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# Two identical charged spheres are suspended by strings of equal length . The strings make an angle of $\;60^{0}\;$ with each other . When suspended in a liquid of density $\;0.5 g/cc,\;$ the angle remains the same . The dielectric constant of the liquid is [density of material of sphere is $\;1.5 g/cc\;$]

$(a)\;2\qquad(b)\;3\qquad(c)\;4\qquad(d)\;1.5$

Explanation :
$tan \theta =\large\frac{qE}{mg}=constant$
$\large\frac{k\;q^2}{d^2\;mg}=tan 30^{0}$
$tan 30^{0}=\large\frac{k\;q^2}{d^2\;mg}=\large\frac{k\;q^2}{1.5 V g\;d^2}\;\quad\;$ [V is volume of sphere]     ---1
When suspended in liquid :
$tan 30^{0}=\large\frac{1}{4 \pi \in}\;\large\frac{q^2}{d^2}\;\large\frac{1}{(mg-\sigma \;V\;g)}$  -----2
$=\large\frac{1}{4 \pi \in_{0} K} \;\large\frac{q^2}{d^2 gV (1.5-0.5)}$
from1 and 2 we get
$\large\frac{kq^2}{d^2 g V K}=\large\frac{kq^2}{1.5 V g d^2}$
$K=1.5$
K -------> dielectric constant of liquid
$k=\large\frac{1}{4 \pi \in_{0}}$
edited Aug 22, 2014 by meena.p

When the angle remains same a quick method is to divide the density of liquid with difference in the densities of the liquid and solid In this case, its simply = 1.5/1.5-0.5=1.5