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Q.

A mercury drop of radius $10^{- 3} m$ is broken into 125 equal size droplets. Surface tension of mercury is $0.45 {Nm}^{- 1}$ . The gain in surface energy is:

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a

$2.26 \times 10^{- 5} J$

b

$28 \times 10^{- 5} J$

c

$5 \times 10^{- 5} J$

d

$17.5 \times 10^{- 5} J$

answer is C.

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Detailed Solution

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From volume conservation,
$\begin{array}{l} (\frac{4}{3} {πr}^{3}) \times 125 = \frac{4}{3} {πR}^{3} \\ R^{3} = 125 r^{3} \\ R = 5 r \Rightarrow r = \frac{R}{5} \end{array}$
gain in surface energy $= [4 {πr}^{2} \times 125 - 4 {πR}^{2}] \times S$
$= 4 {πr}^{2} \times 100 \times 0.45 = 2.26 \times 10^{- 5} J$

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