Question

Sixty four conducting drops each of radius $0.02m$ and each carrying a charge of $5\mu C$ are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be :

• A. $1:4$
• B. $4:1$
• C. $1:8$
• D. $8:1$

Answer 1

Answer: (B)

Let $R=$ radius of combined drop $r=$ radius of smaller drop
Volume will remain same
$\frac{4}{3}\pi R^3=64\times \frac{4}{3}\pi r^3$
$R=4r$
$Q=64q$;
$q:$ charge of smaller drop
$Q$ : Charge of combined drop
$\frac{{\sigma }_{\text{bigger }}}{{\sigma }_{\text{smaller }}}=\frac{\frac{Q}{4\pi R^2}}{\frac{q}{4\pi r^2}}=\frac{Q}{q}\cdot \frac{r^2}{R^2}$
$=64\frac{r^2}{16r^2}=4$
$\frac{{\sigma }_{\text{bigger }}}{{\sigma }_{\text{smaller }}}=\frac{4}{1}$