A water drop breaks into 64 identical droplets of each surface area $10^{- 7} m^{2} .$ If the surface tension of water is $0.07 N m^{- 1},$ the increase in the surface energy in the process is

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$158 \times 10^{- 9} J$

$336 \times 10^{- 9} J$

$216 \times 10^{- 9} J$

$432 \times 10^{- 9} J$

answer is A.

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Detailed Solution

$R = n^{\frac{1}{3}} r \Rightarrow R = {(64)}^{\frac{1}{3}} r = 4 r$

$W = [64 \cdot 4 π r^{2} - 4 π R^{2}] T = 4 π [64 r^{2} - 16 r^{2}] T = 4 π r^{2} \times 48 \times T = 10^{- 7} \times 48 \times 0.07 = 336 \times 10^{- 9} J$

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