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VITEEE 2012 Paper

Section: Physics

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Question : 9 of 120

Marks: +1, -0

The potential of a large liquid drop when eight liquid drops are combined is 20 V. Then the potential of each single drop was

10V
7.5V
5V
2.5V

Solution:

Volume of 8 small drops = Volume of big drop

∴

\left(\frac{4}{3}\pi r^3\right)

× 8 =

\frac{4}{3}\pi R^3

⇒ 2r = R ...(i)

According to charge conservation

8q = Q ... (ii)

Potential of one small drop (V') =

\frac{q}{4\pi\varepsilon_0 r}

Similarly, potential of big drop (V) =

\frac{Q}{4\pi\varepsilon_0 R}

Now,

\frac{V'}{V}

=

\frac{q}{Q} \times \frac{R}{r}

⇒

\frac{V'}{20}

=

\frac{9}{8q} \times \frac{2r}{r}

∴ V' = 5V

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