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Q.

A water drop of radius 1 cm is broken into 729 equal droplets. If surface tension of water is 75 dyne/cm, then the gain in surface energy upto first decimal place will be :
(Given $π = 3.14)$

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a

$8.2 \times 10^{- 4} J$

b

$8.5 \times 10^{- 4} J$

c

$7.5 \times 10^{- 4} J$

d

$5.3 \times 10^{- 4} J$

answer is C.

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Detailed Solution

$\begin{array}{l} 729 \times \frac{4}{3} {πr}^{3} = \frac{4}{3} {πR}^{3} \\ \Rightarrow R = 9 r \\ \Rightarrow r = \frac{R}{9} = \frac{1}{9} c m \\ ΔU = S \times ΔA \\ \Rightarrow ΔU = S \times \{- 4 {πR}^{2} + 729 \times 4 {πr}^{2}\} \\ = S \times 4 π \{729 r^{2} - 81 r^{2}\} \\ = 7 .5 \times 10^{- 4} J \end{array}$

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