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 separation between the balls now become:
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a
2r3
b
2r3
c
122
d
r23
answer is D.
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Detailed Solution
Let m be mass of each ball and q be charge on each ball.Force of repulsion, F=14πε0q2r2In equilibrium Tcosθ=mg .........(i)Tsinθ=F ........(ii)Divide (ii) by (i), we gettanθ=Fmg=14πε0q2r2mgFrom figure (a),r/2y=14πε0q2r2mg −−−(iii)tanθ'=14πε0q2r2mgFrom figure (b)r'/2y/2=14πε0q2r'2mg −−−ivDivide (iv) by (iii), we get2r'r=r2r'2 r'3=r32 r'=r23