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Q.

Two identical charged spheres suspended from a common point by two massless strings of length L are initially a distance d(d<<L) apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result the charges approach each other with a velocity v. Then as a function of distance x between them,

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a

vx1

b

vx1/2

c

vx

d

vx1/2

answer is D.

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Detailed Solution

tanθ=Fmgθ

Kq2x2mg=x2Lx3/2q

Differentiating wrt time, we get

3x2dxdt2qdqdt   where ,dqdt is constant x2(v)q

As x3/2q

We get vx-1/2

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