- Home
- Standard 12
- Physics
2. Electric Potential and Capacitance
medium
Sixty four conducting drops each of radius $0.02 m$ 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
$8: 1$
(JEE MAIN-2022)
Solution
Let $R=$ radius of combined drop
$r=$ radius of smaller drop
Volume will remain same
$\frac{4}{3} \pi R ^{3}=64 \times \frac{4}{3} \pi r ^{3}$
$R =4 r$
$Q =64 q ;$
$q$ : charge of smaller drop
$Q$ : Charge of combined drop
$\frac{\sigma_{\text {bigger }}}{\sigma_{\text {smaller }}}=\frac{\frac{ Q }{4 \pi R ^{2}}}{\frac{ q }{4 \pi r ^{2}}}=\frac{ Q }{ q } \cdot \frac{ r ^{2}}{ R ^{2}}$
$=64 \frac{ r ^{2}}{16 r ^{2}}=4$
$\frac{\sigma_{\text {bigger }}}{\sigma_{\text {smaller }}}=\frac{4}{1}$
Standard 12
Physics
