 65535
Cost for one letter = Rs. 0.50
∴ Hence total cost = \(\frac {4}{3}\) x 65535 x 0.50 = Rs. 43690.

Question 11.
150 workers were engaged to finish a job in a certain number of days, 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed. (NCERT)
Solution:
150, 146, 142, 138, …………, it is a Geometric series.
Let the number of days required complete the work be n.

If the workers not dropped then the work would have complited in (n – 8) days with 150 workers working on each day.
Hence the total workers worked for n days = 150 (n – 8)
= 150n – 1200 ….. (2)
From equation (1) and (2),
150n – 1200 = 152n – 2n₂
2n₂ + 150n – 152n – 1200 = 0
2n₂ – 2n – 1200 = 0
n₂ – n – 600 =0
n₂ – 25n + 24n – 600 = 0
n(n – 25) + 24(n – 25) = 0
(n – 25)(n + 24) = 0
n =25 and n = – 24 (not possible)
∴ Work is completed in 25 days.

MP Board Class 11th Maths Important Questions

MP Board Class 10th Science Solutions Chapter 16 Management of Natural Resources

MP Board Class 10th Science Chapter 16 Intext Questions

Class 10th Science Chapter 16 Intext Questions Page No. 269

Question 1.
What changes can you make in your habits to become more environment friendly?
Answer:

1.  We must refuse to buy products that harm us and the environment.
2.  We should minimise the use of electricity and water.
3.  We should encourage recycling of things.

Question 2.
What would be the advantages of exploiting resources with short-term aims?
Answer:
With the human population increasing at a tremendous rate due to improvement in health-care, thedem and for all resources is increasing at an exponential rate. The management of natural resources requires a long term perspective so that these will last for the generations to come and will not merely be exploited to the hilt for short term gains.

Question 3.
How would these advantages differ from the advantages of using a long term perspective in managing our resources?
Answer:
If resources are used in accordance with short term aims, present generation will be able to utilize the resources properly for overall development. But if we plan to use resources with long term aims, not only the present generation is benefited but also the future generations will also be able to utilize resources for fulfilling its necessities. Thus it would be better to use our natural resources with a long term perspective so that it could be used by the present generation as well as conserved for future use.

Question 4.
Why do you think that there should be equitable distribution of resources? What forces would be working against an equitable distribution of our resources?
Answer:
Nature shows no partiality. Natural resources belong to all and these resources should be used judiciously. Equitable distribution of resources will benefit both poor as well as rich people.
Human greed, corruption, and the lobby of the rich and powerful are the forces working against an equitable distribution of our resources.

Class 10th Science Chapter 16 Intext Questions Page No. 273

Question 1.
Why should we conserve forests and wildlife?
Answer:
We should conserve forests and wildlife to preserve the biodiversity (range of different life-forms) so as to avoid the loss of ecological stability. A large number of tribes are the habitants in and around the forests. If the forests are not conserved,- then it may affect these habitants. Without proper management of forest and wildlife, the quality of soil, the water sources and even the amount of rainfall may be affected. Without forest and wildlife, life would become impossible for human beings.

Question 2.
Suggest some approaches towards the conservation of forests.
Answer:
Some approaches towards the conservation of forests are as follows:

(a) People should show their participation in saving the forest by protesting against the tv ting of trees. For example, Chipko Andolan.
(b) Planting of bees should be increased. Rate of afforestation must be more than that of deforestation.
(c) Some people cut precious trees such as sandalwood to earn money. Government should take legal steps to catch these wood smugglers.
(d) Habitants of forests must not be bothered by the forest officials. Otherwise, this would result in the clash between tribal people and the

government officials, thereby enhancing the naxal activities in forests.

Class 10th Science Chapter 16 Intext Questions Page No. 276

Question 1.
Find out about the traditional systems of water harvesting/ management in your region.
Answer:
We must dug small pits and lakes, put in place simple water shed systems, built small earthen dams, constructed dykes, sand and limestone reservoirs, set up root top water collecting units. These are the traditional systems of water harvesting/management in our region.

Question 2.
Compare the above system with the probable systems in hilly/ mountainous areas or plains or plateau regions.
Answer:
In the above mentioned places check dams are built because here water harvesting is difficult.

Question 3.
Find out the source of water in your region/locality. Is water from this source available to all people living in that area?
Answer:
Tube wells and river water (Tungabhadra) are the water sources available to all people in our area. There are different sources in different places. In some places there is too much shortage of water because of failure of rain recently.

MP Board Class 10th Science Chapter 16 NCERT Textbook Exercises

Question 1.
What changes would you suggest in your home in order to be environment-friendly?
Answer:
Changes that can be undertaken in our homes to be environment friendly are listed below:

1. Switch off the electrical appliances when not in use.
2. Turn the taps off while brushing or bathing and repair the leaking taps.
3. Throw biodegradable and non-biodegradable waste into separate bins.
4. Construct composting pits.
5. Food items such as jam, pickles etc., come packed in plastic bottles. These bottles can later be used for storing things in the kitchen.

Question 2.
Can you suggest some changes in your school which would make it environment-friendly?
Answer:
Changes that can be undertaken in our schools to make it environment friendly are listed below:

1. Electricity can be saved by switching off lights and fans when not required.
2. Turn the taps off when not in use.
3. Biodegradable and non-biodegradable wastes should be thrown into separate bins.

Question 3.
We saw in this chapter that there are four main stakeholders when it comes to forests and wildlife. Which among these should have the authority to decide the management of forest produce? Why do you think so?
Answer:
The forest department of the government should have the authority to decide the management of forest produces. This is because the forest department is the care taker of the forest land and is responsible for any damage to the forest.

Question 4.
How can you as an individual contribute or make a difference to the management of
(a) forests and wildlife
(b) water resources and
(c) coal and petroleum?
Answer:
a) Forests and wild animals.

1. cutting valuable trees should be avoided by destroying forest affects the quality of soil and water resources.
2. Hunting should be prohibited.
3. There should be wild sanctuaries which gives protection for wild animals.

b) Water Resources:
Answer:

1. Water resources should be free from pollution.
2. Excess usage of water should be avoided.

c) Coal and Petroleum:
Answer:
We should minimise the use of coal and petroleum, because these are fossil fuels. By burning these there are ill effects such as air pollution and acid rainfall etc.

Question 5.
What can you as an individual do to reduce your consumption of the various natural resources?
Answer:

1. We must have come across the five R’s to save the environment: Refuse, Reduce, Reuse, Repurpose and Recycle.
2. We should encourage tree plantation programmes.
3. We must reduce the burning of fossil fuels.
4. Encouragement should be given for harvesting the water.

Question 6.
List five things you have done over the last one week to —
(a) conserve our natural resources.
Answer:
We should travel in bus instead of using own vehicles or we should practice walking, we must use LED bulbs or fluorescent tubes in our homes. We must use the lift or taking the stairs, wearing an extra sweater or using a heating device (heater or sign) on cold days.

(b) increase the pressure on our natural resources.
Answer:

1. We should grow Number of trees around our house.
2. Reducing own vehicles by using public transport system or by, walking.
3. There should not be more factories.
4. We must prevent soil erosion.
5. We must reduce the usage of vehicles to avoid air pollution.

Question 7.
On the basis of the issues raised in this chapter, what changes would you incorporate in your lifestyle in a move towards sustainable use of our resources?
Answer:
We need to change our lifestyles so that we can use natural resources on a sustainable basis. The changes which can be brought about are as follows:

• Stop cutting trees and start planting trees.
• Use LED bulbs and fluorescent tubes.
• Take the stairs and avoid using lifts.
• During summers use bamboo made fans avoid air coolers and electricians.
• Use more of public transport.
• Let our conscience be always alert not to pollute the environment from any of our activities.

MP Board Class 10th Science Chapter 16 Additional Important Questions

MP Board Class 10th Science Chapter 16 Multiple Choice Questions

Question 1.
The three R’s to save the environment are:
(a) Reuse
(b) Reduce
(c) Recycle
(d) Reduce, recycle, reuse
Answer:
(d) Reduce, recycle, reuse

Question 2.
Biodiversity is measured by
(a) The number of animals found in an area
(b) The number of mammalian found in an area
(c) The number of species found in an area
(d) The number of insects found in an area
Answer:
(c) The number of species found in an area

Question 3.
What are biodiversity hot spot?
(a) Village
(b) River
(c) Cities
(d) Forests
Answer:
(d) Forests

Question 4.
Harvesting system in hilly areas like Himachal Pradesh uses a local system of irrigation called
(a) Tals
(b) Canals
(c) Kulhs
(d) Ahass
Answer:
(c) Kulhs

Question 5.
Amrita Devi Bishnoi sacrificed her life with 363 people in 1731 to save
(a) Wildlife
(b) Water
(c) Girl child
(d) Trees
Answer:
(d) Trees

Question 6.
Aim of ‘Narmada Bachao Andolan’, was
(a) Stopping dam formation
(b) Stopping ban on dam formation
(c) Stopping water pollution
(d) Garbage maintenance
Answer:
(a) Stopping dam formation

Question 7.
Coli form is
(a) Group ef fungi
(b) Group of viruses
(c) Group of bacteria
(d) All of these
Answer:
(c) Group of bacteria

Question 8.
Main aim of the Chipko movement was:
(a) Water conservation
(b) Ecological conservation
(c) Food conservation
(d) All of the above
Answer:
(b) Ecological conservation

Question 9.
Which gas is a green house gas?
(a) CO₂
(b) CO
(c) SO₂
(d) NO₂
Answer:
(a) CO₂

Question 10.
The Chipko movement started in a village called
(a) Reni in Uttarakhand
(b) Kullu
(c) Delhi
(d) Mumbai
Answer:
(a) Reni in Uttarakhand

Question 11.
Indira Gandhi canal is an example of:
(a) River
(b) Man-made dam
(c) Pond
(d) All of these
Answer:
(b) Man-made dam

Question 12.
Kattas in Karnataka is famous for:
(a) Water harvesting
(b) Solar energy
(c) Biodiversity
(d) None
Answer:
(a) Water harvesting

MP Board Class 10th Science Chapter 16 Very Short Answer Type Questions

Question 1.
Define natural resource.
Answer:
The things available from nature are called natural resource.

Question 2.
Name the 3 R’s to save environment.
Answer:
Reduce, recycle, reuse are the 3 R’s to save environment.

Question 3.
Give some examples of natural resources.
Answer:
Examples of natural resources are water, timber and cotton.

Question 4.
What is biodiversity?
Answer:
Biodiversity is number of species that exists in an area.

Question 5.
What increases demand of resources?
Answer:
Increase in human population increases demand of resources.

Question 6.
Why we should need to manage our resources? Give one example.
Answer:
We need to manage our resources for equitable distribution for every individual.

Question 7.
Give one example of stakeholder of forest.
Answer:
People who live in or around forest are best example of stakeholder of forest.

Question 8.
In which village Chipko movement was started?
Answer:
Chipko movement started in Reni village in Uttarakhand.

Question 9.
What was the main purpose of Chipko andolan?
Answer:
To stop tree cutting indiscriminately.

Question 10.
What are the alternative sources to produce energy without creating pollution and disturbing ecological balance?
Answer:
Sunlight, water, wind etc.

Question 11.
Can we recycle everything?
Answer:
No, because everything do not turn in usable form once being used.

Question 12.
What is an example of biodiversity hot spot?
Answer:
Forest.

Question 13.
Write an importance of conservation of wild life.
Answer:
Preserving biodiversity inherited in time.

Question 14.
Write name of one activist for saving water.
Answer:
Rajendra Kumar.

Question 15.
Give an example of a place famous for water management in India.
Answer:
Kulhs in Himachal Pradesh.

MP Board Class 10th Science Chapter 16 Short Answer Type Questions

Question 1.
What are natural resources? Explain.
Answer:
Natural resources can be broadly categorized into two types: exhaustible and non-exhaustible. Management of natural resources is all about their judicious use in a way that the exhaustible resources can lasfr.for many generations to come and non-exhaustible resources can be maintained in as pristine form as possible.

Question 2.
What are three R’s (Reduce, recycle and reuse)?
Answer:
Reduce: We should reduce the consumption of various resources wherever possible. For example; we can reduce the consumption of electricity and water.

Recycle: There are many items which can be recycled again and again. For example by recycling paper, we reduce the demand for wood and thus, help in saving the forest.

Reuse: Many items can be reused many times. For example; old newspaper, envelopes, plastic bottles.

Question 3.
Who are the stakeholders of forest?
Answer:
The stakeholders are as follows:
People living in or around forests; as they depend on various forest produce for their livelihood, the forest department which is the owner of the forest land, various industrialists who depend on forest for many raw materials. For example: wood is used as raw material in many industries.

So, the people, industry and Government body who are directly or indirectly affected by forest are called stakeholders of forest.

Question 4.
Give one example of saving ecosystem by local communities.
Answer:
Chipko Movement: The Chipko movement began in the early 1980s from a small village; Reni in Garhwal district. The women of the village began hugging the trees to prevent the cutting of trees by the contractors. The Chipko movement later spread to other parts of India.

Question 5.
Give some examples of water harvesting method in ancient India.
Answer:

1. Khadins, tanks and nadis in Rajasthan.
2. Bandharas and tals in Maharashtra.
3. Bundhis in Madhya Pradesh and Uttar Pradesh.
4. Ahars and pynes in Bihar.
5. Kulhs in Himachal Pradesh.
6. Ponds in the Kandi belt of Jammu region and eris (tanks) in Tamil Nadu. Surangams in Kerala, and Kattas in Karnataka.

These are some of the ancient water harvesting examples and still in use at many places.

Question 6.
What is the importance of traditional water harvesting structures?
Answer:
The traditional water harvesting structures usually focus on recharging the groundwater rather making an open reservoir. It has several advantages. Unlike surface water; the groundwater does not evaporate and»thus, loss because of evaporation is prevented. The groundwater does not provide a breeding ground for the mosquitoes and hence is good for public health as well. The groundwater is relatively protected from contamination by human activities.

Question 7.
Why alternate energy sources is required in place of Coal and Petroleum? Give examples of alternate sources.
Answer:
Coal and petroleum are the main energy resources for us. But, since these are exhaustible in nature so, we need to find out alternate sources of energy. Scientists are working on developing some alternate energy sources so that dependency on coal and petroleum can be reduced. Some examples are given below:

1. Solar energy is being used to produce electricity at many places. Although, this technology is still costly.
2. Fuel cell is another development which may help in replacing the internal combustion engines from automobiles.
3. Hydrogen is being used as fuel in buses and cars in many countries. Hydrogen; when used as a fuel produces water as a by-product. Thus, hydrogen can be an environment-friendly fuel.

MP Board Class 10th Science Chapter 16 Long Answer Type Questions

Question 1.
Explain consequences of exploitation of natural resources and sustainable development.
Answer:
There are many consequences of exploitation of natural resources.
Some examples are given below:

1. Burning of fossil fuels creates air pollution. Excess amount of carbon dioxide in the atmosphere leads to global warming.
   Some polluting gases; like oxides of nitrogen and sulphur lead to acid rain, which is harmful for living beings. Acid rain is also harmful for monuments and buildings.
2. Excess exploitation of groundwater leads to a drastic fall in water table. For this reason many places are experiencing acute shortage of drinking water.
3. Overuse of fertilisers and insecticides leads to soil pollution and soil erosion.
4. Many pollutants are directly flown into water bodies. This has resulted in water pollution in many rivers, lakes and even in oceans.

Sustainable Development:

Development is necessary for making all around economic development. But development often comes with a price in the form of environmental damage. Sustainable development means following certain practices which help in saving our environment from damage. This is necessary for maintaining the earth in a good shape so, that future generations can also enjoy bounty of nature.

Question 2.
What are three R’s?
Answer:
1. Reduce: We should reduce the consumption of various resources wherever possible. For example; we can reduce the consumption of electricity by switching off lights and other appliances when they are not required. While leaving the home, one should always check for fans and lights and switch them off. This cannot only help in saving electricity but also in saving the fuels which are utilised in electricity production. We should immediately repair a leaking tap so that precious water can be saved.

2. Recycle: There are many items which can be recycled again and again. Recycling is another way of reducing the demand for natural resources. For example; by recycling paper, we reduce the demand for wood and thus, help in saving the forest.

3. Reuse: Many items can be reused many times. For example; old newspaper can be used for packing many items. Old envelopes can be used, for doing rough work while doing homework. Old plastic bottles can be used for many other purposes.

Question 3.
Explain Arabari’s example of People’s Participation in Forest Management.
Answer:
In 1972, the forest department realized its mistake while reviving the degraded sal forests of Arabari forest range. Arabari forest lies in Midnapore district of West Bengal. The earlier methods of policing and surveillance were a total failure as they often led to frequent clashes with local people. It also led to alienation of people from the conservation programme. Then, came a forest officer; named A.K Baneijee; who was a real visionary. He involved the local people in the revival of 1,272 hectares of forest. In lieu of that the villagers were given employment in silviculture and were given 25% of the harvest. They were also allowed to gather firewood and fodder against a nominal payment. Due to active participation of the local community, there was remarkable revival of the Arabari sal forest. By 1983, the value of the forest rose to ? 12.5 crores.

Question 4.
Explain in brief about people ‘effort for forest conservation’.
Answer:
Bishnoi community: The Bishnoi community of Rajasthan is one such example. Amrita Devi Bishnoi is still remembered with reverence for the way she fought for protecting the khejri trees in Khejrali village. She along with 363 other people sacrificed her life for the protection of khejri trees in 1731. The ‘Amrita Devi Bishnoi National Award for Wildlife Conservation’ has been named in her honour.

Nomadic herders of the Himalayas: The nomadic herders used to graze their animals near the great Himalayan National Park. Every summer, the nomadic people brought their herds down the valley so that the sheep could get plenty of grass to eat. When the National Park was made in that area, the nomadic herders were stopped from grazing their sheep in the protected area. Now, in the absence of grazing by the sheep, the grasses grew very tall in the region. Tall grasses fall over and prevent fresh growth of grass. This shows that by excluding and alienating the local people from forests, proper conservation efforts cannot be carried out.

Chipko movement: The Chipko began in the early 1980s from a small village; Reni in Garhwal district. The women of the village began hugging tree to prevent the cutting of trees by the contractors. The Chipko movement later spread to other parts of India.

Question 5.
Give two examples of forest conservation by local communities.
Answer:
Following are two examples of Forest conservation by local
communities:

1. The Chipko movement began in the early 1980s from a small village; Reni in Garhwal district. The women of the village began hugging a tree to prevent the cutting of trees by the contractors. There are many examples which suggest that involvement of local communities is necessary for any conservation effort. The Bishnoi community of Rajasthan is one such example. Amrita Devi Bishnoi is still remembered with reverence for the way she fought for protecting the khejri trees in Khejrali village. She along with 363 other people, sacrificed her life for the protection of Khejri trees in 1731. The ‘Amrita Devi Bishnoi National Award for Wildlife Conservation’ has been named in her honour.

2. Another example is of the nomadic herders of the Himalayas. The nomadic herders used to graze their animals near the great Himalayan National Park. Every summer, the nomadic people bought their herds down the valley so that the sheep could get plenty of grass to eat. When the National Park was made in that area, the nomadic herders were stopped from grazing their sheep in the protected area. Now in the absence of grazing by the sheep, the grasses grew very tall in the region. Tall grasses fall over and prevent fresh growth of grass. This shows that by excluding and alienating the local people from forests, proper conservation efforts cannot be carried out.

MP Board Class 10th Science Chapter 16 NCERT Textbook Activities

Class 10 Science Activity 16.1 Page No. 266

• Find out about the international norms to regulate the emission of carbon dioxide.
• Have a discussion in class about how we can contribute towards meeting those norms.

Observations:

• There are many laws and norms made to regulate the emission of various gases that creates harm to the environment. One of the norms related to CO₂ emission is based on ‘Kyoto protocol’ where all industrialised countries come forward to minimize the collective emission of CO₂ and other green house gases.

Class 10 Science Activity 16.2 Page No. 267

• There are a number of organisations that seek to spread awareness about our environment and promote activities and attitudes that lead to the conservation of our environment and natural resources. Find out about the organisations(s) active in your neighbourhood/village townicity.
• Find out how you can contribute towards the same cause,

Observations:

• There are many organisations spreading awareness about environment and promoting activities for its benefit. Different states of India have number of organisations. The name of new includes – Ionosphere Social Enterprise, The energy and resonance Institute IRADe, Elnora International, Delhi Greens, Greenpeace India, GVNMAL, India Nateere watch, Kalpvriksh. Nation’s club. National biodiversity Authority etc.

Class 10 Science Activity 16.3 Page No. 266

• Check the pH of the water supplied to your house using universal indicator or litmus paper.
• Also check the pH of the water in the local waterbody (pond, river, lake, stream).
• Can you say whether the water is polluted or not on the basis of your observations?

Observations:

• The normal range for pH in our house may be between 6.5 to 8.5. The pH of local water bodies are around 5.7.
• The water showing little less pH than recommended in water bodies may indicate pollution due to slight acidic substances which make the water acidic showing decrease in pH.
• The optimum range for water bodies in around 6-8.5, the variations in it indicates pollutions.

Class 10 Science Activity 16.4 Page No. 269

• Have you ever visited a town or village after a few years of absence? If so, have you noticed new roads and houses that have come up since you were there last? Where do you think the materials for making these roads and buildings have come from?
• Try and make a list of the materials and their probable sources.
• Discuss the list you have prepared with your classmates. Can you think of ways in which the use of these materials be reduced?

Observations:

• Yes, new roads and houses are coming up very fastly in villages. The materials generally comes from the neighbouring cities and areas where it is build.
• The materials for building roads and buildings includes concrete, composite pavement, asphalt, bituminous, gravel surfaces etc.
• These can be reduced by opting for less toxic and dangerous materials like bituminous and asphalt may be avoided and replaced by sand and naturally made materials.

Class 10 Science Activity 16.5 Page No. 270

• Observe various traditional practices for conservation of nature in your day-to day life. Share within the peer group. Make a report and submit.

Observations:

• Various traditional practices for conservation of nature includes various practices like:
• Religious traditions – temple forest, monastery forests etc.
• Traditional tribal traditions – sacred forests, sacred trees etc.
• Royal traditions – royal hunting practices and preserves, royal gardens etc.
• Livelihood traditions – forests and grows serving as cultural and social space.

Class 10 Science Activity 16.6 Page No. 271

• Make a list of forest produce that you use.
• What do you think a person living near a forest would use?
• What do you think a person living in a forest would use?
• Discuss with your classmates how these needs differ or do not differ and the reasons for the same.

Observations:

• Forest produce that are use includes:
• Wood, sandalwood, rubber, latex, paper, food (fruits and vegetables), Sponges, wood fuel etc.
• A person living near a forest will use wood as fuel, fruits and Vegetables, rubber, sandal cored etc.
• A person leaving in a forest will decrease the storage of forest produce and can used hand in hand as and when required. The use of many things will reduce.
• The person living near will have different requirements as compared to person duriug-inside forest as maximum needs of people living in forest wall increase as he can directly take from there neither than cutting or taking from native and then utilising

Class 10 Science Activity 16.7 Page 272

• Find out about any two forest produce that are the basis for an industry.
• Discuss whether this industry is sustainable in the long run. Or do we viced to control our consumption of these products?

Observations:

• Any two forest produce that are basis for an industry arc wood and coal.
• This industry is not sustainable in the long run as the limited supply for these products are available in nature and we are using them at very high speed. We need to reduce the consumption and wastage of such products. The natural resources should be used judiciously.

Class 10 Science Activity 16.8 Page No. 275

• Debate the damage caused to forests by the following:
  • (a) Building rest houses for tourists in national parks.
  • (b) Grazing domestic animals in national parks.
  • (c) Tourists throwing plastic bottles, covets and other litter in national parks.

Observations:

• The building rest houses for tourists in national parts causes a lot increase in deforestation which disturbs the balance of the nature. The animals and other living organisms living in an ecosystem also gets disturbed and whole area is effected.
• Grazing domestic animals leads to destruction of green grasses and shrubs which destroy the green cover of the park. This also effects the other dependent organisms of the park.
• The plastic bottles/covers and other litter thrown in the national parks makes the park very dirty making it unfit in providing healthy living conditions to the animals. These substances do not decompose and remain there creating pollution for years and leading to destruction of various organisms.

Class 10 Science Activity 16.9 Page No. 275

• Villages suffering from chronic water shortage surround a water theme park in Maharashtra. Debate whether this is the optimum use of the available water.

Observations:

• No, this is not the optimum use of the available water. Water theme park in Maharashtra uses water in large animals for the amusement purposes. This water can be used for various other basic needs. There is acute shortage of water in nearby areas. This water can fulfill the needs there.

Class 10 Science Activity 16.10 Page No. 275

• Study the rainfall patterns in India from an atlas.
• Identify the regions where water is abundant and the regions of water scarcity.

Observations:

• The monsoon affect the most part of India, the amount of rainfall varies from heavy to scanty in different parts. There is great temporal and regional variation in distribution of rainfall. Over 80% of annual rainfall is received in four rainy months of June to September.
• The regions with abundant water includes – most of Canada, Great Lakes, Ireland, Amazon (Brazil), Antarctica.
• The regions of water scarcely includes – Middle East, Sahara Desert, Atacama Desert, India, Gobi Desert

Class 10 Science Activity 16.11 Page No. 279

• Coal is used in thermal power stations and petroleum products like petrol and diesel are used in means of transport like motor vehicles, ships and aeroplanes. We cannot really imagine life without a number of electrical appliances and constant use of transportation. So can you think of ways in which our consumption of coal and petroleum products be reduced?

Observations:

• We can reduce our consumption of coal and petroleum by using other means of energy like solar, hydral, wind and various other natural forms of energy. These sources are renewable and do not cause any harm to the nature.

Class 10 Science Activity 16.12 Page No. 279

• You must have heard of the euro I and Euro II norms for emission from vehicles; Find out how these norms work towards reducing air pollution.

Observations:

• Euro norms refer to the permissible emission levels from both petrol and diesel vehicles. They are available for fuel quality and the method of testing. These norms decides how much a can should smoke and the particles emitted out. It have been in Europe first and then it was also implemented in India as well in the name of Bharat stage I and II.

MP Board Class 10th Science Solutions

MP Board Class 10th Science Solutions Chapter 15 Our Environment

MP Board Class 10th Science Chapter 15 Intext Questions

Class 10th Science Chapter 15 Intext Questions Page No. 257

Question 1.
Why are some substances biodegradable and some non- biodegradable?
Answer:
Some substances can be broken down into simpler substances by the action of enzymes and other physical factors and are returned to the earth. Some substances cannot be degraded into simpler form and exists in nature for very long, deteriorating it and hence, are termed as non- biodegradable.

Question 2.
Give any two ways in which biodegradable substances would affect the environment.
Answer:

1. These substances may cause pollution the environment.
2. They may serve breeding ground for pathogens which may cause diseases

Question 3.
Give two ways in which non-biodegradable substances would affect me environment.
Answer:

1. They do not degrade and pile up in the environment causing harm to the ecosystem.
2. They may lead to bio-magnification in food chain disturbing the various trophic levels.

Class 10th Science Chapter 15 Intext Questions Page No. 261

Question 1.
What are the trophic levels? Give an example of a food chain and state the different trophic levels in it.
Answer:
Each step or level of a food chain is called Trophic levels.
Example for Food chain – Here grass is a producer because it prepares its own food. This grass is eaten by herbivores means secondary, small carnivores (Frog) are tertiary and higher carnivores are in the fourth level.

Question 2.
What is the role of decomposers in the ecosystem?
Answer:
Microorganisms, comprising bacteria and Fungi, break-down the dead remains and waste products of organisms. These microorganisms are the decomposers as they break-down the complex organic substances into simple inorganic substances that go into the soil and are used up once more by the plants.

Class 10th Science Chapter 15 Intext Questions Page No. 264

Question 1.
What is ozone and how does it affect any ecosystem?
Answer:
Ozone at the higher levels of the atmosphere is a product of UV radiations acting on O₂ molecule. The higher energy UV radiations split apart some molecular Oa in free oxygen (O) atoms. These atoms then combine with the molecular O₂ to form Ozone.

Ozone shields the surface of the earth from ultraviolet (UV) radiation from the Sun. This radiation is highly damaging to organisms for example, it is known to cause skin cancer in human beings.

Question 2.
How can you help in reducing the problem of waste disposal? Give any two methods.
Answer:

1. We must minimise the usage of plastics,
2. We can collect wastes and by this we can produce gas which is an alternate source of energy.

Class 10th Science Chapter 15 Ncert Textbook Exercises

Question 1.
Which of the following groups contain only biodegradable items?
(a) Grass, flowers and leather
(b) Grass, wood and plastic
(c) Fruit-peels, cake and lime-juice
(d) Cake, wood and grass
Answer:
(a), (c), (d)

Question 2.
Which of the following constitute a food-chain?
(a) Grass, wheat and mango
(b) Grass, goat and human
(c) Goat, cow and elephant
(d) Grass, fish and goat
Answer:
(b) Grass, goat and human

Question 3.
Which of the following are environment-friendly practices?
(a) Carrying cloth-bags to put purchases in while shopping
(b) Switching off unnecessary lights and fans
(c) Walking to school instead of getting your mother to drop you on her scooter
(d) All of the above
Answer:
(d) All of the above

Question 4.
What will happen if we kill all the organisms in one trophic level?
Answer:
If we kill all the organisms in one trophic level, the population size of organisms in lower level increases uncontrollably and the number of organisms in higher trophic level decreases due to non¬availability of food. This results in an imbalance in ecosystem.

Question 5.
Will the impact of removing all the organisms in a trophic level be different for different trophic levels? Can the organisms of any trophic level be removed without causing any damage to the ecosystem?
Answer:

• Removing producers: All the heterotrophs die.
• Removing herbivores: Carnivores would not get food.
• Removing carnivores: Herbivores would increase to unsustainable levels.
• Removing decomposers: Organic wastes, plant, and animal dead remains would pile up.
• The role of each and every species belonging to every trophic level is unique.
• No, the organisms of any trophic level cannot be removed without damaging the ecosystem.

Question 6.
What is the biological magnification? Will the levels of this magnification be different at different levels of the ecosystem?
Answer:
Some harmful chemicals enter our bodies through the food chain, one of the reasons is the use of several pesticides and other chemicals to protect our crops from disease and pests. These chemicals are either washed down into the soil or into the water bodies. From the soil, these are absorbed by the plants along with water and minerals and from the water bodies these are taken up by aquatic plants and animals.

This is one of the ways in which they enter the food chain. This phenomenon is known as biological magnification. This level of magnification be different at different levels of the ecosystem.

Example; Spraying of DDT will remain for a long time in the environment.

Question 7.
What are the problems caused by the non-biodegradable wastes that we generate?
Answer:

• Non-aesthetic look.
• Death of cattle by ingestion of plastic bags.
• The quality of soil is adversely affected.
• Biomagnification of harmful chemicals like DDT in birds disturb their calcium metabolism.
• Non – biodegradable wastes cause pollution of soil and water.

Question 8.
If all the waste we generate is biodegradable, will this have no impact on the environment?
Answer:
If all the waste we generate is biodegradable, there is a imbalance in nature. Because with the increase of wastes there is decrease in the number of decomposers. These wastes spread every where and microbes are more which causes many diseases to us.

