Class 11

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

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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)

Find the sequence calculator allows to calculate online the terms of the sequence whose index is between two limits.

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) \(\frac {y}{1 – y}\)

Limit of sequence calculator is an online tool that evaluates limits for the given functions and shows all steps.

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

The Infinite Series Calculator an online tool, which shows Infinite Series for the given input.

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 11th Maths Important Questions Chapter 8 Binomial Theorem

Binomial Theorem Important Questions

Binomial Theorem Objective Type Questions

(A) Choose the correct option :

Question 1.
The total number of terms in the expansion of
(a) 7
(b) 12
(c) 13
(d) 6.
Answer:
(c) 13

Question 2.
If y = 3x + 6x² + 10x³ + …………… ∞, then the correct relation will be :
(a) x = 1 – (1 + y)^{–\(\frac { 1 }{ 3 }\)}
(b) x = (1 + y)^{–\(\frac { 1 }{ 3 }\)}
(c) y = 1 – (1 – x)^-3
(d) x = 1 + (1 + y)^{–\(\frac { 1 }{ 3 }\)}
Answer:
(a) x = 1 – (1 + y)^{–\(\frac { 1 }{ 3 }\)}

Question 3.
The total number of terms in the expansion of (1+ x)^-1 will be :
(a) 0
(b) ∞
(c) 2
(d) It can not be expand
Answer:
(b) ∞

Question 4.
The mid – term in the expansion of (x – \(\frac { 1 }{ x }\))¹⁰ will be :
(a) – ¹⁰C₅
(b) ¹⁰C₅
(c) 251
(d) 252
Answer:
(a) – ¹⁰C₅

An online remainder theorem calculator allows you to determine the remainder of given polynomial expressions by remainder theorem.

Question 5.
For all positive integer of n, n(n – 1) is :
(a) Integer
(b) Natural number
(c) Even positive integer
(d) Odd positive integer.
Answer:
(c) Even positive integer

Question 6.
Expansion of (a + x)ⁿ is :
(a) aⁿ + ⁿC₁a^n-1x + ⁿC₂a^n-2x² + ……….. + ⁿC_ra^n-rx^r + ………… + aⁿ
(b) xⁿ + ⁿC₁x^n-1a +ⁿC₂x^n-2a² + ……….. + ⁿC_rx^n-ra^r + ………… + aⁿ
(c) aⁿ – ⁿC₁a^n-1x + ⁿC₂a^n-2x² + ……….. + (-1)^rnC_ra^n-rx^r + ………… + (-1)ⁿaⁿ
(d) xⁿ – ⁿC₁a^n-1a + ⁿC₂a^n-2a² + ……….. + (-1)^rnC_ra^n-rx^r + ………… + (-1)ⁿxⁿ
Answer:
(a) aⁿ + ⁿC₁a^n-1x +ⁿC₂a^n-2x² + ……….. + ⁿC_ra^n-rx^r + ………… + aⁿ

Question 7.
The fifth term in the expansion of ( x – \(\frac { 1 }{ x }\) )¹⁰ from the end, will be :
(a) \(\frac { ^{ 10 }{ C }_{ 6 } }{ x }\)
(b) \(\frac { 105 }{ 32{ x }^{ 2 } }\)
(c) \(\frac { ^{ 10 }{ C }_{ 6 } }{ { x }^{ 2 } }\)
(d) \(\frac { ^{ 10 }{ C }_{ 6 } }{ { x }^{ 10 } }\)
Answer:
(a) \(\frac { ^{ 10 }{ C }_{ 6 } }{ x }\)
[Hint: The total number of terms be 11 in the expansion of it.
∵ The 5^th term from end = (11 – 5)^th = 7^th term from the beginning.]

Question 8.
The number of mid – terms in the expansion of ( x – \(\frac { 1 }{ x }\) )¹⁰
(a) 1
(b) 2
(c) – \(\frac { ^{ 13 }{ C }_{ 7 } }{ x }\)
(d) 1716x
Answer:
(b) 2

Question 9.
The value of ⁿC₀ + ⁿC₁ + ⁿC₂ + …………….. + ⁿC_n :
(a) 2ⁿ + 1
(b) 2^{n – 1}
(c) 2ⁿ – 1
(d) 2ⁿ
Answer:
(d) 2ⁿ

Question 10.
The value of ⁿC₀ + ⁿC₂ + ⁿC₄ + …………….. = ⁿC₁ + ⁿC₃ + ……………. will be :
(a) 2ⁿ + 1
(b) 2^{n – 1}
(c) 2ⁿ – 1
(d) 2ⁿ
Answer:
(b) 2^{n – 1}

The Chebyshev’s Theorem calculator, above, will allow you to enter any value of k greater than 1.

Question 11.
The total number of terms in the expansion of (a + b + c + d)ⁿ will be :
(a) \(\frac { (n+1)(n+2) }{ 2 }\)
(b) \(\frac { n(n+1) }{ 2 }\)
(c) \(\frac { (n+1)(n+2)(n+3) }{ 6 }\)
(d) \(\frac { (n+1)(n+2) }{ 6 }\)
Answer:
(c) \(\frac { (n+1)(n+2)(n+3) }{ 6 }\)

Question 12.
The necessary condition for expansion of (1 + x)^-1 is :
(a) | x | < 1
(b) | x | > 1
(c) | x | = 1
(d) | x | = – 1.
Answer:
(a) | x | < 1

Question 13.
The general term in the expansion of (x + a)ⁿ will be :
(a) r^th
(b) (r+1)^thterm
(c) (r-1)^th
(d) (r+2)^thterm
Answer:
(b) (r+1)^thterm

Question 14.
In the expansion of ( 2x + \(\frac { 1 }{ { 3x }^{ 2 } }\) )⁹, then term independent of x will be :
(a) \(\frac { 8 }{ 127 }\)
(b) \(\frac { 124 }{ 81 }\)
(c) \(\frac { 1792 }{ 9 }\)
(d) \(\frac { 256 }{ 243 }\)
Answer:
(c) \(\frac { 1792 }{ 9 }\)

Question 15.
The coefficient of x3 in the expansion of ( x – \(\frac { 1 }{ x }\) )¹⁵ is :
(a) 14
(b) 21
(c) 28
(d) 35
Answer:
(b) 21

Question 16.
The independent term in the expansion of (x² – \(\frac { 2 }{ { x }^{ 3 } }\))¹⁵ is :
(a) 5^th
(b) 6^th
(c) 7^th
(d) 8^th
Answer:
(c) 7^th

Question 17.
The value of ⁿC₀ + ⁿC₁ + ⁿC₂ + …………….. = ⁿC_n the expansion of (l + x)ⁿ is :
(a) 2ⁿ – 1
(b) 2ⁿ – 2
(c) 2ⁿ
(d) 2^n-1
Answer:
(c) 2ⁿ

Question 18.
The value of ¹⁵C₀ + ¹⁵C₂ + ¹⁵C₄ + ¹⁵C₆ + …………….. = ¹⁵C₁₄ is :
(a) 2¹⁴
(b) 2¹⁵
(c) 2¹⁵ – 1
(d) None of these
Answer:
(a) 2¹⁴

(B) Match the following :

Answer:

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

(C) Fill in the blanks :

1. By binomial theorem the value of (102)⁴ is ……………..
2. The value of second term in the expansion of (1 – x)^-3/2 is ……………..
3. The 5^th term from the end in the expansion of (x – \(\frac { 1 }{ 2x }\) )¹⁰ is ……………..
4. The constant term will be …………….. in the expansion of ( x² – 2 + \(\frac { 1 }{ { x }^{ 2 } }\) )⁶
5. The value of C₁ + 2C₂ + 3C₃ + ………….. + nC_n will be ……………..
6. The value of ⁿC₀ – ⁿC₁ + ⁿC₂ – ⁿC₃ + ………….. will be ……………..
7. The value of is ……………..
8. The value of is ……………..
9. The value of is ……………..
10. (2x + 3y)⁵ = …………….. up to three terms.
11. The coefficient of x⁷ in the expansion of (x² + \(\frac { 1 }{ x }\) )¹¹ will be …………….
12. In the expansion of (1 – x)¹⁰ the value of middle term is ……………..
13. Third (3^rd) term in the expansion of e^-3x will be ……………..
14. If n is odd, in the expansion of (x + a)ⁿ, then number of middle terms are ……………..
15. The middle term in the expansion of (\(\frac { x }{ a }\) + \(\frac { a }{ x }\) )¹⁰
16. The coefficient of xⁿ in the expansion of (1 + x) (1 – x)ⁿ will be ……………..

Answer:

1. 08243216
2. \(\frac { { x }^{ 3 } }{ 16 }\)
3. \(\frac { 105 }{ 32{ x }^{ 2 } }\)
4. 924
5. n.2^n-1
6. 0
7. \(\frac { n(n+1) }{ 2 }\)
8. \(\frac { n(n+1)(2n+1) }{ 6 }\)
9. [ \(\frac { n(n+1) }{ 2 }\) ]²
10. 32x⁵ + 240x⁴y + 720x³y²
11. 462
12. – 252 x^-5
13. \(\frac { 1 }{ 2 }\)
14. Two
15. 252
16. (- 1)ⁿ(1 – n)

(D) Write true / false :

1. The expansion of (1 + x)^-3 is 1 – 3x + 6x² – 10x³ + ………….. + \(\frac { (- 1)(r + 1)(r +2) }{ 2! }\)x^r + ……………..
2. The expansion of (1 – x)^-3 is 1 + 3x + 6x² + 10x³ + ………….. + \(\frac { (- 1)^r(r + 1)(r +2) }{ 2! }\)x^r + ……………..
3. The expansion of (1 – x)^-2 is 1 + 2x + 3x² + (r + 1) x^r+ ………….. +
4. The (r + 1 )^th term in the expansion of (1 – x)^-2 is (- 1)^r(r + 1) x^r+ ………….. +
5. The (r + 1 )^th term in the expansion of (1 – x)ⁿ will be x^r
6. The total number of terms in the expansion of (a + b + c)ⁿ is \(\frac { (n + 1)(n + 1) }{ 2 }\)
7. In the expansion of ( 3x – \(\frac { { x }^{ 3 } }{ 9 } \) )⁹, No . of terms is 9.
8. The number of term in the expansion of ( 3x – \(\frac { { x }^{ 3 } }{ 9 } \) )⁹ is 8.
9. In the expansion of (x + a)ⁿ then sum of powers of x and α in any term is n.
10. The coefficient of x in the expansion of (1 – 2x)^-3 is 6.
11. The second term in the expansion of (2x + 3y)⁵ is 240x⁴y.
12. The value of second term in the expansion of (1 – x)^-3/2 is \(\frac { 3 }{ 2 }\)x.

Answer:

1. True
2. True
3. True
4. True
5. False
6. True
7. False
8. False
9. True
10. False
11. True
12. True.

(E) Write answer in one word / sentence :

1. Find the value of 999³ by the binomial theorem
2. Find the middle term in expansion of (x² – \(\frac { 1 }{ x }\))⁶
3. General term in the expansion of (x + a)ⁿ will be.
4. If in the expansion of (1+x)⁵¹ the coefficient of x^r and x^{r – 5} are equal, then the value of r will be.
5. Find the coefficient of xⁿ in the expansion of (1 + x + x² + …………… ∞)², if | x | < 1.
6. In the expansion of (\(\frac { x }{ 3 }\) – \(\frac { 2 }{ { x }^{ 2 } }\))¹⁰, x⁴ comes in r^th term, then the value of r will be.
7. The 5^th term in the expansion of (1 – 2x)^{– 1} will be.

Answer:

1. 997002999
2. – 20 x⁵
3. ⁿC_r x^{n – r} a^r
4. 28
5. (n + 1)
6. 3
7. 16x²

Binomial Theorem Long Answer Type Questions

Question 1.
Expand : (\(\frac { 2 }{ x }\) – \(\frac { x }{ 2 }\))⁵ (NCERT)
Solution:

Question 2.
Expand : (2x – 3)⁶ (NCERT)
Solution:

Question 3.
Expand : (\(\frac { x }{ 3 }\) + \(\frac { 1 }{ x }\))⁵ (NCERT)
Solution:

Question 4.
Expand : (x + \(\frac { 1 }{ x }\))⁶. (NCERT)
Solution:

Question 5.
find 13^th term in the expansion of ( 9x – \(\frac { 1 }{ 3\sqrt { x } }\) )¹⁸. (NCERT)
Solution:

Question 6.
Find the middle term of (3 – \(\frac { { x }^{ 3 } }{ 6 }\))⁷
Solution:

Question 7.
Find the middle term in the expansion of (\(\frac { x }{ 3 }\) + 9y)¹⁰
Solution:
Here n =10
Total number of terms = n + 1 = 10 + 1 = 11 (odd)
Here, the term will be middle term.

Question 8.
If coefficient of x² and x³ in the expansion of (3 + ax)⁹ are equal, the value of a.
Solution:

Question 9.
Find the coefficient of x⁵ in the expansion of (x + 3)⁸
Solution:
Suppose x⁵ appears in (r + 1)^th term T_r+1 = nC₁x^n-ra^r
Here n = 8, x = x, a = 3
T_r+1 = ⁸C_r(x)^{8 – r}(3)^r
For the coefficient of x⁵,
8 – r = 5
=> r = 3
T₃₊₁ = ⁸C₃(3)³
= \(\frac { 8 × 7 × 6 }{ 3 × 2 × 1}\) × 3 × 3 × 3 × x⁵
= 1512 × x⁵
Hence coefficient of x⁵ is 1512.

Question 10.
Find the coefficient of a⁵b⁷ in the expansion of (a – 2b)¹².
Solution:

Question 11.
If the 17^th and 18^th terms in the expansion of (2 + a)⁵⁰ are equal, then find the value of a. (NCERT)
Solution:
In the expansion of (x + a)ⁿ
T_r+1 = ⁿC_r x^{n – r} a^r
Here n = 50, x = 2, a = a
T₁₇ = T_{16 + 1} = ⁵⁰C₁₆ (2)^{50 – 16} (a)¹⁶
⇒ T₁₇ = ⁵⁰C₁₆ (2)³⁴ (a)¹⁶
and T₁₈ = T_{17 + 1} = ⁵⁰C₁₇ (2)^{50 – 17} (a)¹⁷
= ⁵⁰C₁₇ (2)³³ (a)¹⁷

Question 12.
Prove that the value of the middle term in the expansion of (1+x)²ⁿ is \(\frac { { 1.3.5 …….. (2n – 1)} }{ n! }\).2ⁿ xⁿ
Solution:

Question 13.
In the expansion of (x + 1)ⁿ, the coefficient of the (r – 1)^th, r^th and (r + 1)^th terms are in the ratio 1 : 3 : 5, then find the value of n and r.
Solution:
In the expansion of (x + 1)ⁿ,
T_{r + 1} = ⁿC_rx^{n – r}(1)^r
T_{r – 1} = T_{r – 2 + 1} = ⁿC_{r – 2}(x)^{n – (r – 2)}(1)^{r – 2}
Coefficient of T_{r – 1}^th term = ⁿC_{r – 2}
T_r = T_{r – 1 + 1} = ⁿC_{r – 1}(x)^{n – (r – 1)}(1)^{r – 1}
Coefficient of T_r^thterm = ⁿC_{r – 1}
T_{r + 1} = ⁿC_r x^{n – r} (1)^r
Coefficient of T_r+1^th term = ⁿC_r

Put n = 4r – 5 from equation (1) in equation (2),
3(4r – 5) – 8r = – 3
⇒ 12r – 15 – 8r = – 3
⇒ 4r = 12
∴ r = 3
Put r = 3 in equation (2),
n – 4 x 3 = – 5
⇒ n = 12 – 5
⇒ n = l
n = 7, r = 3

Question 14.
Prove that the coefficient of xⁿ in the expansion of (1 + x)²ⁿ of in the expansion of (1 + x)^{2n – 1}.
Solution:
In the expansion of (x + a)ⁿ
T_r+1 = ⁿC_r x^{n – r} a^r
Here x = 1, a = x, n = 2n
T_r+1 = ²ⁿC_r(1)^{2n – r}(x)^r
For the coefficient of xⁿ, put r = n,
T_n+1 = ²ⁿC_n(a)^{2n – n}(x)ⁿ
and T₁₈ = T_{17 + 1} = ⁵⁰C₁₇ (2)^{50 – 17} (a)¹⁷
= (²ⁿC_n) xⁿ
∴ In the expansion of (1 + x)²ⁿ, the coefficient of xⁿ = ²ⁿC_n …. (1)
and in the expansion of (1 + x)^{2n – 1}, x = 1, a = x, n = 2n – 1
∴ T_r+1 = ²ⁿC_r (1)^{2n – 1  -r} (x)ⁿ
For the coefficient of xⁿ, put r = n, ‘
We get T_n+1 = ^{2n – 1}C_n xⁿ
The coefficient of xⁿ in the expansion of (1 + x)^{2n – 1} = ^{n – 1}C_n

∴ The coefficient of xⁿ in the expansion of (1 + x)²ⁿ
= 2 x The coefficient of xⁿ in the expansion of (1 + x)²ⁿ, [from equation (1) and (2)]

Question 15.
Find the constant term in the expansion of (\(\frac { 3 }{ 2 }\)x² – \(\frac { 1 }{ 3x }\))⁶
Solution:

MP Board Class 11th Maths Important Questions

MP Board Class 11th Chemistry Important Questions Chapter 13 Hydrocarbons

Hydrocarbons Important Questions

Hydrocarbons Very Short Answer Type Questions

Question 1.
Full form of T.E.L?
Answer:
Tetra Ethyl Lead.

Question 2.
What is obtained when ethylene dibromide is heated with Zn dust?
Answer:
Ethylene.

Question 3.
The smelling substance in L.P.G is?
Answer:
Ethyl mercaptan.

Question 4.
What is the name of the method in which by electrolysis of potassium acetate, methane is formed?
Answer:
Kolbe’s reaction.

Question 5.
Write the name of the reaction

Answer:
Sabatier and Senderens reaction

Question 6.
Write the names of the products formed in following reaction:

Answer:
(A) CH E= CH
(B) CH3 – CHO
(C) CH3 – CH₂ – OH

Question 7.
Write the reaction in which alkane is formed by alkyl halide and sodium
Answer:
Wurtz – Fittig reaction

Question 8.
What is number of a and n bonds in ethene?
Answer:
5o and 1 n.

Question 9.
In benzene, carbon has which hybridization?
Answer:
sp² hybridization

Question 10.
In HC = CH, which hybridization is present in C – C?
Answer:
sp – sp² hybridization

Hydrocarbons Short Answer Type Questions – I

Addition to Alkyne and Alkene Reactions Cheat Sheet, Cheat Sheet for Organic Chemistry PDF Download.

Question 1.
Trans – alkene is formed by the reduction of alkyne by liquor ammonia. Is butene show geometrical isomerism obtained by reduction of 2 – butyne?

2 – butene shows geometrical isomerism.

Question 2.
Despite their I – effect, halogens are o – and p – directing in haloarenes. Explain?
Answer:
Halogen have (-1) and (+R) effect, these groups are deactivating due to their (-1) effect and they are ortho and para directing due to (+R) effect.

Question 3.
Alkenes are more reactive than alkanes. Why?
Answer:
In alkanes there is only σ – bond between C – C but in alkenes there is one and one σ – bond between C = C. Due to lateral overlapping the π – bond is weaker than the σ – bond. Hence, alkenes are more reactive than alkanes. (MPBoardSolutions.com) The bond energy of π – bond lower than the bond energy of π – bond. Due to this difference of bond energy alkenes are more reactive as compared to alkanes.

Question 4.
What are asymmetric carbon?
Answer:
The carbon atom in which four different groups are attached is called asymmetric carbon. Due to this property, compound shows optical activity.

Question 5.
What do you mean by Chirality?
Answer:
Those molecules which are not superimposable on their mirror images are chiral molecules and this property is called chirality. They are optically active. The Chirality due to presence of asymmetric carbon in the molecule.

Question 6.
What are Alkanes? Which type of bonds are present in it?
Answer:
Alkanes are saturated hydrocarbons because of their low reactivity, they are also called paraffin. The general formula is C_nH_2n+2, In alkanes each carbon atom is sp³ hybridized. Single σ – bonds are present between C – C and C – H.
Example: Methane CH₄, Ethane C₂H₆.

Question 7.
What are Alkenes? In Alkenes C is present in which hybridized state?
Answer:
A saturated hydrocarbon becomes unsaturated when two hydrogens are less in it. Such hydrocarbons are called olefins. In IUPAC system these olefins are called alkenes. Such alkenes are unsaturated and they contain C = C double bond. General formula is
The hybridization of carbon in C = C is sp².
Example: Ethene CH₂ = CH₂, Propylene CH₃ – CH = CH₂.

Question 8.
What are Alkynes? What type of bonds are present in them?
Answer:
Decrease in four hydrogens in saturated hydrocarbons result in the formation of triple bonds between two carbon atoms. The unsaturated hydrocarbon produced known as alkynes. (MPBoardSolutions.com) Their general formula is C_nH_2n-2. Carbon atom of alkyne is sp³ hybridized. Alkynes are also called acetylenes. They contain carbon – carbon triple bond.
Example: Acetylene CH₂ = CH₂, Propyne CH₃ – C = CH₂.

Question 9.
Why Cyclopropane is more reactive than cyclohexane?
Answer:
In cyclopropane the bond angle in C – C – C is 60° due to which ring feelstrain, and so it is reactive and less stable. Whereas in cyclohexane C – C – C have 109°28 and have less strain in the ring, therefore it is stable and less reactive.

Question 10.
Explain functional isomerism with example?
Answer:
The compound having same molecular formula but different functional groups in the molecule are called functional group isomers.
Example:
1. Alcohols and ethers (C₂H_2n+2 – O)

Question 11.
What is dissociation?
Answer:
The method of separation of dor l image isomers from racemic mixture is called dissociation. This is done through biochemical or chemical methods.

Question 12.
How will you obtain nitrobenzene from acetylene?
Answer:

Question 13.
What is Prototrophy?
Answer:
By hydration of Propyne, enol and keto forms are obtained. It shows tautomerism and this type of isomerism is present in that compounds which have at least one hydrogen. This isomerism is due to the transfer of proton from one place to another. This is called prototrophy.

Question 14.
What is conformational stereioisomerism?
Answer:
The phenomenon of easily interconvertibility at room temperature due to free rotation around the carbon – carbon single bond in alkanes and its derivatives is called conformational stereioisomerism.

Question 15.
Explain the process of polymerisation?
Answer:
In polymerisation many simple molecules of a substance combine together to form a big molecule. The simple molecule is called monomer and the big molecule as polymer or macromolecule. Rubber, nylon, bakelite, P.V.C. are examples of high polymers. (MPBoardSolutions.com) Polymerisation of alkenes takes place in presence of Lewis acid BF₃, AlCl₃ or organic and inorganic peroxides.

Polyethylene is used as electrical insulator and as packing materials.

Hydrocarbons Short Answer Type Questions – II

Question 1.
What is Cracking? Write a note on cracking and its uses?
Answer:
Cracking or Pyrolysis:
Higher hydrocarbons when heated to high temperature decomposes into smaller hydrocarbons (lower carbon atom containing molecule). This type of thermal decomposition is known as cracking or pyrolysis.

Pyrolysis is related to free radical reaction. Manufacture of oil gas or petrol gas is based on the concept of pyrolysis. For example, dodecane when heated to 973 K gives a mixture of heptane and pentene. Platinum, palladium or nickel is used as catalyst.

Steam phase cracking, catalytic cracking and liquid phase cracking are the different methods of cracking.

Uses of alkanes:

1. Alkanes are used for making carbon black which is used for making ink, black paints and polish.
2. As a gaseous fuel in industries and as L.P.G.
3. Higher alkanes like petrol, kerosene, lubricant, oil, paraffin, wax, etc.obtained from petroleum are useful.
4. Some halogen derivatives like chloroform and carbon tetrachloride are useful in laboratories and industries.
5. Catalytic oxidation of alkanes give important compounds like alcohol, aldehyde, acid, etc.

Question 2.
Explain Dehydrohalogenation and Dehalogenation with example?
Answer:
Dehydrohalogenation:
When alkyl halide is heated with alcoholic KOH, molecule of hydrogen halide is eliminated forming alkene. This reaction is known as dehydro – halogenation.
Example:
CH₃ – CH₂ – CH₂ – Cl + KOH CH₃ – CH = CH₂ + KCl + H₂O.

Dehalogenation:
Vicinal dihalogen derivatives of alkanes are compounds in which halogen atoms are present on adjacent carbon atoms. They are also called, 1,2 – dihalogen derivative. They form alkenes when heated with Zn dust.
This reaction is known as dehalogenation:

Question 3.
Write notes on Friedei – Crafts reaction?
Answer:
Alkylation:
When benzene or its higher homologous are heated with alkyl halide in presence of anhydrous AlCl₃ alkyl derivatives of benzene are formed.

Acetylation:
When benzene or its higher homologous are treated with acid chloride in presence of anhydrous AlCl₃, acyl benzene or aromatic ketones are formed.

Question 4.
What is Lindlar’s Catalyst? Write its uses?
Answer:
Lindlar catalyst is a palladium supported mixture of calcium carbonate and poisoned with sulphur or quinoline.

Uses:
Aikynes react with hydrogen in presence of nickel powder or platinum or palladium catalyst to form alkenes. This process is known as catalytic hydrogenation.

Question 5.
Geometrical isomerism is found in which type of compounds? Explain with example?
Answer:
Geometrical isomers and Geometrical isomerism or cis – trans isomerism:
This type of isomerism is shown by those alkene derivatives in which different groups are attached with carbons linked with double bond. (MPBoardSolutions.com) For example: compound like abc = cba. When similar atom or group is on same side of bond, it is called cis – isomer and when they are on opposite side of bond, it is called trans – isomer. This type of isomerism is called cis – trans isomerism.

Following compounds does not show geometrical isomerism.

Question 6.
What are Dienes, write their types? Explain with example?
Answer:
Diene is an unsaturated hydrocarbon. In this, two double bonds are present between carbon – carbon chain. On the basis of position of double bonds dienes are of three types:
1. Isolated dienes:
Dienes in which more than one single bonds are present between two double bonds in C – C chain.
CH₂ = CH – CH₂ – CH = CH₂

2. Conjugated dienes:
In these dienes the double bonds are present in alternate position in carbon – carbon chain.

3. Cumulative dienes:
In these dienes the double bonds are present continuously on two C atoms.
CH₃ = C = CH – CH₃
CH₃ – CH = C = CH₂.

Question 7.
Write Diel’s – Alder reaction with equation?
Answer:
When a conjugated diene is heated with ethene, then a cyclic compound is formed. This is called Diel’s – Alder reaction.

Question 8.
What is Huckel’s Rule?
Answer:
According to it, all those planar cyclic compounds exhibit aromatic character, whose ring contain (4n + 2) n electrons. Where n is an integer. Therefore, planar cyclic compounds in which there are 2 (n = 0), 6(n = 1), 10 (n = 2), 14 (n = 3) π electrons exhibit aromatic property.

Benzene and Naphthalene are aromatic compounds on the basis of Huckel’s rules several heterocyclic compounds should also show aromatic properties.

Question 9.
Write Kolbe’s method for manufacuture of alkanes?
Answer:
When sodium or potassium salt of carboxylic acids are electrolysed, alkanes are obtained at anode.
CH₃COONa ⇄ CH₃COO^– + Na⁺

At anode:
CH₃COO^– – e^– → CH₃COO\(\overset { \bullet }{ C } \)
CH₃ – COO^. → \(\overset { \bullet }{ C } \)H₃ + CO₂
\(\overset { \bullet }{ C } \) H₃ + \(\overset { \bullet }{ C } \) H₃ → C₂H₆

At cathode:
Na⁺ + e^– → Na
2Na + 2H₂O → 2NaOH + H₂

Question 10.
Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitution reactions with difficulty?
Answer:
The orbital structure of benzene shows that the π electrons cloud lying above and below the benzene ring is loosely held and therefore it is likely to be attacked by electrophiles which subsequently bring about substitution. The nucleophiles would be repelled by the n – electron ring and hence benzene ring reacts with nucleophiles with difficulty.

Question 11.
Why do alkenes prefer to undergo electrophilic addition reactions while arenes prefer electrophilic substitution reactions? Explain?
Answer:
Alkenes are source of loosely held π – bonds. Due to which they show electrophilic addition reactions. There occurs a tremendous change in the energy during the electrophilic addition of electrons to alkenes therefore they show electrophilic addition reactions.

In Arenes, during the electrophilic addition reactions the aromatic nature of benzene destroyed, but in electrophilic substitution reactions it remains constant. (MPBoardSolutions.com) So the electrophilic substitution reactions are more stable in order of energy.

Question 12.
What is tautomerism? Explain with example?
Answer:
Tautomerism:
This is a special type of functional isomerism in which the isomers differ in the arrangement of atoms but they exist in dynamic equilibrium with each other. For example, acetaldehyde and vinyl alcohol are tautomers which exist in equilibrium as:

It is of different types but the most common among them is the keto – enol tautomerism. This arises due to 1, 3 – migration of a hydrogen atom from one polyvalent atom to other within the same molecule.

