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

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Answer : (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

$k=\large\frac{1}{4 \pi \in_{0}}$

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