Question

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of ${30}^{\circ }$ with each other. When suspended in a liquid of density $0.8{\text{g /cm}}^3$, the angle remains the same. If density of the marterial of the sphere is $1.6{\text{g /cm}}^3$, then dielectric constant of the liquid is

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is C $2$

From free body diagram,

$F_e=\text{T sin}{15}^{\circ }.....\left(i\right)$
$mg=\text{T cos}{15}^{\circ }.....\left(ii\right)$
from equation $\left(i\right)$ and $\left(ii\right)$,
$\Rightarrow \text{tan}{15}^{\circ }=\frac{Fe}{mg}....\left(iii\right)$

In liquid, $F{e}^{\prime }=T^{\prime }\text{sin}{15}^{\circ }...\left(iv\right)$
$mg=F_B+\text{T' cos}{15}^{\circ }....\left(v\right)$
Where $F_B$ is buoyant force.
$F_B=V(d-\rho )g=V(1.6-0.8)g=0.8Vg$
$[\therefore F_B=V_{\rho g}\text{and}V=\frac{\text{mass}}{\text{density}}]$
$=0.8\frac{m}{d}g=\frac{0.8mg}{1.6}=\frac{mg}{2}$
put the value of $F_B$ in equation (v),
$\Rightarrow \frac{mg}{2}=T^{\prime }\text{cos}{15}^{\circ }....\left(vii\right)$
From equation (iv) and (vii),
$\text{tan}{15}^{\circ }=\frac{2F{e}^{\prime }}{mg}......\left(ii\right)$
from (i) and (ii),
$\frac{Fe}{mg}=\frac{2F{e}^{\prime }}{mg}\Rightarrow Fe=2F{e}^{\prime }$
$F{e}^{\prime }=\frac{Fe}{2}$
By the formula of force,
$\frac{q^2}{4\rho {ϵ}_{0}K}=\frac{q^2}{4\rho {ϵ}_0\times 2}$

$K=2$