Question

$512$ identical drops of mercury are charged to a potential of $2\text{V}$ each. The drops are joined to form a single drop. The potential of this bigger drop is $x\text{V}$. The value of $x$ is _______.

• A. 128.0
• B. 128.00
• C. 128

Answer 1

Answer: (A), (B), (C)

Let charge and radius of each drop be $q$ and $r$ respectively.

The potential of each drop is,

$V=\frac{1}{4\pi {ϵ}_0}\frac{q}{r}$

The charge on the bigger drop, $Q=nq=512q$

Volume of bigger drop $=n\times$ Volume of the smaller drop

$\frac{4}{3}\pi R^3=512\times \frac{4}{3}\pi r^3$

$\Rightarrow R=(512{)}^{1/3}r=8r$

Now, the potential of the bigger drop is,

$V{}^{\prime }=\frac{1}{4\pi {ϵ}_0}\frac{Q}{R}=\frac{1}{4\pi {ϵ}_0}\frac{512q}{8r}$

$=\left(\frac{512}{8}\right)\left(\frac{1}{4\pi {ϵ}_0}\frac{q}{r}\right)$

$=\frac{512}{8}\times V=\frac{512\times 2}{8}$

$\Rightarrow V{}^{\prime }=128\text{V}$

$\therefore x=128$