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 : 1
• B. 1 : 2
• C. 4 : 1
• D. 1 : 4

Answer 1

Answer: (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,$64 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3$
$\Rightarrow 64r^3 = R^3$
$\Rightarrow 4r = R$
$\Rightarrow R = 4r ... (i)$
Now, the capacitance of a spherical conductor is
$C = 4 \pi \varepsilon_0 a$
[a is the radius of the conductor]
Now, the .capacitance of small drop
$C_1 = 4 \pi \varepsilon_0 r... (ii)$
and the capacitance of big drop is
$C_2 = 4 \pi \varepsilon_0 R$
On putting the value of $R$ from Eq. (i), then
$C_2 = 4 \pi \varepsilon_0 (4r)$
$\Rightarrow C_2 = 16 \pi \varepsilon_0 r ... (iii)$
On dividing the Eq. (iii) by Eq. (ii)
$\frac{C_2}{C_1} = \frac{16 \pi \varepsilon_0 r}{4 \pi \varepsilon_0 r}$
$\Rightarrow \frac{C_2}{C_1} = \frac{4}{1}$
$\Rightarrow C_2 : C_1 = 4 : 1$