Conditions for the molecule to show tautomerism:
1. Presence of electron withdrawing groups such as:

2. Presence pf α – hydrogen:

Question 13.
What is Newman projection formula?
Answer:
Newman projection:
Named after M.S. Newman, who first proposed this method of representing the three – dimensional structure on paper, this is easier to visualize than the one described before. In this projection, the molecule is viewed at the C – C bond head on. (MPBoardSolutions.com) In this formula, front carbon atom is shown by dot and rear carbon atom by a circle. Three hydrogen atoms bonded to the carbon atoms are shown by lines making an angle of 120° with each other.

Newman’s projection formula of ethane shown in fig. gives a planar representation for two – dimensional (2 – D) representation of the molecule. (MPBoardSolutions.com) Staggered form changes into eclipsed form when rotated through 60°. Similarly, eclipsed form also changes into staggered form.

Question 14.
What is Markownikoff s rule?
Answer:
During the addition across unsymmetrical double bond, the negative part of the adding molecule attaches itself to the carbon atom carrying less number of hydrogen atoms.

Markownikoffs rule can be explained on the basis of the mechanism of addition reaction. Stability of intermediate decides the yield of the product. Consider the attacks of H⁺ (an electrophile) on the propene molecule. The two inter – mediate carbocations are formed.

Since, a 2° carbocation (I) is more stable than 1° carbocation (II), therefore, carbocation (I) is predominantly formed. (MPBoardSolutions.com) This carbocation then rapidly undergoes nucleo – philic attack by the Br^– ion forming 2 – bromopropane as the major product. Thus, MarkownikofFs addition occurs through the more stable carbocation intermediate. .

Question 15.
Draw the cis and trans structures of hex – 2 – ene? Which isomer will have higher boiling point and why?
Answer:

As cis isomer is more polar than trans therefore magnitude of dipole – dipole interaction is more than trans. Hence boiling point of cis isomer is more than trans.

Question 16.
An alkene, ‘A’ on ozonolysis gives a mixture of ethanal and pentan – 3 – one. Write structure and IUPAC name of ‘A’? Write structure and IUPAC name of ‘A’?
Answer:
The product of ozonolysis are –

Remove the oxygen atom (= 0) and join the two ends by a double bond, the structure of the alkene ‘A’ is

Question 17.
Explain Racemic mixture with example?
Answer:
The mixture of equal amount of two optical isomers is called Racemic mixture. The rotation of these two isomers are in opposite directions. So it is represented by dl or ±.
Example:

Optical isomerism or Enantiomerism:
This type of isomerism is shown by unsymmetrical compounds. Structural, physical and chemical properties of optical isomers are nearly similar but optical properties are different.(MPBoardSolutions.com) Isomer which rotates plane of polarised light in clockwise direction is called dextrorotatory and one which rotates in anti – clockwise direction is called laevo – rotatory. These two forms are represented as d – and l – or (+) and (-) respectively.
Optical isomers have unsymmetrical i.e., chiral carbon in the molecule.

Question 18.
What is B.H.C.? What is its uses?
Answer:
It is prepared by the chlorination of benzene in the presence of ultraviolet light.

Benzene hexachloride is an addition compound and its y isomer is called Gammaxene. It is an important pesticide. It is also called Lindane or 666.

Question 19.
How will you do the following conversion?

1. Methane to Ethane
2. Ethane to Methane
3. Acetylene to Benzene.

Answer:
1. Methane to Ethane:

2. Ethane to Methane:

3. Acetylene to Benzene:

Question 20.
Write the name of the organic compounds which show geometrical isomerism in cyclic compounds?
Answer:
Some of the cyclic compounds also show geometrical isomerism.
Example:

Question 21.
What is Saytzeff’s rule? Explain with example?
Answer:
Saytzeff’s rule:
According to this:
“If an alkyl halide can eliminate the hydrogen in two different ways, that alkene will be formed in excess in which carbon atoms joined by double bond are more alkylated.

Question 22.
What are the essential conditions for a compound to show aromaticity?
Answer:
The property which gives extra stability to benzene and benzene like compounds is called aromaticity.
Aromaticity is decided by Huckel’s rule. According to Huckel’s rule, a compound will be aromatic if it fulfills the following four conditions:

1. Compound shall be cyclic.
2. Compound should be planar or nearly planar (sp² hybridisation)
3. Compound should be conjugated.
4. Compound should have (4n + 2) π electrons. Where n is a whole number and it may be n = 0, 1, 2, 3,4, 5, 6, …

Question 23.
What is cis and trans isomerism? Explain with example?
Answer:
cis and trans isomerism is also called geometrical isomerism. This isomerism is shown by that compounds which have carbon atoms attached to two different atoms. When the two same groups or hydrogen atoms are present at one side of double bond then the compound is called cis isomer. (MPBoardSolutions.com) But when the groups or H – atoms are present in opposite side of the C double bond then such isomers are called trans isomers.
Example:

Question 24.
Arrange benzene, n – hexane and ethyne in decreasing order of their acidic behaviour? What is the reason for this behaviour?
Answer:
The hybridised state of C in the given compounds are:

The acidic character increases with increase of s – character. So the decreasing order of acidty is:
Ethyne > Benzene > n – hexane.

Question 25.
How would you convert the following compounds into benzene:

1. Ethyne
2. Ethene
3. Hexane.

Answer:
Ethyne:

Ethene:

Hexane:

Question 26.
Explain why the following systems are not aromatic?

Answer:
(i)
 = CH₂ does not have six electrons i.e., (4n + 2) π – electrons in the ring. Therefore, it is not an aromatic compound.

(ii)

Reason: Due to sp³ hybridisation the molecule is not planar. It contains 4π – electrons, So, molecule is not aromatic.

(iii)

It is a conjugated system but does not have (4n + 2) π – electrons.

Hydrocarbons Long Answer Type Questions – I

Question 1.
Explain the conformation of n – butane?
Answer:
Conformations of n – butane:
n – Butane can be considered as dimethyl derivative ethane which is produced when terminal hydrogen of each carbon is replaced by methyl group. To assign conformations to n – butane is a difficult task because it has three carbon – carbon single bonds (one in the middle and two at the ends) which undergo free rotation.

Various conformations of n – butane can be obtained by rotating C₂ or C₃ through 360° in six steps (60° each time). Staggered and eclipsed forms are obtained alternately on rotating C₂ or C₃ by 60°. These conformations are shown below. (MPBoardSolutions.com) Fully eclipsed form is shown in (I) and other eclipsed forms are shown in II and III. On rotation of C₂ – C₃ bond by 120°. The completely staggered form is shown in (IV). It is called Antiform also. Other staggered forms shown in V and VI are called skew or gauche forms. The staggered form IV and gauche forms V and VI are termed conformational diastereoisomers.

Question 2.
Give any four points in favour of Kekule’s formula to show its resonance structures?
Answer:
Factors in favour of Kekule’s structure:
1. Benzene reacts with three molecules of hydrogen to form cyclohexane which proves presence of three double bonds.

2. Benzene reacts with three molecules of chlorine to form benzene hexachloride.

3. Three molecules of acetylene polymerize in a red hot tube to form benzene.
3CH ≡ CH → C₆H₆

4. Oxidation of benzene with air in presence of V₂O₅ gives maleic acid which loses a molecule of water to form maleic anhydride.

Question 3.
A hydrocarbon ‘A’ vapour density is 14, makes the Baeyer’s reagent colourless and perform following reaction:

Write the name and formula of A, B, C and D?
Answer:

(A) → Ethylene
(B) → 1, 2 – dibromoethane or ethylene bromide
(C) → Acetylene
(D) → Acetaldehyde.

Question 4.
Explain the laboratory method of formation of alkene with figure?
Answer:
Laboratory preparation of alkene (ethylene):
Ethylene is obtained in laboratory by heating ethyl alcohol with excess of cone. H₂SO₄ at 170°C.
1. Requirements:
Ethyl alcohol, cone. H₂SO₄, sand bath. Thistle funnel, potassiumm hydroxide, gas jar, stand etc.

2. Chemical equation:
C₂H₅OH + H₂SO₄ → C₂H₅HSO₄ + H₂O

3. Method:
50 cc of ethyl alcohol and 100 cc of conc.H₂SO₄ is taken in a flask, then 8 gm anhydrous Al₂ (SO₄)₃ and 50 gm of sand is added. The mixture of Al₂ (SO₄)₃ and sand checks up formation of foam and facilitates the reaction to take place at 140° C.

Now the flask is fitted with a thermometer, an exit tube and a dropping funnel. Flask is placed on a sand bath and fixed with a stand. The other end of exit tube dips in wash bottle containing NaOH. Another tube from wash bottle leads to a behive shelf placed in a trough of water. (MPBoardSolutions.com) A water jar is Inverted over behive shelf. Flask is heated at a temperature of 150° C and simultaneously mixture of alcohol and cone. H₂SO₄ is added dropwise into the flask. Along with ethylene, CO₂ (by oxidation of alcohol) and SO₂ (by reduction of H₂SO₄) are also present as impurities in flask. These impurities get adsorbed in NaOH solution and pure ethylene is collected in gas jars by downward displacement of water. Ethylene prepared by this method is pure.

4. Labelled diagram:

Question 5.
Explain the Laboratory method of preparation of acetylene? Or, Explain the Laboratory method of preparation of acetylene on following points:

1. Method and Chemical reaction
2. Labelled figure.

Answer:
Preparation of acetylene in laboratory:
Acetylene is prepared by dropping water on calcium carbide. Acetylene obtained by this method contains impurities of PH₃ and NH₃. These are removed by passing the gas through CuSO₄ solution.

Precautions:
Before experiment, air in flask is replaced by oil gas because acetylene forms explosive mixture with air.

Reaction of acetylene with water:
In presence of 1% HgSO₄ and 42% H₂SO₄, acetaldehyde is formed.

Reaction with ammoniacal silver nitrate solution:
When acetylene is passed through ammoniacal AgNO₃ solution, white precipitate of silver acetylide is obtained.

Question 6.
Out of benzene, m – dinitrobenzene and toluene which will undergo nitration most easily, why?
Answer:
CH₃ group is electron releasing while NO₂ group is electron withdrawing. Therefore, maximum electron density will be in Toluene followed by in benzene and least in m – nitrobenzene. Therefore, the ease of nitration decreases in the order:
Toluene > Benzene > m – dinitrobenzene.

Question 7.
How will you obtain:

1. B.H.C. from benzene
2. Acetophenone from benzene
3. P.V.C. from chloroethene
4. Teflon from tetrafluoroethene.

Answer:
1. B.H.C. from benzene:

2. Acetophenone from benzene:

3. P.V.C from chloroethene:

4. Teflon from tetrafluoroethene:

Question 8.
Write notes on following:

1. Sabatier and Senderens reaction
2. Wurtz reaction
3. Duma’s reaction,
4. Swart reaction.

Answer:
1. Sabatier and Senderens reaction:
The reaction of alkene with hydrogen in presence of Ni or Pt, after hydrogenation alkane is obtained.

2. Wurtz reaction:
Reaction of two molecules of alkyl halide with sodium in presence of dry ether, alkane is obtained.

3. Duma’s reaction:
Reaction of sodium salt of monocarboxylic acid with soda lime, after decarboxylation alkanes are formed.

4. Swart reaction:
Reaction of alkyl halide with mercuric chloride, chloroalkanes are obtained. During this reaction the substitution of halogen of alkyl halide occur by Cl.
2C₂H₅ – I + HgF₂ → 2C₂H₅ – F + HgI₂

Question 9.
An unsaturated hydrocarbon ‘A’ adds two molecules of H₂ and on reductive ozonolysis gives butane – 1, 4 – dial, ethanal and propanone. Give the structure of ‘A’. Write its IUPAC name and explain the reaction involved?
Answer:

Thus, the structure of compound ‘A’ may be written as:

Hydrocarbons Long Answer Type Questions – II

Question 1.
Write the equations for the following reactions:

1. Reaction of calcium carbide with water.
2. Reaction of bromine water on ethylene.
3. Heating of ethylene with alkaline KMnO₄.
4. Heating benzene with cone. HNO₃ and cone. H₂SO₄.
5. Heating benzene with methyl chloride in presence of anhydrous AlCl₃

Answer:
1. Reaction of calcium carbide with water:
CaC₂ + 2H.OH → CH = CH + Ca(OH)₂

2. Reaction of ethylene with bromine water:

3. Heating ethylene with alkaline KMnO₄:

4. Heating benzene with cone. HNO₃ and cone. H₂SO₄:

5. Heating benzene with methyl chloride in presence of anhydrous AlCl₃:

Question 2.
How will you obtain following:

1. Acetaldehyde from Acetylene.
2. Mustard gas from Ethylene.
3. Ethane from Grignard reagent
4. Cuprous acetylide from Acetylene.
5. Methane from Aluminium carbide.

Answer:
1. Acetaldehyde from Acetylene:

2. Mustard gas from Ethylene:

3. Ethane from Grignard reagent:

4. Cuprous Acetylide from Acetylene:
CH ≡ CH + Cu₂Cl₂ +2NH₄OH → Cu – C ≡ C – Cu + 2NH₄Cl + 2H₂O

5. Methane from Aluminium carbide:
Al₄C₃ + 12H₂O → 3CH₄ + 4Al(OH)₃.

Question 3.
What is conformation? Describe conformation found in ethane?
Answer:
In alkane the C – C bond is formed by the axial overlapping of sp³ hybrid orbitals of the adjacent carbon atoms. The electron distribution in molecular orbitals of sp³ – sp³ sigma bond is cylindrically symmetrical around the inter nuclear axis. This symmetry permits the free rotation about the bond axis without rupture of the molecule. (MPBoardSolutions.com) This result in a large number of different spatial arrangement of atom or group attached to the carbon atom. Thus, “The different spatial arrangement obtained by the free rotation around the bond axis of a C – C cr bond are called conformers and the molecular geometry corresponding to a conformer is known as conformation.

Conformation of ethane:
If the position of one carbon atom of ethane is fixed in space and the other carbon atom is rotated around the C – C bond, then various conformations of ethane are possible. Out of these the conformers which has the lowest energy is called staggered and the one having highest energy is called eclipsed conformation.

1. Staggered conformation:
In staggered conformation, the hydrogen atom of the two carbon atoms are oriented in such a way that they lie far apart from one another. In other words, they are staggered away with respect to one another.

2. Eclipsed conformation:
In eclipsed conformation, the hydrogen atoms of one carbon are lying directly behind the hydrogen atoms of the other. In other words, hydrogen atoms of one carbon are eclipsing the hydrogen atoms of the other.(MPBoardSolutions.com) The conformations of ethane do not have same stability. The staggered conformation is relatively more stable than the other conformation. The difference in the energy content of staggered and eclipsed conformation is 12.5 kJ mol^-1.

Question 4.
An alkane C₈H₁₈ is obtained as the only product on subjecting a primary alkyl halide to Wurtz reaction. On monobromination this alkane yields a single isomer of a tertiary bromide. Write the structure of alkane and tertiary bromide?
Answer:
From Wurtz reaction of an alkyl halide gives an alkane with double the number of carbon atoms present in the alkyl halide. Here, Wurtz reaction of a primary alkyl gives an alkene (C₈H₁₆), therefore, the alkyl halide must contain four carbon atoms. Now the two possible primaiy alkyl halides having four carbon atoms each are, I and II.

Since, alkane C₈H₁₈ on monobromination yields a single isomer of tertiary alkyl bromide, therefore, the alkene must contain tertiary hydrogen. This is possible, only if primary alkyl halide (which undergoes Wurtz reaction) has a tertiary hydrogen.

Question 5.
Explain the free radical mechanism of halogenation of alkane?
Answer:
Mechanism of halogenation of alkane:
Halogenation of alkanes proceed through the formation of free radicals. Therefore, it is also called free radical substitution. The reaction proceeds in the following steps:

1. Chain initiation step:
The first step involves the homolytic fission of chlorine molecule to form chlorine free radicals. This fission takes place in the presence of light or heat.

Note that Cl – Cl bond being weaker than C – H bond and the C – C bonds. Therefore, undergo cleavage first.

2. Chain propagation:
Chlorine free radical is produced in the first step, attacks methane molecule forming methyl free radical and HCl. Methyl free radical reacts with other chlorine molecule forming methyl chloride and chlorine free radical.

When sufficient amount of methyl chloride has been formed, the Cl produced in reaction (ii) has a greater chance of colliding with a molecule of CH₃Cl rather than a molecule of CH₄. If such a collision occurs, a new free radical (CH₂Cl) is produced.
CH₃Cl + \(\overset { \bullet }{ C } \) l → CH₂Cl + HCl
This \(\overset { \bullet }{ C } \)H₂ Cl then reacts with Cl₂ to give CH₂Cl₂ and another \(\overset { \bullet }{ C } \) l free radical.
\(\overset { \bullet }{ C } \)H₂Cl + Cl₂ → CH₂Cl₂ + \(\overset { \bullet }{ C } \) l
This process continues till all the hydrogen is removed.

3. Chain termination:
The chain reaction steps if two or different free radicals combine amongst themselves without producing new free radicals. The possible termination steps are as follows:
\(\overset { \bullet }{ C } \) l + \(\overset { \bullet }{ C } \) l → Cl₂
\(\overset { \bullet }{ C } \) H₃ + \(\overset { \bullet }{ C } \) H₃ → CH₃ – CH₃
\(\overset { \bullet }{ C } \) H₃ + \(\overset { \bullet }{ C } \) l → CH₃Cl

Question 6.
Explain the conformational isomerism in cyclohexane?
Answer:
Conformations of Cyclohexane:
Like alkanes cyclohexane (a cycloalkane) also exhibits conformational isomerism. Sachse (1890) suggested that if cyclohexane has a planar cyclic hexagonal structure, then it will be highly stable due to the presence of angular strain of the ring composed of sp² hybrid carbon atoms. (MPBoardSolutions.com) However, cyclohexane is quite stable. Sachse suggested that two non – planar models are possible which are free from angular strain. These are called chair and boat conformations as shown in Fig.

These are in a staggered way and eclipsed conformation of alkanes. The chain form is more stable that the boat form by an energy equal to about 7 kcal/mol. In the boat form, like the eclipsed form of ethane, the hydrogen atoms being very close repel each other and the system becomes unstable. (MPBoardSolutions.com) In the boat form there is considerable non – bonded interaction between the flagpole hydrogens and also between other eclipsed hydrogens. As a consequence, this form of cyclohexane can flex into what is known as twist boat form (flexible form) which is stable by about 1.8 kcal/mol than regular boat.



In the chain form, the hydrogen atoms are situated quite far apart from one another. As a result the force of repulsion between the nearest (v icinal) hydrogen atoms is minimum. Thus, out of two main conformations, chair and boat, the chair form is more stable and cyclohexane exists mainly in chair form.

Question 7.
Explain the electrophilic substitution reactions in aromatic hydrocarbons giving two examples?
Answer:
Mechanism of monosubstitution in benzene:
Study of many monosubstitution reactions of benzene show that these reactions follow mechanism of electrophilic substitution. In thesi 40% H₂SO₄tie reagent is an electrophile (E⁺). It can be understood in the following steps:

Step I.
Generation of electrophile:
Dissociation of attacking reagent results in the formation of electrophile (E⁺).
E – Nu → E⁺ + Nu^–

Step II.
Attack of the electrophile on the ring:

Step III.
Abstraction of a proton by the base:
Nucleophile displaces proton from the hybrid in fast step forming desired substituted products. In this way elimination takes place in this step.

In this way in electrophilic substitution reaction of benzene, electrophile is added in first step and then H⁺ ion is eliminated.

Example 1.
Mechanism of nitration of benzene:
Nitration of benzene is done in the ahead steps by treating with a nitrating mixture of concentrated nitric acid and concentrated sulphuric acid:

Step I.
Generation of electrophile (NO₂⁺):

Step II.
Attack of the electrophile on the ring:
Electrophile attacks on the benzene ring forming carbocation. It attains stability by ion resonance.

Step III.
Abstraction of H⁺ ion by the base:
Nucleophile HSO₄^– ion substitutes H⁺ ion of the ring in fast step forming nitrobenzene.

Example 2.
Mechanism of halogenation of benzene:
Mechanism ofhalogenation of benzene can be explained by the action of chlorine on benzene in presence of Lewis acid FeCl₃.

Step I.
Generation of electrophile (Cl⁺):

Step II.
Attack of the electrophile on the ring:
Electrophile attacks on the ring forming carbocation intermediate which is resonance stabilised.

Step III.
Abstraction of H⁺ ion by the base:
FeCl₄^– ion is a base here and it displaces H⁺ ion from hybrid forming desired products in the fast step.

MP Board Class 11 Chemistry Important Questions

MP Board Class 11th Chemistry Important Questions Chapter 8 Redox Reactions

Redox Reactions Important Questions

Redox Reactions Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
Metals used in Daniel cell are:
(a) N and Cu
(b) Zn and Ag
(c) Ag and Cu
(d) Zn and Cu
Answer:
(d) Zn and Cu

Question 2.
Which of the following is strongly reducing:
(a) r
(b) cr
(c) Bf
(d) r
Answer:
(d) r

Question 3.
Oxidation in electrolyte occurs:
(a) At anode
(b) At cathode
(c) At both the electrodes
(d) None of these
Answer:
(a) At anode

Question 4.
Which of the following statement is correct? Galvanic cell converts:
(a) Chemical energy to electrical energy
(b) Electrical energy to chemical energy
(c) Electrical state of metal to combined state
(d) Electrolytes to ions
Answer:
(b) Electrical energy to chemical energy

Question 5.
Oxidation number of chlorine in HOCl is:
(a) – 1
(b) 0
(c) +I
(d) +2
Answer:
(c) +I

Question 6.
Oxidation number of Mn in K₂MnO₄ is:
(a) +2
(b) +6
(c) +7
(d) 0
Answer:
(b) +6

Question 7.
Oxidation number of Cr in K₂Cr₂O₇ is:
(a) -6
(b) +6
(c) +2
(d) -2.
Answer:
(b) +6

Question 8.
Oxidation number of Ni in Ni(CO)₄ is:
(a) 0
(b) +2
(c) +1
(d) – 1.
Answer:
(a) 0

Question 9.
Unit of cell constant is:
(a) ohm^-1cm^-1
(b) ohm cm
(c) cm
(d) cm^-1
Answer:
(d) cm^-1

Question 10.
Oxidation state of nitrogen is maximum in:
(a) N₃H
(b) NH₂OH
(c) N₂H₄
(d) NH₃
Answer:
(a) N₃H

Question 11.
Oxidation state of oxygen is zero in:
(a) CO
(b) O₃
(c) SO₂
(d) H₂O₂
Answer:
(b) O₃

Question 12.
Oxidation state of Fe in K₄[Fe(CN)₆] is:
(a) +2
(b) +6
(c) +3
(d) +4
Answer:
(a) +2

Question 13.
Oxidation state of S In H₂S₂O₈ is:
(a) +2
(b) +4
(c) +6
(d) +7
Answer:
(c) +6

Question 14.
Oxidation state of Mn in KMnO₄ is:
(a) +4
(b) +6
(c) +7
(d) +5
Answer:
(c) +7

Question 15.
oxidation state of chlorine is in:
(a) HCl
(b) HClO₄
(c) ICI
(d) Cl₂O
Answer:
(d) Cl₂O

Question 16.
Oxidation state of Cl in ClO₃^– ion is:
(a) +4
(b) +5
(c) +3
(d) +2
Answer:
(b) +5

Question 17.
Halogen which is reduced most easily is:
(a) I₂
(b) Br₂
(c) Cl₂
(d) F₂
Answer:
(a) I₂

Question 18.
Oxidation state of C in CCl₄ is:
(a) +4
(b) -4
(c) +6
(d) – 6
Answer:
(a) +4

Question 19.
Oxidation state of S in S0₄^-2 is:
(a) +6
(b) -6
(c) +5
(d) – 5
Answer:
(a) +6

Question 20.
Oxidizing agent are:
(a) Electron acceptor
(b) Electron donor
(c) Proton acceptor
(d) Neutron acceptor
Answer:
(a) Electron acceptor

Question 2.
Fill in the blanks:

1. Process of loss of electrons is called ……………………..
2. Process of gain of electrons is called …………………….
3. The deterioration of metals in presence of atmospheric gases and moisture is called …………………………
4. A device in which electric energy gets converted to chemical energy is called …………………….. cell
5. In electrochemical series the ability to reduce ………………………… on moving down
6. Most strong reducing element is ………………………….
7. Most strong oxidizing element is ………………………..
8. Theory of dissociation of electrolyte was proposed by ……………………….

Answer:

1. Oxidation
2. Reduction
3. Corrosion
4. Electrochemical cell
5. Decreases
6. Lithium
7. Fluorine
8. Arrhenius

Question 3.
Answer in one word/sentence:

1. What is balanced redox reaction?
2. Why does iron displaced copper from its salt solution?
3. What is the oxidation state of chlorine in Cl₂O?
4. Why does an electrolyte dissociate into ions when dissolved in water?
5. A saturated solution of KNO₃ is used for the formation of salt bridge. Why?
6. In SnCl₂ + 2 FeCl₃ → SnCl₄ + 2 FeCl₂, reaction which element is oxidizing agent?
7. What is oxidation state of Xe in XeO₃?

Answer:

1. Redox reaction in which amount of oxidation and reduction is to the same extent.
2. Because standard eletrode potential value of iron is less then the standard electrode potential of copper.
3. +1
4. Because the electrostate attractive force between the ions break due to water.
5. Because the speed of K⁺ and NO₃ and is almost same.
6. FeCl₃
7. +6

Question 4.
Match the following:

Answer:

1. (c)
2. (d)
3. (a)
4. (b)
5. (e)

Redox Reactions Very Short Answer Type Questions

Question 1.
Why iron displaces copper from its salt solution?
Answer:
Because the standard electrode potential of iron is less than copper.

Question 2.
What is the oxidation number of Cl in Cl₂O?
Answer:
+1.

Question 3.
When an electrolyte is dissolved in water, it dissociates into ions? Why?
Answer:
Because due to water the electrostatic attraction force between the ions breaks.

Question 4.
Why the saturated solution of KNO₃ is used to prepare salt bridge?
Answer:
Because the speed of K⁺ and NO₃^– ions is similar.

Question 5.
In SnCl₂ + 2FeCl₃ → SnCl₄ + 2FeCl₂, which is oxidizing agent?
Answer:
FeCl₃.

Question 6.
What is the oxidation number of Xe in XeO₃?
Answer:
+6.

Question 7.
Which metal is used in Daniel cell?
Answer:
Lithium.

Question 8.
What is the oxidation state of Mn in K₂MnO₄?
Answer:
+6.

Question 9.
Which is the most reducing element?
Answer:
Lithium.

Question 10.
Who maintains the neutrality among the two half cells?
Answer:
Salt bridge.

Question 11.
What is the standard electrode potential of SHE?
Answer:
0.00V (Zero).

Question 12.
Which is the best conducting metal?
Answer:
Silver (Ag).

Question 13.
The reaction 3ClO^–_aq → Cl₃^– + 2Cl^– is an example of?
Answer:
Disproportionation reaction.

Question 14.
Which halogen reduces very easily?
Answer:
F₂.

Question 15.
Which reaction is called electron acceptance reaction?
Answer:
Reduction.

Question 16.
Where oxidation occurs in electrolysis?
Answer:
At anode.

Redox Reactions Short Answer Type Questions – I

Counting atoms calculator in my website step-by-step explanation.

Question 1.
What is oxidation process? Explain redox reaction with example?
Answer:
Oxidation state of an element is defined as “The residual charge left on its atom”. When all the other atoms are removed from the molecule as ion.

2Mg + O₂ → 2MgO [Addition of oxygen]
2FeCl₂ → 2FeCl₃ [Addition of electronegativity element]
H₂S + Cl₂ → S + 2HCl [Removal of hydrogen]
2Ki + H₂O₂ → 2KOH + I₂ [Removal of electropositive element]

Question 2.
What is reduction process? Explain with example?
Answer:
The ability of any substance to accept hydrogen atom or +ve electrolytic element or to release oxygen or – ve element is called reduction process.
CuO + H₂ → Cu + H₂O [Remove of oxygen]
2FeCl₃ + H₂ S → 2FeCl₂ + 2HCl + S [Removal of electromagnetive element]
Cl₂ + H₂S → 2 HCl + S [Addition of H]
S + Fe → Fes [Addition of electropositive element]

Question 3.
AgF₂ is unstable compound, but if it is formed then it will acts as strong oxidising agent Why?
Answer:
In AgF₂, Ag has oxidation state +2, it is highly unstable, it very easily accept electron and forms oxidation state +1. Because Ag⁺ has stable configuration so it is stable.
Ag2⁺ + e^– → Ag⁺
This is the reason why AgF₂ behaves as a strong oxidizing agent.

Question 4.
By the structure of S₄O₆^2-clarify that the oxidation state of S is +5?
Answer:
The oxidation state of two central sulphur atom is 0. As the electron pair remains in the centre cf the bond. So the remaining sulphur atom has oxidation state 4 – 5.

