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MP Board Class 12th Maths Important Questions Chapter 11 Three Dimensional Geometry

Three Dimensional Geometry Important Questions

Three Dimensional Geometry Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
The FZ – plane divides the line segment joining the points P (- 2, 4, 7) and Q (3, -5, 8) in the ratio:
(a) 2 : 3
(b) 1 : 2
(c) 2 : 5
(d) 3 : 4
Answer:
(a) 2 : 3

Question 2.
If a line makes an angle \(\frac { \pi }{ 4 } \) with both the positive direction of the x – axis, and y – axis, then the angle made by the line with positive direction of the Z – axis is:
(a) \(\frac { \pi }{ 6 } \)
(b) \(\frac { \pi }{ 3 } \)
(c) \(\frac { \pi }{ 4 } \)
(d) \(\frac { \pi }{ 2 } \)
Answer:
(d) \(\frac { \pi }{ 2 } \)

Question 3.
Equation of the line passing through the points (2, 3, 4) and (1, -2, 3) are:
(a) \(\frac { x-2 }{ 1 } \) = \(\frac { y-2 }{ -5 } \) = \(\frac { z-4 }{ 1 } \)
(b) \(\frac { x-2 }{ -1 } \) = \(\frac { y-3 }{ -5 } \) = \(\frac { z-4 }{ -1 } \)
(c) \(\frac { x-2 }{ -1 } \) = \(\frac { y-3 }{ 5 } \) = \(\frac { z-4 }{ -1 } \)
(d) \(\frac { x-2 }{ -1 } \) = \(\frac { y-3 }{ -5 } \) = \(\frac { z-4 }{ 1 } \)
Answer:
(b) \(\frac { x-2 }{ -1 } \) = \(\frac { y-3 }{ -5 } \) = \(\frac { z-4 }{ -1 } \)

Question 4.
Angle between the planes x + 2y + z + 7 = 0 and 2x + y – z + 13 = 0 is:
(a) \(\frac { \pi }{ 2 } \)
(b) \(\frac { \pi }{ 3 } \)
(c) \(\frac { 3\pi }{ 2 } \)
(d) π
Answer:
(b) \(\frac { \pi }{ 3 } \)

Question 5.
Equation to the plane which cuts off intercepts 2, 3, -4 on the axis is:
(a) \(\frac{x}{2}\) + \(\frac{y}{3}\) – \(\frac{z}{4}\) = 0
(b) \(\frac{x}{2}\) + \(\frac{y}{3}\) – \(\frac{z}{4}\) = -1
(c) \(\frac{x}{2}\) + \(\frac{y}{3}\) – \(\frac{z}{4}\) = 1
(d) None of these.
Answer:
(c) \(\frac{x}{2}\) + \(\frac{y}{3}\) – \(\frac{z}{4}\) = 1

Question 2.
Fill in the blanks:

1. Direction cosine of unit vector \(\frac { 1 }{ \sqrt { 14 } } \) ( \(\hat { i } \) + 2\(\hat { j } \) + 3\(\hat { k } \) ) are ………………………….
2. Direction cosine of X – axis are ………………………….
3. Angle between the diagonals of cube is ………………………….
4. Angle between the straight lines \(\frac{x}{1}\) = \(\frac{y}{0}\) = \(\frac{z}{-1}\) and \(\frac{x}{3}\) = \(\frac{y}{4}\)
   = \(\frac{z}{5}\) is ………………………….
5. If the lines \(\frac { x-2 }{ 3 } \) = \(\frac { y-3 }{ 4 } \) = \(\frac { z-4 }{ k } \) and \(\frac { x-2 }{ 3 } \) = \(\frac { y-3 }{ 4 } \) = \(\frac { z-4 }{ k } \) are coplaner, then k …………………………..
6. If the line makes an angle α, β, γ with the coordinate axis respectively, then cos²α + cos²β + cos² γ = ……………………………..
7. Intercept cut by the plane 2x + y – z = 5 on axis is ……………………………..

Answer:

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

Question 3.
Write True/False:

1. The points A (1, 2, 3), B (4, 0,4) and C (-2, 4, 2) are collinear.
2. The angle between lines whose direction ratios are (3,4, 5) and (4, -3, 5) is 30°.
3. Angle between the lines 2x = 3y = —z and 6x = -y = -4z is 90°.
4. Shortest distance between two intersecting lines always zero.
5. Angle between the lines \(\frac { x+1 }{ 3 } \) = \(\frac { y+1 }{ 2 } \) = \(\frac { z+2 }{ 4 } \) and plane 2x + y – 3z + 5 = 0 is cos^-1 (\(\frac { 4 }{ \sqrt { 406 } } \) )
6. Straight line \(\frac { x-2 }{ 1 } \) = \(\frac { y+1 }{ -2 } \) = \(\frac { z-4 }{ 1 } \) is parllel to the plane x + 3y + 5z = 4
7. Equation of plane parallel to the X – axis is ax + by + d = 0.

Answer:

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

Question 4.
Match the following:

Answer:

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

Answer:

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

Question 5.
Write the answer in one word/sentence:

1. Find the direction ratios of the normal of the plane x + 2y + 3z + 4 = 0.
2. Find the equation of the plane which intercepts unit length from the coordinate axis.
3. Find the equation of the plane which is perpendicular on FOZ-plane.
4. Find the angle between the planes x + 2y + z + 7 = 0 and 2x + y – z + 13 = 0.
5. Find the distance between the parallel planes 2x – 2y + z + 3= 0 and 4x – 4y + 2z.
6. Find the angle between the lines \(\frac{x}{2}\) = \(\frac{y}{-1}\) = \(\frac{z}{1}\) and \(\frac{x}{1}\) = \(\frac{y}{1}\) = \(\frac{z}{2}\)
7. If a line makes α, β, γ with the positive direction of the axis, then find the value of sin²α +sin² β + sin²γ.

Answer:

1. (1, 2, 3)
2. x + y + z = 1
3. by + cz + d = 0
4. \(\frac { \pi }{ 3 } \)
5. \(\frac{1}{6}\)
6. \(\frac { \pi }{ 3 } \)
7. 2

Three Dimensional Geometry Short Answer Type Questions

Question 1.
Find the direction cosine of the line joining two points (-2, 4, -5) and (1, 2, 3)? (NCERT)
Answer:
Given points A (- 2, 4, – 5) and B (1, 2, 3).
Direction ratio of AB = 1 + 2, 2 – 4, 3 + 5
= 3, -2, 8

Direction cosine of AB

Question 2.
Find direction cosines of the line which makes an angle 90°, 135° and 45° with X, Y and Z axis respectively? (NCERT)
Solution:
Given α = 90°, β = 135°, λ = 45°
cos α = cos 90° = 0
cos β = cos 135° = cos (180° – 45°)
= – cos 45° = – \(\frac { 1 }{ \sqrt { 2 } } \)
cos λ = cos 45° = \(\frac { 1 }{ \sqrt { 2 } } \)
Direction cosines are cos α, cos β, cos λ
i.e; 0, – \(\frac { 1 }{ \sqrt { 2 } } \), \(\frac { 1 }{ \sqrt { 2 } } \)

Question 3.
A line OP, makes the angle 120° and 60° with X – axis and Y – axis respectively find the angle made by line with Z – axis?
Solution:
Given α = 120°, β = 60°.
Let the angle made by the line with Z – axis be γ then
cos²α + cos²β + cos² γ = 1
⇒ cos² 120° + cos² 60° + cos² γ = 1

Hence, the required angle is 45° or 135°.

Question 4.
Show that the points (2, 3, 4), (-1, -2, 1) and (5, 8, 7) are collinear? (NCERT)
Solution:
Given Points are A (2, 3, 4), B (-1,-2, 1), C (5, 8, 7)
Direction ratio of AB x₂ – x₁, y₂ – y₁, z₂ – z₁
= – 1 – 2, – 2 – 3, 1 – 4
= – 3, – 5, – 3 = – (3, 5, 3)
Direction ratio of BC x₂ – x₁, y₂ – y₁, z₂ – z₁
= 5 + 1, 8 + 2, 7 – 1
= 6, 10, 6 = 2 (3, 5, 3)
It is clear that direction ratio of AB and BC are proportional, hence AB is parallel to BC but point B is common to both AB and BC therefore A, B, C are collinear. Proved.

Question 5.
If direction cosine of a line are cos α, cos β and cos γ then prove that:
cos 2α + cos 2β + cos 2γ = – 1.
Solution:
L.H.S. = cos2α + cos2β + cos2γ
= 2cos²α – 1 + 2cos²β – 1 + 2cos²γ – 1
= 2 (cos²α + cos²β + cos²γ) – 3
= 2 × 1 – 3, [∵ cos²α + ²β + cos²γ – 1]
= -1
= R.H.S. Proved.

Question 6.
Prove that the line passing through the points (1, -1, 2) and (3, 4, -2) are perpendicular to the lines passing through the points (0, 3, 2) and (3, 5, 6). (NCERT)
Solution:
Direction ratio’s of the line joining points (1, -1, 2) and (3, 4, -2)
= 3 – 1, 4 + 1, – 2 – 2
= 2, 5, – 4 = a₁, b₁, c₁ (let)
Direction ratios’s of the line joining the points (0, 3, 2) and (3, 5, 6)
= 3 – 0, 5 – 3, 6 – 2
= 3,2,4
= a₂, b₂, c₂ (let)
Lines will be perpendicular if
a₁a₂ + b₁b₂ + c₁c₂ = 0
⇒ 2.3 + 5.2 – 4.4 = 0
⇒ 16 – 16 = 0
⇒ 0 = 0. Proved.

Question 7.
Prove that the line joining the points A (1, 2, 3) and B (2, 3, 5) are parallel to then line joining the points C (-1, 2, -3) and D ( 1, 4, 1)?
Solution:
Direction ratio of line AB
= 2 – 1, 3 – 2, 5 – 3
= 1, 1, 2
= a₁, b₁, C₁
Direction ratio of line CD
= 1 + 1, 4 – 2, 1 + 3
= 2, 2, 4
= a₂, b₂, c₂
Where \(\frac { a_{ 1 } }{ a_{ 2 } } \) = \(\frac { b_{ 1 } }{ b_{ 2 } } \) = \(\frac { c_{ 1 } }{ c_{ 2 } } \)
= \(\frac{1}{2}\) = \(\frac{1}{2}\) = \(\frac{2}{4}\) or \(\frac{1}{2}\)
Hence line AB and CD are parallel. Proved.

Question 8.
Find the angle between the lines \(\frac{x}{2}\) = \(\frac{y}{2}\) = \(\frac{z}{1}\) and \(\frac{x – 5}{4}\) = \(\frac{y – 2}{1}\) = \(\frac{z – 3}{8}\)? (NCERT)
Solution:
Equation of the lines are
\(\frac{x}{2}\) = \(\frac{y}{2}\) = \(\frac{z}{1}\)
and \(\frac{x – 5}{4}\) = \(\frac{y – 2}{1}\) = \(\frac{z – 3}{8}\)
a₁ = 2, b₁ = 2, c₁ = 1
a₂ = 4, b₂ = 1, c₂ = 8
Let the angle between them = θ

Question 9.
Prove that lines
\(\frac{x – 5}{7}\) = \(\frac{y + 2}{- 5}\) = \(\frac{z}{1}\) and \(\frac{x}{1}\) = \(\frac{y}{2}\) = \(\frac{z}{3}\) are perpendicular? (NCERT)
Solution:
\(\frac{x – 5}{7}\) = \(\frac{y + 2}{- 5}\) = \(\frac{z}{1}\)
and \(\frac{x}{1}\) = \(\frac{y}{2}\) = \(\frac{z}{3}\)
Where a₁ = 7, b₁ = -5, c₁ = 1
a₂ = 1, b₂ = 2, c₂ = 3
a₁a₂ + b₁b₂ + c₁c₂ = 7(1) + (- 5)(2) + 1 × 3
= 10 -10 = 0
Hence, the lines are perpendicular. Proved.

Question 10.
Find the cartesion equation of the line which passes through the point (-2, 4, -5) and parallel to the line given by \(\frac { x+3 }{ 3 } \) = \(\frac { y-4 }{ 5 } \) = \(\frac { z+8 }{ 8 } \). (NCERT)
Solution:
Equation of given line
\(\frac { x+3 }{ 3 } \) = \(\frac { y-4 }{ 5 } \) = \(\frac { z+8 }{ 8 } \) ………. (1)
Direction ratio of (1) is 3, 5, 8
Equation of line passing through point (- 2, 4, -5)
\(\frac { x+2 }{ a } \) = \(\frac { y-4 }{ b } \) = \(\frac { z+5 }{ c } \) ……………….. (2)
Direction ratio of eqn. (2) is a, b, c
Eqns. (1) and (2) are parallel
∴\(\frac{a}{3}\) = \(\frac{b}{5}\) = \(\frac{c}{8}\) = k
a = 3k, b = 5k, c = 8k
Putting the value of a, b, c in eqn. (2) we get
\(\frac { x+2 }{ 3k } \) = \(\frac { y-4 }{ 5k } \) = \(\frac { z+5 }{ 8k } \)
\(\frac { x+2 }{ 3 } \) = \(\frac { y-4 }{ 5 } \) = \(\frac { z+5 }{ 8 } \).

Question 11.
Find the eqution of straight line passing through point (1, 2, 3) are parallel to \(\frac{x-6}{12}\) = \(\frac{y-2}{4}\) = \(\frac{z+7}{5}\)?
Solution:
Equation of line passing through point (x₁, y₁, z₁) and direction ratio a, b, c is

Question 12.
Find the angle between the lines \(\frac{x}{1}\) = \(\frac{y}{0}\) = \(\frac{z}{3}\) and \(\frac{x}{4}\) = \(\frac{y}{5}\) = \(\frac{z}{0}\)?
Solution:
Equation of given lines
\(\frac{x}{1}\) = \(\frac{y}{0}\) = \(\frac{z}{3}\) …………………. (1)
and \(\frac{x}{4}\) = \(\frac{y}{5}\) = \(\frac{z}{0}\) ……………………… (2)
Where a₁ = 1, b₁ = 0, c₁ = 3 and a₂ = 4, b₂ = 5, c₂ = 0
Let θ be the angle between eqns. (1) and (2)

Question 13.
Find the condition that the lines x = ay + b, z = cy + d and x = a’y + b’, z = c’y + d’ are perpendicular?
Solution:
x = ay + b and z = cy + 4
⇒ \(\frac { x-b }{ a } \) = \(\frac { y }{ 1 } \) and \(\frac { z-d }{ c } \) = \(\frac { y }{ 1 } \)
Hence the corresponding equation
\(\frac { x-b }{ a } \) = \(\frac { y }{ 1 } \) = \(\frac { z-d }{ c } \) ………………….. (1)
Again x = a’y + b’ and z = c’y + d’ equation is
\(\frac { x-b^{ ‘ } }{ a^{ ‘ } } \) = \(\frac { y }{ 1 } \) = \(\frac { z-d^{ ‘ } }{ c^{ ‘ } } \) ………………… (2)
Eqns. (1) and (2) are perpendicular.
a × a’ + 1 × 1 + c × c’ = 0
⇒ aa’ + cc’ + 1 = 0 is the required condition.

Question 14.
(A) Find the angle between the lines \(\vec { r } \) = ( \(\hat { i } \) + 3\(\hat { j } \) + 5\(\hat { k } \) ) + t ( \(\hat { i } \) – 2\(\hat { j } \) – 3\(\hat { k } \) ) and \(\vec { r } \) = ( \(\hat { i } \) – 2\(\hat { j } \) + 5\(\hat { k } \) ) + s(2\(\hat { i } \) – 2\(\hat { j } \) + \(\hat { k } \) )?
Solution:

(B) Find the angle between lines \(\vec { r } \) = (3\(\hat { i } \) + 4\(\hat { j } \) – 2\(\hat { k } \) ) + t(-\(\hat { i } \) + 2\(\hat { j } \) + \(\hat { k } \) ) and \(\vec { r } \) = ( \(\hat { i } \) – 7\(\hat { j } \) – 2\(\hat { k } \) ) + s( \(\hat { i } \) + 3\(\hat { j } \) + 2\(\hat { k } \) )
Solution:
Solve like Q. 14. (A)

Question 15.
Find angle between two planes 2x – y + z = 6 and x + y + 2z = 7?
Solution:
Given planes are:
2x – y + z = 6 ……………. (1)
and x + y + 2z = 7 ………………… (2)
Direction ratio of eqn. (1) = 2, -1, 1 1 ⇒ A₁, B₁, C₁
Direction ratio of eqn. (2) = 1, 1, 2 ⇒ A₂, B₂, C₂
Let θ be the angle between them

Question 16.
(A) If plane 3x – 6y – 2z = 7 and 2x + y – kz = 5 are perpendicular to each other. Find the value of k?
Solution:
Equation of given planes are
3x – 6y – 2z = 7 ………………………… (1)
and 2x + y – kz = 5 …………………… (2)
Where a₁ = 3, b₁ = -6, c₁ = -2, and a₂ = 2, b₂ = 1, c₂ = -k
Planes (1) and (2) are perpendicular
∴ a₁a₂ + b₁b₂ + c₁c₂ = 0
⇒ (3) (2) + (-6) (1) + (-2) (-k) = 0
⇒ 6 – 6 + 2k = 0
⇒ 2k = 0
⇒ k = 0.

(B) For which value of k the planes 2x + ky + z + 9 = 0 and 5x + 3y – 4z – 6 = 0 are perpendicular?
Solution:
Solve like Q.16 (A)
[Answer: k = -2]

(C) Prove that the planes x + 2y + 3z = 6 and 5x – 3y + z = 1 are perpendicular?
Solution:
Given equation of planes are:
x + 2y + 3z = 6 ……………………. (1)
and 3x – 3y + z = 1 ……………………… (2)
Direction ratio of eqn. (1) = 1, 2, 3
Direction ratio of eqn. (2) = 3,- 3, 1
∴ a₁a₂ + b₁b₂ + c₁c₂ = (1)(3) + 2(-3) + (3)(1)
= 3 – 6 + 3 = 0
Hence planes (1) and (2) are perpendicular. Proved.

Question 17.
A plane meets the co – ordinate axes at points A, B and C. If the centroid of ∆ABC is (2, – 1, 3) then find the equation of the plane?
Solution:
Let A(a, 0, 0), B(0, b, 0), C(0, 0, c)
∵ OA = a, OB = b, OC = c
Centroid (given) is (2,- 1, 3)

Question 18.
A plane meets the coordinate axes in A, B, C such that the centroid of ∆ABC is (a, b, c). Find the equation of the plane is \(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 3?
Solution:

Question 19.
Find the equation of plane which is parallel to the plane 2x + 3y – z = 8 and passes through the point (1,  2, 3)?
Solution:
Any plane parallel to the plane
2x + 3y – z = 8
⇒ 2x + 3y – z + λ = 0 ………………….. (1)
∵ It passes through the point (1,2, 3), then
2(1) + 3(2) – 3 + λ = 0
⇒ 2 + 6 – 3 + λ = 0
⇒ 8 – 3 + λ = 0
⇒ 5 + λ = 0
⇒ λ = -5 ……………………… (2)
∴ From eqns. (1) and (2), we have
2x + 3y – 5 = 0.

Question 20.
Find the direction cosines of normal to the plane 2x + 4y + 4z = 9?
Solution:
Equation of given plane is:
2x + 4y + 4z = 9 …………………… (1)
The direction ratios of normal to the given plane are 2, 4, 4
Hence the direction cosines of the normal are

Question 21.
Find the equation of plane on which the length of perpendicular drawn from origin is 5 unit and the direction cosines of normal to it are \(\frac { 1 }{ \sqrt { 3 } } \), \(\frac { 1 }{ \sqrt { 3 } } \), – \(\frac { 1 }{ \sqrt { 3 } } \)?
Solution:
Equation of plane in normal form
lx + my + nz = p
Given: l = \(\frac { 1 }{ \sqrt { 3 } } \), m = \(\frac { 1 }{ \sqrt { 3 } } \), n = – \(\frac { 1 }{ \sqrt { 3 } } \), p = 5
Putting in eqn. (1),
\(\frac { 1 }{ \sqrt { 3 } } \) x + \(\frac { 1 }{ \sqrt { 3 } } \) y – \(\frac { 1 }{ \sqrt { 3 } } \) z = 5
⇒ x + y – z = 5\(\sqrt{3}\)

Question 22.
Find the equation of plane on which length of perpendicular drawn from origin is 4 units and whose direction cosines are in proportional to 2, -3, 6?
Solution:
Let the equation of plane
lx + my + nz = p
Given:
p = 4, a = 2, b = -3, c = 6

Putting in eqn. (1),
\(\frac{2}{7}\) x – \(\frac{3}{7}\) y + \(\frac{6}{7}\) z = 4
⇒ 2x – 3y + 6z = 28.

Question 23.
Find the equation of plane on which length of perpendicular drawn from origin is 5 unit and the direction ratios of normal to the plane are 2, 3, 6.
Solution:
Solve as Q. No. 22.
[Answer: 2x + 3y + 6z = 35]

Question 24.
Find the equation of the plane which passes through the point (1, – 2, 3) and perpendicular to line whose direction ratios are 2, 1, -1?
Solution:
We know that equation of the plane passing through a point (x₁, y₁, z₁) is
a (x – x₁) + b( y – y₁) + c (z – z₁) = 0
Where, a, b and c are the direction ratios of normal to the plane
Here the point (x₁, y₁, z₁) is (1, -2, 3)
and the direction ratios a, b, c are 2, 1, -1. Therefore by eqn. (1) we get
2. (x – 1) + 1. ( y + 2) -1. (z – 3) = 0
⇒ 2x + y – z – 2 + 2 + 3 = 0
⇒ 2x + y – z + 3 = 0
Which is the required equation of the plane.

Question 25.
(A) Find the perpendicular distance of the plane 6x – 3y + 2z – 14 = 0 from origin?
Solution:
Equation of plane,
6x – 3y + 2z – 14 = 0
The length of perpendicular drawn from (0, 0, 0) to the plane

(B) Find the length of perpendicular drawn from (1, 2, 0) on the plane 4x + 3y + 12z + 16 = 0?
Solution:
Equation of plane,
4x + 3y + 12z + 16 = 0
The length of perpendicular drawn from (1, 2, 0) to the plane

(C) Find the length of perpendicular drawn from (7, 14, 5) to the plane 2x + 4y – z?
Solution:
Solve as Q.No. 25 (B)
Answer:
3\(\sqrt{21}\) unit.

Question 26.
Find the equation of plane which passes through the point (- 1, 2, 3) and parallel to the plane 3x + 4y – 5z = 52?
Solution:
The given plane is
3x + 4y – 5z = 52 …………………… (1)
A plane parallel to the given plane (1),
3x + 4y – 5z = λ ………………(2)
∵ Plane (2) passes through the point (-1, 2, 3)
∴ 3(-1) + 4(2) – 5(3) = λ
⇒ -3 + 8 – 15 = λ
⇒ -18 – 8 = λ
⇒ -10 = λ
∴ From eqn. (2), putting the value of λ
3x + 4y – 5z = -10
∴ The required plane is
3x + 4y – 5z + 10 = 0

Question 27.
(A) Find the intercepts made by the plane 3x + 4y – 7z = 84 from the co – ordinate axes?
Solution:

(B) Find the intercepts made by the plane 3x + 4y – 6z = 72 from the co – ordinate axes?
Solution:
Solve as Q. No. 27 (A).
[Answer: (24, 18, -12).]

Question 28.
Find the equation of plane parallel to the X – axis and cuts intercepts 5 and 7 from Y and Z – axis respectively?
Solution:
Let the equation of plane
\(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 1
The plane (1) is parallel to X – axis
a = ∞
Given b = 5, c = 7
∴ \(\frac{x}{∞}\) + \(\frac{y}{5}\) + \(\frac{z}{7}\) = 1
⇒ \(\frac{y}{5}\) + \(\frac{z}{7}\) = 1 [∵\(\frac{x}{∞}\) = 0]
⇒ 7y + 5z = 35.

Question 29.
Find the equation of the plane passing through the points (0, 0, 0) and perpendicular to the planes x + 2y – z = 1 and 3x – 4y + z = – 5?
Solution:
Equation of the plane passing through point (0,0,0)
a(x – 0) + b(y – 0) + c(z – 0) = 0
⇒ ax + by + cz = 0 ………………………… (1)
Plane (1) is perpendicular to the given planes.
x + 2y – z = 1 and 3x – 4y + z = -5
a + 2b – c = 0 ………………….. (2)
3a – 4b + c = 0 ………………………….. (3)
Solving eqns. (2) and (3),
\(\frac { a }{ 2-4 } \) = \(\frac { -b }{ 1+3 } \) = \(\frac { c }{ -4-6 } \)
⇒ \(\frac { a }{ -2 } \) = \(\frac { -b }{ 4 } \) = \(\frac { c }{ -10 } \)
⇒ \(\frac{a}{1}\) = \(\frac{b}{2}\) = \(\frac{c}{5}\) = k (say)
∴ a = k, b = 2k and c = 5k, where k ≠ 0.
∴ From eqn. (1),
k [x + 2y + 5z] = 0
∴ x + 2y + 5z = 0.

Question 30.
Find the equation of the plane parallel to X – axis and passes through the points (2, 3, -4) and (1, -1, 3)?
Solution:
Let the equation of the plane parallel to X – axis is
by + cz + d= 0 ………………………. (1)
It passes through the points (2, 3, – 4) and (1, -1, 3), then
3b – 4c + d = 0 ……………………….. (2)
and – b + 3c + d = o …………………….. (3)
Solving eqns. (2) and (3), we get
\(\frac { b }{ -4-3 } \) = \(\frac { c }{ -1-3 } \) = \(\frac { d }{ 9-4 } \)
⇒ \(\frac{b}{7}\) = \(\frac{c}{4}\) = \(\frac{d}{-5}\) = k (say)
∴ b = 7k, c = 4k, d = -5k
Put the values of b, c and d in eqn. (1), we get
7ky + 4kz – 5k = 0
⇒ 7y + 4z – 5 = 0
This is the required equation of the plane.

Question 31.
Prove that the distance between two parallel planes 2x – 2y + z + 3 = 0 and 4x – 4y + 2z + 5 = 0 is \(\frac{1}{6}\)?
Solution:
The equations of the plane are 4x – 4y + 2z + 5 = 0
2x – 2y + z + 3 = 0 ……………… (1)
and 4x – 4y + 2z + 5 = 0 ………….. (2)
The length of the perpendicular drawn from O (0, 0, 0) to the plane (1) is

and the length of the perpendicular drawn from O (0, 0, 0) to the plane (2) is

∴ The required distance MN = P₁ – P₂
= 1 – \(\frac{5}{6}\) = \(\frac{6-5}{6}\) = \(\frac{1}{6}\). Proved.

Question 32.
(A) Find the equation of the plane passes through the intersecting line of the planes x + y + z – 6 = 0 and 2x + 3y + 4z + 5 = 0 and also through the point (1, 1, 1)?
Solution:
The given planes are
x + y + z – 6 = 0 …………………. (1)
and 2x + 3y + 4z + 5 = 0 ………………….. (2)
The equation of the plane passes through the intersecting line of the planes (1) and (2), is
(x – y + z – 6) + λ (2x + 3y + 4z + 5) = 0 ………………….. (3)
Because plane (3) passes through the point (1, 1, 1), then
(1 + 1 + 1 – 6) + (2 + 3 + 4 + 5) = 0
⇒ – 3 + λ (14) = 0
⇒ λ = \(\frac{3}{14}\)
Putting the value of λ in eqn.(3),
(x + y + z – 6) + \(\frac{3}{14}\) (2x + 3y + 4z + 5) = 0
⇒ 20x + 23y + 26z – 69 = 0.

(B) Find the equation of the plane passing through the intersecting line of the planes x + 2y + 3z = 5 and 2x – 4y + z = 3 and also through the point (0, 1, 0)?
Solution:
Solve as Q.No. 32. (A).
[Answer: 3x – 2y + 4z + 2 = 0.]

Question 33.
(A) Find the vector equation of a plane passing through the point 2\(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \) and perpendicular to the vector 6\(\hat { i } \) + 2\(\hat { j } \) – 3\(\hat { k } \)?
Solution:
Let the equation of plane be

(B) Find the vector equation of the plane which is at a distance 7 units from origin and perpendicular to the vector 4\(\hat { i } \) + 2\(\hat { j } \) – 3\(\hat { k } \)?
Solution:
Let the equation of plane be

(C) Find the distance of the point (2, – 1, 3) from the plane \(\vec { r } \). (3\(\hat { i } \) + 2\(\hat { j } \) – 6\(\hat { k } \) ) + 15 = 0?
Solution:

(D) Find the distance from the point (2\(\hat { i } \) – \(\hat { j } \) – 4\(\hat { k } \) ) to the plane \(\vec { r } \). (3\(\hat { i } \) – 4\(\hat { j } \) + 12\(\hat { k } \) ) = 19?
Solution:

Question 34.
Find the angle between the planes \(\vec { r } \). (2\(\hat { i } \) – 3\(\hat { j } \) + 4\(\hat { k } \) ) = 1 and \(\vec { r } \) (- \(\hat { i } \) + \(\hat { j } \) ) = 4?
Solution:

Three Dimensional Geometry Long Answer Type Questions – I

Question 1.
Find the perpendicular distance of line \(\frac{x}{1}\) = \(\frac{y-1}{2}\) = \(\frac{z-2}{3}\) from the point (1, 6, 3)?
Solution:

Question 2.
Find the distance between two parallel lines \(\frac { x-1 }{ 2 } \) = \(\frac { y-2 }{ 3 } \) = \(\frac { z-3 }{ 4 } \) and \(\frac { x-2 }{ 4 } \) = \(\frac { y-3 }{ 6 } \) = \(\frac { z-4 }{ 8 } \)?
Solution:
Given lines:
\(\frac { x-1 }{ 2 } \) = \(\frac { y-2 }{ 3 } \) = \(\frac { z-3 }{ 4 } \) …………………… (1)
and \(\frac { x-2 }{ 4 } \) = \(\frac { y-3 }{ 6 } \) = \(\frac { z-4 }{ 8 } \) ………………………… (2)
Any point P (1, 2, 3) lies on line (1). Now we find the length of perpendicular PM drawn from point P (1, 2, 3) to eqn. (2).
∵Eqn. (2) passes through point A (2, 3,4)

Question 3.
Find the equation plane passing through the line \(\frac { x-3 }{ 2 } \) = \(\frac { y+2 }{ 3 } \) = \(\frac { z-4 }{ -1 } \) and point (-6, 3, 2)?
Solution:
Let the equation of plane be
A(x – α) + B(y – β) + C(z – γ) = 0
Passes through line \(\frac { x-3 }{ 2 } \) = \(\frac { y+2 }{ 3 } \) = \(\frac { z-4 }{ -1 } \)
∴ A(x – 3) + B(y + 2) + C(z – 4) = 0 ………………………… (1)
Plane passes through point (-6, 3, 2)
∴ A(-6 – 3) + B(3 + 2) + C(2 – 4) = 0
⇒ -9A + 5B – 2C = 0
⇒ 9A – 5B + 2C = 0 …………………………… (2)
Direction ratio of plane →A, B, C
2A + 3B – C = 0
Solving eqns. (2) and (3)
9A – 5B + 2C = 0
2A + 3B – C = 0
\(\frac { A }{ 5-6 } \) = \(\frac { B }{ 4+9 } \) = \(\frac { C }{ 27+10 } \)
⇒ \(\frac { A }{ -1 } \) = \(\frac { B }{ 13 } \) = \(\frac { C }{ 37 } \)
Putting in eqn. (1)
-1(x – 3) + 13(y + 2) + 37(z – 4) = 0
⇒ -x + 3 + 13y + 26 + 37z – 148 = 0
⇒ -x + 13y + 37z – 119 = 0
⇒ x – 13y – 37z + 119 = 0.

Question 4.
Find the shortest distance between the straight lines \(\frac { x-3 }{ 3 } \) = \(\frac { y-8 }{ -1 } \) = \(\frac { z-3 }{ 1 } \) and \(\frac { x
+3 }{ -3 } \) = \(\frac { y+7 }{ 2 } \) = \(\frac { z-6 }{ 4 } \)?
Solution:
Here, x₁ = 3, y₁ = 8, z₁ = 3, x₂ = -3, y₂ = -7, z₂ = 6
a₁ = 3, b₁ = -1, c₁ = 1, a₂ = -3, b₂ = 2, c₂ = 4
Given lines are
\(\frac { x-3 }{ 3 } \) = \(\frac { y-8 }{ -1 } \) = \(\frac { z-3 }{ 1 } \) and \(\frac { x+3 }{ -3 } \) = \(\frac { y+7 }{ 2 } \) = \(\frac { z-6 }{ 4 } \)
We know that the shortest distance between 2 lines \(\frac { x-x_{ 1 } }{ a_{ 1 } } \) = \(\frac { y-y_{ 1 } }{ b_{ 1 } } \) = \(\frac { z-z_{ 1 } }{ c_{ 1 } } \) and \(\frac { x-x_{ 2 } }{ a_{ 2 } } \) = \(\frac { y-y_{ 2 } }{ b_{ 2 } } \) = \(\frac { z-z_{ 2 } }{ c_{ 2 } } \) is

Question 5.
Find the coordinates of point of intersection of the lines \(\frac { x-1 }{ 2 } \) = \(\frac { y-2 }{ 3 } \) = \(\frac { z-3 }{ 4 } \) and \(\frac { x-2 }{ 3 } \) = \(\frac { y-3 }{ 4 } \) = \(\frac { z-4 }{ 5 } \)?
Solution:
Let \(\frac { x-1 }{ 2 } \) = \(\frac { y-2 }{ 3 } \) = \(\frac { z-3 }{ 4 } \) = r
∴ x = 2r + 1, y = 3r + 2, z = 4r + 3
Suppose the above point is intersection point then it will satisfy line

Put in eqn.(1), we get
x = 2(-1) + 1, y = 3(-1) + 2, z = 4(-1) + 3
x = -1, y = -1, z = -1
∴ Intersection point (-1, -1, -1).

Question 6.
Find the coordinates of point of intersection of the lines x – 3 = \(\frac { y+4 }{ -3 } \) = \(\frac { z-5 }{ 3 } \) and x – 4 = \(\frac { y-5 }{ 3 } \) = \(\frac { z+6 }{ -4 } \)?
Solution:
Given lines are \(\frac { x-3 }{ 1 } \) = \(\frac { y+4 }{ -3 } \) = \(\frac { z-5 }{ 3 } \) ……………….. (1)
and \(\frac { x-4 }{ 1 } \) = \(\frac { y-5 }{ 3 } \) = \(\frac { z+6 }{ -4 } \) ……………………… (2)
x₁ = 3, y₁ = -4, z₁ = 5, l₁ = 1, m₁ = -3, n₁ = 3
x₂ = 4, y₂ = 5, z₂ = -6, l₂ = 1, m₂ = 3, n₂ = -4
If lines (1) and (2) are intersect, then

⇒ 1(12 – 9) -1(-36 + 33) + 1(27 – 33) = 0
⇒ 3 + 3 – 6 = 0
⇒ 0 = 0
Hence the lines (1) and (2) are intersect
Again let \(\frac { x-3 }{ 1 } \) = \(\frac { y+4 }{ 3 } \) = \(\frac { z-5 }{ 3 } \) = r
∴ x = r + 3, y = -3r – 4, z = 3r + 5 ……………………………… (3)
Suppose above point is point of intersection hence it will satisfy eqn. (2)

Putting r = -1 in eqn. (3),
We get x = -1 + 3, y = -3(-1) – 4, z = 3(-1) + 5
x = 2, y = -1, z = 2
∴Intersection point (2, -1, 2).

Question 7.
Find the shortest distance between the lines:
\(\vec { r } \) = \(\hat { i } \) + 2\(\hat { j} \) + 3\(\hat { k } \) + t(2\(\hat { i } \) + 3\(\hat { j } \) + 4\(\hat { k } \) ) and \(\vec { r } \) = 2\(\hat { i } \) + 4\(\hat { j } \) + 5\(\hat { k } \) + s(3\(\hat { i } \) + 4\(\hat { j } \) + 5\(\hat { k } \) )?
Solution:

Question 8.
Prove that the lines \(\vec { r } \) = ( \(\hat { i } \) + \(\hat { j } \) – \(\hat { k } \) ) + λ(3\(\hat { i } \) r =(3\(\hat { i } \) – \(\hat { j } \) ) and \(\vec { r } \)  = (4\(\hat { i } \) – \(\hat { k } \) ) + μ (2\(\hat { i } \) + 3\(\hat { k } \) are intersect to each other. Also And the point of intersection?
Solution:
Given lines are

Equating the above lines (1) and (2)

and two lines intersect each other, if


1 + 3λ = 4 + 2μ …………….. (3)
1 – λ = 0 ………………….. (4)
and -1 = 3μ – 1 ………………… (5)
From eqn. (4) we get
1 – λ = 0 ⇒ λ = 1
From eqn.(5) we get λ and μ in eqns. (1) and (2) respectively, we get
μ = 0
The above values of

Question 9.
(A) Find the shortest distance between the lines:
\(\vec { r } \) = (3 – t) \(\hat { i } \) + (4 + 2t) \(\hat { j } \) + (t – 2) \(\hat { k } \)
and \(\vec { r } \) = (1 + s) \(\hat { i } \) + (3s – 7) \(\hat { j } \) + (2s – 2) \(\hat { k } \)?
Solution:

(B) Find the shortest distance between the lines:
\(\vec { r } \) = (λ – 1) \(\hat { i } \) + (λ + 1) \(\hat { j } \) + (1 + λ) \(\hat { k } \)
\(\vec { r } \) = (1 – µ) \(\hat { i } \) + (2µ – 1) \(\hat { j } \) + (µ + 2) \(\hat { k } \)?
Solution:
Solve as Q.No. 9(A)
[Answer: S.D. = \(\frac { 5 }{ \sqrt { 2 } } \) ]

(C) Find the shortest distance between the lines whose vector equations are
\(\vec { r } \) = (1 + 2λ) \(\hat { i } \) + (2 + 3λ) \(\hat { j } \) + (3 + 4λ) \(\hat { k } \)
\(\vec { r } \) = (2 + 3µ) \(\hat { i } \) + (3 + 4µ) \(\hat { j } \) + (4 + 5µ) \(\hat { k } \)?
Solution:

Question 10.
Find the equation of plane passing through the line of intersection of the planes x + 3y + 4z – 5 = 0 and 3x – 4y + 9z – 10 = 0 and perpendicular to the plane x + 2y = 0?
Solution:
Given equation of planes are
x + 3y + 4z – 5 = 0 ………………….. (1)
and 3x – 4y + 9z – 10 = 0 ………………………. (2)
The plane passing through the line of intersection of the planes (1) and (2)
(x + 3y + 4z – 5) + λ(3x – 4y + 9z – 10) = 0
⇒ (1 + 3λ) x + (3 – 4λ) y + (4 + 9λ) z – 5 – 10λ = 0 ………………… (3)
Again, given plane is
x + 2y = 0 ……………………… (4)
Plane (3) is perpendicular to the plane (4)
(1 + 3λ).1 + (3 – 4λ).2 + (4 + 9λ). 0 = 0
⇒ 1 + 3λ + 6 – 8λ = 0
⇒ λ = \(\frac{7}{5}\)
Putting the value of X in eqn. (3),
(x + 3y + 4z – 5) + \(\frac{7}{5}\) (3x – 4y + 9z – 10) = 0
⇒ 26x – 13y + 83z = 95.

