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Question

Two pith balls carrying equal charges are suspended from a common point by strings of equal length, the equilibrium separation between them is r. Now the strings are rigidly clamped at half the height. The equilibrium seperation between the balls now become

307782_91b7fdb9a6f14a23a221d3e84411e4b1.png

A
(12)2
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B
(r32)
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C
(2r3)
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D
(2r3)
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Solution

The correct option is B (r32)
Let the length of the strings be L and mass of the ball be m and charge be q.
At equilibrium, Fx=0 and Fy=0
Tsinθ=mg ..........(1)
Also Tcosθ=Fe Tcosθ=Kq2r2 .......(2) where K=14πϵo
Dividing (1) and (2) gives r2=mgKq2tanθ
Now as mgKq2=constant=C and tanθ=yr2
r2=C×2yr r(y)13

Thus rr=(y)13y13
Now y=y2 r=r213

479723_307782_ans_05c5b7000f9c4d0ea223943eaef8f4dc.png

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