⇒ 6(-2) + 2x = -2
⇒ -12 + 2x = -2
⇒ 2x = -2 + 12 = 10
⇒ x = \(\frac{10}{2}\) = 5.

Question 5.
What is electrochemical series?
Answer:
Electrochemical series:
When different elements and ions are arranged in increasing order of their standard electrode potential, a series is obtained which is called electrochemical series as activity series.

Question 6.
What is salt bridge? Write its two uses?
Answer:
1. It connects the solution in two half ceil and completes the cell circuit

2. The salt bridge which contains the solution of an electrolyte say K⁺NO₃^– maintains the electrical neutrality by diffusion of ion through it. As positive ion begins to be formed in oxidation half cell negative ion from the salt bridge diffuse out into this half cell. Similarly, as the positive Ion begin to be consumed in the reduction.

Hair cell positive:
On from the salt bridge diffuse out into this half cell. As a result the solution in the two half cell remain electrically neutral and the current continous to flow.

Question 7.
On the basis of electron transfer explain oxidation and reduction reactions?
Answer:
Reactions involving exchange of electrons are called redox reaction. Reaction given below
2FeCl₃ + SnCl₂ → 2FeCl₂ + SnCl₄
It is an example of redox reaction in which FeCl₃ is oxidizing agent while SnCl₂ is a reducing agent.

According to electronic concept during oxidation substance loses one or more electron and its positive oxidation number increases while negative oxidation number decreases. (MPBoardSolutions.com) On the other hand, during reduction substance gains one or more electron and is positive oxidation number decreases while negative oxidation number increases.

Question 8.
What is electrochemical equivalent?
Answer:
According to Faraday’s first law:
The mass of any substance deposited at any electrode is directly proportional to the quantity of electricity passed.

Thus, if W gm substance is deposited on passing Q coulomb electricity, then W ∝ Q or W = ZQ
Where, Z is a constant of proportionality and is called electrochemical equivalent of the substance deposited. If a current of I ampere is passed for t second, then Q = I × t. (MPBoardSolutions.com) So that, W = Z × Q = Z × I × r

Thus, if, Q = 1 coulomb, 1=1 ampere and t = 1 second, then W = Z. Hence, electrochemical equivalent of a substance may be defined as, “The mass of the substance deposited when a current of one ampere is passed for one second”.

As one Faraday (96500 C) deposits one gram equivalent of the substance, hence, electrochemical equivalent can be calculated from the equivalent mass.
\(\frac { Equivalentmassofthesubstance\quad }{ 96500 } \)

Question 9.
What happens when Zn is kept in CuS0₄ solution? Explain with equation?
Answer:
The reaction occurring in the beaker may be written as

Cancelling the common S0₄²⁺ ions
Zn_(s) + Cu²⁺_(eq) → Zn²⁺_(eq) + Cu_(s)
In this case Zinc metal loses electron and gets oxidised to Zn⁺⁺ ions which goes into the solution. Due to this, weight of zinc plate gradually decreases. On the other hand, electrons lost by Zinc atom are taken up by Cu²⁺ ions of CuS0₄ solution.

Question 10.
What do you mean by standard electrode potential?
Answer:
Standard electrode potential: Standard electrode potential (E°) of a half cell (electrode) is the potential difference which is produced when one electrode is dipped in the solution of molar concentration of its ion at 298 K. (MPBoardSolutions.com) If electrode is gaseous, the pressure of gas must be one atmosphere.
In IUPAC system reduction potentials are known as standard electrode potential.

Question 11.
The reaction Cl_2(g) + 2OH^–_(aq) → ClO^–_(aq) + Cl^–_(aq) + H₂O_(l) represents the process of bleaching. Identify and name species that bleaches the substances due to its oxidizing action?
Answer:

In this reaction, O. N. of Cl increases from 0 in Cl₂ to +1 in ClO^– and decreases to -1 in Cl^–. Therefore, Cl₂ is both oxidized to ClO^– and reduced to Cl^– (Disproportionation reaction). Since, Cl^– ion cannot act as oxidizing agent (Because it cannot decrease its O.N. lower than -1) therefore, Cl₂ bleaches substance due to oxidizing action of ClO^– (Hypochlorite) ion.

Question 12.
Define oxidizing agent and reducing agent?
Answer:
Oxidizing agent:
A substance which brings oxidation is called oxidising agent. On the basis of electronic theory, an oxidizing agent may be defined as an electron acceptor. i.e. (MPBoardSolutions.com) in the reaction between Zn atom and Cu²⁺ ion in aqueous solution of Cu²⁺ salt, Cu²⁺ ion is the oxidizing agent as it brings about the oxidation of Zn atom by gaining the electron lost by it.
Zn + Cu²⁺ → Zn² + Cu
(Gains e^–)
(∴ Oxidising agent)
Thus, oxidizing agent or oxidant undergoes gain of electrons during the reaction.
Reducing agent:
A substance which brings about reduction is called a reducing agent. On the basis of electronic theory, a reducing agent may be defined as an electron donor, e.g., in reaction between Zn atom (MPBoardSolutions.com) Cu²⁺ ion in an aqueous salt solution of Cu²⁺, Zn atom is the reducing agent as it brings about reduction of Cu²⁺ ion by losing the electrons.
Zn + Cu²⁺ → Zn²⁺ + Cu
(Loses e^–)
(∴Reducing agent)

Question 13.
If iron rod is dipped in CuS0₄ solution then copper is displaced from its solution by iron but when copper rod is dipped in FeS04 solution. Why iron is not displaced by copper? Explain?
Answer:
Fe occupies higher position in electrochemical series than Cu. So, Fe is more active than Cu. That is why Fe displaces Cu from CuS0₄ solution, but Cu is not able to displace Fe from FeS0₄ solution.

Question 14.
What is oxidation number?
Answer:
Oxidation number of an element is defined as, “The residual charge left on its atom when all the other atoms are removed from the molecule as ions.” Atoms can have positive, negative or zero oxidation state depending upon their state of combination. (MPBoardSolutions.com) In fact, oxidation number is the charge assigned to the atom in a species according to some rules. It is also known as oxidation state.

Question 15.
In electrochemical series, the activity of metals increases or decreases in which sequence?
Answer:
Species present in the right of electrochemical series are reducing agent. Smaller the value of E°, stronger is the reducing agent. The activity of metals decrease on going down the electrochemical series.

Question 16.
What is E° value be + ve shows in Galvanic cell?
Answer:
The positive value of E° shows:

1. Oxidation occurs at anode and
2. The electrode where reduction occurs can be taken as cathode.

Question 17.
What is electrode potential? Its value depends on what factors?
Answer:
The electrical potential difference setup between the metal and its solution is known as electrode potential.
The electrode potential depends upon:

1. Concentration of ions in the solution
2. Temperature.

Question 18.
On the basis of standard electrode potential given below, arrange the metals in their increasing reducing power?
K⁺/ K = – 2.93V, Ag⁺/ Ag = 0.80V, Hg²⁺/ Hg = 0.79V, Mg²⁺/ Mg = -2.37V, Cr³⁺/ Cr = -0.74V?
Answer:
Less the value of E°, the reducing power increases of the metal. So the sequence is Ag < Hg < Cr < Mg < K.

Question 19.
Among MgO, ZnO, CuO and CaO which oxide will reduce by hydrogen?
Answer:
Mg, Zn and Ca are more reactive than H in electrochemical series. So they will not reduced by H. Cu is present below H, so the reactivity of Cu is less than H. So CuO will easily get reduced by H.

Question 20.
The electrode potential (E°) of Ag, Ba, Mg and Au are +0.80, -2.90, -2.37 and +1.42 volt. Among these metals which will displace H from acids and which will not?
Answer:
Those metals which have E° value negative they are more reactive than hydrogen and can easily displace H from weak acids. So, Ba and Mg can displace H from acid solution.

Question 21.
Can be put CuSO₄ solution in silver container? Why?
Answer:
In electrochemical series Ag comes below Cu, so reactivity of Ag is less than Cu. There Ag will not displace Cu from CuSO₄ solution. CuSO₄ can be kept in Ag container.

Redox Reactions Short Answer Type Questions – II

Question 1.
Fluorine reacts with ice as follows:
H₂O_(s) → HF_(g) + HOF_(g)
Justify that, this reaction is a redox reaction.
Answer:

Since, fluorine can undergo oxidation as well as reduction, it is an example of redox reaction.

Question 2.
In acidic medium MnO₄^2- shows disproportionation reaction but MnO₄^– does not show why?
Answer:
In oxidation state of Mn in MnO₄^-2 is +6. It can increase oxidation state (+7) and decrease (+4, +3, +2, 0 etc.). So in acidic medium it shows disproportionation reaction.

In MnO₄^–, Mn has maximum oxidation state (+7). It can only decrease its oxidation state. Due to which it cannot show disproportionation reaction.

Question 3.
In the following reaction whose oxidation and whose reduction occur:
PbS + 4H₂O₂ → PbSO₄ + 4H₂O
Answer:
In this reaction PbS is oxidized as PbSO₄ and H₂O₂ is reduced to form H₂O.

Question 4.
Differentiate between oxidation number and valency?
Answer:
Differences between Oxidation number and Valency:
Oxidation:

1. It is the residual charge left on the atom of an element when all other atoms are removed from molecule or ion.
2. It refers to charge and can be positive, negative or zero, e.g., oxidation number of C in CO₂ is +4 and that of oxygen is -2.
3. It may have fractional value.
4. Elements like C, N, O have variable oxidation number.

Valency:

1. It is the combining capacity of the element expressed as number of H – atom or double the number of oxygen atoms which combine with one atom of the element.
2. It refers to number only and does not carry any sign, e.g., in CO₂ valency of carbon is 4 and that of oxygen is 2.
3. It is always a whole number.
4. Elements like C, N, O exhibit constant valency.

Question 5.
Why fluorine doesn’t show disproportionation reaction?
Answer:
Like chlorine, bromine and iodine also undergo similar disproportionation reactions but fluorine does not. The reason for this anomalous behaviour is that fluorine being the strongest oxidizing agent does not show positive oxidation states. It, therefore, reacts in a different way forming oxygen difluoride (OF2).
2F_2(g) + 2OH^–_(g) → 2F^–_(aq) + OF_2(g) + H₂O_(l)

Question 6.
Show that galvanic cell where following reaction occurs:
Zn_(s) + 2Ag⁺_(aq) → Zn²⁺_(aq) + 2 Ag_(s)
Now explain :

1. Which electrode is -ve?
2. What is the motive of electricity in cell?
3. What is the reaction occurring on each electrode?

Answer:
Galvanic cell Zn_{(s) ∥} Zn²⁺_(aq)Ag(aq) ∥ Ag⁺_(aq) ∥ Ag_(s)

1. Zn electrode is negatively charged. Zn is oxidized into Zn⁺².
2. Electricity flows from Ag electrode to Zn electrode and electrons flow from Zn to Ag electrode.
3. Reactions on electrode:

Cathode — Zn_(s) → Zn⁺²_(s) + 2e^–
Anode — 2Ag⁺_(aq) + 2e^– → 2Ag_(s)

Question 7.
Write the definition of electrochemical cell? Write the chemical reactions of Daniel cell?
Answer:
An electrochemical cell (Galvanic cell) is a device in which the redox reaction takes place indirectly and the decrease in free energy appears in the form of electrical work i.e. chemical energy is converted into electrical energy.

Question 8.
Write the factors affecting electrode potential?
Answer:
The factors affecting electrode potential are:

1. Tendency of metal to donate electrons: More the tendency to release electron more will be electrode potential.
2. Temperature: With increasing temperature electrode potential increases.
3. Concentration of solution: More the concentration of solution more will be electrode potential.

Question 9.
Determine the oxidation number of following:

1. Cr in K₂Cr₂O₇
2. Mn in KMnO₄
3. S in Na₂S₄O₆.

Solution:
1. Cr in K₂Cr₂O₇
2 (+1) + 2x + 7 (-2) = 0
or 2 + 2x – 14 = 0
or 2x = 12
or x = +6.

2. Mn in KMnO₄
1 (+1) + 1x + 4(-2) = 0
or 1 + x – 8 = 0
or x = +7.

3. S in Na₂S₄O₆
2 (+1) + 4x + 6 (-2) = 0
or 2 + 4x – 12 = 0
or 4x = 10
or x = 2.5.

Question 10.
Why does the following reactions occur:
XeO_6(aq)^4- + 2F^–_(aq) + 6H⁺_(aq) → XeO_3(g) + F_2(g) + 3H₂O_(l)?
What conclusion about the compound Na₄XeO₆ (of which XeO₆^4- is a part) K₂MnF₆ can be drawn from the reaction?
Answer:

O.N. of Xe decreases from +8 to +6. This shows that XeO₆^4- is an oxidizing agent. It oxidises F^– to F₂. This reaction shows that Na₂XeO₆ (or XeO₆^–) is a stronger oxidizing agent than F₂.

Question 11.
Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water?
Answer:

Multiplying Cl^– by 2 (As 2 atoms are in Cl₂)
Cl_2(aq) + SO_2(aq) + H₂O_(l) → 2Cl^–_(aq) + SO₄^2-_(aq)
On adding 4H⁺ in left side and multiplying in H₂O by 2,
Cl_2(aq) + SO_2(aq) + 2H₂O_(l) → 2Cl^–_(aq) + SO₄^2- + 4H⁺
It is a balanced disproportionation reaction.

Question 12.
Differentiate between electrochemical cell and electrolytic cell?
Answer:
Differences between electrochemical cell and electrolytic cell:
Electrochemical (Galvanic) cell:

1. In Galvanic cell, chemical energy is changed into electrical energy.
2. Reaction in Galvanic cell is spontaneous.
3. Oxidation and reduction occurs in different half cell.
4. Anode is negative while cathode is positive terminal.
5. The electrons move from anode to cathode in the external circuit.

Electrolytic cell:

1. In electrolytic cell, electrical energy is changed into chemical energy.
2. Reaction in electrolytic cell is non – spontaneous.
3. Oxidation and reduction occurs in same container.
4. Anode is positive while cathode is negative terminal.
5. The electrons are supplied by the external circuit which enter through the cathode and came out through anode.

Question 13.
What is Nernst equation? Derive relation between E and E°?
Answer:
In Standard cell, the concentration of the solution used is 1M and the temperature is maintained at 25°C (For 298K). But the Galvanic cells are often constructed under non standard condition. In such cases the cell voltage (or EMF) is calculated by Nernst equation as follows:
E = E° – \(\frac{RT}{nF}\) log_e \(\frac { M_{ (s) } }{ M^{ n+ }_{ (aq) } } \)
or E = E° + \(\frac{RT}{nF}\) log_e \(\frac { M^{ n+ }_{ (aq) } }{ M_{ (s) } } \)
In changing log_e in log₁₀
or E = E° + \(\frac{2.303}{nF}\) log₁₀ \(\frac { M^{ n+ }_{ (aq) } }{ M_{ (s) } } \) [M_(s) = 1]
or E = E° + \(\frac{2.303RT}{nF}\) log₁₀[Mⁿ⁺_(aq)]
or E = E° + \(\frac{0.0591}{nF}\) log₁₀ [Mⁿ⁺_(aq)].

Question 14.
Consider the reactions:
2S₂O₃^2-_(aq) + I_2(s) → S₄O₆^2-_(aq) + 2I^–_(aq)
S₂O₃^2-_(aq) → 2SO₄^2-_(aq) + 4Br^–_(aq) + 10H⁺_(aq)
Why does the same reductant, thiosulphate reacts differently with iodine and bromine?
Answer:
Br₂ is stronger oxidizing agent than I₂, it oxidises SO₃^2- to SO₄^2- i. e. from +2 state to +6 state of sulphur. However, I₂ being weaker oxidizing agent, oxidises S₂O₃^2- to S₄O₆^2- ion i.e. from + 2 to + 2.5 state of sulphur.

Redox Reactions Long Answer Type Questions – I

Question 1.
What is standard hydrogen electrode? How it is prepared?
Answer:
Standard hydrogen electrode: This consists of gas at 1 atmospheric pressure bubbling over a platinum electrode immersed in 1 M HCl at 25°C (298 K) as shown in figure. The platinum electrode is coated with platinum black to atmospheric increase its surface. (MPBoardSolutions.com) The hydrogen electrode thus Pressure constructed forms a half cell which on coupling with any other half cell begins to work on the principle of oxidation or reduction. Electrode depending upon the 1 M H+solution circumstances works both as anode or cathode.

Cell reaction of standard hydrogen electrode platinum electrode (SHE) when it acts as anode is
H_2(g) → 2H⁺ + 2e^–
It is represented as
H_2(g)(1 atm) Pt | H₃O⁺_(aq)(1.0 M)
When it acts as cathode, the cell reaction is
2H⁺ + 2e^– → H_2(aq)
and it is represented as
H₃O⁺_(aq) (1.0 M) | H_2(g) (1 atm) Pt
Standard hydrogen electrode (SHE) is arbitarily assigned a potential of zero.

Question 2.
On the basis of electrode potential explain which reaction is feasible:
E°_Cu2+/Cu = 0.34 V, E°_Zn2+/Zn = -0.76 V, E°_Mg2+/Mg = 2.37 V, E°_Fe2+/Fe = -0.74V

1. Cu + Zn²⁺ → Cu²⁺ + Zn
2. Mg + Fe²⁺ → Mg²⁺ + Fe.

Solution:
1. Cu + Zn²⁺ → Cu²⁺ + Zn
E°_Cu²⁺/Cu = 0.34 V and E°_Zn2+/Zn = – 0.76 V
In given cell, Cu oxidised to Cu²⁺. So Cu²⁺/Cu will act as anode and Zn²⁺/Zn acts as cathode.
E°_cell = E°_cathode – E°_anode
E°_cell = – 0.74 – (-2.37) = +1.63V
E°_cell = +Ve, shows that the reaction is feasible.

2. Mg + Fe²⁺ → Mg²⁺ + Fe
E°_Mg²⁺_/Mg = 2.37 V and E°_Cu²⁺/Cu = – 0.74 V
Mg oxidised to Mg²⁺ and acts as anode Mg²⁺/Mg.
Fe²⁺ reduced to Fe and acts as cathode Fe²⁺/Fe.
E°_cell = E°_cathode – E°_anode
E°_cell = – 0.74 – (-2.37) = +1.63 V

Question 3.
While sulphur dioxide and hydrogen peroxide can act as oxidizing as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants, why?
Answer:
In sulphur dioxide (SO₂) and hydrogen peroxide (H₂O₂), the oxidation states of sulphur and oxygen are +4 and -1 respectively. Since, they can increase as well as decrease when there compounds take part in chemical reaction, they can act as oxidizing as well as reducing agents. For example,
Increase in O.N.

In ozone (O₃), the oxidation state of oxygen is zero while in nitric acid (HNO₃), the oxidation state of nitrogen is +5. Since, both of them can undergo decrease in oxidation state and not an increase in its value, they can act only as oxidizing agents and not as reducing agents.

In ozone (O₃), the oxidation state of oxygen is zero while in nitric acid (HNO₃), the oxidation state of nitrogen is + 5. Since both them can undergo decrease in oxidation state and not an increase in its value, they can act only as oxidizing agents and not as reducing agents.

Question 4.
How do you count for the following observation though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant, why? Write a balanced redox equation for the reaction?
Answer:
Toluene can be oxidized to benzoic acid in acidic, alkaline and neutral medium by using potassium permanganate.

1. In acidic medium:
[MnO^–_4(aq) + 8H⁺_(aq) + 5e^– → Mn²⁺_(aq) + 4H₂O_(l)] × 6

In the manufacture of benzoic acid alcoholic KMnO₄ is more useful.
Alcohol if used as solvent will help in the formation of a homogeneous mixture between toluene (non – polar) and KMnO₄ (ionic). Actually alcohol has non – polar alkyl group as well as polar OH^– group.

Question 5.
How the redox reactions are balanced by ion – electron method?
Answer:
The redox reactions are balanced by ion – electron method are as follows:
1. Indicate oxidation number of each element and identify the elements which undergoes change in oxidation number.

2. Indicate increase and decrease in oxidation number per atom. Multiply the increase or decrease in oxidation number with number of atoms undergoing the change.

3. Multiply the formula with suitable integer to equalise the increase and decrease in oxidation number.

4. Balance O atoms:
K₂Cr₂O₇ + 6HCl + xHCl → 2KCl + 2CrCl₃ + yCl₂ + H₂O + 6H₂O

5. Balance H atoms:
K₂Cr₂O₇ + 6HCl + 8 HCl → 2KCl + 2CrCl₃ + 3Cl₂ + 7H₂O
Here x = 8 and y = 3 to balance Cl atoms
K₂Cr₂^2-O₇ + 14 HCl → 2KCl + 2rCl₃ + 3Cl₂ + 7H₂O.
Example 2. Cr₂O₇^2- + Fe²⁺ + H⁺ → Cr³⁺ + Fe³⁺ + H₂O.
Solution:
Step 1. (Cr₂O₇)^2- + Fe²⁺ + H⁺ → Cr³⁺ + Fe³⁺ + H₂O.

Step 2.

Step 3.
Multiply Fe by 6 to equalise increase or decrease in O.N.
Cr₂O₇^2- + 6Fe²⁺ + H⁺ → Cr³⁺ + Fe³⁺ + H₂O

Step 4.
Balance atoms other than O and H
Cr₂O₇^2- + 6Fe²⁺ + H⁺ → 2Cr³⁺ + 6Fe³⁺ + H₂O

Step 5.
Balance O atoms
Cr₂O₇^2- + 6Fe²⁺ + H⁺ → 2Cr³⁺ + 6Fe³⁺ + H₂O + 6H₂O

Step 6.
Balance H atoms
Cr₂O₇^2- + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O.

Question 6.
On the basis of electrode potential values given below? Determine reaction between reactants will occur or not?

Answer:
For feasibility of any reaction E° = +ve.
E°_cell = E°_cathode – E°_anode
(a) Fe³⁺_(aq) and I^–_(aq):
2Fe³⁺ + 2I^–_(aq) → 2Fe²⁺_(aq) + 2I_2g
Half cell reaction (oxidation)
2I^–_(aq) → I_2(g) + 2e^– E° = – 0.54 V
Reducion half cell,
2Fe³⁺_(aq) + 2e^– → 2Fe²⁺_(aq) E° = +0.77 V
Complete reaction,
2Fe³⁺_(aq) + 2I^–_(aq) → 2Fe²⁺_(aq) + I_2(g) E°_cell = +0.23 V
E° = +ve, reaction is feasible.

(b) Ag⁺_(aq) and Cu_(s)²⁺:
2Ag_(aq) + Cu_(s) → Cu²⁺_(aq) + 2Ag_(aq)
Oxidation half reaction,
Cu_(s) → Cu²⁺_(aq) + 2e^– E° = – 0.34 V
Reduction half reaction,
2Ag⁺_(aq) + 2e^– → 2Ag_(s) E° = – 0.80 V
Complete reaction,
Cu_(s) + 2Ag⁺_(aq) → Cu²⁺_(aq) + 2Ag_(s) E°_cell = +0.46 V

Question 7.
Give the main characteristics of electrochemical series?
Answer:
The main characteristics of electrochemical series are:

1. Tendency to loose electron becomes greater and the corresponding element or ion behaves as strong reducing agent if it has greater negative value of E°.
2. The substances above hydrogen are weak oxidising agents and substances below it are strong oxidising agents.
3. If the reducing agent of electrode pair is strong reducing agent, its oxidising agent is weak and when reducing agent of electrode pair is weak, its oxidising agent is strong.
4. In series the reducing efficiency decreases from above to downwards.
5. The more the positive value of E°(MPBoardSolutions.com) the stronger is its oxidising properties. In the series the oxidising power of cation decreases as we move upwards from downwards.
6. In series the electropositive character of metals decreases from upwards to downwards.
7. Along the series the electronegative character of non – metal increases from upward to downwards.
8. The metals which come prior to hydrogen displaces hydrogen from acids,
9. The metals which occupy higher position in the series displaces metals which come later in series from their salt solution.

Question 8.
Balance the equation by oxidation number method and identify the oxidising agent and reducing agent?
Cl₂O_7(g) → ClO₂^–_(aq) + O_2(g).
Answer:
Oxidation number method:

Balance the increase and decrease in O.N. multiply H₂O₂ by 4,
Cl₂O₇ + 4H₂O₂ → ClO₂^– + O₂
Balance atom other,
Cl₂O_7(g) + 4H₂O_2(aq) → 2ClO₂^–_(aq) + 4O_2(g)
Balance ‘H’,
Cl₂O_7(g) + 4H₂O_2(aq) + 2OH^–_(aq) → 2ClO₂^–_(aq) + 4O_2(g) + 5H₂O.

Question 9.
Whenever a reaction between an oxidizing agent and a reducing agent is carried out compound of lower oxidation state is formed if the reducing agent is ¡n excess and compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement by giving three illustrations?
Answer:
1. Let us consider the reaction between carbon and oxygen.
Reducing agent + Oxidising agent

2. Let us consider the reaction between white phosphorous (P₄) and Cl_2(g)

3. Let us consider the reaction between sulphur and oxygen.
Reducing agent + Oxidising agent

Redox Reactions Long Answer Type Questions – II

Question 1.
Balance the following reaction by ion – electron method:
Cr₂O₇^2- + Fe²⁺ + H⁺ → Cr⁺³ + Fe⁺³.
Solution:
Cr₂O₇^2- + Fe²⁺ → Cr⁺³ + Fe⁺³ + H₂O

1. Write the oxidation number of each atom in skeleton equation.

Species undergo change is Cr and Fe.

2. Dividing the equation into two half reaction.
Cr₂O₇^2- → Cr⁺³ (Reduction half)
Fe⁺² → Fe⁺³ (Oxidation half)

3. Balancing reduction half:
(a) Balancing Cr atom,
Cr₂O₇^2- → 2Cr⁺³

(b) Adding e to make up the difference in O.N.

(Since each Cr atom gain 3e^– ∴ for 2 Cr atom 6e^–)

(iv) Balancing oxidation half:
(a) Balancing Fe atom (Balanced)
(b) Adding e^– to make up the difference in O.N.
Fe⁺² → Fe⁺³ + e^–

(v) Balancing ‘O’ atom,
Cr₂O₇^2- + 6e^– → 2Cr⁺³ + 7H₂O
Fe⁺² → Fe⁺³ + e^–

(vi) Balancing H by adding H⁺ ions to the side which is deficient on H atoms.
Cr₂O₇^2- + 6e^– + 14H⁺ → 2Cr⁺³ + 7H₂O
Fe²⁺ → Fe⁺³ + e^–

(vii) Equalising e^–,

Question 2.
To determine the oxidising power/reducing power in a solution which method is used? Explain with example?
Answer:
Determination of standard electrode potential ; To determine electrode potential of an electrode, a cell is setup using this electrode and standard hydrogen electrode. The EMF of the cell is measured with the help of a voltameter (or more accurately by potentiometer). (MPBoardSolutions.com) The electrode potential of hydrogen electrode is taken as zero. Therefore, the E.M.F. of such a cell will directly give the value for electrode potential for the given electrode at standard condition (298 K, 1M, 1 atm pressure).

1. If the deflection in galvanometer is towards metal electrode (experimental electrode) then it is anode (-ve terminal) of the cell. It is allotted negative value of standard reduction electrode potential.
For example, in the determination of E° value of zinc, zinc acts as anode and has got negative value of E°.

E°_cell = E°_cathode – E°_anode
∴ 0.76 V = 0.00V – E°_zn⁺²/Zn = -0.76V

Question 3.
What is oxidation number? Write its main rules of determination?
Answer:
Oxidation number of an element is defined as, “The residual charge left on its atom when all the other atoms are removed from the molecule as ions”. Atoms can have positive, negative or zero oxidation state depending upon their state of combination. (MPBoardSolutions.com) In fact, oxidation number is the charge assigned to the atom in a species according to some rules. It is also known as oxidation state.

Rules for assigning oxidation number:
The assignment of oxidation numbers is arbitrary and is usually governed by the following conclusions:

1. The oxidation number of an element in free state i. e., elementary state is regarded as zero. For example,
Oxidation state of hydrogen in H₂ = 0
Oxidation state of helium in He = 0
Oxidation state of sulphur in S₈ = 0
Oxidation state of sodium in Na = 0
Oxidation state of magnesium in Mg = 0

2. In a compound, the more electronegative elements are assigned negative (-) oxidation state and less electronegative elements are assigned positive (+) values. For example, in HCl as chlorine is more electronegative than hydrogen its oxidation state will have negative value while that of hydrogen will be positive.

3. In the formula of a compound, the sum of the negative oxidation states is equal to the sum of the positive oxidation states.

4. The sum of oxidation number of all the atoms in a neutral molecule is taken as zero.

5. Hydrogen has an oxidation state +1 in compounds like H₂S, H₂O, HCl, etc. Exceptionally it has the oxidation number -1 in metallic hydrides, such as NaH, CaH₂, etc.