Question 11.
Find the equation of planes which are parallel to the plane x – 2y + 2z = 3 and whose perpendicular distance from the point (1, 2, 3) is 1?
Solution:
The planes parallel to the given plane x – 2y + 2z = 3 is
x – 2y + 2z + λ = 0 ………………….. (1)
Perpendicular distance from the point (1, 2, 3) to the above plane is

Then substitute the value of 2 in eqn. (1), we have
x – 2y + 2z = 0, x – 2y + 2z = 6

Question 12.
(A) Find the equation of the plane passing through the points (1,-2, 4) and (3, -4, 5) and perpendicular to the plane x + y – 2z = 6?
Solution:
Equation of the plane passing through the point (1, -2, 4) is
A(x – 1) + B(y + 2) + C(z – 4) = 0 ………………………. (1)
It also passes through the point (3, -4, 5)
A(3 – 1) + 5(-4 + 2) + C(5 – 4) = 0
⇒ 2A – 2B + C = 0 …………………………. (2)
The given equation of plane is
x + y – 2z = 6 ……………………. (3)
The planes (1) and (3) are perpendicular
A + B – 2C = 0 ……………………. (4)
On solving by cross multiplication method, we get
\(\frac { A }{ 4-1 } \) = \(\frac { B }{ 1+4 } \) = \(\frac { C }{ 2+2 } \)
⇒ \(\frac{A}{3}\) = \(\frac{B}{5}\) = \(\frac{C}{4}\) = k
∴ A = 3k, B = 5k, C = 4k
Putting these values of A, B and C in eqn. (1), we get
⇒ 3k(x – 1) + 5k(y + 2) + 4k(z – 4) = 0
⇒ 3x – 3 + 5y + 10 + 4z – 16 = 0
⇒ 3x + 5y + 4z – 9 = 0.

(B) Find the equation of plane passing through the points (1, 1,-1) and (1, 1,-1) and perpendicular to the planes x + 2y + 2z = 9?
Solution:
Solve as Q.No. 12.(A)
[Answer: 2x + 2y – 3z + 3 = 0]

Question 13.
(A) Find the equation of plane passing through (1, 1, -1) and perpendicular to the planes x + 2y + 3z = 0 and 2x – 3y + 4z = 0?
Solution:
Let the equation of planes
A(x – x₁) + B(y – y₁) + C(z – z₁) = 0
It passes through (1, 1, -1)
A(x – 1) + B(y – 1) + C(z + 1) = 0 …………………. (1)
Given planes are
x + 2y + 3z = 0 …………………… (2)
2x – 3y + 4z = 0 ………………………. (3)
The plane (1) is perpendicular to eqns. (2) and (3),
∴ 1A + 2B + 3C = 0
and 2A – 3B + 4C = 0
Solving \(\frac { A }{ 8+9 } \) = \(\frac { B }{ 6-4 } \) = \(\frac { C }{ -3-4 } \)
⇒ \(\frac{A}{17}\) = \(\frac{B}{2}\) = \(\frac{C}{-7}\) = k
∴ A = 17k, B = 2k, C = -7k
By eqn. (1),
17k (x – 1) + 2k (y – 1) – 7k (z + 1) = 0
⇒ 17x + 2y – 7z – 26 = 0.

(B) Find the equation of plane passing through (2, 1, 4) and perpendicular to the planes 9x – 7y + 6z + 48 = 0 and x + y – z = 0?
Solution:
Solve as Q.No. 13 (A)
[Answer: x + 15y + 16z = 81]

Question 14.
Find the equation of the plane passing through the points (2, 2, -1), (3, 4, 2) and (7, 0, 6)?
Solution:
Equation of Plane passing through the points (2, 2, -1) is
A(x – 2) + B(y – 2) + C(z + 1) = 0 ………………….. (1)
Since plane given by eqn. (1) passes through the points (3,4, 2) and (7, 0, 6)
Hence it will satisfy eqn. (2),
We get A(3 – 2) + B(4 – 2) + C(2 + 1) = 0
A + 2B + 3C = 0 ………………………. (2)
and A(7 – 2) + B(0 – 2) + C(6 + 1) = 0
5A – 2B + 7C = 0 …………………….. (3)
Solving eqns. (2) and (3),
img 39
\(\frac { A }{ 14+6 } \) = \(\frac { B }{ 15-7 } \) = \(\frac { C }{ -2-10 } \)
\(\frac { A }{ 20 } \) = \(\frac { B }{ 15-7 } \) = \(\frac { C }{ -2-10 } \)
\(\frac { A }{ 20 } \) = \(\frac { B }{ 8 } \) = \(\frac { C }{ -12 } \)
Let \(\frac { A }{ 5 } \) = \(\frac { B }{ 2 } \) = \(\frac { C }{ -3 } \) = k
A = 5k, B = 2k, C = -3k
Let \(\frac{A}{5}\) = \(\frac{B}{2}\) = \(\frac{C}{-3}\) = k
A = 5k, B = 2k, C = -3k
Putting in eqn.(1),
We get 5k(x – 2) + 2k(y – 2) + (-3k) (z + 1) = 0
k (5x + 2y – 3z – 17) = 0
5x + 2y – 3z – 17 = 0.

Question 15.
prove that the points (0, -1, -1), (4, 5, 1), (3, 9, 4) and (-4, 4, 4) are coplanar?
Solution:
Equation of plane passing through the points (0, -1, -1)
A(x – 0) + B(y + l) + C(z + 1) = 0 ………………………… (1)
It passes through the point (4, 5, 1)
A(4 – 0) + B(5 +1) + C(1 +1) = 0
⇒ 4A + 6B + 2C = 0
⇒ 2A + 3B + C = 0 ………………………………. (2)
Eqn. (1) also passes through (3, 9, 4),
A(3 – 0) + B(9 + 1) + C(4 + 1) = 0
⇒ 3A + 10B + 5C = 0 …………………………. (3)
Solving eqns. (2) and (3),
\(\frac { A }{ 15-10 } \) = \(\frac { B }{ 3-10 } \) = \(\frac { C }{ 20-9 } \) = k (let)
⇒\(\frac { A }{ 5 } \) = \(\frac { B }{ -7 } \) = \(\frac { C }{ 11 } \) = k
⇒ A = 5k, B = -7k, C = 11k
Putting in eqn. (1),
5k – 7k (y + 1) + 11k (z + 1) = 0
⇒ k(5x – 7y + 11z + 4) = 0
⇒ 5x – 7y + 11z + 4 = 0 ………………………… (4)
If the plane also passes through the point (-4, 4, 4) will satisfies eqn. (4),
5(- 4) – 7(4) + 11 (4) + 4 = 0
⇒ -20 – 28 + 44 + 4 = 0
⇒ 0 = 0
Hence given points are coplanar. Proved.

Question 16.
A variable plane \(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 1 is at a distance 1 unit from origin. It cuts co – ordinate axes at A, B, C. The centroid (x, y, z) satisfies the equation \(\frac { 1 }{ x^{ 2 } } \) + \(\frac { 1 }{ y^{ 2 } } \) + \(\frac { 1 }{ z^{ 2 } } \) = k? Find the value of k?
Solution:
Given equation of plane \(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 1
OA = a, OB = b, OC = c
The coordinates of the point A, B, C are (a, 0, 0), (0, b, 0) and (0, 0, c).
The length of perpendicular drawn from origin to the plane (1) is 1.

Question 17.
Find the equation of plane passing through the line of intersection of the planes x + 2y + 3z = 4 and 2x + y – z + 5 = 0 and perpendicular to the plane 5x + 3y + 6z + 8 = 0?
Solution:
Given planes
x + 2y + 3z = 4 ………………….. (1)
2x + y – z + 5 = 0 ………………… (2)
Equation of plane passing through the line of intersection of the planes (1) and (2)
x + 2y + 3z – 4 + λ(2x + y – z + 5) = 0 …………………. (3)
⇒ (1 + 2λ)x + (2 + λ)y + (3 – λ)z = 0
⇒ 10λ + 3λ – 6λ + 5 + 6 + 18 = 0
⇒ 7λ + 29 = 0
⇒ λ = \(\frac{-29}{7}\)
Putting the value of A in eqn. (3),
x + 2y + 3z – 4 – \(\frac{-29}{7}\) (2x + y – z + 5) = 0
⇒ 7x + 14y + 21z – 28 – 58x – 29y + 29z – 145 = 0
⇒ -51x – 15y + 50z – 173 = 0
⇒ 51x + 15y – 50z + 173 = 0.

Question 18.
Find the equation of plane passes through the line \(\frac { x-3 }{ 2 } \) = \(\frac { y+2 }{ 9 } \) = \(\frac { z-4 }{ -1 } \) and point (-6, 3, 2)?
Solution:
Point on given line D (3, -2,4)
∴ Eqn. of plane passing through point (3, -2, 4) is
A(x – 3) + B(y + 2) + C(z – 4) = 0 ……………………….. (1)
Plane (1) passes through point (-6, 3, 2),
– 9A + 5B – 2C = 0 …………………………. (2)
Direction ratio’s of given line is 2, 9, -1 and is parallel to plane (1),
2A + 9B – C = 0 ………………………. (3)
Now,
– 9A + 5B – 2C = 0
2A + 9B – C = 0
\(\frac { A }{ -5+18 } \) = \(\frac { B }{ -4-9 } \) = \(\frac { C }{ -81-10 } \)
⇒ \(\frac { A }{ 13 } \) = \(\frac { B }{ -13 } \) = \(\frac { C }{ -91 } \) = k
⇒ \(\frac { A }{ 1 } \) = \(\frac { B }{ -1 } \) = \(\frac { C }{ -7 } \) = k
A = k, B = -k, C = – 7k
Putting in eqn.(1), we get,
k(x – 3) – k (y + 2) – 7k(z – 4) = 0
⇒ x – y – 7z – 3 – 2 + 28 = 0
⇒ x – y – 7z + 23 = 0.

Question 19.
Find the vector equation of the line passing through (1,2,3) and parallel to the planes \(\vec { r } \). ( \(\hat { i } \) – \(\hat { j } \) + 2\(\hat { k } \) = 5 and \(\vec { r } \).(3\(\hat { i } \) + \(\hat { j } \) + \(\hat { k } \) ) = 6? (NCERT)
Solution:
Equation of line is:

Question 20.
Find the vector equation of the line passing through the point (1, 2, -4) and perpendicular to the two lines
\(\frac { x-8 }{ 3 } \) = \(\frac { y+19 }{ -16 } \) = \(\frac { z-10 }{ 7 } \) and \(\frac { x-8 }{ 3 } \) = \(\frac { y-20 }{ 8 } \) = \(\frac { z-5 }{ -5 } \)? (NCERT)
Solution:

Question 21.
A line makes angles α, β, γ and and with four diagonals of a cube then prove that:
cos²α + cos²β + cos² = \(\frac{4}{3}\)
Solution:
Taking three adjacent edges OA, OB, OC of the cube of side a as coordinate the coordinates of vertices of the cube are :
0 (0,0,0), A (a,0,0), B (0,a,0), R (0,0,a),
D (a,a,0), K (a,0,a), L (0,a,a), P (a,a,a)
Direction ratio of diagonal are a – 0, a – 0, a – 0, a, a, a direction cosines of OP are:

MP Board Class 12 Maths Important Questions

MP Board Class 12th Biology Important Questions Chapter 14 Ecosystem

Ecosystem Important Questions

Ecosystem Objective Type Questions 

Question 1.
Choose the correct answer:

Question 1.
The flow of energy in ecosystem: (MP2011)
(a) Unidirection
(b) Bidirectional
(c) Tridirectional
(d) Four directional
Answer:
(a) Unidirection

Question 2.
Chief source of energy in ecosystem :
(a) Solar energy
(b) Green plants
(c) Food substances
(d) All of these.
Answer:
(a) Solar energy

Question 3.
The term ‘ecosystem’ was used for the first time by : (MP 2015)
(a) Tansley
(b) Odum
(c) Reiter
(d) Mishra and Puri.
Answer:
(a) Tansley

Question 4.
The pyramid of number of tree ecosystem is :
(a) Inverted
(b) Upright
(c) Both (a) and (b)
(d) None of these.
Answer:
(a) Inverted

Question 5.
Correct food chain is :
(a) Grass → Grasshopper → Frog → Snake → Hawk
(b) Grass → Frog → Snake → Peacock
(c) Grass → Peacock → Grasshopper → Hawk
(d) Grass → Snake → Rabbit.
Answer:
(a) Grass → Grasshopper → Frog → Snake → Hawk

Question 6.
Pyramid of biomass of lake ecosystem is :
(a) Upright
(b) Inverted
(c) Upright or inverted
(d) None of these.
Answer:
(b) Inverted

Question 7.
Vegetation present between two communities :
(a) Ecad
(b) Ecotype
(c) Ecotone
(d) None of these.
Answer:
(c) Ecotone

Question 8.
Xerosere is started from :
(a) Water
(b) Naked rock
(c) Swamp
(d) All of these.
(b) Naked rock

Question 9.
Serial development of plant community is called :
(a) Attack
(b) Succession
(c) Both (a) and (b)
(d) None of these.
Answer:
(b) Succession

Question 10.
Plants starting succession in any area are called :
(a) Pioneer
(b) Sere
(c) Displacer
(d) All of these.
Answer:
(a) Pioneer

Question 11.
In forest ecosystem, the pyramid of energy is : (MP 2012,17)
(a) Always inverted
(b) Always upright
(c) First upright then inverted
(d) None of these.
Answer:
(b) Always upright

Question 12.
The study of ecology of a species is called :
(a) Ecology
(b) Autecology
(c) Synecology
(d) None of these.
Answer:
(b) Autecology

Question 13.
Food chain starts from :
(a) Respiration
(b) Photosynthesis
(c) Decomposers
(d) Nitrogen fixation.
Answer:
(b) Photosynthesis

Question 14.
Man is:
(a) Autotrophic
(b) Carnivorous
(c) Herbivorous
(d) Omnivorous.
Answer:
(d) Omnivorous.

Question 15.
In any ecosystem, solar energy is conserved by :
(a) By producers
(b) By consumers
(c) By decomposers
(d) All of the above.
Answer:
(a) By producers

Question 16.
The two components of ecosystem are: (MP 2013)
(a) Organism and Plant
(b) Weeds and trees
(c) Frog and man
(d) Biotic and Abiotic.
Answer:
(d) Biotic and Abiotic.

Question 2.
Fill in the blanks:

1. The transitional zone present between to adjacent communities is called ………………….
2. All the plants of a particular area constitute …………………. of that place.
3. Only …………………. % energy is transferred from one trophic level to another.
4. Pyramid of …………………. is always upright. (MP 2013)
5.  …………………. and …………………. are the examples of omnivorous animals.
6. The term ‘ecosystem’ was proposed by ………………….
7. The chief source of energy ecosystems is (MP 2016)
8. N₂ fixing bacteria is called …………………. (MP 2009 Set A)
9. All ecosystems are depended for energy on …………………. (MP 2009 Set D)
10. In forest trees basically function as ………………….
11. Establishment of organisms in new habitat is called ………………….
12. A sereal development of plant community is called ………………….
13. Succession which takes place in baren rock is called ………………….
14. Succession in soil is called ………………….

Answer:

1. Ecotone
2. Flora
3. 10
4. Energy
5. Man, Pig
6. Tansley
7. Sunlight
8. Nitrifying bacteria
9. Sun (solar energy)
10. Producers
11. Ecesis
12. Succession
13. Lithosere
14. Psamosere.

Question 3.
Match the followings:
I.

Answer:

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

II.

Answer:

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

Question 4.
Write the answer in one word/sentances:

1. What is called the study of ecology of a species?
2. What is called the process of soil formation?
3. What type of energy pyramid found in nature?
4. Give one example of an omnivorous animal.
5. Who proposed the term ecosystem?
6. Who gave the 10% rule of ecosystem?
7. In an ecosystem, producer is known as.
8. What is the direction of the flow of energy in an ecosystem?
9. When many food chains operate simultaneously and interlock such patterns is termed as?
10. Give an example of autotrophic componant.
11. Which gas is used by green plants in photosynthesis?
12. Name the special nutrient which is present in metheonine amino acid.
13. Name the cycle in which nitrogen is converted into multiple chemicals and circulated among atmosphere.
14. Due to an ecosystem fungus and bacteria are known as.
15. What is T₂?

Answer:

1. Autecology
2. Pedogenesis
3. Upright
4. Man
5. A.G. Tansley
6. Lindman
7. Biotic componant
8. Leanear
9. Food web
10. Plants
11. CO₂
12. Sulfer
13. Nitrogen cycle
14. Decomposer
15. Herbivorous animal.

Ecosystem Very Short Answer Type Questions

Question 1.
Write the name and ratio of different components of biosphere.
Answer:
Name and ratio of different components of biosphere is:

Some gases are also found in biosphere example Helium, Neyon and Crypton are found in less quantity.

Question 2.
Differentiate between detritivore and decomposer.
Answer:
Detritivore are organisms which feed on detritus and break them into smaller particles, example earth worm. And decomposer are organisms which by secreting enzymes break down complex organic matter into in organic substance example some bacteria and fungi.

Question 3.
Explain consumers of ecosystem
Answer:

1. Producers – All the green plants.

2. Consumers – Depends on others for food.

• Primary consumer: Depends on plants called herbivores.
• Secondary consumers : Depends on herbivores for food.
• Tertiary consumers : Depends on secondary consumers.

3. Decomposers – They decomposed dead organic matter.

Question 4.
Differentiate between biome and ecosystem.
Answer:
An ecosystem is the interaction of living and non – living things in an environment. A biome is a specific geographic area notable for the species living there.

Question 5.
What are transducers according to some ecologists?
Answer:
Some ecologists call green plants as transducers of the ecosystem as they convert solar energy into chemical energy.

Question 6.
Name four submerged plants.
Answer:

1. Hydrilla
2. Vallisneria
3. Elodea
4. Potamogeton.

Question 7.
Name the stages of xerosere.
Answer:
Stages of xerosere:

1. Crustose lichen stage
2. Foliose lichen stage
3. Moss stage
4. Herb stage
5. Shrub stage
6. Climax forest stage.

Question 8.
When many food chain operate simultaneously and interlock such pattern is formed.
Answer:
Food web.

Question 9.
Name the ecosystem which shows most productivity.
Answer:
Tropical ecology.

Question 10.
What are fungi and bacteria called in an ecosystem?
Answer:
Micro – consumer or Decomposer.

Question 11.
Which part of the energy is transferred from one trophic level to other in ecosystem?
Answer:
10%.

Question 12.
Name the type of chemosynthetic bacteria.
Answer:
Autotrophic.

Question 13.
Name the word which is similar to ecosystem given by Prof. R. Mishra.
Answer:
Ecocosm.

Question 14.
Name the trophic level in which green plants are found.
Answer:
Primary trophic level.

Question 15.
Who gave the word transformer for producer?
Answer:
E.J. Kormondy.

Question 16.
Name any two sedimentary cycle.
Answer:

Phosphorus cycle
Sulphur cycle.

Question 17.
The energy pyramids are always.
Answer:
Upright.

Question 18.
Give examples of decomposers.
Answer:
Bacteria and Fungi.

Question 19.
Who gave 10% rule of energy?
Answer:
Lindeman.

Question 20.
Which form of nitrogen is absorbed by plants?
Answer:
In the form of nitrate ion (NO₃^–)

Ecosystem Short Answer Type Questions

Question 1.
Draw a pyramid of energy of grassland ecosystem.

Answer:

Question 2.
Explain nitrogen cycle in nature.

Answer:

Question 3.
Explain sulphur cycle by diagram.
Answer:

Question 4.
Explain the effect of light on plants.
Answer:
Effect of light on plants:
Light is the source of energy. It is essential for life. It is an important factor of an ecosystem. The existence of life on earth is because of light obtained from sun. Sunlight is essential for photosynthesis, help in preparation of food for the whole living world. Light effects biological activities of plants by its intensity, period and duration. Plants are classified into following two categories on the basis of requirement of light intensity:

1. Heliophytes – Plants, which can grow better in bright light are called heliophytes.
2. Sciophytes – Plants, which require relatively less of light and they can grow better in shades are called sciophytes.

Question 5.
Explain the meaning of food web and draw its diagram.
Answer:
Food Web:
In nature, foodchains are not isolated sequences, but are interrelated and interconnected with one another. When many foodchains operate simultaneously and interlock such pattern is termed as food web. Thus, the food web is a description of feeding connections between the organisms which make up a community. Energy passes through one trophic level to next via these food web links, example a rat feeds on various kinds of grains, fruits, stems, roots, etc.

A rat in its turn is consumed by a snake which is eaten by a falcon. The snakes feed on both frogs and rats. Thus, a network of food – chains exists and this is called food web. The food web gets more complicated because of variability in taste and preference availability and compulsion and several other factors at each level. For example, tigers normally do not feed on fishes or crabs but in Sunderbans they are forced to eat them.

Question 6.
Explain calcium cycle with well labelled diagram.
Answer:

Question 7.
Draw ecological pyramid of number of a tree ecosystem and grassland ecosystem.
Answer:

Question 8.
What do you mean by ecosystem? Describe the important components of a pond ecosystem.
Or
Write about role of decomposers in an ecosystem with example.
Answer:
Ecosystem:
The system resulting from the interaction between organisms and their environment is called as ecosystem.

Components of a pond ecosystem:
A pond ecosystem should contain all components of ecosystems like:

1. Producers:
Organism, which can synthesize their own food are included under producers, example Volvox, Pandorina, Oedogonium, Saggitaria, Utricularia, Azolla, Trapa, Lemna, Typha, Nymphaea etc. form the producer class of the pond ecosystem.

2. Consumers:

• Primary consumer – Animals, which feed on producers are included into this category example Daphnia, Cyclops, Paramoecium, Amoeba and small fishes.
• Secondary consumers – Primary consumers also serve as food for water snakes, a few tortoise, few types of fish etc. hence, these are carnivores.
• Tertiary consumers – Secondary consumers also serve as food for aquatic birds like kingfisher, cranes, big fish and these together form a top class carnivorous group and called as tertiary consumers.

3. Decomposers:
All producers and consumers die and accumulate on the floor of the pond. Even the waste material and faeces of these animals get accumulated on the floor of the pond. Similarly, the floor of pond is also occupied by decomposers, which include bacteria and fungi. These decomposers decompose complex organic compounds of their bodies into simpler forms which are finally mixed with soil of floor of ponds. These are again absorbed by the roots of producer plants and thus matter is recycled.

Question 9.
Explain pyramid of biomass of pond ecosystem.
Answer:
The biomass, i.e., the living weight of the organisms in the foodchain present at different trophic levels in an ecosystem forms the pyramid of biomass. When biomass of consumers is greater than biomass of producer then pyramid is called as inverted pyramid of biomass. example pyramid of biomass of pond ecosysyem is always inverted.

Ecosystem:
The system resulting from the interaction between organisms and their environment is called as ecosystem.

Question 10.
Distinguish between:

1. Grazing food chain and Detritus food chain
2. Production and Decomposition
3. Upright and Inverted pyramid
4. Food chain and Food web
5. Litter and Detritus,
6. Primary and Secondary productivity

Answer:
1. Differences between Grazing food chain and Detritus food chain :

Grazing food chain:

• Energy for the food chain comes from the sun.
• First trophic level organisms are producers.

Detritus food chain:

• Energy comes from detritus (organic matter).
• First trophic level organisms are detritivores and decomposers.

2. Differences between Production and Decomposition :

Production:

• It refers to the process of synthesis of organic compounds from inorganic substances utilising sunlight.
• Example: Plants perform the function of production of food.

Decomposition:

• It is the phenomenon of degradation of waste biomass.
• Example : Bacteria and fungi decompose dead organic matter.

3. Differences between Upright pyramid and Inverted pyramid:
Upright pyramid:
When the number of producers or theirbiomass is maximum in an ecosystem and it decreases progressively at each trophic level in a food chain, an uprightpyramid is formed.

Inverted pyramid:
When the number of individuals or their biomass at the producer level is minimum and it Increases progressively at each trophic level in a food chain, an inverted pyramid is formed.

4. Differences between Food chain and Food web:

Food chain:

• A food chain is a single pathway where energy is transferred from producers to successive orders of consumers.
• All food chains start with green plants which are the original source of all food.
• Energy flow is unidirectional.

Food web:

• A food web is a network of various food chains which are interconnected with each other like an interlocking pattern.
• It has many linkages and intercrosscs among producers and consumers.
• Energy flow in multidirectional.

5. Differences between Litter and Detritus:
Litter:
The dead remains of plants (leaves, flowers etc.) and animals excreta which falls on the surface of the earth in terrestrial ecosystems is called litter.

Detritus:
The dead remains of plants and animals constitute detritus. It is differentiated into litter fall (above ground detritus) and below ground detritus.

6. Differences between Primary and Secondary productivity:

Primary productivity:

• It is the rate at which organic matter is built up by producers.
• It is due to photosynthesis.
• Primary productivity is two types :
  Gross Primary Productivity (GPP) and Net Primary Productivity (NPP) GPP – R = NPP (R = loss in Respiration)

Secondary productivity:

• It is the rate of synthesis of organic matter by consumers.
• It is due to herbivory and predation.
• Secondary productivity is two types :
  Gross Secondary Productivity (GSP) and Net Secondary Productivity (NSP) NSP = GSP – R (Loss in Respiration)

Question 11.
What is primary productivity ? Give brief description of factors that affect primary productivity.
Answer:
The rate of biomass production is called productivity.
It is expressed in terms of g^-2yr^-1 or (kcal – m^-2) yr^-1 to compare the productivity of ecosystems. It can be divided into Gross Primary Productivity (GPP) and Net Primary Productivity (NPP).

Gross primary productivity of an ecosystem is the rate of production of organic matter during photosynthesis. A considerable amount of GPP is utilized by plants in respiration. Gross primary productivity minus respiration losses (R), is the Net Primary Productivity (NPP). GPP – R = NPP

Primary productivity depends on:

• The plant species inhabiting a particular area.
• The environmental factors.
• Availability of nutrients.
• Photosynthetic capacity of plants.

Question 12.
Define decomposition and describe the processes and products of decomposition.
Answer:
Decomposition:
Decomposition is the process that involves the breakdown of complex organic matter or biomass from the body of dead plants and animals with the help of decomposers into inorganic raw materials such as carbon dioxide, water, and other nutrients.

The various processes involved in decomposition are as follows :

1. Fragmentation:
It is the first step in the process of decomposition. It involves the breakdown of detritus into smaller pieces by the action of detritivores such as earthworms.

2.Leaching:
It is a process where the water soluble nutrients go down into the soil layers and get locked as unavailable salts.

3. Catabolism:
It is a process in which bacteria and fungi degrade detritus through various enzymes into smaller pieces.

4. Humification:
The next step is humification which leads to the formation of a dark coloured colloidal substance called humus, which acts as reservoir of nutrients for plants.

5. Mineralization:
The humus is further degraded by the action of microbes, which finally leads to the release of inorganic nutrients into the soil. This process of releasing inorganic nutrients from the humus is known as mineralization. Decomposition produces a dark coloured, nutrient rich substance called humus. Humus finally degrades and releases inorganic raw materials such as CO₂, water, and other nutrient in the soil.

Question 13.
Given account of explain energy flow in ecosystem.
Answer:
Energy flow:
In the ecosystem, energy is transferred in an orderly sequence. The flow of solar energy from producers to consumers and to decomposers subsequently in an ecosystem is known as energy flow. Energy flow is an ecosystem is always unidirectional. Sun is the sole source of solar energy in an ecosystem. Green plants utilize this energy in photosynthesis and convert it in the form of chemical energy and store it.

Plants utilize maximum part of this energy to do its biological functions. Some of it is converted into heat and released in the environment. Remaining part of the energy is stored in various components of the body. When a consumer eats these producer plants, the energy is then transferred into its body.

In any food chain energy flows from primary producers to primary consumers, from primary consumers to secondary consumers and secondary consumers to tertiary consumers and so on. Because every organism of a trophic level continuously converts chemical energy into heat, there is always a loss of energy with each step in a food – chain. According to an estimate only 10% of the total energy obtained is transferred from one trophic level to another.

Question 14.
Write important features of a sedimentary cycle in an ecosystem.
Answer:
Sedimentary cycles have their reservoirs in the earth’s crust or rocks. Nutrient elements are found in the sediments of the earth. Elements such as sulphur, phosphorus, potassium, and calcium have sedimentary cycles. Sedimentary cycles are very slow. They take a long – time to complete their circulation and are considered as less perfect cycles. This is because during recycling, nutrient elements may get locked in the reservoir pool, thereby taking a very long – time to come out and continue circulation. Thus, it usually goes out of circulation for a long – time.

Ecosystem Long Answer Type Questions

Question 1.
Describe various components of ecosystem.
Answer:
Ecosystem has following components:

1. Abiotic components:
The physical conditions of the ecosystem depends upon latitude, overall climatic and edaphic factors. The physical deficiencies are overcome by artificial irrigation and use of fertilizers. Thus, like natural ecosystem all organic, inorganic substances and climatic factors together forms abiotic component of the ecosystem.

• Organic substances : Such as carbohydrates, proteins, lipids, etc.
• Inorganic substances : Such as C, H, N, P, K, Ca, I, etc.
• Climatic factors : Such as temperature, light, water, humidity, wind, pH, minerals, soil structure, etc.

2. Biotic factors:
Three types are there

(a) Producers :
This type of crop (dominant species) depends upon the climate, season and the choice of the farmer.

(b) Consumers:

• Primary consumers : Insects, beetles, fishes, etc.
• Secondary consumers : Frog and fishes.
• Tertiary consumers : Snakes and cranes.

(c) Decomposers:
Bacteria, fungi etc.

Question 2.
Explain consumer components ecosystem of a pond in brief.
Answer:
Consumers:
They feed on producers directly or indirectly. It is of following categories:

1. Primary consumers:
Varied forms of zooplanktons are found in the water surface. Most of them are unicellular protists, such as Amoeba, Paramoecium whereas some are multicellular crustaceans, such as Daphnia, Cyclops, etc. Animals which are found under the surface of water are called as benthos, such as many types of fishes, crustaceans, molluscs, insects, beetles, etc. They too feed on producers.

2. Secondary consumers:
They feed on primary consumers, example big fishes, water snakes, etc.

3. Tertiary consumers:
They feed on secondary consumers, example kingfisher, cranes, omnivorous man, etc.

Question 3.
Describe carbon cycle in an ecosystem.
Answer:
Carbon Cycle:
Importance of carbon:
Carbon is considered as the basis of life. Carbon is the most important constituent of proteins, fats and nucleic acids which form the essential constituents of protoplasm.

Sources of carbon – The three sources of carbon in non – living world are:

• The carbon dioxide of the air and that which is dissolved in water.
• The rocks of the earth crust containing carbonates.
• The fossil fuel like coal and petroleum.

Recycling of carbon:
Carbon dioxide (CO₂) is the main source of carbon for the living beings. The carbon of coal, graphite, petroleum are insoluble and carbonates are not available to the organism until they are burnt or chemically changed. Most of the carbon dioxide enters the living world through photosynthesis. In this process green plants trap CO₂ from atmosphere and convert it into carbohydrates by using water and solar energy.

The amount of the carbon fixed by photosynthesis is nearly 7 x 10¹³ kg/year. The organic compounds synthesized in photosynthesis are passed from plants to the herbivores and carnivores. It is estimated that one hectare of a healthy forest produces about 10 tonnes of oxygen and absorbs 30 tonnes of carbon dioxide annually.

Question 4.
What do you mean by trophic levels? What are ecological pyramids ? Explain various types of ecological pyramids.
Answer:
Trophic level:
In an ecosystem, the producer consumer arrangement is a kind of structure known as trophic structure and each food level in the food chain is called as trophic level or energy level. In other words each level of food in food chain is called its trophic level. The first trophic level (T₁) in an ecosystem is occupied by producers. Herbivores (primary consumers) form second trophic level (T₂), secondary consumers form third trophic level (T₃), tertiary consumers form fourth trophic level (T₄) and decomposers form fifth trophic level (T₅) in an ecosystem.

Food or Ecological pyramids:
If we express the organisms of various trophic levels according to their number, biomass and ratio of energy stores within it, then we obtain a cone or pyramid like structure which is known as food or ecological pyramid. Ecological pyramids represent the trophic structure and function of an ecosystem. In base and successive trophic levels the tiers which make up the apex. Ecological pyramids are of the following three types:

1. Pyramid of biomass
2. Pyramid of number
3. Pyramid of energy.

1. Pyramid of Biomass:
Biomass is the dry weight of living organisms per unit of space. The ecological pyramid, which shows the quantitative relationship of the standing crop at each trophic level/The pyramid of biomass shows gradual reduction in biomass at each trophic level from base to apex.

The pyramid of biomass may be :

• Upright – example all terrestrial ecosystems.
• Inverted – example all aquatic ecosystems.

2. Pyramid of Number:
The ecological pyramid which shows the number of individual organisms at each trophic level. It represents numerical relationship between different trophic level of a food chain. In this pyramid more abundant species from the first trophic level and from the base of pyramid and the less abundant species remain near the top. The pyramid of number may be:

• Upright : example grassland, pond, forest ecosystem.
• Inverted : example ecosystem of tree.

3. Pyramid of Energy:
It indicates the total amount of energy at each trophic level of the food chain. At each producer level, the total energy available is relatively more than at the higher trophic levels because of the loss of the energy at each trophic level. Thus, there is a gradual loss of energy at each trophic level. The pyramid of energy of each types of ecosystem is always upright.

Question 5.
What is meant by terrestrial biomes? What are its types? Explain any one biomes in detail.
Answer:
Terrestrial biome:
Large area occupying ecosystems in nature are called biomes. If biomes are on land than they are called terrestrial biomes.
Terrestrial biomes may be :

1. Forest biomes – They may be as below :

• Topical rain forest
• Cold tropical forest
• Taiga forest.

2. Grassland biomes – They may be as below:

• Tropical rain forest
• Cold tropical forest

3. Desert biomes

4. Tundra biomes.

Grassland biome – Grassland biome or ecosystem has long grasses, Its, land is fertile. It receives approximately 25 to 75 cm average rainfall. Its component are:

1. Abiotic component:
All organic, inorganic substances and climatic factors together form abiotic component.

2. Biotic component:

• Producers – Grasses, herb, shrubs.
• Primary consumers – Herbivore like cow, buffalo, goats, sheep, deer, rabbit, rat insect.
• Secondary consumers – Carnivore animals which eat primary consumers, like snake, birds, foxes, jackal etc.
• Tertiary consumers – These organism which eat secondary consumers because no other one eats them, like Hawk, Peacock etc.
• Decomposers – Micro fungus, Bacteria, Actinomycetes are decomposer of grassland biomes and recycle the material back to soil and used by producers.

Question 6.
What are biogeochemical cycles? Write in short sulphur and calcium cycle.
Answer:
Biogeochemical cycles:
All living organisms get matter from the biosphere components i.e., lithosphere, hydrosphere and atmosphere. Essential elements or inorganic substances are provided by earth and are required by organisms for their body building and metabolism, they are known as biogeochemicals or biogenetic nutrients.

Sulphur cycle:
Producers (green plants) need sulphur in the form of sulphates from soil or from water (aquatic plants). The animals get sulphur through food. Some animals get sulphur from water also. Sulphur is found in three amino acids hence, sulphur is component of most proteins, some vitamins and enzymes. Plants pick up sulphur in the form of sulphates. They are converted to organic form mostly as component of some amino acids. It is found in nature as element and also as sulphates in soil, water and rocks. After the death of plants and animals, they are decomposed by microbes like Asperigillus, Neurospora and Escherichia releasing hydrogen sulphide.

