JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Electrostatics

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27 similar drops of mercury are maintained at 10 V each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is ____ times that of a smaller drop.

Answer: 243

Solution

Mercury is incompressible, so

27 \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}

gives

R = 3 r

, while the charges add to

Q = 27 q

. The potential of the large drop is

V_{big} = \frac{k Q}{R} = \frac{k ( 27 q )}{3 r} = 9 \frac{k q}{r} = 9 \times 10 = 90 V

The electrostatic energy of a charged conducting drop is

U = \frac{1}{2} Q V

, so

\frac{U _{big}}{U _{small}} = \frac{\frac{1}{2} ( 27 q ) ( 90 )}{\frac{1}{2} ( q ) ( 10 )} = \frac{2430}{10} = 243

The energy of the big drop is

243

times that of a small drop.

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Integer answer

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