Question

Two small spheres of each charge $q$, mass $m$ and material density $d$ are suspended from a fixed point with the help of inextensible light thread. When the spheres are in air, the angle between the threads is $90^{\circ }$. When the spheres are suspended in a liquid of density $\frac{2}{2}d$, the angle between the threads is $60^{\circ }$. The $3$ value of dielectric constant of the liquid is

• A. $6\sqrt{3}$
• B. $2\sqrt{5}$
• C. $5\sqrt{3}$
• D. $7\sqrt{2}$

Answer 1

Answer: (A)

Key Idea At any instant,
$T\cos \theta =mg$ and $T\sin \theta =F_e$
$\Rightarrow \tan \theta mg=F_e$

$F_e=\frac{k{q}^2}{r^2}$
$r=2l\cos \theta$
[from Fig.]
$\tan \theta =\frac{F_e}{mg}$
So, for $\theta =45^{\circ }$ and $\varphi =45^{\circ }$
$r=\frac{2l}{v^2}=\sqrt{2}l$
$\Rightarrow F_e=\frac{K{q}^2}{2l}$
$\tan 45^{\circ }=\frac{k{q}^2}{2lmg}$
$\left(\because \tan 45^{\circ }=1\right)$
$mg=\frac{k{q}^2}{2l}\dots$(i)
For, $\theta =30^{\circ }$ and $\varphi =60^{\circ }$
$r=2l\times \frac{1}{2}=l$
So , $F_e^{\prime }=\frac{k{q}^2}{l^2}$
$\Rightarrow \tan 30^{\circ }=\frac{k^{\prime }{q}^2}{l^2-m^{\prime }g}$
As, the sphere is suspended in a liquid of density $\frac{2}{3}d$,
then the observed weight of the body.
$m^{\prime }=V\left(d-\frac{2d}{3}\right)=\frac{m}{3}[\because m=V\cdot d]$
$\frac{1}{\sqrt{3}}=\frac{3k^{\prime }{q}^2}{l^{2}mg}$
$\Rightarrow mg=\frac{3\sqrt{3}{k}^{\prime }{q}^2}{l^2}$
So, from Eq. (i) and (ii), we get
$3\sqrt{3}{k}^{\prime }=\frac{k}{2}$
$\Rightarrow k^{\prime }=\frac{1}{4\pi {\epsilon }_0{\epsilon }_r},K=\frac{1}{4\pi {\epsilon }_0}$
$\Rightarrow {\epsilon }_r=6\sqrt{3}$