Calcium cycle:
Calcium is slowly released from the rocks by water and wind action. These are either blown into the air or absorbed by plants through their roots. Animals obtain it directly as compounds and also from plants. Calcium is released from plant and animal bodies by decomposition after death. Molluscs and Corals deposit a large quantity of calcium in their shells and skeletons making it unavailable for quick cycling.

MP Board Class 12th Biology Important Questions

MP Board Class 12th Chemistry Important Questions Chapter 1 The Solid State

The Solid State Important Questions

The Solid State Objective Type Questions

Question 1.
Choose the correct answers:

Question 1.
Due to Frankel defect, density of ionic solids :
(a) Decreases
(b) Increases
(c) Does not change
(d) It changes.
Answer:
(c) Does not change

Question 2.
In CsCl each Cl is surrounded by how many Cs :
(a) 8
(b) 6
(c) 4
(d) 2.
Answer:
(a) 8

Question 3.
Frenkel defect is not shown by :
(a) AgBr
(b) AgCl
(c) KBr
(d) ZnS.
Answer:
(c) KBr

Question 4.
In NaCl crystal number of oppositely charged ions situated at equal distance are:
(a) 8
(b) 6
(c) 4
(d) 2
Answer:
(b) 6

Question 5.
Best conductor of electricity is :
(a) Diamond
(b) Graphite
(c) Silicon
(d) Carbon (non – crystalline).
Answer:
(b) Graphite

Question 6.
Which type of point defect is found in NaCI crystal of KCl crystal: (MP 2009 Set D)
(a) Frenkel defect
(b) Schottky defect
(c) Lattice defect
(d) Impurity defect.
Answer:
(b) Schottky defect

Question 7.
How many space lattices (Bravais lattice) can be obtained from various crystal systems :
(a) 7
(b) 14
(c) 32
(d) 230.
Answer:
(b) 14

Question 8.
Diamond is a :
(a) H – bond solid
(b) Ionic solid
(c) Covalent solid
(d) Glass
Answer:
(c) Covalent solid

Question 9.
The Co – ordination number of Ca²⁺ ions in fluoride structure is :
(a) 4
(b) 6
(c) 8
(d) 3.
Answer:
(c) 8

Question 10.
8 : 8 Co – ordination number is found in which compound :
(a) MgO
(b) A1203
(C) CsCl
(d) All of these
Answer:
(C) CsCl

Question 11.
Co – ordination number of body centred cubic cell is :
(a) 8
(b) 12
(c) 6
(d) 4
Answer:
(a) 8

Question 12.
Density of unit cell is :
(a) \(\frac { ZM }{ { a }^{ 3 }{ N }_{ 0 } } \)
(b) \(\frac { Z{ N }_{ 0 } }{ { a }^{ 3 }M } \)
(c) \(\frac { { N }_{ 0\quad }{ a }^{ 3 } }{ Z } \)
(d) \(\frac { Z }{ M{ N }_{ 0 } } \)
Answer:
(a) \(\frac { ZM }{ { a }^{ 3 }{ N }_{ 0 } } \)

Question 13.
The number of tetrahedral voids in unit cell of cubic close packing :
(a) 4
(b) 8
(c) 6
(d) 2
Answer:
(b) 8

Question 14.
Intra – ionic distance of CsCl will be :
(a) a
(b) \(\frac {a}{2}\)
(c) \(\frac { \sqrt { 3 } a }{ 2 } \)
(d) \(\frac { 2a }{ \sqrt { 3 } } \)
Answer:
(c) \(\frac { \sqrt { 3 } a }{ 2 } \)

Question 15.
Number of atoms in a body centred cubic unit cell is : (MP 2011)
(a) 1
(b) 2
(c) 3
(d) 4.
Answer:
(b) 2

Question 16
Which of the following is Bragg equation :
(a) nλ = 2ϕ sinθ
(b) nλ = 2d sinθ
(c) nλ = sinθ
(d) n\(\frac {θ}{2}\) = \(\frac {d}{2}\) sinθ.
Answer:
(b) nλ = 2d sinθ

Question 17.
Constituents of covalent crystal is :
(a) Atom
(b) Molecule
(c) Ion
(d) All of these.
Answer:
(a) Atom

Question 18.
Number of Na atom present in the unit cell of NaCI crystal is : (MP 2012)
(a) 1
(b) 2
(c) 3
(d) 4.
Answer:
(d) 4.

Question 19.
What type of magnetic substance are Fe, Co, Ni: (MP 2012,18)
(a) Paramagnetic
(b) Ferromagnetic
(c) Diamagnetic
(d) Antiferromagnetic.
Answer:
(b) Ferromagnetic

Question 20.
The correct example of Frenkel defect is : (MP 2012)
(a) NaCI
(b) CsCl
(c) KCl
(d) AgCl.
Answer:
(d) AgCl.

Question 21.
Dry ice (solid CO₂) is a/an : (MP 2012)
(a) Ionic crystal
(b) Covalent crystal
(c) Molecular crystal
(d) Metallic crystal.
Answer:
(c) Molecular crystal

Question 22.
Co – ordination number of Cs in CsCl: (MP2015)
(a) Like Cl i.e., 8
(b) Unlike Cl i.e., 6
(c) Unlike Cl i.e., 8
(d) Like Cl i.e., 6.
Answer:
(a) Like Cl i.e., 8

Question 23.
Structure of NaCI crystal: (MP 2015)
(a) Tetragonal
(b) Cubic
(c) Orthorhombic
(d) Monoclinic.
Answer:
(b) Cubic

Question 24.
Each Na⁺ion in NaCI crystal is surrounded by :
(a) Three Cl^– ions
(b) Eight Cl^– ions
(c) Four Cl^– ions
(d) Six Cl^– ions.
Answer:
(d) Six Cl^– ions.

Question 25.
For increasing of electro conductivity in a solid crystal, mixing of impurities is known as : (MP2016)
(a) Schottky defect
(b) Frenkel defect
(c) Doping
(d) Electronic defect.
Answer:
(c) Doping

Question 26.
Which type of lattice is found in KCl crystal:
(a) Face centred cubic
(b) Body centred cubic
(c) Simple cubic
(d) Simple tetragonal.
Answer:
(a) Face centred cubic

Question 27.
Number of atoms in a body centred cubic unit cell of a monoatomic substance is :
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 28.
Radius ratio limit for tetrahedral symmetry is :
(a) 0155
(b) 0.414
(c) 0.732
(d) 0.225
Answer:
(d) 0.225

Question 29.
The defect produced due to a cation and an anion vacancy in a crystal lattice is known as :
(a) Schottky defect
(b) Frenkel defect
(c) Crystal defect
(d) Ionic defect
Answer:
(a) Schottky defect

Question 30.
If co – ordination number of Cs⁺ is 8 in CsCl then co – ordination number of Cl^– ion is :
(a) 8
(b) 4
(c) 6
(d) 12
Answer:
(a) 8

Question 2.
Answer in one word / sentence :

1. Give two examples of metallic crystal.
2. Give two examples of covalent crystal.
3. Give two examples of ionic crystal.
4. What is the co – ordination number of F⁺ ion in CaF₁?
5. What is the value of co – ordination number of hexagonal close packing structure?
6. What is the type of structure of NaCl crystal?
7. Give an example of body centred cubic cell.
8. Give an example of a compound which has both Schottky and Frenkel type of defect.
9. What types of crystal is SiC? (MP 2011)
10. Write Bragg equation. (MP2017)
11. What is effect on the density of a substance or crystal due to Schottky defect?
12. Write the formula of radius ratio.
13. Give two examples of amorphous or non – crystalline solid.
14. F – centres give colour of crystal due to whose presence? (MP 2018)

Answer:

1. Copper, Nickel
2. Diamond, Graphite
3. NaCl, NaNO₃
4. 4
5. 12
6. Cubic
7. CsCl
8. AgBr
9. Covalent solid
10. nλ = 2d sinθ
11. Due to Schottky defect density of substance decreases
12. Radius ratio = \(\frac { radiusofcation\quad { r }^{ + } }{ radiusofanion\quad { r- } } \)
13. Glass, plastic
14. Due to presence of free electron.

Question 3.
Fill in the blanks :

1. The defect produced due to removal of a cation and an anion from a crystal lattice is called …………………. (MP 2018)
2. If in a crystal lattice a cation leaves its lattice site and occupies a space in the interstitial site then the defect is called ………………….
3. The cause of electric conduction of NaCl in its molten state are its ………………….
4. Due to …………………. defect the density of crystal decreases
5. Total …………………. types of crystal system are there.
6.  …………………. proposed the concept of atom for the first time.
7. The ratio of the cation and anion present in a crystal is known as ………………….
8. The process of adding small amount of impurities in an element or compound is called …………………. (MP 2018)
9. Total 14 types of unit cells are there which are known as ………………….
10. In NaCl crystal structure, co – ordination number of both Na⁺and Cl^– ion is ………………….
11.  …………………. defect is found in ZnS and AgCl crystal.
12. Due to Schottky defect, density of crystal ………………….
13. In metallic solids, conductivity is due to the presence of ………………….
14. Point defects are found in …………………. crystals.
15. Substances which are attracted in magnetic field are called ………………….
16. For a unit cell, if r = \(\frac { a }{ \sqrt { 8 } } \), then it will be …………………. type of unit cell.
17. Conductivity of semiconductor …………………. on increasing temperature.

Answer:

1. Schottky defect
2. Frenkel defect
3. Free ions
4. Schottky
5. Seven
6. Kannad
7. Radius ratio
8. Doping
9. Bravais lattice
10. Six
11. Frenkel
12. Decreases
13. Free electron
14. Ionic
15. Paramagnetic substance
16. Fcc
17. Increases.

Question 4.
Match the following:
I. (MP2014)

Answer:

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

II.

Answer:

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

III.

Answer:

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

IV. (MP2017)

Answer:

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

MP Board Class 12th Chemistry Important Questions

MP Board Class 12th Maths Important Questions Chapter 5A Continuity and Differentiability

Continuity and Differentiability Important Questions

Continuity And Differentiability Objective Type Questions:

Question 1.
Choose the correct answer:

Question 1.
If x = at², y = 2at, then \(\frac{dy}{dx}\) will be:
(a) t
(b) t²
(c) \(\frac{1}{t}\)
(d) \(\frac { 1 }{ t^{ 2 } } \)
Answer:
(c) \(\frac{1}{t}\)

Question 2.
If y = 2\(\sqrt { cot(x^{ 2 }) } \) ,then \(\frac{dy}{dx}\) will be:

Answer:
(a)

Question 3.
The value of \(\frac{d}{dx}\) (x³ + sin x²) is to be:
(a) 3x² + cos x²
(b) 3x² + x sin x²
(c) 3x² + 2x cos x²
(d) 3x² + x cos x²
Answer:
(c) 3x² + 2x cos x²

Question 4.
The value of \(\frac{d}{dx}\) a^x is to be:
(a) a^x
(b) a^x log_a e
(c) a^x log_e a
(d) \(\frac { a^{ x } }{ log_{ e }a } \)
Answer:
(c) a^x log_e a

Question 5.
If y = 500e^7x + 600e^-7x, then the value of \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) will be:
(a) 45 y
(b) 47 y
(c) 49 y
(d) 50 y
Answer:
(c) 49 y

Question 2.
Fill in the blanks:

1. Differential coefficient of cos x⁰ w.r.t x is ……………………………..
2. Differential coefficient of e^log_ea w.r.t x is ……………………………….
3. Differential coefficient of log_e a w.r.t x is ………………………………
4. Differential coefficient of a^x w.r.t x is ……………………………….
5. Differential coefficient of sin 3x w.r.t x is ……………………………….
6. If y = sin^-1 (2x \(\sqrt { (1-x^{ 2 }) } \)), then \(\frac{dy}{dx}\) = ………………………………
7. Differential coefficient of sin x w.r.t. cos x is …………………………………..
8. The value of \(\frac{d}{dx}\) (log tan x) is ………………………………………
9. Differential coefficient of log (log sin x) ………………………………
10. If x = y\(\sqrt { (1-y^{ 2 }) } \), then \(\frac{dy}{dx}\) will be …………………………………
11. n^th differentiation of sin x will be ………………………………………
12. If y = \(\sqrt { x+\sqrt { x+……..\infty } } \), then \(\frac{dy}{dx}\) will be ……………………………….
13. If x = r cos θ, y = r sin θ, then \(\frac{dy}{dx}\) will be ………………………..
14. Differential coefficient of e^x w.r.t \(\sqrt { x } \) will be ………………………….

Answer:

1. – \(\frac { \pi }{ 180 } \) sin x⁰
2. 0
3. 0, 4
4. log_e a.a^x
5. cos 3x
6. \(\frac { 2 }{ \sqrt { 1-x^{ 2 } } } \)
7. – cot x
8. 2 cosec 2x
9. \(\frac { cotx }{ logsinx } \)
10. \(\frac { \sqrt { 1-y^{ 2 } } }{ 1-2y^{ 2 } } \)
11. sin (\(\frac { n\pi }{ 2 } \) + x)
12. \(\frac { 1 }{ 2y-1 } \)
13. – cot θ
14. 2\(\sqrt { x } \)e^x.

Question 3.
Write True/False:

1. Differential coefficient of e^log_ex is \(\frac{1}{x}\)?
2. If f(x) = \(\sqrt { x } \); x>0, then value of f'(2) is \(\frac { 1 }{ 2\sqrt { 2 } } \)?
3. Any function f(x) is said to be differentiatiable at any point x = a when Lf'(a) ≠ Rf'(a)?
4. Differential coefficient of sec^-1a w.r.t x is 0?
5. If y = Ae^mx + Be^-mx, then \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) = – m²y?
6. If y = sin^-1( \(\frac { x-1 }{ x+1 } \) ) + cos^-1 ( \(\frac { x-1 }{ x+1 } \) ), then \(\frac{dy}{dx}\) = 0?
7. Every differentiatiable function is continous?
8. Differential coefficient of a^2x is a^2x log_ea?

Answer:

1. Flase
2. True
3. Flase
4. False
5. True
6. True
7. True
8. Flase

Question 4.
Match the Column:

Answer:

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

Question 5.
Write the answer in one word/sentence:

1. Find differential coefficient of \(\frac { 6^{ x } }{ x^{ 6 } } \) w.r.t. x?
2. Find differential coefficient of y = ^log_etanxw.r.t x?
3. Find n^th derivative of a^x?
4. If y = sin(ax + b), then find the value of \(\frac { d^{ 2 }y }{ dx^{ 2 } } \)?
5. If x² + y² = sin xy, then find the value of \(\frac{dy}{dx}\)?
6. Find differential coefficient of log tan \(\frac{x}{2}\) w.r.t. x?
7. Find differential coefficient of sin^-1 \(\frac { 2x }{ 1+x^{ 2 } } \) w.r.t. x?
8. Find differential coefficient of e^-log_ex w.r.t x?

Answer:

1. \(\frac { 6^{ x } }{ x^{ 6 } } \) [log 6 – \(\frac{6}{x}\) ]
2. sec² x
3. a^x(log_ea)ⁿ
4. -a²y
5. \(\frac { ycosxy-2x }{ 2y-xcosxy } \)
6. cosec x,
7. \(\frac { 2 }{ 1+x^{ 2 } } \)
8. – \(\frac { 1 }{ x^{ 2 } } \)

Continuity And Differentiability Short Answer Type Questions

Question 1.
Find all the points of discontinuity of f, when f is defined as:
f(x) = \(\left\{\begin{array}{lll}
{2 x+3,} & {\text { if }} & {x \leq 2} \\
{2 x-3,} & {\text { if }} & {x>2}
\end{array}\right.\) (NCERT)
Solution:
For x< 2, f(x) = 2x + 3 is polynomial function.
Hence, for x < 2, f(x) is continuous function. For x > 2, f(x) = 2x – 3 is polynomial function.
Hence, x > 2, f(x) is continous.
Now, we shall examine the continuty of f(x) at x = 2 only.
Put x = 2 + h,
When x → 2, then h → 0

= 2(2 + 0) – 3 = 4 – 3 = 1.
Put x = 2 – h,
When x → 2, then h → 0

= 2(2 – 0) + 3 = 7
f(2) = 2(2) + 3 = 7

Hence, f(x) is discontinous at x = 2 only.

Question 2.
Find all the points of discontunity of f, when f is defined as follow:
f(x) = \(\left\{\begin{array}{ccc}
{\frac{|x|}{x},} & {\text { if }} & {x \neq 0} \\
{0} & {\text { if }} & {x=0}
\end{array}\right.\). (NCERT)
Solution:
Hence, we shall examine the continuty of f(x) at x = 0 only,
Put x = 0 + h,
When x → 0, then h → 0

Put x = 0 – h,
When x → 0, then h → 0

= -1
Given: f(0) = 0

Hence, the given function f(x) is discontinous at x = 0.

Question 3.
Examine the continuty of function f(x) at point x = 0?
f(x) = \(\left\{\begin{array}{cc}
{\frac{1-\cos x}{x^{2}},} & {x \neq 0} \\
{\frac{1}{2},} & {x=0}
\end{array}\right.\)
Solution:
f(x) = \(\frac { 1-cosx }{ x^{ 2 } } \), when x ≠ 0.
Put x = 0 + h, when x → 0, then h → 0

Again, put x = 0 – h, when x → 0, then h → 0

Hence, f(x) is continous at x = 0.

Question 4.
Function f is defined as:
f(x) = \(\left\{\begin{aligned}
\frac{|x-4|}{x-4} ; & x \neq 4 \\
0 ; & x = 4
\end{aligned}\right.\)
Then prove that function f is continous function for all points except x = 4?
Solution:

Hence, for x = 4 the function f(x) is dicontinous

When x < 4 then f(x) = -1 which is constant function Hence, it is continous function when x > 4 then f(x) = 1 which is constant function.
Hence, it is continous function
The given functions f is continous at all points except x = 4. Proved.

Question 5.
Find the value of k for which the function
f(x) = \(\left\{\begin{array}{c}
{\frac{k \cos x}{\pi-2 x}, \text { if } x \neq \frac{\pi}{2}} \\
{3, \text { if } x=\frac{\pi}{2}}
\end{array}\right.\)
is contionuous at x = \(\frac { \pi }{ 2 } \). (NCERT)
Solution:
Put x = \(\frac { \pi }{ 2 } \) + h,
When x → \(\frac { \pi }{ 2 } \), then h → 0


\(\frac{k}{2}\) × 1 = \(\frac{k}{2}\)
Put x = \(\frac { \pi }{ 2 } \) – h
When x → \(\frac { \pi }{ 2 } \), then h → 0

\(\frac{k}{2}\) × 1 = \(\frac{k}{2}\)
Given that f ( \(\frac { \pi }{ 2 } \) ) = 3
The given function is continous,

\(\frac{k}{2}\) = \(\frac{k}{2}\) = 3
k = 6.

Question 6.
Find the value of k, if function
f(x) = \(\left\{\begin{array}{lll}
{k x+1,} & {\text { if }} & {x \leq \pi} \\
{\cos x,} & {\text { if }} & {x>\pi}
\end{array}\right.\) is continous at x = π? (NCERT)
Solution:
Put x = π + h,
When x → π, then h → 0

= cos (π + 0)
= -1.
Put x = π – h,
When x → π, then h → 0

= k(π – 0) + 1
= πk + 1
f(π) = kπ + 1
∴ The given function is continous at x = π

-1 = kπ + 1 = kπ + 1
kπ = -2
k =- \(\frac { 2 }{ \pi } \)

Question 7.
Function f is continous at x = 0:
f (x) = \(\left\{\begin{array}{c}
{\frac{1-\cos k x}{x \sin x} ; x \neq 0} \\
{\frac{1}{2} \quad ; x=0}
\end{array}\right.\) Find the value of k?
Solution:
Given:
f(x) = \(\frac { 1-coskx }{ xsinx } \), x ≠ 0


Given: f(0) = \(\frac{1}{2}\)

Question 8.
Find the relationship between a and b so the following function f defined by:
f(x) = \(\left\{\begin{array}{lll}
{a x+1,} & {\text { if }} & {x \leq 3} \\
{b x+3,} & {\text { if }} & {x>3}
\end{array}\right.\) is continous at x = 3. (NCERT; CBSE 2011)
Solution:
Put x = 3 + h,
When x → 3, then h → 0

= b(3 + 0) + 3
= 3b + 3
Put x = 3 – h,
When x → 3, then h → 0

= a(3 – 0) + 1
= 3a + 1
f(3) = 3a + 1
The given function is continous at x = 3.

3b + 3 = 3a + 1 = 3a + 1
3a + 1 = 3b + 3
3a = 3b + 2
a = b + \(\frac{2}{3}\).

Question 9.
Prove that the funcion f(x) = |x – 1|, x ∈ R is not differentiable at x = 1? (NCERT)
Solution:
Given:

f(1) = 1 – 1 = 0
Put x = 1 – h, when x → 1, then h → 0

Put x = 1 + h, when x → 1, then h → 0

Lf'(1) ≠ Rf'(1)

Question 10.
Show that the function:
f(x) = \(\left\{\begin{aligned}
x-1, & \text { if } x<2 \\
2 x-3, & \text { if } x \geq 2
\end{aligned}\right.\), is not differentaible at point x = 2?
Solution:
We know that:
RHD =

From the above, it is clear that,
LHD at x = 2 ≠ RHD at x = 2
∴ f(x) is not differentiable at x = 2. Proved.

Question 11.
Determine the function of defined by
f(x) = \(\left\{\begin{array}{cc}
{x^{2} \sin \frac{1}{x},} & {\text { when } x \neq 0} \\
{0,} & {\text { when } x=0}
\end{array}\right.\) in continous function? (NCERT)
Solution:
Here, f(0) = 0


= 0 × a finite quantiy, [∵ sin \(\frac{1}{h}\) is between -1 and 1]
= 0

Hence, the given function is continous at x = 0.

MP Board Class 12 Maths Important Questions

In this article, we will share MP Board Class 12th Hindi Swati Solutions पद्य Chapter 2 वात्सल्य और स्नेह Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 12th Hindi Swati Solutions पद्य Chapter 2 वात्सल्य और स्नेह

वात्सल्य और स्नेह अभ्यास

वात्सल्य और स्नेह अति लघु उत्तरीय प्रश्न

प्रश्न 1.
कृष्ण ने दही का दोना कहाँ छिपा लिया?
उत्तर:
बालकृष्ण ने दही का दोना अपनी पीठ के पीछे छिपा लिया है।

प्रश्न 2.
दूध पीने से चोटी बढ़ती है, यह सुझाव कृष्ण को किसने दिया था?
उत्तर:
गाय का दूध पीने से चोटी बढ़ती है, यह सुझाव यशोदा माता ने बालकृष्ण को दिया था।

प्रश्न 3.
सूरदास किस भाषा के कवि हैं?
उत्तर:
सूरदास ब्रजभाषा के कवि हैं।

प्रश्न 4.
‘मोसों कहत मोल को लीनो’ यह कथन किसने कहा है? (2016)
उत्तर:
यह कथन बलदाऊ की शिकायत करते हुए कृष्ण यशोदा माता से कहते हैं कि बलदाऊ उन्हें मोल का खरीदा हुआ बताते हैं।

प्रश्न 5.
“घर का पहरे वाला” से कवि का क्या आशय है? (2015)
उत्तर:
“घर का पहरे वाला” से कवि का तात्पर्य घर के रखवाले से है अर्थात् देश की रक्षा करने वाला प्रहरी।

प्रश्न 6.
‘ममता की गोद’ किसे कहा गया है?
उत्तर:
बहन को ‘ममता की गोद’ कहा गया है।

वात्सल्य और स्नेह लघु उत्तरीय प्रश्न

प्रश्न 1.
अपने को निर्दोष सिद्ध करने के लिए कृष्ण ने यशोदा को क्या-क्या तर्क दिए? (2013)
उत्तर:
अपने को निर्दोष सिद्ध करने के लिए कृष्ण ने निम्न तर्क दिए-

1. मैंने दही नहीं खाया है,बल्कि ग्वालवालों ने मेरे मुँह पर लपेट दिया है।
2. तेरा दधि का बर्तन इतने ऊँचे छींके पर लटका हुआ है।
3. मेरे छोटे-छोटे हाथ हैं जो उस छींके तक नहीं पहुँच सकते।

प्रश्न 2.
“गोरे नन्द जसोदा गोरी, तुम कत स्याम सरीर” यह पंक्ति किसने किससे और क्यों कही?
उत्तर:
उक्त पंक्ति बलदाऊ ने कृष्ण से कहीं। उन्होंने यह पंक्ति अपने इस पक्ष को दृढ़ करने के लिए कहीं कि तुझे तो मोल लिया है। यदि तू नन्द और यशोदा माँ का पुत्र होता तो काला क्यों होता ? वे दोनों तो गोरे हैं। यहाँ इस तथ्य को उजागर किया गया है कि गोरे माता-पिता की सन्तान भी गोरी होती है।

प्रश्न 3.
मुँह से मिट्टी निकालने के लिए यशोदा ने कौन-सा उपाय किया?
उत्तर:
कृष्ण के मुँह से मिट्टी निकालने के लिए यशोदा ने कसकर उनकी बाँह पकड़ ली और मारने के लिए एक हाथ में डंडी उठा ली। उनका भाव यह था कि डर के मारे कृष्ण अपने मुख से मिट्टी बाहर निकाल देगा।

प्रश्न 4.
‘मेरा जीवन क्रीड़ा-कौतुक तू प्रत्यक्ष प्रमोद भरी’ से कवि का क्या आशय है?
उत्तर:
कवि कहता है कि भाई का जीवन तो क्रीड़ा-कौतूहल आदि से भरा रहा, लेकिन बहन गम्भीर रहते हुए भी आनन्द देने वाली बनी रही। बहन के होते हुए कभी भी भाई के प्रेम और आनन्द में कमी नहीं आई।.

प्रश्न 5.
बहन को भाई का ध्रुवतारा’ क्यों कहा गया है? (2009, 11, 14, 17)
उत्तर:
भाई तो एक अल्हड़ जीवन या यों कहें कि लापरवाह जीवन बिताता रहा, लेकिन बहन अपने लक्ष्य के लिए ध्रुवतारे की तरह अटल खड़ी हुई दिखाई दी। बहन ने भाई को मुसीबतों के समय उत्साह दिया और उसकी देश-भक्ति को जाग्रत कर देश के प्रति अपने कर्तव्य को पूर्ण करने में सहयोग दिया।

वात्सल्य और स्नेह दीर्घ उत्तरीय प्रश्न

प्रश्न 1.
कृष्ण माता यशोदा से बलदाऊ की क्या-क्या शिकायतें करते हैं?
उत्तर:
कृष्ण माता यशोदा से शिकायत करते हैं कि बलदाऊ मुझे बहुत चिढ़ाते हैं। मझसे यह कहते हैं कि तुझे यशोदा माता ने जन्म नहीं दिया है,बल्कि तुझे किसी से मोल लिया है। में इस गुस्से में खेलने भी नहीं जाता हूँ। बार-बार मुझसे पूछते हैं कि तेरे माता-पिता कौन हैं और कहते हैं कि नन्द बाबा और यशोदा तो दोनों गोरे हैं,यदि तू उनका पुत्र है तो साँवला क्यों है? फिर सभी ग्वालों को सिखा देते हैं और मेरी तरफ देखकर ताली मारकर हँसते हैं और मुझे खिझाते हैं और तू मुझे ही मारना जानती है,बलदाऊ को कभी नहीं मारती।

प्रश्न 2.
‘सूर वात्सल्य के चितेरे हैं’ इस कथन पर अपने विचार व्यक्त कीजिए।
उत्तर:
‘सूर वात्सल्य के चितेरे हैं’ यह कथन अक्षरशः सत्य है। इनका वात्सल्य वर्णन अद्वितीय है। कृष्ण के बाल रूप और उनकी बाल-सुलभ क्रीड़ाओं का जैसे विशद वर्णन सूरदास ने किया है, वैसा सम्पूर्ण विश्व साहित्य में कहीं नहीं मिलता। कृष्ण का घुटनों चलना,मणियों के खम्भ में अपने प्रतिबिम्ब को माखन खिलाना, माँ से जिद करना,खेलते समय खिसिया जाना आदि विविध बाल क्रीड़ाओं का वर्णन सूरसागर में चित्रित है। उदाहरण देखिए-
“मैया मैं तो चंद खिलौना लैहों।
जैहों लोटि धरनि पै अब ही तेरी गोद न ऐहौं।”

बच्चों की पारस्परिक होड़ का चित्रण भी सूरदास ने अनूठे रूप में किया है। माता यशोदा उनसे कहती है कि दूध पीने से चोटी बढ़ती है तो वह इस लालच में रोज दूध पी लेते हैं। जब उन्हें चोटी बढ़ती हुई दिखाई नहीं देती तो वह कहते हैं-
“मैया कबहि बढ़ेगी चोटी।
काचो दूध पियाबति पचि-पचि देत न माखन रोटी ॥
कितनी बार मोहि दूध पिबत भई यह अजहूँ है छोटी॥”

प्रश्न 3.
बहन को ‘चिनगारी’ तथा भाई को ‘ज्वाला’ बताने के पीछे कवि का क्या आशय (2009)
उत्तर:
गोपाल सिंह नेपाली राष्ट्रीय चेतना जाग्रत करने वाले कवि हैं। उन्होंने भाई-बहन के प्रेम को राष्ट्रीय प्रेम के रूप में अभिव्यक्त किया है। अपनी मातृभूमि की रक्षार्थ बहन चिनगारी के रूप में कार्य करेगी तो भाई ज्वाला बन कर देश की रक्षार्थ तत्पर रहेगा। बहन की क्रोध रूपी चिनगारी जलकर भयानक ज्वाला का रूप ले लेगी और मातृभूमि के दुश्मन को जलाकर राख कर देगी। दूसरी ओर भाई का क्रोध तो स्वयं ज्वाला बनकर देश की रक्षा करेगा। भाई बहन को कहता है कि देश की रक्षा के लिए हम दोनों को सजग और सक्रिय रहना है। कवि आश्वस्त है कि भाई और बहन दोनों मिलकर अपने देश की रक्षा के लिए सतर्क रहेंगे और देश के दुश्मन के लिए ज्वाला बनकर उसका सर्वनाश करने में सक्षम होंगे।

प्रश्न 4.
कवि ने भाई-बहन के स्नेह को किन-किन प्रतीकों के माध्यम से अभिव्यक्त किया है?
उत्तर:
कवि ने भाई-बहन के स्नेह को देश-प्रेम में परिवर्तित कर नये-नये प्रतीकों का सहारा लिया है। बहन के प्रेम के प्रतीक हैं-चिनगारी,हहराती गंगा,बसन्ती चोला,कराल क्रान्ति, राधारानी, आँगन की ज्योति,ममता की गोद, बहन की बुद्धि, नदी की धारा, ध्रुवतारा के रूप में बताया है। दूसरी तरफ भाई के प्रेम को ज्वाला, बेहाल झेलम, सजा हुआ लाल, विकराल, वंशीवाला, घर का पहरेदार,प्रेम का पुतला, जीवन का क्रीड़ा कौतुक, क्रियाशीलता, एक लहर बताया है। इन प्रतीकों के माध्यम से कवि ने राष्ट्रीय चेतना के साथ-साथ मानवीय जीवन की ऊष्मापूर्ण अनुभूतियों का प्रभावशाली वर्णन किया है। इस कविता में कवि ने देश-प्रेम का उच्च आदर्श उद्घाटित किया है।

प्रश्न 5.
भाई-बहन’ कविता के माध्यम से कवि क्या संदेश देना चाहता है?
उत्तर:
भाई-बहन’ कविता के माध्यम से कवि ने नये-नये प्रतीकों के द्वारा भाई बहन के प्रेम को राष्ट्रीय प्रेम के परिवेश में व्यक्त किया है। भाई अपनी बहन को सम्बोधित कर कहता है कि हम दोनों को देश की रक्षा के लिए सजग और सक्रिय होना है। बहन की बुद्धि और भाई की क्रियाशीलता मिलकर जीवन और राष्ट्र को आनन्दमय बना देगी। भाई-बहन दोनों मिलकर आजादी के गीत गाकर लोगों में देश की रक्षा के प्रति जागति पैदा कर देश के प्रति अपने कर्तव्य को पूरा कर सकते हैं। भाई-बहन के प्रेम को राष्ट्रीय प्रेम से जोड़कर उदात्त बना दिया गया है। कवि संदेश दे रहा है कि हम अपने आनन्द में ही मस्त होकर देश की रक्षा के अपने कर्त्तव्य को कहीं भूल न जाएँ। यहाँ सभी को सजग रहकर एक सच्चे देशभक्त प्रहरी की तरह अपने देश की रक्षा करनी है।

देश के लिए बसन्ती चोला धारण करना पड़े तो करें और विकराल स्वरूप धारण करना पड़े तो करें तथा अपनी बुद्धि और बल के सहारे देश के लिए अपना बलिदान कर दें। जब देश की आजादी का प्रश्न हो, तो अपनी क्रोध रूपी ज्वाला को जलाकर सामने आये अरि (दुश्मन) को नष्ट कर दें। किसी भी कीमत पर देश की आजादी को बचाना है। यदि मातृभूमि तुम्हें अपनी रक्षा के लिए आवाज दे, तो सभी सुख-सम्पत्ति को भूल कर एक सच्चे देश-भक्त की तरह आगे आयें और अपना सर्वस्व अर्पण करके भी अपनी मातृभूमि की रक्षा करें। कवि देश में एकता बनाये रखने का भी संदेश देता है।

प्रश्न 6.
सन्दर्भ सहित व्याख्या कीजिये
(अ) मैया मैं नाही……. शिव विरंचि बौरायो।
(ब) मैया कबहिं बढ़ेगी……. हरि हलधर की जोटी।
(स) भाई एक लहर ……. भाई का ध्रुवतारा है।
उत्तर:
(अ) सन्दर्भ :
प्रस्तुत पद ‘वात्सल्य और स्नेह’ से सूरदास द्वारा रचित ‘सूर के बालकृष्ण’ नामक शीर्षक से उद्धृत किया गया है।

सन्दर्भ :
इसमें बालकृष्ण अपनी माता यशोदा से अपने प्रति की गई दही खाने की शिकायत को नकारते हुए बड़े ही बुद्धिकौशल से बाल सुलभ उत्तर देते हैं।

व्याख्या :
श्रीकृष्ण माता यशोदा से कहते हैं कि, हे माता! मैंने दही नहीं खाया है। मुझे याद आ रहा है कि इन सभी सखाओं ने मिलकर मेरे मुख पर दही लपेट दिया था। तू जानती है कि इतने ऊँचे टँगे हुए छींके पर दही का बर्तन रखा हुआ है। तू देख सकती है कि मेरे छोटे-छोटे हाथ हैं। इन छोटे हाथों से मैं कैसे उस दही के बर्तन को प्राप्त कर सकता हूँ। फिर नन्दकुमार बालकृष्ण ने दोने को पीठ के पीछे छिपाते हुए और मुख पर लगे हुए दही को पोंछते हुए उपर्युक्त बातें कहीं। यहाँ उनकी बाल सुलभ चतुरता का प्रदर्शन किया गया है। उन्हें भान है कि मुख पर लगे हुए दही से और हाथ में लगे दोने से उनकी चोरी पकड़ी जाएगी, अतः मुख को साफ कर लिया और दोने को पीठ के पीछे छुपा लिया।

यशोदा जी सब समझ गईं, लेकिन पीटने वाली लकड़ी को फेंक कर मुस्कराने लगी और मनमोहन बालकृष्ण को अपने गले से लगा लिया। श्रीकृष्ण के बाल विनोद के आनन्द ने माता यशोदा के मन को मोहित कर लिया। यहाँ श्रीकृष्ण के प्रति उनकी भक्ति का प्रताप दर्शाया गया है। सूरदास कहते हैं कि यशोदा जी और बालकृष्ण के सख को देखकर शिव और ब्रह्मा भी विवेक रहित होकर मोहित हो गए। अर्थात् श्रीकृष्ण यशोदा जी को अपनी बाल लीलाओं का जो सुख दे रहे हैं उससे किसी को भी ईर्ष्या हो सकती है और वह भ्रमित हो सकता है। निरंजन निराकार परब्रह्म साकार रूप में आकर यशोदा जी के आँगन में उनके प्रमोद के लिए जो क्रीड़ाएँ कर रहे हैं, ऐसा सुख शिव-विरंचि को भी दुर्लभ है।

(ब) सन्दर्भ :
पूर्ववत्।

प्रसंग :
प्रस्तुत पंक्तियों में बालकृष्ण अपनी माता यशोदा से यह शिकायत कर रहे हैं कि उनको कितने ही दिन दूध पीते हो गए, लेकिन उनकी चोटी बड़ी नहीं हुई है।

व्याख्या :
बालकृष्ण यह जानने को उत्सुक हैं कि उनकी चोटी कब बढ़ेगी। उनकी माता उन्हें यह भरोसा दिलाकर दूध पिलाती थीं कि दूध पीने से उनकी चोटी बढ़ जाएगी। वह माता से पूछते हैं कि हे माता मुझे कितनी ही बार (बहुत समय) दूध पीते हुए हो गईं, लेकिन यह चोटी अभी तक छोटी है। तेरे कथनानुसार मेरी चोटी बलदाऊ की चोटी के समान लम्बी और मोटी हो जाएगी, काढ़ते में, गुहते में,नहाते समय और सुखाते समय नागिन के समान लोट जाया करेगी। तू मुझे कच्चा दूध अधिक मात्रा में पिलाती है तथा माखन और रोटी नहीं देती है। माखन और रोटी बालकृष्ण को प्रिय हैं, लेकिन वे नहीं मिलते और चोटी बढ़ने की लालसा से उन्हें गाय का कच्चा दूध पीना पड़ता है। सूरदास कहते हैं कि माता बलाएँ लेने लगी और कहने लगी कि हरि हलधर दोनों भाइयों की जोड़ी चिरंजीव हो।