6. Oxygen is usually assigned oxidation number -2, except in H₂O₂ and in oxide of fluorine [F₂O], in which the oxidation number -1 and +2 respectively. In all, oxidation number of oxygen is -1.

7. Fluorine being the most electronegative element is assigned oxidation number -1 in all its compounds. Other halogens also show -1 oxidation number.

8. The number of monoatomic ion in an ionic compound is equal to its electric charge. Thus, the elements of group IA of the periodic table (Li, Na, K, Rb, Cs) all have oxidation number + 1, while the alkaline earth metals of group IIA (Ca, Sr, Ba) have oxidation number + 2.

Question 4.
Balance the following reaction by ion – electron method?
Answer:
Ion – electron method:
Oxidation half equation is:

The oxidation number is balanced by adding 2 electrons as:
H₂O_2(aq) → O_2(g) + 2e^–
The charge is balanced by adding 2OH^– ions as:
H₂O_2(aq) + 2OH^–_(aq) → O_2(g) + 2e^–
The oxygen atoms are balanced by adding 2H₂O as:
H₂O_2(aq) + 20H^–_(aq) → O_2(g) + 2H₂O_(s) + 2e^– ……………… (1)
The reduction half equation is:

The Cl atoms are balanced as:
Cl₂O_7(g) → 2ClO₂^–_(aq)
The oxidation number is balanced by adding 8 electrons as:
Cl₂O_7(g) + 8e^– → 2ClO₂^–_(aq)
The charge is balanced by adding 6OH^– as:
Cl₂O_7(g) + 8e^– → 2ClO₂^–_(aq) + 6OH^–_(aq) ………………………… (2)
The balancing equation can be obtained by multiplying equation (1) with 4 and adding equation (2) to it as:
Cl₂O_7(g) + 4H₂O_2(aq) + 2OH^–_(aq) → 2ClO₂^–_(aq) + 4O_2(g) + 5H₂O_(l)

Question 5.
Write the main uses of electrochemical series?
Answer:
Electrochemical series:
When different elements and ions are arranged in increasing order of their standard electrode potentials, a series is obtained which is called electrochemical series or activity series.
The uses of electrochemical series are:
1. Electropositive character of metal:
Metal which loses electron easily from its outermost shell shows high electropositive character while the atom which loses electron with difficulty shows least electropositive character. In electrochemical series, Li shows maximum electropositive character while fluorine shows least electropositive character.

2. Comparison of reactivity of metals:
Metal having negative value of E° loses electron easily. Hence, greater the negative value of E° more is the reactivity and reducing character of a metal.

3. Knowledge of oxidizing agent and reducing agent:
Substance which gains electron and shows maximum value of standard electrode potential E° is strongest oxidizing agent. On the basis of this F is strongest oxidizing agent while Li is weakest oxidizing agent.

4. Displacement of elements:
Metal having greater tendency to form ion displaces others. Therefore, elements getting priority in the series displace ions of the elements following them. For example. When Cu turnings are added to AgNO₃ solution, Ag gets precipitated.
2AgNO_3(aq) + Cu_(s) → Cu(NO₃)_2(aq) + 2Ag_(s)

5. Electroplating:
Process of depositing layer of gold or silver on copper, brass, iron etc. is called electroplating. It makes articles lustrous and attractive.

6. Metallurgy:
More reactive metal displaces less reactive metal from their aqueous salt solution.

7. Corrosion of metals:
Destruction of metals due to action of chemicals, air and moisture is called corrosion.
e.g., iron rust in moist air. To prevent this a layer of more reactive metal like Sn or Zn is deposited on iron. Coating iron with zinc is called galvanization.

Question 6.
What information did you get from the following reaction:
(CN)_2(g) + 20H^–_(aq) → CN^–_(aq) + CNO^–_(aq) + H₂O_(l)
Answer:
(CN)_2(g) + 2OH^–_(aq) → CNO^–_(aq) + H₂O_(l).
1. Let the O.N. of C in (CN)₂ = x
2x + 2 (-3) = 0
or x = +3.

2. Let the O.N. of C in CN^– = x
x + (-3) = -1
or x = +2.

3. Let the O.N. of C in CNO^– = x
x + (-3) + (-2) = -1 or
x = +4.

Following informations obtained from above equation:

1. It is a disproportionation reaction.
2. The reaction occurs in basic medium.
3. O.N. of N in (CN)₂ is -3 and that in CN^– is -2 and in CNO^– it is -5.
4. Cyanogen (CN)₂ gets simultaneously reduced to CN^– ion as well as oxidized to cyanate, CNO^– ion.

Question 7.
The Mn³⁺ ion is unstable in solution and undergoes disproportionation to give Mn²⁺, MnO₂ and H⁺ ion. Write a balanced ionic equation for the reaction?
Answer:
The reaction is:
Mn³⁺_(aq) → Mn²⁺_(aq) + MnO_2(s) + H⁺_(aq)
Reduction half cell:
Mn³⁺_(aq) + e^– → Mn²⁺ …………. (1)
Oxidation half cell:
Mn³⁺_(aq) → Mn⁺⁴ + O_2(s) + e^–
To balance the equation added 4H⁺ in right and in left 2H₂O is added:
Mn³⁺_(aq) + 2H₂O_(l) → MnO_2(s) + e^– + 4H⁺_(aq) ……………….. (2)
Adding eqns. (1) and (2),
2Mn³⁺_(aq) + 2H₂O_(l) → Mn²⁺ + MnO_2(s) + 4H⁺_(aq)
This is last equation balanced (disproportionation reaction).

MP Board Class 11 Chemistry Important Questions

Here you will find Chapter Wise NCERT MCQ Questions for Class 11 with Answers PDF Free Download based on the important concepts.

MP Board Class 11th Maths Important Questions Chapter 1 Sets

Sets Important Questions

Sets Objective Type Questions

(A) Choose the correct option:

Question 1.
A set is defind as:
(a) Collection of an object
(b) Collection of well defind object
(c) Nothing can be said
(d) None of these.
Answer:
(b) Collection of well defind object

Question 2.
If A = {1, 2, 3},B = {2, 3, 4} and U = {1, 2, 3, 4, 5, 6},then A’ ∩ B’ =
(a) {2, 3}
(b) {1, 5,6}
(c) {5, 6}
(d) {1, 4, 5, 6}.
Answer:
(c) {5, 6}

Question 3.
A and B are two sets, then A ∩ (A ∪ B) =
(a) A
(b) B
(c) ϕ
(d) None of these.
Answer:
(a) A

Question 4.
A set A = {x : x ∈ R, x² = 16 and 2x = 6} =
(a) ϕ
(b) {14,3,4}
(c) {3}
(d) {4}.
Answer:
(a) ϕ

Question 5.
For a non – empty set A:
(a) A∪A’ = A
(b) A∪A’ = A’
(c) A∪A’ = U
(d) A∪A’ = ϕ .
Answer:
(c) A∪A’ = U

Question 6.
If two non – empty sets A and B are not equal, then:
(a) A⊂A∩B
(b) B⊂A∩B
(c) A∩B⊂B
(d) A∩B⊂ϕ.
Answer:
(c) A∩B⊂B

Question 7.
A and B are two sets and n(A) = 70, n(B) = 60 and n(A∪B) = 110, then n(A∩B) =
(a) 240
(b) 50
(c) 40
(d) 20.
Answer:
(d) 20.

Question 8.
If A = {1,2,3,4,5}, then the number of proper subsets:
(a) 120
(b) 30
(c) 31
(d) 32.
Answer:
(d) 32.

Question 9.
If A, B and C are three sets, then A – (B∪C) is equal to:
(a) (B – A)∩C
(b) (A – B) ∪C
(c) (A – B) ∩(A – C)
(d) (A – B) ∪(A – C)
Answer:
(c) (A – B) ∩(A – C)

Question 10.
Let S = {1,2,3,4}. The total numbers of unordered pair of disjoint subsets of S is equal to:
(a) 25
(b) 34
(c) 42
(d) 41.
Answer:
(d) 41.

(B) Match the following:

Answer:

1. (e)
2. (c)
3. (a)
4. (b)
5. (f)
6. (d)

(C) Fill in the blanks:

1. If the elements of A mid B are 3 and 6, then minimum number of elements in A∪B is …………………………….
2. If X and Y are two sets such that n(X) = 17, n(Y) = 23 and n(X∪Y) = 38, then n(X∩Y) = …………………………
3. If sets A = {1, 2, 3, 4, 5, 6} and B = {2, 4, 6, 8}, then B – A = ……………………………..
4. A set which does not contain any element is called ………………………….
5. If set A has n elements, then the number of subsets of A are …………………………..
6. If A = {1, 2} and B = {3, 4}, then (A∪B)∩ϕ = ……………………………..

Answer:

1. 6
2. 2
3. {8}
4. Empty set
5. 2ⁿ
6. ϕ

(D) Write true/false:

1. If A ∩ B = B, then B ⊂ A.
2. If A = {1, 2, 3, 4} and B = {2, 3, 4, 5, 6}, then A ∆ B = {1, 5, 6}.
3. If n(A – B) = n(A) – n(A n B).
4. For a non – empty sets A is A∩A’- A’.
5. If A = {1, 2, 3, 4}, then n[P(A)] = 16.

Answer:

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

Question (E)
Write answer in one word/sentence:

1. If A = {1, 2}, then write all the subsets of set A.
2. If A = {4, 5, 8, 12} and B = {1, 4, 6, 9}, then find the value of A – (B – A).
3. If n(U) = 700, n(A) = 200, n(B) = 300, and n(A∩B) = 100, then find the value of n(A’∩B’). ‘
4. Find the value of n[P{P{P(ϕ}}]
5. If A = {a, b, c, d}, then find all subsets of set A.
6. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6}, then find the value of (B – C)’.

Answer:

1. ϕ, {1}, {2}, {1,2}
2. {4, 5, 8, 12}
3. 300
4. 4
5. 16
6. {1, 3, 4, 5, 6, 7, 8, 9}.

Sets Very Short Answer Type Questions

Question 1.
Write the following sets in set builder form: (NCERT)

1. {3, 6, 9, 12}
2. {2, 4, 8, 16, 32}
3. {5, 25, 125, 625}
4. {2, 4, 6,},
5. {1, 4, 9, ………… 100}.

Solution:

1. A = {x : x is a natural number, multiple of 3 and x < 15}
2. B = {x : x = 2ⁿ, x∈N and n < 6}
3. C = {x : x = 5ⁿ, and x∈N and n ≤ 4}
4. D = {x : x is an even natural number}
5. E = {x : x = n², x ∈ N and n < 11}.

Question 2.
Write the following sets in roster form: (NCERT)

1. A = {x : x is an odd natural number}
2. B = {x : x is an integer – \(\frac{1}{2}\) < x < \(\frac{9}{2}\)
3. C = {x : x is an integer x² ≤ 4}
4. D = {x : x is a letter of a word LOYAL}
5. E = {x : x is a month of year which does not have 31 days}
6. F – {x : x is a consonant of English alphabet which comes before k}

Solution:

1. A = {1, 3, 5, 7, 9}
2. B = {0, 1, 2, 3, 4}
3. C = {-2, -1, 0, 1, 2}
4. D = {L, 0, Y, A}
5. E = {FEBRUARY, APRIL, JUNE, SEPTEMBER, NOVEMBER}
6. F = {b, c, d, f, g, h, j}.

Question 3.
From following find whether A = B. (NCERT)

1. A = {a, b, c, d}, B = {d, c, b, a}.
2. A = {4, 8, 12, 16}, B = {8,4, 16, 18}.
3. A = {2, 4, 6, 8, 10}, B = {x : x is a positive even integer x < 10}.
4. A = {x : x is a multiple of 10}, B = {10, 15, 20, 25, 30}.

Solution:
1. A = {a, b, c, d}, B = {d, c, b, a}
All the elements of set A is in set B
A = B.

2. A = {4, 8, 12, 16}, B = {8, 4, 16, 18}
All the elements of A is not in set B.
∴ A≠B.

3. Here A = {2, 4, 6, 8, 10} and B = { x : x is an even number and x ≤ 10}
B = {2, 4, 6, 8, 10}
i.e., The elements of set A and B are same.
∴ A = B

4. Here A = {x : x is a multiple of 10} = {10, 20, 30, 40}
and B = {10, 15, 20, 25, 30, …}
The elements of set A and B are not same.
∴ A≠B.

Question 4.
If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10} then find the following: (NCERT)

1. A∪B
2. A∪C
3. B∪C
4. B∪D

Solution:

1. A ∪ B = {1, 2, 3, 4} ∪ {3, 4, 5, 6} = {1, 2, 3, 4, 5, 6}.
2. A ∪ C = {1, 2, 3, 4} ∪ {5, 6, 7, 8} = {1, 2, 3, 4, 5, 6, 7, 8}.
3. B ∪ C = {3, 4, 5, 6} ∪ {5, 6, 7, 8} = {3, 4, 5, 6, 7, 8}
4. B ∪ D = {3, 4, 5, 6} ∪ {7, 8, 9, 10} = {3, 4, 5, 6, 7, 8, 9, 10}

Question 5.
If A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C= {11, 13, 15} and D = {15, 17} then, find the following:

1. A ∩ B
2. B ∩ C
3. A ∩ C
4. B ∩ D

Solution:

1. A ∩ B = {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13} = {7, 9, 11}
2. B ∩ C = {7, 9, 11, 13} ∩ {11, 13, 15} = {11, 13}
3. A ∩ C = {3, 5, 7, 9, 11} ∩ {11, 13, 15} = {11}
4. B ∩ D = {7, 9, 11, 13} ∩ {15, 17} = ϕ

Question 6.
If A = {3, 6, 9, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12, 14, 16} and D = {5, 10, 15, 20} then, find the following: (NCERT)

1. A – B
2. A – C
3. A – D
4. B – A
5. C – A
6. D – A
7. B – C
8. B – D
9. C – B
10. D – B
11. C – D
12. D – C?

Solution:
Let A and B are two sets and A – B is a set of those elements which one present in A and not in B.

1. A – B = {Those elements present in A and not in B}
= {3, 6, 9, 12, 15, 18, 21} – {4, 8, 12, 16, 20}
= {3, 6, 9, 15, 18, 21}.

2. A – C = {Those elements present in A and not in C}
= {3, 6, 9, 12, 15, 18, 21} – {2, 4, 6, 8, 10, 12, 14, 16}
= {3, 9, 15, 18, 21}.

3. A – D = {The elements which are in A but not in D}
= {3, 6, 9, 12, 15, 18, 21} – {5, 10, 15, 20}
= {3, 6, 9, 12, 18, 21}.

4. B – A = {The elements which are in B but not in A}
= {4, 8, 12, 16, 20} – {3, 6, 9, 12, 15, 18, 21}
= {4, 8, 16, 20}.

5. C – A = {The elements which are in C but not in A}
= {2, 4, 6, 8, 10, 12, 14, 16} – {3, 6, 9, 12, 15, 18, 21}
= {2, 4, 8, 10, 14, 16}.

6. D – A = {The elements which are in D but not in A}
= {5, 10, 15, 20} – {3, 6,9, 12, 15, 18, 21}
= {5, 10, 20}.

7. B – C = {The elements which are in B but not in C}
= {4, 8, 12, 16, 20} – {2, 4,6, 8, 10, 12, 14, 16}
= {20}.

8. B – D = {The elements which are in B but not in D}
= {4, 8, 12, 16, 20} – {5, 10, 15, 20}
= {4, 8, 12, 16}.

9. C – B = {The elements which are in C but not in B}
= {2, 4, 6, 8, 10, 12, 14, 16} – {4, 8, 12, 16, 20}
= {2, 6, 10, 14}.

10. D – B = {The elements which are in D but not in B}
= {5, 10, 15,20} – {4, 8, 12, 16, 20}
= {5, 10, 15}.

11. C – D = {The elements which are in C but not in D}
= {2, 4, 6, 8, 10, 12, 14, 16} – {5, 10, 15, 20}
= {2, 4, 6, 8, 12, 14, 16}.

12. D – C = {The elements which are in D but not in C}
= {5, 10, 15, 20} – {2, 4, 6, 8, 10, 12, 14, 16}
= {5, 15, 20}.

Question 7.
Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C then find the following:

1. A’
2. B’
3. (A ∪ C)’
4. (A ∩ B)’
5. (A’)’
6. (B – C)’
7. (B – C)’

Solution:
1. A’= U – A
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4}
⇒ A’= {5, 6, 7, 8, 9}.

2. B’ = U – B
⇒ {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 4, 6, 8}
⇒ B’ = {1, 3, 5, 7, 9}.

3. (A ∪ C) = {1, 2, 3, 4} ∪ {3, 4, 5, 6}
⇒ A ∪ C = {1, 2, 3, 4, 5, 6}
(A ∪ C)’= U – (A ∪ C)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4, 5, 6}
⇒ (A ∪ C)’ = {7, 8, 9}

4. A ∪ B = {1, 2, 3, 4} ∪ {2, 4, 6, 8} = {1, 2, 3, 4, 6, 8}
(A ∪ B)’ = ∪ – (A ∪ B)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4, 6, 8}
⇒ (A ∪ B)’ = {5, 7, 9}

5. (A’)’ = ∪ – A’
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {5, 6, 7, 8, 9}
⇒ (A’)’= {1, 2, 3, 4} = A

6. B – C = {2, 4, 6, 8} – {3, 4, 5, 6} = {2, 8}.
(B ∪ C)’ = U – (B – C)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 8}
⇒ (B – C)’= {1, 3, 4, 5, 6, 7, 9}.

Sets Short Answer Type Questions

Question 1.
If X and Fare two sets, n(X) = 17, n(Y) = 23 and n(X ∪ F) = 38, then find n(X ∩ F)? (NCERT)
Solution:
We know that n(X ∪ Y) = n(X) + n(Y) – n(X∩Y)
Given: n(X) = 17, n(Y) = 23, n(X ∪ Y) = 38, n(X ∩ Y) = ?
∴ 38 = 17 + 23 – n(X ∩ Y)
⇒ 38 = 40 – n(X ∩ Y)
⇒ n(X ∩ Y) = 40 – 38
⇒ n(X ∩ Y) = 2.

Question 2.
If X and Y are two sets, such that X∪Y has 18 elements, X has 8 elements and F has 15 elements, then how many elements does X ∩ Y have? (NCERT)
Solution:
Given: n(X ∪ Y) = 18, n(X) = 8, n(Y) = 15.
Applying formula n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
18 = 8 + 15 – n(X ∩ Y)
⇒ n(X ∩ Y) = 23 – 18 = 5.

Question 3.
If S and Tare two sets, such that S has 21 elements, T has 32 elements and S∩T has 11 elements, how many elements does S∪T have? (NCERT)
Solution:
Given: n(S) = 21, n(T) = 32, n(S ∩ T) = 11, n(S ∪ T) = ?
∵ n(S ∪ T) = n(S) + n(T) – n(S ∩ T)
⇒ n(S ∪ T) = 21 + 32 – 11 = 42.

Question 4.
If X and Y are two sets such that X has 40 elements, X∪Y has 60 elements and X∩Y has 10 elements, then how many elements does Y have? (NCERT)
Solution:
Given: n(X) = 40, n(X ∪ Y) = 60, n(X ∩ Y) = 10, n(Y) = ?
Applying formula n(X ∪ Y) = n(X) + n(Y) – n{X ∩ Y)
⇒ 60 = 40 + n(Y) – 10
⇒ 60 = 30 + n(Y)
⇒ n(Y) = 30.

Question 5.
If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10}, then find the following: (NCERT)

1. A ∪ B ∪ C
2. A ∪ B ∪ D
3. B ∪ C ∪ D.

Solution:
1. A ∪ B ∪ C = {1, 2, 3, 4} ∪ {3, 4, 5, 6} ∪ {5, 6, 7, 8}
= {1, 2, 3, 4, 5, 6} ∪ {5, 6, 7, 8}
= {1, 2, 3, 4, 5, 6, 7, 8}.

2. A ∪ B ∪ D = { 1, 2, 3, 4} ∪ {3, 4, 5, 6} ∪{7, 8, 9, 10}
= {1, 2, 3, 4, 5, 6} ∪ {7, 8, 9, 10}
= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

3. B ∪ C ∪ D = {3, 4, 5, 6} ∪ {5, 6, 7, 8} ∪ {7, 8, 9, 10}
= {3, 4, 5, 6} ∪ {5, 6, 7, 8, 9, 10}
= {3, 4, 5, 6, 7, 8, 9, 10}.

Sets Long Answer Type Questions

Question 1.
In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee? (NCERT)
Solution:
Let C and T be the students taking coffee and tea.
Here, n(T) = 150, n(C) = 225, n(C ∩ T) = 100.
Applying formula n(C ∪ T) = n(T) + n(C) – n(C ∩ T)
n(C ∪ T) = 150 + 225 – 100 = 375 – 100
⇒ n(C ∪ T) = 275
Total number of students = 600 = n(U).
Number of students taking neither tea nor coffee = n(C ∪ T)’
n(C ∪ T)’ = n(U) – n(C ∪ T)
= 600 – 275 = 325.

Question 2.
In a group of 70 people, 37 likes coffee, 52 likes tea and each person likes at least one of these two drinks. Find the number of persons who likes both coffee and tea? (NCERT)
Solution:
Let C and T be the persons who likes coffee and tea.
Given: n(C) = 37, n(T) = 52, n(C ∪ T) = 70, n(C ∩ T) = ?
Applying formula n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
⇒ 70 = 37 + 52 – n(C ∩ T)
⇒ 70 = 89 – n(C ∩ T)
⇒ n(C ∩ T) = 89 – 70
⇒ n(C ∩ T) = 19
∴ Number of people who likes both coffee and tea = 19.

Question 3.
In a group of 65 people, 40 likes cricket, 10 likes both cricket and tennis. How many like tennis only and not cricket? How many like tennis? (NCERT)
Solution:
Let C and T denotes the people who likes cricket and tennis respectively.
Given: n(C ∪ T) = 65, n(C) = 40, n(C ∩ T) = 10
We know that
n(C ∪ T) = n(C) + n(T) – (C ∩ T)
65 = 40 + n(T) – 10
⇒ n(T) = 65 – 30 = 35.
∴ Number of people who likes only tennis and not cricket
= n(T ∩ C)’ = n(T) – (C ∩ T)
= 35 – 10 = 25.

Question 4.
In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read newspaper I, 9 read both H and I. 11 read both H and T, 8 read both T and 1.3 read all three newspapers. Find

1. The number of people who read at least one of the newspapers.
2. The number of people who read exactly one newspaper. (NCERT)

Solution:
Given: n (H) = 25, n (T) = 26, n(I) = 26, n(H ∩ I) = 9, n(H ∩ T) = 11, n(T ∩ I) = 8, n(H ∩ T ∩ I) = 3

1. The number of people who reads at least one of the newspaper = n(H∪T∪I)
= n(H) + n(T) + n(I) – n(H ∩ T) – n(T ∩ I) – n(H ∩ I) + n(H ∩ T ∩ I)
= 25 + 26 + 26 – 11 – 8 – 9 + 3
= 77 – 28 + 3 = 80 – 28 = 52.

2. The number of people who reads exactly one newspaper.
= n(H) + n(T) + n(I) – 2n(H ∩ I) – 2n(H ∩ T) – 2n(T ∩ I) + 3n(H ∩ T ∩ I)
= 25 + 26 + 26 – 2 × 9 – 2 × 11 – 2 × 8 + 3 × 3 = 77 – 18 – 22 – 16 + 9
= 86 – 56 = 30.

Question 5.
In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find how many liked product C only? (NCERT)
Solution:
Let A, B and C denotes the people liked the products A, B and C respectively.
Given: n(A) = 21, n(B) = 26, n(C) = 29, n(A ∩ B) = 14, n(C ∩ A) = 12, n(B ∩ C) = 14, n(A ∩ B ∩ C) = 8
n(only C) = n(C) – n(C ∩ A) – n(B ∩ C) + n(A ∩ B ∩ C)
= 29 – 12 – 14 + 8
= 37 – 26 = 11.

MP Board Class 11 Maths Important Questions

MP Board Class 11th Economics Important Questions Unit 3 Statistical Tools and Interpretation

Statistical Tools and Interpretation Important Questions

Statistical Tools and Interpretation Objective Type Questions

Question 1.
Choose the correct option:

Question (a)
The most common measurement of central tendency is:
(a) Median
(b) Multiple
(c) Mean
(d) Weighted mean.
Answer:
(c) Mean

Question 2.
Division of one column in ten equal parts is called:
(a) Decile
(b) Quartile
(c) Percentile
(d) None.
Answer:
(a) Decile

Question 3.
Which of the following equation is correct:
(a) Anova = σ
(b) Anova = σ ^-2
(c) Anova = σ ⁴
(d) Anova = √σ × 2.
Answer:
(b) Anova = σ ^-2

Question 4.
Where is correlation multiple placed:
(a) Between 0 and + 1
(b) Between – 1 and 0
(c) Between – 1 and + 1
(d) None of these.
Answer:
(c) Between – 1 and + 1

Question 5.
Base year is:
(a) Comparison year
(b) Present year
(c) Any year
(d) One year previous to present year.
Answer:
(a) Comparison year

Percentage Difference calculator is a free online tool to find the percent difference between two numbers.

Question 2.
Fill in the blanks:

1. Rule of ……………………….. is supposed to be idle for creation of index.
2. Multiple is the value of column in which ……………………… comes repeatedly.
3. The value which divides the column into more than two parts is called ……………………………
4. ………………………. is the difference between maximum and minimum value.
5. …………………………. is the simple ray that shows relation between two values.

Answer:

1. Fisher
2. Maximum
3. Division
4. Range
5. Analogical relation.

Question 3.
Match the columns:

Answer:

1. (c)
2. (d)
3. (a)
4. (e)
5. (b).

Question 4.
State true or false:

1. Cumulative frequency shows “less than” or “more than” of any sequence.
2. 5 is the mode of the numbers 3, 4, 3, 5, 5, 3, 2.
3. Range is found by adding highest value and smallest value.
4. Money inflation is measured on changes of index number of whole sale market of weekly statement.
5. High income leads high savings.

Answer:

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

Question 5.
Answer in one word:

1. Graphic method of studying dispersion.
2. Definite relation between two or more groups.
3. One which divides column in ten equal parts.
4. In India which index helps to measure average change in prices.
5. Saving is directly related to.

Answer:

1. Lorenz curve
2. Correlation
3. Deciles
4. Consumer Price Index
5. Income.

Statistical Tools and Interpretation Very Short Answer Type Questions

Question 1.
What is measure of central tendency?
Answer:
Measure of central tendency is the single value which represents the characteristics of the entire universe.

Question 2.
Which is the popular measure of central tendency?
Answer:
Arithmetic mean.

Question 3.
Define Arithmetic mean?
Answer:
It is defined as the sum of the values of all observations divided by the number of observations.

Question 4.
Find out mode from the following data:
7, 12, 8, 5, 9, 10, 9, 11, 96
Answer:
9 is the number which is repeated maximum number of times. Hence, mode is 9.

Question 5.
Write the formula of weighted arithmetic mean?
Answer:
Weighted Arithmetic Mean = \(\frac{ΣWx}{ΣW}\)

Free Online Polynomial in Descending Order Calculator determines the polynomial.

Question 6.
Arrange the following series in descending order:
5, 16, 18, 2, 13, 15, 3, 19, 17, 20.
Answer:
20, 19, 18, 17, 16, 15, 13, 5, 3, 2

Question 7.
Name any two positional averages?
Answer:

1. Median and
2. Mode.

Question 8.
What is median?
Answer:
Median is that value which divides a series into two equal parts.

Question 9.
Write the formula of median in discrete series?
Answer:
Median (M) = size of (\(\frac{N+1}{2}\))^th item.

Question 10.
Write the formula of median in continuous series?
Answer:
Median (M) =

Question 11.
What is Quartile?
Answer:
It divides the series into four equal parts. These are three quartiles Q₁, Q₂ and Q₃.

Question 12.
Write the formula of first quartile?
Answer:
First Quartile (Q₁) =

Question 13.
Write the formula of third quartile?
Answer:
Third quartile (Q₃) =

Question 14.
What is a mode?
Answer:
Mode is defined as that value which occurs the most frequently in the distribution.

Question 15.
Write the formula to calculate mode?
Answer:
Mode Z =

Question 16.
Write the names of two measures of central tendency which can be determined from graph paper?
Answer:

1. Median and
2. Mode.

Question 17.
Which is the mostly used measure of central tendency?
Answer:
Mean.

Question 18.
Which measure of central tendency is not affected by extreme values?
Answer:
Median.

Question 19.
What is Percentile?
Answer:
Percentile divides the series into 100 equal parts.

Question 20.
Which measure of central tendency is used to explain qualitative data?
Answer:
Mode.

Question 21.
Name the measure of central tendency which is based on all items?
Answer:
Mean.

Question 22.
Define measure of dispersion?
Answer:
Dispersion measures that extent to which different items tend to disperse away from an average.

Question 23.
What is Range?
Answer:
Range is the difference between the highest value and lowest value in a series.