Question 9.
Why is damage to the ozone layer a cause for concern? What steps are being taken to limit this damage?
Answer:
Damage to the ozone layer causes so many problems. At the higher levels of the atmosphere, ozone performs an essential function. It shields the surface of the earth from ultraviolet radiation from the sun. If ozone layer is damaged no organism can survive. The following are the steps being taken to limit this damage.

1. We should minimize the use of vehicles.
2. We should not encourage the burning of fossilic fuels.
3. It is now mandatory for all the manufacturing companies to make CFC- free refrigerators throughout the world.

MP Board Class 10th Science Chapter 15 Additional Important Questions

MP Board Class 10th Science Chapter 15 Multiple Choice Questions

Question 1.
An environment consists of:
(a) Abiotic components
(b) Biotic components
(c) Both
(d) Not certain
Answer:
(c) Both

Question 2.
Waste could be:
(a) Abiotic components
(b) Biotic components
(c) Both
(d) Not certain
Answer:
(c) Both

Question 3.
Reduction of waste is important to:
(a) Make environmental balance proper.
(b) Make nearby beautiful
(c) waste can be transformed to useful products.
(d) Not clear
Answer:
(d) Not clear

Question 4.
Which one among following is non-biodegradable substance?
(a) Metal
(b) Wood
(c) Water
(d) Urea
Answer:
(a) Metal

Question 5.
Which among the following is a biodegradable waste?
(a) Wood
(b) Teflon pots
(c) Plastic cup
(d) Glass Cups
Answer:
(a) Wood

Question 6.
Most convenient ways of waste management is:
(a) 3 R : Reduce, reuse, recycle principle
(b) Production
(c) Use ban
(d) None
Answer:
(a) 3 R : Reduce, reuse, recycle principle

Question 7.
Which one is not a primary consumer?
(a) Grasshopper
(b) Deer
(c) Ant
(d) Leech
Answer:
(b) Deer

Question 8.
Humans are:
(a) Primary consumers
(b) Secondary consumers
(c) Top consumers
(d) All
Answer:
(d) All

Question 9.
Energy while going up in a trophic level is:
(a) Increased
(b) Decreased
(c) Remain same
(d) Can’t predict
Answer:
(b) Decreased

Question 10.
Pesticides are used to:
(a) Develop new varieties of crops.
(b) Kill unwanted plants.
(c) Kill insects and enrobes attacking crops.
(d) Save crops from birds.
Answer:
(c) Kill insects and enrobes attacking crops.

Question 11.
In a marine ecosystem producers are:
(a) Plants
(b) Sand
(c) Water
(d) Fishes
Answer:
(d) Fishes

Question 12.
Top consumer in a crop field is:
(a) Rat
(b) Hawk
(c) Snake
(d) (a) and (b)
Answer:
(b) Hawk

Question 13.
Effect of bio-magnification is maximum in:
(a) Primary consumers
(b) Secondary consumers
(c) Top consumers
(d) All of these
Answer:
(c) Top consumers

Question 14.
Which one of the following is artificial ecosystem?
(a) Jungle
(b) Town
(c) A pond
(d) Mountain
Answer:
(b) Town

Question 15.
Cockroach is a:
(a) Producer
(b) Primary consumer
(c) Secondary consumer
(d) Decomposer
Answer:
(d) Decomposer

Question 16.
Energy source of an ecosystem is:
(a) Producers
(b) Sunlight
(c) Top consumer
(d) Atmospheric gases
Answer:
(b) Sunlight

Question 17.
Link between primary and secondary consumers are:
(a) Autotrophs
(b) Omnivorous
(c) Carnivorous
(d) Herbivorous
Answer:
(a) Autotrophs

Question 18.
What is CFC?
(a) A waste
(b) A coolant gas
(c) A bakery product
(d) An organization
Answer:
(b) A coolant gas

Question 19.
UNEP stands for:
(a) United Nations Environment Programme.
(b) United Nations Entertainment Programme.
(c) United Nations Excellence Programme.
(d) Unlimited Nations Excellence Programme.
Answer:
(a) United Nations Environment Programme.

Question 20.
How many atoms of oxygen compose an ozone molecule?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

MP Board Class 10th Science Chapter 15 Very Short Answer Type Questions

Question 1.
Which gas of our environment help in formation of energy which we get from various sources?
Answer:
Oxygen.

Question 2.
How non-biodegradable objects effect environment?
Answer:
They pollute the environment.

Question 3.
Name two substances you think as most non-biodegradable.
Answer:
Plastic and chemicals like pesticides.

Question 4.
Which organisms help in biodegradation of a substance?
Answer:
Decomposers.

Question 5.
Write three common waste produced by our daily use.
Answer:
Soap and detergent, consumed food materials used paper and plastic garbage.

Question 6.
Can a big tree be treated as an isolated small ecosystem?
Answer:
Yes.

Question 7.
Give examples of natural ecosystem.
Answer:
Forests, pond, river etc.

Question 8.
Give examples of artificial ecosystem.
Answer:
Aquariums, Garden, town etc.

Question 9.
Can a ecosystem survive without autotrophs?
Answer:
No.

Question 10.
At which category parasite come in an ecosystem?
Answer:
Decomposers.

Question 11.
At which level rabbit and mole come in trophic level?
Answer:
Rabbit – Primary level.
Mole – Primary and decomposer level.

Question 12.
How much percentage of sunlight is converted into chemical energy by all autotrophs?
Answer:
1 %.

Question 13.
What percentage of average organic matter is present at each step of trophic level?
Answer:
10%.

Question 14.
What kind of plants comes at primary consumers level?
Answer:
Carnivorous plants.

Question 15.
How many minimum food chains can be observed in a food web?
Answer:
2 – 3.

MP Board Class 10th Science Chapter 15 Short Answer Type Questions

Question 1.
What is ecosystem? What are components of ecosystem?
Answer:
An ecosystem includes all of the living things in a given area, interacting with each other and also with their non-living environments.

Components of ecosystem:
An ecosystem has two types of components – biotic component (plants, animals and organisms) and abiotic component (weather, earth, sun, soil, climate, atmosphere).

Question 2.
What is abiotic component?
Answer:
All the non-living things make the abiotic component of an ecosystem. Air, water and soil are the abiotic components.

Question 3.
What is the importance of abiotic component?
Answer:
Air provides oxygen (for respiration), carbon dioxide (for photosynthesis), water (for metabolic activities) and soil is the reservoir of various nutrients which are utilized by plants. Through plants; these nutrients reach other living beings.

Question 4.
What is a biotic component?
Answer:
Ail living beings make the biotic component of an ecosystem. Examples: Green plants, animals and other living beings. Bacteria and fungi are examples of biotic component.

Question 5.
What is a food chain?
Answer:
A food chain is a simple representation of transfer of energy from the sun to different biotic components of an ecosystem. Sun is the ultimate source of energy. Green plants convert solar energy into chemical energy during photosynthesis. When an animal takes food, this energy is supplied to the animal and the process goes on.

Question 6.
What is food web and trophic level?
Answer:
Food web: In an ecosystem, there can be many food chains which are interlinked at various levels. Thus, many food chains form a network which is called food web.

Trophic level: Transfer of energy occurs through a food chain. Different levels in the food chain are called trophic level.

Question 7.
What are biodegradable substances?
Answer:
Substances which can be decomposed by microorganisms are called biodegradable substances. All the organic substances are biodegradable.

Question 8.
What are non-biodegradable substances?
Answer:
Substances which cannot be decomposed by microorganisms are non biodegradable. All inorganic substances are non-biodegradable. Many synthetic substances are also non biodegradable.

Question 9.
Give two differences between biodegradable and non bio-degradable.
Answer:

Question 10.
What is ozone layer? How it is protected from ultraviolet radiations?
Answer:
Ozone layer is also known as stratosphere. When ultraviolet radiations act on oxygen, the oxygen gets converted into ozone. Ozone layer works like a protective shield for living beings. The ozone layers guards from harmful ultraviolet rays of the sun.

MP Board Class 10th Science Chapter 15 Long Answer Type Questions

Question 1.
What is ecosystem? Explain in detail the components of ecosystem.
Answer:
An ecosystem includes all of living things in a given area, interacting with each other and also with their non living environments.

Components of ecosystem:

1. Biotic component (plants, animals and organisms) and abiotic components (weather, sun, soil, climate, atmosphere).
All living beings make the biotic component of an ecosystem. Examples: Green plants, animals and other living beings. Bacteria and fungi are examples of biotic component.

Green plants play the role of producers: Because they prepare the food by photosynthesis.

Animals and other living beings play the role of consumers; because they take food (directly or indirectly) from plants.

Bacteria and fungi play the role of decomposers; as they decompose dead remains of plants and animals so that raw materials of organisms can be channelized back to the environment.

2. Abiotic component: All the non-living things make the abiotic component of an ecosystem. Air, water and soil are the abiotic components.

Air provides oxygen (for respiration), carbon dioxide (for photosynthesis) and other gases for various needs of the living beings.

Water is essential for all living beings because all the metabolic activities happen in the presence of water.

Soil is the reservoir of various nutrients which .are utilized by plants. Through plants, these nutrients reach other living beings.

Question 2.
Explain in briefly about food chain.
Answer:
A food chain is a simple representation of transfer of energy from the sun to different biotic components of an ecosystem. Sun is the ultimate source of energy. Green plants convert solar energy into chemical energy during photosynthesis. When an animal takes food, this energy is supplied to the animal and the process goes on. A simple food chain can be shown as follows:

Producer → Primary consumer → Secondary consumer

Real life cannot be as simple as a food chain shown above. In any ecosystem, there can be many food chains which are interlinked at various levels. Thus, many food chains form a network which is called food web.

Transfer of energy occurs through a food chain. Different levels in the food chain are called trophic level. Out of the energy consumed by an organism at a particular trophic level, 90% is utilized for its own need and rest 10% is left for the organism of the next trophic level. So. very little energy is tell for the organism which is at the tertiary level. Letus assume that a green plant makes 100% energy in the form of chemical energy, 90% of this energy would be utilized for its own purpose. This would leave just 10% energy for the primary consumer. Now, primary consumer shall also utilize 90% of energy which was consumed by it. This would leave just 1% energy for (10% of 100%) for the secondary consumer. By this logic, the tertiary consumer would get just 0.1% of energy which was originally made by the green plant. This is the reason, there can be just one or two organisms at the top of the food pyramid.

This explains why the population of producers is always the largest in an ecosystem; followed by the population of herbivores and then that of carnivores. Moreover, herbivores needs to eat many plants in its lifetime to fulfill its energy need. Similarly, carnivores needs to eat many herbivores in its lifetime.

Question 3.
What is ozone layer depletion?
Answer:
Ozone layer is also known as stratosphere. When ultraviolet radiations act on oxygen, the oxygen gets converted into ozone.

Ozone layer works like a protective shield for living beings. The ozone layers guards from harmful ultraviolet radiations from the sun.

Effect of CFCs: Use of CFCs (Chlorofluorocarbon) has damaged the ozone layer. As a result, the ozone layer has become thinner at certain parts. In 1987, the UNEP (United Nations Environment Programme) succeeded in forging an agreement among different nations to freeze the CFC production at 1986 level. Later, an agreement was signed among different nations to phase out CFCs. It is important to note that CFC is used in refrigerators and aerosol sprays. India is also a signatory of that agreement and thanks to the efforts by the United Nations and different environmentalists, the CFC emission has been put under some control.

Problems of waste disposal: During our day to day activities, we produce lot of waste. While some of the waste is biodegradable, a large chunk is composed of non-biodegradable substances. Plastic waste is a serious concern because plastic is non-biodegradable. We need to respect our environment and find out ways to reduce the burden on our environment.

MP Board Class 10th Science Chapter 15 NCERT Textbook Activities

Class 10 Science Activity 15.1 Pages No. 256,257

• You might have seen an aquarium. Let us try to design one.
• What are the things that we need to keep in mind when we create an aquarium? The fish would need a free space for swimming (it could be a large jar), water, oxygen and food.
• We can provide oxygen through an oxygen through an oxygen pump (aerator) and fish food which is available in the market.
• If we add a few aquatic plants and animals it can become a self- sustaining SySieni.’ Can you think how this happens? An aquarium is an example of a human-made ecosy stem.
• Can we leave the aquarium as such after we set it up?
• Why does it have to be cleaned once in a while? Do we have to clean ponds or lakes in the same manner? Why or why not?

Observations:

• Pesticides are the chemicals sprayed to kill the pests infecting crop and harming them. These are very harmful to living organisms. These get biologically magnified in the food chains and in the bodies, once entered. Organic farming methods and other natural methods can be adopted to reduce dependence on chemicals.

Class 10 Science Activity 15.2 Page No. 257

• While creating an aquarium did you take care not to put an aquatic animal which would eat others? What would have happened otherwise?
• Make groups and discuss how each of the above groups of organisms are dependent on each other.
• Write the aquatic organisms in order of who eats whom and form a chain of at least three steps.
  
• Would you consider any one group of organisms to be of primary importance? Why or why not?

Observations:

• CFC’s are responsible for depletion of ozone layer.
• These chloroflouro carbons were banned in various countries and since, then ozone hole has decreased in size.

FIg. 15.1: Food chain ¡n nature (a) in forest, (b) in grassland and (c) in pond.

Class 10 Science Activity 15.3 Pages No. 244,245

• Newspaper reports about pesticide levels in ready made food items are often seen these days and some states have banned these products. Debate in groups the need for such bans.
• What do you think would be the source of pesticides in these food items? Could pesticides get into our bodies from this source through other food products too?
• Discuss what methods could be applied to reduce our intake of pesticides.

Observations:

• The materials that are non-biodegradable do not degrade easily. The hard materials that are not organic in nature take more time to decompose. The biodegradable materials turns soft and start decomposing and mixing in the soil. They completely change their form and structure.
• The waste from plants – fruits and vegetables change the fastest.

Class 10 Science Activity 15.4 Page No. 248

• Find out from the library, internet or newspaper reports, Which chemicals are responsible for the depletion of the ozone layer.
• Find out if the regulations put in place to control the emission of these chemicals have succeeded in reducing the damage to the ozone layer. Has the size of the hole in the ozone layer changed in recent years?

Observations:

• CFCs, SO₂ other pollutant gases are responsible for the depletion of ozone layer.
• Non-biodegradable materials take years to decompose. Plastics are generally non-biodegradable and not decompose. The materials like paper, jute etc. easily degrade and do not harm the environment.

Class 10 Science Activity 15.5 Page No. 249

• Collect waste material from your homes. This could include all the waste generated during a day, like kitchen waste (spoils food, vegetable peels, used tea leaves, milk packets and empty cartons), waste paper, empty medicine bottles/strips/ bubble packs, old and tom clothes and broken footwear.
• Bury this material in a pit in the school garden or it there is no space available, you can collect the material in an old bucket/flower pot and cover with at least 15cm of soil.
• Keep this material moist and observe at 15-day intervals,
• What are the materials which change their form and structure over time?
• Of these materials that are changed, which ones change the fastest.1?”

Observations:

• We cannot leave the aquarium as it is as waste is generated in it by fishes which need to be cleaned up frequently so that it does not become toxic and harm the living organisms. The water bodies should also be cleaned up properly to ensure better living conditions. The aquatic life is affected by such kind of pollutions.

Class 10 Science Activity 15.6 Pages No. 249-250

• Use the library or Internet to find out more about biodegradable and non-biodegradable substances.
• How long are various non-biodegradable substances expected to last in our environment?
• These days, new types of plastics which are said to be biodegradable are available. Find out more about such materials and whether they do or do not harm the environment.

Observations:
Phytoplankton → Zoo planktons → Small fishes → Large fishes Aquatic Food Chain

• Phytoplankton are the autotrophs present in the water body and are of prime importance.
• The removal of any one group from the food chain will disturb the whole aquatic ecosystem.

Class 10 Science Activity 15.7 Page No. 252

• Find out what happens to the waste generated at home. Is there a system in place to collect this waste?
• Find out how the local body (Panchayat, municipal corporation, resident welfare association) deals with the waste. Are there mechanisms in place to treat the biodegradable and non-biodegradable wastes separately?
• Calculate how much waste is generated at home in a day.
• How much of this waste is biodegradable?
• Suggest ways of dealing with this waste.

Observations:

• The waste generated at home are differentiated into biodegradable and non-biodegradable materials. They are placed separately in the bins and thereafter, processed.

Class 10 Science Activity 15.8 Page No. 253

• Find out how the sewage in your locality is treated. Are there mechanisms in place to ensure that local water bodies are not polluted by untreated sewage?
• Find out how the local industries in your locality treat their wastes. Are there mechanisms in place to ensure that the soil and water are not polluted by this waste?

Observations:

• The amount of waste generated in homes and in the class-rooms is very high. At the end of the day. number of dustbins are fully filled which are thrown away in dump yard which too have become problematic. The wastes should be reused and amount generated should be reduced.

Class 10 Science Activity 15.9 Page No. 253

• Search the Internet or library to find out what hazardous materials have to be dealt with while disposing of electronic items. How would these materials affect the environment?
• Find out how plastics are recycled. Does the recycling process have any impact on the environment?

Observations:

• ETP (Effluent treatment plant) or STP (Sewage treatment plant) are used to treats the waste generated before it enters the water body. All the harmful treatments are removed and level of the toxic materials are reduced.

Class 10 Science Activity 15.10 Page No. 253

• Search the Internet or library to find out what hazardous materials have to be dealt with while disposing of electronic items. How would these materials affect the environment?
• Find out how plastics are recycled. Does the recycling process have any impact on the environment?

Observations:

• Electronic items are very hazardous as they are non-biodegradable and increase the mars on the earth.
• Plastic are recycled after melting again and-reforming into new shapes and products. This process create lots of pollution by emitting dangerous fumes and harms the environment.

MP Board Class 10th Science Solutions

MP Board Class 10th Science Solutions Chapter 11 Human Eye and Colourful World

MP Board Class 10th Science Chapter 11 Intext Questions

Class 10th Science Chapter 11 Intext Questions Page No. 190

Question 1.
What is meant by power of accommodation of the eye?
Answer:
The ability of the eye to focus the distant objects as well as the nearby objects on the retina by changing focal length or converging power of its lens is called accommodation. The normal eye has a power of accommodation which enables the object as close as 25cm & as far as infinity to be focused on its retina.

Question 2.
A person with a myopic eye can not see objects beyond 1.2 m distinctly. What should be the type of the corrective lens used to restore proper vision?
Answer:
The person is able to see nearby objects clearly, but he is unable to see objects beyond 1.2m. This happens because the image of an object beyond 1.2 m is formed in front of the retina and not at the retina, as shown in the figure.

To correct this defect of vision, he must use a concave lens. The concave lens will bring the image back to the retina is shown in the given figure.

Fig. 11.1 Myopic eyes and their correction.

Question 3.
What is the far point and near point of the human eye with normal vision?
Answer:
Far point of the human eye with normal vision is near than infinity and near point is 1.2 m.

Question 4.
A student has difficulty reading the blackboard while sitting in the last row. What could be the defect the child is suffering from? How can it be corrected?
Answer:
A student has difficulty reading the black board while sitting in the last row means he is suffering from myopia. This defect can be corrected using concave lens of suitable power.

MP Board Class 10th Science Chapter 11 NCERT Textbook Exercises

Question 1.
The human eye can focus objects at different distances by adjusting the focal length of the eye lens. This is due to:
(a) presbyopia
(b) accommodation
(c) near-sightednes
(d) far-sightedness
Answer:
(b) Human eye can change the focal length of the eye lens to see the objects situated at various distances from the eye. This is possible due to the power of accommodation of the eye lens.

Question 2.
The human eye forms the image of an object at its:
(a) cornea
(b) iris
(c) pupil
(d) retina
Answer:
(d) The human eye forms the image of an object at its retina.

Question 3.
The least distance of distinct vision for a young adult with normal vision is about
(a) 25 m.
(b) 2.5 cm
(c) 25 cm
(d) 2.5 m
Answer:
(c) 25 cm

Question 4.
The change in focal length of an eye lens is caused by the action of the
(a) pupil.
(b) retina
(c) ciliary muscles
(d) iris
Answer:
(c) ciliary muscles

Question 5.
A person needs a lens of power – 5.5 dioptres for correcting his distant vision. For correcting his distant vision. For correcting his near vision he needs a lens of power + 1.5 dioptre. What is the focal length of the lens required for correcting (i) distinct vision, and (ii) near vision?
Answer:
i) Lens required for correcting
distant vision = – 5.5
Focal length of lens F \(=\frac{1}{P}\)
\(\mathrm{F}=\frac{1}{-5.5}=0.181 \mathrm{m}\)
Lens required for correcting this defect =-0.181 M
ii) Lens required for correcting near vision = + 1.5 D
Focal length of lens F \(=\frac{1}{P}\)
\(F=\frac{1}{1.5}=0.667 \mathrm{m}\)

Question 6.
The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?
Answer:
The person is suffering from an eye defect called myopia. In this defect, the image is formed in front of the retina. Hence, a concave lens is used to correct this defect of vision.
Object-distance, u = infinity
Image-distance, v = -80 cm
Focal length = f
According to the lens formula,

A concave lens of Power – 1.25 D is required by the person to correct his defect.

Question 7.
Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.
Answer:
A person suffering from hypermetropia can see distinct objects clearly but faces difficulty in seeing nearby objects clearly. It happens because . the eye lens focuses the incoming divergent rays beyond the retina. This defect of vision is corrected by using a convex lens. A convex lens of suitable power converges the incoming light in such a way that the image is formed on the retina, as shown in the following figure.

Fig. 11.2: Correction for hypermetropic eye.

The convex lens actually creates a virtual image of a nearby object (N’ in the figure) at the near point of vision (N) of the person suffering from hypermetropia. The given person will be able to clearly see the object kept at 25 cm (Near point of the normal eye), if the image of the object is formed at his near point, which is given as 1 m.

Object-distance, u = -25 cm
Image-distance, v = -1m = -100m
Focal length, f = ?

Using the lens formula,
\(\frac { 1 }{ v } \) – \(\frac { 1 }{ u } \) = \(\frac { 1 }{ f } \)
\(\frac { 1 }{ -100 } \) – \(\frac { 1 }{ -25 } \) = \(\frac { 1 }{ f } \)
\(\frac { 1 }{ f } \) = \(\frac { 1 }{ 25 } \) = \(\frac { 1 }{ f } \)
\(\frac { 1 }{ f } \) = \(\frac { 1 }{ 25 } \) – \(\frac { 1 }{ 100 } \)
\(\frac { 1 }{ f } \) = \(\frac { 4-1 }{ 100 } \)
f = \(\frac { 100 }{ 3 } \) = 33.3 cm = 0.33cm

A convex lens of power + 3.0 D is required to correct the defect.

Question 8.
Why is a normal eye not able to see clearly the objects placed closer than 25 cm?
Answer:
The maximum accommodation of a normal eye is reached when the object is at a distance of 25 cm from the eye. The focal length of the eye lens cannot be decreased below this minimum limit. Thus an object placed closer than 25cm [or very close to eye] cannot be seen clearly by a normal eye.

Question 9.
What happens to the image distance in the eye when we increase the distance of an object from the eye?
Answer:
The distance eye lens and retina is the image distance inside the eye. The image distance is fixed. It cannot be changed at all. Therefore, when we increase the distance of an object from the eye, there is no change in the image distance inside the eye.

Question 10.
Why do stars twinkle?
Answer:
Stars emit their own light and they twinkle due to the atmospheric refraction of light. Stars are very far away from the earth. Hence, they are considered as point sources of light. When the light coming from stars enters the earth’s atmosphere, it gets refracted at different levels because of the variation in the air density at different levels of the atmosphere. When the star’s light refracted by the atmosphere comes more towards us, it appears brighter than when it comes less towards us. Therefore, it appears as if the stars are twinkling at night.

Question 11.
Explain why the planets do not twinkle.
Answer:
Planets do not twinkle because they appear larger in size than the stars as they are relatively closer to earth. Planets can be considered as a collection of a large number of point-size sources of light. The different parts of these planets produce either brighter or dimmer effect in such a way that the average of brighter and dimmer effect is zero. Hence, the twinkling effects of the planets are nullified and they do not twinkle.

Question 12.
Why does the Sun appear reddish early in the morning?
Answer:
Light from the Sun near the horizon passes through thicker layers of air and larger distance in the earth’s atmosphere before reaching our eyes. However, light from the Sun overhead would travel relatively shorter distance. At noon, the Sun appears white as only a little of the blue and violet colours are scattered. Near the horizon, most of the blue light and shorter wavelengths are scattered away by the particles. Therefore, the light that reaches our eyes is of longer wavelengths. This gives rise to the reddish appearance of the Sun.

Question 13.
Why does the sky appear dark instead of blue to an astronaut?
Answer:
When sunlight passes through the atmosphere, the fine particles in air scatter the blue colour (shorter wavelength more strongly than red. The scattered blue light enters our eyes. If the earth had no atmosphere, there would not have been any scattering. Then the sky would have looked dark. The sky appears dark to astronaut flying at very high attitude, as scattering is not prominent at such heights.

MP Board Class 10th Science Chapter 11 Additional Important Questions

MP Board Class 10th Science Chapter 11 Multiple Choice Questions

Question 1.
When our eye see a near object, its lens?
(a) Expand
(b) Remain same
(c) Contract
(d) None
Answer:
(a) Expand

Question 2.
What transfer information from eyes to brain?
(a) Neuro trans
(b) Iris
(c) Optical nerve
(d) Retina
Answer:
(c) Optical nerve

Question 3.
What controls the size of pupil?
(a) Iris
(b) Cornea
(c) Lens
(d) Aqueous humour
Answer:
(a) Iris

Question 4.
Near point of human eye is approximately:
(a) 50 cm
(b) 1 m
(c) 25 cm
(d) 25 m
Answer:
(c) 25 cm

Question 5.
Far point of human eye is approximately:
(a) 10 km
(b) infinity
(c) 5 km
(d) 25 cm
Answer:
(a) 10 km

Question 6.
Cataract is caused mostly to
(a) Farm workers
(b) Factory workers
(c) Kids watching TV and Mobiles
(d) Old age people
Answer:
(d) Old age people

Question 7.
In which kind of defect, image of a distant object is formed in front of retina?
(a) Myopia
(b) Cataract
(c) Hypermetropia
(d) Presbyopia
Answer:
(a) Myopia

Question 8.
A concave lens with correct power can help a person with
(a) Hypermetropia
(b) Myopia
(c) Presbyopia
(d) None of these
Answer:
(b) Myopia

Question 9.
Adjusting “cording to object’s distance from eye is called
(a) Vision power
(b) Motivation
(c) Accommodation power
(d) Presbyopia
Answer:
(c) Accommodation power

Question 10.
For an eye donation, eyes must be removed within ……… after death.
(a) 8 hours
(b) 25 hours
(c) 4-6 hours
(d) None.
Answer:
(c) 4-6 hours

Question 11.
What type of lens are required to correct a vision of hypermetropic eyes?
(a) Concave
(b) Convex
(c) both
(d) Plane mirror is required ;
Answer:
(b) Convex

Question 12.
When we pass a light through a prism, it is
(a) Refracted
(b) Scattered
(c) Reflected
(d) Inverted
Answer:
(a) Refracted

Question 13.
Which colour lies in centre of a rainbow?
(a) Red
(b) Violet
(c) Yellow
(d) Green
Answer:
(d) Green

Question 14.
If two prism are placed inverted to each other and white light is passed from one prism, what will we get refracted by other prism?
(a) VIBGYO
(b) white light
(c) All light will be absorbed
(d) Black spot
Answer:
(b) white light

Question 15.
We see twinkling of stars due to:
(a) Reflection of light.
(b) Refraction of light
(c) Scattering of light
(d) Polarization of light
Answer:
(b) Refraction of light

Question 16.
Particle looks moving in smoke filled place, when a light beam coming from a whole passes through it, because:
(a) Tyndall effect
(b) atmospheric refraction
(c) White light spectrum
(d) reflection
Answer:
(a) Tyndall effect

MP Board Class 10th Science Chapter 11 Very Short Answer Type Questions

Question 1.
Where image is formed in our eyes?
Answer:
Retina.

Question 2.
Which part of eye helps in adjustment of lens according to distance?
Answer:
Lens.

Question 3.
Which nerve transfer information regarding image to brain?
Answer:
Optic nerve.

Question 4.
What is power of accommodation?
Answer:
The ability of the eye to focus both near and distant objects, by adjusting its focal length is called power of accommodation of eyes.

Question 5.
How many points of accommodation are set to observe clarity of a person’s vision?
Answer:
Two-far point and near point.

Question 6.
What is the maximum distance till which a person with normal eye can see?
Answer:
No limit, infinite distance.

Question 7.
Name common refractive defects of eyes.
Answer:
Myopia, Hypermetropia and Presbyopia.

Question 8.
What kind of lens is used to correct a myopic defect in a person?
Answer:
Concave.

Question 9.
What kind of treatment is available to correct a cataract?
Answer:
Surgery.

Question 10.
What is bifocal lens?
Answer:
Lens made by both concave and convex lens is called bifocal lens.

Question 11.
What happens to the light ray entering a prism?
Answer:
The light ray get refracted.

Question 12.
What happens if white light is passed through a prism?
Answer:
It Scatters to seven colours spectral pattern.

Question 13.
What is an emergent ray?
Answer:
Ray which comes out after being refracted from a prism is called emergent ray.

Question 14.
Why sky looks red at sunrise and sunset?
Answer:
Scattering of white light.

Question 15.
Which colour light bend most after coming out of a triangular prism?
Answer:
Violet.

Question 16.
Why we never see a shadow without light?
Answer:
When light passes through an object, it get partially deviated or absorbed so, disappear from that very way which forms a shadow.

Question 17.
Why light ray in a dusty room look sparkling and moving?
Answer:
Due to tyndall effect.

Question 18.
Why we can observe a star twinkling?
Answer:
Due to atmospheric refraction.

Question 19.
What is the time difference between actual and observed sunrise and sunset?
Answer:
2 minutes.

Question 20.
Why sky looks blue to all of us?
Answer:
Due to scattering of light.

MP Board Class 10th Science Chapter 11 Short Answer Type Questions

Question 1.
Why does the Sun appear reddish early in the morning?
Answer:
Early in the morning, the sun is near the horzion. Sunlight reaches us after covering a large thickness of the atmosphere. So, shorter waves of blue region are almost completely scattered away by the air molecules. Red waves of longer wavelength are least scattered and reach our eyes. The sun appears red.

Question 2.
Why does the sky appear dark instead of blue to an astronaut?
Answer:
The atmosphere is quite thin at very high altitudes. There is almost no scattering of sunlight. So, the sky appears dark to an astronaut.

Question 3.
Why is a concave lens used to correct myopia or short-sightedness?
Answer:
A concave lens of suitable focal length diverges the parallel rays from the distant object as if they are coming from the far point F of the myopic eye. This helps the eye lens to form a clear image at the retina.

Question 4.
Dispersion is caused by refraction not by reflection. Why?
Answer:
The reason is that for a given angle of incidence, the angle of reflection is same for all the wavelengths of white light while the angle of refraction is different for different wavelengths.

Question 5.
A hypermetropic person prefers to remove his spectacles, while driving. Give reason.
Answer:
When a hypermetropic person wearing the spectacles looks at a distant object, the parallel rays from the distant object get converged in front of the retina. The image appears blurred. In order to avoid this, the person prefers to remove his spectacles.