(स) सन्दर्भ :
पूर्ववत्।

प्रसंग :
भाई और बहन एक-दूसरे के सहायक हैं। हर उन्माद में बहन को भाई का ही सहारा है। दोनों मिलकर आजादी के गीत गाकर मातृभूमि के प्रति अपने कर्तव्य को पूर्ण कर सकते हैं।

व्याख्या :
कवि गोपाल सिंह नेपाली कहते हैं कि यहाँ पर भाई एक लहर के रूप में है, तो बहन नदी की एक धारा है। दोनों साथ-साथ रहकर कल्याण के मार्ग पर चलकर सभी का सहारा बनेंगे। गंगा-यमुना के संगम में पानी की अधिकता होने से कभी बाढ़ जैसी स्थिति हो जाती है और किनारे डूबने लगते हैं। ऐसे पागलपन भरे अवसर पर बहन को भाई का ही एकमात्र सहारा है। बेफिक्री के समय में भी भाई अपनी बहन के लिए ध्रुवतारे की तरह अडिग है। कुछ समय ऐसे आते हैं जब हमें अपने संयम को दृढ़ रखना है। भाई और बहन दोनों मिलकर आजादी के गीत गाकर ही मातृभूमि के प्रति अपने कर्तव्य को पूरा कर सकते हैं। अपनी मुसीबतों को झेलकर और बलिदानों के माध्यम से पत्थर-हृदय देश के दुश्मनों को चेतावनी देते हैं कि हम हर हाल में देश की रक्षा के लिए तत्पर हैं।

वात्सल्य और स्नेह काव्य सौन्दर्य

प्रश्न 1.
निम्नलिखित शब्दों के मानक रूप लिखिए
उत्तर:

प्रश्न 2.
निम्नलिखित पंक्तियों में अलंकार पहचान कर लिखिए
(अ) काढ़त गुहत न्हवावत ओछत, नागिन सी भुई लोटी।
(ब) मेरा जीवन क्रीड़ा-कौतुक तू प्रत्यक्ष प्रमोद भरी।
(स) काचो दूध पिआवत पचि-पचि, देत न माखन रोटी।
उत्तर:
(अ) उपमा अलंकार
(ब) अनुप्रास अलंकार
(स) पुनरुक्तिप्रकाश अलंकार।

प्रश्न 3.
निम्नलिखित शब्दों में से तत्सम और तद्भव शब्द छाँटकर लिखिए
ज्योति, उन्माद, बहन, कलंक, जननी,माटी,मैया, पूत,तात, पत्थर।
उत्तर:

प्रश्न 4.
गोद राखि चचुकारि दुलारति पुन पालति हलरावति।
आँचर ढाँकि बदन विधु सुंदर थन पय पान करावति॥
उपर्युक्त पंक्तियों में प्रयुक्त रस बताइये।
उत्तर:
वात्सल्य रस।

प्रश्न 5.
इस पाठ में से पुनरुक्ति प्रकाश और अनुप्रास अलंकार के उदाहरण छाँट कर लिखिए।
उत्तर:
(i) पुनरुक्तिप्रकाश-तू चिनगारी बनकर उड़ री,जाग-जाग मैं ज्वाल बनूँ।
(ii) अनुप्रास अलंकार-तू भगिनी बन क्रान्ति कराली,मैं भाई विकराल बनूँ।

सूर के बालकृष्ण भाव सारांश

‘सूर के बालकृष्ण’ नामक पदों के रचयिता भक्त कवि ‘सूरदास’ हैं। सूरदास ने शिशु सुलभ मुद्राओं, क्रीड़ाओं और शिशु के स्वभावगत सौन्दर्य को भी अपनी उदात्त और आकर्षक छवियों में प्रकट किया है।

वात्सल्य प्रेम में प्रेम का निश्छल,उदार और शिशु सुलभ स्वभाव प्राप्त होता है। शिशु के सुन्दर स्वरूप, आकर्षक मुद्राओं और उसके अबोध व्यवहारों के प्रदर्शन से यह वात्सल्य भाव दर्शक के हृदय में संचरित होता है। हिन्दी साहित्य में वात्सल्य भाव की रचनाओं की कमी है, किन्तु जितनी भी रचनाएँ प्राप्त होती हैं, वे वत्सलता के प्रभावी स्वरूप को प्रकट करती हैं। सूरदास ने शिशु सुलभ भुद्राओं,क्रीड़ाओं को तो अपने काव्य में स्थान दिया ही है साथ ही शिश के स्वभावगत सौन्दर्य को भी अपनी आकर्षक छवियों में प्रस्तुत किया है। शिशु कृष्ण अपने वाक्यचातुर्य से अपने दही खाने वाली बात को काट देते हैं। शिशु सहज विश्वासी होता है। शिशु कृष्ण से यदि माँ ने कह दिया कि गाय का दूध पीने से चोटी बढ़ती है तो कृष्ण खूब दूध पीते हैं और रोज-रोज माँ से पूछते हैं कि चोटी क्यों नहीं बढ़ रही? यहाँ शिशु की अधीरता दिखाई देती है। बच्चों में पारस्परिक चिढ़ाने का भाव है। कृष्ण अपनी माँ से बलराम के चिढ़ाने की शिकायत करते हैं। बाल सुलभ चेष्टाओं में मिट्टी खाने की प्रवृत्ति भी समाहित है। सूरदास ने इस तथ्य का आकर्षक वर्णन अपने पदों में किया है। सूरदास की भाषा में चमत्कृत कर देने वाला प्रवाह इन पदों में उपलब्ध है।

सूर के बालकृष्ण संदर्भ-प्रसंग सहित व्याख्या

(1) मैया मैं नाहीं दधि खायो।
ख्याल परे ये सखा सबै मिलि, मेरे मुख लपटायो।
देखि तही सीके पर भाजन, ऊँचे धर लटकायो।
तुही निरखि नान्हे कर अपने, मैं कैसे करि पायो।
मुख दधि पोंछि कहत नंदनंदन, दोना पीठि दुरायो।
डारि सॉट मुसुकाई तबहि, गहि सुत को कंठ लगायो।
बाल विनोद मोद मन मोह्यो, भक्ति प्रताप दिखायो।
सूरदास प्रभु जसुमति के सुख, शिव विरंचि बौरायो।

शब्दार्थ :
ख्याल परे = याद आया; लपटायो = लपेट दिया है; भाजन = बर्तन; निरखि = देख; नान्हे = छोटे,कर = हाथ; दोना = पत्तों का बना पात्र; साँटि = पीटने की डंडी; सुत = पुत्र; कंठ = गला; मोद = प्रसन्नता; मोह्यो = मोहित हो गया; विरंचि = ब्रह्माजी।

सन्दर्भ :
प्रस्तुत पद ‘वात्सल्य और स्नेह’ से सूरदास द्वारा रचित ‘सूर के बालकृष्ण’ नामक शीर्षक से उद्धृत किया गया है।

सन्दर्भ :
इसमें बालकृष्ण अपनी माता यशोदा से अपने प्रति की गई दही खाने की शिकायत को नकारते हुए बड़े ही बुद्धिकौशल से बाल सुलभ उत्तर देते हैं।

व्याख्या :
श्रीकृष्ण माता यशोदा से कहते हैं कि, हे माता! मैंने दही नहीं खाया है। मुझे याद आ रहा है कि इन सभी सखाओं ने मिलकर मेरे मुख पर दही लपेट दिया था। तू जानती है कि इतने ऊँचे टँगे हुए छींके पर दही का बर्तन रखा हुआ है। तू देख सकती है कि मेरे छोटे-छोटे हाथ हैं। इन छोटे हाथों से मैं कैसे उस दही के बर्तन को प्राप्त कर सकता हूँ। फिर नन्दकुमार बालकृष्ण ने दोने को पीठ के पीछे छिपाते हुए और मुख पर लगे हुए दही को पोंछते हुए उपर्युक्त बातें कहीं। यहाँ उनकी बाल सुलभ चतुरता का प्रदर्शन किया गया है। उन्हें भान है कि मुख पर लगे हुए दही से और हाथ में लगे दोने से उनकी चोरी पकड़ी जाएगी, अतः मुख को साफ कर लिया और दोने को पीठ के पीछे छुपा लिया।

यशोदा जी सब समझ गईं, लेकिन पीटने वाली लकड़ी को फेंक कर मुस्कराने लगी और मनमोहन बालकृष्ण को अपने गले से लगा लिया। श्रीकृष्ण के बाल विनोद के आनन्द ने माता यशोदा के मन को मोहित कर लिया। यहाँ श्रीकृष्ण के प्रति उनकी भक्ति का प्रताप दर्शाया गया है। सूरदास कहते हैं कि यशोदा जी और बालकृष्ण के सख को देखकर शिव और ब्रह्मा भी विवेक रहित होकर मोहित हो गए। अर्थात् श्रीकृष्ण यशोदा जी को अपनी बाल लीलाओं का जो सुख दे रहे हैं उससे किसी को भी ईर्ष्या हो सकती है और वह भ्रमित हो सकता है। निरंजन निराकार परब्रह्म साकार रूप में आकर यशोदा जी के आँगन में उनके प्रमोद के लिए जो क्रीड़ाएँ कर रहे हैं, ऐसा सुख शिव-विरंचि को भी दुर्लभ है।

काव्य सौन्दर्य :

1. बालकृष्ण की बाल सुलभ चतुरता दर्शनीय है।
2. ब्रजभाषा का प्रयोग।
3. अनुप्रास अलंकार का प्रयोग।
4. वात्सल्य रस है।

(2) मैया कबहिं बढ़ेगी चोटी।
किती बार मोहि दूध पिअत भई, यह अजहूँ है छोटी।
तू जो कहति बल की बेनी, ज्यौं, है है लॉबी मोटी।
काढ़त गुहत न्हवावत ओछत, नागिन सी भुइँ लोटी।
काचो दूध पिआवत पचि-पचि देत न माखन रोटी।
सूर श्याम चिरजीवौ दोऊ भैया, हरि हलधर की जोटी।। (2010, 15)

शब्दार्थ :
किती बार = कितनी बार; बल = बलदाऊ; बेनी = चोटी; काढ़त गुहत = काढ़ते में और गुहते में; न्हवावत = नहाने में; ओछत = सुखाने में; पचि-पचि = खूब; हरि हलधर = कृष्ण बलराम; जोटी = जोड़ी।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
प्रस्तुत पंक्तियों में बालकृष्ण अपनी माता यशोदा से यह शिकायत कर रहे हैं कि उनको कितने ही दिन दूध पीते हो गए, लेकिन उनकी चोटी बड़ी नहीं हुई है।

व्याख्या :
बालकृष्ण यह जानने को उत्सुक हैं कि उनकी चोटी कब बढ़ेगी। उनकी माता उन्हें यह भरोसा दिलाकर दूध पिलाती थीं कि दूध पीने से उनकी चोटी बढ़ जाएगी। वह माता से पूछते हैं कि हे माता मुझे कितनी ही बार (बहुत समय) दूध पीते हुए हो गईं, लेकिन यह चोटी अभी तक छोटी है। तेरे कथनानुसार मेरी चोटी बलदाऊ की चोटी के समान लम्बी और मोटी हो जाएगी, काढ़ते में, गुहते में,नहाते समय और सुखाते समय नागिन के समान लोट जाया करेगी। तू मुझे कच्चा दूध अधिक मात्रा में पिलाती है तथा माखन और रोटी नहीं देती है। माखन और रोटी बालकृष्ण को प्रिय हैं, लेकिन वे नहीं मिलते और चोटी बढ़ने की लालसा से उन्हें गाय का कच्चा दूध पीना पड़ता है। सूरदास कहते हैं कि माता बलाएँ लेने लगी और कहने लगी कि हरि हलधर दोनों भाइयों की जोड़ी चिरंजीव हो।

काव्य सौन्दर्य :

1. बच्चों को बहाने से दूध पिलाने के तथ्य को उजागर किया गया है।
2. ब्रजभाषा का सुन्दर प्रयोग हुआ है।
3. पुनरुक्तिप्रकाश,उपमा अलंकार का प्रयोग।

(3) मैया, मोहिं दाऊ बहुत खिझायो।
मोसों कहत मोल को लीनो, तोहि जसुमति कब जायो।
कहा कहौं यहि रिस के मारे, खेलन हौं नहिं जात।
पुनि-पुनि कहत कौन है माता, को है तुमरो तात।।
गोरे नन्द जसोदा गोरी, तुम कत स्याम सरीर।
चुटकी दै दै हँसत ग्वाल सब, सिखे देत बलवीर।।
तू मोही को मारन सीखी, दाऊ कबहुँ न खीझै।
मोहन को मुख रिस समेत लखि, जसुमति सुनि-सुनि रीझै।।
सुनहु कान्ह बलभद्र चबाई, जनमत ही को धूत।
सुर-श्याम मो गोधन की सौं, हों माता तू पूत।।

शब्दार्थ :
खिझायो = चिढ़ाना; जायो = पैदा किया; लखि = देखकर, रिसके = गुस्से में; तात = पिता; चबाई = चुगलखोर; धूत = धूर्त; सौं = सौगन्ध।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
बलदाऊ और कृष्ण अन्य ग्वाल-बालों के साथ बाहर खेलने जाते हैं तो बलदाऊ भिन्न-भिन्न प्रकार से उन्हें चिढ़ाते हैं। यहाँ यही शिकायत कृष्ण माता यशोदा जी से कर रहे हैं।

व्याख्या:
श्रीकृष्ण अपने बड़े भाई बलभद्र की शिकायत करते हुए अपनी माता से कहते हैं कि हे माता बलभद्र भाई मुझे बहुत चिढ़ाते हैं। मुझसे कहते हैं कि तुझे यशोदा जी ने जन्म नहीं दिया है, तुझे तो किसी से मोल लिया है। मैं क्या बताऊँ इस गुस्से के कारण मैं खेलने भी नहीं जाता। मुझसे बार-बार पूछते हैं कि तेरे माता-पिता कौन हैं। तू नन्द-यशोदा का पुत्र तो हो नहीं सकता क्योंकि तू साँवले रंग का है जबकि नन्द और यशोदा दोनों गोरे हैं। ऐसी मान्यता है कि गोरे माता-पिता की सन्तान भी गोरी होती है। यह तर्क बलदाऊ ने इसलिए दिया ताकि कृष्ण इसको काट न सके। इस बात पर सभी ग्वाल-बाल ताली दे-देकर हँसते हैं।

सभी ग्वाल-बालों को बलदेव सिखा देते हैं और सभी हँसते हैं, तब मैं खीझ कर रह जाता हूँ। तू सिर्फ मुझे ही मारना सीखी है दाऊ से कभी कुछ भी नहीं कहती कृष्ण के गुस्से से भरे मुख को बार-बार देखकर उनकी रिस भरी बातें सुन-सुनकर यशोदा जी अत्यन्त प्रसन्न होती हैं। मोद भरे मुख से यशोदा जी बोलीं, हे कृष्ण! बलदाऊ तो चुगलखोर है और जनम से ही धूर्त है। सूरदास जी कहते हैं, यशोदाजी कहने लगीं-मुझे गोधन (अपनी गायों) की सौगन्ध है,मैं माता हूँ और तू मेरा पुत्र है। इस कथन ने कृष्ण के गुस्से को दूर कर दिया।

काव्य सौन्दर्य :

1. ब्रजभाषा में अनूठा माधुर्य भर दिया है।
2. माता और पुत्र के प्रश्नोत्तर तार्किक दृष्टि से उत्तम हैं।
3. अनुप्रास व पुनरुक्तिप्रकाश अलंकार का प्रयोग है।

4. मो देखत, जसुमति तेरे ढोटा, अबहिं माटी खाई।
यह सुनिकै रिसि करि उठि धाई, बांह पकरि लै आई।।
इन कर सों भज गहि गाढ़े, करि इक कर लीने सांटी।
मारति हौ तोहिं अबहिं कन्हैया, वेग न उगिलौ माटी॥
ब्रज लरिका सब तेरे आगे, झूठी कहत बनाई।
मेरे कहे नहीं तू मानति, दिखरावौं मुख बाई॥ (2016)

शब्दार्थ :
ढोटा = पुत्र; भुइँ = भूमि; मुख बाई = मुख फैलाकर; गहि गाढ़े = जोर से पकड़कर।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
ग्वाल-बालों ने यशोदा माँ से यह शिकायत की है कि कान्हा ने मिट्टी खाई है। उनका विश्वास कर माँ उसे पकड़ लेती है।

व्याख्या :
कृष्ण के साथ खेलने वाले बालकों ने माता यशोदा से शिकायत की कि कान्हा ने मिट्टी खाई है। यह सुनकर माता को बड़ा क्रोध आया और कृष्ण को बाँह से पकड़ लिया। उनके हाथ से कहीं छूट कर भाग न जाय इसलिए कसकर बाँह पकड़ ली और एक हाथ में मारने के लिए एक डंडी ले ली। माता यशोदा ने कृष्ण से कहा हे कृष्ण ! मैं तुझे अभी मारूँगी नहीं तो जल्दी से मुँह से मिट्टी उगल दो। यह सुनकर बालकृष्ण ने डरते हुए अपनी माँ से कहा-ये सब ब्रज के ग्वाल-वाल तेरे पास आकर मेरी झूठी शिकायत लगाते हैं। मैं जानता हूँ कि तू मेरा कहना तो मानेगी नहीं इसलिए में अपना मुंह खोलकर दिखाता हूँ कि मुँह में माटी कहाँ है।

काव्य सौन्दर्य :

1. बाल सुलभ चेष्टाओं का सुन्दर वर्णन है।
2. ब्रजभाषा का सुन्दर प्रयोग किया है।
3. बच्चों के माटी खाने के तथ्य को उजागर किया गया है।
4. अनुप्रास अलंकार प्रयुक्त हुआ है।

भाई-बहन भाव सारांश

प्रस्तुत कविता ‘भाई-बहन’ सुप्रसिद्ध कवि ‘गोपाल सिंह नेपाली’ द्वारा रचित है। इसमें कवि ने भाई बहन के प्रेम को राष्ट्रीय प्रेम के परिवेश में व्यक्त किया है।

गोपाल सिंह नेपाली ने राष्ट्रीय चेतना के साथ-साथ जीवन की अनुभूतियों का वर्णन अपनी कविताओं में किया है। प्रस्तुत कविता में भाई-बहिन के प्रेम को राष्ट्र-प्रेम के रूप में व्यक्त किया है। यहाँ भाई अपनी बहन को सम्बोधित कर राष्ट्र की रक्षा हेतु समर्पित होने का आग्रह कर रहा है। इस कविता में अनेक प्रतीकों के माध्यम से राष्ट्र-प्रेम को उजागर किया गया है। बहन की बुद्धि और भाई की क्रियाशीलता मिलकर ही जीवन और राष्ट्र को आनन्दमय बना सकेगी। दोनों मिलकर आजादी के गीत गाकर अपने राष्ट्र के प्रति अपने प्रेम को प्रकट कर सकते हैं। कविता में दिए गए प्रतीक और बिम्ब मौलिक हैं।

भाई-बहन संदर्भ-प्रसंग सहित व्याख्या

1. तू चिनगारी बनकर उड़ री, जाग-जाग मैं ज्वाल बनें,
तू बन जा हहराती गंगा, मैं झेलम बेहाल बून,
आज बसन्ती चोला तेरा, मैं भी सज लूं लाल बनूँ
तू भगिनी बन क्रान्ति कराली, मैं भाई विकराल बनूँ
यहाँ न कोई राधारानी, वृन्दावन, वंशीवाला;
तू आँगन की ज्योति बहन री, मैं घर का पहरेवाला।

शब्दार्थ :
हहराती गंगा = कलकल ध्वनि करती गंगा; चोला = वेश; भगिनी = बहन; विकराल = भयानक।

सन्दर्भ :
प्रस्तुत पंक्तियाँ ‘वात्सल्य और स्नेह’ पाठ के ‘भाई-बहन’ शीर्षक कविता से ली गई हैं। इसके रचयिता गोपाल सिंह नेपाली हैं।

प्रसंग :
यहाँ पर भाई अपनी बहन से अपने देश की रक्षा के लिए बसन्ती चोला पहनने और अपने हृदय के अन्दर की ज्वाला को जलाए रखने को प्रेरित कर रहा है।

व्याख्या :
कविवर गोपाल सिंह नेपाली कहते हैं कि एक भाई अपनी बहन से देश की रक्षा के लिए जाग्रत रहने का आह्वान कर रहा है। भाई कहता है कि हे बहन! तू चिनगारी बनकर आकाश में उड़ और मैं ज्वाला बनकर देश-रक्षा के लिए जाग्रत रहूँ। तू कल-कल ध्वनि करती हुई गंगा बनकर बह और मैं बेहाल की तरह चलने वाली झेलम नदी बन कर बहूँ तू देश की रक्षा के लिए बसन्ती रंग के वस्त्र पहन ले और मैं भी देश के वीरों के रूप में सज जाऊँ। तू दुश्मन के लिए भयंकर क्रान्ति बन जा और मैं भयानक वीर बन जाऊँ ताकि उसकी दृष्टि हमारे देश पर न पड़े। यहाँ सिर्फ देश की रक्षा का सवाल है, सभी की एकता का प्रश्न है। यहाँ न तो वृन्दावन अलग है,न राधारानी अलग है और न नन्दलाल अलग है। सभी देश की रक्षा के लिए उद्यत हैं। भाई अपनी बहन से कहता है कि वह आँगन की ज्योति बनकर प्रकाश करे और वह घर का पहरेदार बनेगा। हम दोनों अपने प्रेम को देश-प्रेम पर न्यौछावर करते हैं।

काव्य सौन्दर्य :

1. भाषा साहित्यिक होने के साथ-साथ व्यावहारिक है।
2. देश-भक्ति का अनूठा समर्पण है।
3. अनुप्रास अलंकार है।
4. कविता में मधुरता है।

(2) बहन प्रेम का पुतला हूँ मैं, तू ममता की गोद बनी;
मेरा जीवन क्रीड़ा-कौतुक तू प्रत्यक्ष प्रमोद भरी;
मैं भाई फूलों में भूला, मेरी बहन विनोद बनी;
भाई की गति, मति भगिनी की दोनों मंगल-मोद बनी
यह अपराध कलंक सुशीले, सारे फूल जला देना।
जननी की जंजीर बज रही, चल तबियत बहला देना।

शब्दार्थ :
पुतला = मूर्ति; विनोद = मनोरंजन; प्रमोद = आनन्द; भगिनी = बहन।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
यहाँ पर भाई अपनी बहन को इस बात के लिए प्रेरित करता है कि यदि अपने देश की परतन्त्रता पुकार रही हो तो अपने सारे सुख भुलाकर उसकी रक्षा के लिए तैयार हो जाना।

व्याख्या :
भाई कहता है कि मैं तो प्रेम की मूर्ति बनकर रहा हूँ और तू ममता की गोद बनकर रही है। अर्थात् तूने हर किसी को अपनी ममता की छाँव में बैठाया है। मेरा जीवन क्रीड़ा और कौतूहल बनकर रहा है और तू साक्षात् आनन्दमयी बन कर रही है। मैं अपने जीवन के सुखों में खोया रहा और मेरी बहन मनोरंजन में डूबी रही। भाई की क्रियाशीलता और बहन की बुद्धि दोनों मिलकर आनन्द का साधन बनीं। भाई अपनी बहन को चेतावनी देते हुए कहता है कि हे बहन! हमारे सुख और आनन्द कहीं अपराध और कलंक न बन जायँ, इसलिए सभी सुखों को तिलांजलि देकर भारतमाता की परतन्त्रता की बेड़ियों की आवाज को सुन और चलकर उसे धैर्य दिला कि हम सब मिलकर उसे स्वतन्त्र करेंगे। हम सभी सुखों को अपने देश की रक्षा के लिए अर्पण कर देंगे।

काव्य सौन्दर्य :

1. अनेक प्रतीकों के माध्यम से कवि ने इस कथ्य को प्रकट किया है।
2. अनुप्रास अलंकार की छटा दृष्टव्य है।
3. वीर रस का पुट दिया गया है।

(3) भाई एक लहर बन आया, बहन नदी की धारा है।
संगम है, गंगा उमड़ी है, डूबा कूल-किनारा है;
यह उन्माद बहन को अपना, भाई एक सहारा है;
यह अलमस्ती, एक बहन ही भाई का ध्रुवतारा हैं;
पागल घड़ी, बहन-भाई है, वह आजाद तराना है।
मुसीबतों से बलिदानों से पत्थर को समझाना है। (2009)

शब्दार्थ :
तराना = ताल स्वर; अलमस्त = मतवाला, बेफ्रिक; उन्माद = पागलपन।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
भाई और बहन एक-दूसरे के सहायक हैं। हर उन्माद में बहन को भाई का ही सहारा है। दोनों मिलकर आजादी के गीत गाकर मातृभूमि के प्रति अपने कर्तव्य को पूर्ण कर सकते हैं।

व्याख्या :
कवि गोपाल सिंह नेपाली कहते हैं कि यहाँ पर भाई एक लहर के रूप में है, तो बहन नदी की एक धारा है। दोनों साथ-साथ रहकर कल्याण के मार्ग पर चलकर सभी का सहारा बनेंगे। गंगा-यमुना के संगम में पानी की अधिकता होने से कभी बाढ़ जैसी स्थिति हो जाती है और किनारे डूबने लगते हैं। ऐसे पागलपन भरे अवसर पर बहन को भाई का ही एकमात्र सहारा है। बेफिक्री के समय में भी भाई अपनी बहन के लिए ध्रुवतारे की तरह अडिग है। कुछ समय ऐसे आते हैं जब हमें अपने संयम को दृढ़ रखना है। भाई और बहन दोनों मिलकर आजादी के गीत गाकर ही मातृभूमि के प्रति अपने कर्तव्य को पूरा कर सकते हैं। अपनी मुसीबतों को झेलकर और बलिदानों के माध्यम से पत्थर-हृदय देश के दुश्मनों को चेतावनी देते हैं कि हम हर हाल में देश की रक्षा के लिए तत्पर हैं।

काव्य सौन्दर्य :

1. भाषा सरल और मधुर है।
2. भाई-बहन का स्नेह देश-प्रेम से मिलकर उदात्त बन गया है।
3. बिम्ब और प्रतीक नवीन रूप में दर्शाए गए हैं।

MP Board Class 12th Hindi Solutions

MP Board Class 11th Chemistry Important Questions Chapter 3 Classification of Elements and Periodicity in Properties

 Classification of Elements and Periodicity in Properties Important Questions

Classification of Elements and Periodicity in Properties Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
Variable valency is represented by:
(a) Metallic elements
(b) Normal elements
(c) Transitional elements
(d) Non – metallic elements.
Answer:
(c) Transitional elements

Question 2.
Of the following, whose size is the largest:
(a) Al
(b) Al⁺
(c) Al²⁺
(d) Al³⁺
Answer:
(a) Al

Question 3.
Electronic structure of the last shell of highly electronegative element is:
(a) ns²p³
(b) ns² np⁴
(c) ns²p⁵
(d) ns² np⁶
Answer:
(c) ns²p⁵

Question 4.
Electron affinity enthalpy depends on:
(a) On atomic size
(b) On nuclear charge
(c) On atomic number
(d) On both atomic size and nuclear charge
Answer:
(d) On both atomic size and nuclear charge

Question 5.
Electron gain enthalpy of noble gases is:
(a) Less
(b) Nearly zero
(c) High
(d) Very high.
Answer:
(b) Nearly zero

Question 6.
Alkali metals behave:
(a) As a good oxidising agent
(b) As a good reducing agent
(c) As a good hydrolyser
(d) None of these.
Answer:
(b) As a good reducing agent

Question 2.
Fill in the blanks:

1. Electron affinity of noble gases is …………………..
2. Electron affinity of Be and Mg of second group is …………………….
3. In a period, on moving from left to right, ionisation energy …………………………
4. Periodic law made by taking atomic mass as the basis is called …………………………….
5. …………………. is called Eka Boran
6. Na⁺ is …………………….. than Na.
7. Li shows diagonal relationship with ………………………..

Answer:

1. Zero
2. Zero
3. Increases
4. Mendeleev’s periodic law
5. Aluminium
6. Smaller
7. Mg.

Question 3.
Answer in one word/sentence:

1. Which scientist gives Modem periodic law?
2. Which element have highest electronegativity?
3. An element having atomic number 20, will be kept in which group in periodic table?
4. What is electron affinity of Noble gases?
5. What is the change in ionization energy on moving left to right in periodic table?
6. How many periods and groups in modem periodic table?
7. What is the other name of Eka – silicon?
8. How many elements are their in first period of modem periodic table?
9. What is the reason for the similarity in properties of Li and Mg?
10. Which oxidation state of A1 is most stable?

Answer:

1. Mosle (1913)
2. Fluorine
3. In second group
4. 0 (Zero)
5. Increases
6. Period = 7 and Group = 18,
7. Germanium
8. 2 (Two)
9. Diagonal relationship
10. +3.

Question 4.
Match the following:

Answer:

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

Classification of Elements and Periodicity in Properties Short Answer Type Questions – I

Question 1.
What is Mendeleev’s periodic law?
Answer:
Mendeleev’s periodic law:
“The physical and chemical properties of the elements are a function of their atomic masses”. This means that the physical and chemical properties of the elements get repeated after a fixed interval.

Question 2.
What is Modern periodic law?
Answer:
Modern periodic law: “Physical and chemical properties of elements are periodic functions of their atomic number”. This means when the elements are arranged in their increasing atomic numbers then after a fixed interval the elements having similar properties are repeated.

Question 3.
What is Doebereiner’s triad?
Answer:
This law was given by Doebereiner. According to this law atomic mass of the middle element was an arithmetic mean of the atomic masses of the other two.
Li₇ Na₂₃ K₃₉
That is \(\frac{39 + 7}{2}\) = 23.

Question 4.
Prove on the basis of quantum numbers that the 6th period of periodic table should have 32 elements?
Answer:
The sixth period begins with the filling of principal quantum number n = 6. In this period, the electrons enters in 6s, Af 5 d and 6p. In this sub – shells total 16 (1 + 7 + 5 + 3) orbitals are present. According to Pauli’s exclusion principle every’ orbital have maximum two electrons. So 16 orbitals should have only 32 electrons. So, in 6^th period 32 elements are present.

Question 5.
How many periods and groups were present in Mendeleev’s periodic table? What are the number of elements in each period?
Answer:
In Mendeleev’s periodic table 7 periods and 9 groups are present. In first period 2 elements, in second and third period 8 – 8 elements, fourth and fifth period, 18 – 18 elements, in sixth period 32 elements and seventh period is incomplete. In this period elements of atomic number 90 to 103, including actinides were present.

Question 6.
What change takes place on moving from left to right in period?
Answer:
On moving from left to right in period, ionization energy and electron affinity increases, but metallic property, basicity of oxides and atomic radii decreases.

Question 7.
What change occur on moving from top to bottom in a group?
Answer:
On moving top to bottom in a group atomic radii and ionic radii increases but ionization potential, electron affinity and melting point decreases.

Question 8.
What is diagonal relationship? Explain giving example?
Answer:
Diagonal relationship:
It has been observed that some of the elements of second period show similarities with the elements of third period present diagonally to each other, though belonging to different groups. For example, lithium resembles with magnesium of (group 2) and beryllium resembles with aluminium (of group 3) and so on. This similarity in properties of elements present diagonal is called diagonally relationship.

Question 9.
What is the difference between electron gain enthalpy and electro negativity?
Answer:
Electron gain enthalpy is the amount of energy released when an electron is added to the neutral gaseous atom. But electronegativity is the power of an element to attract covalent electron towards it.

Question 10.
Tell in the form of periods and group that Z = 114 will be placed?
Answer:
By writing the electronic configuration the maximum value of n will show the period.
₁₁₄Z = ₈₆[Rn]7s², 5f¹⁴, 6d¹⁰, 7p².
(∵ n = 7) period = 7
Group = 14(10 + 2 + 2).

Question 11.
Why the ionization potential of noble gases are high?
Answer:
Due to fully filled orbitals and complete octet, the electronic configuration of noble gases are stable. So to remove an electron from outer orbit high energy is required.

Question 12.
Why the electron affinity of halogen elements are high?
Answer:
Halogens have the large (-ve) value of (∆_egH) electron affinity because of their largest effective nuclear charge (Z_eff) and smallest size in their respective period. Also they have strong tendency to gain one electron to attain the stable noble gas configuration i.e., from ns², np⁵ to ns², np⁶ configuration. Thus, the electron gain process is highly exothermic for halogens. However, as we move from Cl to I, the electron enthalpies becomes less and less negative due to corresponding increase in their atomic size.

Question 13.
Whose electron affinity is more among F and Cl and why?
Answer:
Electron affinity of F is less than Cl because of small size of F, there is more repulsion among the electrons of 2p orbitals. So additive electron does not stabilize the atom.

Question 14.
Why the electron affinity of noble gases is zero?
Answer:
Noble gases have zero electron gain enthalpy. All noble gases have fully – filled valence shells (ns², np⁶). Due to their highly stable (ns², np⁶). Configurations, noble gases have absolutely no tendency to take an additional electron. Hence, noble gases have zero electron gain enthalpy.

Question 15.
Differentiate between Atomic radii and Ionic radii?
Answer:
Differences between Atomic radii and Ionic radii:
Atomic radii:

1. Atomic radius gives us idea about size of the atom. Atomic radius may be taken as the distance between the centre of the nucleus and the outermost shell of the atom. It can be measured either by X – ray or by spectroscopic methods.
2. Atomic radii of atom is more than ionic radii.

Ionic radii:

1. Ionic radius tells us about size of the ion. It is defined as the distance between centre of the nucleus and the point upto which the ion has influence in the ionic bond.
2. Ionic radii of anion is more than atomic radii.

Question 16.
Why the value of second IE₂ is more than IE₁ for any element?
Answer:
IE₁ value is the energy needed to remove first electron from the gaseous atom. With the loss of the electron. Therefore, Z _eff increases for the remaining electron. Therefore, greater energy is needed to remove the second electron from the gaseous atom. Therefore, IE₂ value of an element is always more than its IE₁ value.

Question 17.
Ionic radii of cation is less than atomic radii. Why?
Or, Size of cation is smaller than the related atom. Why?
Answer:
Cations are positively charged ions and are formed when a neutral atom loses ‘ one or more valency electrons. Thus a cation possesses the same nuclear charge but less number of electrons as compared to the parent atom. Therefore effective nuclear charge on remaining electrons increases. This causes the decrease in size. Hence, the radius of a cation is always smaller than the radius of the atom from which it is formed.

Question 18.
Ionic radii of anion is less than atomic radii. Why?
Answer:
Radii of anion:
Anions are negatively charged ions and are formed when one or more electrons are added to the valence shell of a neutral atom. Thus, an anion contains the same nuclear charge but more electrons as compared to its parent atom. This decreases the effective nuclear charge i.e., nucleus exert less influence on the valence electrons. The valence electrons move away from the nucleus and the size increases due to expansion of electron cloud. Hence, the radius of an anion is always larger than that of parent atom.

Question 19.
Why the value of electron affinity of N and Be is near zero?
Answer:
The 2s orbital of Be is complete so it is stable, so it does not allow the entry of any electron. Similarly the p orbitals of N is half filled and so stable and does not accept any electron. That is why the value of electron affinity for Be and N is near zero.

Question 20.
The size of Mg⁺² ion less than O^-2 ion whereas the electron configuration of both is same. Explain?
Answer:
The electronic configuration of Mg⁺² ion and O^2- ion is same. Both have eight electrons in valence shell but the nuclear charge of Mg⁺²2 ion is +12 whereas that of O^2- ion is +8. That is why the nuclear charge for outer electron in Mg⁺² is more than O^2-, so the size of Mg⁺² ion is less than O^2- ion.

Question 21.
The value of electron affinity increases on moving left to right in a period. Why?
Answer:
On moving across a period, the size of the atom decreases and nuclear charge 1 increases. Due to this the attraction for incoming electron increases. That is why the value of electron affinity increases in a period.

Question 22.
Explain the trend of metallic character of elements in periodic table?
Answer:
Metallic and non – metallic character:
Metallic character of an element is measured in terms of tendency of that element to lose electron i.e., electropositivity. Similarly, non – metallic character of an element is measured in terms of tendency of that element to gain electron i.e., electronegativity.

We know that ionisation enthalpies as well as electronegativities of the element increases along a period from left to right. This implies that the metallic character is maximum on extreme left (alkali metals) while the non – metallic character is the maximum on extreme right (halogens).

Question 23.
Among the O and N whose electron gain enthalpy is more?
Answer:
Among O and N, the electron gain enthalpy of O is more than N, because in N the valence orbital 2s²2p³ is half filled and so stable. So the tendency to accept the incoming electron is more in O than in N. Other than this the size of O atom is smaller than N atom so due to high nuclear charge the incoming electron added fastly.

Classification of Elements and Periodicity in Properties Short Answer Type Questions – II

Question 1.
The first IE₁ of Al is less than Mg, Why?
Answer:
The electronic configuration of A1 and Mg is:
Al₁₃ – 1s², 2s², 2p⁶, 3s², 3p¹
Mg₁₂ – 1s², 2s², 2p⁶, 3s²
The atom will be stable if it have full – filled or half – filled orbitals. In Al the p – orbital is vacant whereas in Mg the s orbital is fully filled. That is the orbital of Al is less stable than Mg. So, the energy required to remove an electron from Al atom is less in compare on to Mg atom, that is why the IE₁ of Al is less than Mg.