Question 24.
Write the formula of range?
Answer:
Range = Highest value in series – Lowest value in series.

Question 25.
Define Quartile deviation?
Answer:
Quartile deviation is the half of inter quartile range.

Question 26.
Write the formula of quartile deviation?
Answer:
Quartile deviation = \(\frac { Q_{ 3 }-Q_{ 1 } }{ 2 } \)

Question 27.
Write the formula of coefficient of quartile deviation?
Answer:
Coefficient of quartile deviation = \(\frac{Q_{3}-Q_{1}}{Q_{3}+Q_{1}}\)

Question 28.
Define mean deviation?
Answer:
It is the arithmetic mean of the deviations of various items from their average.

Question 29.
Write the formula of coefficient of mean deviation?
Answer:
Coefficient of mean deviation = \(\frac { MD }{ \overline { XorMorZ } } \)

Question 30.
Define standard deviation?
Answer:
Standard deviation is the square root of the arithmetic mean of the squares of all deviation.

Question 31.
Write the formula of standard deviation?
Answer:
Standard deviation =

Question 33.
What is coefficient of variation?
Answer:
It is the percentage variation in the mean, the standard deviation being considered as the total variation in the mean.

Question 34.
Write the formula of coefficient of variation?
Answer:
Coefficient of variation = \(\frac { \sigma }{ \overline { X } } \) × 100.

Actual yield calculator of products formed in reactions that involve limiting reagents.

Question 35.
What does Lorenz curve show?
Answer:
Lorenz curve is the graphic representation of dispersion in a distribution. It shows the actual distribution of items.

Question 36.
What is coefficient of correlation?
Answer:
Correlation is an analysis of covariation between two or more variables.

Question 37.
What is positive correlation?
Answer:
Correlation is said to be positive when two variables move together in the same direction.

Question 38.
What is negative correlation?
Answer:
Correlation is said to be negative when two variables move together in the opposite direction.

Question 39.
If coefficient of correlation is zero then what does it mean?
Answer:
There is no correlation between two variable.

Question 40.
Write the formula of Karl Pearson to calculate correlation?
Answer:
\(r=\frac{\sum x y}{N \sigma_{1} \sigma_{2}}\)

Question 41.
When is Spearman’s rank difference method used?
Answer:
When variables are qualitative like Beauty, Bravery, Ability, Intelligence etc.

Question 42.
Write the formula of Spearman’s rank difference method?
Answer:

Question 44.
What is the measure of Index in base year?
Answer:
100.

Question 45.
What is base year?
Answer:
Base year is the year of reference with which the extent of change in the current year is measured.

Question 46.
Write two types of Index number?
Answer:

1. Simple Index number and
2. Weighted Index number.

Question 47.
What is simple index number?
Answer:
These are the index numbers in which all items of the series are accorded equal importance.

Question 48.
What is weighted index number?
Answer:
These are the index numbers in which different items of the series are accorded different weightage, depending upon their relative importance.

Question 49.
Who publishes wholesale price index in India?
Answer:
Central statistics organization (CSO).

Question 50.
What does wholesale price index show?
Answer:
It measures the relative changes in the prices of commodity traded in the wholesale markets.

Question 51.
On which basis wholesale price index is available in India?
Answer:
Weekly basis.

Question 52.
What is the other name of consumer price index?
Answer:
Cost of living index.

Question 53.
Define consumer price index?
Answer:
Consumer price index is the index number which measures the average change in prices paid by the specific class of consumers for goods and services consumed by them in the current year in comparison with base year.

Question 54.
What is base year of Index number of Industrial production?
Answer:
1993 – 1994.

Question 55.
What is Assumed mean?
Answer:
Assumed mean is the value of a series which helps in calculating the arithmetic mean.

Question 56.
What is senses?
Answer:
Sensex is an index representing the movement in share price of major companies listed in Bombay stock exchange.

Question 57.
What is the base year of sensex?
Answer:
1978 – 1979.

Question 58.
How many companies are included in sensex?
Answer:
30.

Question 59.
What does it mean when sensex goes up?
Answer:
If sensex goes up it means that the prices of the stocks of most of the companies under BSE sensex have gone up.

Statistical Tools and Interpretation Short Answer Type Questions

Question 1.
A measure of dispersion is a good supplement to the central value in under – standing a frequency distribution? Comment?
Answer:
Dispersion is the extent to which values in a distribution differ from the average of the distribution. Knowledge of only average is insufficient as it does not reflect the quantum of variation in values. Measures of dispersion enhance the understanding of a distribution considerably by providing information about how much the actual value of items in a series deviate from the central value, e.g., per capita income. Thus, a measure of dispersion is a good supplement to the central value in understanding a frequency distribution.

Question 2.
Which measure of dispersion is the best and how?
Answer:
Standard deviation is considered to be the best measure of dispersion and is therefore the most widely used measure of dispersion.

1. It is based on all values and thus provides information about the complete series.
2. It is independent of origin but not of scale.
3. It is useful in advanced statistical calculations like comparison of variability in two data sets.
4. It can be used in testing of hypothesis.
5. It is capable of further algebraic treatment.

Question 3.
Some measures of dispersion depend upon the spread of values whereas some calculate the variation of values from a central value. Do you agree?
Answer:
Yes, it is true that some measures of dispersion depend upon the spread of values, whereas some calculate the variation of values from the central value. Range and Quartile Deviation measure the dispersion by calculating the spread within which the value lie. Mean deviation and standard deviation calculate the extent to which the values differ from the average or the central value.

Question 4.
In town, 25% of the persons earned more than? 45,000 whereas 75% earned more than 18,000. Calculate the absolute and relative values of dispersion?
Answer:
25% of the persons earned more than 45,000. This implies that upper quartile Q₃ = 45,000 75% earned more than 18,000. This implies that lower quartile Q₁ = 18,000. Absolute measure of dispersion = Q₃ – Q₁ = 45,000 – 18,000 = 27,000
Relative measure of Dispersion:
Co – efficient of Quartile Deviation = \(\frac { Q_{ 3 }-Q_{ 1 } }{ Q_{ 3 }+Q_{ 1 } } \) where Q₃ = 3rd Quartile Q₁ = 1st Quartile
= \(\frac{45,000-18,000}{45,000+18,000}\) = \(\frac{27,000}{63,000}\)

Question 5.
When is rank correlation more precise than simple correlation coefficient?
Answer:
Rank correlation is more precise than simple correlation coefficient in the following situations:

1. When the measurement of the variables are suspect e.g., in a remote village where measuring rods or weighing scales are not available, height and weight of people cannot be measured precisely but the people can be easily ranked in terms of height and weight.

2. When data is qualitative: It is difficult to quantify qualities such as fairness, honesty etc.

3. When data has extreme values: Sometimes the correlation coefficient between two variables with extreme values may be quite different form the coefficient without the extreme values. Under these circumstances rank correlation provides a better alternative to simple correlation.

Question 6.
Does zero correlation mean independence?
Answer:
No, zero correlation does not mean independence. If there is zero correlation (r_XY= 0), it means the two variables are uncorrelated and there is no linear relation between them. However, other types of relation may be there and they may not be independent.

Question 7.
Why do we need on Index number?
Answer:
Index number are needed due to the following reasons:

1. Measurement of change in the price level or the value of money.
2. Information of foreign trade.
3. Calculating real wages.
4. Measuring and comparing output.
5. Policy making of government.

Question 8.
What are the desirable properties of the base period?
Answer:
Base period should have the following properties:

1. The base year should be a normal period and periods in which extraordinary events have occurred should not be taken as base periods as they are not appropriate for general comparisons.
2. Extreme value should not be selected as base period.
3. The period should not be too far in the past as comparison with current period cannot be done with such base year as policies, economic and social conditions change with time.
4. It should be updated periodically.

Question 9.
Why is it essential to have different CPI for categories of consumers?
Answer:
The Consumer Price Index (CPI) in India is calculated for different categories as under:

1. CPI for industrial workers.
2. CPI for urban non – manual employees.
3. CPI for agricultural labourers.

Question 10.
What is the difference between a price index and a quantity index?
Answer:
The difference between a Price index and a Quantity index:
Price Index:

1. It measure and allow for comparison of the prices of certain goods.
2. Its widely used.
3. It is known as unweighted index numbers.

Quantity Index:

1. It measure the changes in the physical volume of production construction or employment.
2. Its narrowly used.
3. It is known as weighted index numbers.

Question 11.
“Index numbers are Economic Barometers”? Explain?
Answer:
Index numbers are used to measure the level of economic activities like barometers. They are called as barometers due to following reasons:

1. Wholesale price index number shows the changes in general price level which can be used for any consumer category.
2. Index number of agricultural production is a weighted average of quantity relatives.
3. Index number is a useful guide to help in decision making while investing in the stock market.

Question 12.
What is “mode”? Explain its merits?
Answer:
The word “mode” has been derived from the French word “La mode” which means fashion. In statistics, mode means the value, which occurs most frequently in the series. Following are the merits of mode:

1. Simple calculation:
Calculation of mode is simple. Sometimes in individual and discrete series it can be found out by inspection only.

2. Least effect of extreme values:
Mode is not affected by extreme values. If the groups are unequal even then mode can be calculated.

3. Representative measure:
Mode is a representative and real measure of the series because this is one of the values from the group itself.

4. Use in quantitative facts:
Mode is also used for describing the quantitative data.

Question 13.
Explain the merits and demerits of Karl Pearson’s coefficient of correlation?
Answer:
Merits:

1. This method gives us precise magnitude of the correlation due to which interpretation of result becomes easy.
2. It is the most popular method of measuring relationship between the two variables.
3. It expresses the direction and degree of change in two variables.

Demerits:

1. It cannot be used to quantity qualities such as honesty fairness, etc.
2. The value of correlation are largely affected by the values of extreme items.

Question 14.
Explain the merits and demerits of Spearman rank difference method?
Answer:
Merits:

1. Simple:
Spearman’s ranking method is very simple compare to Pearson’s method. It can be easily calculated and understood than Pearsonian coefficient of correlation. In Karl Pearson’s method more calculation are required

2. Compulsory calculations:
In case ranks of values in the series are given rank methods is the only way to compute coefficient of correlation.

3. Useful for data of qualitative nature:
In case the data is related to ability, performance, beauty, honesty, intelligence and honesty etc. So this is the most suitable method to calculate coefficient of correlation.

4. Same result:
The answer obtained by rank method equals the answers determined through Pearsonian method, if values are not repeated.

5. Convenient:
This method can be used convenient by even if numbers in series are given in irregular way.

6. Best method:
This method is the best method in cases where the correlation is found by personal data.

7. This method is appropriate when number of elements is only 20 or 30.

Demerits:
Though the method is good but still there are certain drawbacks in it. They are as follows:

1. Unsuitable of grouped frequency distribution:
This method cannot be used for finding out correlation in grouped frequency distribution. This method is applicable only in individual observation.

2. Unsuitable if the values of the series exceed 30:
In such cases, obtaining ranks their difference and square becomes difficult so Pearsonian coefficient of correlation method is recommended. When N becomes more than thirty the counting becomes complicated. The scope of this method is limited.

3. Lack in precision:
This method is based on ranks. All the information related to data is not used, so it lacks precision as compared to Pearsonian method of coefficient of correlation.

Question 15.
The monthly expenditure of some families in a village are as follows:

Calculate range and its coefficient?
Answer:
Range = H – L
= 5,000 – 500 = 4,500 Rs.
Coefficient of range = \(\frac{H – L}{H + L}\) = \(\frac{5000 – 500}{5000 + 500}\) =\(\frac { 4500 }{ 5500 }\) = 0.82

Question 16.
Two factories A and B are established in one industry? Following information are available about them:

In which of factory wages rates are more?
Answer:
To find out the difference in wage rates we will have to find out coefficient of variation.
Coefficient of variation = \(\frac{S D}{\bar{X}} \times 100\)
Coefficient of variation of factory A = \(\frac{3}{120}\) × 100 = 2.5 %
Coefficient of variation of factory B = \(\frac{4}{85}\) × 100 = 4.7 %
Wage rates of factory B is more.

Question 17.
Explain the importance of correlation in statistics?
Answer:
Some importance features about the correlation:

1. The correlation of a sample is represented by the letter r.
2. The range of possible values for a correlation is between – 1 to +1.
3. A positive correlation indicates a positive linear association. The strength of the positive linear association increases as the correlation becomes closer to +1.
4. A negative correlation indicates a negative linear association. The strength of the negative linear association increases as the correlation becomes closer to -1.
5. A correlation of either +1 or -1 indicates a perfect linear relationship. This is hard to find with real data.

Question 18.
Explain the utility of index numbers?
Answer:

1. Index numbers help in framing suitable policies:
Index numbers provide various type of information on the basis of which various policies suitable to our economy are framed. On the basis of index number the insurance companies fix their premiums and banks fix their interest rates. Railways fixed their fare on this index only. It gives knowledge about the purchasing power, industrial and agricultural production about other countries.

2. Make the difficult facts easy:
By index number, sometimes we make some facts and events easy which is not possible by any other method. For example, measurement of trading activities is not possible by studying single fact. But by analysing the progress of industrial production, Banking and transportation the trading index can be found out. Many emotional facts take the concrete form with the help of index. Thus, they are made simple and easier for general people.

3. Comparative study is possible:
By index number comparative study is possible. This is because they express changes in relative form so it is not difficult to compare two things.

4. Indication towards future economic tendency:
Index number expresses not only the present conditions but on the basis of it we can find out the conclusions of the facts of future also. Thus, we reach to the present facts with the help of future inferences and thus we get the indication of future economic tendency.

Question 19.
Define base year? What are the guidelines to be observed while determining the base year?
Answer:
Base year is the period with respect to which change is measured. Following guidelines are to be observed while determining the base year:

1. Base year should be a normal year i.e. it should be free form abnormalities like wars, earthquakes, floods, etc.
2. Base year should not be too old.
3. Base year should be clearly defined.

Question 20.
Construct price index for 1990 taking 1989 as base year from the following information:

Answer:
Calculation of Price Index by simple aggregative method:

Price Index for year 1990 (P₀₁) = \(\frac { \Sigma P_{ 1 } }{ \Sigma P_{ 0 } } \) x 100
= \(\frac{360}{300}\) x 100
= \(\frac{36000}{300}\) = 120
These is a net increase of (120 – 100) = 20% in prices in 1989 as compared to 1990.

Question 21.
Calculate Index for 1989 taking base year 1989 by average of price relatives method?

Answer:
Calculation of Price Index by average of price relatives method:

Question 22.
Calculation of Index by Laspeyre’s method from the following:

Answer:
Calculate of Index by Laspeyre’s method:

Question 23.
Calculate index by Paasche’s method the following:

Answer:
Calculation of index by Paasche’s method:

Statistical Tools and Interpretation Long Answer Type Questions

Question 1.
If the arithmetic mean of the data given below is 28. Find:

1. Missing frequency
2. Median of the series.

Answer:
Calculation of missing frequency:

\(\overline { X } \) = \(\frac{Σfm}{Σf}\)
or 28 = \(\frac{2100+35x}{80+x}\)
or or 2240+ 28x = 2100 + 35x
or 2240 – 2100 = 35x – 28x
or 7x = 140
x = \(\frac{140}{7}\)
Hence, missing frequncy is 20.

Question 2.
Calculate median from the following:

Answer:
Calculation of median:

Median = Size of (\(\frac{N}{2}\))^th items.
= \(\frac{100}{2}\)^th item.
= 50^th item.
50th median class is 20 – 30.

= 20 + \(\frac{20}{27}\) x 10 = 20 + 7.41 = 27.41

Question 3.
Calculate arithmetic mean by shortcut method:

Answer:
Calculate of mean by shortcut method:

Formula: \(\overline { X } \) = A + \(\frac { \Sigma fdx }{ N } \)
Here, \(\overline { X } \) = ?, N = 100, A = 25, Σfdx = 620.
∴ \(\overline { X } \) = 25 + \(\frac{620}{100}\) = 25 + 6.2 = 31.2.

Question 4.
Calculate mode from the following data:

Answer:
From the following table modal class is 20 – 30 as its frequency is the highest.

Question 5.
Calculate quartiles from the followings:

Answer:
Calculate of quartiles:

Q₁ = \(\frac{N+1}{4}\) th item = \(\frac{100+1}{4}\) th item.
= 25.25 th item
= 25 th item + 0.25 (26 th item – 25 th item)
= 7 + 0.25 (7-7)
= 7 + 0.25(0) = 7 + 0 = 7.
Q₂ = 2 (\(\frac{N+1}{4}\)) th item = 50.50 th item.
= 50 th item + 0.5 (51th item – 50th item)
= 9 + 0.5 (9 – 9) = 9 + 0.5(0) = 9 + 0 = 9.
Q₃ = 3 (\(\frac{N+1}{4}\)) th item = 75.75 th item
75th item + 0.75 (76 th item – 75 th item)
= 16 + 0.75 (16 – 16)
= 16 + 0.75(0) = 16 + 0
= 16.

Question 6.
Calculate dispersion and relative measures of dispersion:

Answer:
Calculating relative measure of dispersion:

Here, R = ?, L = 35.5, S = 3.5.
Formula: R = L – S = 35.5 – 3.5 = 32
∴ Relative measure of dispersion = \(\frac{L-S}{L+S}\)
= \(\frac{35.5-3.5}{35.5+3.5}\)
= \(\frac{32}{39}\) = 0.8205 (Approx).

Question 7.
Calculate quartile deviation and its coefficient from the following data:

Answer:
Calculation of quartile deviation and its coefficient:

Q₁ = Size of \(\frac{N+1}{4}\) th item = Size of \(\frac{87+1}{4}\) th item.
= Size of \(\frac{88}{4}\) th item.
= Size of 22 th item, which lies in C.F. 38
∴ Q₁ =9

∴ Coefficient of Q.D. = 0.5.

Question 8.
The yield of wheat and rice per acre for 10 districts of a state:

Calculate mean deviation about mean.
Answer:

Question 9.
The yield of wheat and rice of 10 districts of a state are:

Calculate mean deviation from median.
Answer:

Question 10.
The yield of rice and wheat of 10 districts of a state are:

Calculate standard deviation and its coefficient?
Answer:

Question 11.
A batsman is to be selected for a cricket team? The choice is between X and Y on the basis of their scores in five previous scores which are:

Which batsman should be selected if we want?

1. A higher run gether or
2. A more reliable batsman is the team

Answer:

Question 12.
Calculate the correlation coefficient between the heights of fathers in inches (x) and their son (y):?

Answer:

Question 13.
Calculate coefficient of correlation between x and y, and comment on their relationship?

Answer:

These are perfectly positive correlation between x and y.

Question 14.
There are 80 students in a class. Average marks of statistics in class is 65 marks. There are two sections in the class, first have 50 students whose average marks is 60. Calculate the average marks of other section?
Answer:
Total students N = 80
Average of marks \(\bar{X}_{1}\) = 65
Students of 1st section N₁ = 50
Students of second section N₂ = 80 – 50 = 30
Average marks of students of 1st section \(\bar{X}_{1}\) = 60
N \(\bar{X}_{1}\) = + N₁ \(\overline { X } _{ 1 }\) + N₂ \(\overline { X } _{ 2 }\)
80 x 65 = 50 x 60 + 30 \(\overline { X } _{ 2 }\)
5,200 – 3,000 = 30 \(\overline { X } _{ 2 }\)
30 \(\overline { X } _{ 2 }\) = 2,200
Average marks of students of 2nd section \(\overline { X } _{ 2 }\) = \(\frac { 1 }{ 2 }\)
\(\overline { X } _{ 2 }\) = 73.33.

Question 15.
Calculate mode from the following data:

Answer:
Modal class is 30 – 40 as from observation highest frequency is 32?

Question 16.
The median in following table is 42? Determine the missing frequency?

Answer:
Let the frequency of class 20 – 30 be f₂ Unknown frequency of class 50 – 60 f₂.

N = 288 = 106 + f_{1 +} f₂
160 + f₁ + f₂ = 288
f₁ + f₂ = 288
f₁ + f₂ = 288 – 106 = 182
Median 42 is in class 40 – 50.

= 5 (42 – 40) = 94 – f₁
= 210 – 200 = 94 – f₁
= 10 – 94 = – f₁
f₁ = 84
f₁ + f₂ = 182
f₂ = 182 – f₁ = 182 – 84
f₂ = 98

Question 17.
Compare the arithmetic mean, median and mode as measures of central tendency?
Answer:
Following table shows the comparison of arithmetic mean, median and mode as measures of central tendency:

Question 18.
Distinguish between mean deviation and standard deviation?
Answer:
Following are differences between mean deviation and standard deviation:

1. Algebraic sign like plus (+) and minus (-) are ignored while calculating mean deviation, they are taken into account in the calculation of standard deviation.
2. Mean deviation can be calculated by any of the three averages i.e., mean, median and mode standard deviation can be calculated only by the Arithmetic mean.
3. Mean deviation cannot be used for further algebraic treatment while standard deviation is capable of further algebraic treatment.
4. Standard deviation is considered to be superior to mean deviation.
5. Mean deviation is easy to calculated and to understand as compared to standard deviation.

Question 19.
Calculate standard deviation from the following data:

Answer:

Question 20.
Calculate coefficient of correlation by Karl Pearson method from following data:

Answer:

Question 21.
Calculate coefficient of correlation by Karl Pearson method from following data:

Answer:

Question 22.
Calculate coefficient of correlation by Spearman rank difference method?

Answer:

Question 23.
In a drama competition two judges gave the following marks to 10 participants? Calculate rank coefficient?

Answer:

Question 24.
What is index number? What is its importance in statistics?
Answer:
An index number is a specialised average designed to measure the net change in a group of related variables over a period of time. Following are the main uses of index number which show its importance in statistics:

1. To simplify complicated matters:
Index numbers present the given information in such a manner that it can be easily understood.

2. To measure comparative changes:
Index numbers facilitate comparison of changes from time to time, among different places and in series expressed in different units. The changes in price level, cost of living etc. which are not capable to measurement directly are measured with the help of index numbers.

3. To frame suitable policies:
Index numbers guide a lot in framing suitable economic policies. For example, wholesale and retail price index number help in economic and business policy – making regarding, price, output, demand, sales etc. The indices of consumption of various commodities help in the planning of their future production. Index numbers are applied with advantage for formulating and revising their policies from time to time.

4. To measure the purchasing power of money:
Index numbers are helpful in measuring the purchasing power, i. e., value of the money. This helps in fixing proper wage policy in the country.

5. To study trends and to make forecast:
Index number are most widely used for measuring changes over a period of time. On the basis of present indices, the forecast for the future can be made.

Question 25.
What is the purpose of constructing an index number of price and quantities?
Answer:
The purpose of constructing and index number of prices is to enable us to compare prices of certain goods over a period of time. It is used to measure change in the price level or the value of money during different periods of time.

The purpose of constructing an index number of quantities is to enable us to compare physical quantity of goods produced, consumed or distributed over a period of time. It is used to show whether the level of agricultural and industrial production in the economy is increasing or decreasing.

Question 26.
What are the limitations of index number?
Answer:
The limitation of index numbers are as follows:

1. Since index numbers are based on sample data, they only provide approximate results which may not exactly represent the changes in relative levels.
2. There are more chances of errors in construction of index numbers. It may be any where from level of selection of commodities to choice of the formula.
3. Index numbers often find difficulty to record change in the quality of a variable.
4. There does not exist a unique index number method which is acceptable to all.
5. Index numbers are also misused by dishonest persons to draw desirable conclusion for their selfish motives.

Question 27.
Throw light on importance of consumer price index?
Answer:
The importance of consumer price index is:

• It is used to determine the purchasing power of money and real wages.
• It is also used to analyze the market of specific commodities.
• It helps to formulate government policies.
• It helps in estimation of national income.
• Cost of living index numbers are used as basis for the wage adjustments.

Question 28.
Calculate Price Index by Fisher’s method from following data:

Answer:

Question 29.
What are the points to be kept in mind while constructing index number?
Answer:
The points to be kept in mind while constructing Index number are:

1. Purpose:
The purpose of construction of Index numbers should be clearly defined because every index number has a specific use.

2. Selection of a base years:
Selection of base year is important for construction of index number. It should be selected carefully.

3. Selection of number of Items:
It refers to collection of data which should be determined by the purpose for which the index is constructed.

4. Selection of an average:
As Index number are specialized averages. We need to decide on the average to be used. We need to select from various averages such as
arithematic mean, geometric mean, etc.

5. Selection system of weighing:
Weighing is having an important place in construction of index number. In this various goods are weighed according to their weights.

MP Board Class 11th Economics Important Questions

In this article, we will share MP Board Class 11th Hindi Solutions शुद्ध वाक्य रचना सम्बन्धी नियम Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 11th Special Hindi शुद्ध वाक्य रचना सम्बन्धी नियम

वाक्य रचना की सामान्य अशुद्धियाँ

वाक्य रचना में होने वाली सामान्य अशुद्धियाँ प्रायः इस प्रकार हैं

1. वचन सम्बन्धी अशुद्धियाँ-आदि, अनेक अथवा बहुवचन संज्ञा की क्रियाओं में प्रायः वचन सम्बन्धी त्रुटियाँ हो हैं। जैसे-
अशुद्ध वाक्य – शुद्ध वाक्य
(1) मैंने आज फूल आदि खरीदा। – (1) मैंने आज फूल आदि खरीदे।
(2)मैंने अनेकों स्थान देखे। – (2) मैंने अनेक स्थान देखे।
(3) बार-बार एक ही बात सुनते-सुनते मेरा कान पक गया। – (3) बार-बार एक ही बात सुनते-सुनते मेरे कान पक गये।

2. लिंग सम्बन्धी अशुद्धियाँ-विशेषण और विशेष्य का लिंग एक जैसा हो, समस्त पदों के सर्वनाम का लिंग एवं क्रिया का लिंग अन्तिम पद के अनुसार होना चाहिए। उदाहरण
अशुद्ध वाक्य – शुद्ध वाक्य
(1) सीता एक विद्वान् छात्रा है। – (1) सीता एक विदुषी छात्रा है।
(2) मैं अपनी देवी-देवताओं को मानता हूँ। – (2) मैं अपने देवी-देवताओं को मानता हैं।
(3) रीता ने आज मिष्ठान्न और नमकीन खरीदी। – (3) रीता ने आज मिष्ठान्न और नमकीन खरीदे।

3. अर्थ सम्बन्धी अशुद्धियाँ
अशुद्ध वाक्य – शुद्ध वाक्य
(1) आपकी सौजन्यता से मुझे कार्य मिला। – (1) आपके सौजन्य से मुझे कार्य मिला। [2009]
(2) निरपराधी को दण्ड देना अनुचित है। – (2) निरपराध को दण्ड देना अनुचित है। [2009]
(3) तुम अपने लड़के को मेरे यहाँ भेजो। – (3) तुम अपने पुत्र को मेरे यहाँ भेजो।
(4) शोक है कि तुम अनुत्तीर्ण हो गए। – (4) खेद है कि तुम अनुत्तीर्ण हो गए।

4. काल सम्बन्धी अशुद्धियाँ
अशुद्ध वाक्य – शुद्ध वाक्य
(1) मैंने कल पुस्तकें खरीदूंगा। – (1) मैं कल पुस्तकें खरीदूंगा।
(2) वह अगले साल दिल्ली गया था। – (2) वह पिछले साल दिल्ली गया था।
(3) मैं अगले सप्ताह विदेश जाता हूँ। – (3) मैं अगले सप्ताह विदेश जाऊँगा।

5. पुनरुक्ति सम्बन्धी अशुद्धियाँ
अशुद्ध वाक्य – शुद्ध वाक्य
(1) वह सदैव हमेशा सच बोलता है। – (1) वह सदैव सच बोलता है।
(2) वे सज्जन पुरुष आए हैं। – (2) वे सज्जन आए हैं।
(3) पूरे संसार जगत् में ईश्वर व्याप्त है। – (3) पूरे संसार में ईश्वर व्याप्त है।

6. मात्रा सम्बन्धी अशुद्धियाँ
अशुद्ध वाक्य – शुद्ध वाक्य
(1) राना सांगा एक ऐतिहासिक पुरुष थे। – (1) राणा सांगा एक ऐतिहासिक पुरुष थे।
(2) सौन्दर्यता सबको मोह लेती है। [2009] – (2) सुन्दरता सबको मोह लेती है।
(3) उसे रास्ते में सात पुरुष मिलें। – (3) उसे रास्ते में सात पुरुष मिले।

7. अन्य अशुद्धियाँ
अशुद्ध वाक्य – शुद्ध वाक्य
(1) घर में चीनी केवल नाममात्र को है। – (1) घर में चीनी नाममात्र को है।
(2) राम अथवा श्याम गए हैं। – (2) राम अथवा श्याम गया है।
(3) वह एक फूलों की माला लाया। – (3) वह फूलों की एक माला लाया।
(4) राम ने कहा कि राम का भाई आया हैं। – (4) राम ने कहा कि उसका भाई आया है।
(5) आज मोहन मीरा के साथ आया है। – (5) मोहन आज मीरा के साथ आया है।