Question 6.
Give reasons for the following:

1. The stars appear to twinkie.
2. The planets do not twinkle.

Answer:

1. Stars appear to twinkle due to atmospheric refraction of starlight and physical conditions of the earth’s atmosphere are not being stationary.
2. Planets are much closer to the earth and are seen as extended sources. The fluctuations caused in the amount of light due to atmospheric refraction are negligible. Hence, planets do not twinkle.

MP Board Class 10th Science Chapter 11 Long Answer Type Questions

Question 1.
Why does the Sun seem to rise two minutes before the actual sunrise and set two minutes after the actual sunset? Explain with the help of labelled diagram.
Answer:
Advance sunrise and delayed sunset. Apparent shift in the position of sun at sunrise and sunset:
The sun is visible before actual sunrise and after actual sunset, because of atmospheric refraction. With altitude, the density and hence, refractive index of air-layer decreases. As shown in Fig. given the light rays starting from the sun travel from rarer to denser layers. They bend more and more towards the normal. To an observer on the earth, light rays appear to come from position S. The sun which is actually in position S below the horizon appears in position S above the horizon.

Fig. 11.6: Atmospheric refraction effect at sun rise.

Thus, the sun appears to rise early by about two minutes and for the same reason, it appears to set late by about two minutes. This increases the length of the day by about four minutes.

Question 2.
“Stars appear higher than they actually are.” Give reason.
Or
Is the position of a star as seen by us its true position? Justify your answer.
Answer:
Since, the atmosphere bends starlight towards the normal, the apparent position of the star is slightly different from its actual position. The stars appear slightly higher (above) than their actual position when viewed near the horizon.

Question 3.
Why do we observe the apparent random wavering or flickering of objects when seen through a turbulent stream of hot air rising above fire, a stove or radiator?
Answer:
This is due to atmospheric refraction i.e., refraction of light by the earth’s atmosphere. The air just above the fire becomes hotter than air further up. Hotter air is lighter (less denser) than the cooler air (denser) above it. This causes refraction of light due to decrease of refractive index with decreasing density or increasing temperature. Since, the physical conditions of the refracting medium (air) are not stationary, the apparent positions of the objects, as seen through the hot air, fluctuate. Consequently, the objects seen through such air show a wavering effect.

Question 4.
Why do we see stars appear twinkling whereas planets do not twinkle?
Or
A star sometimes appear brighter and some other times fainter. What is this defect called, state reason for this effect.
Answer:
Twinkling of stars: Differential the apparent position of a star is slightly different from the actual position due to refraction of starlight by the atmosphere. Further, this apparent position is not stationary but keeps on changing due to the change in atmospheric conditions like density, temperature etc. The path of the rays of light coming from the star goes on varying slightly. The amount of light entering our eyes from a particular star increases or decreases randomly with time.

Fig. 11.7: Apparent star position due to atmospheric refraction.

Sometimes, the star appears bright and other times, it appears fainter. This gives rise to the twinkling effect of the star.

The planets do not show twinkling effect. As the planets are much closer to the earth, the amount of light received from them is much greater and the fluctuations caused in the amount of light due to atmospheric refraction are negligible as compared to the amount of light received from them.

Question 5.
What is tyndall effect? What is its cause? Name two phenomena observed in daily life which are based on Tyndall effect.
Or
A beam of light is allowed to pass through two beakers A and B containing a true solution and colloidal solution respectively. What do you observe? Name the phenomenon responsible for your observation.
Answer:
Tyndall effect: When a beam of light is passed through a colloidal solution, placed in a dark room, the path of beam becomes illuminated (or visible), when observed through a microscope placed perpendicular to the path of light. This effect is called Tyndall effect.

On the other hand, the path of a beam of light is not visible through a true solution, as shown in figure 11.8.

Fig. 11.8: Tyndall effect.

Cause of Tyndall effect: The size of the colloidal particle is relatively larger than the solute particle of a true solution. The colloidal particles first absorb energy from the incident light and then scatter a part of this energy from their surfaces. Thus, Tyndall effect is due to scattering of light by the colloidal particles and the colloidal particles are seen as points of light moving against a dark background.

Some daily life phenomena based on Tyndall effect are as follows:

1. When a fine beam of sunlight enters a smoke filled room through a small hole, the smoke particles become visible due to the scattering of light.
2. When sunlight passes through a canopy of a dense forest, the tiny water droplets in the mist scatter light and become visible.

MP Board Class 10th Science Chapter 11 NCERT Textbook Activities

Class 10 Science Activity 11.1 Page No. 192

• Fix a sheet cf white paper on a drawing board using drawing pins.
• Fife a glass prism on it in such a way that it rests on its triangular fee. Trace the outline of the prism using a pencil.
• Draw a straight line PE inclined to one of the refracting surfaces, say AB, of the prism.
• Fix two pins, say at points P and Q, on the line PE as shown in Fig. 11.3.
• Look for the images of the pins, fixed at P and Q, through the other face AC.
• Fix two more pins, at points R and S, such that the pins at R and S and the images of the pins at P and Q lie on the same straight line.
• Remove the pins and the glass prism.
• The line PE meets the boundary of the prism at point E (see Fig. 11.3). Similarly, join and produce the points R and S. Let these lines meet the boundary of the prism at E and F, respectively. Join E and F.
• Draw perpendiculars to the refracting surfaces AB and AC of the prism at points E and F, respectively.
• Mark the angle of incidence (∠i), the angle of refraction (∠r) and the angle of emergence (∠e) as shown in Fig. 11.3.

PE – Incident ray – ∠i – Angle of incidence
EF – Refracted ray ∠r – Angle of refraction
FS – Emergent ray ∠e – Angle of emergence
∠A – Angle of the prism – ∠D – Angle of deviation

Fig. 11.3 : Refraction of tight through a triangular glass prism

Observations:

• The light rays enter the prism and emerge out as the emergent ray. The light ray bent towards the normal upon refraction. At the second surface, the light ray enters from glass to air. The emergent ray appears to be coming along the rays FRS, the angle between the incident ray produced forward and the emergent ray produced backward, is called the angle of deviation.

Class 10 Science Activity 11.2 Page No. 193

• Take a thick sheet of cardboard and make a small hole or narrow slit in its middle.
• Allow sunlight to fall on the narrow slit. This gives a narrow beam of white light.
• Now, take a glass prism and allow the light from the slit to fall on one of its faces as shown in Fig. 11.4.
  
  Fig. 11.4: Dispersion of white light by the glass prism.
• Turn the prism slowly until the light that comes out of it appears on a nearby screen.
• What do you observe? You will find a beautiful band of colours. Why does this happen?

Observations:

• The white light gets dispersed into seven colour components by a prism. The band of the coloured components of a light beam is called spectrum. The splitting of light into its component colours is called dispersion. The colours bends at different angles as they pass through prism and are observed separately as VIBGYOR.

Class 10 Science Activity 11.3 Page No. 196

• Place a strong source (S) of white light at the focus of a converging lens (L). This lens provides a parallel beam of light.
• Allow the light beam to pass through a transparent glass tank containing clear water.
• Allow the beam of light to pass through a circular hole (e) made in a cardboard. Obtain a shaip image of the circulai hole on a screen (MN) using a second converging lens (L₂), as shown in Fig. 11.5
  
  Fig. 11.5: An arrangement for observing scattering of light in colloidal solution.
• Dissolve about 200 g of sodium thiosulphate (hypo) in about 2 L of clean water taken in the tank. Add about 1 to 2 mL of concentrated sulphuric acid to the water. What do you observe?

Observations:

• The microscopic particles of sulphur precipitate, in about 2 to 3 minutes. As they precipitate, the blue light from the three sides of the glass tank is seen. This happens due to scattering of short wavelengths by minute colloidal sulphur particles.

MP Board Class 10th Science Solutions

MP Board Class 10th Science Solutions Chapter 9 Heredity and Evolution

MP Board Class 10th Science Chapter 9 Intext Questions

Class 10th Science Chapter 9 Intext Questions Page No.143

Question 1.
If a trait Aexists in 10% of a population of an asexually reproducing species and trait B exists in 60% of the same population, which trait is likely to have arisen earlier?
Answer:
Trait ‘B’ is likely to have arisen earlier. Because in asexually reproducing species, small differences are seen due to DNA replication.

Question 2.
How does the creation of variations in a species promote survival?
Answer:
Depending on the nature of variations, different individuals would have different kinds of advantages. Bacteria that can withstand heat will better in a heat wave, as selection of variants by environmental factors forms the basis for evolutionary process.

Class 10th Science Chapter 9 Intext Questions Page No. 147

Question 1.
How do Mendel’s experiments show that traits may be dominant or recessive?
Answer:
Mendel used a number 0 contrasting visible characters of garden peas-round/wrinkled seeds, tall/short plants, white/violet flowers and so on. He took pea plants with different characteristics a tall plant and a shot plant, produced progeny by crossing then, and calculated the percentages of tall or short progeny.

In the first place, there were no halfway characteristics in this first generation, or F₁ progeny no medium-height plants. All plants were tall. This means that only one of the parental traits was seen, not some mixture of the two, so the next question was were the tall plants in the F₁ generation exactly the same as the tall plants of the parent generation? Mendelian experiments test this by getting both the parental plans and these F₁ tall plants to reproduce by self-pollination.

The progeny of the parental plants are of course, all tall. However the second generation, or F₂ progeny of the F₁ tall plants are not all tall. Instead, one quarter of them are short. This indicates that both the tallness and shortness traits were inherited in the F₁ plants, but only the tallness trait was expressed. This led Mendel to propose that two copies of factor (now called genes) controlling traits are present in sexually reproducing organism. These two may be identified or may be different, depending on the percentage. A pattern of inheritance can be worked out with this assumption as shown in figure.

In this explanation, both TT and Tt are tall plants, while only it is a short plant. Traits like T are called dominant traits, while those that behave like ‘t’ are called recessive traits.

Question 2.

How do Mendel’s experiments show that traits are inherited independently?
Answer:
Mendel crossed

1. pure breeding tall plants having round seeds and
2. pure breeding short plants having wrinkled seeds.

The plants of F1 generation were all tall with round seeds indicating that the traits of tallness and round seeds were dominant.

While self breeding, of F1 yielded plants with characters of 9 tall round seeded, 3 tall wrinkled seeded, 3 short round seeded and one short wrinkled seeded. Tall wrinkled and short round seeded plants are new combinations which can develop only when the traits are inherited independently.

Question 3.
A man with blood group A marries a woman with blood group C and their daughter has blood group O. Is this information enougl to tell you which of the traits – blood group A or O – is dominant? Why or why not?
Answer:
From this information, it is not possible to tell which of the traits-blood group A or O is dominant. AA group becomes AO. So this information is complete.

Question 4.
How is the sex of the child determined in human beings?
Answer:
Pair of sex chromosomes determines the particular sex of a child. In human, the males have one X and one Y chromosome and the females have two X chromosomes therefore, the females are XX and the males are XY. The gametes receive half of the chromosomes. The child gametes have 22 autosomes and either X or Y sex chromosome in males while X in females.
Type of male gametes: 22 + X Or 22 + Y.
Type of female gamete: 22 + X.
This is the basis of sex determination in human beings.

Class 10th Science Chapter 9 Intext questions Page No. 150

Question 1.
What are the different ways in which individuals with a particular trait may increase in a population?
Answer:
A big reason could be accurate copying of DNA and limited variations. Variation causes in generation of new traits and non preserving of existing parental traits. Individuals with a particular trait may also increase due to natural selection that is trait offers some survival advantage. Genetic drift which is caused by genes governing that trait become common in a population.

Question 2.
Why are traits acquired during the life-time of an individual not inherited?
Answer:
Variation is not hereditary from generation to generation. In case of Asexual reproduction DNA will not transfer to germ cells. So experiences of an individual during its lifetime cannot be passed on to its progeny and cannot direct evolution.

Question 3.
Why are the small numbers of survivingjigers a cause of worry from the point of view of genetics?
Answer:
Tigers are adopted to their environment as per genes. If tigers number is decreasing number of genes also decrease. So its generation is becoming less.

Class 10th Science Chapter 9 Intext Questions Page No. 151

Question 1.
What factors could lead to the rise of a new species?
Answer:
Variations, non copying of DNA, natural selection, genetic drift and acquisition of traits during the life time of an individual can give rise to new species.

Question 2.
Will geographical isolation be a major factor in the speciation of a self-pollinating plant species? Why or why not?
Answer:
Geographical isolation will not a major factor in the speciation of a self pollinating plant species, because self pollination is taking place in one plant. In cross pollination it is a major factor.

Question 3.
Will geographical isolation be a major factor in the speciation of an organism that reproduces asexually? Why or why not?
Answer:
Geographical isolation will not be a major factor in the speciation of an organism that reproduces asexually because it is a major factor in organisms that reproduces sexually.

Class 10th Science Chapter 9 Intext Questions Page No. 156

Question 1.
Give an example of characteristics being used to determine how close two species are in evolutionary terms.
Answer:
The characteristics in different organisms would be similar because they are inherited from a common ancestor. As an example, consider the fact that mammals have four limbs, as do birds, reptiles and amphibians. The basic structure of the limbs is similar through it has been modified to perform different functions in various vertebrates. Such a homologous characteristic helps to identify an evolutionary relationship between apparently different species.

Question 2.
Can the wing of a butterfly and the wing of a bat be considered homologous organs? Why or why not?
Answer:
The wing of a butterfly and the wing of a bat be is not considered homologous organs because the designs of the two wings, their structure and components are very different.

Question 3.
What are fossils? What do they tell us about the process of evolution?
Answer:
Fossils are the remains of organisms that once existed on earth.

We can gather information about the development of the structures from simple structured to complex structured organisms. They tell us about the phases of evolutions through which they must have undergone in order to sustain themselves in the competitive environment.

Class 10th Science Chapter 9 Intext Questions Page No. 158

Question 1.
Why are human beings who look so different from each other in terms of size, colour and looks said to belong to the same species?
Ans.
A species is a group of organisms that are capable of interbreeding to produce a fertile offspring. Skin colour, looks and size are all variety of features present in human beings. These features are genetic but also environmentally controlled. Various human races are formed based on these features. All human races have more than enough similarities to be classified as same species. Therefore, all human beings are a single species as humans of different colour, size and looks are capable of reproduction and can produce a fertile off spring.

Question 2.
In evolutionary terms, can we say which among bacteria, spiders, fish and chimpanzees have a ‘better’ body design? Why or why not?
Answer:
Bacteria is a better body design, because even though it a simple organism, it can survive in hot springs, at the bottom of sea and even in coldest ice covered place such as Antarctica.

MP Board Class 10th Science Chapter 9 NCERT Textbook Exercises

Question 1.
A Mendelian experiment consisted of breeding tall pea plants bearing violet flowers with short pea plants bearing white flowers. The progeny all bore violet flowers, but almost half of them were short. This suggests that the genetic make-up of the tall parent can be depicted as:
(a) TTWW
(b) TTww
(c) TtWW
(d) TtWw
Answer:
(c) TtWW. Genetic make up of tall plant.

Question 2.
An example of homologous organs is:
(a) our arm and a dog’s fore-leg.
(b) our teeth and an elephant’s tusks.
(c) potato and runners of grass.
(d) all of the above.
Answer:
(d) Both organs in all option have same basic structural design but have different functions and appearance.

Question 3.
In evolutionary terms, we have more in common with:
(a) a Chinese school-boy
(b) a chimpanzee
(c) a spider
(d) a bacterium
Answer:
(a) a Chinese school-boy.

Question 4.
A study found that children with light-coloured eyes are likely to have parents with light-coloured eyes. On this basis, can we say anything about whether the light eye colour trait is dominant or recessive? Why or why not?
Answer:
No, we cannot say anything about whether the light eye colour trait is dominant or recessive. As this information is not sufficient. For considering a trait as dominant or recessive, we need data of at least last three generations.

Question 5.
How are the areas of study – evolution and classification – interlinked?
Answer:
An example will help with this A brother and a sister are closely related. They have common ancestors in the first generation before them, namely their parents. A girl and her first cousin are also related, but less than the girl and her brother. This is because cousins have common ancestors, their grandparents, in the second generation before them, not in the first one. We can now appreciate that classification of species is in fact a reflection of their evolutionary relationship.

Question 6.
Explain the terms analogous and homologous organs with examples.
Answer:

Consider the fact that mammals have four limbs, as do birds, reptiles and amphibians. The basic structure of the limbs is similar though it has been modified to perform different functions in various vertebrates. This is an example of homologous characteristic.

We find that the wings of bats are skin folds stretched mainly between elongated fingers. But the wings of birds are feathery covering all along the arm. The design of the two wings, their structure and components, are thus very different. They look similar because they have a common use for flying, but their origins are not common. This makes them analogous characteristics, rather than homologous characteristics.

Question 7.
Outline a project which aims to find the dominant coat colour in dogs.
Answer:
There are variety of genes that govern coat colour of a dog. At least eleven identified gene series (A, B, C, D, E, F, G, M, P, S, T) that influence coat colour in dog.

A dog inherits one gene from each of its parents. The dominant gene gets expressed in the phenotype. For example, in the B series, a dog can be genetically black or brown.

Let us assume that one parent is homozygous black (BB), while the other parent is homozygous brown (bb)

In this case, all the off springs will be heterozygous (Bb).
Since black (B) is dominant, all the offsprings will be black. However, they will have both B and b alleles.
If such heterozygous pups are crossed, they will produce 25% homozygous black (BB), 50% heterozygous black (Bb), and 25% homozygous brown (bb) offsprings.

Question 8.
Explain the importance of fossils in deciding evolutionary relationships.
Answer:
Analysis of the organ structure in fossils allows us to make estimates of how far back evolutionary relationships go. The wild cabbage plant is a good example. Humans have over more two thousand years, cultivated wild cabbage as a food plant, and generated different vegetables from it by selection. This is of course, artificial selection rather than natural selection. Kale, cauliflower. Broccoli, cabbage, Red cabbage and Kohl rabi all these have same ancestor.

Question 9.
What evidence do we have for the origin of life from inanimate matter?
Answer:
As we all know, that there occurs a time when our planet was lifeless, at that time intial matter to develop life was water, sand and some atmospheric gases as CO₂, CH4 and nitrogen. The evidence for the origin of life from inanimate matter, was provided through an experiment, conducted in 1953, by Stanley L. Miller and Harold C. Urey. In experiment, they assembled an atmosphere containing molecules like ammonia, methane and hydrogen sulphide, but not oxygen.

This was similar to atmosphere that existed on early earth . This was maintained at a temperature just below 100°C and sparks were passed through the mixture of gases to simulate lightning. At the end of a week, 15% of the carbon from methane, had been converted to simple compounds of carbon including amino acids which make up protein molecules and support the life in basic form.

Question 10.
Explain how sexual reproduction gives rise to more viable variations than asexual reproduction. How does this affect the evolution of those organisms that reproduce sexually?
Answer:

Change in on-reproductive tissues cannot be passed on to the DNA of the germ cell. Therefore the experiences of an individual during its lifetime cannot be passed on to its progeny and cannot direct evolution.

Ex: If we breed a group of mice, all their progeny will have tails, as expected. Now, if the tails of these mice are removed by surgery in each generation, do these tailless mice have tailless progeny? The answer is no and it makes sense because removal of the tail cannot change the genes of the germ cells of the mice. Hence sexual reproduction gives rise to more viable variations than asexual reproduction.

Question 11.
How is the equal genetic contribution of male and female parents ensured in the progeny?
Answer:
Genetic inheritance begins at the time of conception, progeny inherited 23 chromosomes from female parent and 23 from male parent. Together it form 22 pairs of autosomal chromosomes and a pair of sex chromosomes (either XX in case of female, or XY in male). Homologous chromosomes have the same genes in the same positions, but may have different alleles (varieties) of those genes. An individual has two copies of alleles, and that can be homozygous (both copies the same) or heterozygous (the two copies are different) for given gene.

Hence, in human beings, equal genetic contribution of male and female parents is ensured in the progeny through inheritance of equal number of chromosomes from both parents. Females have a equal pair of two X sex chromosomes and males have a pair of one X and one Y sex chromosome. As fertilisation takes place, the male gamete (haploid) fuses with the female gamete (haploid) resulting in formation of the diploid zygote. The zygote in the progeny receive an equal contribution of genetic traits from the parental generations.

Question 12.
Only variations that confer an advantage to an individual organism will survive in a population. Do you agree with this statement? Why or why not?
Answer:
Only variations that confer an advantage to an individual organism will survive in a population we agree to this statement variation is convenient for survival. This provides diversity for organisms.

MP Board Class 10th Science Chapter 9 Additional Important Questions

MP Board Class 10th Science Chapter 9 Multiple Choice Questions

Question 1.
Fossil archaeopteryx exhibits connection between:
(a) Amphibian and fish
(b) Reptiles and fish
(c) Reptile and birds
(d) Birds and mammals
Answer:
(c) Reptile and birds

Question 2.
The sex of the human child depends on the sex chromosome present in the:
(a) Egg
(b) Spenn
(c) Both (a) and (b)
(d) None of these
Answer:
(b) Spenn

Question 3.
Genetic information is carried out by long chain of molecules made up of:
(a) Enzymes
(b) DNA
(c) Amino acids
(d) Proteins
Answer:
(b) DNA

Question 4.
Which one of the following represents a ratio of monohybrid cross?
(a) 9 : 7
(b) 3 : 1
(c) 1 : 1 : 1 : 1
(d) 9 : 3 : 3 : 1
Answer:
(b) 3 : 1

Question 5.
On which plant Mendel carried his experiments of inheritance?
(a) Cow pea
(b) Wild pea
(c) Garden pea
(d) Pigeon pea
Answer:
(c) Garden pea

Question 6.
A gamete certains which of the following?
(a) Both alleles of a gene
(b) Only one allele of a gene
(c) All alleles of a gene
(d) No allele of a gene
Answer:
(b) Only one allele of a gene

Question 7.
Chromosomes are made up of
(a) Proteins
(b) DNA
(c) RNA
(d) All of these.
Answer:
(d) All of these.

Question 8.
Pea plants were more suitable than cats for Mendel’s experiments because:
(a) Cats have many genetic traits
(b) No pedigree record of cats
(c) Pea plants can be self-pollinated or fertilised
(d) Pea plants favour cross pollination.
Answer:
(c) Pea plants can be self-pollinated or fertilised

Question 9.
In a cross Tt × Tt, the percentage of offsprings produced having same phenotype as the parents would be:
(a) 50%
(b) 100%
(c) 25%
(d) 0%
Answer:
(a) 50%

Question 10.
Who proposed the laws of heredity?
(a) Darwin
(b) Mendel
(c) Morgan
(d) Dalton
Answer:
(b) Mendel

MP Board Class 10th Science Chapter 9 Very Short Answer Type Questions

Question 1.
Define heredity.
Answer:
The process by which traits and characteristics are reliably inherited or passed from the parents to the offspring is called heredity.

Question 2.
What is a gene?
Answer:
Gene is a functional segment of DNA on a chromosome occupying specific position, which carries out a specific biological function.

Question 3.
Name the plant on which Mendel performed his experiments.
Answer:
Garden pea (Pisum sativum).

Question 4.
Define the term variation.
Answer:
Variation: There are differences found in structure, function, behaviour and genetic make up of different individuals of the same parentage, variety, race and species. These differences refer to variation.

Question 5.
Write the expanded form of DNA.
Answer:
Deoxyribonucleic acid (DNA).

Question 6.
Define genetics.
Answer:
The branch of biology which deals with heredity and variations, is known as genetics.

Question 7.
Define the term offspring.
Answer:
Offspring is an individual formed as a result of sexual reproduction involving the formation and fusion of two gametes. The genotype of an offspring is different from either of the parents due to shuffling of chromosomes and their genes.

Question 8.
What are reciprocal crosses?
Answer:
They are two types of crosses involving two groups of individuals where the male of one group is crossed with the female of the other and vice versa.

Question 9.
Where are the genes located? What is the chemical nature of gene?
Answer:
Genes are located at a specific position on a chromosome. Chemical Nature of Gene: Chemically, gene is a segment of deoxyribonucleic acid (DNA) consisting of specific sequence of the nucleotides. The sequence of the constituent nucleotides determines the functional property of a gene.

MP Board Class 10th Science Chapter 9 Short Answer Type Questions

Question 1.
Define genetics. What is the contribution of Mendel in this branch of Biology?
Answer:
Genetics is the branch of science of heredity and variations which deals with the study of the transmission of traits from parents to the offsprings and the occurrence of differences among the individuals.

Contribution of Mendel: Mendel did his experiments on garden pea (Pisum sativum) and discovered the scientific principles, which govern patterns of inheritance i.e., the principle of inheritance. He explained that contrasting characters are controlled by units which he called ‘Factors Today, these factors are called genes.

Question 2.
Differentiate between inherited and acquired traits.
Answer:
Inherited traits:

1. The traits which are inherited from the parents (Father and Mother) to the offsprings (progeny) are called inherited traits.
2. These traits are due to genetic make up of the progeny.

Acquired characters

1. These traits cannot be passed on to their future generations.
2. These traits develop in response to the environment.

Question 3.
What are Mendel’s laws of inheritance?
Answer:
Law of dominance: When two homozygous individuals with one or more sets of contrasting characters are crossed the characters that appear in the F, hybrids are dominant characters.

Law of segregation: Contrasting characters brought together in hybrid remain together without being contaminated and when gametes are formed from the hybrid, the two separate out from each other and only one enters each gamete.

Law of independent assortment: In inheritance of more than one pair of contrasting characters simultaneously, the factors for each pair of characters assort independently of other pair’s.

Question 4.
How did life originate on earth?
Answer:
Life originated on earth from inorganic elements and compounds under extreme atmospheric conditions (such as very high temperature, electric discharges, reducing atmosphere etc.) by formation of complex organic compounds such as amino acids.

Question 5.
Why did Mendel choose garden pea for his experiments?
Answer:
Due to the following reasons, Mendel selected garden pea for his experiment:

1. Garden pea flowers are normally self-pollinated but can be easily cross-pollinated.
2. Many varieties with distinguished contrasting characters e.g., smooth seed coat, wrinkled seed coat are available.
3. A large number of progeny can be produced in a short duration.
4. Its flowers can be easily handled for experimentation.

Question 6.
What are the factors which help in speciation?

1. Genetic drift: Due to genetic drift, there will be accumulation of # different changes in each sub-populations. The levels of gene flow ’ between them will decrease if they are further isolated, it will be more on a small sub-population.
2. Over generations, genetic drift will accumulate, causing different changes in the populations.
3. Natural selection may also operate differently in the different geographical location.
4. Together, genetic and natural selection will make the population more and more different from each other. As a result, members will be incapable of reproducing with each other. Changes may be due to change in DNA or number of chromosomes.

Question 7.
Does geographical isolation of individuals of a species lead to formation of a new species? Provide a suitable explanation.
Answer:
Yes, geographical isolation of sub-populations of a population of a species leads to genetic drift. This may impose limitations to.sexual reproduction of the separated population. Slowly, the separated individuals will reproduce among themselves and generate new variations. Continuous accumulation of those variations through a few generations may ultimately lead to the formation of a new species.

Question 8.
What tools have been used to study human evolution?
Answer:
The tools used for tracing evolutionary line are:

1. Excavating time – dating and study of fossils.
2. Determining DNA sequences.

MP Board Class 10th Science Chapter 9 Long Answer Type Questions

Question 1.
A husband have 46 chromosomes, his wife has 46 chromosomes. Then, why don’t their offspring have 46 pairs of chromosomes, which is obtained by the fusion of male and female gametes? Support your answer with a neat illustration.
Answer:
At the stage of gamete formation, meiosis division (reduction division) occurs. As a result, each gamete receives half number of chromosomes of the parent. So, when male gamete (sperm) fuses with egg, original number of chromosomes of the parent is received by the zygote.

Fertilisation

Question 2.
Name the characters studied by Mendel in garden pea.
Answer:

Question 3.
Explain the terms:
Monohybrid cross, dihybrid cross, monohybrid ratio and dihybrid ratio.
Answer:
Monohybrid cross: Monohybrid cross is that cross which is made to study the inheritance of a single pair of genes or factors of a character.

Dihybrid cross: It is a cross which is made to study the inheritance of two pairs of genes or two characters.

Monohybrid ratio: It is the ratio which is obtained in the F2 generation when a monohybrid cross is made. It is usually 3 : 1 (Phenotypic ratio) or 1 : 2 : 1 (genotypic ratio).

Dihybrid ratio: It is the ratio, which is obtained in the F2 generation when a dihybrid cross is studied. It is usually 9 : 3 : 3 : 1 (phenotypic ratio).

Question 4.
Describe any three methods of tracing evolutionary relationships among organisms.
Answer:
The following methods help us in tracing evolutionary relationships:

(i) Study of homologous organs: Organs which have similar structure and origin are called homologous organs. For example: limbs of birds, frog, human may look different but they have similar structure and origin. Such homologous organs help to identify an evolutionary relationship between apparently diffejent species.

(ii) Study of analogous organs: Analogous organs are similar in function but differ in structure and origin. For example’. forelimbs of birds and bats are used for flying but their origins and components are not common. Thus, study of analogous organs reveals difference in their ancestry and their evolutionary relationship.

(iii) Study of fossils: All impressions, casting of body or hard remains of ancient life in the sedimentary rocks are called fossils. Study of fossils helps in finding out:
(a) Interrelationship of ancient life.
(b) Correlation of forms of life existing today and their line of evolution from ancient life.

Question 5.
A tall pea plant bearing violet flowers is crossed with short pea plant bearing white flowers. Work out the F1 and F2 generations. Give F2 ratio.
Answer:
Parents: Tall pea plant with violet flower × Short pea plant with white flower

Question 6.
Given below is the experiment carried out by Mendel to study inheritance of two traits in garden pea:
(а) What do A, B, C, D, E, F and G represent in these boxes?
(b) State the objective for which Mendel performed this experiment.

Independent inheritance of two separate traits, shape, and colour of seeds.
Answer:
(a) A = gamete (Ry) of round green plant.
B = gamete (rY) of wrinkled yellow plant.
C = (RrYy).
D = 9, E = 3, F = 3, G = 1.
(b) To show independent inheritance of traits or to prove law of independent assortment.

MP Board Class 10th Science Chapter 9 NCERT Textbook Activities

Class 10 Science Activity 9.1 Page No. 143

Observe the ears of all the students in the class. Prepare a list of students having free or attached earlobes and calculate the percentage of students having each (Fig. 9.1). Find out about the earlobes of the parents of each student in the class. Correlate the earlobe type of each student with that of their parents. Based on this evidence, suggest a possible rule for the inheritance of earlobe types.

(a) Free and (b) attached earlobes. The lowest part of the ear, called the earlobe, is closely attached to the side of the head in some of us, and not in others. Free and attached earlobes are two variants found in human populations.

Observations:

• Earlobes can be free or attached. The genes for earlobe inheritance consists of two alleles. Both the alleles for attached and free earlobes can be present in a single human being, the one which is dominant shows and the recessive one do not express itself,

Class 10 Science Activity 9.2 Page No. 144

• In Fig. 9.2, what experiment would we do to confirm that the F2 generation did in fact have a 1 : 2 : 1 ratio of TT, Tt and tt trait combinations?

Inheritance of traits over two generations.

Observations:
The cross fertilisation of pea plants showing different traits can be done at F2 stage and the number of plants for particular trait (height here) can be studied and used to confirm 1 : 2 : 1 ratio of TT, Tt and it

MP Board Class 10th Science Solutions

MP Board Class 10th Science Solutions Chapter 14 Sources of Energy

MP Board Class 10th Science Chapter 14 Intext Questions

Class 10th Science Chapter 14 Intext Questions Page No. 243

Question 1.
What is a good source of energy?
Answer:
We could then say that a good source of energy would be one.