Question 2.
What do you mean by Shielding effect?
Answer:
Electrons have negative charge and they repel each other and this force of repulsion decreases the attractive force from nucleus to outer shell, due to this the electrons present in valency shell bounded loosely with the nucleus. In this way, the electrons present between nucleus and outer electrons causes shield and this effect is called Shielding effect.

Question 3.
Discuss the trend in ionization potential from moving left to right in a periodic table?
Answer:
On moving left to right in a periodic table the nuclear charge increases and atomic radius decreases, due to this the outer electrons are attracted more towards the nucleus and to remove an electron from outer shell more energy is required. So the ionization energy increases on moving left to right.

Question 4.
What do you think that the second electron gain enthalpy of O will be positive, more negative or less negative than first electron gain enthalpy? Give reason for your answer:
Answer:
O(g) + e^–_g → O^–_g; ∆_egH = – 141 kJ mol^-1
O^–(g) + e^–_g → O^2-_g; ∆_eg H = + 780 kJ mol^-1
When, on gaining one electron O atom forms O^– ion than energy is released. So first electron gain enthalpy is negative. But when an electron added in O^– ion than O^2- ion forms and it feels a strong repulsive force and so energy is required for addition of electron. Due to this the second electron gain enthalpy of O is positive.

Question 5.
The electron gain enthalpy of F is less negative than Cl, Why?
Answer:
Electron gain enthalpy of F is less – ve than that of Cl because when an electron is added to F, the added electron goes to the smaller n = 2 quantum level and suffers repulsion from other electrons present in this level. In case of Cl, the added electron goes to the larger n = 3 quantum level and suffer much less repulsion from other electrons.

Question 6.
Why the long form of periodic table is better than Mendeleev’s periodic table?
Answer:
The long form of periodic table is better than Mendeleev’s becasuse it has several advantages over Mendeleev’s table. The important advantages are:
1. The arrangement of elements is easy to remember and reproduce.

2. The elements have been classified on the basis of atomic number which is more fundamental than atomic weight of the elements.

3. This periodic table is closely connected to the electronic configuration of ele-ments. Therefore, the position of an element in the periodic table can easily be justified. The electronic configuration of an element can be predicted if its position in the periodic table is known.

4. The resemblances and differences of the properties of elements in periods and groups are explained on the basis of electronic arrangement in various shell i.e., electronic configuration. Thus it reflects trends in physical and chemical properties of the elements.

Question 7.
Write the characteristics of Modern periodic table?
Answer:
Characteristics of modern periodic table:

1. This periodic table is based on the electronic configuration of elements.
2. It reflects the sequence of filling the electrons in order of sub – energy levels s, p, d an Af
3. It gives clear division of elements, transitional elements and inner transition elements.
4. Attempt has been made to separate metals from non – metals. The elements on the left – hand side of the periodic table are metals and more non – metallic elements have been placed on right side.

Question 8.
Explain, why all transition elements are d-block elements but all d – block elements are not transition elements?
Answer:
Those elements whose outer electron goes in d – orbital are known as J – block elements or transition elements. The general outer electronic configuration of these elements is
(n – 1)d^1-10ns^0-2. The electronic configuration of Zn, Cd and Hg is (n – 1 )d¹⁰nsd², but they do not show the properties of transition elements. In ground state and in general oxidation state the d – orbitals of these elements are completely filled, therefore they will not be considered as transition elements. So on the basis of properties all transition elements are d – block elements, but on the basis of electronic configuration all d – block elements are not transition elements.

Question 9.
Among the pairs whose ionization energy is less. Why?

1. Cl or F
2. Cl or S
3. K or Ar
4. Kr or Xe.

Answer:

1. Among Cl and F ionization energy of Cl is less than F because size of F is small.
2. Among Cl and S, the ionization energy of S is less as the size of S is bigger than Cl.
3. Among K and Ar the ionization energy of K is less as in the valence shell of K only one electron is present which can be easily donated and K⁺ is formed but Ar has complete octet and so stable and so more energy is required for removal of electron.
4. Among Kr and Xe, the ionization energy of Xe is less because the size of Xe is bigger than Kr.

Question 10.
What are s – block elements? Write their main properties?
Answer:
Elements in which the last electron enters into s – orbital are known as s – block elements. It includes Alkali metals of group – I and Alkaline earth metals of group – II.
General properties of s – block elements:

1. In outermost orbit the electronic configuration is ns¹ or ns².
2. These elements have definite positive oxidation number which is +1 and +2 in group 1 and group 2.
3. Except hydrogen, all are metals, ionization potentials are low. Thus, these are strong electropositive, strong reducing agent and strongly metallic in nature.
4. Elements of 1st group are alkali metals and elements of 2nd group are called alkaline earth metals.
5. Oxides of these elements are basic in nature. Oxides of 1st group dissolve in water to give alkalis.
6. These elements form electrovalent compounds.
7. These elements provide colour on heating with flame.
   
8. These elements are very reactive. Reacts with water and acids displacing their hydrogen as H₂ gas.

Question 11.
What are p – block elements? Write their main properties?
Answer:
The elements in which the last electron outers in p – block are called p – block. General characteristic of p – block elements:

1. These elements have 2 electrons in 5 sub – shell and 1 to 6 electrons in p sub-shell in outermost orbit. Only in zero group elements, outer shell is completed.
2. Definite positive or negative oxidation numbers are represented by the elements. Some of the elements show variable valencies.
3. These elements form simple ions as well as complex ions of CO₃^2-, NO₃^–.
4. These are generally non – metals and metalloids. Some of the elements are heavy elements. e,g., Pb, Bi, etc.
5. Oxides are acidic in nature. Some oxides are amphoteric, e.g., PbO, SnO etc.
6. These elements form covalent compounds with each other but with the elements of s – block form electrovalent compounds.

Question 12.
Which elements are called transition elements? Throw light on their properties?
Answer:
Transition elements:
The elements which are in between s – block and p – block are called as transition elements or elements in which d – orbitals are partially filled are called transition elements. In this, last electron goes to d – orbital of orbit inner to outer so called d – block elements. Boiling point, melting point and densities of these elements are high.
Example: Cr, Mn, Fe, Cu, etc.
Properties:

1. In this two outer orbits are incomplete
2. Electronic configuration is (n – 1) d^{1 to 10} ns^{1 to 2}
3. It has metallic properties
4. These elements show variable oxidation state
5. These form coloured ions
6. These form complex salt
7. These are good catalysts
8. These form complex compounds
9. These are generally diamagnetic and
10. These also form with non-metal compounds.

Question 13.
What are inner transition elements? Write their general properties?
Answer:
In these elements three outer orbitals are incomplete. In all these elements, the s – orbital of the last shell (n) is completely filled, the d – orbitals the the penultimate (n – 1) shell invariably contains zero or one electron but the f – orbital of the antipenultimate (n – 2) shell (being lower in energy than d – orbitals of the penultimate shell) gets progressively filled. Hence the general configuration of f – block elements is (n – 2) f^1-14 (n – 1) d^{0 – 1}ns²
Lanthanides: The last electron enters in 4f subshell.
Actinides: The last electron enters in 5f subshell.

Question 14.
Explain ionization energy and electron affinity?
Answer:
Ionization Energy:
The minimum amount of energy which is needed to remove the most loosely bound electron from a neutral isolated gaseous atom in its ground state to form a gaseous cation.
M_g + Energy → M⁺_g + e^–_g
Electron Affinity:
It is equal to the change in enthalpy when an isolated gaseous atom accepts an electron to form a monovalent anion. It is denoted by ∆_egH.
In this way, if energy is released by addition of an electron than electron affinity is positive. But if energy is required to add an electron to a negative ion than electron affinity will be negative.

Question 15.
Explain atomic radii and ionic radii?
Answer:
Atomic radius:
The size of an atom is usually expressed in terms of its radius called atomic radius. It is very important property because several physical and chemical Cation properties are related to it. If the atom is assumed to be spherical than the term atomic radius means the distance from the centre of the nucleus to the outermost shell of electrons. According to quantum mechanical model, the atomic radius is defined as, the distance from the centre of the nucleus to the point up to which the density of electron cloud (i. e., probability of finding the electron) is maximum.

Ionic radius:
The ionic radius of an ion may be defined as the distance, from its nucleus to the point up to which the nucleus has influence on the electron cloud of the ion.

Question 16.
In a periodic table how the size of the atom changes? Explain?
Answer:
The distance from the centre of the nucleus to the outermost electron is called atomic radius.
In a period:
As we move left to right in a period atomic number increases by one unit in each successive element. As the addition of electron takes place in the same principal shell, they do not screen each other from the nucleus. It means the increased nuclear charge is not neutralize by the extra valency electron, cause a decrease in the size of atom.

In a group:
In moving down the group, nuclear charge is increasing with the increase in atomic number and we expect that the size of atom should decrease. But at the same time while going from one atom to another, there is increase in the number of electron shells. The effect of increased nuclear charge is reduced by shielding effect (screening effect) of the electrons present in the inner shells. Therefore, the effect of increase in the electron shell is more pronounced than the effect of increase in nuclear charge. Consequently, the atomic size or atomic radius increases down the group.

Question 17.
What is the change in reduction and oxidation properties of elements in periodic table?
Answer:
Reducing agents:
Those elements which form positive ion by removal of electron, are called reducing agents. The reducing power of any element depends upon the tendency of it to donate electron. The ionization potential of alkali elements are less due to their large size. Therefore they can easily donate electrons, so alkali elements are strong reducing agents. On moving from left to right in a periodic table in a period the reducing power decreases and on moving top to bottom in a group reducing power increases.

Oxidizing agents:
Those elements which accept electrons will work as oxidizing agents. The oxidizing power of any element depends upon its electron accepting property. Halogen easily accept electron so they are strong oxidizing agent. On moving left to right in a period the oxidizing power increases and on moving top to bottom the oxidizing power decreases.

Question 18.
Differenctiate the Modern periodic table and Mendeleev’s periodic table?
Answer:
Differences between Modem periodic table and Mendeleev’s periodic table:
Modem Periodic Table:

1. In Modem periodic table the elements are arranged in increasing order of atomic number.
2. The elements are kept in 18 groups.
3. The metals and non – metals are separately kept.

Mendeleev’s Periodic Table:

1. In Mendeleev’s periodic table the elements are arranged in increasing order of atomic mass.
2. The elements are kept in 9 groups and 7 sub – groups.
3. The metals and non-metals are not separated.

Classification of Elements and Periodicity in Properties Long Answer Type Questions – I

Question 1.
How many types of elements are there on the basis of electronic configuration? Give example of each?
Or Describe the different types of element on the basis of electronic arrangement?
Answer:
The elements have been divided into four main classes as follows:

1. Inert gases or Noble gases or Aerogens:
The elements which belongs to the group 18 of the periodic table are known as noble gases or inert gases. They have 8 electrons in their outermost shell (ns2 np6). Therefore their combining capacity or valence is zero. Hence they are inert in nature. All the members are gaseous in nature.

2. Representative elements:
All the elements of s and p – block with exception of noble gases are called representative elements or normal elements. Representative elements includes

These elements have the outer electronic configuration from ns¹ to ns²np⁵. Elements of this type use only thier outer shell electrons in the bonding with the other atom.

3. Transition elements (d – block elements):
These are the elements in which the last shell is not completely filled. These elements have configuration (n – 1)d^1-10ns^1-2 i.e., these elements contain 1 – 9 electrons in the penultimate shell and 1 – 2 electrons in the valence shell. The elements having partially filled d – orbitals are named transition elements because these represent a transition (change) from the most electropositive elements to the most electronegative elements. Strictly speaking, elements of group 12 (Zn, Cd, Hg) cannot be included in transition elements, although these elements chemically resemble the transition elements in many respects. All these elements closely resemble each other due to the presence of the same number of electrons in the outermost shell.

4. Inner transition elements (f – block elements):
Lanthanides (₅₈Ce – ₇₁Lu) and actinides (₉₀Th – ₁₀₃Lw) are collectively known as inner transition elements. In these elements, three outermost shells are not completely filled. The last electron in those enters in the (n – 2) subshell. These elements generally have the same number of electrons in the last two shells, (n -1) and n – orbital. The properties of these elements are so close to each other that it is not easy to separate lanthanides from one another.

Question 2.
Write the applications of Mendeleev’s periodic table?
Answer:
The important contributions of Mendeleev’s periodic table to chemistry are as follows:
1. Systematic study of chemistry:
The Mendeleev’s periodic table categorised the elements for the first time in a systematic way. This helped immensely in the study of the chemistry of elements and their compounds and made the study much easier. If the properties of a group are known, one can easily guess the properties of all the elements (and their compounds) placed in that group.

2. Prediction of new elements:
At the time of Mendeleev’s 56 elements were known. While arranging these elements, he left some gaps. These gaps represented the undiscovered elements. Further Mendeleev predicted the properties of these undiscovered elements f on the basis of their positions. For example, both gallium (Ga) and germanium (Ge) were not known when Mendeleev gave the periodic table. On the basis of their expected positions, he named these elements as Eka – aluminium and Eka – silicon (from the Sanskrit word eka meaning next) because he believed that these would be similar to aluminium and silicon respectively. When these elements were actually found, their properties were the same as predicted by him.

3. Correction of atomic masses:
Mendeleev also corrected the atomic masses of certain elements with the help of their expected positions and properties. For example, be – ryllium was assigned an atomic weight of 13 – 5 on the basis of its equivalent weight (4 – 5) and valency (wrongly calculated as 3). As such, it should have been placed between carbon (atomic mass 12) and nitrogen (atomic mass 14).

But no vacant place was available in between C and N and further more properties of beryllium did not justify such a position. Therefore valency 2 was assigned to beryllium which gave it an atomic weight of 4.5 × 2 = 9, as atomic weight = equivalent weight × valency and it was placed in proper position between lithium (atomic weight 7) and boron (atomic weight 11). In a similar way atomic masses of many other elements were corrected.

4. Useful in research:
Mendeleev’s periodic table is very useful to further research and study of properties of different elements.

Question 3.
What do you mean by electronegativity of an element? How it is different from electron affinity? How electronegativity changes in periodic table?
Answer:
1. Electronegativity:
It is defined as the relactive tendency of an element to attract the shared pair of electrons in a covalent bond towards itself.

2. Electron affinity:
Electron affinity is equal to the change in enthalpy when an isolated gaseous atom accepts an electron to form a monovalent anion.
The value of first electron affinity is negative as the process is exothermic but in second electron affinity energy is absorbed because the negative ion repels the upcoming electron.

3. Difference:
Electron affinity is the property to bind extra electron whereas electro negativity is the ability of an atom of a molecule to attract electron.
Trend of Electronegativity:

4. In a period:
As we move left to right in a period electronegativity increases due to decrease in size and a corresponding increase in effective nuclear charge.

5. In a group:
While moving down a group the increased nuclear charge is neutralized by screening effect. In general, the electronegativity decreases down the group due to in-creasing size.

Question 4.
The atomic number of an element is 17. By giving the electronic configuration give its place in periodic table?
Answer:
The electronic configuration of atom having atomic number 17 = Is², 2s² 2p⁶, 3s², 3p⁵ = 2, 8, 7.
1. Its outer configuration is 3s²3p⁵, so it is a p – block elements.

2. Group:
If one orbital is vacant, than their present in outer shell show the group i.e., the element is 7th group.

3. Sub – group:
If the last electron enter in s or p-sub-shell than the element is of a sub – group.

4. Period:
No. of shells show the no. of periods. This element is member of third period. group – 7, sub – group – A, period – 3.

Question 5.
What do you mean by isoelectronic species? Write the name of any one species which is isoelectronic with following atoms and ions:

1. F^–
2. Ar
3. Mg²⁺
4. Rb⁺

Answer:
In isoelectronic species the number of electrons are same but the nuclear charge is different. In such species by increasing nuclear charge, the size of the atom increases.

1. Number of e^– in F^– = 9 + 1 = 10
2. Number of e^– in Ar = 18
3. Number of e^– in Mg⁺² = 12 – 2 = 10
4. Number of e^– in Rb⁺ = 37 – 1 = 36

N^3-, O^2-, Ne, Na⁺ and Al⁺³ species, F^– and Mg²⁺ are isoelectronic species. P^3-, S^2-, Cl^–, K⁺ and Ca²⁺ species, Ar is isoelectronic species.
Similarly Br^–, Kr and Sr²⁺ species and Rb⁺ is isoelectronic species.

Classification of Elements and Periodicity in Properties Long Answer Type Questions – II

Question 1.
What is Modern Periodic law? Give the description of periodic table based on this law?
Answer:
Modern Periodic Law:
“The physical and chemical properties of elements are the periodic functions of their atomic numbers”.

Description of Periodic table:
Characteristics of Modern Periodic table:

1. Metals and non – metals are kept separate.

2. Strongly electropositive elements (s – block elements) are kept on the left side of transitional element (d – block elements) in groups 1 and 2, transitional elements are kept in the middle in groups 3 to 12 and non-metallic elements (p – block elements) are kept on the right side of transitional element (d – block elements) in groups 13 to 17.
Thus, s – block
1 – 2 groups
d – block
3 – 12 groups
p – block
13 – 17 groups

3. On the basis of filling of electrons, elements are classified into s – block, p – block, d – block and f – block elements.

4. Transitional elements are kept apart from normal elements. These elements lie in between s – and p – block. In them the last electron enters into (n – 1) d orbital therefore, these elements show similarly in properties rather than gradation.

5. Rare earth elements (Lanthanides and Actinides) are kept away from the main table in a suitable place.

6. Sub – groups of a group are eliminated.

7. Fe, Co and Ni are kept with transitional elements which is the suitable place.

8. Hydrogen is kept in group 1.

9. This classification relates the atomic configuration of elements with their position in the periodic table due to which their study is very easy.

Question 2.
What are the defects of Mendeleev’s periodic table ? How they can be remove by Modern Periodic Law?
Answer:
Defects of Mendeleev’s Periodic table:
Inspite of remarkable contribution made to chemistry, Mendeleev’s periodic table had certain defects. These defects are as follows:

1. Position of Hydrogen:
Hydrogen is a unique element and resembles with the elements of group IA (alkali metals) as well as with those of VII A group (Halogens) in its properties. Therefore, it should have been placed in both IA and VII A groups in Mendeleev’s periodic table. Therefore, the position of hydrogen in the periodic table is anomalous or controversial.

2. Position of Isotopes:
Isotopes are the atoms of same element having same atomic number but different atomic masses. Therefore, according to Mendeleev’s classification, these should be placed at different places depending upon their atomic masses. However, isotopes have not been given separate places in the periodic table.

3. Position of Lanthanides and Actinides:
Position of lanthanides (14 elements following lanthanum, atomic number 58 – 71) and actinides (14 elements following actinium, 1 atomic number 90 – 103) is also anomalous in Mendeleev’s periodic table. In this periodic table, all these elements are supposed to be placed together in III group which is not in accordance to the periodic law.

4. Similar elements placed in different group:
The elements like silver and thallium, barium and lead, copper and mercury show similar properties, yet they are placed in . different groups in Mendeleev’s periodic table.

5. Eighth group:
This group is full of anomalies. Except osmium, no other elements of this group shows the group valency i.e., no other element is octavalent. Further they all are arranged in three triads without any justification.

6. Cause of periodicity:
Mendeleev could not explain the cause of periodicity among the elements.

Modern periodic table removes a lot of defects of Mendeleev’s periodic table like:

1. Since, this table is based on electronic configuration therefore hydrogen is placed in group 1.

2. To put element of high atomic mass before the element of low atomic mass:
In Mendeleev’s periodic table, elements of higher atomic mass precedes elements of lower atomic mass. But their position is justified in Modem periodic table because element with higher atomic mass has lower atomic number, e.g., cobalt (atomic mass 58.93, atomic number 28).

3. Position of Isotopes:
Isotopes of an element possess same atomic number, therefore, there is no necessity of giving them separate positions in the periodic table.

4. To place elements with different properties at the same place: In Mendeleev’s periodic table, elements of different properties are placed at same place. For example, elements of sub – group IA and IB. But in Modem periodic table, this defect is removed by separating the sub – groups.

5. Position of Noble gas:
Noble gases have been placed between VIIIB electromotive elements and IA electropositive elements.

6. Diagonal relationship:
It can be explained on the basis of electronic configuration and atomic radius.

Question 3.
Compare the Mendeleev’s periodic table and Modern periodic table?
Answer:
Comparison between Mendeleev’s periodic table and Modem periodic table:
Mendeleev’s periodic table:

1. Elements are arranged in order of their increasing atomic weight.
2. There are 9 horizontal blocks known as groups.
3. Zero group is added later on.
4. Elements of different properties kept in same group.
5. Every isotope don’t have different place.

Modem periodic table:

1. Elements are arranged in order of their increasing atomic number.
2. There are 18 horizontal blocks called groups.
3. Nobel gases are at the end of each period.
4. Elements of different properties kept in different groups.
5. There is no need to give different place to isotopes.

Question 4.
What is ionisation energy? Explain the factors affecting ionisation energy?
Answer:
Ionization energy:
The amount of energy which is required to separate the outermost electron from atom is called ionic energy or ionic potential i.e., to convert atom into positive ion the necessary energy required is called as ionic energy. Its unit is kJ mole^-1.
So, M_(g) + Energy → M⁺_(g) + e^–
M_(g) – Electron → M⁺_(g) + Ionisation energy.
Factors affecting ionization energy:
1. Size of atom or ion:
Greater the size of atom or ion, weaker are the forces of attra-ction and lower is the value of ionization energy.

2. Nuclear charge:
Greater the nuclear charge, more is the attraction for electrons and hence, greater is the value of ionization energy.

3. Shielding effect:
In multielectron atom with increase in atomic number, shielding effect increases due to which valence shell electron feels lesser attraction and hence, value of ionization energy is lower.

4. Penetration effect:
Simple the shape of orbitals, more is the penetration of it for the nucleus. That is it experience greater attraction. Thus, value of ionization energy is higher. It follows the order s > p > d > f

5. Electronic configuration:
Completely filled and half-filled orbitals are more stable than any other arrangement. Thus, value of ionization energy is higher for it.

6. Trend in periodic table:

(a) In period:
Ionisation energy generally increases from left to right in a period. This is due to gradual increase in nuclear charge and decrease in atomic size of the elements.

(b) In a group:
There is a gradual decrease in ionisation energy moving top to bottom. This is due to increase in the number of the main energy shell i.e., size of the atom increases.

Question 5.
What do you mean by Electron affinity? Explain the factors affecting it?
Answer:
Electron affinity:
The electron affinity of an atom is the energy released, when an electron is added to a neutral atom. It is expressed in kcal/mol. Electronegativity is a relative number on an arbitrary scale while electron affinity is expressed in the units of energy (kcal/mol).
A_g + e^– → A^–_g + E₁ (Exothermic)
A^–_g + e^– → A^2- – E₂ (Endothermic)
A^2-_g + e^– → A^3-_g – E₃ (Endothermics)
This way in the addition of an electron to neutral isolated atom energy is released and electron affinity is positive, but in an anion energy is required to add an electron because the anion opposes the intrance of an electron. Thus, electron affinity for charged ions is negative.

Factors affecting electron gain enthalpy: Some important factors affecting electron gain enthalpy are:

1. Atomic size
2. Nuclear charge
3. Electronic configuration.

1. Atomic size:
As the atomic size increases, the distance between the nucleus and the incoming electron increases. This results in lesser attraction. Consequently, electron gain enthalpy becomes less negative.

2. Nuclear charge:
Greater the nuclear charge, greater will be attraction for the incoming electron and as a result, AH becomes more negative.

3. Electronic configuration:
Atoms having stable electronic configuration have lesser tendency to accept the electron. Due to this the value of electron gain enthalpy becomes less negative.

4. Half – filled and ful – filled orbitals:
Half – filled and ful – filled orbitals are stable and in this condition they cannot easily accept the electron. So the value of electron affinity for such elements are zero.

5. Periodicity:

1. In periods: On moving left to right in a period the electron affinity increases as the nuclear charge increases.
2. In group: On moving down the group the electron affinity decreases due to increase in size of atom.

MP Board Class 11 Chemistry Important Questions

MP Board Class 12th Physics Important Questions Chapter 5 Magnetism and Matter

Magnetism and Matter Important Questions

Magnetism and Matter Objective Type Questions

Question 1.
Choose the correct answer of the following:

Question 1.
If a bar magnet is placed with its north pole pointing towards geographical north and south pole pointing towards geographical south, then the neutral point will be:
(a) Situated on the axial line
(b) Situated on the equatorial line
(c) Situated at a place where is neither axial line nor equatorial line
(d) Not formed at all.
Answer:
(b) Situated on the equatorial line

Question 2.
Magnetic moment is which type of physical quantity :
(a) Scalar
(b) Vector
(c) Neutral
(d) None of these.
Answer:
(b) Vector

Question 3.
If a bar magnet of magnetic moment M is divided in two equal parts, the magnetic moment of each part will be :
(a) 2M
(b) \(\frac {M}{2}\)
(c) M
(d) Zero
Answer:
(b) \(\frac {M}{2}\)

Question 5.
‘weber’ is the unit of:
(a) Magnetic moment
(b) Magnetic induction
(c) Magnetic field
(d) Magnetic flux
Answer:
(d) Magnetic flux

Question 6.
The two magnetic lines of force :
(a) Meet each other at the poles
(b) Meet at the neutral point
(c) Never meet
(d) All the above statements are correct.
Answer:
(c) Never meet

Question 7.
The intensity of magnetic field is defined as :
(a) Magnetic moment per unit volume
(b) Magnetic induction force acting on unit magnetic pole
(c) Number of magnetic lines of force passing per unit area
(d) Number of lines of force passing through unit volume.
Answer:
(c) Number of magnetic lines of force passing per unit area

Question 8.
A magnetic needle is placed in a non – uniform magnetic field. The needle will experience :
(a) A force without any torque
(b) A torque without any force
(c) A force and a torque
(d) Neither a torque nor a force.
Answer:
(c) A force and a torque

Question 9.
The ratio of magnetic field intensities at equal distance in end on position and broadside on position of a small bar magnet is :
(a) 1 : 4
(b) 1 : 2
(c) 1 : 1
(d) 2 : 1.
Answer:
(d) 2 : 1.

Question 10.
A magnetic dipole of magnetic moment M is placed in a magnetic field of intensity B with its axis along the magnetic field. The work done in rotating it by 180° is :
(a) -MB
(b) MB
(c) Zero
(d) +2 MB
Answer:
(d) +2 MB

Question 11.
If the net magnetic moment of individual atom of a substance is zero, the substance is :
(a) Diamagnetic
(b) Paramagnetic
(c) Ferromagnetic
(d) Non – magnetic.
Answer:
(a) Diamagnetic

Question 12.
Electromagnets are made up of:
(a) Paramagnetic substances
(b) Soft iron
(c) Steel
(d) Diamagnetic substances.
Answer:
(b) Soft iron

Question 13.
At equator the total intensity of earth’s magnetic field is equal to :
(a) V
(b) H
(c) Both
(d) None of these.
Answer:
(b) H

Question 14.
The resultant intensity of earth’s magnetic field at a place is given by :
(a) \(\frac {H}{V}\)
(b) \(\frac {V}{H}\)
(c) \(\sqrt { { H }^{ 2 }+V^{ 2 } }\)
(d) \(\sqrt { { H }^{ 2 }-V^{ 2 } }\)
Answer:
(c) \(\sqrt { { H }^{ 2 }+V^{ 2 } }\)

Question 15.
The value of angle of dip near the magnetic poles is :
(a) 90°
(b) 45°
(c) 30°
(d) Zero.
Answer:
(a) 90°

Question 16.
The south pole of earth’s magnet is :
(a) Near the geographic north pole
(b) Near the geographic south pole
(c) In geographic east
(d) In geographic west.
Answer:
(a) Near the geographic north pole

Question 17.
The angle of dip at equator is :
(a) 90°
(b) 30°
(c) 0°
(d) 45°
Answer:
(c) 0°

Question 18.
In a plane perpendicular to the magnetic meridian, a dip needle :
(a) Will be horizontal
(b) Will be vertical
(c) Will be inclined at angle of dip at that place
(d) Will be inclined at any angle.
Answer:
(b) Will be vertical

Question 2.
Fill in the blanks :

1. The SI unit of pole strength is ……………………….
2. The direction of magnetic moment of a magnet is always from ………………………. to ………………………. pole.
3. The SI unit of magnetic moment is ……………………….
4. The magnetic lines of force are ………………………. curve.
5. The tangent drawn at any point of a magnetic line of force gives ……………………….
6. The magnetic field produced due to a solenoid is same as that produced by a ……………………….
7. A magnet is also called a ……………………….
8. At same distance, magnetic field intensity in broadside on position is ………………………. the intensity in end on position.
9. Nowadays magnetic lines of force are called ……………………….
10. At ………………………. point the resultant intensity of magnetic field is zero.
11. The temperature at which ferromagnetic substance is converted into paramagnetic substance is known as ……………………….
12. The strength of ………………………. magnet can be changed.
13. The vertical component of earth’s magnetic field at a place becomes zero where angle of dip is ……………………….
14. The angle of dip from equator to poles lies between ………………………..
15. ………………………. substances can easily be magnetized.

Answer:

1. ampere x metre
2. south; north
3. ampere x metre²
4. Closed
5. Direction of magnetic field
6. Bar magnet
7. Magnetic dipole
8. Half
9. Magnetic field line
10. Neutral
11. Curie temperature
12. Electro
13. Zero
14. Zero to 90°
15. Ferro – magnetic.

Question 3.
Match the Columns :
I.

Answer:

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

II.

Answer:

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

III.

Answer:

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

Question 4.
Write the answer in one word / sentence :

1. Name the elements or parameters of earth’s magnetic field.
2. What is the value of angle of dip at poles and equator?
3. How is the relative permeability (μ_r) ofthe material related to susceptibility (x_m)?
4. Give two examples of diamagnetic substance.
5. Name any two paramagnetic substances.

Answer:

1.

• Declination
• Angle of dip
• Horizontal component of earth’s magnetic field

2. Angle of dip at poles is 90° and at equator it is 0°

3. μ_r = 1 + x_m

4. Zinc and Bismuth

5. Aluminium and Manganese.

Magnetism and Matter Very Short Answer Type Questions

Question 1.
Define effective length of a magnet?
Answer:
The distance between two poles of magnet is called its effective length.

Question 2.
What do you mean by intensity of magnetic field? Write its SI unit. Is it scalar or vector?
Answer:
Intensity of magnetic field:
The intensity of field at a point is defined by the force experienced by a unit north pole, placed-at that point.
Its SI unit is tesla or weber metre^-2. Magnetic field is a vector.

Question 3.
Define magnetic lines of force.
Answer:
1st definition:
The magnetic lines of force are the curves in the magnetic field, on which if a unit north pole is placed, then it will follow the imaginary curve drawn.

2nd definition:
“A magnetic line of force is a smooth curve in a magnetic field such that the tangent at any point on it gives the direction of the magnetic field at that point.”

Question 4.
Can two magnetic lines of force intersect?
Or
Magnetic lines of force do not intersect each other, why?
Answer:
No. If the two magnetic lines of force intersect, then there will be two tangents and hence two directions of magnetic field at the point of intersection. This is impossible.

Question 5.
Define magnetic moment. Write its SI unit. Is it a scalar or a vector?
Answer:
The product of pole strength (m) and effective length (2l) of the magnet is called magnetic moment (M).
If m be the pole strength and 2l be the effective length, then
M = m x 2l
SI unit of magnetic moment is weber x metre. It is a vector quantity having a direction from south pole to north pole.

Question 6.
What is a diamagnetic substance?
Answer:
A substance which when placed in a magnetizing field develops very weak magnetization in the opposite direction of the applied field is called diamagnetic substance.

Question 7.
What is paramagnetic substance?
Answer:
A substance which when placed in a magnetizing field develops weak magnetism in the direction of the applied field is called paramagnetic substance.

Question 8.
What is paramagnetism?
Answer:
The atoms or molecules of some materials (e.g., Al, CuCl₂) have non – zero magnetic moment. When such a substance is placed in a magnetic field \(\vec { B }\), the individual magnetic dipoles align in the direction of \(\vec { B }\). There is net magnetization in the direction of \(\vec { B }\) and proportional to \(\vec { B }\) This is called paramagnetism.

Question 9.
What are ferromagnetics or ferromagnetic substances?
Answer:
Ferromagnetics are the substances, which when placed in a magnetic field are strongly magnetized in the direction of the magnetizing field. Example Fe. Ni, Co etc.

Question 10.
Write about the number of electrons in diamagnetic and paramagnetic substances.
Answer:
The number of electrons in diamagnetic substances are in even number and in paramagnetic substances electrons are in odd numbers.

Question 11.
Does the magnetism of paramagnetic salts depend upon temperature? Give reason.
Answer:
Yes, with the increase of temperature its magnetism decreases. When a paramagnetic salt is placed in a magnetic field then on each elementary magnet, a torque acts which tends to bring them in the direction of magnetic field. When the temperature is increased the thermal agitation opposes this tendency, hence the paramagnetism is decreased.

Question 12.
Define magnetic intensity. Give its SI unit.
Answer:
Magnetic intensity is the ability of a magnetizing field to magnetize a material and is defined as the number of ampere turns flowing around unit length of solenoid required to produce magnetic induction B0 inside it.
H = \(\frac { { B }_{ 0 } }{ { \mu }_{ 0 } }\)
SI unit of H is Am^-1

Question 13.
Define magnetic permeability. State its SI unit.
Answer:
Magnetic permeability is defined as the ratio of magnetic induction B to the magnetizing field intensity H ie., μ = \(\frac {B}{H}\)
SI unit is TmA^-1.

Question 14.
Why the magnetic property increases in paramagnetic substances with cooling?
Answer:
When a paramagnetic substances is kept in an external magnetic field, then on each elementary magnet a torque acts which tries to bring them parallel to the direction of magnetic field. The thermal vibrations opposes it. If the temperature is decreased, then thermal vibrations decreases, hence the magnetic property increases.

Question 15.
Why is diamagnetism independent of temperature?
Answer:
The induced magnetic moment in diamagnetic sample is always opposite to the magnetizing field, no matter what the internal motion of atom is.

Question 16.
What is Curie point?
Answer:
Curie point is the temperature above which a ferromagnetic substance becomes paramagnetic.

Question 17.
At any point on the surface of earth, horizontal component of earth magnetic field and vertical component of it are equal. What will be the angle of dip at that point?
Solution:
According to question H = V
Or B_H = B_v
But tanθ = \(\frac { { B }_{ v } }{ { B }_{ H } }\)
Or tanθ = \(\frac { { B }_{ v } }{ { B }_{ v } }\)= 1 =tan45°
θ = 45°
Angle of dip will be 45°.

Question 18.
When a bar magnet is cut into two equal pieces perpendicular to its axis, then what will be its charge in magnetic moments.
Answer:
In this position, magnetic moment of each pieces will be M’ = m’ x 2l
But m’= \(\frac {M}{2}\)
∴ M’ = \(\frac {m}{2}\) x 2l
= \(\frac {M}{2}\)
Therefore magnetic moment will become half of its initial value.

Magnetism and Matter Short Answer Type Questions

Question 1.
Write Coulomb’s law of magnetism and define the unit magnetic pole with its help.
Answer:
Coulomb’s law:
The force of attraction or repulsion between two magnetic poles is directly proportional to the product of pole strength and inversely proportional to the square of the distance between them and acts along the line joining the roles.
Let m₁ and m₂ be the pole strengths and d be the distance between them, then

Unit pole:
If F = 10^-7N, d = 1m and m₁=m₂ = m, then putting the values in eqn. (2),
we get
10^-7 = 10^-7\(\frac { { m }^{ 2 } }{ 1 } \) ⇒ m = ±1
Thus, if two similar poles are kept 1m apart in vacuum and repei each other by a force of 10^-7N, then the poles are called unit poles.

Question 2.
What is end – on – position or axial position? Derive an expression for the intensity of field at a point on the axis of a bar magnet. What is the direction of resultant field?
Or
Derive an expression for the intensity of field at a point on the axial position of a bar magnet.
Answer:
End – on – position:
The point where the intensry of the magnetic field is to be found, is on the magnetic axis, then this point is called end – on – position.

Let NS be a bar magnet of pole strength m and effective length 2l. Consider a point P on its axis at a distance d from the centre of the magnet O. Magnetic field at P has to be found out.
Now, the intensity of field at P due to N – pole :
B₁ = \(\frac { { \mu }_{ o } }{ 4\pi } .\frac { m }{ N{ P }^{ 2 } }\),(along\(\vec { NP } \))
NP = OP – ON =d – l
∴ B₁ =\(\frac { { \mu }_{ o } }{ 4\pi } .\frac { m }{ (d-1)^{ 2 } }\) … (1)
Similarly, the intensity of field at P due to S – pole :
B₂ = \(\frac { { \mu }_{ o } }{ 4\pi } .\frac { m }{ S{ P }^{ 2 } } \),(along\(\vec { PS }\))
or ∴ B₁ =\(\frac { { \mu }_{ o } }{ 4\pi } .\frac { m }{ (d+1)^{ 2 } }\) … (2)
Since, B₁ and B₂ are acting in opposite direction and B₁ > B₂
∴Resultant field B = B₁ – B₂ (along \(\vec { NP }\))
Putting the values from eqns. (1) and (2),

This is the required expression for the intensity of field on the axis.
Again, if the magnet is small i.e., I << d
By neglecting l.
\(\frac { { \mu }_{ o } }{ 4\pi } .\frac { 2m }{ ({ d }^{ 2 })^{ 2 } }\)
or \(\frac { { \mu }_{ o } }{ 4\pi } .\frac { 2M }{ ({ d })^{ 3 } }\)
In CGS units, B = \(\frac { 2M }{ ({ d })^{ 3 } }\)
The direction of resultant field is along the magnetic axis from south pole to north pole.