शब्द-बोध
1. वर्तनी की दृष्टि से शुद्ध शब्दों का प्रयोग

वर्तनी (Spelling) का तात्पर्य अक्षर-विन्यास से है। पहले इसके लिए ‘हिज्जे’ या ‘अक्षरी’ नाम प्रचलित थे। उच्चारण की शुद्धता के अनुसार एवं व्याकरण की दृष्टि के अनुरूप शब्दों को लिखना ही, वर्तनी की दृष्टि से शुद्ध शब्दों का प्रयोग कहलाता है। – इन अशुद्धियों के कारण होते हैं-अज्ञान, भ्रान्ति, असावधानी, बनकर बोलने की दुष्प्रवृत्ति और विदेशी प्रभाव।

शुद्ध लेखन के लिए शुद्ध वर्तनी अपेक्षित होती है। वर्तनी ही भाषा की प्रभाव-क्षमता बढ़ाती है। अत: वर्तनी के सुधार या शब्द-शुद्धिकरण के लिए नीचे कुछ अशुद्ध शब्दों के शुद्ध रूप दिये हैं-

2. अर्थ की दृष्टि से सही शब्दों का चयन

हिन्दी में कतिपय ऐसे शब्द हैं जिन्हें ऊपर से देखने पर कोई अर्थ-भेद नहीं दिखाई देता है, पर सूक्ष्मतः उनमें अन्तर होता है। ऐसे शब्दों का ज्ञान आवश्यक है।

कुछ शब्दों का अर्थ तो एक ही रहता है, पर प्रयोग में उनमें कुछ अन्तर रहता है। कभी-कभी सम्बद्ध शब्दों के प्रयोग से भी वाक्य अशुद्ध हो जाता है। जैसे
अशुद्ध वाक्य – शुद्ध वाक्य
(1) उसे व्यापार में हानि होने की आशा है। – (1) उसे व्यापार में हानि होने की आशंका है।
(2) विनय को परीक्षा में उत्तीर्ण होने की आशा है। – (2) विनय को परीक्षा में उत्तीर्ण होने की आशंका है।
(3) राम बुरी तरह प्रसिद्ध है। – (3) राम बुरी तरह बदनाम है।
(4) शोक है कि आपने मेरे पत्रों का उत्तर नहीं दिया है। – (4) खेद है कि आपने मेरे पत्रों का उत्तर नहीं दिया।
(5) मैंने एक वर्ष तक उसकी प्रतीक्षा देखी। – (5) मैंने एक वर्ष तक उसकी प्रतीक्षा की।
(6) रमा एक मोतियों का हार गले में पहने है। – (6) रमा मोतियों का एक हार गले में [2009] पहने है।
(7) मीना चाचाजी के ऊपर विश्वास करती है। – (7) मीना चाचाजी पर विश्वास करती है।
(8) शब्द केवल संकेत मात्र होते हैं। – (8) शब्द संकेत मात्र होते हैं।
(9) आपके एक-एक शब्द तुले हुए होते हैं। – (9) आपका एक-एक शब्द तुला हुआ होता है।
(10) उसे अनेकों व्यक्तियों ने आशीवाद दिया। – (10) उसे अनेक व्यक्तियों ने आशीर्वाद दिया।

(11) विंध्याचल पर्वत के दक्खिन में नर्मदा स्थित है। – (11) विंध्याचल के दक्षिण में नर्मदा स्थित है।
(12) मैं अपनी बात का स्पष्टीकरण करने के लिए तैयार हूँ। [2009]- (12) मैं अपनी बात का स्पष्टीकरण देने के लिए तैयार हूँ।
(13) मैं अपने गुरु के ऊपर बहुत श्रद्धा करता है। – (13) मैं अपने गुरु पर बहुत श्रद्धा रखता हूँ।
(14) सौन्दर्यता सबको मोह लेती है। [2009] – (14) सुन्दरता सबको मोह लेती है।
(15) हवा, चल रही है। – (15) हवा बह रही है।
(16) महादेवी वर्मा विद्वान महिला थी। – (16) महादेवी वर्मा एक विदुषी महिला थी।
(17) श्रीकृष्ण के अनेकों नाम हैं। [2009] – (17) श्रीकृष्ण के अनेक नाम हैं।
(18) यह घी की शुद्ध दुकान है। – (18) यह शुद्ध घी की दुकान है।
(19) मोहन आटा पिसाकर लाया। – (19) मोहन अनाज पिसाकर लाया।
(20) वे सज्जन पुरुष कौन हैं? – (20) वे सज्जन कौन हैं?
(21) प्रत्येक वृक्ष फल देते हैं।। – (21) प्रत्येक वृक्ष फल देता है।

अर्थ की दृष्टि से सही शब्दों का चयन करने के लिए सूक्ष्म अन्तर वाले शब्दों को कान लेना चाहिए जिससे सही अर्थ का ज्ञान हो तो हम सभी जगह उसका उपयोग कर सकें। जैसे
(1) अनुराग = स्त्री-पुरुष के बीच होने वाला स्नेह। प्रेम = स्नेह सम्बन्ध मात्र। वात्सल्य = छोटों के प्रति बड़ों का स्नेह। भक्ति = बड़ों के प्रति छोटों का श्रद्धायुक्त स्नेह।
(2) अस्त्र = जो फेंक कर मारा जाय, जैसे-तीर, भाला, बम। शस्त्र = जिससे काटा जाय; जैसे-तलवार, छुरा। आयुध = लड़ाई के सभी साधन, आयुध कहलाते हैं। इसमें अस्त्र-शस्त्र के अतिरिक्त लाठी, गदा, छड़ आदि को भी ले सकते हैं। हथियार = हाथ से चलाये जाने वाले आयुध।
(3) अशुद्धि = अपवित्रता, मिलावट, लिखने अथवा बोलने में कमी, इन सभी को अशुद्धि कहते हैं। भूल = विस्मृति, जिसमें कर्ता की असावधानी प्रकट होती है। त्रुटि = वस्तु में किसी कमी का होना।
(4) अमूल्य = वस्तु जो मूल्य देकर प्राप्त न की जा सके, जैसे-विद्या, ज्ञान, प्रेम, सफलता, सुख, विश्वास। बहुमूल्य = जिसका अधिक मूल्य देना पड़े, जो वस्तु साधारण मूल्य की न हो। जैसे-सोना, प्लेटिनम, हीरा। महँगा = जिस वस्तु को बदलते बाजार भाव के कारण अधिक मूल्य देकर पाया जाये।
(5) अलभ्य = जिसे किसी उपाय से न पाया जाय। दुर्लभ = जिसे अधिक कष्ट उठाकर पाया जाय।
(6) अलौकिक = जो संसार में इन्द्रियों के द्वारा सुलभ न हो। असाधारण = जो सांसारिक होकर भी अधिकता से न मिलता हो। अस्वाभाविक = अप्राकृतिक, जो प्रकृति के नियमों से बाहर का प्रतीत हो।
(7) अर्पण = शरणगति का भाव लिए अर्पण। प्रदान = धार्मिक भावना के साथ दूसरे का स्वत्व स्थापित करना, जैसे-जलदान, दीपदान, विद्यादान, गोदान, अन्नदान। प्रदान = सामान्य रूप से बड़ों के द्वारा छोटों को देना। वितरण = समाज में उत्पादित वस्तुओं के विभाजन की प्रणाली।
(8) अनुग्रह बड़े की ओर से छोटों के अभीष्ट सम्पादन की स्नेहपूर्ण क्रिया अनुग्रह है। अनुकम्पा = दुःख दूर करने की चेष्टा से युक्त कृपा। दया = दुःख देखकर द्रवित होना। कृपा = दुःख-निवारण के सामर्थ्य से युक्त दया। सहानुभूति = दूसरे के दुःख में समभागी होने का भाव। करुणा = दुःख निवारण की व्याकुलता लिए हुए द्रवित होना। ममता = माता के समान किसी पर वत्सल भाव।
(9) अंग = (अवयव) भौतिक समष्टि का भाग, जैसे-शरीर का अंग, वृक्ष का अंग, शाखा आदि। देह = शरीर। आकृति = रेखाओं से बनी कोई समष्टि, चाहे वह सजीव हो या निर्जीव।
(10) अमर्ष = द्वेष या दु:ख जो तिरस्कार करने वाले के कारण होता है। क्रोध = प्रतिकूल आचरण के बदले में प्रतिकूल आचरण की प्रवृत्ति। कलह = क्रोध को जब सक्रिय रूप दिया जाता है तो उससे होने वाला वाचिक या शारीरिक आचरण।

(11) असुर = दैत्य या दानव-एक जाति विशेष, जो देवों के समकक्ष थी। राक्षस = यह अनुचित आचरण के कारण मनुष्य जाति के लिए आता है।
(12) अनुपम = बेजोड़-जिसकी तुलना न हो सके। अपूर्व = जिसे पहले भनुभव न किया गया हो। अद्भुत = जो विस्मयजनक हो।
(13) आयु = जीवन के सम्पूर्ण समय को आयु कहते हैं। वयस् = जीवन के शरीर विकास सूचक भाग को वयस् कहा जाता है, जैसे-शैशव। अवस्था = दशा-जीवन के प्रसंग में बीते हुए जीवन का भाग। वयस् के अर्थ में भी चलता है, जैसे-युवावस्था, वृद्धावस्था।
(14) अबला = स्त्री की संज्ञा है जिससे उसकी कोमलता एवं निर्बलता का संकेत मिलता है। नारी = केवल नर जाति की स्त्री या मानवी। कान्ता = प्रिया, किसी पुरुष के स्नेह-सम्बन्ध से दिया हुआ स्त्री का नाम-जो चमक को भी प्रकट करता है। भार्या = वह स्त्री जो किसी पुरुष से अपने भरण-पोषण की प्राप्ति का अधिकार रखती हो। पत्नी = पति से सम्बद्ध धार्मिक कार्यों में पति का साथ देने वाली, जिसका धार्मिक रीति से विवाह सम्पन्न हुआ हो। वधू = विवाह के बाद नारी की संज्ञा। अंगना = अंगों की कोमलता तथा सुन्दरता की सूचक स्त्री की संज्ञा।
(15) आनन्द = समस्त मन, प्राण, शरीर, आत्मा के सम्मिलित सुखानुभूति को आनन्द कहते हैं, हर्षित, प्रसन्न। सुख = मन के अनुकूल किसी भी प्रतीति को सुख कहते हैं। हर्ष = सुख या आनन्द के कारण शरीर की रोमांचमयी अवस्था हर्ष है। उल्लास-उमंग = मन में सुख की वेगयुक्त अल्पकालिक क्रिया।
(16) भिज्ञ = विषय के ज्ञान से सम्पन्न। विज्ञ = विषय का विशेष ज्ञान रखने वाला। बहुज्ञ = अनेक विषयों का ज्ञाता।।
(17) अंश = मूर्त, अमूर्त तत्त्व का भाग। खण्ड = मूर्त पदार्थ का विभाजन।
(18) अन्न = मुख्य भोजन का धान्य विशेष। दाना = किसी भी वस्तु का गोलाई लिए छोटी इकाई का नाम, जैसे-माला का दाना, मोती का दाना, अनाज का दाना। शस्य = फसल; जो खेत में काटने हेतु लगी हो। खाद्य = सभी प्रकार का भोजन योग्य पदार्थ। पकवान = पका हुआ भोजन।

(19) आज्ञा = बड़ों द्वारा छोटों के प्रति काम की प्रेरणा। अनुज्ञा = श्रोता की इच्छा के समर्थन में स्वीकृतिसूचक कथन। अनुमति = बड़ों द्वारा सहमति अथवा किसी काम करने की इच्छा का समर्थन। आदेश = प्रायः कार्याधिकारी द्वारा दी गई आज्ञा।
(20) अनुनय = किसी बात पर सहमत होने की प्रार्थना। विनय = निवेदन, शिष्टता, अनुशासन। प्रार्थना = किसी कार्यसिद्धि या वस्तु प्राप्ति के लिए विनययुक्त कथन। निवेदन = अपनी बात विनयपूर्वक रखना। आवेदन = अपनी योग्यता आदि के कथन द्वारा किसी पद या कार्य हेतु प्रस्तुत होना।
(21) आमन्त्रण = किसी अवसर पर सम्मिलित होने का बुलावा। निमन्त्रण = उत्सव के अवसर पर बुलाना। आह्वान = ललकारना, संघर्ष के लिए बुलाना।
(22) इच्छा = किसी वस्तु के प्रति मन का राग। आशा = इच्छा के साथ प्राप्ति की सम्भावना का सुख। उत्कण्ठा = जिज्ञासा। स्पृहा = उत्कृष्ट इच्छा। मनोरथ = अभिलाषा।
(23) उत्साह = बड़े कार्यों के प्रति प्रेरित करने वाला आनन्ददायक भाव। साहस = सहने की शक्ति को साहस कहते हैं।
(24) तेज = बाहरी प्रकाश। कान्ति = चमक। प्रकाश = किरण समूह।
(25) ईर्ष्या = दूसरे की उन्नति न सह पाना। स्पर्धा = दूसरे को उन्नत देखकर स्वयं उन्नति की होड़। असूया = दूसरे के गुणों में दोष ढूँढ़ना। द्वे ष- दूसरे की बुराई की कामना। बैर = बहुत दिनों का संचित क्रोध का रूप।
(26) खेद = अनिष्ट सूचना। अवसाद= मन-शरीर की निश्चेष्टता। क्लेश = मन-शरीर दोनों का पीड़ित होना। कष्ट = दुःखजनक स्थिति। व्यथा = पीड़ा। वेदना = पीड़ा का मानसिक अनुभव। यातना = दुःख भोगने की मनोदशा। यन्त्रणा = दुःख की दीर्घकालिक जकड़न।
(27) ऋतु = छः ऋतुएँ होती हैं। मौसम = पारिस्थितिक भेद।

MP Board Class 11th Maths Important Questions Chapter 6 Linear Inequalities

Linear Inequalities Important Questions

Linear Inequalities Objective Type Questions

(A) Choose the correct option :

Question 1.
If x is a real number, then the solution of inequality 3x + 1 < 5x + 7 is :
(a) (- ∞, 3)
(b) (- 3,∞)
(c) (3, ∞)
(d) None of these.
Answer:
(b) (- 3,∞)

Question 2.
If x is a real number, then the solution of inequality \(\frac { 1 }{ 2 }\) (3x – 1) ≥ \(\frac { 1 }{ 3 }\)(4x + 3) – 1 is :
(a) (- ∞, 3)
(b) (3, ∞)
(c) (3, ∞)
(d) None of these.
Answer:
(c) (3, ∞)

Question 3.
The graph of x ≤ 2 and y ≥ 2 is situated in the :
(a) First and second quadrant
(b) Second and Third quadrant
(c) First and third quadrant
(d) None of these.
Answer:
(a) First and second quadrant

Question 4.
In linear programming problems, the objective function will be :
(a) One of the constraints
(b) A linear function which gives optimum solution
(c) Relation between varivable
(d) None of these.
Answer:
(b) A linear function which gives optimum solution

Question 5.
In linear constraints the maximum value of objective function will be :
(a) At the centre of feasible region
(b) At (0, 0)
(c) At one of the vertices of the feasible region
(d) At the vertex which is situated at maximum distance from (0, 0)
Answer:
(c) At one of the vertices of the feasible region

Question 6.
Which word is not used in linear programming problems :
(a) Slack variable
(b) Objective function
(c) Concave region
(d) Feasible solution.
Answer:
(a) Slack variable

Question 7.
The minimum value of p = 6x + 16y when constraints are x ≤ 40 and y ≥ 20 and x, y ≥ 0 is :
(a) 240
(b) 320
(c) 0
(d) None of these.
Answer:
(b) 320

Question 8.
At which point the value of 3x + 2y is maximum under the constraints x + y ≤ 2, x ≥ 0 y ≥ 0 :
(a) (0, 0)
(b) (1.5, 1.5)
(c) (2, 0)
(d) (0, 2).
Answer:
(c) (2, 0)

Question 9.
Under constraints x ≥ 0, y ≥ 0, x + y ≤ 4, maximum value of P = 3x + y is :
(a) 8
(b) 12
(c) 6
(d) 10.
Answer:
(b) 12

Question 10.
Under constraints x – 2y ≥ 6, x + 2y ≥ 0, x ≤ 6 , maximum value of P = x + 3y is :
(a) 16
(b) 17
(c) 18
(d) 19.
Answer:
(c) 18

Question 11.
Feasible solution of linear programming is in :
(a) Second quadrant only
(b) First and third quadrant
(c) First and second quadrant
(d) First quadrant only.
Answer:
(d) First quadrant only.

Question 12.
If x is a real number and |x| < 4, then :
(a) x ≥ 4
(b) – 4 < x < 4
(c) x ≤ – 4
(d) – 4 ≤ x ≤ 4
Answer:
(b) – 4 < x < 4

Question 13.
Solution of inequality 3x – 2 < 0 will be :
(a) [3, ∞]
(b) [- ∞, \(\frac { 2 }{ 3 }\)]
(c) [3, 2]
(d) [2, 3]
Answer:
(b) [- ∞, \(\frac { 2 }{ 3 }\)]

Welcome to our step-by-step math inequality solver.

Question 14.
The set of solutions of inequality 4x – 12 ≥ 0 is :
(a) (4, 2)
(b) [4, 12]
(c) [3, ∞]
(d) [3, ∞]
Answer:
(c) [3, ∞]

Question 15.
The solution of inequality |4x – 3| < 27 is :
(a) (- 6, \(\frac { 15 }{ 2 }\))
(b) [ – 6, \(\frac { 15 }{ 2 }\) ]
(c) [ – 6, \(\frac { 15 }{ 2 }\) )
(d) ( – 6, \(\frac { 15 }{ 2 }\) ]
Answer:
(a) (- 6, \(\frac { 15 }{ 2 }\))

Question 16.
Solution of inequality x > 2 is :

Answer:

Question 17.
Sawant has a space to store at most 13 boxes of apples and oranges. If he bought x boxes of oranges and y boxes of apples, then correct inequality will be :
(a) x + y < 13
(b) x + y > 13
(c) x + y < 13
(d) x + y = 13.
Answer:
(c) x + y < 13

Question 18.
In which point the maximum of objective function P = 10x + 6y in the following points:
(a) (0, 0)
(b) (8, 4)
(c) (10, 0)
(d) (0, 0).
Answer:
(b) (8, 4)

Question 19.
The graph of inequality x +y > 2 is :

Answer:

Question 20.
The solution of inequality 3x – 7 ≥ x + 1 is :
(a) [4, ∞]
(b) (4, ∞]
(c) (4, ∞)
(d) [4, ∞).
Answer:
(d) [4, ∞).

Question 21.
The solution of inequality 3x – 6 > 0; 2x – 6 > 0 is :
(a) [3, ∞]
(b) (3, ∞)
(c) [2, ∞)
(d) (2, ∞).
Answer:
(b) (3, ∞)

Question 22.
The solution of inequality 2x + 6 = 0; 4x – 8 < 0 is :
(a) [- 3, ∞)
(b) (3, 2)
(c) [- 3, 2)
(d)[- 3, 2].
Answer:
(c) [- 3, 2)

Question 23.
The set of solution of inequality x ≥ 0 is :
(a) All points of X – axis and first, fourth quadrants
(b) All points of first and fourth quadrants
(c) All points of first and second quadrants
(d) All points of Y – axis and first, fourth quadrants.
Answer:
(d) All points of Y – axis and first, fourth quadrants.

Question 24.
Variables of the objective function of linear Programming problem are :
(a) Negative
(b) Zero or Negative
(c) Zero
(d) Zero or Positive.
Answer:
(d) Zero or Positive.

(B) Match the following :

Answer:

1. (c)
2. (d)
3. (a)
4. (f)
5. (e)

(C) Fill in the blanks :

1. The maximum or minimum value of objective function is called ……………….
2. The graph of x ≥ 0 is situated in the ………………. quadrant.
3. The graph of y ≤ 0 is situated in the ………………. quadrant.
4. Solution of inequality 3x + 1 < 5x + 7 is (where x is an integer) ……………….
5. Solution of inequality 3x – 4 < 5 is (where x is an integer) ……………….
6. Solution of inequality 4x + 3 > – 13 is (where x is a real number) ……………….
7. Solution of inequality 7x – 2 < 5x + 4 is (where x is a real number) ……………….
8. Solution of inequality 20x < 90 is (where x is a natural number) ……………….
9. Solution of inequality 3x – 2 < x + 4 is ……………….
10. The inequality of one variable is ……………….
11. The set of solutions of y > 0 are in the ………………. and ………………. quadrant.
12. x ≥ 2 and y ≥ 2 will be situated in the graph ……………….
13. The domain of inequality 3x – 15 ≤ 0 is ……………….
14. A function whose maximum and minimum value is to be found subject to the given constrains is known as ……………….

Answer:

1. Optimum value
2. First and fourth
3. Third and fourth
4. (- 3, ∞)
5. {… – 3, – 2, -1, 0, 1, 2}
6. (- 4, ∞)
7. (- ∞, 3)
8. {1, 2, 3, 4}
9. (- ∞, 3)
10. ax + b ≥ c or ax + b ≤ c; where a ≠ 0
11. First and second
12. First and second quadrant
13. (- ∞, \(\frac { 15 }{ 3 }\))
14. Objective function

(D) Write true / false :

1. If the variable x is such that its value lies between two fixed point a and b, then {x : a < x < b} is called a closed interval.
2. The function whose maximum or minimum value is to be found is called objective function.
3. The set of values of the variable satisfying all constraint is called feasible solution of the problem.
4. The process of doing certain specified steps in a given order is called programming.
5. The set {x : a < x < b} which consists of both a and b is called open interval.
6. The solution of inequality 6x – 30 ≥ 0 will be x ≤ 5
7. The solution of inequality – 2x + 7 < – 13 will be (10, ∞).
8. The linear inequality of one variable is ax + b = 0; a ≠ 0, ∀a, b ∈
9. R. The line inequality of two variables is ax + by + c = 0; a ≠ 0, b ≠ 0, ∀a, b, c ∈ R.
10. The graph of inequality x > 0 is :
    
11. The graph of inequality y ≤ 0 is :
    
12. The solution of inequality 5x – 30 ≤ 0 is :
    

Answer:

1. False
2. True
3. True
4. True
5. False
6. False
7. True
8. True
9. True
10. True
11. True
12. False.

(E) Write answer in one word / sentence :

1. In which quadrant lies the solution of x ≥ 0 and y ≥ 0?
2. In which quadrant lies the solution of x ≤ 2 and y ≥ 2?
3. Write the solution of 3(2 – x) ≥ 2(1 – x) .
4. Write the solution of \(\frac { x }{ 3 }\) > \(\frac { x }{ 2 }\) + 1.
5. Write the solution of \(\frac { x – 4 }{ x + 2 }\) ≤ 2.

Answer:

1. First quadrant
2. First and second quadrant
3. (- ∞, 4]
4. (- ∞, – 6)
5. (- ∞, – 8]∪(- 2, ∞)

Linear Inequalities Long Answer Type Questions

Question 1.
Draw the graph of inequality 2x + y ≥ 6 in two – dimensional plane. (NCERT)
Solution:
Given inequality is 2 x + y ≥ 6
Consider this as an equation 2x + y = 6.
For this equation following values of x and y are given in the table :

Plot the graph using the above table. Take a point (0, 0) and put in given inequality,
2x + y ≥ 6
⇒ 2 x 0 + 0 ≥ 6
⇒ 0 ≥ 6 Which is false.
Hence, shaded portion will be opposite of origin.

∴ The shaded region represents the inequality 2x + y ≥ 6.

Question 2.
Draw the graph of inequality 3x + 4y ≤ 12 in two – dimensional plane. (NCERT)
Solution:
Given inequality : 3x + 4y = 12
Consider this as an equation 3x + 4y = 12.
For this equation the following values of x and y are given in the table :

Plot the above points in a graph paper and join them.
Put (0, 0) in given inequality,
⇒ 3x + 4y ≤ 12
⇒ 3 x 0 + 4 x 0 ≤ 12
⇒ 0 ≤ 12
Which is true.
Hence, shaded portion will be towards origin as shown in figure.

∴ The shaded region represents the inequality 3x + 4y ≤ 12

Question 3.
Draw the graph of inequality y + 8 ≥ 2x in two – dimensional plane. (NCERT)
Solution:
Given inequality : y+8 > 2x Consider this as an equation 2x – y = 8
For this equation following values of x and y are given in the table :

Plot the above points in graph paper and join them.
Put (0, 0) in the given inequality,
y + 8 ≥ 2x
⇒ 0 + 8 ≥ 2 x 0
⇒ 8 ≥ 0 Which is true.
Hence, the shaded portion will be towards origin.

∴ Shaded portion represents the inequality y + 8 ≥ 2x.

Draw the graph of following inequality in two – dimensional plane :

Question 4.
2x – 3y > 6.
Solution:
Given inequality is :
2x – 3y > 6
Consider this as an equation 2x – 3y = 6
Following table is prepared for different values of x and y :

Plot the above points on xy – plane and join them.
Put x = 0, y = 0 in given inequality,
2x – 3y > 6
⇒ 2 x 0 – 3 x 0 > 6
⇒ 0 > 6 Which is false.
Hence, the shaded portion will be opposite of origin.

∴The shaded region represents the inequality 2x – 3y > 6

Question 5.
– 3x + 2y ≥ – 6.
Solution:
Given inequality is : – 3x + 2y ≥ – 6.
Consider this as an equation – 3x + 2y = – 6.
Following table is prepared for different values of x and y :

Plotting the above points on xy – plane and join them. Put x = 0, y = 0 in given inequality,
– 3x + 2y ≥ – 6
⇒ – 3(0) + 2(0) ≥ – 6
⇒ 0 ≥ – 6 Which is true.
∴ Shaded portion will be towards origin.

Hence, the shaded portion represents the inequality – 3x + 2y ≥ – 6.

Question 6.
3y – 5x < 30.
Solution:
Given inequality: y – 5x < 30.
Consider this as an equation 3y – 5x = 30.
Following table is prepared for different values of x and y :

Ploting the above points on xy – plane and join them.
Put x = 0, y = 0 in given inequality,
3y – 5x < 30
⇒ 3(0) – 5(0) < 30
⇒ 0 < 30
Which is true.
Hence, the shaded region will be towards origin.

∴ Shaded region represents the inequality 3y – 5x < 30

Question 7.
2x + y ≥ 6, 3x + 4y ≤ 12. (NCERT)
Solution:
Given inequalities are :
2x + y ≥ 6
and 3x + 4y ≤ 12
For the equation 2x+y = 6.
Following table is prepared for different values of x and y :

Plotting the above points on xy – plane to get straight line.
Put x = 0, y = 0 in the inequality,
2x + y ≥ 6
⇒ 2 x 0 + 0 ≥ 6
⇒ 0 ≥ 6
Which is false.
∴ The shaded region will be opposite of origin.
For the equation 3x + 4y = 12.
Following table is prepared for different values of x and y :

Plotting the above points on xy – plane to get straight line.
Put x = 0, y = 0 in given inequality,
3x + 4y ≤ 12
⇒ 3 x (0) + 4(0) ≤ 12
⇒ 0 ≤ 12
Which is true.
Hence, the shaded portion will be towards origin.

∴ The common shaded region is the required solution of the given inequalities.

Question 8.
Solve the following inequalities graphically in two – dimensional plane : x + y ≥ 4, 2x – y >0. (NCERT)
Solution:
Given inequalities are :
x + y ≥ 4
and 2x – y > 0
For the equation x + y = 4, following table is prepared for different values of x and y:

Plotting the above points on xy – plane to get straight line.
Put x = 0, y = 0 in the inequality,
x + y ≥ 4
⇒ 0 + 0 ≥ 4
⇒ 0 ≥ 4
Which is false.
Hence, the shaded region will be opposite of origin.
Preparing table for different value of x and y of the equation 2x – y = 0 :

Plotting the above points in xy – plane to get straight line.
Put x = 1 and y = 0 in given inequality,
2x – y > 0
⇒ 2(1) – 0 > 0
⇒ 2 > 0
Which is true.
Hence, the shaded region will be towards of point (1,0).

The common shaded region is the required solution of the given inequalities.

Question 9.
Solve the following inequalities graphically in two – dimensional plane :
5x + 4y ≤ 20, x ≥ 1, y ≥ 2 (NCERT)
Solution:
The given inequalities are :
5x + 4y ≤ 20
x ≥ 1
y ≥ 2
Prepared table for different values of x and y of the equation 5x +4y = 20 :

Plotting the points of the above table on xy – plane to get straight line.
Then, put x = 0, y = 0 in the inequality,
5x + 4y ≤ 20
⇒ 0 + 0 ≤ 20
⇒ 0 < 20
Which is true.
∴Shaded region will be towards origin.
For equation x + O.y = 1 to get different values of x and y are in the table :

Plotting the points from table on xy plane to get straight line.
Putting x = 0, y = 0 in the inequality,
x ≥ 1
⇒ 0 ≥ 1
Which is false.
Hence, the shaded region will be opposite side of origin.
For equation 0.x + y = 2, the different values of x and y are given in the table :

Plotting the points on xy plane to get straight line.
Then, put x = 0, y = 0 (0,0) in the inequality,
y ≥ 2
⇒ 0 ≥ 2
Which is false.
Hence, the shaded region will be opposite side of origin.

