Question

$n$ identical drops, each of capacitance $C$ and charged to a potential $V$, coalesce to form a bigger drop. Then the ratio of the energy stored in the big drop to that in each small drop is

• A. $ {{n}^{5/3}}:1 $
• B. $ {{n}^{4/3}}:1 $
• C. $ n:1 $
• D. $ {{n}^{3}}:1 $

Answer 1

The Correct Option is A

Volume of big drop $ =n\times $ volume of small drop $ \frac{4}{3}\pi {{R}^{3}}=n\times \frac{4}{3}\pi {{r}^{3}} $ $ R={{n}^{1/3}}r $ Capacitance of small drop, $ C=4\pi {{\varepsilon }_{0}}r $ Capacitance of big drop, $ C=4\pi {{\varepsilon }_{0}}R $ $ =4\pi {{\varepsilon }_{0}}{{n}^{1/3}}r $ $ C={{n}^{1/3}}C $ The potential of small drop $ V=\frac{q}{C}=\frac{q}{4\pi {{\varepsilon }_{0}}r} $ The potential of big drop $ V=\frac{nq}{(4\pi {{\varepsilon }_{0}}){{n}^{1/3}}r} $ $ V={{n}^{2/3}}V $ $ \therefore $ Energy of small drop $ =\frac{1}{2}C{{V}^{2}} $ Energy of big drop $ =\frac{1}{2}CV{{}^{2}} $ $ =\frac{1}{2}{{n}^{1/3}}C{{({{n}^{2/3}}V)}^{2}} $ $ ={{n}^{5/3}}\frac{1}{2}C{{V}^{2}} $ $ \therefore $ $ \frac{Energ{{y}_{(big\,drop)}}}{Energ{{y}_{(small\,drop)}}}=\frac{{{n}^{5/3}}}{1} $