Question 3.
What is broad – side – on position or equatorial position? Derive an expression for the intensity at a point on broad – side – on position of a bar magnet. What will be the direction of resultant field?
Or
Determine the force on a unit north pole, kept on the broad-side-on position of a small bar magnet.
Answer:
Broad – side – on position:
When the point where the intensity of the magnetic field has to be found lies on the perpendicularbisector of magnetic axis, i.e., on the neutral axis, then it is called broad – side – on position.

Magnetic field is the force experienced by unit north pole placed at that point.
Hence, B = \(\frac {F}{m}\)
If m = 1, then B = F.
Let NS be a bar magnet of pole strength m and effective length 2l and magnetic moment M = m2l.
Consider a point P at a distance d from the centre O of a magnet on its neutral axis. Let unit north pole be placed at P.
Now, the intensity of field at P, due to N – pole will be :

Resolving B₁ and B₂ into its components, we have B₁, cosθ along NS and B₂sinθ⊥ to NS along OP. Also B₂ cosθ along NS and B₂ sinθ⊥ to NS along PO.
But B₁ = B₂
⇒ B₁ sinθ= B₂ sinθ
Since, their directions are opposite and their magnitudes are equal, hence they cancel each other.
The resultant field is therefore

Question 4.
What are magnetic lines of force ? Write down its properties.
Answer:
Magnetic lines of force :
1st definition:
The magnetic lines of force are the curves in the magnetic field, on which if a unit north pole is placed, then it will follow the imaginary curve drawn.

2nd definition:
“A magnetic line of force is a smooth curve in a magnetic field such that the tangent at any point on it gives the direction of the magnetic field at that point.”

Properties of lines of force :

1. They are closed and continuous curves.
2. Outside the magnet the direction is from north to south and inside the magnet the direction is from south to north.
3. The tangent drawn at any point on the curve gives the direction of the resultant field at that point.
4. They do not intersect each other. If two lines of force intersect at a point, then there would be two tangents at that point hence, the resultant force would have two directions; which is not possible, therefore the lines of force do not intersect.
5. They are dense near the poles where the magnetic field is strong and get separated where the magnetic field is weak.
6. They repel each other in the direction, perpendicular to it, therefore the like poles repel each other.
7. They experience tension along the lines of force therefore, unlike poles attract each other.
8. They behave just like a stretched elastic string.

Question 5.
Derive an expression for the torque acting on a bar magnet placed in a uniform magnetic field, making angle 0with the field and hence define magnetic moments with its help.
Answer:
Let NS be a bar magnet, placed in a uniform magnetic field of intensity B, making an angle θ with the field. Suppose m be the pole strength and 2l be the effective length, Force acting on each pole will be mB.

On the N – pole this force will be along the direction of field, whereas on S – pole this will be opposite to the direction of field. As two equal and opposite forces are acting on it along different line of action, hence a couple acts on it which tries to bring the magnet along the direction of magnetic field. This couple is called ‘restoring couple’ or ‘restoring torque’.

Restoring torque is defined as the product of magnitude of any one of the forces and the perpendicular distance between them.
∴ τ = Force x Perpendicular distance
or Torque, τ = mB x SP … (1)

Also, in ANPS, we get sinθ = \(\frac {SP}{NS}\)
or SP = NSsinθ = 2lsinθ
Putting the value of SP in eqn. (1),
τ = mB x 2l sinθ
But m x 2l = M (magnetic moment)
∴ τ = mB sinθ
In vector form :
\(\vec { τ }\) = \(\vec { M }\) x \(\vec { B }\)
and the direction of \(\vec { τ }\) will be perpendicular to the plane containing \(\vec { M }\) and \(\vec { B }\).

Definition of magnetic moment :
As τ = MBsinθ
If the magnet is held perpendicular to the field, then 0= 90° or sinθ = 1 then the torque acting on the magnet will be maximum, if the strength of the applied field is 1 i.e., B = 1,then
τ_max = M
Hence, magnetic moment is numerically equal to the maximum torque acting on the bar magnet when it is held perpendicular in a uniform magnetic field of unit intensity.

Question 6.
Compare to a bar magnet and a current – carrying solenoid.
Answer:
Comparison of a bar magnet and a solenoid :
Bar magnet:

• It attracts magnetic substances.
• When it is suspended freely it rests in the direction of N – S.
• It has two poles.
• Like poles of magnet repel and unlike poles attract.

Solenoid:

• It also attracts magnetic substances.
• It also rests in N – S direction if suspended freely.
• It has also two poles.
• Like poles of solenoid also repel and unlike poles attract.

Question 7.
Explain how does an atom behave as a magnetic dipole. Derive an expression for the magnetic dipole moment of the atom. Also define Bohr magneton.
Or
Deduce the expression for the magnetic dipole moment of an electron orbiting around the central nucleus.
Answer:
Magnetic dipole moment of a revolving electron:
In hydrogen – like atoms, an electron revolves around the nucleus. Its motion is equivalent to a current loop which possesses a magnetic dipole moment = IA. As shown in Fig., consider an electron revolving anticlockwise around a nucleus in an orbit of radius r with speed v and time – period T.

Equivalent current,

According to right hand thumb rule, the direction of the magnetic dipole moment of the revolving electron will be perpendicular to the plane of its orbit and it the downward direction, as shown in Fig.
Also, the angular momentum of the electron due to its orbital motion is
I = m_evr  … (2)
The direction of / is normal to the plane of the electron orbit and in the upward direction, as shown in Fig.
Dividing equation. (1) by equation. (2), we get

The above ratio is a constant called gyromagnetic ratio. Its value is 8.8 x 10¹⁰Ckg^-1.
So

The negative sign shows that the direction of \(\vec { l }\) is opposite to that of \(\vec { { \mu }_{ 1 } }\) According to Bohr’s quantization condition, the angular momentum of an electron in any permissible orbit is,
l = \(\frac { nh }{ 2π }\) , where n =1,2 ,3, ………….
∴ µ₁ = n(\(\frac { eh }{ 4\pi { m }_{ e } }\))
This equation gives orbital magnetic moment of an electron revolving in nth orbit.

Bohr magneton:
It is defined as the magnetic moment associated with an electron due to its orbital motion in the first orbit of hydrogen atom. It is the minimum value of µ₁, which can be obtained by putting n = 1 in the above equation. Thus Bohr magneton is given by
µ_B= (µ₁)_min = \(\frac { eh }{ 4\pi { m }_{ e } }\) = 9.27 x 10^-24Am².

Question 8.
Derive an expression for work done in rotating a bar magnet in uniform magnetic field through 8 angle.
Answer:
Let a bar magnet of effective length 2l and of magnetic moment M be kept in uniform magnetic field B. When a bar magnet is rotated through some angle in the magnetic field then some work has to be done against moment of restoring couple.
If magnet is rotated through dQ angle then work done, dW = τ dθ,
(where τ is moment of restoring couple)
or dW = MB sinθ dθ … (1)
When magnet is rotated from θ₁ to θ₂, then work done is given by :

This is the required expression.

Question 9.
Establish the expression for potential energy of a bar magnet placed in a uniform magnetic field.
Answer:
The potential energy of the bar magnet in any orientation is the work done by the external agent to turn the dipole from its zero position (θ = 90°) to that orientation (θ = θ°)
dW = MB sinθ dθ … (1)
Amount of work done to rotate the bar magnet from zero position (θ = π/2) to an arbitrary position (θ = θ) will be obtained by integrating equation  (1) under proper limit.

Question 10.
Define the magnetic elements of earth’s magnetic field at a place.
Or
Establish relation between element of earth’s magnetic field?
Answer:
Elements of earth’s magnetic field:
The earth’s magnetic field at a place can be completely described by three parameters which are called elements of earth’s magnetic field. They are declination, dip and horizontal component of earth’s magnetic field.

1. Magnetic declination:
The angle between the geographical meridian and the magnetic meridian at a place is called the magnetic declination (α) at that place, Or, it is the angle which a compass needle (free to swing in a horizontal planb) makes with the geographic north – south direction.

2. Angle of dip or magnetic inclination:
The angle made by the earth’s total magnetic field \(\vec { B }\) with the horizontal direction in the magnetic meridian is called angle of dip (δ) at any place. Or, it is the angle which a dip needle (free to swing in the plane of the magnetic meridian) makes with the horizontal.

At the magnetic equator, the dip needle rests horizontally so that the angle of dip is zero at the magnetic equator. The dip needle rests vertically at the magnetic poles so that the angle of dip is 90° at the magnetic poles. At all other places, the dip angle lies between 0° and 90°.

3. Horizontal component of earth’s magnetic field:
It is the component of the earth’s total magnetic field \(\vec { B }\) in the horizontal direction in the magnetic meridian. If δ is the angle of dip at any place, then the horizontal component of earth’s field \(\vec { B }\) at that place is given by
B_H = Bcosδ
At the magnetic equator,
δ = 0°,B_H = Bcos0°= B
At the magnetic poles,
δ = 90°,B_H =5cos90°= 0
Thus the value of BH is different at different places on the surface of the earth.

Question 11.
Prove tanδ = \(\frac { { B }_{ v } }{ { B }_{ H } }\) and B = \(\sqrt { { B }_{ H }^{ 2 }+{ B }_{ v }^{ 2 } }\) where symbol have there usual meaning.
Answer:
Relations between elements of earth’s magnetic field:
Fig. shows the three elements of earth’s magnetic field. If 8 is the angle of dip at any place, then the horizontal and vertical components of earth’s magnetic field B at that place will be

B_H = Bcosδ .. (1)
and B_v = Bsinδ
\(\frac { { B }_{ v } }{ { B }_{ H } }\) = \(\frac { Bsinδ }{ Bcosδ}\)
\(\frac { { B }_{ v } }{ { B }_{ H } }\) =tanδ .. (2)
Also
B²_H + B²_v = B²(cos²δ + sin²δ) = B²
or B = \(\sqrt { { B }_{ H }^{ 2 }+{ B }_{ v }^{ 2 } }\) .. (3)
Equations (1), (2) and (3) are the different relations between the elements of earth’s magnetic field.

Question 12.
Compare the magnetic properties of soft iron and steel.
Answer:
Comparison of magnetic properties of soft iron and steel:
Soft iron:

• In soft iron, greater magnetism can be produced, than steel. Its magnetic nature is greater than steel.
• Soft iron does not retain magnetism for longer time. Its retaintivity is less.
• The magnetization and demagnetization of soft iron are easy.
• Temporary magnets are made by soft iron.

Steel:

• In steel, less magnetism can be produced than soft iron, its magnetic nature is less than soft iron.
• Steel retains magnetism for longer time. Its retaintivity is greater than soft iron.
• The magnetization and demagnetization of steel are difficult.
• Permanent magnets are made by soft steels.

Magnetism and Matter Long Answer Type Questions

Question 1.
Answer the following regarding terrestrial magnetism quantities :

1. Three quantities are required to express a vector completely write the name of that there independent quantity.
2. At which place of south India the angle of dip in 18° will you expect more value of angle of dip at britain ?
3. If you draw lines of forces at Melbourne city of Austrilia. This lines of forces will go inside the earth or outside.
4. The magnetic needle which is free to revolve in vertical plane. If it is kept at geographical north or south pole then in which direction it will revolve?

Answer:
1.

• Angle of declination
• Angle of dip
• Horizontal component of Earth magnetic field.

2. Britain is near magnetic pole of earth, therefore angle of dip at Britain is more than the angle of dip at south India (approx. 70°).

3. The magnetic lines of force at Melbourne city of Austrilia will go outside.

4. Geographical North pole or South pole is just situated vertically to the direction of earth magnetic field. Therefore the magnetic needle will be independent to revolve in vertical plane

Question 2.
Compare the magnetic properties of paramagnetic substance and ferromagnetic substance on any three points.
Answer:
Comparison :

Question 3.
What do you mean by magnetic field intensity. Derive an expression for magnetic field due to a bar magnet in general position. How is this formula used to find magnetic field in

1. Axial position
2. Equatorial position.

Answer:
The force experienced by unit north pole at any point in the magnetic field is known as magnetic field intensity. NS is a bar magnet of magnetic moment \(\vec { M }\), we have to find out magnetic field intensity at P, which is situated at θangle from axis of magnet.

Now, we divide M into two components

• Mcosθ
• Msinθ

For Mcosθ point ‘P’ lies in axial position, therefore magnetic field at P due to M cos G component is :
B₁ = \(\frac { { \mu }_{ 0 } }{ 4\pi } \frac { 2Msin\theta }{ { d }^{ 3 } }\) … (1)

For M sinθ point P lies on equatorial position :
B₂ = \(\frac { { \mu }_{ 0 } }{ 4\pi } \frac { Msin\theta }{ { d }^{ 3 } }\) … (2)
∵ B₁is perpendicular to B₂ , the resultant of B₁ and B₂ is given by :

This is the required expression.
(i) For axial position θ = 0° => cos 0°= 1
∴From eqn. (3),
B = \(\frac { { \mu }_{ 0 } }{ 4\pi } \frac { 2M }{ { d }^{ 3 } } \)
From equatorial position θ = 90° ⇒ cos90° = 0
B = \(\frac { { \mu }_{ 0 } }{ 4\pi } \frac { M }{ { d }^{ 3 } } \)

Magnetism and Matter Numerical Questions

Question 1.
Magnetic wire of magnetic moment ‘M’ is bent in the shape of L, at one third of its length. What will be the new magnetic moment.
Solution:

Question 2.
The length of magnetic wire is L and its magnetic moment is M. If it is bent in the form of semi – circle then what will be its new magnetic moment?
Solution:
Initial magnetic moment of magnetic wire M = mL
If it is bent in the form of semi – circle then
L = πr ⇒ r = \(\frac {L}{π}\)
New magnetic moment M’ = m x 2r
M’ = m x \(\frac {2L}{π}\)
or M’ = \(\frac {2M}{π}\)

Question 3.
The distance between two magnetic poles of pole strengths ‘m’ and ‘4m’ is 3m. Find the distance of point in between them at which magnetic field intensity is zero.
Solution:

Question 4.
If the pole strength of each pole of two similar magnetic poles is made two times and distance between them becomes half of its initial value then how will magnetic force acting between them change?
Solution:

Force becomes 16 times its initial value.

Question 5.
Obtain the earth’s magnetization. Assume that the earth’s field can be approximated by a giant bar magnet of magnetic moment 8.0 x 10²²Am². The earth’s radius is 6400 km. [NCERT]
Solution:
Here magnetic moment m = 8.0 x 10²² Am²
Radius of the Earth R = 6400 km = 6.4 x 10⁶m

Question 6.
A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per metre, calculate (a) H, (b) M, (c) B and (d) the magnetizing current IM. [NCERT]
Solution:
Here n = 1000 tums/m, I = 2A, μ_r = 400
1. H = nI = 1000 x 2 = 2x 10³ Am^-1

2. M = x_mH = (μ_r -1)H
= (400 – 1) x 2 x 10³ ≈ 8 x 10⁵Am^-1

3. B = μH = μ_rμ₀H
= 400 x 4π x 10^-7 x 2 x 10³ T = 1.0T .

4. As M = nI_M
∴ I_M = \(\frac {M}{n}\)
= \(\frac { 8\times { 10 }^{ 5 } }{ 1000 }\)
= 8 x 10² A.

Question 7.
A short bar magnet placed with its axis at 30° experiences a torque of 0.016 Nm in an external field of 800G

1. What is the magnetic moment of the magnet?
2. What is the work done by an external force in moving it from its most stable to most unstable position?
3. What is the work done by the force due to the external magnetic field in the process mentioned in part (b)?
4. The bar magnet is replaced by a solenoid of cross – sectional area 2 x 10^-4 and 1000 turns, but the same magnetic moment Determine the current flowing through the solenoid. [NCERT]

Solution:
1. Here θ = 30°, B = 800G = 800 x 10^-4T, τ = 0.016Nm
Magnetic moment,
m = \(\frac {τ }{B sinθ}\) =\(\frac { 0.016 }{ 800\times { 10 }^{ -4 }\times sin30° }\)
= 0.40 Am².

2. For most stable position θ = 0°and for most unstable position θ = 180°. So the required work done by the external force,
W = mB (cos 180°- cos0°) = 2mB
= 2 x 0.40 x 800 x 10^-4
=0.064J

3. Here the displacement and the torque due to the magnetic field are in opposition. So the work done by the magnetic fied due to the external magnetic field is,
W_B = 0.064J

4. Here A =2 x 10^-4m², N = 1000
Magnetic moment of solenoid,
m_s = m = 0.40 Am²
But m_s = NIA.
∴ current, I = \(\frac { { m }_{ s } }{ NA }\) = \(\frac { 0.40 }{ 1000\times 2\times { 10 }^{ -4 } }\)

Question 8.
In the magnetic meridian of a certain place, the horizontal component of the earth’s magnetic field is 0.26G and the dip angle is 60°. What is the magnetic field of the Earth in this location?
Solution:
Here B_H= 0.26G, δ = 60°
As B_H = Bcosδ
∴ B = \(\frac { { B }_{ H } }{ cos\delta }\)
= \(\frac {0.26}{cos60°}\)
= \(\frac {0.26}{0.5}\)
= 0.52G

Question 9.
What is the magnitude of the equatorial and axial fields due to a bar magnet of length 5 cm at a distance of 50 cm from its mid – point? The magnetic moment of the bar magnet is 0.40 Am². [NCERT]
Sol. Here m = 0.40 Am²
r = 50 cm = 0-50 m, 21 = 5-0 cm
Clearly, the magnet is a short magnet (l<<r)

Question 10.
A planar loop of irregular shape encloses an area of 7.5 x 10^-4 m² and carries a current of 12 A. The sense of flow of current appears to be clockwise to an observer. What is the magnitude and direction of the magnetic moment vector associated with the current loop? [NCERT]
Solution:
Here A = 7.5 x 10^-4m², l = 12A
Magnetic moment associated with the loop is
m = IA = 12 x 7.510^-4 = 9.0 x 10^-3 JT^-1
Applying right hand rule, the direction of magnetic moment is along the normal to the plane of the loop away from the observer.

MP Board Class 12th Physics Important Questions

MP Board Class 12th Biology Important Questions Chapter 6 Molecular Basis of Inheritance

Molecular Basis of Inheritance Important Questions

Molecular Basis of Inheritance Objective Type Questions

Question 1.
Choose the correct answers:

Question 1.
The process of transfer of enetic information from DNA to RNA is :
(a) Transversion
(b) Transcription
(c) Translation
(d) Translocation.
Answer:
(b) Transcription

Question 2.
Transcription involves :
(a) Synthesis of RNA over DNA
(b) Joining of amino acids over polypeptides
(c) Synthesis of RNA over ribosomes
(d) Synthesis of DNA.
Answer:
(a) Synthesis of RNA over DNA

Question 3.
In operon model, RNA polymerase binds to :
(a) Structural gene
(b) Promotor gene
(c) Operator gene
(d) Regulator gene.
Answer:
(b) Promotor gene

Question 4.
The process of translation is :
(a) Ribosome synthesis
(b) Protein synthesis
(c) DNA synthesis
(d) RNA synthesis.
Answer:
(b) Protein synthesis

Question 5.
Operon model of gene expression is prokaryotes was proposed by :
(a) Meselson and stahl
(b) Wilkins and Franklin
(c) Beadle and Tatum
(d) Jacob and Monod.
Answer:
(d) Jacob and Monod.

Question 6.
Reverse transcription was discovered by :
(a) Beadle and Tatum
(b) Temin and Baltimore
(c) Watson and Crick
(d) Khorana.
Answer:
(b) Temin and Baltimore

Question 7.
Nitrogen base present in a codon are :
(a) 4
(b) 3
(c) 2
(d) 1.
Answer:
(a) 4

Question 8.
A non – sense/termination codon is :
(a) UUU
(b) GCG
(c) UAA
(d) CCC.
Answer:
(a) UUU

Question 9.
A unit of lac – operon which in the absence of lactose, suppresses the activity of operator gene is :
(a) Structural gene
(b) Regulator gene
(c) Repressor gene
(d) Promoter gene.
Answer:
(d) Promoter gene.

Question 10.
Sequence of nitrogenous bases on t RNA is known as :
(a) Anti codon.
(b) Terminating coon
(c) Repressor codon
(d) Initiate codon.
Answer:
(a) Anti codon.

Question 11.
Which is found in nucleosome :
(a) Histone molecule
(b) Luxary genes
(c) Nucleoplasmine
(d) House keeping genes
Answer:
(a) Histone molecule

Question 12.
Functional gene is :
(a) Gene battery
(b) Luxary genes
(c) Mylud gene
(d) House keeping genes.
Answer:
(d) House keeping genes.

Question 13.
Which RNA has very short life span :
(a) m RNA
(b) t RNA
(c) r RNA
(d) Sn RNA.
Answer:
(a) m RNA

Question 14.
Who terminate the transcription :
(a) Co – protein
(b) Sigma factor
(c) Raw Protein
(d) Omega factor.
Answer:
(c) Raw Protein

Question 15.
A Person which has a trisomy on 21 st chromosome is called :
(a) Klinefelter syndrome
(b) Down’s syndrome
(c) Turner syndrome
(d) None of these.
Answer:
(b) Down’s syndrome

Question 2.
Fill in the blanks:

1. ……………… is found in RNA, in the place of DNA’s thymine.
2. Formation of ……………… from DN A is called transcription.
3. In ……………… ribosomes are of 70s type.
4. ……………… is an exogenous gene that has been introduced into the genome of other organism.
5. The joining of DNA strands together by ………………
6. ……………… is the vector of genes.
7. ……………… enzyme is necessary for transcription.
8. ……………… discovered polytene chromosome.
9. Intron is known as ………………
10. ……………… nitrogenous protein which bind the DNA molecule with chromosome.
11. Highly condensed chromatin which is not available for transcription is called ………………
12. ……………… vims infects the bacteria.
13. ……………… is the outer covering of viruses.
14. ……………… type of RNA is found in ribosome.
15. ……………… is made by joining of peptide bond.

Answer:

1. Uracil
2. mRNA
3. Prokaryotes
4. Transgenic
5. Ligase
6. Plasmid
7. RNA Polymerase
8. Balbiani
9. Junk DNA
10. Histone
11. Hetero chromatin
12. Bacteriophage
13. Capsid
14. Ribosomal RNA
15. Polypeptide.

Question 3.
Match the followings :
I.

Answer:

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

II.

Answer:

(b)
(a)
(d)
(c)

III.

Answer:

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

Question 4.
Write the answer in one word/sentences:

1. Who discovered nucleic acid for the first time?
2. How many codons are coded to 20 types of amino acid?
3. Which codon is known as the starting of codon?
4. Name the RNA which function as enzyme.
5. Who proposed the operon model of gene regulation?
6. Name the organism which contain single stranded DNA.
7. Name the process in which to make R&IA from DNA.
8. Which bond is produce when sugar and phosphoric acid combine in DNA?
9. Name the enzyme which help in transcription.
10. How many nucleotides are found in a gene?
11. Name any one termination codon.
12. Genes which are not expressed their characters.
13. Who developed the DNA finger printing technique?
14. Name the gene which is control the activity of other gene?
15. Write the name of amino acid which starts the protein synthesis.

Answer:

1. Friedrich Meischer
2. 64
3. AUG
4. Ribozyme
5. Jacob and Monod
6. ϕ x 174
7. Transcription
8. Phosphodiester
9. RNApolymerase
10. 1000
11. UAA
12. Silent gene
13. Alec Jaffreys
14. Regulator gene
15. Methionine.

Molecular Basis of Inheritance Very Short Answer Type Questions

Question 1.
Structure formed by regulation + structural + operator + promoter gene.
Answer:
Operon.

Question 2.
What are the animals that have a foreign gene deliberately inserted into their genome?
Answer:
Transgenic animals.

Question 3.
What are the group of cells or organisms which have same hereditary characters?
Answer:
Clone.

Question 4.
By which the instructions of our DNA are converted into a functional product?
Answer:
Gene expression.

Question 5.
Write the name of sugar found in RNA.
Answer:
Ribose sugar.

Question 6.
Which codon is AUG?
Answer:
Anticodon.

Question 7.
Name the enzyme which takes part in transcription.
Answer:
RNA Polymerase.

Question 8.
Who tell that DNA is a heredity material?
Answer:
Alfred Hershey and Martha Chase.

Question 9.
Which bond is made in DNA when join the sugar and phosphoric acid?
Answer:
Phosphodiester bond.

Question 10.
Name the segment in which any nucleotide sequence within a gene that is removed by RNA splicing during maturation of the final RNA products.
Answer:
Intron.

Question 11.
Who gave the operon model?
Answer:
Jacob and Monod.

Question 12.
What do you mean by commaless genetic code?
Answer:
Between two codon has no internal punctuation.

Question 13.
Write the full name of Sn RNP.
Answer:
Small nuclear Ribonucleo Proteins.

Molecular Basis of Inheritance Short Answer Type Questions

Question 1.
Group the following as nitrogenous bases and nucleosides :
Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.
Answer:
Adenine, Guanosine, Thymine, Uracil and Cytosine are nitrogenous bases. (Adenine and Guanosine → Purine, Thymine, Uracil and Cytosine → Pyrimidine) Cytidine is a nucleoside.

Question 2.
If a double stranded DNA has 20 % of cytosine, calculate the percent of adenine in the DNA.
Answer:
According to Chargaff’s rule, the DNA molecule should has an equal ratio; Cytosine = 20 % therefore, Guanine = 20%
A + T = 100 – (G – C)
A + T = 100 – 40 since, both Adenine and Thymine are in equal amounts.
Thymine = Adenine = \(\frac { 60 }{ 2 }\) = 30%
So, quantity of Adenine is 30% in DNA helix.

Question 3.
What are oncogenes?
Answer:
Genes which are responsible for production of cancer in host by uncontrolled mitotic cell division are called as oncogenes.

Question 4.
What are Okazaki fragments and leadings strands?
Answer:
Okazaki fragments:
On second parental DNA template new complementary DNA strands are formed in smaller fragments starting from RNA primer. These short fragments are called Okazaki fragments.

Leading strands:
Second strand is formed on 5’ → 3’ strand of parental DNA in a continuous stretch in reverse direction 3’ → 5’ and is called as leading strand.

Question 5.
DNA nucleotides are formed by which molecule?
Answer:
Components of DNA Nucleotides:

Question 6.
What is peptide bond?
Answer:
The bond formed between the carboxylic group (- COOH) of one amino acid and amino group (- NH₂) of another amino acid is called as peptide bond. A molecule of water is released during the formation of peptide bond.

Question 7.
Write five characters of gene hypothesis.
Answer:
Sutton, Bridges, Muller and Morgan suggest these theory. The characters of gene of this theory are as follows:

1. Genes are situated on the chromosome.
2. They make the physiological character of organisms.
3. These are called functional unit of specific characters.
4. Genes have the capacity of self-transcription.
5. They perform mutation.
6. Characters go to one generation to other by parents.

Question 8.
What is gene expression? Explain by different methods of gene expression in animals.
Answer:
The mechanism at molecular level by which a gene is able to express itself in the phenotype of an organism is called gene expression.
Different methods of gene expression in animals are:

1. Transduction:
It is the process in which bacteriophages pick up pieces of DNA from one bacterial cell and transfer the same to another on infection.

2. Transformation:
It is the process by which DNA isolated from one type of cell when introduced into another, is able to bestow some of the properties of the former to the later.

Question 9.
What is proof reading and repair of DNA?
Answer:
Variety of environmental factors such as radiations, chemicals etc. may cause damage in DNA of a cell. The bacterial DNA polymerase III can do proof reading, in the sense that it can go back and remove the wrong base before it proceeds to add new bases in the 5′ → 3′ direction. It is called proof reading. Obviously, the survival of the cell depends on its availability of damages:

1. Monoadduct:
Which involve alterations in a single nitrogenous base.

2. Diadducts :
They are the alterations involving more than one nitrogenous base. Number of nucleases have been found to be involved in repair replication such as Exonucleases (defined as phosphodiesterases which require a terminus for hydrolysis and cut’off terminal nucleotides), Endonucleases (which are also phosphodiesterases which do not require a terminus for hydrolysis and break internal bonds). The endonucleases which act on the damaged DNA and cause repair or correction of this molecule are referred to as correctional nucleases. The following steps are said to be involved in the repair replication i.e., Incision, Excision, Reinsertion and joining of newly formed strands.

Question 10.
Write any four differences between DNA and RNA.
Answer:
Differences between DNA and RNA:
DNA:

• It contains deoxyribose sugar.
• It has adenine, thymine, cytosine and guanine as nitrogenous bases.
• It consists of two polynucleotide chains coiled into a double helix.
• It is main constituent of chromosome which is found in nucleus.

RNA:

• It contains ribose sugar.
• It has adenine, uracil, guanine and cytosine as nitrogenous bases.
• It consists of single polynucleotide chains which may get folded on it self to form double helix.
• It is main constituent of ribosome and generally found in cytoplasm.

Question 11.
Write the names of enzymes used in DNA replication.
Answer:
The names of enzymes used in DNA replication are as follows:

1. DNA helicase : For unwinding of two strands.
2. DNA gyrase : For relieving tension.
3. Primase : For formation of primer.
4. DNA polymerase : For DNA synthesis.
5. RNA primer : For initiation of the synthesis of DNA segments.
6. DNA ligase : For joining of DNA Okazaki segments.

Question 12.
What is transcription? Name the enzyme catalysing it.
Answer:
Transcription:
Formation of wRNA from DNA in the presence of enzyme is called transcription. It is the first stage of protein synthesis which is catalysed by RNA polymerase enzyme. The process of transcription involves in the following steps:

1. Exposing of the bases of DNA:
The two strands of DNA are separated due to presence of an unwinding protein and thus, their bases are exposed. The exposed chain of DNA functions as template for the synthesis of oiRNA in the presence of RNA polymerase enzyme.

2. Base pairing:
The ribonucleotides are jointed in a definite fashion on the exposed strand of DNA. G is bonded with ‘C’ ‘C’ bonded with ‘G’, ‘T’ bonded with ‘A’ and ‘A’ bonded with ‘T’ respectively.

3. Synthesis of RNA chain:
The new ribonucleotide bonded on DNA template are jointed with the help of RNA polymerase and thus, forming a new chain of RNA. Then this mRNA is separated from DNA and reaches the cytoplasm where it combines with ribosomes and thus, initiating the synthesis of protein.

Question 13.
What is translation? Explain it.
Answer:
Translation:
The translation step of protein synthesis involves translation of the language of nucleic acids (available in the form of mRNA) into language of protein. The sequence of bases in wRNA, decides the sequence of amino acids in proteins. Each amino acid is programmed by a triplet code. It consists of a sequence of three bases in the DNA and the complementary bases in /wRNA. The synthesis of protein occurs in three steps, initiation, elongation and termination. After the final step i.e., termination, the proteins are transported out of the cell or translocated within the cell. Thus, the transformation of nucleotides chain of RNA into polypeptide chain of protein is called as translation.

It is completed in following steps:

1. Activation of amino acids.
2. Binding of activated amino acids with?RNA.
3. Binding of mRNA with smaller unit of ribosome.
4. Initiation of polypeptide chain.
5. Elongation of polypeptide chain.
6. Termination of polypeptide chain.

Question 14.
Describe the evidence given by Griffith in support of DNA as genetic material. Explain it along with suitable diagram.
Answer:
Griffith had done transformation experiments in mice to prove that DNA is the genetic material. He took virulent strain of Diplococcus pneumonae (S – III) which causes pneumonia in mice and injected it into mice which resulted in the production of pneumonia in mice. He also injected a non – virulent strain of that bacteria in the body of mice and found that all the mice were unaffected. In third experiment he injected heat killed (S – III) strain and non-virulent strain R – II strain together in the body of mice and found that all the mice suffered from pneumonia and became dead.

After analysis it was found that these mice contained both the strains of Diplococcus pneumonae. Thus, this experiment proved that any substance of S – III strain is transferred into R – II strain due to which R – II strain become virulent. Later, McLeod, Avery and McCarty observed that DNA molecules are transferred from S – III to R – II strain and make virulent; Thus, it is proved that DNA is the genetic material.

Question 15.
Explain the semi-conservative method of replication of DNA.
Or
Explain the method of DNA duplication.
Answer:
Synthesis of new DNA strands:
The DNA polymerase plays an important role in adding the building blocks to the primer in a sequence as influenced by the template. Replication of DNA is not continuous. It takes place by semi – conservative method. The parent DNA unwinds sequentially in local areas. A nick in one strand of the helix provides two free ends in one strand and a swivel in the other to absorb the twist that occur in the unwinding process.

When the double stranded DNA gets unwound up to a point it will represent a Y – shaped replication fork. This unwinding exposes the internal bases for replication.The enzyme DNA polymerase (discovered by Kornberg in 1957) now starts adding the nucleotides complementary to the DNA templates in the direction 5′ → 3′. Since, the two strands run in anti – parallel manner, the synthesis of new strands will be in opposite direction. As the synthesis of new strand progresses the point of divergence of the fork will be seen moving further due to unwinding of the strands of parental DNA.

Second strand is formed on 5′ → 3′ strand of parental DNA in a continuous stretch in reverse direction 3′ 5′ and is called the leading strand. However, on the second parental DNA template, the new complementary DNA strands are formed in smaller fragments starting from the RNA primer.

These short fragments are called Okazaki fragments after the name of scientist who discovered them. Since these fragments are joined later to form the complete strand it is called the lagging strand. The enzyme which joins the Okazaki fragments with the help of the RNA primer to form the lagging strand is called polynucleotide ligase or the joining enzyme.

Question 16.
Give the functions of nucleotides.
Answer:
Functions of Nucleotides:

1. It works as a activated precursors of DNA and RNA.
2. They are perform the storage and conduction of energy to the form of ATR
3. It required for activation of intermediates in many biosynthetic pathway.
4. It works as Carrier of methyl group in the form of SAM.
5. It components of co – enzyme : NAD, FAD and Co A.
6. Some functions are as a vitamin.
7. They are control and coordinates different activities in our body.

Question 17.
Write four features of genetic code.
Answer:
According to Nirenberg, Khorana and Holley, genetic code is that sequence of nitrogenous bases of DNA in which genetic informations for the synthesis of protein are coded.

Characteristic features of genetic code:

1. The code is triplet:
The codon is a specific sequence of three nitrogenous bases of mRNA.

2. The code is commaless:
The sequence of bases read in blocks of three at a time form a particular position. There is no gap between two subsequent codons.

3. Code is degenerating:
Presence of more than one codon for one amino acid is called as degeneracy of codons, example Serine having three codons UCU, UCA, AGU.

4. Codes are universal:
Codons are similar in all organisms, example serine is coded by UCU codon in all the living beings.

5. Codes are non – ambiguous:
The position of genetic code in cellular medium is nonambiguous because a codon always codes only one amino acid. Sometimes a codor codes more than one amino acid, example in E. coli. UUV codon generally code phenylalanine, after treatment of their ribosome with streptomycin. It can also code isoleucine, leucine and serine.

6. Initiation and termination codon:
Codons responsible for the initiation of polypeptide chain are called as initiation codon, example AUG. Likewise codons responsible for the termination of polypeptide chain are called as chain termination codon, example UAA, U AG, UGA.

Question 18.
Define Codon and Anticodon.
Answer:
Codon:
A specific sequence of three consecutive nucleotides that is a part of the genetic code and that specifies a paticular amino acid in a protein or starts or stops protein synthesis example AUG codon which is situated on the wRNA, code methionine amino acid.

Anticodon:
A sequence of three adjacent nucleotides located on one end of transfer RNA. It bounds to the complementary coding triplet of nucleotides in wRNA during translation phase of protein synthesis. For example, the anticodon for Glycine is ccc that binds to the codon (which is GGE) ofwRNA.

Question 19.
Explain DNA duplication in short.
Answer:
Watson and Crick after giving the double helix model of DNA, also postulated the mechanism of DNA duplication, also known as replication. According to them, during duplication, the weak hydrogen bonds between the nitrogenous base of the nucleotides get separated, so that two polynucleotide chains of DNA also separate and uncoil. The chains thus, separated are complementary to one another. These strands act as template and because of the specificity of base pairing each nucleotide of separated chain attracts its complementary nucleotide from the cell cytoplasm.

Once the nucleotides are attached by their hydrogen bonds their sugar radicals write through their phosphate components completing the formation of a new polynucleotide chain. This results in the formation of two double helixes of DNA where each molecule has one old strand contributed by parent DNA and one synthesized new. This method of DNA duplication is known as semi-conservative method.

Question 20.
Describe the functions of nucleic acids.
Or
Explain the utility of nucleic acids.
Answer:
Utility of Nucleic acids:

1. Nucleic acids are the hereditary materials of organisms which involve in the transfer of hereditary characters from one generation to the next.
2. DNA controls the synthesis of enzymes which control the various activities of the body.
3. Nucleic acids also control protein synthesis.
4. Nucleic acids form maximum portion of chromatin network.
5. It causes mutation in living beings.
6. They form enzymes.

Question 21.
Explain the structure of RNA.
Answer:
RNA molecules are single stranded nucleic acids composed of nucleotides. Four types of bases are present in RNA. These nitrogenous bases joint in different manner and form the ribonucleoside. Ribonucleoside joins together and make a polyribonucleotide chain.
All four types of nucleoside and nucleotide are as follows:

Question 22.
Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acid synthesized from it (DNA or RNA) list the types of nucleic acid polymerases.
Answer:
These are two different types of nucleic acid polymerases:

1. DNA – dependent DNA polymerases
2. DNA – dependent RNA polymerases

The DNA dependent DNA polymerases use a DNA template for synthesizing a new strand of DNA, whereas DNA dependent RNA polymerases use a DNA template strand for synthesizing RNA.

