Class 12th

92. Kindly Consider the following

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Class 12th physics ncert exemplar solution class 12th chapter one

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Kindly go through the solution

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A positive charge Qis uniformly distributed along a circular ring of radius R. A small test charge q is placed at the centre of the ring as shown in figure. Then

(a) If q>0 and is displaced away from the centre in the plane of the ring it will be pushed back towards the centre

(b) If q<0 and is displaced away from the centre in the plane of the ring, it will never return to the centre an will continue moving till it hits the ring

(c) If q<0 , it will perform SHM for small displacement along the axis.

(d) Q at the centre of the ring is in an unstable equilibrium within the plalne of the ring for q>O

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This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a, b, c

Explanation- the positive charge Q is uniformly distributed at the outer surface of the enclosed sphere thus electric field inside the sphere is zero. So the effect of electric field on charge q due to positive charge Q is zero.

Now the only attractive and repulsive force between Q and q

Case 1 q>0 this creates repulsive force

Case 2 q<0 this creates attractive force

If q is shifted from the centre then the positive charges nearer to this charge will attract it towards itself and charge q will never return to its centre.

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Kindly Consider the following

91. $\frac{x^{2}}{1 - x^{6}}$

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Class 12th physics ncert exemplar solution class 12th chapter one

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$\begin{array}{l} \begin{array}{l} L e t x^{3} = t \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} 3 x^{2} d x = d t \end{array} \\ I = \int \frac{x^{2}}{1 - x^{6}} d x = \frac{1}{3} \int \frac{d t}{1 - t^{2}} \\ = \frac{1}{3} [\frac{1}{2} l o g | \frac{1 + t}{1 - t} |] + C \\ = \frac{1}{6} l o g | \frac{1 + x^{3}}{1 - x^{3}} | + C \end{array}$

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A positive charge Q is uniformly distributed along a circular ring of radius R. A small test charge q is placed at the centre of the ring figure. Then,

(a) If q > 0 and is displaced away from the centre in the plane of the ring, it will be pushed back towards the centre

(b) If q < 0 and is displaced away from the centre in the plane of the ring, it will never return to the centre and will continue moving till it hits the ring

(c) If q < 0, it will perform SHM for small displacement along the axis

(d) Q at the centre of the ring is in an unstable equilibrium within the plane of the ring for q > 0

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This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer. (a), (b), (c), (d)

Explanation- The positive charge Q is uniformly distributed along the circular ring then electric field at the centre of ring will be zero, hence no force is experienced by the charge if it is placed at the centre of the ring.
Now the charge is displaced away from the centre in the plane of the ring. There will be net electric field opposite to displacement will push back the charge towards the centre of the ring if the charge is positive. If charge is negative, it will experience net force in the direction of displacement and the charge w

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Class 12th Integrals

Kindly Consider the following

90. $\frac{3 x}{1 + 2 x^{4}}$

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$\begin{array}{l} L e t√2 x^{2} = t \\ \Rightarrow 2√2 x d x = d t \\ I = \int \frac{3 x}{1 + 2 x^{4}} d x \\ = \frac{3}{2√2} \int \frac{d t}{1 + t^{2}} \\ = \frac{3}{2√2} [t a n^{- 1} t] + C \\ = \frac{3}{2√2} [t a n^{- 1} (\sqrt2 x^{2})] + C \end{array}$

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89. Kindly Consider the following

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Class 12th physics ncert exemplar solution class 12th chapter one

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Kindly go through the solution

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Refer to the arrangement of arrangement of charges in figure and a gaussian surface of radius R with Q charge at the center. Then

(a) Total flux through the surface of the sphere is $\frac{- Q}{ε 0}$ .

(b) Field on the surface of the sphere of the sphere is -Q/4 $π ε$ ₀R².

(c) Flux through the surface of sphere due to 5Q is zero.

(d) Field on the surface of sphere due to -2Q is same everywhere

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This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (a, c)

Explanation- Gauss's law states that the total electric flux of an enclosed surface is given by q/? _0,where q is the charge enclosed by the surface.

So total charge inside the surface is = Q-2Q = -Q

Therefore total flux through the surface of the sphere = -Q/? ₀

Now, charge 5Q is lies outside the surface, thus it makes no contribution to electric flux through the given surface. So both option a and c are true.

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88. Kindly Consider the following

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Class 12th physics ncert exemplar solution class 12th chapter one

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Kindly go through the solution

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Consider a region inside which there are various types of charges but the total charge is zero. At points outside the region,
(a) The electric field is necessarily zero
(b) The electric field is due to the dipole moment of the charge distribution only
(c) The dominant electric field is ∝ 1/r³, for large r, where r is the distance from an origin in this regions r
(d) The work done to move a charged particle along a closed path, away from the region, will be zero

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This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer. (c), (d)

Explanation- electric field is not necessarily zero may it become zero by their algeabric sum. For dipole the electric field is always ∝ 1/r³and it is also conservative.

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