1. Which would do a large amount of work per unit volume or mass.
2. be easily accessible.
3. be easy to store and transport, and
4. perhaps most importantly, be economical.

Question 2.
What is a good fuel?
Answer:
A good fuel is one which

• produces more heat per unit mass. It has high calorific value.
• produces less harmful gases on combustion.
• is cheap and easily available.
• is every to handle safe to transport and convenient to store.

Question 3.
If you could use any source of energy for heating your food, which one would you use and why?
Answer:
We should select which is easily available and it should be cheaper. Bio-gas is an excellent fuel as it contains. It burns without some. Its heating capacity is high. This gas is convenient for consumption and transportation.

Class 10th Science Chapter 14 Intext questions Page No. 248

Question 1.
What are the disadvantages of fossil fuels?
Answer:
Disadvantages of fossil fuels are as following:

1. Fossil fuels are limited source and we can not use them for more than 110-120 years. From not so designing or making machines dependent on them will be a big failure after their loss.
2. Fossil fuel generates big amount of pollution which will destroy our atmosphere and increase temperature of earth which lead to destruction of our ecosystem.

Question 2.
Why are we looking at alternate sources of energy?
Answer:
Fossil fuels are a non-renewable source of energy. So we need to conserve them. If we were to continue consuming these sources at such alarming rates, we would soon run out of energy. In order to avoid this, alternate sources of energy were explored.

Question 3.
How has the traditional use of wind and water energy been modified for our convenience?
Answer:
(1) Wind energy: The kinetic energy of the wind can be used to do work. This energy was harnessed by wind mills in the past to do mechanical work. For example in a water lifting pump, the rotatory motion of windmill is utilized to lift water from a well. Today wind energy is also used to generate electricity. A windmill essentially consists of a structure similar to a large electric fan that is erected at some height on a rigid support.

A number of windmills are erected over a large area, which is known as wind energy farm. The energy output of each windmill in a farm is coupled together to get electricity on a commercial scale wind energy farms can be established only at those places where wind blows for the greater part of a year. The wind speed should also be higher than 15 km/h to maintain the required speed of the turbine, since, the tower and blades are exposed to the vagaries of nature like rain, sun, storm and cyclone, they need a high level of maintenance.

(2) Water energy: In order to produce hydel electricity, high rise dams are constructed on the river to obstruct the flow of water and thereby collect water in larger reservoirs. The water level rises and in this process the kinetic energy of flowing water gets transformed into potential energy. The water from the high level in the dam is carried through pipes, to the turbine, at the bottom of the dam. Sine the water in the reservoir would be refilled each time. It rains (hydropower is a renewable source of energy) we would not have to worry about hydro electricity sources getting used up the way fossil fuels would get finished one day.

Class 10th Science Chapter 14 Intext Questions Page No. 253

Question 1.
Can any source of energy be pollution free? Why or why not?
Answer:
Yes, nature show itself some examples of energy conversion which arg pollution free as photosynthesis, as we know in this process photo energy is converted to chemical energy. Hence, solar energy is best way to produce pollution free energy.

Question 2.
Hydrogen has been used as rocket fuel. Would you consider it a cleaner fuel than CNG? Why or why not?
Answer:
Yes, it is cleaner because it does not create any residual hazardous product or chemicals which pollute environment or disbalance the ecosystem. But, handling such big amount of energy properly is required.

Class 10th Science Chapter 14 Intext Questions Page No. 254

Question 1.
Name two energy sources that you would consider to be renewable. Give reasons for your choices.
Answer:
Energy derived from water, wind, sun and ocean all are renewable. All these energies can be harnessed into usable form as long as the solar system exists.

Question 2.
Give the names of two energy sources that you would consider to be exhaustible. Give reasons for your choices.
Answer:

• fossil fuels
• Nuclear fuels

Fossil fuels are present in a limited. amount in the earth. Once exhausted, they will not be available to us again. It takes millions of years for fossil fuel to be formed. The nuclear materials which can be conveniently extracted from earth 7. are limited and hence they will get exhausted one day.

Class 10th Science Chapter 14 NCERT Textbook Exercises

Question 1.
A solar water heater cannot be used to get hot water on:
(a) A sunny day
(b) A cloudy day
(c) A hot day
(d) A windy day
Answer:
(b) A cloudy day

Question 2.
Which of the following is not an example of a bio-mass energy source?
(a) Wood
(b) Gobar-gas
(c) Nuclear energy
(d) Coal
Answer:
(c) Nuclear energy

Question 3.
Most of the sources of energy we use represent stored solar energy. Which of the following is not ultimately derived from the sun’s energy?
(a) Geothermal energy
(b) Wind energy
(c) Nuclear energy
(d) Bio-mass
Answer:
(c) Nuclear energy

Question 4.
Compare and contrast fossil fuels and the sun as direct sources of energy.
Answer:

Question 5.
Compare and contrast bio-mass and hydro electricity as sources of energy.
Answer:

Question 6.
What are the limitations of extracting energy from—
(a) the wind?
(b) waves?
(c) tides?
Answer:
(a) The wind:

1. Wind energy farms can be established only at those places where wind blows for the greater part of a year.
2. The wind speed should also be higher than 15 km/h to maintain the required speed of the turbine.
3. Establishment of wind energy farms require large are of land.

These are the limitations of extracting energy from the wind.

(b) Limitations of extracting waves energy: The waves are generated by strong winds blowing across the sea. Wave energy would be a viable proposition only where waves are very strong.

(c) Limitations of extracting tidal energy: The locations where such dams can be built are limited.

Question 7.
On what basis would you classify energy sources as
(a) renewable and non-renewable?
(b) exhaustible and inexhaustible?
Are the options given in (a) and (b) the same?
Answer:
The options given in (a) and (b) are the same.

Question 8.
What are the qualities of an ideal source of energy?
Answer:
The qualities of an ideal source of energy are as follows:

1. Which would do a large amount of work per unit volume or mass.
2. Be easily accessible.
3. Be easy to store and transport, and
4. Perhaps most importantly, be economical.

Question 9.
What are the advantages and disadvantages of using a solar cooker? Are there places where solar cookers would have limited utility?
Answer:
Solar cooker is a device used to trap solar energy and utilize it to cook food. It consists of a box painted black from inside to absorb heat of the Sun (black colour is the best absorber of heat). A thick glass lid is placed over the box to trap heat energy of the sun. A plane mirror reflector is also attached to the box so that a strong beam of sunlight falls over the cooker’s top.

The reflector of the solar cooker sends strong beams of sunlight over the top of the cooker. Sunlight consists of about 1/3 rd infra-red rays which have a heating effect. These infra-red rays are of shorter wavelength as these are produced by a very hot source of heat. The glass lid over the cooker allows these infra-red rays of short wavelength into the cooker but does not allow the infra-red rays which are emitted by the black surface of the cooker to escape as these are of longer wavelength. Thus, heat energy of the sun gets trapped in the black box of the solar cooker. This heat cooks the food material kept in the black box.

Fig. 14.1: Solar heating device (solar cooker).

Advantages of solar cooker:

1. It is used to cook food and saves precious fossil fuel.
2. It does not cause any pollution.
3. No smoke is produced during the working of a solar cooker.
4. Nutrients of food material, which is to be cooked in the solar cooker, do not get destroyed.
5. Four food items can be cooked at the same time.

Disadvantages:

1. It cannot be used to cook food during the night.
2. It cannot be used to cook food on a cloudy day.
3. The direction of the reflector has to be changed after a small interval of time with the change of position of the sun.

Question 10.
What are the environmental consequences of the increasing demand for energy? What steps would you suggest to reduce energy consumption?
Answer:
They are:

• Burning of fossil fuels to meet the increasing demand for energy causes air pollution.
• Construction of dams and rivers to generate hydroelectricity destroys large ecosystems which get submerged underwater in the dams further, a large amount of methane [which is a green house gas] is produced when submerged vegetation rots under anaerobic conditions.

In order to reduce energy consumption

• Fossil fuel should be used with care and caution to derive maximum benefit out of them.
• Fuel saving devices such as pressure cookers etc should be used.
• Efficiency of energy sources should be maintained be getting them regularly serviced.
• And last of all, we should be economical in our energy consumption as energy saved is energy produced.

MP Board Class 10th Science Chapter 14 Additional Important Questions

MP Board Class 10th Science Chapter 14 Multiple Choice Questions

Question 1.
Can we convert any form of energy to its other form?
(a) Yes
(b) No
(c) May be Yes
(d) None
Answer:
(a) Yes

Question 2.
Photosynthesis represents what kind of conversion of energy:
(a) Conversion of photo-energy to chemical energy
(b) Conversion of chemical energy to photo-energy
(c) Conversion of physical energy to photo-energy
(d) Conversion of chemical energy to electrical energy
Answer:
(a) Conversion of photo-energy to chemical energy

Question 3.
Which one of the following is good source of energy?
(a) Burning wood
(b) Burning cooking oil
(c) Burning wax
(d) Burning petroleum
Answer:
(d) Burning petroleum

Question 4.
Fuel not used for cooking is:
(a) LPG
(b) Coal
(c) Petrol
(d) PNG
Answer:
(c) Petrol

Question 5.
Renewable source of energy is:
(a) Diesel
(b) Water
(c) Coal
(d) Petrol
Answer:
(b) Water

Question 6.
Physical energy is changed to electrical energy in:
(a) Thermal power plant
(b) Hydro power plant
(c) Photosynthesis
(d) None
Answer:
(a) Thermal power plant

Question 7.
Which one among following is not a property of biomass fuel?
(a) It contains 75% methane
(b) Bum without smoke
(c) Leaves no residue
(d) Easy to install
Answer:
(c) Leaves no residue

Question 8.
What kind of energy is used in wind energy form?
(a) Potential energy
(b) Kinetic energy
(c) Chemical energy
(d) Photo-energy
Answer:
(b) Kinetic energy

Question 9.
Which one is among alternative non-conventional source of energy?
(a) Fossil fuel
(b) Biomass
(c) Solar energy
(d) None of these
Answer:
(c) Solar energy

Question 10.
What percent of total of solar energy is absorbed on earth:
(a) 10%
(b) 50%
(c) 100%
(d) Can’t calculate
Answer:
(a) 10%

Question 11.
Ultimate source of energy on earth is:
(a) Electricity
(b) Nuclear energy
(c) Fossil fuel
(d) Sun
Answer:
(d) Sun

Question 12.
To maintain a wind energy farm, the wind speed should be:
(a) 1-2 km/h
(b) 180-200 km/h
(c) 15-20 km/h
(d) Can’t be calculated
Answer:
(c) 15-20 km/h

Question 13.
Which of the following element is chosen for generating nuclear energy?
(a) Uranium
(b) Silicon
(c) Germanium
(d) Carbon
Answer:
(c) Germanium

Question 14.
How bio gas is generated in biomass plant?
(a) By distillation
(b) By anaerobic fermenting
(c) By reduction
(d) By simple burning
Answer:
(b) By anaerobic fermenting

Question 15.
Natural gas contains:
(a) CO₂
(b) H₂O
(c) CH₄
(d) NH₃
Answer:
(c) CH₄

Question 16
…………. is used in a solar cell for storing energy.
(a) Gold
(b) Silver
(c) Germanium
(d) Silicon
Answer:
(d) Silicon

Question 17.
The stored heat in the earth is harnessed as:
(a) Fuel
(b) Geothermal energy
(c) Solar energy
(d) Biomass
Answer:
(b) Geothermal energy

Question 18.
Main constituent of LPG is:
(a) Butane
(b) Methane
(c) Propane
(d) Ethane
Answer:
(c) Propane

Question 19.
High calorific value of a material represents:
(a) A good manure
(b) A good fuel
(c) Protein rich material
(d) A good conductor of electricity
Answer:
(b) A good fuel

Question 20.
…………………… is a coal with highest carbon content
(a) Bituminous
(b) Peat
(c) Anthracite
(d) None
Answer:
(c) Anthracite

MP Board Class 10th Science Chapter 14 Very Short Answer Type Questions

Question 1.
Name two things which are considered as good source of energy.
Answer:
Coal and petrol.

Question 2.
Write a main characteristic of a good fuel.
Answer:
It should release a large amount of energy per unit volume.

Question 3.
Which food product can be used to produce light energy?
Answer:
Cooking oil and fodder.

Question 4.
What kind of energy source can be used with the help of concave mirror?
Answer:
Solar energy when being used in solar appliances.

Question 5.
What kind of energy is heat energy of molten rock present inside earth core?
Answer:
Geothermal energy.

Question 6.
Name two energy sources which are conventional and renewable in nature.
Answer:
Biomass and hydroelectricity.

Question 7.
Write one limitation of use of fossil fuels.
Answer:
Pollution.

Question 8.
How much percent of CH₄ is present in Bio gas?
Answer:
75%.

Question 9.
Write two kind of solar energy manifestation to oceans.
Answer:
Wave energy and ocean thermal energy.

Question 10.
What percentage of nuclear energy is contributed in total energy production in India?
Answer:
Approximately 3-4%.

Question 11.
What is main source of solar energy?
Answer:
Sun is the main source of energy.

Question 12.
Give one cause of limitation of solar energy.
Answer:
Solar energy is costly.

Question 13.
What is nuclear energy?
Answer:
Nuclear fission is the process during which two nucleus fuse to form one nucleus. The energy which is produced in this process called nuclear energy.

Question 14.
Give three examples of energy which is produced from sea.
Answer:
Tidal Energy, wave energy and Ocean thermal energy.

MP Board Class 10th Science Chapter 14 Short Answer Type Questions

Question 1.
Why it is important to consume energy wisely?
Answer:
When we use energy in its usable form we convert the form of energy and get our work done during the process. Since, we cannot reverse . the change involved in this process so we cannot get back the original usable form of energy. Due to this, it becomes important to think about energy shortage and the related energy crisis.

Question 2.
Write characteristics of a good source of energy.
Answer:
Characteristics of a good source of energy are:

• It should be effective i.e., able to do large amount of work in mass or volume.
• It should be easily accessible.
• It should be easily transported from one place to other.
• It should be economical.

Question 3.
What are conventional sources of energy?
Answer:
The sources of energy which have been in knowledge since a long time are conventional sources of energy. Firewood, coal, petroleum, natural gas, hydel energy, wind energy and nuclear energy are considered to be the conventional sources of energy.

Question 4.
Give two examples of fossil fuels.
Answer:
Two examples of fossil fuels used commercially for producing energy are:

Coal: The plants buried under swamps after death for very long time and due to high pressure and high temperature inside the earth; they are converted into coal. Coal is the highest used energy source in India.

Petroleum: The animals body after death get buried under the ocean surface and due to high pressure and high tempera lure inside the water were converted into petroleum; in due course of time Petroleum is among the major source of energy. Petroleum products are used as automobile fire! and also in the chemical industries.

Question 5.
Differentiate between non-renewable and renewable sources of Energy.
Answer:
Non-renewable sources of Energy: It takes millions of years for the formation of fossils fuels. Since, they cannot be replenished in the foreseeable future, they are known as non-renewable sources of energy.

Renewable sources of Energy: Those sources of energy which can be replenished quickly are called renewable sources of energy. Hydel energy, wind energy and solar energy are examples of renewable sources of energy.

Question 6.
What is hydel energy?
Answer:
Hydel Energy: Hydel energy is produced by utilizing the kinetic energy of flowing water. Huge dams are built over a source of water. Water is collected behind ‘the dam and released. When the water falls on the turbine; the turbine moves; because of kinetic energy of water. Electricity is generated by the turbine. Electricity; thus generated is called hydel energy of hydroelectricity. Water in the reservoir is replenished with rainwater and so, availability of water is, not a problem for hydroelectricity.

Question 7.
How electricity is generated in thermal power plant?
Answer:
In a thermal power plant, coal or petroleum is used for converting water into steam. The steam is used to run the turbine and thus, electricity in generated.

Question 8.
Define biomass.
Answer:
Biomass: The plants and animals constitute the biomass. Farm waste; such as stalks of harvested plants and dung of cattle; can be used to generate methane. The decomposition of biomass produces methane; which can be channelized for useful purposes.

Question 9.
What is bio-gas plant? How it is channelized?
Answer:
Bio-gas plant: Bio-gas plant can be very useful in solving the energy need of rural areas. A bio gas plant is a dome like structure which is usually built from bricks and concrete. In the mixing tank; the slurry is made from cow dung and water. The slurry then goes to the digester; which is a closed chamber. Since oxygen is absent in the digester, the anaerobes carry on their work of decomposition. The process of decomposition produces bio gas. Bio gas has about 70% of methane and the rest is composed of other gases.

The bio gas is channelized through a pipe and can be utilitzed as kitchen fuel and also as fuel for getting light. The slurry; left behind; is removed. It is used as manure, once it dries.

Question 10.
What is wind energy?
Answer:
Wind energy has been in use since ages. The sail boats of the pre-industrialisation era used to run on wind power. Windmills have been in use; especially in Holland; since the medieval period. Nowadays, windmills are being used to generate electricity. The kinetic energy of wind is utilized to run the turbines; which generate electricity.

Question 11.
What is non-conventional sources of energy?
Answer:
Energy sources which are relatively new are called non-conventional sources of energy, e.g., nuclear power and solar energy.

Question 12.
Explain solar energy.
Answer:
The sun is the main source of energy for all living beings on this earth. Even the energy in the fossil fuels has come from the sun. The sun is an endless reservoir of energy which would be available as long as the solar system is in existence. Technologies for harnessing the solar energy have been developed in recent times.

Question 13.
What is solar cooker and how food is cooked in it?
Answer:
Solar cooker is very simple in design and mode of function. It is usually made from mirrors. Plane mirrors are placed inside a rectangular box. The light reflected from the plane mirrors concentrates the solar energy inside the solar cooker which generates enough heat to cook food.

Question 14.
What are solar cells?
Answer:
Solar cells: Solar cells are made from silicon. The solar panel converts solar energy into electrical energy which is stored in a battery; for later use.

Question 15.
What is tidal energy?
Answer:
Due to the gravitational pull of the moon, tides happen near seashores. Water rushes up near the seashore during a high tide and goes down during a low tide. Dams are built near seashores to collect the water which comes during a high tide. When the water runs back to the ocean, the flow of water can be utilized to generate electricity.

Question 16.
How energy is generated from molten rocks?
Answer:
The molten rocks from the inside the earth are pushed in certain regions of the earth. Such regions are called the hot spots of the earth. When groundwater comes in contact with such hot spots, lot of steam is generated. This steam can be harnessed to produce energy which is called Geothermal Energy.

Question 17.
What is nuclear energy?
Answer:
Nuclear fission is the process during which two nucleus fuse to form one . nucleus. The process generates a huge amount of energy. This phenomenon is utilized in nuclear power plants. Nuclear power is safest for the environment but the risk of damage due to accidental leaks of radiation is pretty high. Further, storage of nuclear waste is a big problem because of potential risk of radiation involved. Nonetheless, many countries are using nuclear power in a big way.

MP Board Class 10th Science Chapter 14 Long Answer Type Questions

Question 1.
Define biomass. What is Bio gas plant? How it is channelized?
Answer:
Biomass: The plants and animals constitute the biomass. Farm waste; such as stalks of harvested plants and dung of cattle; can be used to generate methane. The decomposition of biomass produces methane; which can be channelized for useful purposes.

Bio gas plant: Bio gas plant can be very useful in solving the energy need of rural areas. A bio gas plant is a dome like structure which is usually built from bricks and concrete. In the mixing tank; the slurry is made from cow dung and water. The slurry then goes to the digester; which is a closed chamber. Since oxygen is absent in the digester, the anaerobes carry on their work of decomposition. The process of decomposition produces bio gas, has about 70% of methane and the rest ‘is composed of other gases. The bio gas is channelized through a pipe and can be utilized as kitchen fuel and also as fuel for getting light. The slurry; left behind; is removed. It is used as manure, once it dries.

Question 2.
What is wind energy? Write limitations of wind energy.
Answer:
Wind energy has been in use since ages. The sail boats of the re-industrialization era used to run on wind power. Windmills have been in use; especially in Holland; since the medieval period.

Nowadays, windmills are being used to generate electricity. The kinetic energy of wind is utilized to run the turbines; which generate electricity.

Limitations of wind Energy: Wind farms can only be established at those places where the wind speed is high enough and is more than 15 km/hr for most parts of the year. Wind farms need to be established on large tracts of land. The fan of the windmill has many moving parts; so cost of maintenance and repair is quite high. The fact, that it has to suffer the vagaries of the nature, further compounds the problem. Initial cost of establishing a wind farm is very high.

Question 3.
Briefly explain energy from sea.
Answer:
Tidal energy: Due to the gravitational pull of the moon, tides happen at seashores. Water rushes up near the seashore during a high tide and goes down during a low tide. Dams are built near seashores to collect the water which comes during a high tide. When the water runs back to the ocean, the flow of water can be utilized to generate electricity.

Wave Energy: Waves can also be a good source of energy. Many devices are being designed and tested to produce wave energy. For example; a hollow tower is built near the seashore. When water gushes in the tube because of wave, it forces the air upwards. The kinetic energy of air in the tube is used to run a turbine. When the wave goes down; air from up goes down the tube which is also used in running the turbine.

Ocean Thermal Energy. The water at sea surface is hot during daytime, while the water at lower level is cold. The temperature differential in water levels can be utilized to generate energy. If the temperature differential is more than 20°C, then ocean thermal energy can be utilized from that place. For this, a volatile liquid; like ammonia; is boiled using the heat from the hot water at the surface. The steam of the volatile liquid is utilized to run the turbine to generate electricity. Colder water from the surface below is utilized to condense ammonia vapour which is then channelized to the surface to repeat the cycle.

Question 4.
Explain Solar Energy. How is food cooked with help of solar energy and also explain its limitations.
Answer:
The sun is the main source of energy for all living beings on this earth. Even the energy in the fossil fuels has come from the sun. The sun has an endless reservoir of energy which would be available as long as the solar system is in existence. Technologies for harnessing the solar energy hav been developed in recent times.

Solar cooker is very simple in design and mode of function. It is usually made from mirrors. Plane mirrors are placed inside a rectangular box. The light reflected from the plane mirrors concentrates the solar energy inside the solar cooker which generates enough heat to cook food.

Limitations of solar Energy: The technologies for harnessing solar energy are at a nascent stage. At present, the cost benefit ratio for using solar energy is not conducive. Using solar energy is exhorbitantly costly.

MP Board Class 10th Science Chapter 14 NCERT Textbook Activities

Class 10 Science Activity 14.1 Page No. 242

• List four forms of energy that you use from morning, when you wake up, till you reach the school.
• From where do we get these different forms of energy?
• Can we call these ‘sources’ of energy? Why or why not?

Observations:

• Forms of energy that we use are electrical energy, mechanical energy, chemical energy for vehicles, chemical energy from food.
• Energy can neither created nor destroyed. It is just transferred from one form to the other.
• The ‘sources’ of energy are the one which releases energy to be used in such forms.

Class 10 Science Activity 14.2 Page No. 243

• Consider the various options we have when we choose a fuel for cooking our food.
• What are the criteria you would consider when trying to categories something as a good fuel?
• Would your choice be different if you lived:

(a) in a forest?
(b) in a remote mountain village or small island?
(c) in New Delhi?
(d) lived five centuries ago?

• How are the factors different in each case?

Observations:

• The criteria for good fuel inducts amount of energy released upon combustion, smoke produced or not, easily accessible, economical and easy be store and transport.
• In forest or remote village, wood from forests can be used as fuel.
• In new Delhi, electrical energy would be a choice.
• Five centuries ago, may be mechanical energy would have been used.
• Factors includes availability of the sunrises and case in utilizing them.

Class 10 Science Activity 14.3 Pages No. 244-245

• Take a table-tennis ball and make three slits into it.
• put semicircular fins cut out a metal sheet into these slits.
• Pivot the tennis ball on an axle through its centre with a straight metal wire fixed to a rigid support. Ensure that the tennis ball rotates freely about the axle.
• Now connect a cycle dynamo to this,
• Connect a bulb in series.
• Direct a jet of water or steam produced in a pressure cooker at the fins (Fig. 14.2). What do you observe?

Fig. 14.2: A model to demonstrate the process – of thermo – electric production.

Observations:

• We observe that fan starts morning. The rotor blade also moves with speed that turn the shaft of the dynamo and concert the mechanical energy into electrical energy.

Class 10 Science Activity 14.4 Page No. 248

• Find out from your grand-parents or other elders:
  • (a) how did they go to school?
  • (b) how did they get water for their daily needs when they were young?
  • (c) what means of entertainment did they use?
• Compare the above answers with how you do these tasks now.
• Is there a difference? If yes, in which case more energy from external sources is consumed? _

Observations:

• Earlier, people used to go school on-foot.
• The water were drawn from wells from far off places and carried on head in pots to the home.
• Means of entertainment includes folk dances, songs etc.
• The scenario has totally changed, nowadays, people are being separately and generally, not in noses as were done decades ago. More energy from internal sources is consumed these days to fulfill the demands of energy which is inversed a lot.

Class 10 Science Activity 14.5 Page No. 249

• Take two conical flasks and paint one white and the other black. Fill both with water.
• Place the conical flasks in direct sunlight for half an hour to one hour.
• Touch the conical flasks. Which one is hotter? You could also measure the temperature of the water in the two conical flasks with a thermometer.
• Can you think of ways in which this finding could be used in your daily life?

Observations:

• The one with black colour is hotter as black absorbs more heat as compared to white one.
• In daily life, we apply this on the colour we wear. The days which are that, we avoid black colours as they absorb more heat. Sincerity for cooling effect, white colours used.

Class 10 Science Activity 14.6 Pages No. 249-250

• Study the structure and working of a solar cooker and/or a solar water- heater, particularly with regard to how it is insulated and maximum heat absorption is ensured.
• Design and build a solar cooker or water-heater using low-cost material available and check what temperatures are achieved in vour system.
• Discuss what would be the advantages and limitations of using the solar cooker or water-heater.

Observations:

• The minimum heat absorption is ensured by painting it black in colour. The glass used on the lop traps the infrared rays from the sun and do not allow them the escape.
• Advantages includes no wastage of energy as solar energy is a trapped to be used further. It is a renewable some of energy that do not create any pollution.
• Limitations includes plausibility of solar rays at certain times of day only. It will not work on sunny days.

Class 10 Science Activity 14.7 Page No. 252

• Discuss in class the question of what is the ultimate source of energy for bio-mass, wind and ocean thermal energy.
• Is geothermal energy and nuclear energy different in this respect? Why ?
• Where would you place hydro electricity and wave energy?

Observations:

• The cleaner source of energy is the sun. Yes, cleaner energy is obtained from fusion or fission of molecules whereas geothermal energy is the energy present in the earth. The energy from geological changes is harnessed to be used in various ways.
• Hydro electricity and wave energy are also form of renewable energy.

Class 10 Science Activity 14.8 Page No. 253

• Gather information about various energy sources and how each one affects the environment.
• Debate the merits and demerits of each source and select the best source of energy on this basis.

Observations:

• Among the various source of energy, renewable sources of energy are the best as they do not harm the environment and this energy can be reused and when required. Non-renewable resources of’energy are harmful to the environment and causes pollution.
• Best sources of energy arc renewable sources of energy like solar, hydro, tidal, wave energy etc.

Class 10 Science Activity 14.9 Page No. 254

• Debate the following two issues in class.

(a) The estimated coal reserves are said to be enough to last us for another two hundred years. Do you think we need to worry about coal getting depleted in this case? Why or why not?
(b) It is estimated that the Sun will last for another five billion years. Do we have to worry about solar energy getting exhausted? Why or why not?

• One the basis of the debate, decide which energy sources can be considered

1. exhaustible
2. inexhaustible
3. renewable
4. non-renewable.

Give your reasons for each choice.

Observations:

• Yes, we need to worry about availability of coal as resources are limited they are made from fossil fuels which take millions of years to be formed. Also, they cause pollution and harm the environment.
• Sun will least for another five billion years but we should use this to the maximum as there are still so many years in which it can be utilized, harness eat and stored.
• Exhaustible are those energy sources which will get exhausted soon from fossil fuel inexhaustible are the one like solar, tidal etc. Renewable can be used whereas non-renewable are not used again and again.

MP Board Class 10th Science Solutions

MP Board Class 10th Science Solutions Chapter 10 Light Reflection and Refraction

MP Board Class 10th Science Chapter 10 Intext Questions

Class 10th Science Chapter 10 Intext Questions Page No. 168

Question 1.
Define the principal focus of a concave mirror.
Answer:
The number of rays parallel to the principal axis are falling on a concave mirror which meat at a point is called principal focus of the concave mirror.

(or)

Light rays that are parallel to the principal axis of a concave mirror converge at a specific point on its principal axis after reflecting from the mirror. This point is known as the principal focus of concave mirror.

Question 2.
The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Answer:
R = 2f Here R = 20 cm
20 = 2f
∴ \(f=\frac { 20 }{ 2 } =10\)
∴ Focal length = 10 cm.

Question 3.
Name a mirror that can give an erect and enlarged image of an object.
Answer:
Concave mirror can give an erect and enlarged image of an object when object is placed between the pole and principal focus.

Question 4.
Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Answer:
A convex mirror when fitted at rear-view position of vehicles, it gives a wider field of view, with which driver can see most of the traffic behind him. Convex mirrors give a virtual, erect and diminished image of the objects in front of it. So, we prefer a convex mirror as a rear-view mirror in vehicles.

Class 10th Science Chapter 10 Intext Questions Page No. 171

Question 1.
Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer:
Radius of curvature, R = 32 cm
Radius of curvature = 2f
\(R=2f=\frac { R }{ 2 } =\frac { 32 }{ 2 } =16\)
∴ Convex mirror focal length is = 16cm

Question 2.
A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?
Answer:

Let the height of the object = h₀ = h
Then, height of the image, h₁ = -3h (Image formed is real)
= \(\frac { -3h }{ h } \) = \(\frac { -v }{ u } \)
Object-distance, u = – 10 cm
v = 3 × (- 10)
= – 30 cm
Here, the negative sign indicates that an inverted image is formed at a distance of 30 cm in front of the given concave mirror.

Class 10th Science Chapter 10 Intext Questions Page No. 176

Question 1.
A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Answer:
Lightray bend towards normal. Because when a ray of light enters from rearer medium to denser medium, it changes its direction in the second medium.

Question 2.
Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is 3 × 108 ms-1.
Answer:
Refractive index of a medium:
µm = Speed of light in vacuum/Speed of light in the medium
Speed of light in vacuum, c = 3 × 10⁸ ms^-1
Refractive index of glass, µ_g = 1.50
Speed of light in the glass,
v = Speed of light in vacuum / Refractive index of glass
= c/µ_g
= 3 × 10⁸/1.50
= 2 × 10⁸ ms^-1.

Question 3.
Find out, from Table the medium having highest optical density. Also find the medium w ith lowest optical density.
Table:

Answer:
Highest optical density = Diamond.
Lowest optical density = Air.

Optical density of a medium is proportional to the refractive index. Hence, medium with highest refractive index will have the highest optical density and vice-versa. It can be observed from table that diamond and air respectively have the highest and lowest refractive index. Therefore, diamond has the highest optical density and air has the lowest optical density.

Question 4.
You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in table.

Answer:
Light travel faster in water when compared to kerosene and turpentine, since the refractive index of water is lower than kerosene and turpentine. The speed of light is inversely proportional to the refractive index.