Question 23.
List two essential roles of ribosome during translation.
Answer:
Two essential roles of ribosome during translation are:

1. One of the RNA acts as a peptidyl transferase ribozyme for formation of peptide bonds.
2. Ribosome provides sites for attachment of TMRNA and charged fRNA for polypetide synthesis.

Molecular Basis of Inheritance Long Answer Type Questions

Question 1.
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
Answer:
Hershey and Chase experiment:

1. They grew some bacteriophages on a medium that contained radioactive phosphorus and some in another medium that contained radioactive sulphur.

2. Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein as phosphorus is present only in DNA.

3. Viruses grown on radioactive sulphur contained radioactive protein but not radioactive DNA because DNA does not contain sulphur.

4. It was found that bacteria which were infected with bacteriophages that had radioactive DNA were radioactive, indicating that DNA was the material that passed from the virus to the bacteria.

5. Bacteria that were infected with viruses that had radioactive proteins were not radioactive. This indicated that proteins did not enter the bacteria from the viruses.

6. This was a clear cut proof that DNA is the genetic material that is passed from virus to bacteria.

Question 2.
Differentiate between the followings:

1. Repetitive DNA and Satellite DNA.
2. OTRNA and IRNA.
3. Template strand and Coding strand.

Answer:
1. Differences between Repetitive DNA and Satellite DNA :

Repetitive DNA :

• DNA in which certain base sequences are repeated many times are called repetitive DNA.
• Repetitive DNA sequences are transcribed.

Satellite DNA:

• DNA in which large protein of the gene is randomly repeated is called satellite DNA.
• Satellite DNA sequences are not transcribed.

2. Differences between mRNA and tRNA:

mRNA:

• It is linear.
• It carries coded information.
• mRNA undergoes additional processing, i.e., capping and tailing splicing
• Nitrogen bases are unmodified.

tRNA:

• It is clover – leaf shaped.
• It carries information for association with an amino acid and a anticodon for its in corporation in a polypeptide.
• It does not require any processing.
• Nitrogen bases may be modified.

3. Differences between Template strand and Coding strand:

• Template strand:
• It is the strand of DNA which takes part in transcription.
• The polarity is 3 ’ → 5′
• Nucleotide sequence is complementary to the one present in mRNA.

Coding strand:

• It is the stand that does not take part in transcription.
• The polarity is 5’ → 3’.
• The nucleotide sequence is same as the one present in mRNA except for presence of Thymine instead of Uracil.

Question 3.
Explain (in one or two lines) the function of the followings:

1. Promoter
2. tRNA
3. Exons.

Answer:
1. Promoter is an essential component of the transcription unit. It is located at the beginning of 5’-end. It provides a site for the attachment of transcription factors and RNA polymerase.

2. tRNA is a small sized RNA molecule that takes part in transcription. It physically picks up activated amino acids from the cytoplasm and carries (transfers) them to ribosomes, where they join together through peptide bonds and leave the /RNA to fetch more amino acids.

3. Exons are the coding sequences of DNA that are transcribed and translated.

Question 4.
Why is the human genome project called a mega project?
Answer:
Human genome project is called a mega project because:

1. Its aim was to determine the nucleotide sequence of complete human genome which was a task of enormous magnitude.
2. A total of 3 x 10⁹ base pairs were to be sequenced and the cost was about 9 billion US dollars.
3. It requires bioinformatics data base techniques and other contemporary devices for the analysis, storage and retrieval of information.
4. Many countries worked jointly to complete this timed project.

Question 5.
What is DNA Fingerprinting? Mention its application.
Answer:
DNA Fingerprinting:
Every human individual is characterised by unique print at the fingertips. The study of fingers, palm and sole print is called ‘dermatoglyphics’. Like prints of the fingertips, each individual has unique DNA fingerprint. Unlike the prints of finger, the DNA fingerprints can not be altered by surgery. The later is exactly similar in all the cells and tissues of an individual. It can not be changed by medical treatment.

The distinction of individuals on the basis of DNA fingerprint is due to sequence of nucleotides in whole genomic DNA. The technique to identify a person on the basis of his/her DNA specificity is called DNA fingerprinting. This was invented by Sir Alec Jeffreys in 1984 at Leicester University, U.K. In India, Dr. V. K. Kashyap and Dr. Lalji Singh started this technique at CCMB, Hyderabad.

DNA fingerprinting involves following steps:

1. The DNA of the organism to be tested is isolated, it is called host DNA
2. Host DNA is cleaved with the help of specific restriction enzymes into several fragments.
3. Double stranded DNA fragments are denatured to produce single stranded DNA by alkali treatment.
4. DNA segments are separated by electrophoresis.

Question 6.
Briefly describe the following:

1. Transcription
2. Polymorphism
3. Translation
4. Bioinformatics.

Answer:
1. Transcription:
It is the formation of RNA over the template of DNA. It forms single-stranded RNA which has a coded information similar to the sense or coding strand of DNA with the exception that thymine is replaced by uracil. One strand of DNA is used as template strand for the synthesis of a complementary strand of RNA called mRNA.

2. Polymorphism:
Genetic polymorphism means occurrence of genetic material in more than one form. It is of three major types, i.e., Allelic, SNP and RFLP.

(a) Allelic polymorphism:
Allelic polymorphism occurs due to multiple alleles of a gene. Allele possess different mutations which alter the structure and function of a protein formed by them as a result, change in phenotype may occur.

(b) SNP or Single Nucleotide Polymorphism:
Over 1 – 4 million single base DNA differences have been observed in human beings. According to SNP, every human being is unique. SNP is very useful for locating alleles, identifying disease associated sequence and tracing human history.

3. Translation:
The translation step of protein synthesis involves translation of the language of nucleic acids (available in the form of mRNA) into language of protein. The sequence of bases in mRNA, decides the sequence of amino acids in proteins. Each amino acid is programmed by a triplet code. It consists of a sequence of three bases in the DNA and the complementary bases in mRNA. The synthesis of protein occurs in three steps, initiation, elongation and termination. After the final step i.e., termination, the proteins are transported out of the cell or translocated within the cell. Thus, the transformation of nucleotides chain of RNA into polypeptide chain of protein is called as translation.

It is completed in following steps:

• Activation of amino acids.
• Binding of activated amino acids with tRNA.
• Binding of mRNA with smaller unit of ribosome.
• Initiation of polypeptide chain.
• Elongation of polypeptide chain.
• Termination of polypeptide chain.

4. Bioinformatics:
The science which deals with handling storing of huge information of genomics as databases, analysing, modelling and providing various aspects of biological information, especially the molecules connected with genomics and proteomics is called bioinformatics.

Question 7.
Explain the Watson and Crick model of DNA.
Answer:
The structure of DNA was proposed by Watson and Crick. It is twisted ladder like structure. It has got two coiled polynucleotides which are joined together by nitrogen bases with hydrogen bond in the centre. The longitudinal strands of DNA are made of sugars and phosphates of nucleotides. The horizontally placed nitrogen bases are of two types, purine and pyrimidine. Purines are adenine and guanine whereas pyrimidines are cytosine and thymine.

MP Board Class 12th Biology Important Questions

MP Board Class 12th Biology Important Questions Chapter 7 Evolution

Evolution Important Questions

Evolution Objective Type Questions

Question 1.
Choose the correct answers:

Question 1.
Atmosphere of earth just before the origin of life consisted of:
(a) Water vapour, CH₄, NH₃and O₂
(b) CO₂, NH₃and CH₄
(c) CH₄, NH₃, H₂ and water vapour
(d) CH₄ O₃, O₂ and water vapour.
Answer:
(a) Water vapour, CH₄, NH₃and O₂

Question 2.
Primitive atmosphere is made up of mixture of these gases :
(a) CH₄
(b) NH₃
(C) Water vapour
(d) None of these.
Answer:
(d) None of these.

Question 3.
What is the meaning of Abiogenesis :
(a) Life is originated from non – living things
(b) Life is originated from living things
(c) Origin of virus and microorganisms
(d) None of these.
Answer:
(a) Life is originated from non – living things

Question 4.
Primitive atmosphere of earth did not posses :
(a) CH₄
(b) NH₃
(C) H₂
(d) O₂
Answer:
(a) CH₄

Question 5.
Which of the following was formed in Miller’s experiment:
(a) Microspheres
(b) Nucleic acid
(c) Amino acid
(d) UV – radiation.
Answer:
(d) UV – radiation.

Question 6.
What is the unit of evolution :
(a) Community
(b) Genus
(c) Order
(d) Species.
Answer:
(a) Community

Question 7.
Whith of the following has not life :
(a) Azoic era
(b) Mesozoic era
(c) Paleozoic era
(d) Cenozoic era.
Answer:
(a) Azoic era

Question 8.
Which of the following are not an analogous organ :
(a) Wings of birds and butterfly
(b) Eye of octopus and mammals
(c) Thom of Baugainvilla and tendril of Cucurbita
(d) Tuberous roots of sweet potato and tuber of potato.
Answer:
(d) Tuberous roots of sweet potato and tuber of potato.

Question 9.
The earliest era is :
(a) Cenozoic
(b) Mesozoic
(c) Paleozoic
(d) Precambrian.
Answer:
(c) Paleozoic

Question 10.
Which of the following is analogous organ :
(a) Which has structural symmetry
(b) Which has functional and structural symmetry
(c) Similar in functional structure
(d) Mostly inactive.
Answer:
(c) Similar in functional structure

Question 11.
Who gave the exact proof of evolution :
(a) Fossils
(b) Vestigeal organ
(c) Embryo
(d) Morphology.
Answer:
(a) Fossils

Question 12.
Which era is called of mammals and angiosperm’s era :
(a) Mesozoic
(b) Cenozoic
(c) Paleozoic
(d) Archaeozoic.
Answer:
(b) Cenozoic

Question 13.
Whose dental structure is similar to human :
(a) Homo erectus erectus
(b) Homo erectus pekinensis
(c) Homo habillis
(d) Homo sapiens neanderthalensis.
Answer:
(c) Homo habillis

Question 14.
Which of these cells are immortal:
(a) Germ cells
(b) Liver cells
(c) Kidney cells
(d) Neurons.
Answer:
(a) Germ cells

Question 2.
Fill in the blanks:

1. Earth is a member of ……………….
2. The great explosion that resulted into the formation of universe is ……………….
3. ‘Philosophic Zoologique’ is written by ……………….
4. ………………. is a link between reptiles and birds.
5. ………………. is the ancient ancestor of modem horses.
6. At present marsupiales (kangaroos) are found only in ……………….
7. ………………. is the first photosynthetic organism that originate on the earth.
8. ………………. is the process by which changes in the genetic composition of organism.
9. ………………. constructed geological time scale.
10. ………………. is the first prehistoric human.
11. ………………. was among the first fossils to be recognized as belonging to Homo sapiens.
12. ………………. is found since the 19th century in the Siwalik Hills of the Indian sub – continent.
13. Discoveries related to the genetic code and its function in protein synthesis is given by ……………….
14. ………………. was the father of the Paleontology.
15. ………………. is the name of Darwin’s ship.

Answer:

1. Solar system
2. Big bang
3. Lamarck
4. Archaeopteryx
5. Eohippus
6. Australia
7. Cyanobacteria
8. Organic evolution
9. Geovani Abduna
10. Homo habilis
11. Cro – magnon
12. Ramapithecus
13. Dr. Hargovind khorana
14. Leonardo da Vinci
15. Beagle.

Question 3.
Match the followings :
I.

Answer:

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

II.

Answer:

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

Question 4.
Write the answer in one word/sentences:

1. The matter which is a link between acellular and cellular system.
2. The island at which Darwin studies the organisms for his theory of evolution.
3. When the earth is originate?
4. Which is known as the link between living and non – living?
5. Where are originate the first animal on the earth?
6. Name the scientist who disapproved Lamarkism.
7. The whole process of the development at changes from embryo to adult organism.
8. Other name of natural selection.
9. Who proposed the Mutaion Theory of Evolution?
10. The organs which are different in their origin but have similar function.
11. Who proposed the theory of inheritance of acquired characters?
12. Who proposed the theory of natural selection ?
13. The remains of ancient organisms.

Answer:

1. Coacervates
2. Galapagos
3. 4.6 million years ago
4. Virus
5. In water
6. Weismann
7. Ontogeny
8. Darwinism
9. Hugo de Vries
10. Homologous organ
11. Lamarck
12. Darwin
13. Fossils.

Evolution Very Short Answer Type Questions

Question 1.
Who proposed the theory of natural selection?
Answer:
Darwin.

Question 2.
Who proposed the Recapitulation theory or Biogenetic law?
Answer:
Haeckel.

Question 3.
Astronomical distance measured in?
Answer:
Astronomical distance measured in Light years.

Question 4.
Name the theory by which earth is said to originate.
Answer;
The big – bang theory.

Question 5.
What is fossil?
Answer:
Fossils are the remains or impressions of ancient organisms preserved in sedimentary rocks or other media.

Question 6.
What is mutation?
Answer:
New species originate due to changes of hereditary characters are called mutation.

Question 7.
Name the scientist who tell the spontaneous theory is wrong.
Answer:
Louis Pasteurs.

Question 8.
Which era is called golden period of Dinosaurs?
Answer:
Mesozoic period is billed golden period of Dinosaurs.

Question 9.
In which ship Darwin studied the nature?
Answer:
Beagle.

Question 10.
Name any two vertibrates body organ which are homologus organs of human forelimb.
Answer:

1. Flipper of whale
2. Wing of birds.

Question 11.
What is the scientific name of modern man?
Answer:
Homo sapiens.

Question 12.
Who is the early man of the modern human?
Answer:
Cro – Magnon peoples are early human of modem human.

Question 13.
Who are the early human and sub – human.
Answer:
Ramapithecus is early human and Australopithecus is early human.

Question 14.
Which human form first started to walk on two legs?
Answer:
Australopithecus form first started to walk on two legs.

Question 15.
Which type of human was ‘Cro – Magnon’ on the basis of food intake?
Answer:
Cro – Magnon was carnivorous.

Question 16.
On the basis of evolution which human had brain size of 1400cc?
Answer:
Neanderthal.

Question 17.
Give the name of ape-like ancestors of humans.
Answer:
Ape – like ancestors of human’s are Dryopithecus.

Question 18.
Differenciate between Dryopithecus and Ramapithecus.
Answer:
Dryopithecus were apelike but Ramapithecus were mostly human like.

Evolution Short Answer Type Questions

Question 1.
Explain antibiotic resistance observed in bacteria in light of Darwinism selection theory.
Answer:
Darwinism theory of natural selection states that environment selects organisms with favourable variations and these organisms thus, survive and reproduce. It is observed when bacterial populations are exposed to certain antibiotic, the sensitive bacteria could not tolerate and hence, died due to the adverse environment. Whereas some bacteria that devel¬oped mutation became resistant to the particular antibiotic and survived.

As a result such resistant bacteria survive and multiply quickly as compared to other sensitive bacteria. So, the whole population is regained by multiplication of resistant variety and antibiotic resistant gene becomes widespread in the bacterial population.

Question 2.
Find out from newspapers and popular science articles any new fossil dis – coveries or controversies about evolution.
Answer:
Fossils of dinosaurs have revealed the evolution of reptiles in Jurassic period. As a result of this evolution of other animals such as birds and mammals has also been discovered. However, two unusual fossils recently unearthed in China have ignited a controversy over the evolution of birds confuciusomis is one such genus of primitive birds that were crow sized and lived during the Cretaceous period in China.

Question 3.
Attempt giving a clear definition of the term species.
Answer:
A species generally includes similar organism. Members of this group can show interbreeding. Similar group of genes are found in the members of same species and this group has capacity to produce new species. Every species has some cause of isolation which intruped the interbreeding with nearest reactional species which is refer as reproduptively isolated.

Question 4.
What is virus? Why is it treated as a link between living and non – iiving ?
Answer:
Viruses are simplest organism,of the earth, which consist of nucleic acid (DNA or RNA) surrounded by protein cover. It shows characteristics of living as well as non – living organisms.

(A) Living characters of virus:

1. Virus shows structural differentiations.
2. They contain hereditary material.
3. They exhibit mutation.
4. They spread plant and animal diseases.
5. Growth and development present.
6. They exhibit adaptaion.
7. They possess sensitivity.

(B) Non – living characters of virus:

1. Lack protoplasm and cell organelles.
2. Can be crystallized.
3. No metabolic activities seen.
4. Cannot reproduce outside living cells.
5. They lack enzymes.
6. Due to above reason, viruses are considered as link between living and non-living organisms, thus, it is the first life
7. originated in the earth.

Question 5.
What is oxygen revolution? Explain.
Answer:
Oxygen revolution:
Evolution of O₂ in photosynthesis during primitive environmental conditions is very important because, it is required in the evolution of organism and conversion of reducing environment into oxidizing environment hence, it is called oxygen revolution. Oxygen evolution should cause the following changes in the environment:

1. Oxygen evolution should cause the conversion of reducing envronment into oxidizing environment.
2. Ozone layer is formed 15 miles above from the earth which absorbs the ultraviolet light of the sunlight and thus, prevents
3. the entry of uv light in the atmosphere.
4. O₂ present in the environment dissociates methane (CH₄) into CO₂ and O₂. This CO₂ is used in photosynthesis.
5. NH₃ of the primitive environment is dissociated into H₂O and nitrogen.
   CH₄ + 2O₂ → CO₂ + 2H₂O
   4NH₃ + 3O₂ → 2N₂ + 6H₂O.

Question 6.
Explain the origin of the earth.
Answer:
Origin of the earth:

1. Earth was formed 4 – 5 billion years back.
2. Initially, the surface was covered with water vapour, methane, CO₂ and NH₃.
3. The UV rays of the sun broke water into hydrogen and oxygen.
4. Hydrogen escaped and oxygen combined with NH₃ and CH₄ to form water, CO₂ and other gases, also forming the ozone layer.
5. Cooling of water vapour led to rain which filled the depressions on earth’s surface, forming water bodies.

Question 7.
Mention the names of discoveries and principles given by the following scientists:

1. Louis Pasteur
2. A. I. Oparin
3. Urey and Miller
4. Francesco Redi
5. Faux.

Answer:

1. Louis Pasteur – He proved that air contains spores of microorganism and Biogenesis theory was supported by him.
2. A. I. Oparin – He presented the biochemical explanation of origin of life in his book “The Origin of Life on Earth”.
3. Urey and Miller – They supported the evidence of Oparin – Haldane theory of Origin of life.
4. Francesco Redi – He by conducting experiments proved that abiogenesis cannot exist but bioginesis theory can exist i.e., Life arises from pre – existing life.
5. Faux – He has been experimentally supporting the organic substances as described by Oparin.

Question 8.
What are homologous organs?
Or
What is homology?
Answer:
Organs which are similar in structure and origin but different in appearance and functions are called homologous organs and the phenomenon is called homology.

Examples:
Forelimbs of bat, wings of bat, hands of man, forelimbs of horse. These are the examples of homologous organs because, they are made up of similar bones, humerus, radius – ulna, carpals, metacarpals and fingers.

Question 9.
What do you mean by analogous?
Answer:
Analogous organs:
Organs which are different in origin and structure but performing similar functions are known as analogous organs and the phenomenon is called as analogy. Analogous organs do not indicate phylogeny.

Examples:
Wings of butterflies are made up of chitin, wings of birds made by produc¬tion of feathers on forelimbs and skin present between the fingers of bat are the examples of analogous orgAnswer:

Question 10.
What is the difference between homologous and analogous organs ? Give two examples of each of them.
Answer:
Differences between Homologous and Analogous organs :

Homologous organ:

• Organs which are similar in structure and origin but different in function are known as homologous organs.
• These structures indicate that organisms bearing them are evolutionary related.
• These organs are differ in outlook.
• They exhibit similarity in internal structure.
  Example: Forelimbs of frog, hand of human, wings of bat and forelimb of horse.

Analogous organ:

• Organs which are differing in structure and origin but similar in function are known as analogous organs.
• These structures indicate that organisms bearing them are evolutionary different.
• These organs are similar in outlook.
• They do not exhibit similarity in internal structure.
  Example: Wings of butterflies, bats and birds.

Question 11.
What do you mean by vestigial organs?
Or
Write the two names of vestigial organs of man.
Or
What are vestigial organs? Explain. Write four vestigial organs of the human body.
Answer:
Vestigial organs:
Organs that are reduced and have become functionless in an organism but were functional in their ancestors are called as vestigial organs.
Examples:

1. Vermiform appendix.
2. Coccygial vertebrae.
3. Nictitating membrane in the eyes of human.
4. Muscles of external ear (Pinna).

Question 12.
Write down the demerits of Darwinism.
Answer:
Some of the demerits of Darwinism are:

1. Darwinism stresses upon small fluctuating variations which has no role in evolution.
2. Does not satisfactorily explain effect of use and disused and presence of vestigial organs.
3. It did not differentiate somatic and germinal variations.
4. It explains survival of the fittest but not arrival of the fittest.

Question 13.
Differentiate between lamarckism and darwinism.
Answer:
Differences between Lamarckism and Darwinism:

Lamarckism:

• It is based on the theory of inheritance of acquired characters.
• According to this theory the length of neck of Giraffe increased only to facilitate it in eating leaves of higher trees.
• Increase in length of neck of Giraffe resulted in evolution of Giraffe with long necks.
• The theory is based on the use or misuse of the organs.

Darwinism:

• It is based on the theory of natural selection.
• Ancestors of Giraffe were both with long and short necks.
• Giraffe with long neck only could survive in struggle for existence.
• This theory is based on the inheritance of the characters in next generation.

Question 14.
Explain Lamarckism in short.
Or
Explain Lamarckism of organic evolution in brief.
Answer:
Lamarckism:
In 1809, Lamarck has proposed a theory to explain organic evolution, which is known, as the “Theory of inheritance of acquired characters.” According to Lamarck, organisms acquire certain characters during their lifetime due to changes in environment and these acquired characters are heritable. According to this theory, new species are originated as follows:

1. New requirements and wills are produced in the organisms due to the effect of changed environment.
2. New requirements and wills of organisms resulting in the production of new habits.
3. Changes in the habit bringing about modifications of the organ.
4. New habits resulting in the use or disuse of the organs.
5. Use of organ resulting in the development of acquired characters.
6. These acquired characters are heritable.
7. Inheritance of acquired characters resulting in the development of new species.

Question 15.
Write two incidences which influenced Darwin to set out theory of evolution.
Answer:
Two incidences which influenced Darwin are:

1. During the voyage of Beagle Darwin noticed distribution of a kind of bird called as finches and other organisms by looking at his observation of Galapagos island. He found variation in the beaks of finches adapted according to the environment.

2. Pegion keepers kills weak varieties of pegion generation after generation to obtain better variety of pegion.

3. Above two incidences influenced Darwin to set out theory of evolution (Natural selection).

Evolution Long Answer Type Questions

Question 1.
Draw a well – labelled diagram of Miller and Urey’s experiment.
Answer:

Experimental evidence of Chemical evolution or Miller’s experiment:

1. Experiment was performed by S.L. Miller and H.C. Urey in 1953.

2. Experimental set – up:
In a closed flask containing CH₄, H₂, NH₃and water vapour at 800°C, electric discharge was created. The conditions were similar to those in primitive atmosphere.

3. Observations:
After a week, they observed presence of amino acids and complex molecules like sugars, nitrogen bases, pigments and fats in the flask.

4. Conclusions:

• It provides experimental evidence for the theory of chemical origin.
• It showed that the first non-cellular form of life was created about 3 billion years ago.
• It showed that non-cellular biomolecules exist in the form of DNA, RNA, polysaccharides and protein.

Question 2.
Write an essay on modern concepts of origin of life.
Or
Explain the role of non – living in origin of life.
Answer:
Modern concept of origin of life:
The modem concept of origin of life was postulated by a Russian biochemist A.I. Oparin in 1936. According to this theory, after the formation of earth various chemicals played important role in the formation of atmosphere. Life originated and first organism came into existence from certain molecules when atmospheric conditions became suitable. According to Oparin, life originated in the following steps:

1. Formation of earth and its atmosphere:
Earth is believed to be originated some 4,500 million years ago by the condensation and cooling of the clouds of cosmic dust and gases called ylem. The heavier elements collected at the core and lighter elements around the core. Outermost layer contains H, C, O and N. Oxygen was found only in combination of other elements. These four elements reacted with each other forming H₂, H₂O, CH₄, NH₃, CO₂ and HCN.

2. Formation of small organic molecules:
The mixture of methane, ammonia, water and hydrogen comes in contact of solar energy. Cosmic rays and electric discharge could produce some simple organic compounds. These simple organic compounds formed in such a way and accumulated in primitive atmosphere and oceans were responsible for synthesis of complex micro-molecules as follows :

1. CH₄, H₂O → Sugars
→ Fatty acids
→ Glycerine

2. CH₄, H₂O, NH₃ → Amino acids

3. CH₄, HCN, H₂O, NH₃ → Nitrogenous bases (Purines and Pyrimidines)

3. Formation of polymers:
It is clearly understood from the above description that a large number of micro molecules such as hydrocarbons, amino acids, fatty acids, purine and pyrimidines and simple sugars accumulated in the oceans. When atmospheric water condensed on further cooling, the inorganic precursors collided, reacted and aggregated to form new molecules of increasing size and complexity. Thus, by polymerization macromolecules were formed. The chemical reactions for the formation of macro – molecules can be summarized as follows:

• Sugar + Sugar → Polysaccharides
• Fatty acid + Glycerine → Lipids
• Amino acid + Amino acid → Protein
• Nitrogenous base (Adenine) + Sugar + Phosphate → Adenosine phosphate
• Nitrogenous base + Sugar → Nucleoside
• Nucleoside + Phosphate → Nucleotide
• Nucleotide + Nucleotide → Nucleic acid.

4. Formation of molecular aggregates and primitive cells:
Over a vast of time, these molecules became associated with one temporary complex. Ultimately, it leads to the formation of a coacervate. A coacervate is a solution of high molecular weight of chemicals, i.e., proteins and carbohydrates, which become bounded by lipid membrane, which is selectively permeable.

The coacervate grows by absorbing molecules from their environment. The substances which got accumulated in the coacervates underwent reactions and resulted in the molecular reorganization of some proteins into enzymes. A coacervate having nucleoprotein surrounded by various nutritive organic substances and covered by surface membrane is considered to be the precell, which got later transformed into first living cell. The coacervate can reproduce by budding.

5. Evolution of complex biochemical reactions:
Primitive organism utilize chemical substances present in the environment as food hence, they are :

(a) Heterotrophic, chemosynthetic organisms appeared due to mutation and natural selection in heterotrophs.

(b) Blue – green algae evolved from chemosynthetic organisms by mutation and natural selection.

(c) The liberation of free oxygen into the atmosphere produced by the blue – green algae due to the process of photosynthesis. It finally changed the reducing atmosphere into an oxidizing one and therefore, all possibilities of further chemical evolution were finished.

Free living eukaryotes originated in the ocean from blue – green algae.

6. The origin of well – developed organisms:
From the simple eukaryotes which were like unicellular organisms of today various forms of life evolved during passage of time.

Question 3.
Write the process of formation of organic molecules in sea water on earth with the help of Miller and Urey’s experiment.
Answer:
The work of A.I. Oparin (1938 – 1965), H.Urey and Stanley Miller (1959) provided evidences in the favour of biochemical origin of life. They had prepared the atmosphere like that of primitive earth and as described by Oparin, they made the synthesis of organic compounds by the following methods:

1. Four elements H, C, O (not free O₂) and N react with each other to form H₂O, CH₄, NH₃, CO₂ and HCN on primitive earth.

2. From these four elements following organic molecules were formed in sea water of earth:
(i)

(ii) CH₄, H₂O, NH₃ → Amino acids.

(iii) CH₄, HCN, H₂O, NH₃ → Nitrogenous bases.

3. Macro – molecules of organic compounds were synthesized by these above prepared organic compounds.

• Sugar + Sugar → Polysaccharides (Carbohydrate).
• Fatty acids + Glycerol → Fats.
• Amino acid + Amino acid → Protein.
• Nitrogenous base + Sugar → Nucleoside.
• Nucleoside + Phosphoric acid → Nucleotide.
• Nucleotide + Nucleotide → Nucleic acid.
• Nitrogenous base (Adenine) + Sugar + Phosphate → Adenosine phosphate.

4. The above organic compounds and salts together constituted the first living being.

Question 4.
Name connecting link of reptiles and birds. Also write their characters.
Answer:
Archaeopteryx is the connecting link of birds and reptiles. Archaeopteryx was a bird. It is regarded as the connecting link between reptiles and birds, which suggests the path of evolution of the latter from the former. It is found as fossils. They are found during Jurassic period 140 million years. Archaeopteryx exhibits both reptiles and birds like characters.

1. Reptiles like characters:

• Bones were similar to that of reptiles in which air sacs were absent.
• Tail bearing vertebra.
• It had teeth in jaws, scales were present on the body.
• Metacarpals were free.
• Pelvic girdle recombines with the pelvic girdle of reptiles.

2. Birds like characters:

• Presence of feathers on body.
• Forelimbs were modified in wings.
• Skull large and monocondylar.
• Jaws were modified into beak.
• Hallux was backwarded and pointed.

Question 5.
Give a detailed account of theory of natural selection.
Or
Describe the Darwin’s opinion about the origin of new species of organisms.
Answer:
Charles Darwin (1809 – 1882) explained the theory of evolution in his book “Origin of species by natural selection.” Darwin undertook a long voyage for five years in the capacity of a naturalist on a British warship ‘Beagle’. He travelled to islands of Galapagos and collected evidences to explain evolution.

To explain origin of species he gave theory of natural selection as a mechanism for evolution.

The main points of Darwinism are given below:

1. Over production of offsprings:
Every living being has an inherent tendency to produce more offspring than that can survive.

2. Struggle for existence:
Though the offsprings are produced in large number yet their production remain almost constant. This is because of struggle for existence. There is struggle for food, space, breeding, etc. Moreover death of individuals due to diseases and predators, population are kept under control.

3. Survival of the fittest:
Darwin believed that any individual is successful in struggle for existence if it survives long enough to produces offsprings. Individuals who are fit in a particular environment can only survive.

4. Variation:
Due to constant struggle, the organisms change themselves in accordance with the new needs.

5. Natural selection:
In the struggle, for existence organisms having variations favourable to the environment would have more chance to survive and reproduce their own kind. Those which do not possess favourable variations would die or fail to reproduce.

6. Origin of species:
Any changes in environmental conditions cause natural selection to act upon the population and select the well adapted individuals. It results in changes of characters of the populations. By the inheritance of these changes in successive generations, new species are formed.

Question 6.
Organic Evolution is a continuous process, explain it. And giving any three evidences.
Answer:
Evolution is a complex phenomenon accounting for the present day diversity among organisms. But it has clearly maintained the basic unity among them since it occurred over a period of millions of years, no one would have seen/recorded evolution and hence scientists have provided various evidences to prove evolution. Some of the evidences of organic evolution are described below:

I. Evidences from Embryology:
Important activities that occur various animals are:

1. For the survival, all animals get energy and various substances from environment.
2. In all organisms energy is produced from ATP.
3. In all organisms, the duplication of DNA is similar.
4. In all organisms, protein synthesis is same and it is produced from ribosomes.
5. In all organisms respiration and steps of respiration is same.
6. All organisms multiplicate and reproduce, due to which they have basic similarities.
7. All organisms conduct hereditary characters on similar principles.

II. Evidence from Anatomy:
Anatomy of living organisms will be explained with different examples:

1. Homologous organs : Organs which are similar in structure and origin but different in function and appearance is known as homologous organs.

2. Analogous organs : Organs which are different in origin and structure but performing similar functions are known as analogous organs.

3. Vestigial organs : Organs that are reduced and have become functionless in an organism, but were functional in their ancestors are called vestigial organ.

III. Evidences from vestigial organs:
Organs of the body which are non – functional but they are functional in some other organisms are called vestigial organs: Morphological evidence of evolution is provided by the presence of vestigial organs of body which are often undesired, degenerated and non-functional. These might have been large and functional in some other animals or in ancestors of those which now possess it in rudimentary forms, example vermiform appendix in man, muscles of external ear (pinna) in man, nictitating membrane or plica, semilunaris in human eye, wisdom teeth (third pair of molars) tail bone (coccyx) in man, wings of ostrich, hindlimbs in snakes, etc.

Question 7.
What do you mean by organic evolution? How do fossils exhibit evidence to prove organic evolution?
Answer:
Descent with modification in organism is known as organic evolution.

Evidences of organic evolution from fossils record:
Fossils are treated as significant evidence of organic evolution. Fossils are the remains or impressions of ancient organisms preserved in the layers of rock and soil. Fossils only do not prove the theory of organic evolution, yet it evidently prove that gradually complexity increased in body organization. The complexity in the body of organization can be noticed as we study the upper layers. Thus, it can be concluded from above observations :

1. The crust of the earth and the organisms living on it underwent change in the course of time.
2. The organisms with simple structural organization originated earlier than the complex ones.
3. Some of the organisms lived on the earth for short time and became extinct. This was a result of drastic changes in the climate on the earth.

Hence, forth fossils produce bonafide record of such plants and animals which had shown their existence once upon a time and now are extinct or not present exactly in the same form, thus, producing strong evidence in favour of organic evolution.

Question 8.
Try to trace the various components of human evolution (Hint: Brain size and function, skeletal structure, dietary preference, etc.)
Answer:
The various components of human evolution are as follows:

1. Brain capacity.
2. Posture.
3. Food/Dietary preference and other important features.

Name, brain capacity, posture and food features:

1. Dryopithecus Africans:
Knuckle walker, walked similar to Gorillas and Chimpanzees (was more apelike), soft fruit and leaves; canines large, arm and legs are of equal size.

2. Ramapithecus:
Semi – erect (more manlike) ate seeds and nuts, canines were small while molars were large.

3. Australopithecus africanus:
Brain capacity 450 cm³, full erect prosture, height 1.05m, herbivorous (ate fruits), hunted with stone weapons, lived as trees, canines and incisors were small.

4. Homo habilus:
Brain capacity 735 cm³, fully erect posture, height, 1.5m, carnivorous, canines were small. They were first tool makers.

5. Homo erectus:
Brain capacity 800 – 1100 cm³, fully erect posture, height 1.5 – 1.8m, omnivorous. They used stone and bone tools for hunting games.

6. Homo neanderthalensis :
Brain capacity 1300 – 1600 cm³, fully erect posture, height 1.5 – 1.6m, omnivorous cave dwellers, used hiles to protect their bodies and buried their dead.

7. Homo sapiens fossils:
Brain capacity 1650 cm3, fully erect posture with height 1.8m, omnivorous. They had strong jaw with teeth close together. They were cave dwellers, made painting and carvings in the caves. They developed a culture and were called first modern man.

8. Homo sapiens sapiens:
Brain capacity 1200 – 1600 cm³, fully erect posture, height 1.5 – 1.8m, omnivorous. They are the living modem men with high intelligence. They developed art, culture, language, speech, etc. They cultivated crops and domesticated animals.

Question 9.
Find out through internet and popular science articles whether animals other than man has self – consciousness.
Answer:
There are many animals other than humans, which have self – consciousness. An example of an animal being self-conscious is dolphins. They are highly intelligent. They have a sense of self and, they also recognize others among themselves and others. They communicate with each other by whistles, tail – slapping, and other body movements. Not dolphins, there are certain other animals such as Crow, Parrot, Chimpanzee, Gorilla, Orangutan, etc., which exhibit self – consciousness.

Question 10.
List 10 modern day animals and using the internet resources link it to a corresponding ancient fossil. Name both.
Answer:
Modern and Ancient corresponding animals:

Question 11.
Practise drawing various animals and plants.
Answer:

Question 12.
Using various resources such as your school Library or the internet and discussions with your teacher, trace the evolutionary stages of any one animal, say horse.
Answer:
The evolution of horse started with Eohippus during Eocene period. It involved the following evolutionary stages:

1. Gradual increase in body size.
2. Elongation of head and neck region.
3. Increase in the length of limbs and feet.
4. Gradual reduction of lateral digits.
5. Enlargement of third functional toe.
6. Strengthening of the back.
7. Development of brain and sensory organs.
8. Increase in the complexity of teeth for feeding on grass.

1. Eohippus:
It had a short head and neck. It had four functional toes and a splint of 1 and 5 on each hind limb and a splint of 1 and 3 in each forelimb. The molars were short crowned that were adapted for grinding the plant diet.

2. Mesohippus:
It was slightly taller than Eohippus. It had three toes in each foot.

3. Merychippus:
It had the size of approximately 100 cm. Although it still had three toes in each foot, but it could run on one toe. The side toe did not touch the ground. The molars were adapted for chewing the grass.

4. Pliohippus:
It resembled the modern horse and was around 108 cm tall. It had a single functional toe with splint of 2nd and 4th in each limb.

Equus:
Pliohippus gave rise to Equus or the modern horse with one toe in each foot. They have incisors for cutting grass and molars for grinding food.

Question 13.
How far inter relationship among? The living organisms is helpful to understand the process of evolution?
Answer:
Organisms which are externally different shows some similarities. This is known as interrelationship. It proves that they might have evolved from the same ancestor. Inter relationship among the living organisms can be understood by following example:

1. All living organisms obtain energy and various materials from the environment.
2. All living organisms multiply and reproduce.
3. All living organisms show transmission of genetic information according to same principle.
4. All living organisms synthesize protein in the ribosome and steps of protein synthesis are same in all living organisms.
5. All organisms respire and steps of respiration in them are same.
6. In all organism method of DNA replication are same.
7. In all living organisms flow of information occurs by the help of nucleic acids present in the nucleus.
8. In all living organisms energy is obtained from ATP.