∴ The common shaded region is the required solution of the given inequalities.

Question 10.
Solve the following inequalities graphically :
3x + 4y ≤ 60, x + 3y ≤ 30, x ≥ 0, y ≥ 0. (NCERT)
Solution:
The given inequalities are :
3x + 4y ≤ 60
x + 3y ≤ 30
x ≥ 0
y ≥ 0
Prepared table for values of x and y of equation 3x + 4y = 60 :

Plotting the above points on xy – plane to get straight line.
Then, put x = 0, y = 0 in the inequality,
3x + 4y ≤ 60
⇒ 0 + 0 ≤ 60
⇒ 0 ≤ 60
Which is true.
Hence, the shaded portion will be towards origin.
Prepared table for values of x and y of equation x + 3y = 30:

Plotting the above points on xy – plane to get straight line.
Then, put x = 0, y = 0 in the inequality,
x + 3y ≤ 30
⇒ 0 + 0 ≤ 30
⇒ 0 ≤ 30
Which is true.
Hence, the shaded region will be towards origin.
The graph of x = 0 is y – axis and graph of y = 0 is x – axis.

The common shaded region is the required solution set of the given inequalities.

MP Board Class 11th Maths Important Questions

MP Board Class 11th Maths Important Questions Chapter 15 Statistics

Statistics Important Questions

Statistics Long Answer Type Questions

Question 1.
Find the mean deviation about the mean for the data in following table: (NCERT)

Solution:

Mean \(\bar { x }\) = \(\frac{\sum f_{i} x_{i}}{\sum f_{i}}\) = \(\frac { 4000 }{ 80 }\) = 50
Mean deviation about mean = \(\frac{\sum f_{i}\left|x_{i}-\bar{x}\right|}{\sum f_{i}}\)
= \(\frac { 1280 }{ 80 }\) = 16.

You can use this Standard Deviation Calculator to calculate the standard deviation, variance, mean, and the coefficient of variance for a given set .

Question 2.
Find the mean deviation about the median for the data in following table: (NCERT)

Solution:

\(\frac { N }{ 2 }\) = \(\frac { 26 }{ 2 }\) = 13
Which lies on cumulative frequency (C.F.) 14.
∴ Median = M_d = 7.
Mean deviation about mean = \(\frac{\sum f_{i}\left|x_{i}-M_{d}\right|}{\sum f_{i}}\)
= \(\frac { 1128.8 }{ 100 }\)

Question 3.
Find the mean deviation about the mean for the data in following table: (NCERT)

Solution:
Let A = 130

Mean \(\bar { x }\) = A + \(\frac { Σfd }{ Σf }\) × i = 130 + \(\frac { (- 47) }{ 10 }\) = 10
\(\bar { x }\) = 130 – 4.7 = 125.3
Mean deviation about mean = \(\frac{\sum f_{i}\left|x_{i}-\bar{x}\right|}{\sum f_{i}}\)
= \(\frac { 1128.8 }{ 100 }\) = 11.228.

Question 4.
Find the mean deviation about the median for the data in following table: (NCERT)

Solution:

\(\frac { N }{ 2 }\) = \(\frac { 50 }{ 2 }\) = 25
20 – 30 is median class.
Here F = 14, f = 14, l = 20, i = 10

Question 5.
Calculate the mean deviation about median for the age distribution of 100 persons given below :

Solution:
First make the class interval uniform:

\(\frac { N }{ 2 }\) = \(\frac { 100 }{ 2 }\) = 50
Median class = 35.5 – 40.5
where l = 35.5, F = 37, f = 26, i = 5

Question 6.
Find the mean and variance for the following frequency distribution in following table:
(A)

Solution:

Let assumed mean A = 105

(B)

Solution:

Let assumed mean A = 25

Question 7.
Find the mean and variance and standard deviation using short cut method:

Solution:

Let assumed mean A = 92.5

Standard deviation σ = \(\sqrt {variance}\)
= \(\sqrt {105.58}\) = 10.27

Question 8.
The diameter of circle (in mm) drawn in design are given below :

Calculate the standard deviation and mean diameter of the circles. (NCERT)
Solution:
First make the class interval uniform :

Let assume mean A = 42.5

Question 9.
From the prices of shares X and Y below, find out which is more stable in value:

Solution:
For first share X : Let assumed mean A = 58, n = 10

Foe second share Y : Let assumed mean A = 107, n = 10

Coefficient of variation for X :
Mean M = A + \(\frac { Σd }{ n }\)
= 58 + \(\frac { ( – 70) }{ 10 }\) = 58 – 7
= 51
∴ Coefficient of variance = \(\frac { σ }{ M }\) x 100
= \(\frac { 6 }{ 51 }\) x 100 = 11.8
Coefficient of variance of Y :
Mean M = A + \(\frac { Σd }{ n }\)
= 107 + \(\frac { – 20 }{ 10 }\) = 107 – 2
= 105
∴ Coefficient of variation = \(\frac { σ }{ M }\) x 100
= \(\frac { 2 }{ 105 }\) x 100 = 1.9
Coefficient of variance of Y is less than the coefficient of variance of X.
Share Y is more stable than share X.

Question 10.
An analysis of monthly wages paid to the workers in two firms A and B belonging to the same industry. Give the following results: (NCERT)

1. Which firm A or B pays out larger amount as monthly wages.
2. Which firm A or B shows greater variability in individual wages.

Solution:
For firm A :
Number of wages earners = 586
Mean of monthly wages x = Rs. 5,253
Amount paid by firm A = 586 x 5,253 = Rs. 30, 78, 258
Variance of distribution of wages = 100
Standard deviation σ = \(\sqrt {100}\) = 10
Coefficient of variation = \(\frac{\sigma}{\bar{x}}\) x 100 = \(\frac { 10 }{ 5253 }\) = 0.19

For firm B :
Number of wages earners = 648
Mean \(\bar { x }\) = Rs. 5253
Amount paid by firm B = 648 x 5253 = Rs. 34, 03, 944
Coefficient of variation = 121
Standard deviation σ = \(\sqrt {121}\) = 11
Coefficient of variation = \(\frac{\sigma}{\bar{x}}\) x 100 = \(\frac { 11 }{ 5253 }\) x 100 = 0.21
Thus, firm B pays out larger amount as monthly wages.
∵ C.V. of firm B > C.V. of firm A.
∴ Firm B shows greater variability in individual wages.

Question 11.
The following is the record of goals scored by team A in a football session:

For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent? (NCERT)
Solution:

For team B :

C.V of team A < C.V of team B. Hence team A is more consistent.

Question 12.
The mean and standard deviation of marks obtained by 50 students of three subjects Mathematics, Physics and Chemistry are given below :

Which of these three subjects shows the highest variability in marks and which shows lowest? (NCERT)
Solution:
C.V. of Mathematics (C.V.) = \(\frac{\sigma}{\bar{x}}\) x 100
Where σ = 12, \(\bar { x }\) = 42
∴ C.V. of Mathematics = \(\frac { 12 }{ 42 }\) x 100 = 28.57
C.V. of Physics = \(\frac{\sigma}{\bar{x}}\) x 100
Where σ = 15, \(\bar { x }\) = 32
∴ C.V. of Physics = \(\frac { 15 }{ 32 }\) x 100 = 46.87
C.V. of Chemistry = \(\frac{\sigma}{\bar{x}}\) x 100
Where σ = 20, \(\bar { x }\) = 40.9
∴ C.V. of Chemistry = \(\frac { 20 }{ 40.9 }\) x 100
= 48.89
∵ C.V. of Chemistry > C.V. of Physics > C.V. of Mathematics.
∴ Chemistry shows the highest variability and Mathematics shows the least variability.

Question 13.
The mean and standard deviation of a group of 100 observations were found to be 20 and 3 respectively. Later on it was found that three observations were incorrect Which are recorded by 21,21, and 18. Find the mean and standard deviation, if the incorrect observations are omitted. (NCERT)
Solution:
Given: n = 100, \(\bar { x }\) = 20, σ = 3
\(\bar { x }\) = \(\frac{\sum x}{n}\) ⇒ 20 = \(\frac{\sum x}{n}\)
Σx = 20 x 100 = 2000
If incorrect observations 21,21 and 18 are omitted, then correct sum
Σx = 2000 – 21 – 21 – 18 = 2000 – 60 = 1940
Now correct mean of remaining 97 observations are
\(\bar { x }\) = \(\frac{\sum x}{n}\) = \(\frac { 1940 }{ 97 }\) = 20
Given : σ = 3

correct Σx² = 40900 – (21)² – (21)² – (18)²
= 40900 – 441 – 441 – 324 = 39694
Now correct S.D. of remaining 97 observations are

Question 14.
The mean and variance of eight observations are 9 and 9.25 respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, then And the remaining two observations. (NCERT)
Solution:
Let the two remaining observations are x₁ and x₂
Mean M = \(\frac{6 + 7 + 10 + 12 + 12 + 13 + x_{1} + x_{2}}{8}\)
9 = \(\frac{60 + x_{1} + x_{2}}{8}\)
⇒ x₁ + x₂ + 60 = 72
⇒ x₁ + x₂ = 12 …. (1)

⇒ x²₁ + x²₂ = 722 – 642
⇒ x²₁ + x²₂ = 80 …. (2)
From eqn. (1),
x₂ = 12 – x₁
Put the value of x₂ in equation (2),
x²₁ + (12 – x₁)² = 80
⇒ x² + 144 + x²₁ – 24x₁ = 80
⇒ 2x²₁ – 24x₁ + 64 = 0
⇒ x²₁ – 12x₁ + 32 = 0
⇒ x²₁ – 4x₁ – 8x₁ + 32 = 0
⇒ x₁(x¹ – 4) – 8(x₁ – 4) = 0
⇒ (x₁ – 8)(x₁ – 4) = 0
⇒ x₁ = 4, 8
When x₁ = 4, then x₂ = 12 – 4 = 8
When x₁ = 8, then x₂ = 12 – 8 = 4
Hence remaining observations are 4 and 8.

MP Board Class 11th Maths Important Questions

In this article, we will share MP Board Class 11th Biology Solutions Chapter 10 कोशिका चक्र और कोशिका विभाजन Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 11th Biology Solutions Chapter 10 कोशिका चक्र और कोशिका विभाजन

कोशिका चक्र और कोशिका विभाजन NCERT प्रश्नोत्तर

प्रश्न 1.
स्तनधारियों की कोशिकाओं की औसत कोशिका-चक्र अवधि कितनी होती है ?
उत्तर:
स्तनधारियों की कोशिकाओं की औसत कोशिका-चक्र की अवधि 24-25 घण्टे होती है।

प्रश्न 2.
जीवद्रव्य विभाजन व केन्द्रक विभाजन में क्या अंतर है ?
उत्तर:
जीवद्रव्य विभाजन (Cytokinesis) में कोशिका-द्रव्य का विभाजन होता है जबकि केन्द्रक विभाजन (Karyokinesis) में कोशिका के केन्द्रक का विभाजन होता है।

प्रश्न 3.
अंतरावस्था में होने वाली घटनाओं का वर्णन कीजिए।
उत्तर:
अंतरावस्था(इण्टरफेज) दो विभाजनों के बीच की अवस्था है, जिसमें निम्नलिखित तीन अवस्थाएँ पायी जाती हैं –
(i)G₁फेज-इस अवस्था में प्रोटीन एवं RNA का संश्लेषण किया जाता है।
(ii) S फेज-इस अवस्था में DNA एवं हिस्टोन प्रोटीन का संश्लेषण होता है।
(ii) G₂फेज-इस अवस्था में आवश्यक प्रोटीन तथा RNA का संश्लेषण किया जाता है तथा विभिन्न कोशिकांगों का निर्माण होता है।

प्रश्न 4.
कोशिका-चक्र की G₀(प्रशांत अवस्था) क्या है ?
उत्तर:
वे कोशिकाएँ जो आगे विभाजित नहीं होती हैं तथा निष्क्रिय अवस्था में पहुँचती हैं, जिसे कोशिका, चक्र की प्रशांत अवस्था (G₀) कहा जाता है। इस अवस्था की कोशिका उपापचयी रूप से सक्रिय होती है, लेकिन विभाजित नहीं होती, इनमें विभाजन जीव की आवश्यकता के अनुसार होता है।

प्रश्न 5.
सूत्री विभाजन को सम-विभाजन क्यों कहते हैं ?
उत्तर:
सूत्री विभाजन को सम-विभाजन कहा जाता है, क्योंकि विभाजन के अंत में दो ऐसी कोशिकाएँ बनती हैं जिसमें गुणसूत्रों की संख्या जनक कोशिका के बराबर होती है।
इस विभाजन में जनक कोशिका के आनुवंशिक गुण पुत्री कोशिका में पहुँचते हैं । अतः इस विभाजन से आनुवंशिक समानता बनी रहती है।

प्रश्न 6.
कोशिका-चक्र की उस अवस्था का नाम बताइए, जिसमें निम्न घटनाएँ सम्पन्न होती हैं –

1. गुणसूत्र तर्कु मध्य रेखा की ओर गति करते हैं।
2. गुणसूत्र बिंदु का टूटना व अर्द्धगुणसूत्र का पृथक् होना।
3. समजात गुणसूत्रों का आपस में युग्मन होना।
4. समजात गुणसूत्रों के बीच विनिमय का होना।

उत्तर:

1. मध्यावस्था (Metaphase)
2. पश्चावस्था (Anaphase)
3. अर्धसूत्री विभाजन-I (Meiosis-I) की युग्मपट्ट (Zygotene) अवस्था
4. पैकीटीन (Pachytene) अवस्था।

प्रश्न 7.
निम्न के बारे में वर्णन कीजिए –

1. सूत्रयुग्मन
2. युगली
3. काएज्मेटा।

उत्तर:
1. सूत्रयुग्मन (Synapsis):
कोशिका विभाजन की जायगोटीन अवस्था में समजात गुणसूत्रों के जोड़ा बनाने की प्रक्रिया को सूत्रयुग्मन या अन्तर्ग्रथन या सिनैप्सिस कहते हैं। सिनैप्सिस बनाने वाले गुणसूत्र अलग – अलग जनकों के होते हैं।

2. युगली (Bivalent):
जाइगोटीन अवस्था में समजात गुणसूत्रों द्वारा युग्मन करके सिनैप्सिस का निर्माण करने वाले गुणसूत्रों को बाइवैलेण्ट या डायड कहते हैं, क्योंकि उस समय गुणसूत्र दो की संख्या में दिखाई देते हैं।

3. काएज्मेटा (Chaismata):
अर्द्धसूत्री विभाजन के प्रोफेज-I के पैकीटीन अवस्था में समजात गुणसूत्रों के युग्मन के समय अर्द्ध गुणसूत्र जिस स्थान पर एक-दूसरे को स्पर्श कर जीन विनिमय करते हैं, वह स्थान डिप्लोटीन अवस्था में गुणसूत्रों के विलगाव (Terminalization) के क्रम में ‘X’ आकृतिकी संरचनाओं के रूप में परिलक्षित होता है, जिसे किएज्मेटा (Chiasmata) कहते हैं।

प्रश्न 8.
पादप व प्राणी कोशिका के कोशिकाद्रव्य विभाजन में क्या अंतर है ?
उत्तर:
कोशिका द्रव्य विभाजन-कोशिका विभाजन के समय केन्द्रक विभाजन के बाद कोशिकाद्रव्य विभाजन को साइटोकाइनेसिस या कोशिकाद्रव्य विभाजन कहते हैं, यह पादप एवं प्राणियों में दो अलग-अलग विधियों द्वारा होता है –

(a) कोशिका खाँच द्वारा इस कोशिकाद्रव्य विभाजन में कोशिका के मध्य में एक खाँच बनती है, जो बढ़कर कोशिका के कोशिकाद्रव्य को दो भागों में बाँट देती है। प्रत्येक भाग में एक केन्द्रक होता है, जन्तुओं में इसी प्रकार का विभाजन पाया जाता है।

(b) कोशिका प्लेट द्वारा इस कोशिकाद्रव्य विभाजन में कोशिका के मध्य गॉल्गीकाय तथा E.R. एकत्रित होकर एक पट बना देते हैं जो बाद में कोशिका भित्ति बन जाती है और कोशिका को दो भागों में बाँट देती हैं। प्रत्येक भाग में एक केन्द्रक होता है। यह विभाजन पादप कोशिकाओं में पाया जाता है।

प्रश्न 9.
अर्द्धसूत्री विभाजन के बाद बनने वाली चार संतति कोशिकाएँ कहाँ आकार में समान व कहाँ भिन्न आकार की होती हैं ?
उत्तर:
अर्द्धसूत्री विभाजन (Meiosis) के बाद बनने वाली चार संतति कोशिकाएँ (daughter cells) अर्द्धसूत्री विभाजन II (Meiosis-II) के अंत्यावस्था II (Telophase-II) के अंत में आकार में कहीं पर समान और कहीं पर असमान होती हैं।

प्रश्न 10.
सूत्री विभाजन की पश्चावस्था तथा अर्द्धसूत्री विभाजन की पश्चावस्था-I में क्या अंतर है ?
उत्तर:
सूत्री विभाजन एवं अर्द्धसूत्री विभाजन की पश्चावस्था-I में अंतर –

1. सूत्री विभाजन (Mitosis division):
इसमें एक गुणसूत्र का एक अर्द्ध-गुणसूत्र। एक ध्रुव की ओर तथा दूसरा, दूसरे ध्रुव की ओर चला जाता है।

2. अर्द्धसूत्री विभाजन (Meiosis division):
इसमें एनाफेज-I में पूरा गुणसूत्र ध्रुवों की ओर जाता है, जबकि ऐनाफेज-II में गति समसूत्री ऐनाफेज के समान होती हैं।

प्रश्न 11.
सूत्री एवं अर्द्धसूत्री विभाजन में प्रमुख अंतरों को सूचीबद्ध कीजिए।
उत्तर:
दोनों ही कोशिका विभाजनों में कोशिकाओं की संख्या बढ़ती है तथा दोनों में गुणसूत्रों का विभाजन होता है। इन समानताओं के बावजूद दोनों विभाजनों में निम्नलिखित अन्तर पाये जाते हैं –

प्रश्न 12.
अर्द्धसूत्री विभाजन का क्या महत्व है?
उत्तर:
अर्द्धसूत्री विभाजन का महत्व –

1. इसके कारण पीढ़ी-दर-पीढ़ी कोशिकाओं में गुणसूत्रों की संख्या एक समान बनी रहती है।
2. इस विभाजन के कारण जनक के समान ही कोशिकाएँ पैदा होती हैं।
3. इस विभाजन में जीन विनिमय होने के कारण यह नये गुणों के बनने में सहायता करता है।
4. इसके कारण विभिन्नता पैदा होती है, जो जैव विकास के लिए आवश्यक है।
5. इसके कारण एक द्विगुणित कोशिका से चार अगुणित कोशिकाएँ बनती हैं।

प्रश्न 13.
अपने शिक्षक के साथ निम्न के बारे में चर्चा कीजिए –

1. अगुणित कीटों व निम्न श्रेणी के पादपों में कोशिका विभाजन कहाँ संपन्न होती है ?
2. उच्च श्रेणी के पादपों की कुछ अगुणित कोशिकाओं में कोशिका विभाजन कहाँ नहीं होता है?

उत्तर:

1. निम्न श्रेणी के पादपों (क्लैमाइडोमोनास, स्पाइरोगायरा) में अगुणित बीज (Spores) युग्मकोद्भिद (Gametophyte) के निर्माण की प्रक्रिया में समसूत्री (Mitotic) कोशिका विभाजन पाया जाता है। जबकि अर्द्धसूत्री (Meiotic) विभाजन, इनके युग्मनज (Zygote) में अगुणित बीजाणु (Spores) के निर्माण के समय होता है।

2. उच्च श्रेणी के पादपों (Angiospores) में बीजाण्ड (Ovule) के भ्रूणपोष (Embryo sac) में पाये जाने वाले सखि (Synergids) एवं प्रतिध्रुवीय कोशिकाओं (Antipodal cell) में कोई भी कोशिका विभाजन नहीं पाया जाता है। अंत में ये कोशिकाएँ नष्ट हो जाती हैं।

प्रश्न 14.
क्या S – अवस्था में बिना डी. एन. ए. प्रतिकृति के सूत्री विभाजन हो सकता है ?
उत्तर:
नहीं, S प्रावस्था में बिना डी. एन. ए. प्रतिकृति (DNA replication) के सूत्री विभाजन संभव नहीं है। सूत्री विभाजन के लिए डी. एन. ए. की मात्रा का दुगुना होना आवश्यक है।

प्रश्न 15.
क्या बिना कोशिका विभाजन के डी. एन. ए. प्रतिकृति हो सकती है?
उत्तर:
नहीं, क्योंकि कोशिका विभाजन के दौरान ही डी. एन. ए. प्रतिकृति व कोशिका वृद्धि होती है।

प्रश्न 16.
कोशिका विभाजन की प्रत्येक अवस्थाओं के दौरान होने वाली घटनाओं का विश्लेषण कीजिए और ध्यान दीजिए कि निम्नलिखित दो प्राचलों में कैसे परिवर्तन होता है –

1. प्रत्येक कोशिका की गुणसूत्र संख्या (N)
2. प्रत्येक कोशिका में DNA की मात्रा (C)।

उत्तर:

1. किसी भी जीव में कोशिका विभाजन की पूर्वावस्था (Prophase), मध्यावस्था (Metaphase) एवं पश्चावस्था (Anaphase) में गुणसूत्र की संख्या (N) दुगुनी हो जाती है। अंत्यावस्था (Telophase) में पुत्री कोशिकाओं (Daughter cell) के निर्माण के समय गुणसूत्रों की संख्या (N) आधी हो जाती है।

2. कोशिका विभाजन के दौरान विभिन्न अवस्थाओं में प्रत्येक कोशिका में DNA की मात्रा (C) में परिवर्तन होता है। कोशिका की पूर्वावस्था, मध्यावस्था एवं पश्चावस्था में DNA की मात्रा (C) दुगुनी होती है, लेकिन अंत्यावस्था में पुत्री कोशिका के निर्माण के समय इनकी मात्रा आधी हो जाती है।

कोशिका चक्र और कोशिका विभाजन अन्य महत्वपूर्ण प्रश्नोत्तर

कोशिका चक्र और कोशिका विभाजन वस्तुनिष्ठ प्रश्न

प्रश्न 1.
सही विकल्प का चयन कीजिए –
1. किएज्मेटा निर्मित होते हैं –
(a) डिप्लोटीन अवस्था में
(b) लेप्टोटीन अवस्था में
(c) पैकीटीन अवस्था में
(d) डाइकाइनेसिस अवस्था में।
उत्तर:
(c) पैकीटीन अवस्था में

2. सूत्री विभाजन में गुणसूत्रों का द्विगुणन होता है –
(a) प्रारम्भिक पूर्वावस्था में
(b) पश्च पूर्वावस्था में
(c) विभाजनान्तराल अवस्था में
(d) पश्च अन्त्यावस्था में।
उत्तर:
(b) पश्च पूर्वावस्था में

3. अर्द्धसूत्री विभाजन की प्रथम मध्यावस्था में सेण्ट्रोमियर –
(a) विभाजित होते हैं
(b) विभाजित नहीं होते
(c) विभाजित होकर पृथक् नहीं होते
(d) समान नहीं होते।
उत्तर:
(c) विभाजित होकर पृथक् नहीं होते

4. गुणसूत्र का वह भाग जहाँ पर गुणसूत्र विनिमय होता है उसे कहते हैं –
(a) क्रोमोमियर्स
(b) बाइवैलेन्ट
(c) किएज्मेटा
(d) सेण्ट्रोमियर।
उत्तर:
(c) किएज्मेटा

5. गुणसूत्रों की संख्या कब आधी हो जाती है –
(a) प्रोफेज-I
(b) ऐनाफेज-I
(c) मेटाफेज-I
(d) मेटाफेज-II.
उत्तर:
(b) ऐनाफेज-I

6. अर्द्धसूत्री विभाजन किसमें होता है –
(a) परागकण
(b) परागनलिका
(c) पराग मातृ कोशिका
(d) जनन कोशिका।
उत्तर:
(c) पराग मातृ कोशिका

7. तर्क तन्तु किसके बने होते हैं –
(a) प्रोटीन
(b) लिपिड
(c) सेल्यूलोज
(d) पेक्टिन।
उत्तर:
(a) प्रोटीन

8. युग्मन के समय गुणसूत्रों के मध्य युग्मन होता है –
(a) समान गुणसूत्रों के मध्य
(b) समजात गुणसूत्रों के मध्य
(c) विषमजात गुणसूत्रों में
(d) इनमें से कोई नहीं।
उत्तर:
(b) समजात गुणसूत्रों के मध्य

9. क्रोमोसोम्स का द्विगुणन किस अवस्था में होता है –
(a) S अवस्था
(b) G अवस्था
(c) G₂अवस्था
(d) M अवस्था।
उत्तर:
(a) S अवस्था

10. किस प्रकार के कोशिका विभाजन में गुणसूत्रों की संख्या में कमी होती है –
(a) मियोसिस
(b) माइटोसिस
(c) द्विविभाजन
(d) विदलन।
उत्तर:
(a) मियोसिस

11. माइटोसिस के समय कोशिका का कौन-सा अंगक लुप्त हो जाता है –
(a) प्लास्टिड
(b) प्लाज्मा झिल्ली
(c) केन्द्रिका
(d) इनमें से कोई नहीं।
उत्तर:
(c) केन्द्रिका

12. समजात गुणसूत्रों की एक जोड़ी में चार क्रोमैटिड किस अवस्था में पाई जाती है –
(a) डिप्लोटिन
(b) पैकीटिन
(c) जाइगोटीन
(d) डायकाइनेसिस।
उत्तर:
(b) पैकीटिन

13. किएज्मेटा किसके स्थान को दर्शाते हैं –
(a) सिनैप्सिस
(b) डिस्जंक्शन
(c) क्रॉसिंग ओवर
(d) टर्मिनेलाइजेशन।
उत्तर:
(c) क्रॉसिंग ओवर

14. लैंगिक जनन में कोशिका विभाजन किस प्रकार का होता है –
(a) एमाइटोटिक
(b) माइटोटिक
(c) मियोटिक
(d) मियोटिक एवं माइटोटिक।
उत्तर:
(c) मियोटिक

15. कोशिका विभाजन को रोकने वाला रसायन किस पौधे से प्राप्त किया जाता है –
(a) क्राइसेन्थेमम
(b) कॉल्चिकम
(c) डल्बर्जिया
(d) क्रोकस।
उत्तर:
(b) कॉल्चिकम

16. स्पिण्डल तंतुओं का निर्माण किससे होता है –
(a) सेन्ट्रीओल
(b) केन्द्रक
(c) माइटोकॉन्ड्रिया
(d) सेन्ट्रोमियर।
उत्तर:
(d) सेन्ट्रोमियर।

17. कोशिका विभेदन होने तक कोशिका चक्र की कौन-सी अवस्था आ चुकी होती है –
(a)G₀प्रावस्था
(b)G₁ प्रावस्था
(c)G₂प्रावस्था
(d) S-प्रावस्था।
उत्तर:
(a)G₀प्रावस्था

18. कोशिका विभाजन का प्रेरण किसके द्वारा होता है –
(a) साइटोकाइनिन
(b) ऑक्सिन
(c) जिबरेलिन
(d) ए.बी.ए.।
उत्तर:
(a) साइटोकाइनिन

19. कोशिका विभाजन में कोशिका पट्टिका का निर्माण किस अवस्था में होता है –
(a) ऐनाफेज
(b) मेटाफेज
(c) टीलोफेज
(d) साइटोकाइनेसिस।
उत्तर:
(c) टीलोफेज

20. माइटोसिस में सेन्ट्रोमियर का विभाजन किस अवस्था में होता है –
(a) प्रोफेज
(b) मेटाफेज
(c) ऐनाफेज
(d) टीलोफेज।
उत्तर:
(c) ऐनाफेज

21. गुणसूत्रों के किस भाग से तर्कु तन्तु जुड़े रहते हैं –
(a) सेन्ट्रोमियर
(b) क्रोमोमियर
(c) क्रोमानिमा
(d) काइनेटोफोर।
उत्तर:
(d) काइनेटोफोर।

22. वह अवस्था जिसमें किएज्मेटा को देखा जा सकता है –
(a) लेप्टोटीन
(b) जाइगोटीन
(c) पैकीटीन
(d) डाइकाइनेसिस।
उत्तर:
(c) पैकीटीन

23. मियोसिस में किन कारणों से विभिन्नता उत्पन्न होती है –
(a) स्वतंत्र अपव्यूहन
(b) क्रॉसिंग ओवर
(c) (a) एवं (b) दोनों
(d) सहलग्नता।
उत्तर:
(c) (a) एवं (b) दोनों

24. गुणसूत्रों पर पाये जाने वाले क्रोमोमियर्स की संख्या होती है –
(a)250
(b)300
(c) 150
(d) 300 से अधिक।
उत्तर:
(a)250

25. टेरिडोफाइटा में रिडक्शन विभाजन होता है –
(a) गैमीट बनने के समय
(b) स्पोर बनने के बाद
(c) स्पोर बनने के समय
(d) गैमीट बनने के बाद।
उत्तर:
(c) स्पोर बनने के समय

प्रश्न 2.
रिक्त स्थानों की पूर्ति कीजिए

1. अर्द्धसूत्री विभाजन …………….. में होता है।
2. सिनैप्टिकल जटिल का निर्माण ……………. अवस्था में होता है।
3. अर्धसूत्री विभाजन के गुणसूत्र …………….. अवस्था में विभाजित होते हैं।
4. डिप्लोटीन अवस्था में …………….. होता है।
5. कैरियोकाइनेसिस में …………… का विभाजन होता है।
6. जनक एवं संतति कोशिका में गुणसूत्रों की संख्या बराबर होती है इसलिए इसे …………….. कहते हैं।
7. मध्यावस्था में जिस तल पर गुणसूत्र पंक्तिबद्ध हो जाते हैं उसे …………………. कहते हैं।

उत्तर:

1. जनन कोशिकाओं
2. जाइगोटीन
3. द्वितीय पश्चावस्था
4. जीन विनिमय
5. केन्द्रक
6. समसूत्री विभाजन
7. मध्यावस्था पट्टिका।

प्रश्न 3.
एक शब्द में उत्तर दीजिए –

1. अर्द्धसूत्री विभाजन में दो गुणसूत्र के बीच बनने वाली रचना का नाम लिखिए।
2. कोशिका झिल्ली में गर्त बनने से किस कोशिका में कोशिकाद्रव्य विभाजन होता है ?
3. कोशिका में जीन्स की स्थिति कहाँ होती है ?
4. किस कोशिका विभाजन द्वारा पौधों में युग्मक निर्माण होता है ?
5. कोशिका विभाजन की किस अवस्था में गुणसूत्र मध्य रेखा पर स्थित हो जाते हैं ?

