Question

64 small drops of water having same charge and same radius are combined to form one big drop. The ratio of capacitance of big drop to small drop is

• A. 2:01
• B. 1:02
• C. 4:01
• D. 1:04

Answer 1

The Correct Option is C

Let the radius of each small drop is r and the radius of big drop
is R. When 64 small drops of water are combined to form one
big drop, then the volume remains constant. So, the voiume of
64 small drops = the volume of big drop
ie,$\hspace30mm 64 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3$
$\Rightarrow \hspace40mm 64r^3 = R^3$
$\Rightarrow \hspace40mm 4r = R$
$\Rightarrow \hspace40mm R = 4r \hspace35mm ... (i)$
Now, the capacitance of a spherical conductor is
$\hspace40mm C = 4 \pi \varepsilon_0 a$
$\hspace50mm$ [ a is the radius of the conductor]
Now, the .capacitance of small drop
$\hspace40mm C_1 = 4 \pi \varepsilon_0 r \hspace35mm ... (ii)$
and the capacitance of big drop is
$\hspace40mm C_2 = 4 \pi \varepsilon_0 R$
On putting the value of R from E (i), then
$\hspace40mm C_2 = 4 \pi \varepsilon_0 (4r)$
$\Rightarrow \hspace30mm C_2 = 16 \pi \varepsilon_0 r \hspace35mm ... (iii)$
On dividing the E (iii) by E (ii)
$\hspace40mm \frac{C_2}{C_1} = \frac{16 \pi \varepsilon_0 r}{4 \pi \varepsilon_0 r}$
$\Rightarrow \hspace30mm \frac{C_2}{C_1} = \frac{4}{1}$
$\Rightarrow \hspace30mm C_2 : C_1 = 4 : 1$