Question

$N$ identical drops of mercury are charged simultaneously to $10V$ . When combined to form one large drop, the potential is found to be $40V$ , the value of $N$ is

• A. 4
• B. 6
• C. 8
• D. 10
• E. 12

Answer 1

Answer: (C)

After combining, the volume remains same ie, volume of bigger drop
$=N\times$ volume of smaller drop or $\frac{4}{3}\pi R^3=N\times \frac{4}{3}\pi r^3$ or $N={\left(\frac{R}{r}\right)}^3$ .... (i)
As charge is conserved, hence $Q=Nq$ ...(ii)
Capacity of bigger drop $=4\pi {\epsilon }_{0}R$ Capacity of smaller drop $=4\pi {\epsilon }_{0}r$
From Eq. (ii), we have $(4\pi {\epsilon }_{0}R)V_{big}=N(4\pi {\epsilon }_{0}r)V_{small}$
Or $(4\pi {\epsilon }_{0}R)\times 40=N(4\pi {\epsilon }_{0}r)\times 10$ Or $4R=Nr$
Or $\frac{R}{r}=\frac{N}{4}$ .... (iii)
From Eqs. (i) and (iii),
$N={\left(\frac{N}{4}\right)}^3$
$N=\frac{N^3}{64}$ Or
$N=\frac{N^3}{64}$ Or
$N^2=64$ Or $N=8$