MP Board Class 12th Biology Important Questions

MP Board Class 12th Hindi Swati Solutions गद्य Chapter 11 गीता और स्वधर्म (निबंध, आचार्य विनोबा भावे)

गीता और स्वधर्म अभ्यास

गीता और स्वधर्म अति लघु उत्तरीय प्रश्न

प्रश्न 1.
अर्जुन के स्वभाव में कौन-सी वृत्ति विद्यमान थी?
उत्तर:
कृत-निश्चय होकर और कर्तव्य भाव से समर-भूमि में खड़े अर्जुन के स्वभाव में क्षात्रवृत्ति विद्यमान थी।

प्रश्न 2.
किस सजा से अपराधी के सुधरने की आशा नष्ट हो जाती है?
उत्तर:
फाँसी की सजा अमानुषी होने के कारण मनुष्य को शोभा नहीं देती क्योंकि इससे अपराधी के सुधरने की आशा नष्ट हो जाती है।

प्रश्न 3.
लेखक किस सागर में डुबकी लगाकर बैठ जाता है?
उत्तर:
लेखक विनोबा भावे जब अकेले होते हैं तब गीता के अमृत सागर में गहरी डुबकी लगाकर बैठ जाते हैं।

प्रश्न 4.
किसने श्रीकृष्ण से अपना सारथ्य स्वीकार कराया? (2016)
उत्तर:
दुर्योधन के द्वारा संधि प्रस्ताव ठुकराने पर अर्जुन ने अनेक देशों के राजाओं को एकत्र करके श्रीकृष्ण से अपना सारथ्य स्वीकार कराया।

प्रश्न 5.
युद्ध में आप्त स्वजन सम्बन्धियों की कितनी पीढ़ियाँ एकत्र हुई थी?
उत्तर:
युद्ध भूमि के मध्य खड़े होकर अर्जुन ने देखा कि दादा, बाप, बेटे, पोते, आप्त-स्वजन सम्बन्धियों की चार पीढ़ियाँ एकत्र थीं।

प्रश्न 6.
अर्जुन के मन में कौन-सा भाव उत्पन्न हो गया था?
उत्तर:
युद्ध के लिए तत्पर स्वजन समूह को देखकर सदैव विजयी रहने वाले अर्जुन के मन में अहिंसा का भाव उत्पन्न हो गया था।

गीता और स्वधर्म लघु उत्तरीय प्रश्न

प्रश्न 1.
गीता के गगन में लेखक किन उपकरणों से उड़ान भरता है?
उत्तर:
जिस प्रकार एक पक्षी अपने दो पंखों की सहायता से नीले खुले गगन में स्वछन्द होकर यथाशक्ति दूर तक उड़ान भरता है। ठीक उसी प्रकार लेखक तर्क को त्यागकर श्रद्धा और प्रयोग,इन दो उपकरणों से गीता के गगन में यथाशक्ति उड़ान भरता रहता है।

प्रश्न 2.
अर्जुन ने अकेले ही किन-किन के दाँत खट्टे कर दिए थे? (2017)
उत्तर:
जिस समय पांडव अज्ञातवास में थे उस समय कौरवों की सेना ने उत्तर कुमार की गायों का हरण करने का प्रयास किया तब अर्जुन ने उत्तर कुमार का सारथी बनकर अकेले ही धीर भीष्म पितामह, आचार्य द्रोण और सूर्यपुत्र कर्ण के दाँत खट्टे कर दिए थे।

प्रश्न 3.
युद्ध टालने के लिए कौन-कौन सी दोनों बातें बेकार हो चुकी थीं?
उत्तर:
अर्जुन के स्वभाव में क्षात्रवृत्ति थी। वह कृत निश्चय होकर और कर्त्तव्य भाव से समर-भूमि में खड़ा था। लेकिन युद्ध के विनाशकारी परिणामों का उसे आभास था। अत: युद्ध को टालने के लिए अर्जुन ने पहली, कौरवों के पास कम से कम माँग का प्रस्ताव भेजा-दूसरी, कृष्ण को मध्यस्थ बना कर भेजा। लेकिन विनाशकारी बुद्धि वाले दुर्योधन ने एक भी बात को स्वीकार नहीं किया।

प्रश्न 4.
लेखक के अनुसार गीता के जन्म का उद्देश्य क्या था?
उत्तर:
युद्धभूमि में अर्जुन धर्म-सम्मूढ़ हो गया था, इस कारण उसके मन में स्वधर्म के विषय में मोह पैदा हो गया था। इस मोह को अर्जुन स्वयं स्वीकार करता है। अतः स्पष्ट है कि गीता का जन्म स्वधर्म में बाधक जो मोह है, उसके निवारणार्थ हुआ है। गीता के जन्म का प्रधान उद्देश्य यही था। गीता सुनने के बाद अर्जुन कृष्ण से कहता है-भगवन्, मेरा मोह नष्ट हो गया है, मुझे स्वधर्म का भान हो गया है।

प्रश्न 5.
लेखक के अनुसार गीता का मुख्य काम क्या-क्या है?
उत्तर:
यदि गीता के उपक्रम और उपसंहार को मिलाकर देखें तो स्वधर्म में बाधक मोह-ममत्व तथा स्वजनासक्ति दूर करना गीता का मुख्य काम है,मोह का निवारण करना ही गीता का काम है। गीता ही नहीं,सम्पूर्ण महाभारत का काम लोक हृदय के मोहावरण को दूर करना है। यही काम गीता ने किया जिसके परिणामस्वरूप अर्जुन का मोह नष्ट हो गया और उसे क्षात्रधर्म का ज्ञान हो गया।

गीता और स्वधर्म दीर्घ उत्तरीय प्रश्न

प्रश्न 1.
विनोबा जी का गीता के साथ किस प्रकार का सम्बन्ध था?
उत्तर:
विनोबा जी का गीता के साथ जो सम्बन्ध था उसे बताना कठिन है क्योंकि वह अलौकिक था जिसे तर्क द्वारा नहीं बताया जा सकता है। विनोबा जी कहते हैं कि उनके शरीर का निर्माण तो माँ के दूध से हुआ किन्तु उनके हृदय और बुद्धि का पोषण गीता के दूध अर्थात् गीता के संदेशों से हुआ। इस सम्बन्ध को हम हार्दिक सम्बन्ध कह सकते हैं। हार्दिक सम्बन्धों में तर्क बहस,दलील या चर्चा का कोई महत्व नहीं होता। तर्क के स्थान पर श्रद्धा और प्रयोग इन दो बातों के आधार पर गीता को समझने का प्रयास किया जाता है।

गीता उनका प्राणतत्व है, आत्मा है। जब वह गीता के मोह निवारण पर चर्चा करते हैं तो मानो, वह गीता रूपी सागर पर तैर रहे होते हैं अर्थात् गीता के श्लोकों का शब्दार्थ कर रहे हैं परन्तु जब वह एकाकी होकर मोह निवारण का माप श्रद्धा व प्रयोग के आधार पर हृदय में मनन करते हैं तो उन्हें ऐसा लगता है, मानो वे अमृत-सागर में डुबकी (गोता) लगा रहे हैं जिससे उनकी आत्मा को परम सुख की प्राप्ति होती है। इसी आत्मा के परम सुख की प्राप्ति के रूप में ही विनोबा जी का सम्बन्ध गीता से है। यह सम्बन्ध अलौकिक होने के साथ-साथ हृदय के तारों से मधुर ईश्वरीय संगीत को सुनाने वाला है।

प्रश्न 2.
अर्जुन को महावीर क्यों कहा गया?
उत्तर:
विनोबा जी ने अर्जुन को महावीर कहा है, उनका यह कथन बिल्कुल सत्य है, क्योंकि अर्जुन में अठारह अक्षौहिणी सेना से अधिक शक्ति थी। वह सैकड़ों लड़ाइयों में अपना जौहर दिखा चुका था। उत्तर-गो ग्रहण के समय उसने अकेले ही भीष्म, द्रोण और कर्ण के दाँत खट्टे कर दिए थे। वह सदा विजय प्राप्त करने वाला और सब नरों में एक ही सच्चा नर था। वीरवृत्ति उसके रोम-रोम में भरी थी। उसके विचार कृत-निश्चय और कर्त्तव्य भाव से पूर्ण थे। क्षात्रवृत्ति उसके स्वभाव में थी। वीरवृत्ति का उत्साह उसकी प्रेरणा थी। युद्ध क्षेत्र में उसने असंख्य वीरों का संहार किया था।

उसका गांडीव हाथ से कभी नहीं गिरा था। उसके शरीर में कभी कम्पन नहीं हुआ था न आँखें गीली हुई थीं। वह सदैव युद्ध-प्रवृत्त ही था। युद्ध उसकी दृष्टि से उसका स्वभाव प्राप्त और अपरिहार्य रूप से निश्चित कर्त्तव्य था। पलभर के लिए उसके मन में परिवार के प्रति आसक्ति तथा मोह जाग्रत हुआ था। परन्तु कृष्ण ने उसके इस मोह को गीता सुनाकर नष्ट कर दिया था। गीता कर्मयोग तथा युद्धयोग दोनों का प्रतिपादन करती है जिसे अर्जुन ने पूर्ण रूप से अपनाया था। इन्हीं सब कारणों से अर्जुन को महावीर कहा गया।

प्रश्न 3.
युद्ध क्षेत्र में स्वजनों को देखकर अर्जुन के मन में कौन-कौन से भाव उत्पन्न होते है? (2014)
उत्तर:
पाण्डवों ने महाभारत के युद्ध को टालने के लिए दुर्योधन के पास कम-से-कम माँग का प्रस्ताव और श्रीकृष्ण जैसे मध्यस्थ को भेजा परन्तु दुर्योधन नहीं माना तो अर्जुन कृष्ण को अपना सारथी बनाकर रणांगन में पहुँचा और अपने रथ को दोनों सेनाओं के बीच खड़ा करने को कहा। सेनाओं के बीच खड़े होकर उसने सारे स्वजन समूह को देखा तो उसके हृदय में उथल-पुथल मच गई। उसे बहुत बुरा लगा। अर्जुन के मन में भाव उठे कि दादा, बाप, बेटे, पोते-आप्त,स्वजन-सम्बन्धियों की चार पीढ़ियाँ मरने-मारने के अन्तिम निश्चय से वहाँ एकत्र हुई हैं। इस प्रत्यक्ष दर्शन का उसके हृदय पर हृदय को विचलित कर देने वाला प्रभाव पड़ा।

आज तक उसने अनेक युद्धों में असंख्य वीरों का संहार किया है, परन्तु उस समय उसे बुरा नहीं लगा था। उसका गांडीव हाथ से नहीं गिरा था। शरीर में कंपन नहीं हुआ था, उसकी आँखें गीली नहीं हुई थीं। तो इसी समय ऐसा क्यों हुआ? उसके मन में स्वजनासक्ति के भाव उत्पन्न हुए। कर्त्तव्यविमुख होने के भावों के साथ तत्व ज्ञान (दर्शन) के भाव जाग्रत हुए। वह युद्ध को व्यर्थ समझकर पाप समझने लगा। युद्ध से कुल क्षय होगा, धर्म का लोप होगा। स्वैराचार मचेगा, व्यभिचार फैलेगा, अकाल आ पड़ेगा,समाज पर तरह-तरह के संकट आयेंगे। इस प्रकार के अनेक भाव अर्जुन के मन में उत्पन्न होते हैं।

प्रश्न 4.
न्यायाधीश के उदाहरण से लेखक क्या सीख देना चाहता है?
उत्तर:
एक न्यायाधीश ने सैकड़ों अपराधियों को फाँसी की सजा दी थी। परन्तु जब अपने पुत्र को खून के जुर्म में मृत्युदण्ड देने का समय आया तो पुत्र-स्नेह के कारण वह हिचकने लगा तथा बुद्धिवाद बघारने लगा-फाँसी की सजा बड़ी अमानुषी है, लज्जा की बात है, बड़ा कलंक है। उसकी यह बात आंतरिक नहीं थी। वह आसक्ति-जनित थी। इस प्रकार महावीर अर्जुन भी स्वजन आसक्ति में आकर अहिंसावादी तथा बुद्धिवादी बन गया और युद्ध में दोष बताने लगा। अतः लेखक बताना चाहता था कि यह अर्जुन का तत्व ज्ञान नहीं कोरा प्रज्ञावाद था। मनुष्य परिजनों के स्नेह में पड़कर कर्त्तव्य को भूल जाता है, वह मोह में पड़ जाता है। इसी मोह को दूर करना गीता का धर्म है। मनुष्य को मोह में न पड़कर अपना कर्तव्य पालन करना चाहिए। मोह मनुष्य को नीचे गिरा देता है। अर्जुन की गति न्यायाधीश जैसी हो गई थी।

प्रश्न 5.
श्रीकृष्ण ने अर्जुन की जिज्ञासाओं का कैसे समाधान किया?
उत्तर:
श्रीकृष्ण द्वारा मध्यस्थता करने तथा कम-से-कम मांग के प्रस्ताव को दुर्योधन ने स्वीकार नहीं किया तो युद्ध अनिवार्य हो गया। दोनों पक्षों की सेनाएँ युद्धभूमि में खड़ी थीं। तब अर्जुन के मन में अचानक जिज्ञासा जागी कि वह दोनों सेनाओं के चेहरे देखे। अर्जुन ने कृष्ण को अपनी जिज्ञासा बताई, कृष्ण ने अर्जुन का रथ दोनों सेनाओं के मध्य खड़ा कर दिया। अर्जुन ने अपनी चार पीढ़ियों को देखा और मोहग्रस्त हो युद्ध से विरक्त हो गया। इसी प्रकार स्वधर्म के विषय में उनके मन में मोह पैदा हो गया था। श्रीकृष्ण ने अर्जुन की जिज्ञासा व मोह को दूर किया। इसके लिए श्री कृष्ण ने अर्जुन को गीता सुनाई जो गीता महाभारत के मध्यभाग में एक ऊँचे दीपक के समान है तथा समस्त जिज्ञासाओं को हल करने में पूर्णतः सक्षम है।

प्रश्न 6.
गीता का निष्कर्ष क्या है?
उत्तर:
गीता की योजना महाभारत में की गई थी जो महाभारत के मध्य भाग में एक ऊँचे दीपक की तरह पूर्ण महाभारत को प्रकाशित करती है। गीता का निष्कर्ष केवल कर्म-योग ही नहीं, बल्कि युद्ध-योग का भी प्रतिपादन करता है। युद्ध-भूमि के मध्य खड़ा अर्जुन स्वजन आसक्ति के कारण क्षात्रवृत्ति से विमुख होकर युद्ध के दुष्परिणाम बघारने लगा क्योंकि सारे जन समूह को देखकर उसके हृदय में उथल-पुथल मच गई थी। कृष्ण अर्जुन के आसक्तिजनित मोह को जानते थे, इसलिए उन्होंने अर्जुन के मोह-पाश का उपाय शुरू किया। स्वधर्म में बाधक जो मोह है उसका निवारण करना ही गीता का उद्देश्य है। मोह, ममत्व, आसक्ति को दूर करके स्वधर्म का भान कराना ही गीता का निष्कर्ष है। गीता ही नहीं सम्पूर्ण महाभारत में लोक हृदय के मोहावरण को दूर करने के लिए यह इतिहास-प्रदीप जलाया गया है। मुख्य निष्कर्ष मोह निरसन ही है।

गीता और स्वधर्म भाषा-अध्ययन

प्रश्न 1.
दिए गए मुहावरों का वाक्यों में प्रयोग कीजिए

1. दाँत खट्टे कर देना
2. उथल-पुथल मचाना
3. आँखें गीली होना
4. जौहर दिखाना
5. उड़ान भरना
6.  डुबकी लगाना
7. जुनून उतर जाना।

उत्तर:

1. दाँत खट्टे करना – भारतीय सैनिकों ने 1971 के युद्ध में पाकिस्तानी सैनिकों के दाँत खट्टे कर दिए थे।
2. उथल-पुथल मचना – कलिंग युद्ध के मैदान में अनेक सैनिकों की कराहट को सुनकर सम्राट अशोक के हृदय में उथल-पुथल मच गई।
3. आँखें गीली होना – रेल दुर्घटना के वीभत्स दृश्य को देखकर उपस्थित जनसमूह की आँखें गीली हो गईं।
4. जौहर दिखाना – झाँसी वाली रानी ने अकेले ही ब्रिटिश सैनिकों के बीच जौहर दिखाया।
5. उड़ान भरना – जीवन में सफलता पाने के लिए कल्पना की उड़ान भरने के साथ परिश्रम करना भी आवश्यक है।
6. डुबकी लगाना – श्री विनोबा भावे ने गीता के अमृत सागर में डुबकी लगाकर जीवन। का लक्ष्य प्राप्त कर लिया।
7. जुनून उतर जाना – स्वजनों के मोह का जूनून उतरने पर ही अर्जुन महाभारत के युद्ध को जीत सका।

प्रश्न 2.
निम्नलिखित वाक्यों का भाव-विस्तार कीजिए-
(अ) जहाँ हार्दिक सम्बन्ध होता है, वहाँ तर्क की गुंजाइश नहीं होती। (2017)
उत्तर:
भाव विस्तार :
‘गीता और स्वधर्म’ शीर्षक में ‘श्री विनोबा भावे’ने बताया है कि गीता के साथ उनका हार्दिक सम्बन्ध है। इस कारण इस सम्बन्ध को साबित करने के लिए किसी भी प्रकार की दलील नहीं दी जा सकती और न ही बहस की जा सकती है। हार्दिक सम्बन्ध भावना से जुड़े होते हैं जिन्हें अनुभव किया जाता है, परन्तु दलीलों द्वारा नहीं बताया जा सकता। साथ ही यह सम्बन्ध सूक्ष्म होता है। इस सम्बन्ध में बड़ी शक्ति होती है,जो पाषाण हृदय को भी कोमल बना देती है। इस सम्बन्ध के द्वारा आत्मिक प्रसन्नता व शान्ति मिलती है।

(ब) इस आसक्ति जनित मोह ने उसकी कर्त्तव्य निष्ठा को ग्रस लिया और तब उसे तत्त्व ज्ञान याद आया।
उत्तर:
भाव विस्तार :
‘श्री विनोबा भावे’ जी ने अपने निबन्ध ‘गीता और स्वधर्म’ में अर्जुन के मोह व कृष्ण की गीता के विषय में बताया है। अर्जुन युद्धभूमि के मध्य रथ पर सवार था,तब उसने स्वजन समूह को देखा तो उसके हृदय में उथल-पुथल मच गई। उसके हृदय में परिजनों के प्रति मोह उत्पन्न हुआ, इस कारण वह अपने क्षात्रधर्म अर्थात् युद्ध कर्म को भूल कर प्रज्ञावादी बन गया। युद्ध को पाप व मानव जाति का कलंक कहने लगा। जो अर्जुन युद्ध प्रवृत्त था, युद्ध करना उसका स्वभाव था, अपरिहार्य रूप से उसका कर्त्तव्य था, उस कर्त्तव्य को अर्जुन भूल गया। युद्ध के दुष्परिणामों को बताने लगा यद्यपि जो सत्य थे परन्तु अर्जुन के मुख से सुशोभित नहीं हो रहे थे क्योंकि अर्जुन का कर्तव्य युद्ध में निहित था। उसका तत्व ज्ञान आत्मिक न होकर बौद्धिक था। कहीं न कहीं उस तत्त्व ज्ञान की जड़ में मोह था जिसने अर्जुन को कर्तव्य से गिरा दिया था।

गीता और स्वधर्म पाठ का सारांश

भूदान आन्दोलन के प्रणेता तथा किसानों के मसीहा ‘आचार्य विनोबा भावे’ द्वारा लिखित प्रस्तुत निबन्ध ‘गीता और स्वधर्म’ में आचार्य जी ने गीता के माध्यम से मनुष्य को अपनी बुद्धि के आधार पर न्यायपूर्ण कार्य करने का संदेश दिया है।

लेखक का गीता के साथ तर्कपूर्ण नहीं, आत्मिक रिश्ता है। उनके हृदय और बुद्धि का पोषण गीता के दूध से हुआ है। गीता पर उनकी अपार श्रद्धा है। गीता उनके लिए अमृत का सागर है। यह गीता श्रीकृष्ण ने युद्ध स्थल में युद्ध से विमुख हुए अर्जुन को कर्तव्य का ज्ञान कराने के लिए सुनाई थी। इसीलिए गीता महाभारत में दीपक के समान प्रकाश देती है।

माना जाता है कि कृष्ण ने अर्जुन को गीता दो कारणों से सुनाई होगी। एक, अर्जुन की नपुंसकता को दूर करने तथा दूसरी अर्जुन की अहिंसा-वृत्ति को दूर करने के लिए। लेकिन ये दोनों बातें ही ठीक नहीं हैं.क्योंकि अर्जन ‘महावीर’ थे,उन्होंने उत्तर-गो ग्रहण के समय अकेले ही भीष्म, द्रोण और कर्ण को पराजित किया था। वीर-वृत्ति उनके रोम-रोम में बसी थी। क्षात्रवृत्ति उनके स्वभाव में थी। युद्ध को टालने के लिए कम से कम माँग का प्रस्ताव और कृष्ण ने मध्यस्थता की थी परन्तु दुर्योधन सहमत नहीं था। इस कारण कृष्ण को अपना सारथी बनाकर वीर अर्जुन कुरुक्षेत्र में पहुँचे। वहाँ दोनों पक्षों की सेनाओं के मध्य अपने रथ में बैठे अपनी चार पीढ़ियों के मरण का उत्साह देखकर वह स्वजनाशक्ति में आ गए थे। वीरता का शौर्य दिखाने वाले उस महावीर अर्जुन का गांडीव कभी झुका नहीं था,न शरीर में कम्पन हुआ था और न कभी आँखें गीली हुई थीं।

वहीं अर्जुन जिसमें कभी अहिंसा-वृत्ति ने जन्म नहीं लिया था, उसके आसक्ति जनित मोह ने कर्तव्य को त्याग कर तत्व ज्ञान की शरण ली। कहने लगे कि युद्ध मानव जाति के विनाश का कारण होगा। उस समय अर्जुन की दशा उस न्यायाधीश जैसी थी जो अनेकों को मृत्यु दण्ड दे चुका था परन्तु जब अपने बेटे को खून के जुर्म में मृत्यु दण्ड देने का समय आया तो कहने लगा कि मृत्यु दण्ड अमानुषी सजा है, बड़ा कलंक है। यह कथन आसक्ति जनित है। अर्जुन के इस तत्व ज्ञान को कृष्ण जानते थे इसलिए अर्जुन के मोहपाश को नष्ट करने का उपाय किया और गीता में अर्जुन के इसी मोहपाश पर गदा प्रहार किया है।

अतः गीता का जन्म,स्वधर्म में बाधक जो मोह है,उसके निवारणार्थ हुआ है। गीता सुनाने के बाद कृष्ण ने अर्जुन से पूछा कि अर्जुन तेरा मोह गया? तब अर्जुन ने उत्तर दिया-हाँ भगवन! मेरा मोह नष्ट हो गया, मुझे स्वधर्म का भान हो गया। गीता ही नहीं सम्पूर्ण महाभारत का यही उद्देश्य है। व्यास जी ने महाभारत के आरम्भ में कहा है कि लोक हृदय के मोहावरण को दूर करने के लिए मैं यह इतिहास प्रदीप जला रहा हूँ।

गीता और स्वधर्म कठिन शब्दार्थ

तर्क = दलील। क्लैष्य = नपुंसकता। प्रवृत्त = संलग्न। परावृत्त = दूर भागना। जौहर = वीरता। ख्याति = यश। समर-भूमि = युद्ध-भूमि। क्षात्रवृत्ति = क्षत्रिय धर्म। रणांगण = युद्ध भूमि। स्वजनासक्ति = पारिवारिक जनों के प्रति मोह। प्रतिपादन = सिद्ध करना। क्षय = नष्ट। स्वैराचार = स्वेच्छाचार। नौबत = अवसर। अमानुषी = जंगलीपन। जुनून = धुन। आसक्ति = मोह। प्रज्ञावेद = बुद्धिवाद। अपरिहार्य = जो छोड़ा न जा सके। प्रहार = चोट। निरसन = निराकरण। सारथ्य = सारथी का कार्य। कर्त्तव्यच्युति = अपने कर्त्तव्य से विमुख। अवान्तर = अन्य।

गीता और स्वधर्म संदर्भ-प्रसंग सहित व्याख्या

(1) गीता का और मेरा सम्बन्ध तर्क से परे है। मेरा शरीर माँ के दूध पर जितना पला है, उससे कहीं अधिक मेरे हृदय और बुद्धि का पोषण गीता के दूध पर हुआ है। जहाँ हार्दिक सम्बन्ध होता है, वहाँ तर्क की गुंजाइश नहीं रहती। तर्क को काटकर श्रद्धा और प्रयोग, इन दो पंखों से ही मैं गीता-गगन में यथाशक्ति उड़ान भरता रहता हूँ।

संदर्भ :
प्रस्तुत गद्य अवतरण हमारी पाठ्य-पुस्तक के निबन्ध ‘गीता और स्वधर्म’ नामक पाठ से अवतरित है। इसके लेखक ‘आचार्य विनोबा भावे’ हैं।

प्रसंग :
इन पंक्तियों में गीता तथा विनोबा भावे के सम्बन्ध को तर्कहीन बताकर गीता के प्रति उनकी अपार श्रद्धा को व्यक्त किया गया है।

व्याख्या :
विनोबा भावे गीता पर प्रवचन देते हुए कहते हैं कि उनका और गीता का सम्बन्ध अलौकिक है जिसे शब्दों के द्वारा नहीं बताया जा सकता, न उस सम्बन्ध के विषय में कोई दलील या बहस की जा सकती है। गीता के साथ अपने सम्बन्ध को स्पष्ट करते हुए वे कहते हैं कि माँ के दूध व वात्सल्य से उनके शरीर का निर्माण हुआ परन्तु उनके हृदय व बुद्धि का विकास गीता के ज्ञान से हुआ। गीता के रहस्य ने ही उनके हृदय को पवित्र तथा बुद्धि को परिष्कृत किया है। जब किसी से हार्दिक सम्बन्ध हो जाते हैं तो वहाँ बहस या दलील व्यर्थ हो जाती है।

हृदय दलीलों को नहीं मानता, वह तो सत्य व कोमलता को स्वीकार करता है। जिस प्रकार एक पक्षी अपने दोनों पंखों की सहायता से स्वछन्द नीले आकाश में अपनी पूर्ण शक्ति के साथ उड़ान भरता हुआ प्रसन्न होता है ठीक उसी प्रकार विनोबा जी भी गीता में श्रद्धा रखते हुए तथा उसे अपने जीवन में साक्षात प्रयोग करते हुए गीता के अर्थ को या उसके रहस्य को समझते हैं और सुख का अनुभव करते हैं। श्रद्धा और प्रयोग उनके गहन अध्ययन के दो आधार रूपी पंख हैं।

विशेष :

1. विनोबा जी की गीता के प्रति श्रद्धा अभिव्यक्त की गई है।
2. संस्कृतनिष्ठ शुद्ध साहित्यिक खड़ी बोली का प्रयोग है।
3. गीता-गगन में रूपक अलंकार का प्रयोग है।
4. उदाहरण व गवेषणात्मक शैली का प्रयोग है।

(2) मैं प्राय: गीता के ही वातावरण में रहता हूँ। गीता मेरा प्राणतत्व है। जब मैं गीता के सम्बन्ध में किसी से बात करता हूँ, तब गीता-सागर पर तैरता हूँ और जब अकेला रहता हूँ तब उस अमृत-सागर में गहरी डुबकी लगाकर बैठ जाता हूँ।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
प्रस्तुत अंश में विनोबा जी गीता के प्रति अपनी श्रद्धा का वर्णन करते हैं। गीता के प्रति उनकी श्रद्धा तर्करहित हार्दिक है। उनके जीवन का हर पल गीता के मूल से जुड़ा है।

व्याख्या :
विनोबा भावे, आसक्ति रहित कर्म करने का संदेश देने वाली गीता से, आजीवन जुड़े रहे। गीता ने उन्हें जीवन प्रदान किया। गीता उनके प्राण-आत्मा थी। गीता की चर्चा करते समय वह उसके बाह्य रूप के सम्पर्क में रहते थे लेकिन एकान्त की अवस्था में वह उसके गूढ़ रहस्य को समझने में लगे रहते थे। परन्तु प्रतिपल वे गीता के सम्पर्क में रहते थे। एकाकी अवस्था में उन्होंने गीता रूपी सागर के निष्काम कर्म रूपी अमृत को पीया। गीता-सागर में डुबकी लगाने अर्थात् उसका गहन अध्ययन करने पर उन्हें सुख मिलता था। जैसे अमृत पीकर मनुष्य अमर हो जाते हैं, उसी प्रकार गीता के रहस्य को समझने पर वे अमर हो गये या कहें कि वे सुख-दुःख से परे हो गए। इस प्रकार गीता व विनोबा जी दो अलग रूप होते हुए एक हो गए थे। गीता ने उन्हें आनन्दपूर्ण जीवन प्रदान किया।

विशेष :

1. गीता के द्वारा ही मनुष्य मोक्ष पा सकता है।
2. प्राणतत्व जैसे गूढ़ शब्दों के कारण भाव कठिन हो गए हैं।
3. अलंकारिक व मुहावरेदार भाषा।
4. संस्कृत शब्दावली का प्रयोग।
5. साहित्यिक खड़ी बोली।

(3) अर्जुन, तो लड़ाई से परावृत्त हो रहा था, सो भय के कारण नहीं। सैकड़ों लड़ाइयों में अपना जौहर दिखाने वाला वह महावीर था। उत्तर-गो ग्रहण के समय उसने अकेले ही भीष्म, द्रोण और कर्ण के दाँत खट्टे कर दिये थे। सदा विजय प्राप्त करने वाला और सब नरों में एक ही सच्चा नर, ऐसी उसकी ख्याति थी। वीरवृत्ति उसके रोम-रोम में भरी थी।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
प्रस्तुत पंक्तियों में बताया गया है कि जब अर्जुन कुरुक्षेत्र की युद्ध-भूमि में खड़ी अट्ठारह अक्षोहिणी सेना के मध्य अपनी चार पीढ़ियों को खड़ा देखते हैं तो उनके मन में स्वजनासक्ति आ जाती है और वह युद्ध न करने की बात कृष्ण से कहते हैं। तब कृष्ण गीता के माध्यम से क्षात्रधर्म का सन्देश देकर उनमें उत्साह का संचार करते हैं।

व्याख्या :
श्री विनोबा जी बताते हैं कि अर्जुन ने युद्ध करना अस्वीकार किया। इसका कारण विशाल सेना की शक्ति को देखकर वह भयभीत नहीं था। अर्जुन महावीर था, उसने अनेक युद्धों में अपनी वीरता दिखाई थी तथा वह अनेक लड़ाइयाँ जीत चुका था। जिस समय पाण्डव अज्ञातवास में थे तब कौरव सेना, उत्तर कुमार की गायों का अपहरण करने आई थी। जिनमें भीष्म, द्रोण और सूर्यपुत्र कर्ण जैसे अजेय वीर थे, उस समय अर्जुन ने उत्तर कुमार का सारथी बनकर कौरवों की सेना को पराजित किया था।

यह अर्जुन की वीरता और निडरता का उदाहरण है। अत: अर्जुन शत्रु की सेना की शक्ति से नहीं डरा था। अर्जुन की प्रत्येक लड़ाई में विजय हुई थी, वह नर-श्रेष्ठ था। उसकी वीरता का यश दसों दिशाओं में फैला था। अर्जुन के सम्पूर्ण शरीर व रोम-रोम में वीरता की भावना भरी हुई थी। लेखक ने अर्जुन द्वारा युद्ध न किए जाने का कारण स्वजनासक्ति बताया है। उस आसक्ति को कृष्ण ने गीता सुना कर समाप्त किया था।

विशेष :

1. शुद्ध खड़ी बोली का प्रयोग।
2. ‘दाँत खट्टे करना’ तथा ‘रोम-रोम में भरा’ जैसे मुहावरों का प्रयोग।
3. उद्धरण शैली का प्रयोग।
4. ‘परावृत्त’ जैसे संस्कृत शब्दों का प्रयोग हुआ है।
5. अर्जुन की वीरता का परिचय दिया गया है।

(4) क्या अशोक की तरह उसके मन में अहिंसा-वृत्ति का उदय हुआ था? नहीं, यह तो केवल स्वजनासक्ति थी। इस समय भी यदि गुरु बंधु और आप्त सामने न होते, तो उसने शत्रुओं के मुण्ड गेंद की तरह उड़ा दिये होते। परन्तु इस आसक्ति जनित मोह ने उसकी कर्त्तव्यनिष्ठा को ग्रस लिया और तब उसे तत्वज्ञान याद आया।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
विनोबा जी ने बताया है कि युद्धभूमि में अपने परिजनों को देखकर अर्जुन मोह में पड़ गया, इस कारण वह अपने क्षात्रधर्म को भूल गया।

व्याख्या :
कुरुक्षेत्र के मैदान में अपने स्वजन समूह को देखकर अर्जुन के हृदय में हलचल मच गई। वह अचानक अहिंसावादी बन गया । अर्जुन की अहिंसा अशोक जैसी अहिंसा नहीं थी, उसकी अहिंसा का कारण परिवार के प्रति मोह था। अशोक युद्ध के वीभत्स परिणामों के कारण अहिंसावादी बना था। अशोक और अर्जुन की अहिंसा में अन्तर था। अर्जुन के सामने यदि उसके गुरु,भाईबन्धु, मित्र आदि न होकर शत्रु पक्ष की सेना होती तो वह विपक्ष को नष्ट कर देता। इस समय परिजनों के प्रति आसक्ति ने उसे मोह में डुबो दिया जिसका परिणाम था कि वह अपने क्षात्रधर्म अर्थात् कर्त्तव्य को भूल गया। वह अहिंसा को श्रेष्ठ मानने लगा तथा अपना तत्त्व-ज्ञान अर्थात् दर्शन बघारने लगा कि युद्ध पाप है,समाज का कलंक है, धर्म का नाश है, व्यभिचार की जड़ है, आदि। इसी मोह को नष्ट करने व क्षात्रधर्म याद कराने के लिए कृष्ण ने अर्जुन को युद्धभूमि में गीता सुनाई थी।

विशेष :

1. अर्जुन मोहवश अपने कर्तव्य-युद्ध को भूल गया।
2. शुद्ध साहित्यिक खड़ी बोली का प्रयोग है।
3. संस्कृत शब्दों की अधिकता है।
4. अशोक और अर्जुन की अहिंसा में अन्तर बताया है।
5. शैली में संक्षिप्तता तथा क्लिष्टता है।

(5) परन्तु सारी गीता में इस मुद्दे का कहीं भी जवाब नहीं दिया है, फिर भी अर्जुन का समाधान हुआ है। यह सब कहने का अर्थ इतना ही है कि अर्जुन में अहिंसा-वृत्ति नहीं थी, वह युद्ध प्रवृत्त ही था। युद्ध उसकी दृष्टि से उसका स्वभावप्राप्त और अपरिहार्य रूप से निश्चित कर्त्तव्य था। उसे वह मोह के वश होकर टालना चाहता था। और गीता का मुख्यत: इस मोह पर ही गदा प्रहार है।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
निबन्ध की अन्तिम पंक्तियों में बताया गया है कि युद्ध न करना अर्जुन का दर्शन नहीं बुद्धिवाद था। अतः कृष्ण ने अर्जुन के बुद्धिवाद पर ध्यान न देकर अर्जुन के मोह को तोड़ा।

व्याख्या :
कृष्ण अर्जुन की परिवार में निहित आसक्ति को जानते थे, वह यह भी जानते थे कि अर्जुन अहिंसावादी नहीं है, उसका स्वभाव युद्ध है, लेकिन इन तथ्यों को गीता में नहीं बताया गया है। कृष्ण ने तो केवल अर्जुन के मोह को नष्ट किया है, उसके कर्त्तव्य की याद दिलाई है। कृष्ण ने ऐसा इसलिए किया क्योंकि वे जानते थे कि अर्जुन क्षत्रिय है, युद्ध करना उसका स्वभाव है और एक क्षत्रिय अहिंसा को कभी नहीं अपनाता। युद्ध एक क्षत्रिय का कभी न त्यागने वाला कर्त्तव्य होता है। अर्जुन सच्चे अर्थों में एक वीर योद्धा था। युद्ध के परिणामों को जानते हुए भी वह युद्ध को नहीं त्याग सकता था। लेकिन परिवार के मोह ने उसे युद्ध से विमुख कर दिया था। इसी मोह को श्रीकृष्ण ने गीता के सन्देश के द्वारा नष्ट कर दिया। अन्त में, अर्जुन ने कहा कि उसका भ्रम अर्थात् मोह नष्ट हो गया है और वह अपने धर्म के लिए तैयार है।

विशेष :

1. कृष्ण के द्वारा अर्जुन का मोह भंग हुआ है।
2. संस्कृत के शब्दों के प्रयोग के साथ भावों की सरलता है।
3. लघु वाक्यों ने विचारों को स्पष्ट किया है।
4. शुद्ध साहित्यिक खड़ी बोली।
5. ‘मुद्दे’ व ‘जवाब’ जैसे उर्दू के शब्दों का प्रयोग।

MP Board Class 12th Hindi Solutions