Question 5.
The refractive index of diamond is 2.42. What is the meaning of this statement?
Answer:
It means Ratio of velocity of light in air and velocity of air in diamond is 2.42.

Class 10th Science Chapter 10 Intext Questions Page No. 184

Question 1.
Define 1 dioptre of power of a lens.
Answer:
1 dioptre is the power of lens whose focal length is 1 metre 1 D = 1 m^-1

Question 2.
A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.
Answer:
Image of Needle is real and inverted means this is real image it is 2f
Image is at a distance of 50 cm
Hence needle is kept 50 cm in front of convex lens.
Distance of object, u = – 50 cm.
Distance of image v = 50 cm
Focal length f = ?
As per lens formula.

f = 25 cm = 0.25 m
Power of the lens

Power of the lens P = + 4D.

Question 3.
Find the power of a concave lens of focal length 2 m.
Answer:
Focal length of concave lens, f = 2 m
Power of lens, P = \(\frac { 1 }{ f } \) = \(\frac { 1 }{ (-2) } \) = -0.5D

MP Board Class 10th Science Chapter 10 NCERT Textbook Exercises

Question 1.
Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Answer:
(d) A lens allows light to pass through it, but clay does not have that property.

Question 2.
The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?
(a) Between the principal focus and the centre of curvature
(b) At the centre of curvature
(c) Beyond the centre of curvature
(d) Between the pole of the mirror and its principal focus.
Answer:
(d) Between the pole of the mirror and its principal focus.

Question 3.
Where should an object be placed in front of a convex lens to get a real image of the size of the object?
(a) At the principal focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre of the lens and its principal focus.
Answer:
(b) When an object is placed at the centre of curvature in front of a convex lens, its image is formed at the centre of curvature on the other side of the lens.

Question 4.
A spherical mirror and a thin spherical lens have each a focal length of -15 cm. The mirror and the lens are likely to be:
(а) both concave
(b) both convex
(c) the mirror is concave and the lens is convex
(d) the mirror is convex, but the lens is concave
Answer:
(а) both concave

Question 5.
No matter how far you stand from a mirror, your image appears erpct. The mirror is likely to be
(a) plane
(b) concave
(c) convex
(d) either plane or convex
Answer:
(d) either plane or convex

Question 6.
Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
(a) A convex lens of focal length 50 cm
(b) A concave lens of focal length 50 cm
(c) A convex lens of focal length 5 cm
(d) A concave lens of focal length 5 cm
Answer:
(c)

Question 7.
We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.
Answer:
Range of the distance of the object = 0 cm to 15 cm.
Nature of the image = virtual, erect and larger than the object.

Question 8.
Name the type of mirror used in the following situations.
(a) Headlights of a car.
(b) Side/rear-view mirror of a vehicle.
(c) Solar furnace.
Support your answer with reason.
Answer:
(a) Concave mirror: Concave mirrors can produce powerful parallel ’ beam of light when the light source is placed at their principal focus. Hence, we can visualize ways easily in little light.

(b) Convex mirror: A convex mirror when fitted at rear view position of vehicles, it gives a wider field of view, with which driver can see most of the traffic behind him.

(c) Concave mirror: They are converging mirrors. This is because it concentrates the parallel rays of sun at principal focus and increase intensity of light falling on it.

Question 9.
One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.
Answer:
Yes, the convex lens will form complete image of the object, even if its one half is covered with black paper. Following two cases can better explain it:

Case I : When the upper half of the lens is covered.
In this case, a ray of light coming from the object is being refracted by the lower half of the lens. These rays meet at the other side of the lens to form the image of the given object.

Case II : If the lower half of the lens is covered.
In this case, a ray of light coming from the object is being refracted by the upper half of the lens. These rays meet at the other side of the lens to form the image of the given object.

Question 10.
An object 5 cm in length.is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Answer:
Converging lens means a convex lens. As the distances given in the question are large, so we choose a scale of 1 : 5, i.e., 1 cm represents 5 cm. Therefore, on this scale 5 cm high object, object distance of 25 cm and focal length of 10 cm can be represented by 1 cm high, 5 cm and 2 cm lines respectively. Now, we draw the ray diagram as follows;

1. Draw a horizontal line to represent ‘die principal axis of the convex lens.
2. Centre line is shown by DE.
3. Mark two foci F and F’ on two sides of the lens, each at a distance of 2 cm from the lens.
4. Draw an arrow AB of height 1 cm on the left side of lens at a distance of 5 cm from the lens.
5. Draw a line AD parallel to principal axis and then allow it to pass straight through the focus (F’) on the right side of the lens.
   
6. Draw a line from A to C (centre of the lens), which goes straight without deviation.
7. Let the two lines starting from A meet at A’.
8. Draw A’B’, perpendicular to the principal axis from A’.
9. Now A’B represents the real, but inverted image of the object AB.
10. Then, measure CB’ and A’B’. It is found that CB’ = 3.3 cm and A’B’ = 0.7 cm.
11. Thus the final position, nature and size of the image A’B’ are
    • (a) Position of image A’B’ = 3.3 cm × 5 = 16.5 cm from the lens on opposite side.
    • (b) Nature of image A’B’: Real and inverted.
    • (c) Height of image A’B’: 0.7 × 5 = 3.5 cm, i.e., image is smaller than the object.

Question 11.
A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.

Answer:
Focal length of concave lens f = – 15 cm
Image distance, v = – 10 cm
According to the lens formula,
\(\frac { 1 }{ v } \) – \(\frac { 1 }{ u } \) = \(\frac { 1 }{ f } \)

On solving we get, u = – 30 cm
The negative value of u indicates that the object is placed 30 cm in front of the lens. This is shown in the above ray diagram.

Question 12.
An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Answer:
Focal length of convex mirror,
f = +15 cm
Object distance, u = -10 cm
As per lens formula

Magnification \(=\frac{v}{u}=\frac{-6}{-10}=0.6\)
Virtual image is formed at the distance of 6 cm and it is erect.

Question 13.
The magnification produced by a plane mirror is +1. What does this mean?
Answer:
The positive sign means image formed by a plane mirror is virtual and erect. Since, the magnification is I, it means that the size of the image is equal to the size of the object.

Question 14.
An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Answer:
Object distance, u = – 20 cm
Object height, h = 5 cm
Radius of curvature, R = 30 cm
Radius of curvature = 2 × focal length
R = 2f ⇒ f = 15 cm
According to the mirror formula,
\(\frac { 1 }{ v } \) + \(\frac { 1 }{ u } \) = \(\frac { 1 }{ f } \)
The positive value indicates that the image is formed behind the mirror.

The positive value of image height indicates image is virtual, erect and smaller in size.

Question 15
An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.
Answer:
Object-distance, u = – 27 cm
Object-height, h = 7 cm
Focal length, f = – 18 cm
According to the mirror formula,
\(\frac { 1 }{ v } \) + \(\frac { 1 }{ u } \) = \(\frac { 1 }{ f } \)
Putting values, \(\frac { 1 }{ v } \) + \(\frac { 1 }{ (-27) } \) = \(\frac { 1 }{ (-18) } \)
So, v = – 54 cm
The screen should be placed at a distance of 54 cm in front of the given mirror
and \(\frac{h_{2}}{h_{1}}=\frac{-v}{u}\)
h₂ = -14 cm
The negative value of image indicates that the image is inverted.

Question 16.
Find the focal length of a lens of power – 2.0 D. What type of lens is this?
Answer:
Given, P = -2D
Power of lens. p = \(\frac { 1 }{ f } \)
and, f = – \(\frac { 1 }{ 2 } \)
\(\frac{h_{2}}{h_{1}}=\frac{-v}{u}\)
= – 0.5 m
A concave lens, because it has a negative value of focal length.

Question 17.
A doctor has prescribed a corrective lens of power + 1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
Answer:
Given, P = 1.5.D
Power of lens, P = \(\frac { 1 }{ f } \)
and, focal length f = \(\frac { 1 }{ 1.5 } \)
= \(\frac { 10 }{ 15 } \) = 0.66 m
A convex lens, because it has a positive focal length. Lens is converging.

MP Board Class 10th Science Chapter 10 Additional Important Questions

MP Board Class 10th Science Chapter 10 Multiple Choice Questions

Question 1.
The image formed by a convex lens is virtual, erect and larger than the object. The position of the object must be:
(a) Between the lens and its focus
(b) At the focus
(c) At twice the focal length
(d) At infinity
Answer:
(a) Between the lens and its focus

Question 2.
A real image formed by a convex lens is always:
(a) On the same side of the lens as the object
(b) Erect
(c) Inverted
(d) Smaller than the object
Answer:
(c) Inverted

Question 3.
If an object is moved towards a convex lens, the size of its image:
(a) Decreases
(b) Increases
(c) First decreases and then increases
(d) Remains the same
Ans.
(b) Increases

Question 4.
An object is placed at a distance of 30 cm from a concave mirror of focal length 15 cm. The image will be:
(a) Real and of same size
(b) Real and magnified
(c) Real and diminished
(d) virtual and magnified
Answer:
(a) Real and of same size

Question 5.
A concave mirror always forms real and inverted image except when the object is placed:
(a) At infinity
(b) Between F and C
(c) At F
(d) Between F and pole of the mirror
Answer:
(b) Between F and C

Question 6.
The mirror which has a wide field of view must be:
(a) Concave
(b) Convex
(c) Plane
(d) None of these
Ans.
(b) Convex

Question 7.
The image formed by a concave mirror:
(a) Is always real
(b) Is always virtual
(c) Can be both real and virtual
(d) None of these
Ans.
(c) Can be both real and virtual

Question 8.
An object is placed 20 cm from a convex lens of focal length 10 cm. The image must be:
(a) Real and diminished
(b) Real and of same size
(c) Real and enlarged
(d) Virtual and enlarged
Answer:
(b) Real and of same size

Question 9.
The ratio of the focal length of spherical mirror to its radius of curvature is:
(a) 0.5
(b) 1
(c) 2
(d) 3
Answer:
(a) 0.5

Question 10.
A real and inverted image of the same size is formed by a concave mirror when the object is placed:
(a) Between the mirror and its focus.
(b) Between the focus and the centre of curvature.
(c) At the centre of curvature.
(d) Beyond the centre of curvature.
Ans.
(c) At the centre of curvature.

MP Board Class 10th Science Chapter 10 Very Short Answer Type Questions

Question 1.
What is a mirror? Mention the different types of mirrors commonly used.
Answer:
Mirror: A highly polished surface which is smooth enough to reflect a good fraction of light incident on it is called a mirror. The mirror may be a highly polished metal surface or an ordinary glass plate coated with a thin silver layer.

Question 2.
What is the number of images of an object held between two plane parallel mirrors?
Answer:
Infinity.

Question 3.
Does the refractive index for a given pair of media depend on the angle of incidence?
Answer:
No, it is independent of the angle of incidence.

Question 4.
The refractive index of water with respect to air is \(\frac { 4 }{ 3 } \). What is the refractive index of air with respect to water?
Answer:
Refractive index of air with respect to water = \(\frac { 3 }{ 4 } \)

Question 5.
Can absolute refractive index of a medium exceed unity?
Answer:
No, because speed of light is maximum in vacuum.

Question 6.
Why does a ray of light bend when it travels from one medium to another?
Answer:
The bending of light or refraction occurs due to the change in the speed of light as it passes from one medium to another due to change in the density of the medium.

Question 7.
What happens when a ray of light strikes the surface of separation between the two media at right angle?
Ans.
The ray of light passes undeflected from one medium to another.
Here, ∠i = ∠r = 0°

Question 8.
What do you mean by a magnification less than unity?
Answer:
It means that the size of the image is smaller than the size of the object.

Question 9.
Which spherical mirror has

1. a real focus and
2. a virtual focus?

Answer:

1. A concave mirror has a real focus.
2. A convex mirror has a virtual focus.

Question 10.
State the position of the object for which a concave mirror produces virtual magnified image.
Answer:
The object should be placed between F and P of the concave mirror.

MP Board Class 10th Science Chapter 10 Short Answer Type Questions

Question 1.
Name the type of mirror(s) that should be used to obtain:
(i) a magnified and virtual image.
(ii) a diminished and virtual image of an object.
Draw labelled diagrams to show the formation of the required image in each of the above two cases. Which of these mirrors could also form a magnified and real image of the object? State the position of object for which this could happen.
Answer:
(i) Concave mirror.

Fig. 10.3 : Concave mirror with the object between F and P.

(ii) Convex mirror.

Fig. 10.4: Convex mirror with the object between pole and infinity.

Question 2.
Explain the uses of concave and convex mirrors.
Answer:
Uses of concave mirrors:
1. Shaving mirror : A concave mirror is used as a shaving or make-up mirror because it forms erect and enlarged image of the face when it is held closer to the face.

2. As head mirror : E.N.T. specialists use a concave mirror on their forehead. The light from a lamp after reflection from the mirror is focussed into the throat, ear or nose of the patient making the affected part more visible.

3. In ophthalmoscope : It consists of a concave mirror with a small hole at its centre. The doctor looks through the hole from behind the mirror while a beam of light from a lamp reflected from it, is directed into the pupil of patient’s eye which makes the retina visible.

4. In headlights : Concave mirrors are used as reflectors in headlights of motor vehicles railway engines, torch lights etc. The source is placed at the focus of the concave mirror. The light rays after reflection travel over a. large distance as a parallel beam of high intensity.

5. In astronomical telescopes : A concave mirror of large diameter (5 m or more) is used as objective in an astronomical telescope. It collects light from the sky and makes visible even those faint stars which cannot be seen with naked eye.

6. In solar furnaces : Large concave mirrors are used to concentrate sunlight to produce heat in solar furnace.

Uses of convex mirrors : Drivers use convex mirror as a rear-view mirror in automobiles because of the following two reasons:

1. A convex mirror always forms an erect, virtual and diminished image of an object placed anywhere in front of it.
2. A convex mirror has a wider field of view than a plane mirror of the same size as shown in Fig. 10.5.

Fig. 10.5: Field of view of (a) a plane mirror, (b) a convex mirror.

Thus, convex mirrors enable the driver to view much larger traffic behind him than would be possible with a plane mirror. The main disadvantage of a convex mirror is that it does not give the correct distance and the speed of the vehicle approaching from behind.

Question 3.
State the characteristics of the image formed by a convex mirror. What is the value of angle of incidence and angle of reflection when a ray of light retraces its path after reflection from a convex mirror? Illustrate with the help of a ray diagram.
Answer:
Properties of the image formed by a convex mirror:

(a) The image is always virtual and erect.
(b) The image is highly diminished or point sized.
(c) It is always formed between F and P.
(d) As the object is moved towards the pole of a convex mirror, image also moves towards its pole and gradually increases in size till its size becomes almost equal to that of the object.

When array of light retraces its path, ∠i = ∠r = 0°.

Fig. 10.6: A ray directed towards C is reflected back along same path after reflection from a convex mirror.

Question 4.
State the new Cartesian sign convention followed for reflection of light by spherical mirrors.
Answer:
According to this convention:

1. The object is on the left of the mirror. So all the ray diagrams are drawn with the incident light travelling from left to right.
2. All the distances parallel to the principal axis are measured from the pole of the mirror.
3. All distances measured in the direction of incident light are taken as positive.
4. All distances measured in the opposite direction of incident light are taken as negative.
5. Heights measured upwards and perpendicular to the principal axis are taken positive.
6. Heights measured downwards and perpendicular to the principal axis are taken negative.

Fig. 10.7: New Cartesian sign convention for reflection of light by spherical mirrors.

Question 5.
State the type of mirror preferred as
(i) rear view mirror in vehicles
(ii) shaving mirror. Justify your answer giving two reasons in each case.
Answer:
(i) A convex mirror is preferred as a rear-view mirror because:

(a) It always forms an erect, virtual and diminished image of an object placed anywhere in front of it.
(b) It has wider field of view.

(ii) A concave mirror is preferred as a shaving mirror because when it is held closer to the face, it forms:

(a) an enlarged image of the face.
(b) an erect image of the face.

Question 6.
State the laws of refraction of light.
Answer:
Laws of refraction of light: The refraction of light obeys the following two laws:
1st law: The incident ray, the refracted ray and normal to the interface of two transparent media at the point of incidence, all lie in the same plane.

2nd law: The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a given pair of media.
Mathematically,

The ratio µ21 is called refractive index of the second medium with respect to the first medium. The second law of refraction is also called Snell’s law of refraction.

Question 7.
What is the physical significance of refractive index?
Answer:
The refractive index of any medium gives the ratio of the speed of light in vacuum to the speed of light in that medium. For example, the refractive index of water, µ_w = 1.33. This means that the ratio of the speed of light in vacuum or air to the speed of light in water is 1.33.

Question 8.
What do you mean by optically denser and optically rarer media? How is the speed of light related to optical density?
Answer:
The optical density of a medium represents its ability to refract light. A medium having larger refractive index is called optically denser medium than the other. The other medium having lower refractive index is called optically rarer medium.

The speed of light is higher in a rarer medium than a denser medium. Thus, a ray of light travelling from a rarer medium to a denser medium slows down and bends towards the normal.

When it travels from a denser medium to a rarer medium, it speeds up and bends away from the normal.

Table:
Refractive indices of some material media (with respect to vacuum)

MP Board Class 10th Science Chapter 10 Long Answer Type Questions

Question 1.
With the help of a ray diagram, state and explain the laws of reflection of light at a plane mirror. Mark the angles of incidence and reflection clearly on the diagram.
Answer:
As shown in Fig. 10.8, when a ray of light is incident on a mirror, it gets reflected in accordance with the following laws of reflection.
1st law: The incident ray, the reflected ray and the normal at the point of incidence all lie in the same plane.
2nd law: The angle of incidence (i) is equal to the angle of reflection (r) i.e.∠i = ∠r

Fig.10.8: Reflection in a plane mirror.

Question 2.
What is lateral inversion of an image? What is the cause of lateral inversion?
Answer:
Lateral inversion: If we stand before a plane mirror and move our right hand, our image appears to move its left hand. In fact, our entire image is reversed sideways. This sideways reversal of the image is known as lateral inversion.

Cause of lateral inversion: Lateral inversion is due to the fact that in a plane mirror the image is as far behind the mirror as the object is in front of it, and that the front of the image and the front of the object face each other. The laterally inverted image of the word PAPYRUS is as shown in Fig. 10.9. The images of symmetrical letters like A, H, I, M, O, T, U, V, W, X, Y, 8 are not affected by lateral inversion.

Fig.10.9: Lateral inversion before a mirror.

Question 3.
Define the following terms in connection with spherical mirrors:
(i) Angular aperture
(ii) Centre of curvature
(iii) Radius of curvature
(iv) Principal axis
(v) Linear aperture
(vi) Pole
(vii) Principal force
(viii) Focal length
(ix) Principle focus Focal plane.
Answer:
Definition in connection with spherical mirrors: In Fig 10.10, let APB be a principal section of a spherical mirror, i.e., the section cut by a plane passing through pole and centre of curvature of the mirror
(i) Angular aperture : It is the angle ACB subtended by the boundary of the spherical mirror at its centre of curvature.
(ii) Centre of curvature : It is the centre C of the sphere of which the mirror forms a part.
(iii) Radius of curvature : It is the radius R (= AC or BC) of the sphere of which the mirror forms a part.
(iv) Principal axis : The line passing through the pole and the centre of curvature of mirror is called its principal axis.
(v) Linear aperture : It is the diameter AB of the circular boundary of the spherical mirror.
(vi) Pole: It is the middle point P of the spherical mirror.

Fig. 10.10: Characteristics of a concave mirror.

Question 4.
Deduce a relation between focal length (f) and radius of curvature (R) for a concave mirror.
Answer:
Relation between f and R for a concave mirror: As shown in Fig. 10.11, consider a ray AB parallel to the principal axis and incident at the point B of a concave mirror. After reflection from the mirror, this ray passes through its focus F, obeying the laws of reflection. If C is the centre of curvature, then CP = R, is the radius of curvature and CB is normal to the mirror at point B.

Fig. 10.11: Relation betweenTand R for a concave mirror.

According to the law of reflection, ∠i = ∠r
As AB is parallel to CP, so ∠a = ∠i (Alternate angles)
∠a = ∠r
Thus, ∆ BCF is isosceles,
Hence, CF = FB.
If the aperture (or size) of the mirror is small, then B lies close to P, so that,
FB = FP
FP = CF = \(\frac { 1 }{ 2 } \) CP
or f = \(\frac { R }{ 2 } \)
or Focal lenght = \(\frac { 1 }{ 2 } \) × Radius of curvature
Thus, the principal focus of a spherical mirror lies midway between the pole and the centre of curvature.

Question 5.
What happens to the size of the image formed by a convex mirror, when an object is gradually moved towards the mirror?
Answer:
When the object is at position A1B1, its virtual image is at a1b1 When the object is at position A2B2, its virtual image is at a2b2. So, when an object is gradually moved towards the pole of a convex mirror, its image also moves towards its pole and gradually increases in size till it has a size almost equal to that of the object. However, the image is always formed between F and P.

MP Board Class 10th Science Chapter 10 NCERT Text Book Activities

Class 10 Science Activity 10.1 Page No. 161

• Take a large shining spoon. Try to view your face in its curved surface.
• Do you get the image? Is it smaller or larger?
• Move the spoon slowly away from your face. Observe the image. How docs it change?
• Reverse the spoon and repeat the Activity. How does the image look like now?
• Compare the characteristics of the image on the tw o surfaces.

Observations:

• The image is formed that is smaller in size.
• On moving the spoon away, the size of the image gradually decreases with increasing field of view.
• On reversing the spoon, spoon w hen dose forms elect and magnified image on the inner curved ..urface. As we move away, image gets inverted and gradually decrease in Its size.
• The image on outer surface of spoon is erect and gradually decreases in size on moving away. The image of spoon on inner surface is erect and gets inverted on moving away. The size also decreases when moved away.

Class 10 Science Activity 10.2 Page No. 162

Caution: Do not look at the Sun directly or even into a mirror reflecting sunlight. It may damage your eyes.

• Hold a concave mirror in your hand and direct its reflecting surface towards the Sun.
• Direct the light reflected by the mirror on to a sheet of paper held close to the mirror.
• Move the sheet of paper back and forth gradually until you find on the paper sheet a bright, sharp spot of light.
• Hold the mirror and the paper in the same position for a few minutes. What do you observe? Why?

Observations:

• The paper starts burning when the mirror and paper are held in the same position for a few minutes as light rays from the sun, sharply focuses on this point and due to which heat concentrates at a point resulting in burning because of intense heating.

Class 10 Science Activity 10.3 Pages No. 163,164

You have already learnt a way of determining the focal length of a concave mirror. In activity 10.2, you have seen that the sharp bright spot of light you got on the paper is, in fact, the image of the Sun. It was a tiny, real, inverted image. You got the approximate focal length of the concave mirror by measuring the distance of the image from the mirror:

• Take a concave mirror. Find out its approximate focal length in the w ay described above. Note down the value of focal length. (You can also find it out by obtaining image of a distant object on a sheet of paper.)
• Mark a line on a table with a chalk. Place the concave mirror on a stand. Place the stand over the line such that its pole lies over the line.
• Draw with a chalk two more lines parallel to the previous line such that the distance between any two successive lines it equal to the focal length of the minor. These lines will now correspond to the positions of the points P, F and C, respectively. Remember – For a spherical mirror of small aperture, the principal focus F lies midway between the pole P and the centre of curvature C.
• Keep a bright object, say a burning candle, at a position far beyond C. Place a paper screen and move it in front of the mirror till you obtain a sharp bright image of the candle flame on it.
• Observe the image carefully. Note down its nature, position and relative size with respect to the object size.
• Repeat the activity by placing the candle-(a) just beyond C, (b) at C, (c) between F and C. (d) at F, and (c) between P and F.
• In one of the cases, you may not get the image on the screen. Identify the position of the object in such a case, Then, look for its virtual image in the mirror itself.
• Note down and tabulate your observations.

Observations:

Class 10 Science Activity 10.4 Page No. 166

• Draw neat ray diagrams for each position of the object shown in previous activity 10.3 observations.
• You may take any two of the rays mentioned in the previous section for locating the image.
• Compare your diagram with those given in Fig. 10.1.
• Describe the nature, position and relative size of the image formed in each case.
• Tabulate the results in a convenient format.

Observations:

Fig. 10.1: Ray diagrams for the image formation by a concave mirror.

Position of the object	Position of the. image	Size of the image	Nature of the image
At infinity	At the focus F, behind the mirror	Highly diminished, point-sized	Virtual and erect
Between infinity and the pole P of the mirror	Between P and F, behind the mirror	Diminished	Virtual and erect

Class 10 Science Activity 10.5 Page No. 167

• Take a convex mirror. Hold it in one hand.
• I Iold a pencil in the upright position in the other hand.
• Observe the image of the pencil in the mirror. Is the image erect or inverted? Is it diminished or enlarged?
• Move the pencil away from the mirror slowly. Does the image become smaller or larger?
• Repeat this Activity carefully. State whether the image will move closer to or farther away from the focus as the object is moved away from the min or?

Observations:

• The image is erect and diminished. The image becomes smaller or moving pencil away.
• The image moves closer to the focus as the object is moved away from the mirror.

Class 10 Science Activity 10.6 Page No 167

• Observe the image of a distant object, say a distant tree, in a plane mirror.
• Could you see a full-length image?
• Try with plane minors of different sizes. Did you see the entire object in the image?
• Repeat this Activity with a concave mirror. Did the mirror show full length image of the object?
• Now try using a convex mirror. Did you succeed? Explain your observations with reason.

Observations:

• No, full – length image of a distant object is not seen in a plane mirror.
• The entire images of the objects were not seen.
• No, the mirror do not show full length image of the object.
• Yes, with the convex mirror we can see full length image of distant object with under field of view’ this is because the convex mirror are used as rear or side view mirrors in vehicles. The image formed is diminished, erect and virtual.

Class 10 Science Activity 10.7 Page No. 172

• Place a coin at the bottom of a bucket filled with water.
• With your eye to a side above water, try to pick up the coin in one go. Did you succeed in picking up the coin?
• Repeat the Activity. Why did you not succeed in doing it in one go?
• Ask your friends to do this. Compare your experience with theirs.

Observations:

• No, we cannot succeed in picking up the coin.
• This happens due to refraction, the coin appears to he at some other place from where it is actually present. The light rays coming out from water tends to bend creating this problem.

Class 10 Science Activity 10.8 Page No. 172

• Place a large shallow bowl on a Table and put a coin in it.
• Move away slowly from the bowl. Stop when the coin just disappears from your sight.
• Ask a friend to pour water gently into the bowl without disturbing the coin.
• Keep looking for the coin from your position. Does the coin becomes visible again from your position? How could this happen?

Observations:

• Yes, on pouring water, it again becomes visible and little raised due to refraction.

Class 10 Science Activity 10.9 Page No. 172

• Draw a thick straight line in ink, over a sheet of white paper placed on a Table.
• Place a glass slab over the line in such a way that one of its edges make an angle with the line.
• Look at the portion of the line under the slab from the sides. What do you observe? Does the line under the glass slab appear to be bent at the edges ?
• Next, place the glass slab such that it is normal to the line. What do you observe now’ Does the part of the line under the glass slab appear bent?
• Look at the line from the top of the glass slab. Does the part of the line, beneath the slab, appear to be raised? Why does this happen?

Observations:

• The line under the glass slab appear bent at the edges due to refraction.
• No, now it does not appear bend as a ray of light perpendicular to the plane of a refracting medium does not change angle during refraction.
• Yes, this is also due to refraction that apparent position of image of object seems nearer than its actual position.

Class 10 Science Activity 10.10 Page No. 173

• Fix a sheet of white paper on a drawing board using drawing pins.
• Place a rectangular glass slab over the sheet in the middle.
• Draw the outline of the slab with a pencil. Let us name the outline as ABCD.
• Take four identical pins.
• Fix two pins, say E and F, vertically such that the line joining the pins is inclined to the edge AB.
• Look for the images of the pins E and F through the opposite edge. Fix two other pins, say G and H, such that these pins and the images of E and F lie on a straight line.
• Remove the pins and the slab.
• Join the positions of tip of the pins E and F and produce the line up to AB. Let EF meet AB at O. Similarly, join the positions of tip of the pins G and H and produce it up to the edge CD. Let HG meet CD at O’.
• Join O and O’. Also produce EF up to P, as shown by a dotted line in Fig. 10.2.

Observations:

Fig. 10.2: Refraction of light through a rectangular glass slab.

Class 10 Science Activity 10.11 Page No. 177

Caution: Do not look at the Sun directly or through a lens while doing this Activity or otherwise. You may damage your eyes if you do so.

• Hold a convex lens in your hand. Direct it towards the Sun.
• Focus the light from the Sun on a sheet of paper. Obtain a sharp bright image of the Sun.
• Hold the paper and the lens in the same position for a while. Keep observing the paper. What happened? Why? Recall your experience in Activity 10.2.

Observations:

• This is due to sharp focusing of all rays at a single point after passing through the lens. The concentration of light rays increases at a point resulting in burning of paper.

Class 10 Science Activity 10.12 Page No. 178

• Take a convex lens. Find its approximate focal length in a way described in Activity 10.11.
• Draw five parallel straight lines, using chalk, on a long Table such that the distance between the successive lines is equal to the focal length of the lens.
• Place the lens on a lens stand. Place it on the central line such that the optical centre of the lens lies just over the line.
• The two lines on cither side of the lens correspond to F and 2F of the lens respectively. Mark them with appropriate letters such as 2F₁, F₁, F₂ and 2F₂, respectively.
• Place a burning candle, far beyond 2F, to the left. Obtain a clear sharp image on a screen on the opposite side of the lens.
• Note down the nature, position and relative size of the image.
• Repeat this Activity by placing object just behind 2F₁ between F₁ and 2F₁ at F₁, between F₁ and O. Note down and tabulate your observations.

Observations:

Class 10 Science Activity 10.13 Page No. 179

• Take a concave lens. Place it on a lens stand.
• Place a burning candle on one side of the lens.
• Look through the lens from the other side and observe the image. Try to get the image on a screen, if possible. If not, observe the image directly through the lens.
• Note down the nature, relative size and approximate position of the image.
• Move the candle away from the lens. Note the change in the size of the image. What happens to the size of the image when the candle is placed too far away from the lens.

Observations:

MP Board Class 10th Science Solutions

MP Board Class 10th Science Solutions Chapter 8 How do Organisms Reproduce?

MP Board Class 10th Science Chapter 8 Intext Questions

Class 10th Science Chapter 8 Intext Questions Page No. 128

Question 1.
What is the importance of DNA copying in reproduction?
Answer:
The chromosomes in the nucleus of a cell contain information for inheritance of features from parents to next generation in the form of DNA (Deoxyribo Nucleic Acid) molecules. The DNA in the cell nucleus is the information source for making proteins. Hence DNA copying is important in reproduction.

Question 2.
Why is variation beneficial to the species but not necessarily for the individual?
Answer:

If a population of reproducing organisms were suited to particular niche and if the niche were drastically altered, the population could be wiped out. However, if some variations were to be present in a few individuals in these populations, there would be some chance for them to survive.

Thus, if there were a population of bacteria living in temperature waters and if the water temperature were to be increased by global warming, most of these bacteria would die, but the few variants resistant to heat would survive and grow further. Variation is thus useful for the survival of species over time. Variation is not useful for all organisms.