उत्तर:

1. किएज्मा (किएज्मेटा)
2. जन्तु कोशिका
3. केन्द्रक के गुणसूत्र में
4. अर्द्धसूत्री
5. मेटाफेज।

प्रश्न 4.
उचित संबंध जोडिए –

उत्तर:

1. (d) G₂ उपावस्था
2. (a) पूर्वावस्था
3. (e) समसूत्री विभाजन
4. (c) दो कोशिकाएँ
5. (b) अगुणित संतति कोशिका

प्रश्न 5.
सत्य / असत्य बताइए –

1. कोशिकीय विभाजन ही जीवन की सत्यता का आधार है।
2. तंत्रिका कोशिका विभाजित होती है।
3. कोशिकीय चक्र के M – चरण में वास्तविक केंद्रकीय विभाजन होता है।
4. जीवाणु कोशिका का कोशिका चक्र 20 घंटे में पूरा होता है।
5. तर्कुतन्तु गुणसूत्रों की गति को अनियंत्रित करते हैं।

उत्तर:

1. सत्य
2. असत्य
3. सत्य
4. असत्य
5. असत्य।

कोशिका चक्र और कोशिका विभाजन अति लघु उत्तरीय प्रश्न

प्रश्न 1.
कोशिका विभाजन की किस अवस्था में क्रॉसिंग ओवर होती है ?
उत्तर:
अर्द्धसूत्री विभाजन के प्रोफेज की पैकीटीन अवस्था में क्रॉसिंग ओवर होती है।

प्रश्न 2.
ऐसे रसायन का नाम बताइए जो कोशिका विभाजन के अध्ययन हेतु गुणसूत्रों के अभिरंजन के लिए इस्तेमाल किया जाता है।
उत्तर:
ऐसीटोकार्मिन (Acetocarmine) या ऐसीटोऑर्सिन नामक अभिरंजन का उपयोग कोशिका विभाजन · में गुणसूत्रों के अभिरंजन हेतु किया जाता है। .

प्रश्न 3.
समसूत्री विभाजन किन कोशिकाओं का लक्षण है ? उत्तर-समसूत्री विभाजन कायिक कोशिकाओं का लक्षण है। प्रश्न 4. अर्द्धसूत्री विभाजन किन कोशिकाओं में होता है ?
उत्तर:
यह विभाजन जनन कोशिकाओं में होता है, जिसके फलस्वरूप युग्मकों का निर्माण होता है और पीढ़ी-दर-पीढ़ी गुणसूत्रों की संख्या समान बनी रहती है।

प्रश्न 5.
प्रथम अर्द्धसूत्री विभाजन में पायी जाने वाली अवस्थाओं का नाम लिखिए।
उत्तर:
प्रथम अर्द्धसूत्री विभाजन में निम्नलिखित अवस्थाएँ पायी जाती हैं –
(A) प्रोफेज प्रथम-इसमें पाँच उप-अवस्थाएँ होती हैं –

1. लेप्टोटीन
2. जाइगोटीन
3. पैकीटीन
4. डिप्लोटीन
5. डायकाइनेसिस।

(B) मेटाफेज प्रथम
(C) ऐनाफेज प्रथम
(D) टिलोफेज प्रथम।

प्रश्न 6.
उस विधि का नाम बताइए, जिसके कारण आनुवंशिक स्थिरता, वृद्धि, अलैंगिक प्रजनन, पुनरावृत्ति तथा कोशिकाओं का प्रतिस्थापन होता है।
उत्तर:
समसूत्री विभाजन (Mitotic division)

प्रश्न 7. समसूत्री विष किसे कहते हैं ?
उत्तर:
कुछ रसायन ऐसे होते हैं, जो समसूत्री विभाजन को रोक देते हैं, इन्हें समसूत्री विष कहते हैं। कुछ प्रमुख समसूत्री विष निम्नलिखित हैं –

• कोल्चिसीन-यह विभाजन के समय त’ बनने की क्रिया को रोककर मध्यावस्था को स्थिर कर देता है।
• राइबोन्यूक्लिएज-यह प्रोफेज अवस्था को रोक देता है।
•  मस्टर्ड गैस-गुणसूत्रों को खण्डित कर देता है।

प्रश्न 8.
एक पुष्पीय पादप के उन अंगों के नाम बताइए जहाँ अर्द्धसूत्री विभाजन सम्भव है।
उत्तर:
पुष्पीय पादप के पुंकेसर के परागकोष तथा स्त्रीकेसर के अण्डाशय में अर्द्धसूत्री विभाजन होता है।

कोशिका चक्र और कोशिका विभाजन लघु उत्तरीय प्रश्न

प्रश्न 1.
जीन विनिमय क्या है ? इसके महत्व पर प्रकाश डालिए।
उत्तर:
अर्द्धसूत्री विभाजन की डिप्लोटीन अवस्था में होने वाली समजात गुणसूत्र खण्डों की अदलाबदली को जीन विनिमय या क्रॉसिंग ओवर कहते हैं। जब डिप्लोटीन अवस्था में विकर्षण के पैदा होने के कारण समजात गुणसूत्र अलग-अलग होना प्रारम्भ करते हैं तब ये आपस में कुछ स्थानों पर संलग्न रह जाते हैं इन स्थानों को किएज्मा कहते हैं।

इन स्थानों पर सिस्टर क्रोमैटिड टूटकर फिर से क्रॉस के रूप में जुड़ जाते हैं, लेकिन गुणसूत्रों के फिर से जुड़ने की इस क्रिया में क्रोमोनिमा की अदला-बदली (पुनर्योजन) हो जाती है। गुणसूत्रों के क्रोमोनिमा की इसी अदला-बदली को परस्पर जीन विनिमय (Crossing over) कहते हैं। अतः इस प्रक्रिया के कारण समजात गुणसूत्रों के बीच जीन विनिमय होता है।

महत्व:

1. इसके कारण जीवों में विभिन्नता पैदा होती है।
2. इसके कारण जीवों में अनुकूलन पैदा होता है।
3. इसके कारण जीवों में विकासात्मक लक्षण बनते हैं।
4. इसकी सहायता से गुणसूत्रों के आनुवंशिक मानचित्र बनाये जाते हैं।
5. इसके कारण जीवों में नये लक्षण बनते हैं।

प्रश्न 2.
समसूत्री विभाजन के महत्व को लिखिए।
उत्तर:
समसूत्री विभाजन का महत्व –

1. इस विभाजन के कारण जीवों में वृद्धि तथा विकास होता है।
2. इस विभाजन के कारण जनक कोशिकाओं के समान ही सन्तति कोशिकाएँ बनती हैं।
3. इस विभाजन के द्वारा घाव भरते हैं तथा मृत कोशिकाओं का प्रतिस्थापन भी इसी विभाजन के द्वारा होता है।
4. इस विभाजन के द्वारा सूचनाओं का मातृ कोशिका से सन्तति कोशिका में प्रवाह होता है।

प्रश्न 3.
मियोसिस की ऐनाफेज प्रथम, माइटोटिक ऐनाफेज से किस बात में भिन्न है ? इसका पूरी प्रक्रिया पर क्या प्रभाव पड़ता है ?
उत्तर:
मियोसिस के ऐनाफेज – I में कोशिका के गुणसूत्रों की द्विगुणित संख्या में से आधे गुणसूत्र एक ध्रुव पर तथा आधे गुणसूत्र दूसरे ध्रुव पर जाते हैं, जबकि माइटोसिस में एक ही गुणसूत्र के अर्द्ध-गुणसूत्र सेण्ट्रोमियर से अलग होकर एक अर्द्ध-गुणसूत्र एक ध्रुव पर तथा दूसरा दूसरे ध्रुव पर जाता है। पूरी प्रक्रिया पर पड़ने वाला प्रभाव-मियोसिस के ऐनाफेज में आधे-आधे गुणसूत्र ध्रुवों पर जाने के कारण विभाजन पूर्ण होने पर दो ऐसी सन्तति कोशिकाएँ बनती हैं, जिनमें गुणसूत्रों की संख्या मूल संख्या की आधी हो जाती है, फलत: पीढ़ी-दर-पीढ़ी गुणसूत्रों की संख्या एक समान बनी रहती है।

प्रश्न 4.
किसी भी बहुकोशिकीय जीवधारी में दो प्रकार के कोशिका विभाजनों की आवश्यकता एवं महत्व पर संक्षिप्त टिप्पणी लिखिए।
उत्तर:
दो कोशिकीय विभाजनों की आवश्यकता-बहुकोशिकीय जीवों की संरचना तथा जैविक क्रिया जटिलता का. प्रदर्शन करती हैं। इस कारण इनमें दो प्रकार के विभाजनों की आवश्यकता पड़ती है, जिससे एक विभाजन (अर्द्धसूत्री) जनन कोशिकाओं में प्रजनन के समय हो जिससे गुणसूत्रों की संख्या में स्थिरता बनी रहे, जबकि दूसरा विभाजन ऐसा हो जो सामान्य कोशिकाओं की मरम्मत कर सके। दो प्रकार का विभाजन इन आवश्यकताओं को पूरा करता है।

महत्व:
इन दो प्रकार के विभाजनों के कारण ही कोशिकीय संलयन (निषेचन) के बाद भी जीवों तथा कोशिकाओं में पीढ़ी-दर-पीढ़ी गुणसूत्रों की संख्या समान बनी रहती है और टूट-फूट की मरम्मत तथा वृद्धि एवं विकास की क्रियाएँ भी संचालित होती हैं।

प्रश्न 5.
समसूत्री विभाजन की प्रोफेज तथा ऐनाफेज को चित्र सहित समझाइए।
उत्तर:
समसूत्री प्रोफेज:

1. गुणसूत्र लम्बे, पतले धागे के समान पाये जाते हैं, जिसे क्रोमैटिन जाल कहते हैं।
2. सेण्ट्रिओल गति करके ध्रुवों पर जाने लगते हैं।
3. गुणसूत्र सेण्ट्रोमियर से जुड़े हुए दिखाई देने लगते हैं।
4. केन्द्रकीय झिल्ली समाप्त हो जाती है।

समसूत्री ऐनाफेज:

1. सेण्ट्रोमियर्स के विभाजन से दोनों क्रोमैटिड्स अलग हो जाते हैं एवं दो गुणसूत्र बना देते हैं।
2. गुणसूत्र ध्रुवों की ओर गति करते हैं। 3. गुणसूत्र की संख्या एवं प्रकार स्पष्ट हो जाते हैं।

प्रश्न 6.
समसूत्री विभाजन की विभिन्न अवस्थाओं को चित्रित कीजिए।
उत्तर:
समसूत्री विभाजन (Mitosis cell division):

प्रश्न 7.
समसूत्री विभाजन की क्या विशेषताएँ हैं ? मेटाफेज का नामांकित चित्र बनाकर वर्णन कीजिए।
उत्तर:
विशेषताएँ:

1. नई बनी कोशिकाओं में गुणसूत्रों की संख्या जनक कोशिका के समान होती है।
2. एक कोशिका विभाजित होकर दो समान कोशिकाएँ बनाती हैं।
3. यह विभाजन कायिक कोशिका में होता है तथा इसके कारण जीवों में वृद्धि होती है तथा घाव भरता है।
4. जनक कोशिका के आनुवंशिक गुण पौत्रिक कोशिका में पहुँचते हैं। इस विभाजन से आनुवंशिक समानता बनी रहती है।

समसूत्री मेटाफेज:

1. इस अवस्था में केन्द्रक कला तथा केन्द्रिका लुप्त हो जाती है।
2. गुणसूत्र कोशिका की मध्यरेखा पर एकत्रित हो जाते हैं।
3. गुणसूत्रों के अर्द्ध गुणसूत्र स्पष्ट एवं अलग हो जाते हैं।
4. तर्क का निर्माण पूर्ण हो जाता है। टीप-चित्र के लिए उपर्युक्त प्रश्न क्रमांक 6 के चित्र D को देखिए।

प्रश्न 8.
निम्नलिखित में अन्तर स्पष्ट कीजिए –

1. समसूत्री एवं अर्द्धसूत्री विभाजन
2.  गुणसूत्र एवं अर्द्ध-गुणसूत्र
3. सेण्ट्रोमियर एवं क्रोमोमियर
4.  सेण्ट्रोसोम एवं सेण्ट्रिओल
5. मेटाफेज-I एवं मेटाफेज-II
6.  जायगोटीन एवं पैकीटीन
7. कोशिका खाँच एवं कोशिका प्लेट।

उत्तर:

1. समसूत्री एवं अर्द्धसूत्री विभाजन में अन्तर:
दोनों ही कोशिका विभाजनों में कोशिकाओं की संख्या बढ़ती है तथा दोनों में गुणसूत्रों का विभाजन होता है। इन समानताओं के बावजूद दोनों विभाजनों में निम्नलिखित अन्तर पाये जाते हैं –
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2. गुणसूत्र एवं अर्द्ध-गुणसूत्र में अन्तर:
क्रोमैटिन जाल ही कोशिका विभाजन के समय संघनित होकर एक विशिष्ट रचना बनाता है, जिसे गुणसूत्र कहते हैं, जबकि गुणसूत्र स्वयं दो कुण्डलित अर्धांशों का बना होता है, जिसे अर्द्ध-गुणसूत्र कहते हैं।

3. सेण्ट्रोमियर एवं क्रोमोमियर में अन्तर:
गुणसूत्रों में पाये जाने वाले प्राथमिक संकीर्णन को सेण्ट्रोमियर कहते हैं। यह गुणसूत्र को दो भुजाओं में बाँटकर उनके आकार को निर्धारित करता है, जबकि गुणसूत्र के अन्दर क्रोमैटीन तन्तु (क्रोमोनिमा) पर पायी जाने वाली गाँठ सदृश रचनाओं को क्रोमोमियर कहते हैं । सम्भवत: DNA अणु इन्हीं स्थानों पर क्रोमैटीन तन्तु से जुड़े होते हैं।

4. सेण्ट्रोसोम एवं सेण्ट्रिओल में अन्तर:
कोशिका के अन्दर तथा केन्द्रक के पास एक कलाविहीन कणीय कोशिकांग पाया जाता है, जिसे सेण्ट्रोसोम कहते हैं। सेण्ट्रिओल दो उप-इकाइयों का बना होता है, जो एक-दूसरे से समकोण पर स्थित होती है।

5. मेटाफेज-I एवं मेटाफेज-II में अन्तर –
मेटाफेज – I:
1. समजात गुणसूत्र जोड़े में कोशिका के मध्य में स्थित होते हैं।
2. इसमें सेण्ट्रोमियर विभाजित नहीं होता पूरा गुणसूत्र ध्रुवों की ओर जाता है।

मेटाफेज – II
1. इकहरे गुणसूत्र कोशिका के मध्य में स्थित होते हैं।
2. इसमें सेण्ट्रोमियर विभाजित होकर दो भाग बना देता है तथा अर्द्ध – गुणसूत्र ध्रुवों की ओर जाता है।

6. जायगोटीन एवं पैकीटीन में अन्तर:
जायगोटीन अवस्था में गुणसूत्र जोड़े में रहकर बाइवैलेण्ट अवस्था में रहता है, जबकि पैकीटीन में गुणसूत्रों के अर्द्ध-गुणसूत्र अलग होकर टेट्राड अवस्था में रहते हैं।

7. कोशिका खाँच एवं कोशिका प्लेट:
कोशिका खाँच वह रचना है, जिसमें कोशिका संकीर्णित होकर दो भागों में बँटती हैं, जबकि कोशिका प्लेट वह रचना है, जिसमें कोशिका के मध्य में कोशिकांग व्यवस्थित होकर एक प्लेट बनाते हैं फलतः कोशिका दो भागों में बँट जाती है।

प्रश्न 9.
निम्नलिखित का अर्थ स्पष्ट कीजिए –

1. समजात गुणसूत्र
2. बाइवैलेण्ट
3. टेट्राड
4. G₂फेज।

उत्तर:
(i) समजात गुणसूत्र – एक समान गुणसूत्रों को समजात (Homologous) गुणसूत्र कहते हैं अगर ये गुणसूत्र एक साथ स्थित हों, तो उनकी सभी रचनाएँ हूबहू एकसमान होती हैं।

(ii) बाइवैलेण्ट-NCERT प्रश्न क्रमांक 7 का उत्तर देखिए।

(iii) टेट्राड-पैकीटीन अवस्था में समजात गुणसूत्रों के अर्द्ध – गुणसूत्र विभाजित होकर प्रत्येक गुणसूत्र को दो अर्द्धाशों अर्थात् तन्तुओं में बाँट देते हैं। इस कारण जोड़े के गुणसूत्र चार गुणसूत्रों के रूप में प्रतीत होने लगते हैं, तब समजात गुणसूत्रों के जोड़े को टेट्राड कहते हैं।

(iv)G₂;फेज या द्वितीय वृद्धि उपअवस्था – कोशिकीय चक्र की वह उप-अवस्था है, जिसमें कोशिकीय संश्लेषण फिर से होता है। इस उप-अवस्था में माइटोकॉण्ड्रिया व क्लोरोप्लास्ट स्वविभाजित हो जाते हैं। सेण्ट्रिओल का विभाजन होकर दो सेट बन जाते हैं तथा स्पिण्डिल निर्माण प्रारम्भ हो जाता है। इस अवस्था में कोशिकीय पदार्थों का संश्लेषण फिर से होता है, इस कारण इसे G₂ फेज कहते हैं।

प्रश्न 10.
निम्नलिखित में से कौन-से कथन समसूत्री विभाजन की निम्नलिखित अवस्थाओं से सम्बन्धित हैं –
(A) प्रोफेज
(B) मेटाफेज
(C) ऐनाफेज
(D) टीलोफेज
(E) इण्टरफेज।

1. केन्द्रक झिल्ली का पुनः निर्माण
2. गुणसूत्र सर्वाधिक छोटे एवं मोटे
3. गुणसूत्र कुण्डलित होने लगते हैं
4. सेण्ट्रोमियर का दो भागों में विभाजन
5. केन्द्रक सक्रिय होता है, किन्तु क्रोमोसोम स्पष्ट दिखाई नहीं देते हैं
6. साइटोकाइनेसिस के बाद की अवस्था
7. प्रत्येक गुणसूत्र दो अर्द्ध-गुणसूत्रों का बना दिखाई देता है।

उत्तर:

1. टीलोफेज
2. मेटाफेज
3. टीलोफेज
4. प्रोफेज
5. प्रोफेज
6. इण्टरफेज
7. प्रोफेज।

कोशिका चक्र और कोशिका विभाजन दीर्घ उत्तरीय प्रश्न

प्रश्न 1.
अर्द्धसूत्री विभाजन की क्रिया का नामांकित चित्रों सहित विवरण दीजिए तथा इसका महत्व लिखिए।
उत्तर:
अर्द्धसूत्री विभाजन-वह विभाजन है, जिसमें विभाजन के बाद चार ऐसी कोशिकाएँ बनती हैं, जिसमें गुणसूत्रों की संख्या मूल संख्या की आधी रह जाती है। यह विभाजन दो चरणों में पूरा होता है –

(A) अर्द्धसूत्री विभाजन प्रथम:
इसके द्वारा दो ऐसी कोशिकाएँ बनती हैं, जिनमें गुणसूत्रों की संख्या मूल कोशिका की आधी होती हैं। यह चार अवस्थाओं में पूरा होता है –

1. प्रोफेज-I:
इसमें पाँच उप अवस्थाएँ होती हैं –

1. लेप्टोटीन – सेण्ट्रिओल ध्रुवों पर चले जाते हैं तथा गुणसूत्र स्पष्ट हो जाते हैं।
2. जायगोटीन – समजात गुणसूत्र जोड़ा बनाते हैं।
3. पैकीटीन – गुणसूत्रों के अर्द्ध-गुणसूत्र अलग होकर टेट्राड बना देते हैं, और उनमें अन्तर्ग्रथन हो जाता है तथा जीन विनिमय की क्रिया होती है।
4. डिप्लोटीन-समजात गुणसूत्र विकर्षण के कारण अलग हो जाते हैं तथा उनमें जीन विनिमय पूर्ण हो जाता है।
5. डायकाइनेसिस-गुणसूत्र दूर जाने लगते हैं । केन्द्रक तथा केन्द्रिका विलुप्त हो जाती हैं तथा तर्कु बन जाते हैं।

2. मेटाफेज-I:
समजात गुणसूत्र कोशिका के मध्य में आ जाते हैं, त’ पूर्ण विकसित हो जाते हैं।

3. ऐनाफेज-I:
समजात गुणसूत्रों में से एक, एक ध्रुव पर तथा दूसरा, दूसरे ध्रुव पर चला जाता है। अतः आधे गुणसूत्र एक ध्रुव पर तथा आधे दूसरे ध्रुव पर चले जाते हैं।

4. टीलोफेज:
केन्द्रिका तथा केन्द्रक फिर से बन जाते हैं अन्त में दो ऐसी कोशिकाएँ बनती हैं, जिनमें गुणसूत्रों की संख्या आधी रह जाती है।
कोशिकाद्रव्य विभाजन-दो केन्द्रक बनने के बाद कोशिका का कोशिकाद्रव्य भी खाँच या पट्ट निर्माण द्वारा दो भागों में बँट जाता है।

(B) अर्द्धसूत्री विभाजन द्वितीय:
यह समसूत्री विभाजन के समान होता है, जिसमें अर्द्धसूत्री विभाजन से बनी कोशिकाएँ दो-दो कोशिकाओं में विभाजित हो जाती हैं –
इसमें निम्नलिखित चार अवस्थाएँ होती हैं –

1. प्रोफेज – II:
प्रथम विभाजन की सन्तति कोशिकाओं के गुणसूत्र स्पष्ट हो जाते हैं। केन्द्रक तथा केन्द्रिका विलुप्त हो जाती हैं तथा तर्कु बन जाते हैं।

2. मेटाफेज-II:
गुणसूत्र के दोनों अर्द्ध-गुणसूत्र अलग होकर कोशिका के मध्य में आ जाते हैं।

3. ऐनाफेज-II:
अर्द्ध-गुणसूत्र अलग-अलग ध्रुवों पर चले जाते हैं और गुणसूत्र का रूप ले लेते हैं।

4. टीलोफेज-II:
प्रत्येक ध्रुव पर केन्द्रक बन जाता है। कोशिकाद्रव्य विभाजन-प्रथम विभाजन से बनी दोनों कोशिकाओं के दोनों केन्द्रकों के चारों तरफ कोशिकाद्रव्य भी विभाजित हो जाता है। इस प्रकार चार ऐसी कोशिकाएँ बन जाती हैं, जिनमें गुणसूत्रों की संख्या आधी हो जाती है।

अर्द्धसूत्री विभाजन का महत्व:

1. इसके कारण पीढ़ी-दर-पीढ़ी कोशिकाओं में गुणसूत्रों की संख्या एक समान बनी रहती है।
2. इस विभाजन के कारण जनक के समान ही कोशिकाएँ पैदा होती हैं।
3. इस विभाजन में जीन विनिमय होने के कारण यह नये गुणों के बनने में सहायता करता है।
4. इसके कारण विभिन्नता पैदा होती है, जो जैव विकास के लिए आवश्यक है।
5. इसके कारण एक द्विगुणित कोशिका से चार अगुणित कोशिकाएँ बनती हैं।

प्रश्न 2.
कोशिका चक्र की आवश्यकता एवं महत्व को दर्शाते हुए उसकी विभिन्न अवस्थाओं में होने वाली घटनाओं का वर्णन कीजिए।
उत्तर:
कोशिका चक्र की आवश्यकता एवं महत्व-एक कोशिका के अस्तित्व में आने से लेकर उसका विभाजन होने तक की क्रियाएँ संयुक्त रूप से कोशिका चक्र कहलाती हैं। कोशिका चक्र के कारण ही नई कोशिकाओं का निर्माण होता है, जिससे नई कोशिकाएँ बनकर घाव को भरती हैं।

इसके अलावा इसी के कारण ही पीढ़ी-दर-पीढ़ी गुणसूत्रों की संख्या एकसमान बनी रहती है। इसी के कारण जीवों में वृद्धि तथा विकास सम्भव हो पाता है एवं पुरानी कोशिकाओं की पुनर्स्थापना होती है अर्थात् कोशिकीय चक्र के बिना जीव अस्तित्व में नहीं रह पाएँगे। कोशिका चक्र की अवस्थाएँ-कोशिकीय चक्र में निम्नलिखित अवस्थाएँ पायी जाती हैं –

(A) अन्तरावस्था या विभाजनान्तराल या इण्टरफेज:
अंतरावस्था(इण्टरफेज) दो विभाजनों के बीच की अवस्था है, जिसमें निम्नलिखित तीन अवस्थाएँ पायी जाती हैं –

(i)G₂ फेज-इस अवस्था में प्रोटीन एवं RNA का संश्लेषण किया जाता है।

(ii) S₂ फेज-इस अवस्था में DNA एवं हिस्टोन प्रोटीन का संश्लेषण होता है।

(ii) G₂फेज-इस अवस्था में आवश्यक प्रोटीन तथा RNA का संश्लेषण किया जाता है तथा विभिन्न कोशिकांगों का निर्माण होता है।

(B) विभाजन अवस्था या m – अवस्था:
इसमें कोशिका का मूल विभाजन होता है, जिससे 2 या 4 कोशिकाएँ बनती हैं। इसमें निम्नलिखित अवस्थाएँ पायी जाती हैं –
1. केन्द्रकीय विभाजन-इस प्रावस्था में विभाजित होने वाली कोशिका का केन्द्र विभाजित होकर दो अथवा चार केन्द्र बना देता है। यह विभाजन चार अवस्थाओं प्रोफेज, मेटाफेज, ऐनाफेज तथा टीलोफेज में पूर्ण होता है।

2. कोशिकाद्रव्य विभाजन या साइटोकाइनेसिस:
कोशिका द्रव्य विभाजन-कोशिका विभाजन के समय केन्द्रक विभाजन के बाद कोशिकाद्रव्य विभाजन को साइटोकाइनेसिस या कोशिकाद्रव्य विभाजन कहते हैं, यह पादप एवं प्राणियों में दो अलग-अलग विधियों द्वारा होता है –

(a) कोशिका खाँच द्वारा इस कोशिकाद्रव्य विभाजन में कोशिका के मध्य में एक खाँच बनती है, जो बढ़कर कोशिका के कोशिकाद्रव्य को दो भागों में बाँट देती है। प्रत्येक भाग में एक केन्द्रक होता है, जन्तुओं में इसी प्रकार का विभाजन पाया जाता है।

(b) कोशिका प्लेट द्वारा इस कोशिकाद्रव्य विभाजन में कोशिका के मध्य गॉल्गीकाय तथा E.R. एकत्रित होकर एक पट बना देते हैं जो बाद में कोशिका भित्ति बन जाती है और कोशिका को दो भागों में बाँट देती हैं। प्रत्येक भाग में एक केन्द्रक होता है। यह विभाजन पादप कोशिकाओं में पाया जाता है।

MP Board Class 11th Biology Solutions