JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Electrostatics

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512 identical drops of mercury are charged to a potential of 2 V each. The drops are joined to form a single drop. The potential of this drop is ____ V.

Answer: 128

Solution

A drop of radius

r

held at potential

V

carries charge

q = 4 π ε_{0} r V

. Mercury is incompressible, so the volume is conserved on merging:

n (\frac{4}{3} π r^{3}) = \frac{4}{3} π R^{3} \Rightarrow R = n^{1/3} r

The big drop carries the whole charge

n q

, so

V^{'} = \frac{n q}{4 π ε _{0} R} = \frac{n ( 4 π ε _{0} r V )}{4 π ε _{0} n ^{1/3} r} = n^{2/3} V

With

n = 512

and

V = 2

V, the potential of the single drop is

(512)^{2/3} \times 2 = 64 \times 2 = 128

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Integer answer

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