Class 10th Science Chapter 8 Intext Questions Page No. 133

Question 1.
How does binary fission differ from multiple fission?
Answer:
Binary fission: It is a simple kind of division which formate new individual. In binary fission, a single cell divides into two equal halves but it is possible only with very simple single cell kind. Amoeba and Bacteria divide by binary fission.

Multiple fission: Another type of simple division is multiple fission, in this, a single cell divides into many daughter, cells, e.g., Plasmodium divide by multiple fission.

Binary fission	Multiple fission
In this fission, one cell split into two equal halves during cell division. Eg: Bacteria.	Here one organism divide into many daughter cells simultaneously. Eg: yeast.

Question 2.
How will an organism be benefited if it reproduces through spores?
Answer:
The spores are covered by thick walls that protect them until they come into contact with another moist surface and can begin to grow. Thus organism be benefited if it reproduces through spores.

Question 3.
Can you think of reasons why more complex organisms cannot give rise to new individuals through regeneration?
Answer:
Multicellular organisms are not simply a random mass of cells but a carefully organized entity of tissues and organs are placed at definite positions in the body to form organ systems. These systems are well coordinated to perform specific functions. Hence complex organisms cannot reproduce through fragmentation.

Question 4.
Why is vegetative propagation practised for growing some types of plants?
Answer:
Advantages of vegetative propagation:

• Used in methods such as layering or grafting, to grow many plants like sugarcane, roses or grapes for agricultural purposes.
• Plants raised can bear more flowers and fruits in comparison to plants produced from seeds.
• Plants such as banana, orange, rose and jasmine which have lost the capacity to produce seeds can be propagated.
• All plants produced by vegetative propagation are genetically similar enough to the parent plant.

Question 5.
Why is DN Acopying an essential part of the process of reproduction?
Answer:
The consistency of DNA copying during reproduction is important for the maintenance of body design features that allow the organism to use that particular niche. Because of this DNA copying is an essential part of the process of reproduction.

Class 10th Science Chapter 8 Intext Questions Page No. 140

Question 1.
How is the process of pollination different from fertilization?
Answer:
Pollination is movement of pollens from one plant to another plant’s or its own plant’s stigma. It may require certain agents called pollinators such as air, water birds or some insects to perform. Fertilization, is a complex process, it involves the fusion of the male and female gametes. It occurs inside the ovule and leads to the formation of zygote.

Question 2.
What is the role of the seminal vesicles and the prostate gland?
Answer:
Along the path of the vas deferens, gland like the prostrate and the seminal vesicles add their secretions so that the sperms are now in a fluid which makes their transport easier and this fluid also provides nutrition.

Question 3.
What are the changes seen in girls at the time of puberty?
Answer:
The changes seen in girls at the time of puberty are:

1. Development of secondary sexual characteristics.
2. Growth in breast size and darkening of skin of the nipples.
3. Growth of hair in the genital area and other areas of skin like underarms, face, hands and legs.
4. Growth in the size of uterus and ovary hence, start of menstrual cycle periodically.

Question 4.
How does the embryo get nourishment inside the mother’s body?
Answer:
The embryo gets nutrition form the mother’s blood with the help of a special tissue called placenta. This is a disc which is embedded in the uterine wall. It contains villi on the embryo’s side of the tissue on the mother’s side are blood spaces, which surround the villi. This provides a large surface area for glucose and oxygen to pass from the mother to the embryo. The developing embryo will also generate waste substances which can be removed by transferring them into the mother’s blood through the placenta.

Question 5.
If a woman is using a copper-T, will it help in protecting her from sexually transmitted diseases?
Answer:
Copper-T will helps in protecting her from sexually transmitted diseases by helping to prevent infections of diseases.

MP Board Class 10th Science Chapter 8 NCERT Textbook Exercises

Question 1.
Asexual reproduction takes place through budding in:
(a) amoeba
(b) yeast
(c) plasmodium
(d) leishmania
Answer:
(b) yeast

Question 2.
Which of the following is not system in human beings? a part of the female reproductive
(a) ovary
(b) uterus
(c) vas deferens
(d) fallopian tube
Answer:
(c) vas deferens

Question 3.
The anther contains:
(a) sepal
(b) ovules
(c) carpel
(d) pollen grains
Answer:
(d) pollen grains

Question 4.
What are the advantages of sexual reproduction over asexual reproduction?
Answer:
In case of asexual reproduction, new generations are produced by one organism. But in sexual reproduction, new generations are produced by two organisms (male and female). In case of sexual reproduction germ cells are produced in testes and these secrete a hormone testosterone. In human beings also develop special tissues for this purpose.

Question 5.
What are the functions performed by the testis in human beings?
Answer:
They are the glands where sperm and testosterone are generated and present in male body. The testes are contained in the scrotum and are composed of dense connective tissue. Functions of testes are as follows:

• It produces sperms, which contain haploid set of chromosomes of
• It produces testosterone, which initiate secondary sexual characteristics

Question 6.
Why does menstruation occur?
Answer:
Since the ovary releases one egg every month, the uterus also prepares itself every month to receive a fertilised egg. Thus its lining becomes thick and spongy. This would be required for nourishing the embryo if fertilisation had taken place. Now, however, this lining is not needed any longer. So the lining slowly breaks and comes out through the vagina as blood and mucous. This cycle takes place roughly every month and is known a menstruation. It usually lasts for about two to eight days.

Question 7.
Draw a labelled diagram of the longitudinal section of a flower.
Answer:

Longitudinal section flower.

Question 8.
What are the different methods of contraception?
Answer:

Many ways have been devised to avoid pregnancy. These contraceptive methods fall in a number of categories. One category is the creation of a mechanical barrier so that sperm does not reach the egg. Condoms on the penis or similar coverings worn in the vagina can serve this purpose.

Another category of contraceptives acts by changing the hormonal balance of the body so that eggs are not released and fertilisation cannot occur. These drugs commonly need to be taken orally as pills. However, Since they change hormonal balances, they can cause side effects too. Other contraceptive devices such as the loop or the copper-T are placed in the uterus to prevent pregnancy. Again, they can cause side effects due to irritation of the uterus. Surgery can also be used for removed of unwanted pregnancies.

Question 9.
How are the modes for reproduction different in unicellular and multicellular organisms?
Answer:
In unicellular organisms, reproduction occurs by the division of the entire cell. The modes of reproduction in unicellular organisms can be fission, budding etc. whereas in multi cellular organisms, specialised reproductive organs are present. Therefore, they can be reproduced by complex reproductive methods such as vegetative propagation, spore formation etc. In more complex multicellular organisms such as human beings and plants, the mode of reproduction is sexual reproduction.

Question 10.
How does reproduction help in providing stability to populations of species?
Answer:
Reproduction is the process of producing new individuals of the same species by existing organisms of a species. So, it helps in providing stability to population of species by giving birth to new individuals as the rate of birth must be at par with the rate of death to provide stability to population of a species.

Question 11.
What could be the reasons for adopting contraceptive methods?
Answer:
Contraceptive methods are mainly adopted because of the following reasons:

• It prevent unwanted pregnancies.
• It control rise in population and birth rate.
• It prevent sexually transmitted diseases.

MP Board Class 10th Science Chapter 8 Additional Important Questions

MP Board Class 10th Science Chapter 8 Multiple Choice Questions

Question 1.
All individuals produced by an organism are:
(a) Genetically similar
(b) Non-identical
(c) Fission
(d) Moneociuos
Answer:
(a) Genetically similar

Question 2.
Sexual reproduction is completed by _______ division:
(a) Mitotic
(b) Meiotic and mitotic both
(c) Meiosis
(d) Mitotic at some stages .
Answer:
(c) Meiosis

Question 3.
In yeast cell, division results in:
(a) Offspring
(b) Bud
(c) Clone
(d) Branch
Answer:
(b) Bud

Question 4.
Which of the following organisms undergo multiple fission?
(a) Paramecium
(b) Plasmodium
(c) Amoeba
(d) All of the above
Answer:
(b) Plasmodium

Question 5.
Hydra reproduces asexually through:
(a) Budding
(b) Binary fission
(c) Multiple fission
(d) Vegetative propagation
Answer:
(a) Budding

Question 6.
In which plant, the site of origin of new plants is node?
(a) Potato tuber
(b) Onion bulb
(c) Rhizome ginger
(d) All of the above
Answer:
(d) All of the above

Question 7.
In which of the following, asexual reproduction takes place through binary fission?
(a) Amoeba
(b) Yeast
(c) Plasmodium
(d) Leishmania
Answer:
(b) Yeast

Question 8.
Which of the following is in human beings?
(a) Ovary
(b) Uterus
(c) (a) and (b)
(d) Fallopian tube
(e) (a), (b) and (d)
Answer:
(e) (a), (b) and (d)

Question 9.
The anther, a part of male flower have:
(a) Sepals
(b) Ovules
(c) Carpel
(d) Pollen grains
Answer:
(d) Pollen grains

Question 10.
The information for making proteins is provided by:
(a) Rough endoplasmic reticulum
(b) DNA
(c) Hormones
(d) Enzymes
Answer:
(b) DNA

Question 11.
Nature of gametes are usually:
(a) Haploid
(b) Diploid
(c) Both (a) and (b)
(d) None of the above
Answer:
(a) Haploid

Question 12.
With the help of which tissues embryo gets nutrition from the mother’s blood?
(a) Zygote
(b) Uterus only
(c) Placenta
(d) None of these
Answer:
(c) Placenta

Question 13.
Which of the following is not a part of the male reproductive system in human beings?
(a) Testes
(b) Uterus
(c) Vas deferens
(d) Urethra
Answer:
(b) Uterus

Question 14.
Binary fission in some organisms occurs in definite orientation in relation to the cell structures. One such organisms is:
(a) Leishmania
(b) Plasmodium
(c) Amoeba
(d) Bacteria
Answer:
(c) Amoeba

Question 15.
Plants that have lost their capacity to produce seeds, reproduce by:
(a) Spores
(b) Vegetative propagation
(c) Fission
(d) Regeneration
Answer:
(a) Spores

Question 16.
A stamen consists of two parts namely:
(a) Anther and style
(b) Anther and filament
(c) Stigma and style
(d) Filament and style
Answer:
(b) Anther and filament

Question 17.
A bisexual flower contains:
(a) Stamens only
(b) Carpels only
(c) Either stamens or carpels
(d) Both stamens and carpels
Answer:
(d) Both stamens and carpels

Question 18.
Germinated seeds do not contains:
(a) Sepals
(b) Cotyledon
(c) Plumule
(d) Radicle
Answer:
(a) Sepals

Question 19.
A feature of reproduction that is common to amoeba, spirogyra and yeast is that:
(a) they reproduce asexually
(b) they are all unicellular
(c) they reproduce only sexually
(d) they are all multicellular
Answer:
(a) they reproduce asexually

Question 20.
Which of the part of flower ripens to form a fruit?
(a) Ovule
(b) Ovary
(c) Carpel
(d) Egg cell
Answer:
(b) Ovary

Question 21.
The testes perforin the following function/functions:
(a) Produce testosterone
(b) Produce sperms
(c) Produce male gametes and hormone
(d) Produce sperms and urine
Answer:
(b) Produce sperms

Question 22.
Where does fertilisation take place in human beings?
(a) Uterus
(b) Vagina
(c) Cervix
(d) Fallopian Tube
Answer:
(d) Fallopian Tube

Question 23.
Condom is a method of control that falls under the following category:
(a) Surgical method
(b) Hormonal method
(c) Mechanical method
(d) Chemical method
Answer:
(c) Mechanical method

Question 24.
The common passage for sperms and urine in the male reproductive system is:
(a) Ureter
(b) Seminal vesicle
(c) Urethra
(d) Vas deferens
Answer:
(c) Urethra

Question 25.
In sperm, which part dissociates after fertilization?
(a) Acrosome
(b) Tail
(c) Head
(d) Middle piece
Answer:
(b) Tail

MP Board Class 10th Science Chapter 8 Very Short Answer Type Questions

Question 1.
Which life process is not essential to maintain the life of an individual organism but important for the survival of species?
Answer:
Reproduction.

Question 2.
How a species can get a danger of being extinct?
Answer:
If individuals of any species stops reproducing, then that species can get a danger of being extinct.

Question 3.
How an individual is able to make a copy of itself?
Answer:
DNA copying is a process at cellular level which enables an individual to make copy of it self.

Question 4.
Write the name of process by which Hydra reproduces.
Answer:
Budding only.

Question 5.
Generally, how many individuals are involved in asexual reproduction?
Answer:
One.

Question 6.
Write the name of some common method of asexual reproduction.
Answer:
Vegetative propagation, budding, fragmentation and spore formation.

Question 7.
Which type of flower is called unisexual flowers?
Answer:
A flower which have either male or female reproductive parts is called unisexual flowers.

Question 8.
What is pollination?
Answer:
The transfer of pollen grains from the anther to the stigma of the same or of another flower of the same kind is known as pollination.

Question 9.
What do you understand by term fertilisation?
Answer:
The fusion of male and female gametes is termed as fertilisation.

Question 10.
How seed is dispersed?
Answer:
Seed dispersal takes place by means of wind, water and animals.

MP Board Class 10th Science Chapter 8 Short Answer Type Questions

Question 1.
How does plasmodium undergo fission?
Ans.
Plasmodium divides into many daughter cells through multiple fission.

Question 2.
How spirogyra reproduces by fragmentation?
Answer:
An individual spirogyra breaks up into many smaller pieces, each fragment grows into new individual.

Question 3.
Which cells are responsible for budding in hydra?
Answer:
Regenerative cells.

Question 4.
Name the structure into which following develops: the plumule and radicle?
Answer:
Plumule develops to shoot while radicle form root of a plant.

Question 5.
On which plant can you find buds on its leaves?
Answer:
Bryophyllum.

Question 6.
Write the scientific name of the bread mould.
Answer:
Rhizopus.

Question 7.
Where are the testes located in human beings?
Answer:
In abdominal cavity, in scrotum.

Question 8.
For what specific reason have the testes specific location?
Answer:
As testes, requires lesser temperature, to produce sperm than of abdominal cavity.

Question 9.
Correlate the rate of general body growth and maturation of reproductive tissue during puberty.
Answer:
When reproductive tissues (organs) begin to mature, body growth rate slows down.

Question 10.
Where does the zygote get implanted in human beings?
Answer:
In the wall of uterus.

Question 11.
Which two important substances are delivered to developing embryo through placenta?
Answer:
Glucose and oxygen.

Question 12.
How change in hormonal balance prevents pregnancy?
Answer:
It prevents the release of eggs.

Question 13.
Name the tissue in mother’s body that provides nutrition to developing embryo?
Answer:
Placenta provides nutrition to developing embryo.

Question 14.
Write one side effect of loop placed in uterus.
Answer:
It may cause permanent irritation and excessive and prolong bleeding in uterus.

Question 15.
Which structures need to be blocked in males and females respectively to prevent pregnancy?
Answer:
Vas deferens in male (vasectomy), fallopian tube in female (tubectomy).

Question 16.
Why is children sex ratio alarmingly declining in our country.
Answer:
Abortions based on sex selections.

Question 17.
Name the chemical methods of preventing pregnancy.
Answer:
Morning over oral pills.

Question 18.
Name some of the devices used as mechanical method for preventing pregnancy.
Answer:
Loop, copper T, condoms.

Question 19.
Name the only mammal(s) which lays eggs.
Answer:
Echidna and duck-billed platypus.

Question 20.
What is parthenogenesis?
Answer:
Parthenogenesis is a type of asexual reproduction. In this case, embryo development takes places without fertilisation. A few species of insects, bees, wasps, birds and lizards (e.gKomodo dragon lizard) reproduce this way.

Question 21.
Give an example of an organism which reproduces by:
(a) Fragmentation
(b) Spore formation
(c) Stems
Answer:
(i) Spirogyra.
(ii) Bacteria, fungi (rhizopus), moss, algae.
(iii) Plants like potato (tuber), onion (bulb) reproduce by vegetative propagation of stems.

Question 22.
Discuss various artificial vegetative propagation techniques.
Answer:
Various artificial vegetative propagation techniques are:

1. Cutting
2. Layering
3. Grafting
4. Tissue culture

Question 23.
What is grafting? What are different types of grafting techniques?
Answer:
In grafting, one part of a plant is inserted into another plant in a way that both of them will unite and grow together as a single plant. Different methods of grafting are:

• Approach grafting
• Cleft grafting
• Bud grafting
• Tongue grafting

Question 24.
Name some:

1. Plants which are reproduced by vegetative propagation.
2. Plants which have unisexual flowers.
3. Plants which have bisexual flowers.
4. Plants with self-pollination.
5. Plants that do cross-pollination.

Answer:

1. Rose, sweet potatoes, bryophyllum.
2. Coconut, papaya, watermelon.
3. Lily, rose, sunflower.
4. Beans, peas, tomatoes.
5. Grasses, catkins, maple trees.

Question 25.
What is germination?
Answer:
The seed contains the future plant or embryo which develops into a seedling under appropriate conditions. This process is known as germination.

Question 26.
What is cross-pollination?
Answer:
Cross-pollination is the process of transfer of pollen from the anther of a flower to stigma of a flower of another plant of the same species or closely related species.

Question 27.
Explain hormonal pills of contraception.
Answer:
Oral contraceptives: In this method, tablets or drugs are taken orally by females to check pregnancy These contain small doses of hormones in forms of pills that prevent the release of eggs and thus, fortilisation cannot occur.

MP Board Class 10th Science Chapter 8 Long Answer Type Questions

Question 1.
Why simply copying of DNA in a dividing cells not enough to maintain continuity of life?
Answer:
Copying of DNA preserve and pass specific characters of a generation to next generation offsprings. In reproduction, it is very important to create DNA copy. It determines the body design of an individual. But variation in genotype is also important,, because sometimes existing genotype don’t find its potential to survive in changing surroundings. So, genotype must have some alterations which are caused by variations only. Hence, simply copying of DNA in a dividing cells is not enough to maintain continuity of life.

Question 2.
Describe in brief the fragmentation mode of asexual reproduction.
Answer:
Fragmentation: Many lower organisms, use fragmentation mode of asexual reproduction for its growth e.g., algae. When water and nutrients are available in sufficient amount algae grow and multiply rapidly by fragmentation. An algae breaks up to multiple fragments. These fragments or pieces grow into new individuals.

Question 3.
Explain budding in yeast.
Answer:
The yeast is a single-celled organism. The small bulb-like projection come out from the yeast cell in favourable time and is called a bud. The bud gradually grows and gets dettached from the parent cell and forms a new yeast cell. The new yeast cell grows, matures and produces more yeast cells.

Question 4.
Describe the process of implantation.
Answer:
A week after the sperm fertilizes the egg, the fertilized egg (zygote) undergo development and become a multicelled blastocyst. The blastocyst fix itself into the lining of the uterus, called the endometrium. The hormone estrogen causes the endometrium to become thick and rich with blood. Progesterone and other hormone released by the ovaries, keeps the endometrium thick with blood so that the blastocyst can absorb nutrients from uterus. This process is called implantation.

Question 5.
Explain the following.

1. Hermaphrodites
2. Unisexual
3. Syngamy

Answer:

1. Hermaphrodites are bisexual organisms which possess both male and female reproductive organs. Examples: earthworm, leech, starfish.
2. Animals which have different male and female individuals as birds, mammals etc.
3. The process of fusion of male gamete with female gamete is called syngamy.

Question 6.
What is contraception? Discuss natural and barrier method of contraception.
Answer:
Contraception or birth control methods include: condoms, the diaphragm, the contraceptive pill, implants, IUDs (intrauterine devices), sterilization and the morning after pill and many more some of best methods are given below:

• Natural method: It involves avoiding the chances of meeting of sperms and ovum. In this method, the sexual intercourse is avoided by the couple from day 10th to 17th of the menstrual cycle of female as in this period, ovulation is expected and therefore, the chances of fertilisation are very high.
• Barrier method: In this method, the fertilisation of ovum and sperm is checked out with the help of artificially developed barriers. Barriers are developed for both males and females. Most common barrier available in market are condoms.

Question 7.
Describe implants and surgical methods of contraception
Answer:
Contraceptive devices are also developed as the loop or copper-T to prevent pregnancy. Surgical methods are also used to block the gamete transfer. It includes the blocking of vas deferens to prevent the transfer of Sperms known as vasectomy. Similarly, tubectomy in the fallopian tubes of the female can be blocked so that the egg will not reach the uterus.

Question 8.
Discuss fertilization in flowering plants.
Answer:
There are two main procedures of completing fertilization in flowering plants, which are:
(i) Pollination
(ii) Fertilisation

(i) Pollination: Pollination is a very important part of the life cycle of a flowering plant which results in seeds that grow into new plants. It is part of the sexual reproduction process of flowering plants. Flowers are the structures of flowering plants that contain all the specialized parts needed for sexual reproduction. Plants have gametes, which contain half the normal number of chromosomes for that plant species. Male gametes are found inside tiny pollen grains on the anthers of flowers. Female gametes are found in the ovules of a flower. Pollination is the process that brings these male and female gametes together. The wind or animals, especially insects and birds, pick up pollen from the male anthers and carry it to the female stigma. Flowers have to encourage animals to pollinate them.

(ii) Fertilisation: After pollination, when pollen has landed on the stigma of a suitable flower of the same species, various process occurs in the making of seeds. A pollen grain on the stigma grows a tiny tube, all the way down the style to the ovary. This pollen tube carries a male gamete to meet a female gamete in an ovule. In a process called fertilization, the two gametes join. The fertilised ovule form a seed, which contains a food store and an embryo that grow into a new plant. The ovary develops into a fruit to protect the seed.

Question 9.
Inside womb, how does a child receive food, oxygen and water? Discuss.
Answer:
As a mother eats something the nutrient like glucose, proteins, fats, vitamins, etc. are absorbed into the mother’s blood by the small intestine. The nutrients flow to the placenta, and then transferred to the baby’s bloodstream through the umbilical cord. The baby’s waste products (like CO2) are disposed of in the mother’s blood stream as well. In the placenta, the mothers blood flows into a network of blood Vessels and capillaries. Molecules in the mother’s blood like glucose, proteins, fats, oxygen etc. flow out of the mother’s blood supply and are absorbed into another network of blood vessels and capillaries containing the baby’s blood supply. The baby’s blood then flows through the umbilical cord back to the baby. It is the complete process of baby’s nutrition inside womb.

Question 10.
Discuss the advantages and disadvantage of autogamy or self¬pollination.
Answer:
Advantages of autogamy:

It is a sure method of seed formation. Scent and Nectar are not needed by the flower to attract insects. Parent characteristics are preserved in off spring’s. Small quantity of pollen is required for pollination. Flowers need not be large or attractive. Disadvantages of autogamy plants lose their vigor in their future generations due to repeated self-pollination. Since, there is no variation, no genetic improvement occurs in offsprings. Weak characteristics of the plant are inherited by the next generations.

MP Board Class 10th Science Chapter 8 Textbook Activities

Class 10 Science Activity 8.1 Page No. 129

• Dissolve about 10 gin of sugar in 100 mL of water.
• lake 20 mL of this solution in a test tube and add a pinch of yeasl granules to it.
• Put a cotton plug on the mouth of the test tube and keep it in a warm place.
• Alter 1 or 2 hours, put a small drop of yeast culture from the test tube on a slide and cover it with a coverslip.
• Observe the slide under a microscope.

Observations:

• Formation of yeast cells can be seen. Some of them, shows chain budding.

Class 10 Science Activity 8.2 Page No. 129

• Wet a slice of bread, and keep it in a cool, moist and dark place.
• Observe the surface of the slice with a magnifying glass.
• Record your observation for a week.

Observations:

• A layer of while cottony mass is seen over the surface of slice. These inercase in size and number and after a week, the layer turns black show ing formation of spores or sporangia.

Class 10 Science Activity 8.3 Page No. 129

• Observ e a permanent slide of Amoeba under a microscope.
• Similarly observe another permanent slide of Amoeba show-ing binary fission.
• Now, compare the observation of both the slides.

Observations:

• The permanent slide of amoeba shows normal cytoplasm and nucleus. Nucleus can be seen dividing and construction in cytoplasm can also be seen. The binary fission with two daughter cells is observed in the other slide.

Class 10 Science Activity 8.4 Page No. 129

• Collect water from a lake or pond that appears dark green and contains filamentous structures.
• Put one or two filaments on a slide.
• Put a drop of glycerin on these filaments and cover it with a coverslip
• Observe the slide under a microscope.
• Can you identify different tissues in the Spimgyra filaments.

Observations:

• Spirogyra filament consists of many cells which are attached linearly to form a filament.

Class 10 Science Activity 8.5 Page No. 132

• Take a potato and observe its surface. Can notches be seen?
• Cut the potato into small pieces such that some pieces contain a notch or bud and some do not.
• Spread some cotton on a tray and wet it. Place the potato pieces on this cotton. Note where the pieces with the buds are placed.
• Observe changes taking place in these potato pieces over the next few days. Make sure that the cotton is, kept moistened.
• Which arc the potato pieces that give rise to fresh green shoots and roots.

Observations:

• The,potato undergoes various changes in few days. The buds in notches show growth of young shoots and roots. The pieces which do not have eye buds do not show any growth.

Class 10 Science Activity 8.6 Page No. 132

• Select a money-plant.
• Cut some pieces such that they contain at least one leaf.
• Cut out some other portions between two leaves.
• Dip one end of all the pieces in water and observe over the next few days.
• Which ones grow’ and give rise to fresh leaves?
• What can you conclude from your observations ?

Observations:

• The leaves at the nodes show formation of fresh leaves. The formation of branch from axillary buds axil of leaf is also observ ed.
• The leaves that undergo photosynthesis show tendency to grow into a new plant through vegetative propagation.

Class 10 Science Activity 8.7 Page No. 135

• Soak a few seeds of Bengal gram (chana) and keep them overnight.
• Drain the excess water and cover the seeds with a wet cloth and leave them for a day. Make sure that the seeds do not become dry.
• Cut open the seeds carefully and observe the different parts.
• Compare your observations with the Fig. 8.2 and sec if you can identify all the parts.

Observations:

• The parts identified includes- cotyledon which stores food, plumule which is a future shoot radicle that is a future root.

Germination.

MP Board Class 10th Science Solutions

MP Board Class 10th Science Solutions Chapter 13 Magnetic Effects of Electric Current

MP Board Class 10th Science Chapter 13 Intext Questions

Class 10th Science Chapter 13 Intext Question Page No. 224

Question 1.
Why does a compass needle get deflected when brought near a bar magnet?
Answer:
A compass needle get deflected when brought near a bar magnet because a compass needle is in fact, a small bar magnet. The ends of the compass needle point approximately towards north and south directions.

Class 10th Science Chapter 13 Intext Questions Page No. 228

Question 1.
Draw magnetic field lines around a bar magnet.
Answer:
Magnetic field lines of a bar magnet emerge from the north pole and terminate at the south pole. Inside the magnet, the field lines emerge from the south pole and terminate at the north pole.

Question 2.
List the properties of magnetic field lines.
Answer:
The properties are:

• They travel from north pole to south pole outside the magnet and south pole to north pole inside the magnet.
• They are closed and continuous curves.
• Two magnetic field lines never intersect each other. If the lines intersect, then at the point of intersection there would be two directions [the needle would point towards two directions] for the same magnetic fields which is not possible.
• The number of field lines per unit area is the measure of the strength of magnetic field, which is maximum at poles. The magnetic field is strong, where the field lines are close together and weak where the lines are far apart.

Question 3.
Why don’t two magnetic field lines intersect each other?
Answer:
Two magnetic fields lines of force never intersect each other. If the lines intersect, then at [the point of intersection there would be two directions [the needle would point towards two directions] for the same magnetic field, which is not possible.

Class 10th Science Chapter 13 Intext Questions Pages No. 229. 230

Question 1.
Consider a circular loop of wire lying in the plane of the table. Let the current pass through the loop clockwise. Apply the right-hand rule to find out the direction of the magnetic field inside and outside the loop.
Answer:
Inside the loop = Pierce inside the table.

Outside the loop = Appear to emerge out from the table.

For downward direction of current flowing in the circular loop, the direction of magnetic field lines will be as if they are emerging from the table outside the loop and merging in the table inside the loop. Similarly, for upward direction of current flowing in the circular loop, the direction of magnetic field lines will be as if they are emerging from the table outside the loop and merging in the table inside the loop, as shown in the given figure.

Question 2.
The magnetic field in a given region is uniform. Draw a diagram to represent it.
Answer:
The magnetic field lines are parallel and equidistant.

Question 3.
Choose the correct option.
The magnetic field inside a long, straight, solenoid-carrying current
(a) is zero
(b) decreases as we move towards its end
(c) increases as we move towards its end
(d) is the same at all points
Answer:
(d) The magnetic field inside a long, straight, current-carrying solenoid is uniform. It is the same at all points inside the solenoid.

Class 10th Science Chapter 13 Intext Questions Pages No. 231,232

Question 1.
Which of the following property of a proton can change while it moves freely in a magnetic field? (There may be more than one correct answer.)
(a) mass
(b) speed
(c) velocity
(d) momentum
Answer:
(c) velocity
(d) momentum.

Question 2.
In Activity 13.7, how do we think the displacement of rod AB will be affected if

1. current in rod AB is increased;
2. a stronger horse-shoe magnet is used; and
3. length of the rod AB is increased?

Answer:

1. displacement of A is increased.
2. If a stronger horse-shoe magnet is used magnetic field is increasing.
3. current flows is more.

Question 3.
A positively-charged particle (alpha-particle) projected towards west is deflected towards north by a magnetic field. The direction of magnetic field is
(a) towards south
(b) towards east
(c) downward
(d) upward
Answer:
(d) upward.
Since the positively charged particle alpha particle projected towards west, so the direction of current is towards west. Now the deflection is towards north, so the force is towards north. Now hold the forefinger, centre finger and thumb of our left – hand at right angles to one another. Let us adjust the hand in such a way that our centre finger points towards west and thumb points towards north. If we look at our forefinger, it will be pointing, upward. Thus, the magnetic field is in the upward direction. So, the correct answer is (d).

Class 10th Science Chapter 13 Intext Questions Page No. 233

Question 1.
State Fleming’s left-hand rule.
Answer:
According to this rule, stretch the thumb, forefinger, and middle finger of your left hand such that they are mutually perpendicular. If the first finger points in the direction of the Magnetic field and the second finger in the direction of current, then the thumb will point in the direction of motion or the force acting on the conductor.

Question 2.
What is the principle of an electric motor?
Answer:
A current-carrying conductor when placed in a magnetic field experiences a force. This is the principle of an electric motor.

Question 3.
What is the role of the split ring in an electric motor?
Answer:
The split ring reverse the direction of current in the armature coil after every half rotation i.e., it acts as a commutator. The reversal of current reverses, the direction of the forces acting on the two arms of the armature after every half rotation. This allows the armature coil to rotate continuously in the same direction.

Class 10th Science Chapter 13 Intext Question Page No. 236

Question 1.
Explain different ways to induce current in a coil.
Answer:
The different ways to induce current in a coil are as follows:

(a) If a coil is moved rapidly between the two poles of a horse-shoe magnet, then an electric current is induced in the coil.
(b) If a magnet is moved relative to coil, then an electric current is induced in the coil.

Class 10th Science Chapter 13 Intext Questions Page No. 237

Question 1.
State the principle of an electric generator.
Answer:
An electric generator works on the principle of electromagnetic induction. It generates electricity by rotating a coil in a magnetic field.

Question 2.
Name some sources of direct current.
Answer:
Some sources of direct current are cell, Dc generator, etc.

