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JEE Advanced 2020 Paper 2

Section: Physics

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Question : 5 of 54

Marks: +1, -0

A point charge

q

m

is suspended vertically by a string of length

l.

A point dipole of dipolemoment

\overset{\rightarrow}{p}

is now brought towards

q

from infinity so that the charge moves away. The finalequilibrium position of the system including the direction of the dipole, the angles and distances is shown in the figure below. If the work done in bringing the dipole to this position is

N \times (mgh),

g

is the acceleration due to gravity, then the value of

N

is ________. (Note that for three coplanar forces keeping a point mass in equilibrium,

\frac{F}{\sin \theta}

is the same for all forces, where

F

is any one ofthe forces and

\theta

is the angle between the other two forces)

[JEE Adv 2020 P2]

Your Answer:

Solution:

Clearly

\frac{mg}{\sin \left(90 + \frac{\alpha}{2}\right)} = \frac{T}{\sin \left(90 + \frac{\alpha}{2}\right)}

= \frac{F_d}{\sin (180 - \alpha)}

\therefore

Aslo

F_d = E q = \frac{q^2 p}{4 \pi \epsilon_0 \left(2l \sin \frac{\alpha}{2}\right)^3}

Initial PE = (0)

Final

P E = \frac{p q}{4 \pi \epsilon_0 \left(2l \sin \frac{\alpha}{2}\right)^2}

+ m g h = W

Also

2l \sin \frac{\alpha}{2} . \sin \frac{\alpha}{2}

= 2 \sin^2 \frac{\alpha}{2} = h

\frac{F_d}{\sin \alpha} = \frac{m g}{\cos \frac{\alpha}{2}}

\Rightarrow \frac{2 p q}{4 \pi \epsilon_0 \left(2l \sin \frac{\alpha}{2}\right)^3} = \frac{m g \sin \alpha}{\cos \frac{\alpha}{2}}

\therefore \frac{pq}{4 \pi \epsilon_0} = \frac{mg \left(2l \sin \frac{\alpha}{2}\right)^3 \cdot \sin \alpha}{2 \cdot \cos \frac{\alpha}{2}}

W = \frac{mg \left(2l \sin \frac{\alpha}{2}\right)^3 \sin \alpha}{2 \cos \frac{\alpha}{2} \left(2l \sin \frac{\alpha}{2}\right)^2}

+ mgh

\Rightarrow W = \frac{mg \sin \alpha \cdot 2l \sin \frac{\alpha}{2}}{2 \cos \frac{\alpha}{2}} + mgh

\Rightarrow W = \frac{mg \cdot 2 \sin \frac{\alpha}{2} \cdot \cos \frac{\alpha}{2} \cdot l \sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}

+ mgh

\Rightarrow W = mg \left(2l \sin^2 \frac{\alpha}{2}\right) + mgh

= 2 mgh

\therefore N = 2

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