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Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of $30^o$ with each other. When suspended in a liquid of density $1\, g\, cm^{-3}$, the angle remains the same. If density of the material of the sphere is $4/3\, g\, cm^{-3}$, the dielectric constant of the liquid is
$4$
$3$
$2$
$1$
Solution
From $\mathrm{FBD}$ of sphere, using theorem,
$\frac{\mathrm{F}}{\mathrm{mg}}=\tan \theta$ ……….$(i)$
when suspended in liquid, as $\theta$ remains same,
$\frac{\mathrm{F}^{\prime}}{\operatorname{mg}\left(1-\frac{\rho}{\mathrm{d}}\right)}=\tan \theta$ ………$(ii)$
Using eqns. $(i)$ and $(ii),$
$\frac{\mathrm{F}}{\mathrm{mg}}=\frac{\mathrm{F}^{\prime}}{\mathrm{mg}\left(1-\frac{\rho}{\mathrm{d}}\right)}$
where, $\quad \mathrm{F}^{\prime}=\frac{\mathrm{F}}{\mathrm{K}}$
$\therefore \frac{\mathrm{F}}{\mathrm{mg}}=\frac{\mathrm{F}^{\prime}}{\operatorname{mg} \mathrm{K}\left(1-\frac{\rho}{\mathrm{d}}\right)}$
or $\quad \mathrm{K}=\frac{1}{1-\frac{\rho}{\mathrm{d}}}=2$