Question 3.
Which sources produce alternating current?
Answer:
AC generators, power plants etc., produce alternating current.

Question 4.
Choose the correct option:
A rectangular coil of copper wires is rotated in a magnetic field. The direction of the induced current changes once in each
(a) two revolutions
(b) one revolution
(c) half revolution
(d) one-fourth revolution
Answer:
(c) When a rectangular coil of copper is rotated in a magnetic field, the direction of the induced current in the coil changes once in each half revolution. As a result, the direction of current in the coil remains the same.

Class 10th Science Chapter 13 Intext Questions Pages No. 238

Question 1.
Name two safety measures commonly used in electric circuits and appliances.
Answer:

1. Electric fuse
2. Earthing wire.

Question 2.
An electric oven of 2 kW power rating is operated in a domestic electric circuit (220 V) that has a current rating of 5 A. What result do you expect? Explain.
Answer:
P = VI
Here P = 2 KW = 2000 W
V = 220

The current drawn by this electric oven is 9 A whereas the fuse in the circuit is ( only 5 A capacity. When a high current of 9 A flows through the 5 A fuse, the fuse wire will get heated too much, melt and break, the circuit. Therefore, when a 2 kW power rating electric oven is operated in a circuit having a 5 A fuse will blow off cutting off the power supply in this circuit.

Question 3.
What precaution should be taken to avoid the overloading of domestic electric circuits?
Answer:

1. Each appliance has a separate switch to ON/OFF the flow of current through it.
2. The use of an electric fuse prevents the electric circuit and the appliance from possible damage by stopping the flow of unduly high electric current.
3. We should not connect too many appliances to a single socket to prevent overloading.

MP Board Class 10th Science Chapter 13 NCERT Textbook Exercises

Question 1.
Which of the following correctly describes the magnetic field near a long straight wire?
(a) The field consists of straight lines perpendicular to the wire
(b) The field consists of straight lines parallel to the wire
(c) The field consists of radial lines originating from the wire
(d) The field consists of concentric circles centered on the wire.
Answer:
(d) The magnetic field lines, produced around a straight current-carrying conductor are concentric circles. Their centres lie on the wire.

Question 2.
The phenomenon of electromagnetic induction is
(a) the process of charging a body
(b) the process of generating magnetic field due to a current passing through a coil
(c) producing induced current in a coil due to relative motion between a magnet and the coil
(d) the process of rotating a coil of an electric motor.
Answer:
(c) When a straight coil and a magnet are moved relative to each other, a current is induced in the coil. This phenomenon is known as electromagnetic induction.

Question 3.
The device used for producing electric current is called a
(a) generator
(b) galvanometer
(c) ammeter
(d) motor.
Answer:
(a) generator.

Question 4.
The essential difference between an AC generator and a DC generator is that
(a) AC generator has an electro-magnet while a DC generator has permanent magnet.
(b) DC generator will generate a higher voltage.
(c) AC generator will generate a higher voltage.
(d) AC generator has slip rings while the DC generator has a commutator.
Answer:
(c) AC generator will generate a higher voltage.

Question 5.
At the time of short circuit, the current in the circuit
(a) reduces substantially.
(b) does not change.
(c) increases heavily.
(d) vary continuously.
Answer:
(c) increases heavily.

Question 6.
State whether the following statements are true or false.
(a) An electric motor converts mechanical energy into electrical energy.
(b) An electric generator works on the principle of electromagnetic induction.
(c) The field at the centre of a long circular coil carrying current will be parallel straight lines.
(d) A wire with a green insulation is usually the live wire of an electric supply.
Answer:
(a) False
(b) true
(c) true
(d) False.

Question 7.
List two methods of producing magnetic fields.
Answer:

1. Permanent magnet
2. Electromagnet.

Question 8.
How does a solenoid behave like a magnet? Can you determine the north and south poles of a current-carrying solenoid with the help of a bar magnet? Explain.
Answer:
A solenoid is a long coil of circular loops of insulated copper wire. Magnetic field lines are produced around the solenoid when a current is allowed to flow through it. The magnetic field produced by it is similar to the magnetic field of a bar magnet. The field lines produced in a current¬carrying solenoid is shown in the following figure.

Fig. 13.4: Field lines of the magnetic field through and around a current carrying solenoid.

In the above figure, when the north pole of a bar magnet is brought near the end connected to the negative terminal of the battery, the solenoid repels the bar magnet. Since like poles repel each other, the end connected to the negative terminal of the battery behaves as the north pole of the solenoid and the other end behaves as a south pole. Hence, one end of the solenoid behaves as a north pole and the other end behaves as a south pole.

Question 9.
When is the force experienced by a current-carrying conductor placed in a magnetic field largest?
Answer:
The force experienced by a current -carrying conductor is the maximum when the direction of current is perpendicular to the direction of the magnetic field.

Question 10.
Imagine that you are sitting in a chamber with your back to one wall. An electron beam, moving horizontally from back wall towards the front wall, is deflected by a strong magnetic field to your right side. What is the direction of magnetic field?
Answer:
According to Fleming’s left-hand rule, the magnetic field acts in the vertically downward direction.
Note that the direction of current will be opposite to that of the electron beam.

Question 11.
Draw a labelled diagram of an electric motor. Explain its principle and working. What is the function of a split ring in an electric motor?
Answer:
An electric motor converts electrical energy into mechanical energy. It works on the principle of the magnetic effect of current. A current-carrying coil rotates in a magnetic field. The following figure shows a simple electric motor. When a current is allowed to flow through the coil MNST by closing the switch, the coil starts rotating anti-clockwise. This happens because a downward force

acts on length MN and at the same time, an upward force acts on length ST. As a result, the coil rotates anti-clockwise.

Current in the length MN flows from M to N and the magnetic field acts from left to right, normal to length MN. Therefore, according to Fleming’s left hand rule, a downward force acts on the length MN. Similarly, current is the length ST. flows from S to T and the magnetic fields acts from left to light, normal to the flow of current. Therefore, an upward force acts on the length ST. These two forces cause the coil to rotate anti-clockwise.

After half a rotation, the position of MN and ST interchange. The half¬ring D comes in contact with brush A and half-ring C comes in contact with brush B. Hence, the direction of current in the coil MNST gets reversed.

The current flow’s through the coil in the direction TSNM. The reversal of current through the coil MNST repeats after each half rotation. As a result, the coil rotate unidirectional ly. The split rings help to reverse the direction of current in the circuit. These are called the commutator.

Question 12.
Name some devices in which electric motors are used.
Answer:
Electric motor is used as an important component in electric fans, refrigerators, mixers, washing machines, computers, MP3 players etc.

Question 13.
A coil of insulated copper wire is connected to a galvanometer. What will happen if a bar magnet is

1. pushed into the coil
2. withdrawn from inside the coil
3. held stationary inside the coil?

Answer:

1. There is a momentary deflection in the needle of the galvanometer.
2. Now the galvanometer is deflected towards the left showing that the current is now set up in the direction opposite to the first.
3. When the coil is kept stationary with respect to the magnet, the deflection of the galvanometer drops to zero.

Question 14.
Two circular coils A and B are placed closed to each other. If the current in the coil A is changed, will some current be induced in coil B? Give reason.
Answer:
If the current in the coil A is changed there is a change in its magnetic field. By this electricity is induced in B. This is called Electromagnetic induction.

Question 15.
State the rule to determine the direction of a

1. magnetic field produced around a straight conductor carrying current,
2. force experienced by a current-carrying straight conductor placed in a magnetic field which is perpendicular to it, and
3. current induced in a coil due to its rotation in a magnetic field.

Answer:
(i) Right-hand thumb rule: If the current-carrying conductor is held in the right hand such that the thumb points in the direction of the current, then the direction of the curl of the fingers will be given the direction of the magnetic field.

(ii) Fleming’slefthandrule: Stretch the forefinger, the central finger of the right hand mutually perpendicular to each other. If the forefinger points in the direction of the magnetic field, the central finger in the direction of the current, then the thumb points in the direction of a force in the conductor.

(iii) Fleming’s right-hand rule: Stretch the thumb/ forefinger and the central finger of the right hand mutually perpendicular to each other. If the forefinger points in the direction of the magnetic field, thumb in the direction conductor, then the central finger points in the direction of current induced in the conductor.

Question 16.
Explain the underlying principle and working of an electric generator by drawing a labelled diagram. What is the function of brushes?
Answer:
An electric generator converts mechanical energy into electrical energy. The principle of working of an electric generator is that when a loop is moved in a magnetic field, an electric current is induced in the coil, ft generates electricity by rotating a coil in a magnetic field. The following figure shows a simple AC generator.

• MNST → Rectangular coil
• C and D → Two slip rings
• A and B → Brushes
• X → Axle, G → Galvanometer

If axle X is rotated clockwise, then the length MN moves upwards while length ST moves downwards. Since the lengths MN and ST are moving in a magnetic field, a current will be induced in both of them due to electromagnetic induction. Length MN is moving upwards and the magnetic field acts from left to right. Hence, according to Fleming’s right hand rule, the direction of induced current will be from M to N. Similarly, the direction of induced current in the length ST will be from S to T.

The direction of current in the coil is MNST. Hence, the galvanometer shows a deflection in a particular direction. After half a rotation, length MN starts moving down whereas length ST starts moving upward. The direction of the induced current in the coil gets reversed as TSNM. As the direction of current gets reversed after each half rotation the produced current is called an alternating current (AC).

To get a unidirectional current, instead of two slip rings, two split rings are used, as shown in the following figure.

In this arrangement, brush ‘A’ always remains in contact with the length of the coil that is moving up whereas brush B always remains in contact with the length that is moving down. The split rings C and D act as a commutator.

The direction of current induced in the coil will be MNST for the first rotation and TSNM in the second half of the rotation. Hence, unidirectional current is produced from the generator called DC generator. The current is called AC current.

Question 17.
When does an electric short circuit occur?
Answer:
If the resistance of an electric circuit becomes very low. Then the current flowing through the circuit becomes very high. This is caused by connecting too many appliances to a single socket or connecting – high power rating appliances to the light circuits. This results in a short circuit when the insulation of live and neutral wires undergoes wear and tear and then touches each other, the current flowing in the circuit increase abruptly. Hence, a short circuit occurs.

Question 18.
What is the function of an earth wire? Why is it necessary to earth metallic appliances?
Answer:
The metallic body of electric appliances is connected to the earth by means of earth wire so that any leakage of electric current is transferred to the ground. This prevents any electric shock to the user. That is why earthing of the electrical appliances is necessary.

MP Board Class 10th Science Chapter 13 Additional Important Questions

MP Board Class 10th Science Chapter 13 Multiple Choice Questions

Question 1.
Two magnet when come closer:
(a) Attract each other
(b) Repei each other
(c) Sometimes attract and sometimes repel
(d) No reaction
Answer:
(c) Sometimes attract and sometimes repel

Question 2.
Permanent magnet can be made by:
(a) Gold
(b) Carbon
(c) Alnico
(d) Wood
Answer:
(c) Alnico

Question 3.
When we draw magnetic field lines of any magnet it:
(a) Begin from N pole and end at S pole
(b) Begin from S pole and end at N pole
(c) Form circles around a magnet
(d) Form box around a magnet
Answer:
(a) Begin from N pole and end at S pole

Question 4.
Strongest magnetic pole around a magnet is
(a) Near North pole
(b) Near south pole
(c) In center of magnet
(d) At both poles
Answer:
(d) At both poles

Question 5.
Electro magnetism was discovered by
(a) Newton
(b) Oersted
(c) Ohm
(d) Joule
Answer:
(b) Oersted

Question 6.
Electro magnets are used in
(a) AC
(b) Fridge
(c) Radio
(d) All of these
Answer:
(d) All of these

Question 7.
Magnetic field are lines forming
(a) Straight lines
(b) Closed curves
(c) Dotted lines
(d) Dotted triangles
Answer:
(b) Closed curves

Question 8.
Magnetic field of a straight conductor form:
(a) Straight lines
(b) Concentric lines
(c) Points
(d) None of above
Answer:
(b) Concentric lines

Question 9.
According to right hand thumb rule, current is generated in a system:
(a) Parallel to magnetic field
(b) Perpendicular to magnetic field
(c) Just above the magnetic field
(d) All of these.
Answer:
(b) Perpendicular to magnetic field

Question 10.
If a circular loop conductor has 5 turn, magnetic field produced by it will be to than single loop conductor.
(a) 5 times bigger
(b) 5 times lesser
(c) Similar
(d) 10 times lesser
Answer:
(a) 5 times bigger

Question 11.
Magnetic field generated due to a passing current in a solenoid forms pattern:
(a) Similar to circular loops
(b) Similar to straight conductor
(c) Similar to Bar magnet
(d) Which is unique
Answer:
(c) Similar to Bar magnet

Question 12.
According to Fleming’s left hand rule
(a) Field is perpendicular to current
(b) Field is perpendicular to force generated
(c) Current is perpendicular to force generated .
(d) All of these
Answer:
(d) All of these

MP Board Class 10th Science Chapter 13 Very Short Answer Type Questions

Question 1.
What we call the end, where north end of freely hanged magnet stops?
Answer:
North pole.

Question 2.
Where force of a magnet can be detected?
Answer:
Near a magnet, in its magnetic field.

Question 3.
How we mark a magnetic force?
Answer:
With the help of magnetic field lines.

Question 4.
What do closer field lines of a magnet presents?
Answer:
Closer lines present stronger magnetic strength.

Question 5.
What is reason for different kinds of pattern of a magnetic pattern around a conductor generated magnetic field?
Answer:
Shape of conductor.

Question 6.
Give some examples of daily use where electro magnet is being used.
Answer:
Radio and Television.

Question 7.
How an electromagnet is formed?
Answer:
Electro magnet is formed by wrapping coil of insulated copper wire over core of soft iron.

Question 8.
What is experienced by a current carrying conductor when placed in a magnetic field?
Answer:
Force.

Question 9.
Who gave the theory of induced electromagnetic induction?
Answer:
Michael Faraday.

Question 10.
What is a galvanometer?
Answer:
An instrument which detects current in a circuit.

Question 11.
Give examples of appliances using electric motor.
Answer:
Electric pan, mixer, washing machines.

Question 12.
Which rule of electromagnetism suits best to express working of an electric motor?
Answer:
Fleming’s left hand rule.

Question 13.
What is the basic principle used in electric generator?
Answer:
It is based on electromagnetic induction.

Question 14.
What is the permissible value of electricity use in India?
Answer:
220 V, 50Hz.

Question 15.
How short-circuit can be prevented?
Answer:
By using fuse.

MP Board Class 10th Science Chapter 13 Short Answer Type Questions

Question 1.
Write two properties of magnet.
Answer:

1. A magnet always points north or south when suspended freely.
2. Like pole repel each other while opposite pole attract each other.

Question 2.
Define magnetism.
Answer:
A magnet influence its near by object and attracts towards itself, if magnetic in nature. This phenomenon is called magnetism.

Question 3.
Define magnetic field.
Answer:
Area under which a magnet can influence other magnetic objects is called its magnetic field.

Question 4.
Where we find closer line and what does it indicate?
Answer:
Closer lines show high strength of magnetic field while wide lines represents weak strength of magnet.

Question 5.
State Right hand thumb rule.
Answer:
Right hand thumb rule is represented by a down thumb fist in which fist represents a magnetic field while thumb represents current movement.

MP Board Class 10th Science Chapter 13 Long Answer Type Questions

Question 1.
An electric heater rated 800 W operates 6h/day. Find the cost of energy to operate it for 30 days at ₹ 3.00 per unit.
Solution:
Power of the heater, P = 800 W Time, t = 6 hour /day
No; of days, n = 30 Cost per unit = ₹ 3.00
Total cost of its usage = ? Energy, E = P × t
Consumed in 1 day = 800 × 6 = 4800 Wh
Energy consumed in 30 days = 4800 × 30 = 144000 Wh

∴ Cost of 144 units = 3 × 144 = ₹ 432.

Question 2.
(a) Draw magnetic field lines produced around a current carrying straight conductor passing through cardboard. How will the strength of the magnetic field change, when the point where magnetic field is to be determined, is moved away from the straight wire carrying constant current ? Justify your answer.
(b) Two circular coils A and B are placed close to each other. If the current in the coil A is changed, will some current be induced in the coil B? Give reason.
Answer:
(a) (i) The magnetic field lines around a straight conductor carrying current are concentric circles whose center lies on the wire.
(ii) When a point where magnetic field is to be determined is moved away from the straight wire, the strength of the magnetic field decreases because as we move away from a current carrying straight conductor, the concentric circles around it representing magnetic field lines become larger and larger indicating the decreasing strength of magnetic field.

Fig. 13.18: A pattern of concentric circles indicating the field lines of a magnetic field a straight conducting wire.

(b) Yes, current is induced in the coil B. Because as the current in the coil A changes, the magnetic field lines around the coil B also change. Therefore, the change in magnetic field lines associated with the coil B is the cause of induced electric current in it.

Question 3.
(a) Draw magnetic field lines of a bar magnet. “Two magnetic field lines never intersect each other.” Why?
(b) An electric oven of 1.5 kW is operated in a domestic circuit (220 V) that has a current rating of 5 A. What result do you expect in this case? Explain.
Answer:
(a)

Two magnetic field lines do not intersect one another. The direction of magnetic field lines is always from north pole to south pole. If the two magnetic field line do intersect, it means at the point of intersection the compass needle is showing two different directions which is not possible.

(b) Power, P = 1.5 kW = 1.5 × 1,000 = 1,500 W
Voltage, V = 220 V, I = ?
P = V × I I = \(\frac { P }{ V } \) = \(\frac { 1,500 }{ 220 } \) = 6.8 A
Now, the current drawn by the oven is 6.8 A which is very high but the fuse in this circuit is only 5 A capacity. When a very high current of 6.8. A flows through 5 A fuse, the fuse wire will get heated too much, melt and break the circuit, cutting off the power supply.

Question 4.
A circuit has a line of 5 A. How many lamps of rating 40W; 220V can simultaneously run on this line safely?
Answer:

Question 5.
A bulb is rated at 200 V, 100 W. Calculate its resistance. Five such bulbs burn for 4 hours daily. Calculate the units of electrical energy consumed per day. What would be the cost of using these bulbs per day at the rate of ₹ 4.00 per unit?
Solution:

Electrical energy consumed, E = P × t
Energy consumed by 1 bulb = 0.1 × 4 = 0.4 kWh
∴ Energy consumed by 5 bulb = 5 × 0.4 = 2 kWh = 2 units
Cost of electrical energy:
Cost of 1 unit of electricity = ₹ 4
∴ Cost of 2 units of electricity = 4 × 2 = ₹ 8

Question 6.
(a) Describe an activity to show with the help of a compass that magnetic field is strongest near poles of a magnet.
(b) Mention the direction of magnetic field lines (i) inside a bar magnet and (ii) outside a bar magnet.
Answer:
(a) A bar magnet is placed on a sheet of paper and its boundary is marked with a pencil. A magnetic compass is brought near the N-pole of the bar magnet. It is observed that N-pole of magnet repels the N-pole of compass needle due to which the tip of the compass needle moves away from the N-pole. Thus, a magnetic field pattern is obtained around a bar magnet.

(b) Each magnetic field line is directed from the north pole of a magnet to its south pole. The field lines are closest together at the two poles of the bar magnet.

1. Inside a bar magnet, the lines of forces start from south pole and end on north pole.
2. Outside a bar magnet, magnetic lines of forces start from north pole and end on south pole.

Question 7.
(a) With the help of a labelled diagram, describe an activity to show that a current carrying conductor experiences a force when placed in a magnetic field. Mention the position when this force is maximum.
(b) Name and state the rule which gives the direction of force acting on the conductor.
Answer:
(a) Activity:

• A small aluminium rod (AB) about 5 cm is suspended with two connecting wires horizontally from a stand.
• A strong horse-shoe magnet is placed in such a way that the rod lies between the two poles with the magnetic field directed upwards, the north pole of the magnet vertically below and south pole vertically above the aluminium rod.
  
  Fig. 13.20: A current-carrying conductor experiences force in magnetic field.

• The aluminium rod is connected in series with a battery, a key and a rheostat.
• When a current is allowed to pass through aluminium rod. Form end B to end A, it is observed that the rod is displaced towards the left.
• When the direction of the current is reversed from A to B, it is observed that the direction of displacement- of the rod is towards the right.

This activity shows that when a current carrying conductor is placed in a magnetic field, a mechanical force is exerted on conductor which makes it move.

The maximum force is exerted on a current carrying conductor only when it is perpendicular to the-direction of magnetic field.

(b) The direction of force acting on the current carrying conductor can be found out by using Fleming’s left-land rule.

According to Fleming’s left-hand rule, hold the fore finger, the center finger and the thumb of your left hand at right angles to one another. If the first finger of your left hand points in the direction of magnetic field and sound in the direction of current, then the thumb will point in the direction of motion or the force acting on conductor.

Question 8.
Study the following current-time graphs from two different sources:

1. Use above graphs to list two differences between the current in the two cases.
2. Name the type of current in the two cases.
3. Identify one source each for these currents.
4. What is meant by the statement that “ the frequency of current in India is 50 Hz”?

Answer:

1. D.C. – Direct Current.
2. A.C.- Alternating Current.
3. Source of D.C. → a cell, battery, solar cell, D.C. generator.
   Source of A.C. → A.C. Generator.
4. The frequency of current in India is 50 Hz means the direction of current in India changes 50 times in 1 second.

Question 9.
Explain two disadvantages of series arrangement for household circuit.
Answer:
Disadvantages of series circuits for domestic wiring:

1. In series circuit, if one electrical appliance stops working due to some defect then all other appliances also stop working because the whole circuit is broken.
2. In series circuit, all the electrical appliances have only one switch , due to which they cannot be turned off or turned on separately.

Question 10.
(i) State Maxwell’s right-hand thumb rule.
(ii) PQ is a current carrying conductor in the plane of the paper as shown in the figure. Mention the direction of magnetic fields produced by it at points A and B.
Given r₁ < r₂ where will the strength of the magnetic field be larger?
Answer:
(i) Maxwell’s right hand thumb rule: The direction of the current is given by Maxwell’s right hand thdmb rule, “If the current carrying conductor is gripped with the right hand in such a way that the thumb gives the direction of the current, then the direction of the fingers gives the direction of the magnetic field produced around the conductor”.

(ii) Since the direction of current in the straight conductor is from Q to P, then according to Maxwell’s right hand thumb rule, magnetic field at point A is inside the paper and at point B is outside the paper. Since r₁ < r₂ the strength of the magnetic field at A is more than at B because greater the
distance of a point from the current carrying wire, weaker will be the magnetic field produced at that point.

MP Board Class 10th Science Chapter 13 NCERT textbook activities

Class 10 Science Activity 13.1 Page No. 223

• Take a straight thick copper wire and place it between the points X and Y in an electric circuit, as shown in Fig. 13.8. The wire XY is kept perpendicular to the plane of paper.
  
  Compass needle is deflected on passing an electric current through a metallic conductor.
• Horizontally place a small compass near to this copper wire. See the position of its needle.
• Pass the current through the circuit by inserting the key into the plug.
• Observe the change in the position of the compass needle.

Observations:

• The needle is deflected showing that electric current through the copper wire has produced a magnetic effect.

Class 10 Science Activity 13.2 Page No. 224

• Fix a sheet of white paper on a drawing board using some adhesive material.
• Place a bar magnet in the centre of it.
• Sprinkle some iron filings uniformly around the bar magnet (Fig. 13.9). A salt – sprinkler may be used for this purpose.
• Now tap the board gently.
• What do you observe?
  
  Fig. 13.9: Iron filings near the bar magnet align themselves along the field lines

Observations:

• It is observed that iron filings arrange themselves in a pattern of concentric circles. This shows that iron filings experience a force due to magnetic effect of a magnetic. The lines along which the iron filings align themselves are magnetic field lines.

Class 10 Science Activity 13.3 Pages No. 224,225

• Take a small compass and a bar magnet
• Place the magnet on a sheet of white paper fixed on a drawing board, using some adhesive material.
• Mark the boundary of the magnet.
  
  Fig. 13.10: Drawing a magnetic field line with the help of a compass needle.
• Place the compass near the north pole of the magnet. How does it behave? The south pole Fig 13 10: Drawing a magnetic field line of the needle points with the help of a compass needle, towards the north pole of the magnet. The north pole of the compass is directed away from the north pole of the magnet.
• Mark the position of two ends of the needle.
• Now move the needle to a new position such that its south pole occupies the position previously occupied by its north pole.
• In this way, proceed step by step till you reach the south pole of the magnet as shown in Fig. 13.10.
  
  Fig. 13.11: Field lines around a bar magnet.
• Join the points marked on the paper by a smooth curve. This curve represents a field line.
• Repeat the above procedure and draw as many lines as you can. You will get a pattern shown in Fig. 13.11. These lines represent the magnetic field around the magnet. These are known as magnetic field lines.
• Observe the deflection in the compass needle as you move it along a field line. The deflection increases as the needle is moved towards the poles.

Observations:

• The magnetic field is strong at the poles due to which deflection increases at the poles as the needle move towards it.

Class 10 Science Activity 13.4 Page No. 226

• Take a long straight copper wire, two or three cells ofl .5 V each, and a plug key. Connect all of them in scries as shown in Fig. 13.12 (a)
• Place the straight wire parallel to and over a compass needle.
• Plug the key in the circuit.
• Observe the direction of deflection of the north pole of the needle. If the current flows from north to south, as shown in Fig. 13.12 (a), the north pole of the compass needle would move towards the east.
• Replace the cell connections in the circuit as shown in Fig. 13.12 (b). This would result in the change of the direction of current through the copper wire, that is, from south to north.
• Observe the change in the direction of deflection of the needle You will see that now the needle moves in opposite direction, that is, towards the west (Fig. 13.12 (b)). It means that the direction of magnetic field produced by the electric current is also reversed.
  Fig. 13.12: A simple electric circuit in which a straight copper wire is placed parallel to and over a compass needle. The deflection in the needle becomes opposite when the direction of the current is reversed.

Observations:

• As current flow changes its direction from south to north, the needle in the compass moves in a opposite direction that is towards the west. This shows the direction of magnetic filed produced by the electric current is also reversed.

Class 10 Science Activity 13.5 Pages No. 226, 227

• Take a battery (12 V), a variable resistance (or a rheostat), an ammeter (0 5 A), a plug key, connecting wires and a long straight thick copper wire.
• Insert the thick wire through the centre, normal to the plane of a rectangular cardboard. Take care that the cardboard is fixed and does not slide up or down.
  Fig. 13.13: (a) A pattern of concentric circles indicating the field lines of a magnetic field around a straight conducting wire. The arrows in the circles show the direction of the field lines. (b) A close up of the pattern obtained.

• Connect the copper wire vertically between the points X and Y, as shown in Fig. 13.13 (a), in series with the battery, a plug and key.
• Sprinkle some iron filings uniformly on the cardboard. (You may use a salt sprinkler for this purpose).
• Keep the variable of the rheostat at a fixed position and note the . current through the ammeter.
• Close the key so that a current flows through the wire. Ensure that the copper wire placed between the points X and Y remains vertically straight.
• Gently tap the cardboard a few times. Observe the pattern of the iron filings. You would find that the iron filings align themselves showing a pattern of concentric circles around the copper wire (Fig. 13.13).
• What do these concentric circles represent? They represent the magnetic field lines.
• How can the direction of the magnetic field be found? Place a compass at a point (say P) over a circle. Observe the direction of the needle. The direction of the north pole of the compass needle would give the direction of the field lines produced by the electric current through the straight wire at point P. Show the direction by an arrow.
• Does the direction of magnetic field lines get reversed if the direction of current through the straight copper wire is reversed? Check it.

Observations:

• The deflection is the needle changes. If the current is increased, the deflection also increases. It indicates the magnitude of the magnetic field produced at a given point increases as the current through the wire increases.

Class 10 Science Activity 13.6 Page No. 229

• Take a rectangular cardboard having two holes. Insert a circular coil having large number of turns through them, normal to the plane of the cardboard.
• Connect the ends of the coil in series with a battery, a key and a rheostat, as shown in Fig. 13.14.
• Sprinkle iron filings uniformly on the cardboard.
• Plug the key.
• Tap the cardboard gently a few times. Note the pattern of the iron filings that emerges on the cardboard.

Observations:

Fig. 13.14: Magnetic field produced by a current carrying circular coil.

Class 10 Science Activity 13.7 Page No. 230

• Take a small aluminium rod AB (of about 5 cm). Using two connecting wires suspend it horizontally from a stand, as shown in Fig. 13.15.
• Place a strong horse-shoe magnet in such a way that the rod lies between the two poles with the magnetic field directed upwards. For this put the north pole of the magnet vertically below and south pole vertically above the aluminium rod (Fig. 13.15).
  
  Fig. 13.15: A current carrying rod, AB, experiences a force perpendicular to its length and the magnetic field.
• Connect the aluminium rod in series with a battery, a key and a rheostat.
• Now pass a current through the aluminium rod from end B to end A.
• What do you observe? It is observed that the rod is displaced towards the left. Yr- will notice that the rod gets displaced.
• Reverse the direction of current flowing through the rod and observe the direction of its displacement. It is now towards the right.
  Why does the rod get displaced?

Observations:

• The rod is displaced due to force exerted on the current – carrying aluminium rod when placed in a magnetic field.
• The direction of force is recedes as the direction of anent through the induction is reversed.

Class 10 Science Activity 13.8 Pages No. 233,234

• Take a coil of wire AB having a large number of turns.
• Connect the ends of the coil to a galvanometer as shown in Fig. 13.16.
• Take a strong bar magnet and move its north pole towards the end B of the coil. Do you find any change in the galvanometer needle?
• There is a momentary deflection in the needle of the galvanometer, say to the right. This indicates the presence of a current in the coil AB. The deflection becomes zero the moment the motion of the magnet stops.
• Now withdraw the north pole of the magnet away from the coil. Now the galvanometer is deflected toward the left, showing that the current is now set up in the direction opposite to the first.
• Place the magnet stationary at a point near to the coil, keeping its north pole towards the end B of the coil. Wc sec that the galvanometer needle deflects toward the right when the coil is moved towards the north pole of the magnet. Similarly the needle moves toward left when the coil is moved aways.
  
  Fig. 13.16: Moving a magnet towards a coil sets up a current in the coil circuit, as indicated by deflection in the galvanometer needle.

• When the coil is kept stationary with respect to the magnet, the deflection of the galvanometer drops to zero. What do you conclude from this activity?

Observations:

• From this activity, it can be concluded that the motion of a magnet with respect to the coil produces an induced potential difference, which sets up an induced electric current in the circuit.

Class 10 Science Activity 13.9 Page No. 235

• Take two different coils of copper wire having large number of turns (say 50 and 100 turns respectively). Insert them over a non-conducting cylindrical roll, as shown in Fig. 13.17. (You may use a thick paper roll for this purpose.)
• Connect the coil-1, having larger number of turns, in series with a battery and a plug key. Also connect the other coil-2 with a galvanometer as shown.
• Plug in the key. Observe the galvanometer. Is there a deflection in its needle? You will observe that the needle of the galvanometer instantly jumps to one side and just as quickly returns to zero, indicating a momentary current in coil-2.
• Disconnect coil-1 from the battery. You will observe that the needle momentarily moves, but to the opposite side. It means that now the current flows in the opposite direction in coil-2
  Fig. 13.17: Current is induced in coil-2 when current in coil-1 is changed.

