Question

$n$ identical charged drops having charge $q_0$, potential $V_0$, capacity $C_0$ and energy $E_0$, coalesce to form a big drop. The new potential, charge, capacity, energy stored will be:

• A. $n^{\frac{1}{3}}{V}_0;n{q}_0;n^{\frac{2}{3}}{C}_0;n^{\frac{3}{5}}{U}_0$
• B. $n^{\frac{2}{3}}{V}_0;\frac{q_0}{n_0};n^{\frac{1}{3}}{C}_0;n^{\frac{3}{5}}{U}_0$
• C. $n^{\frac{2}{3}}{V}_0;n{q}_0;n^{\frac{1}{3}}{C}_0;n^{\frac{5}{3}}{U}_0$
• D. $n^{\frac{2}{3}}{V}_0;\frac{q_0}{n};n^{\frac{2}{3}}{C}_0;n^{\frac{5}{3}}{U}_0$

Answer 1

The correct option is C $n^{\frac{2}{3}}{V}_0;n{q}_0;n^{\frac{1}{3}}{C}_0;n^{\frac{5}{3}}{U}_0$

charge on each drop $=q_0$

total charge must $\mu$ a constant $\therefore$ final charge on big drop $=n{q}_0$

We know potential of small drop $V_0=\frac{k{q}_0}{r_0}$

Volumes must be equal

$\Rightarrow n\cdot \frac{4}{3}\pi r_0^3=\frac{4}{3}\pi R^3$

$\Rightarrow R=n^{\frac{1}{3}}{r}_0$

$\therefore \text{final voltage}=\frac{k\left(n{q}_0\right)}{n^{\frac{1}{3}}{r}_0}=n^{\frac{2}{3}}{V}_0.$

We know $C_0=\frac{q_0}{V_0}\Rightarrow \text{final C}=\frac{n{q}_0}{n^{\frac{2}{3}}{V}_0}=n^{\frac{1}{3}}{C}_0,$

Initial energy $U_0=\frac{1}{2}{C}_0{V}_0^2,$

final energy$=\frac{1}{2}\left(n^{\frac{1}{3}}{C}_0\right)(n^{\frac{2}{3}}{V}_0{)}^2$

$=\frac{1}{2}{C}_0{V}_0^2{n}^{\frac{5}{3}}=n^{\frac{5}{3}}{U}_0.$