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STD 12 - 1. Electric charges and fields — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 1. Electric charges and fields4 Marks
MCQ
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
  • ✓
    $4$
  • B
    $3$
  • C
    $2$
  • D
    $1$
✓

Answer

Correct option: A.
$4$
a
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 $\quad \mathrm{K}=\frac{1}{1-\frac{\rho}{\mathrm{d}}}=2$

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