Observations:

• We will observe that the needle of th e galvanometer instantly jumps to one side and just quickly returns to its initial position.

MP Board Class 10th Science Solutions

MP Board Class 10th Science Solutions Chapter 6 Life Processes

MP Board Class 10th Science Chapter 6 Intext Questions

Class 10th Science Chapter 6 Intext Questions Page No. 95

Question 1.
Why is diffusion insufficient to meet the oxygen requirements of multi cellular organisms like humans?
Answer:
Diffusion is insufficient to meet the oxygen requirement of multicellular organisms like humans because multicellular organisms have complex body designs. Moreover, all their cells may not be in direct contact with the surrounding environment.

Question 2.
What criteria do we use to decide whether something is alive?
Answer:
We tend to think of some sort of movement, either growth-related or not, as common evidence for being alive.

Question 3.
What are outside raw materials used for by an organism?
Answer:
Food, water and oxygen are outside raw materials mostly used by an organism. Depending on the complexity of the organism its requirement varies from organism to organism.

Question 4.
What processes would you consider essential for maintaining life?
Answer:
Nutrition, Respiration, Transportation of materials into the body and Excretion are the processes we consider essential for maintaining life.

Class 10th Science Chapter 6 Intext Questions Page No. 101

Question 1.
What are the differences between autotrophic nutrition and heterotrophic nutrition?
Answer:

Autotrophic Nutrition	Heterotrophic Nutrition
It is the process by which autotrophs take in substances from the outside and convert them into stored forms of energy. This mate­rial is taken in the form of carbon dioxide and water which is converted into carbohydrates in the presence of sunlight and chlorophyll	This involves the intake of complex material prepared by other convert organisms
Chlorophyll is necessary.	Chlorophyll is absent.

Question 2.
Where do plants get each of the raw materials required for photosynthesis?
Answer:
The following raw materials are required for photosynthesis:

• The CO₂ enters from the atmosphere through stomata.
• Water is absorbed from the soil by the plant roots.
• Sunlight, an important component for manufacture food, is absorbed by the chlorophyll and other green parts of the plants.

Question 3.
What is the role of the acid in our stomach?
Answer:

1. The hydrochloric acid reacts an acidic medium which facilitates the action of the enzyme pepsin.
2. Acid in the stomach kills micro organisms.

Question 4.
What is the function of digestive enzymes?
Answer:
Digestive enzymes such as amylase, lipase, pepsin, trypsin etc. help in the breaking down of complex food particles into simpler ones. These simple particles can be easily absorbed by the blood and thus, transported to all the cells of the body.

Question 5.
How is the small intestine designed to absorb digested food?
Answer:
The small intestine is designed in such a way that the digested end products are easily absorbed into the body. The innermost lining of the small intestine has many finger-like foldings called villi which increase the surface area for absorption. The villi are richly supplied with blood capillaries and a large lymph vessel which takes the absorbed food to all the cells of the body.

Class 10th Science Chapter 6 Intext Questions Page No. 105

Question 1.
What advantage over an aquatic organism does a terrestrial organism have with regard to obtaining oxygen for respiration?
Answer:
Terrestrial organisms take up oxygen from the atmosphere whereas aquatic animals need to utilize oxygen present in the water. Air contains more O₂ as compared to water. Since, the content of O₂ in air is high, the terrestrial animals do not have to breathe faster to get more oxygen. Therefore, unlike aquatic animals, terrestrial animals do not show various adaptations for better gaseous exchange.

Question 2.
What are the different ways in which glucose is oxidised to provide energy in various organisms?
Answer:
Glucose is first broken down in the cell cytoplasm into a three carbon molecules called pyruvate. Pyruvate is further broken down by different ways to provide energy. The breakdown of glucose by different pathways can be illustrated as follows,

Break-down of glucose by various pathways.
In yeast and human muscle cells, the breakdown of pyruvate occurs in the absence of oxygen whereas in mitochondria, the breakdown of pyruvate occurs in the presence of oxygen.

Question 3.
How is oxygen and carbon dioxide transported in human beings?
Answer:
In human beings, respiratory pigments take up oxygen from the air in the lungs and carry it to tissues which are deficient in oxygen before releasing it. The respiratory pigment is haemoglobin which has a very high affinity for oxygen. This pigment is present in the red blood corpuscles. Carbon dioxide is more soluble in water than oxygen is and hence is mostly transport the dissolved from in our blood.

Question 4.
How are the lungs designed in human beings to maximize the area for exchange of gases?
Answer:
Within the lungs, the passage divides into smaller and smaller tubes which finally terminate in ballon like structures which are called alveoli. The alveoli provide a surface where the exchange of gases can take place.

Class 10th Science Chapter 6 Intext Questions Page No. 110

Question 1.
What are the components of the transport system in human beings? What are the functions of these components?
Answer:
The components of the transport system in human beings are the heart, blood and blood vessels and lymph:

• Heart pumps oxygenated blood throughout the body. It takes deoxygenated blood from the various body parts and sends this impure blood to the lungs for oxygenation.
• Being a fluid connective tissue, blood helps in the transport of oxygen, nutrients, CO₂ and nitrogenous wastes.
• The blood vessels (arteries, veins and capillaries) carry blood either away from the heart to various organs or from various organs, back to the heart.

Question 2.
Why is it necessary to separate oxygenated and deoxygenated blood in mammals and birds?
Answer:
Because mammals and birds require more energy. Hence there must be separation of oxygenated blood and deoxygenated blood. By this these organisms get sufficient oxygen and helps to maintain their body temperature.

Question 3.
What are the components of the transport system in highly organised plants?
Answer:
Xylem and Phloem are the two main types of conducting tissues. In highly organised plants, xylem conducts water and minerals obtained from the soil (via roots) to the rest of the plant. Phloem transports food materials from the leaves to different parts of the plant body.

Question 4.
How are water and minerals transported in plants?
Answer:
In xylem tissue, vessels and tracheids of the roots, stems and leaves are inter connected to form a continuous system of water conducting channels reaching all parts of the plant. At the roots, cells in contact with the soil actively take up ions. This creates a difference in the concentration of these ions between the root and the soil. Water therefore moves into the root from the soil to eliminate this difference. This means that there is steady movement of water into root xylem, creating a column of water that is steadily pushed upwards.

Question 5.
How is food transported in plants?
Answer:
The transport of soluble products of photosynthesis is called translocation and it occurs in the part of the vascular tissue known as phloem. Besides the products of photosynthesis, the phloem transports amino acids and other substances. These substances are specially delivered to the storage organs of roots, fruits and seeds and to growing organs. The translocation of food and other substances takes place in the sieve tubes with the help of adjacent companion cells both in upward and downward directions.

Class 10th Science Chapter 6 Intext Questions Page No. 112

Question 1.
Describe the structure and functioning of nephrons.
Answer:
Nephrons are the basic units of kidneys. Each kidney possesses large number of nephrons, approximately 1-1.5 million. The main components of the nephron are glomerulus, Bowman’s capsule and a long renal tubule.

Structure of a nephron:

Structure of a nephron

Functioning of a nephron:

• The blood with metabolic waste enters the kidney through the renal artery, which branches into many capillaries associated with glomerulus.
• The water and solute are drained to the nephron at Bowman’s capsule.
• In the proximal tubule, some substances such as amino acids, glucose and salts are selectively reabsorbed and unwanted molecules are added in the urine.
• The filtrate then moves down into the loop of Henle, where more water is absorbed. The amount of water reabsorbed depends on how much excess water is present in the body and on, how much of dissolved waste is to be excreted.
• From here, the filtrate moves into the distal tubule and finally reach to the collecting duct Collecting duct collects urine from many nephrons.
• The urine formed in each kidney enters a long tube called ureter. From ureter, it gets transported to the urinary bladder and then into the urethra.

Question 2.
What are the methods used by plants to get rid of excretory products?
Answer:
Plants can get rid of excess water by transpiration. For other wastes, plants use the fact that many of their tissues consist of dead cells, and that they can even lose some parts such as leaves. Many plant waste products are stored incellular vacuoles waste products may be stored in leaves that fall off. Other waste products are stored as resins and gums especially in old xylem. Plants also excrete some waste substances into the soil around them.

Question 3.
How is the amount of urine produced regulated?
Answer:
The amount of urine produced depends on the amount of excess water and dissolved wastes present in the body. Some other factors such as habitat of an organism and hormone such as Antidiuretic hormone (ADH) also regulates the amount of urine produced.

MP Board Class 10th Science Chapter 6 NCERT Textbook Exercises

Question 1.
The kidneys in human beings are a part of the system for:
(a) nutrition
(b) respiration
(c) excretion
(d) transportation
Answer:
(c) In human beings, the kidneys are a part of the system for excretion.

Question 2.
The xylem in plants are responsible for:
(a) transport of water
(b) transport of food
(c) transport of amino acids
(d) transport of oxygen
Answer:
(a) In a plant, the xylem is responsible for transport of water.

Question 3.
The autotrophic mode of nutrition requires:
(a) carbon dioxide and water
(b) chlorophyll
(c) sunlight
(d) all of the above.
Answer:
(d) The autotrophic mode of nutrition requires carbon dioxide, water, chlorophyll and sunlight, all the four components.

Question 4.
The breakdown of pyruvate to give carbon dioxide, water and energy takes place in:
(a) cytoplasm
(b) mitochondria
(c) chloroplast
(d) nucleus
Answer:
(b) The breakdown of pyruvate to give carbon dioxide, water and energy takes place in mitochondria.

Question 5.
How are fats digested in our bodies? Where does this process take place?
Answer:
Fats are present in the intestine in the form of large globules which makes it difficult for enzymes to act on them. Bile salts break them down into smaller globules increasing the efficiency of enzyme action. The enzymes present in the wall of the small intestine converts fats into fatty acids and glycerol.

Question 6.
What is the role of saliva in the digestion of food?
Answer:
Saliva is secreted by the salivary Glands, located around tongue. It moistens the food for easy swallowing. It contains a digestive enzyme – amyjase, which breaks down bulky starch into sugar. So, sometimes it is advised to consume less water during a meal.

Question 7.
What are the necessary conditions for autotrophic nutrition and what are its by-products?
Answer:
Autotrophic nutrition is accomplished by the process of photosynthesis. Carbon dioxide, water, chlorophyll pigment and sunlight are the necessary conditions required for autotrophiq nutrition. Carbohydrates (food) and 02 are the by-products of photosynthesis.

Question 8.
What are the differences between aerobic and anaerobic respiration? Name some organisms that use the anaerobic mode of respiration.
Answer:

Aerobic respiration	Anaerobic respiration
i) This take place in presence of oxygen	No oxygen is
ii) It takes place in cytoplasm and mitochondria.	it takes place only in cytoplasm
iii) By this process more energy is released.	Less energy is released by this process.

Question 9.

How are the alveoli designed to maximise the exchange of gases?
Answer:
The alveoli are the small hollow structures present in the lungs. The walls of the alveoli consist of extensive network of blood vessels. Each lung contains 300-350 million alveoli, making it a total of approximately 700 millions in both the lungs. The alveolar surface when spread out covers about 80m² area. This large surface area makes the gaseous exchange more efficient in alveoli and capillaries.

Question 10.
What would be the consequences of a deficiency of haemoglobin in our bodies?
Answer:
Haemoglobin is the respiratory pigment that transports oxygen to the body cells for cellular respiration. Therefore, deficiency of haemoglobin in blood can affect the oxygen supplying capacity of blood. This can lead to deficiency of oxygen in the body cells. It can also lead to a disease called anemia.

Question 11.
Describe double circulation in human beings. Why is it necessary?
Answer:
Oxygen rich-blood from the lungs comes to the thin walled upper chamber of the heart on the left, the left atrium. The left atrium relaxes when it is collecting this blood. It then contracts while the next chamber, the left ventricle, relaxes, so that the blood is transferred to it. When the muscular left ventricle contracts in its turn, the blood is pumped out to th body. De-oxygenated blood comes from the body to the upper chamber on the right, the right atrium, as it relaxes. As the right atrium contracts, the corresponding lower chamber, the right ventricle, dilates. This transfer blood to the right ventricle, which in turn pumps it to the lungs for oxygenation.

Blood goes through the heart twice during each cycle in other vertebrates. This is known as double circulation. The separation of the right side and the left side of the heart is useful to keep oxygenated and deoxygenated blood from mixing.

Importance of double circulation:

The separation of oxygenated and de-oxygenated blood allows a more efficient supply of oxygen to every single cells. This efficient system of oxygen supply is very useful in warm-blooded animals such as human beings. As we know, warm-blooded animals have to maintain a constant body temperature. Thus, the circulatory system of humans becomes more efficient because of the double circulation.

Schematic sectional view of the human heart.

Question 12.
What are the differences between the transport of materials in xylem and phloem?
Answer:
(i) Xylem tissue helps in the transport of water and minerals.

• Phloem tissue helps in the transport of food.

(ii) Water is transported upwards from roots to all other plant parts.

• Food is transported in both upward and downward directions. Transport in xylem occurs with the help of simple physical forces such as transpiration pull.

(iii) Transport of water and minerals do not require energy in the form of ATP.

• Transport of food in phloem requires energy in the form of ATP.

Question 13.
Compare the functioning of alveoli in the lungs and nephrons in the kidneys with respect to their structure and functioning.
Answer:
Structure of Alveoli

1. Alveoli are tiny balloon-like structures present inside the lungs.
2. The walls of the alveoli are one cell thick and it contains an extensive network of blood capillaries.

Functions:

1. The exchange of O₂ and CO₂ takes place between the blood of the capillaries that surround the alveoli and the gases present in the alveoli.
2. Alveoli are the site of gaseous exchange.

Structure Nephrons

1. Nephrons are tubular structures present inside the kidneys.
2. Nephrons are made of glomerulus, bowman’s capsule, and a long renal tube. It also contains a cluster of thin walled capillaries.

Functions:

(i) The blood enters the kidneys through the renal artery which branches into many capillaries in the glomerulus. The water and solute are transferred to the nephron at Bowman’s capsule. Then, the filtrate moves through the proximal tubule and then down into the loop of henle. From henle’s loop, filtrate passes into the distal tubule and then to the collecting duct. The collecting duct collects the urine from many nephrons and passes it to the ureter. During the ’ flow of filtrate some substances such as glucose, amino acid and water are selectively re-absorbed.

MP Board Class 10th Science Chapter 6 Additional Questions

MP Board Class 10th Science Chapter 6 Multiple Choice Questions

Question 1.
Which out of the following can prepare their own food:
(a) Carnivores
(b) Omnivores
(c) Herbivores
(d) Autotrophs
Answer:
(d) Autotrophs

Question 2.
Which of the following type includes plant as majority of its population?
(a) Carnivores
(b) Omnivores
(c) Herbivores
(d) Autotrophs
Answer:
(d) Autotrophs

Question 3.
Which of the given organisms can feed on plants and animals as well?
(a) Carnivores
(b) Omnivores
(c) Herbivores
(d) Autotrophs
Answer:
(b) Omnivores

Question 4.
Amylase is secreated by:
(a) Pancreas
(b) Stomach
(c) Small intestine
(d) Oesophagus
Answer:
(a) Pancreas

Question 5.
of the following metal is associated with chlorophyll?
(a) Aluminium
(b) Iron
(c) Potassium
(d) Calcium
Answer:
(b) Iron

Question 6.
Which of the following metals are helpful for cellular level transport?
(a) AI – Zn
(b) Fe
(c) Na – K
(d) Calcium
Answer:
(c) Na – K

Question 7.
Where does water get absorbed in body?
(a) Stomach
(b) Food canal
(c) Large intestine
(d) Small intestine
Answer:
(c) Large intestine

Question 8.
The lungs in human beings helps in:
(a) Excretion
(b) Nutrition
(c) Respiration
(d) Transportation
Answer:
(c) Respiration

Question 9.
The liver in human beings helps in:
(a) Excretion
(b) Digestion
(c) Respiration
(d) Transportation
Answer:
(b) Digestion

Question 10.
Bile is originated from in human digestive system.
(a) Pancreas
(b) Liver
(c) Kidney
(d) Stomach
Answer:
(b) Liver

Question 11.
Villi are find inside
(a) Brain
(b) Stomach
(c) Small intestine
(d) Oesophagus
Ans.
(c) Small intestine

Question 12.
The decomposition of carbohydrate is a process of:
(a) Esterification
(b) Hydrogenation
(c) Oxidation
(d) Emulsification
Answer:
(c) Oxidation

Question 13.
Which one of the following organism use air dissolved in water for respiration?
(a) Amoeba
(b) Sheep
(c) Lion
(d) Leech
Answer:
(a) Amoeba

Question 14.
Energy is released and stored in the form of ATP during:
(a) Excretion
(b) Nutrition
(c) Respiration
(d) Transportation
Answer:
(c) Respiration

Question 15.
Nephron are unit of kidney, it works for:
(a) Nutrition
(b) Respiration
(c) Excretion
(d) Transportation
Answer:
(c) Excretion

Question 16.
The phloem in plants are responsible for:
(a) Transport of water
(b) Transport of food
(c) Transport of amino acids
(d) Transport of oxygen
Answer:
(b) Transport of food

Question 17.
The autotrophic mode of nutrition requires:
(а) Carbon dioxide and water
(b) Chlorophyll and sunlight
(c) Carbohydrate
(d) (a) and (b)
Answer:
(d) (a) and (b)

Question 18.
Energy generated during cellular level of metabolism is stored in:
(a) Cytoplasm
(b) Mitochondria
(c) Chloroplast
(d) Nucleus
Answer:
(b) Mitochondria

Question 19.
During expiration, the lungs are:
(a) Arched
(b) Flattened s
(c) Perforated
(d) None of these
Answer:
(a) Arched

Question 20.
The correct pathway of blood in double circulatory system is:
(a) atria → ventricles → arteries → veins
(b) atria → veins → arteries
(c) veins → arteries → atria
(d) veins → ventricles → atria → arteries
Answer:
(a) atria → ventricles → arteries → veins

MP Board Class 10th Science Chapter 6 Very Short Answer Type Questions

Question 1.
Name one process which cannot be seen in non living things in comparison to living beings.
Answer:
Reproduction.

Question 2.
Najne two pigments which absorb sunlight.
Answer:
Carotenoids and chlorophyll.

Question 3.
Which life process synthesise chemical energy and turn it into heat energy?
Answer:
Nutrition and respiration.

Question 4.
Name some carbohydrate rich food.
Answer:
Rice and wheat.

Question 5.
Name raw material required for photosynthesis.
Answer:
Carbon dioxide, water, chlorophyll rich living cell, sunlight.

Question 6.
In which form, the food is stored in leaves?
Answer:
Starch.

Question 7.
What is the colour of iodine-starch complex?
Answer:
Blue-Purple.

Question 8.
During blood circulation in humans, in one cycle how many times blood moves to heart?
Answer:
Two times.

Question 9.
How many chambers do a human heart have?
Answer:
Four.

Question 10.
What is the term for process of taking food in the body?
Answer:
Ingestion.

Question 11.
Which gland secretes amylase enzyme?
Answer:
Salivary gland.

Question 12.
What is the function of large intestine during digestion?
Answer:
Absorption of water.

Question 13.
How do aquatic plants and animal get oxygen for photosynthesis?
Answer:
Aquatic plants and animals obtain oxygen through the process of diffusion.

Question 14.
Where does light reaction and dark reaction of photosynthesis occur?
Answer:
In the grana thylakoids of chloroplasts, light reaction occurs while dark reaction occur in the stroma of chloroplasts.

Question 15.
Which wavelength of light is best absorbed by chlorophyll?
Answer:
Red colour wavelength.

Question 16.
Which products formed during light reaction in photosynthesis process are used by dark reaction?
Answer:
NADPH and ATP.

Question 17.
What is the function of thylakoid membranes in chloroplast?
Answer:
It provides large surface area for light absorption.

Question 18.
Write down the full form of the following:

(a) ATP
(b) NADP

Answer:

(a) ATP : Adenosine triphosphate.
(b) NADP : Nicotinamide adenine dinucleotide phosphate.

Question 19.
Define excretion.
Answer:
Excretion is a biological process by which an organism gets rid of metabolic toxic wastes from its body.

Question 20.
Name the toxic wastes obtained as by products of metabolism.
Answer:
These wastes are nitrogenous materials i.e., ammonia, urea and uric acid, carbon dioxide, inorganic salts, excess of water.

Question 21.
What is osmoregulation?
Answer:
It is a process that maintains the amount of water and ionic balance in the body.

Question 22.
Where urea is formed in humans?
Answer:
Urea is made in the liver via the ornithine cycle by combining the ammonia made by deamination with carbon dioxide made from respiration.

Question 23.
What are the major excretory products of plants?
Answer:
Oxygen, water, carbon dioxide, latex, gums, resins, excessive salts, calcium oxalate and other toxic substances are excretory organs of plants.

Question 24.
Name the excretory organs of earthworm.
Answer:
Nephridia.

MP Board Class 10th Science Chapter 6 Short Answer Type Questions

Question 1.
What is gout? How it affects life?
Answer:
In humans, the high concentrations of uric acid in the blood cause uric acid crystals to precipitate in the kidneys and joints. This can cause huge pain and swelling in the joints, particularly in the big toes and disturb the movement of body. .

Question 2.
List the excretory system organs in human beings.
Answer:
The excretory system in humans consists of:

• a pair of kidneys
• a pair of ureter
• a urinary bladder
• a urethra

Question 3.
Explain unit of kidney.
Answer:
Nephron is the structural and functional unit of kidney. Each kidney of the pair contains millions of nephrons.

Question 4.
Which organ in human beings is related to homeostasis?
Answer:
Kidney is also concerned with homeostasis since, it carries out osmoregulation of body fluids and controls the pH of the blood.

Question 5.
How does excretion takes place in segmented worms, like earthworms?
Answer:
Segmented worms, such as earthworms, produce urea that is excreted through long tubules called nephridia.

Question 6.
What are the excretory organs of insects like grasshoppers?
Answer:
Malpighian tubules.

Question 7.
Why does uric acid is the best nitrogenous waste product for insects, reptiles and birds?
Answer:
Uric acid has low solubility and does conserve water in insects, reptiles and birds.

Question 8.
Name the two possible treatments during chronic renal (kidney) failure.
Answer:
Hemodialysis and kidney transplant.

Question 9.
What is the main excretory product in Amoeba and jellyfish?
Answer:
Ammonia (NH₃).

Question 10.
What are the two main functions of kidneys?
Answer:
The two main functions of kidneys are:

1. To remove toxic metabolic waste products (for example, urea, uric acid, ammonia, salts etc.) from the blood.
2. To regulate the blood pH, blood water and salt content, blood osmotic pressure, blood pressure (homeostasis).

Question 11.
Why is photosynthesis important to the global world?
Answer:

(a) In photosynthesis, solar energy (sunlight) is converted to chemical energy and is stored in plants as starch.
(b) Plants also store excess sugar by synthesis of starch.
(c) Heterotrophs, including humans, may completely or partially consume plants for fuel and raw materials.
(d) Photosynthesis is responsible for the presence of oxygen in our atmosphere.
(e) Each year, photosynthesis synthesizes approx. 160 billion metric tons of carbohydrate.

MP Board Class 10th Science Chapter 6 Long Answer Type Questions

Question 1.
What is photosynthesis? Explain in detail.
Answer:
The process by which green plants make their own food from carbon- dioxide and water in the presence of sunlight and chlorophyll is called photosynthesis.

During this process, oxygen gas is released. This process can be represented as:

The green plants convert energy of sunlight into chemical energy by making glucose.

The extra glucose formed changes into starch which is stored in leaves. The oxygen released comes from the water.

Question 2.
(a) Define translocation.
Answer:
Transportation of food from photosynthetic parts of the plant to the non-green part of the plant through phloem is known as translocation.

(b) Name the correct substrates for the following enzymes.
(i) Trypsin
(ii) Amylase
(iii) Pepsin
(iv) Lipase
Answer:

S.No.	Enzyme	Substrate
(i)	Trypsin	Proteins (Peptides)
(ii)	Amylase	Starch
(iii)	Pepsin	Proteins
(iv)	Lipase	Emulsified fats

Question 3.
What are the important events that occur during photosynthesis process?
Answer:
The following events occur during this process:

(a) Absorption of light energy by chlorophyll.
(b) Conversion of light energy to chemical energy and splitting of water molecules into hydrogen and oxygen. (Light reaction)
(c) Reduction of carbon dioxide to carbohydrates. (Dark reaction by Calvin cycle)

Question 4.
Mention the conditions necessary for photosynthesis. Also, mention the process involved in each of these steps.
Answer:

1. Sunlight: Chlorophyll of the leaves of the plant trap sunlight and converts them into chemical energy during photosynthesis. Plant utilizes visible light only which is made up of 7 colours and green . colour is least absorbed (it reflects green the most – that’s why the leaves appear green).
2. Chlorophyll: Chlorophyll is a green pigment mainly present in the leaves of the plant. There are many types of chlorophyll named a, b, c, d, e and bacteria – chlorophyll. Chlorophyll ‘a’ and chlorophyll ‘b’ are most abundantly present in the nature.
3. Carbon Dioxide: Plants take carbon dioxide through stomata to make glucose (food for the plants).
4. Water: Water is absorbed by the roots of the plants through osmosis. Through xylem, water is transported to all parts of the plant.

MP Board Class 10th Science Chapter 6 Textbook Activities

Class 10 Science Activity 6.1 Page No. 96

1. Take a potted plant with variegated leaves – for example, money plant or crotons.
2. Keep the plant in a dark room for three days so that all the starch gets used up.
3. Now keep the plant in sunlight for about six hours.
4. Pluck a leaf from the plant. Mark the green areas in it and trace them on a sheet of paper.
5. Dip the leaf in boiling water for a few minutes.
6. After this, immerse it in a beaker containing alcohol.
7. Carefully place the above beaker in a water-bath and heat till the alcohol begins to boil.
8. What happens to the colour of the leaf? What is the colour of the solution?
9. Now dip the leaf in a dilute solution of iodine for a few minutes,
10. Take out the leaf and rinse off the iodine solution.
11. Observe the colour of the leaf and compare this with the tracing of the leaf done in the beginning Fig. 6.4.
12. What can you conclude about the presence of starch in various areas of the leaf?

Observations:

1. The green coloured leaf become colourless. The solution of alcohol becomes green in colour as chlorophyll of leaf gets dissolved it upon immersing.
2. leaf On dipping in iodine solution, green areas of leaf turns dark blue whereas colourless part of leaf show no formation of starch.

Class 10 Science Activity 6.2 Page No. 97

1. Take two healthy potted plants which are nearly the same size.
2. Keep them in a dark room for three days.
3. Now place each plant on separate glass plates. Place a watch-glass containing potassium hydroxide by the side of one of the plants. The potassium hydroxide is used to absorb carbon dioxide.
4. Cover both plants with separate bell-jars as shown in Fig. 6.5.

1. Use vaseline to seal the bottom of the jars to the glass plates so that the set-up is air-tight.
2. Keep the plants in sunlight for about two hours.
3. Pluck a leaf from each plant and check for the presence of starch as in the above activity.
4. Do both the leaves show the presence of the same amount of starch?
5. What can you conclude from this activity?

Observations:

1. No. both leaves do not show the presence of the same amount of starch as starch is produced during the process of photosynthesis utilizing sunlight, chlorophyll and CO₂ In first set up, availability of CO₂, will be less as potassium hydroxide present absorbs CO₂, In the second set up, leaves will have more amount of starch.
2. From this activity, we can conclude that amount of carbon dioxide affects the process and outcome of photosynthesis.

Class 10 Science Activity 6.3 Page No. 99

1. Take 1 mL starch solution (1%) in two test tubes (A and B).
2. Add 1 mL saliva to test tube A and leave both test tubes undisturbed for 20-30 minutes.
3. Now add a few drops of dilute iodine solution to the test tubes.
4. In which test tube do you observe a colour change?
5. What does this indicate about the presence or absence of starch in the two test tubes?
6. What does this tell us about the action of saliva on starch.

Observations:

1. The colour change is observed in test tube B, showing presence of starch. Whereas test tube A will show no colour change as saliva present converts sugar into starch.
2. This tells us about the action of salivary amylase enzymes present in starch. The salivary amylase acts on starch and breakdown into sugar.
   

Class 10 Science Activity 6.4 Page No. 101

1. Take some freshly prepared lime water in a test tube.
2. Blow air through this lime water.
3. Note how long it takes for the lime water to turn milky.
4. Use a syringe or pichkari to pass air through some fresh lime water taken in another test tube Fig. 6.6.

1. Note how long it takes for this lime water to turn milky.
2. What does this tell us about the amount of carbon dioxide in the air that we breathe out?

Observations:

1. Immediately, on blowing air in turns milky,
2. On using syringe / pichkari, much time is taken as amount of CO₂ entering now is restricted.
3. It Shows, the air we breathe out contains higher amount of CO₂.

Class 10 Science Activity 6.5 Page No. 101

1. Take some fruit juice or sugar solution and add some yeast to this. Take this mixture in a test tube fitted with a one-holed cork.
2. Fit the cork with a bent glass tube. Dip the free end of the glass tube into a test tube containing freshly prepared lime water.
3. What change is observed in the lime water arid how long does it take, for this change to occur?
4. What’does this tell us about the pradabts of fermentation?

Observations:

1. Lime water turns milky as CO₂ produced gets mixed with yeast, sugar and afeohol,
2. The products of fermentation.are-CO₂ and alcohol.

Class 10 Science Activity 6.6 Page No. 103

1. Observe fish in an aquarium. They open and close their mouths and the gill-slits (or the operculum which covers the gill-slits) behind their eyes also open and close. Are the timings of the opening and closing of the mouth and gill-slits coordinated In some mariner?
2. Count the number of times the fish opens and closes its mouth in a minute.
3. Compare this to the number of times you breathe in and out in a minute.

Observations:

1. Yes, the timings of opening and closing of mouth and gill slits are coordinated. They open and close alternatively.
2. In dissolved water, availability of oxygen is less so fish breathes at higher rate.

Class 10 Science Activity 6.7 Page No. 105

1. Visit a health centre in your locality and find out what is the normal range of haemoglobin content in human beings.
2. Is it the same for children and adults?
3. Is there any difference In the haemoglobin levels for men and women?
4. Visit a veterinary clinic in your locality. Find out what is the normal range of haemoglobin content in an animal like the buffalo or cow.
5. Is this content different in calves, male and female animals?
6. Compare the difference seen in male and female human beings and animals.
7. How would the difference. If any, be explained?

Observations:

1. The normal haemoglobin content for human male is 13.8-17.2 g/ decilitre and for female is 12.1 – 15.1 g/dl.
2. No. it is not same for children and adults.
3. Male has higher level of Hb level as compared to females.
4. Normal buffalo or low Hb level is 10.4-16.4 g/dl.
5. Yes, Hb content in calves is higher than male and female animals.

Class 10 Science Activity 6.8 Page No. 109

1. Take two small pots of approximately the same size and having the same amount of soil. One should have a plant in it. Place a stick of the same height as the plant in the other pot.
2. Cover the soil in both pots with a plastic sheet so that moisture cannot escape by evaporation.
3. Cover both sets, one with the plant and the other with the stick, with plastic sheets and place in bright sunlight for half an hour.
4. Do you observe any difference in the two cases?

Observations:

1. Yes, water droplets can be seen on the plastic sheet covering the pot with plant. This is due to condensation of water vapours released during transpiration.

MP Board Class 10th Science Solutions