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)

$6 \sqrt{3}$

Answer 2

Key Idea At any instant, $T\, \cos\, \theta=m g$ and $T \,\sin \,\theta=F_{e}$ $\Rightarrow \tan \,\theta \,m g=F_{e}$ $F_{e} =\frac{k q^{2}}{r^{2}}$ $r =2 l \cos \theta$ [from Fig.] $\tan \,\theta=\frac{F_{e}}{m g}$ So, for $\theta=45^{\circ}$ and $\phi=45^{\circ}$ $r=\frac{2 l}{v^{2}}=\sqrt{2} l$ $\Rightarrow F_{e}=\frac{K q^{2}}{2 l}$ $\tan 45^{\circ}=\frac{k q^{2}}{2 l\, m g}$ $\left(\because \tan 45^{\circ}=1\right)$ $m g=\frac{k q^{2}}{2 l} \dots$(i) For, $\theta=30^{\circ}$ and $\phi=60^{\circ}$ $r=2 l \times \frac{1}{2}=l$ So , $F_{e}'=\frac{k q^{2}}{l^{2}}$ $\Rightarrow \tan 30^{\circ}=\frac{k' q^{2}}{l^{2}-m' g}$ As, the sphere is suspended in a liquid of density $\frac{2}{3} d$, then the observed weight of the body. $m'=V\left(d-\frac{2 d}{3}\right)=\frac{m}{3} [\because m=V \cdot d]$ $\frac{1}{\sqrt{3}}=\frac{3 k' q^{2}}{l^{2} m g}$ $\Rightarrow m g=\frac{3 \sqrt{3} k'q^{2}}{l^{2}}$ So, from E (i) and (ii), we get $3 \sqrt{3} k' =\frac{k}{2}$ $\Rightarrow k'=\frac{1}{4 \pi \varepsilon_{0} \varepsilon_{r}}, K=\frac{1}{4 \pi \varepsilon_{0}}$ $\Rightarrow \varepsilon_{r} =6 \sqrt{3}$