If 1000 droplets of water of surface tension 0.07 N/m, having same radius 1 mm each, combine to from a single drop. In the process the released surface energy is –
$(Take π = \frac{22}{7})$

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$7.92 \times 10^{- 4} J$

$7.92 \times 10^{- 6} J$

$9.68 \times 10^{- 4} J$

$8.8 \times 10^{- 5} J$

answer is D.

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Detailed Solution

$R = n^{1 / 3} r \Rightarrow R = 10 \times 1 mm$

Surface energy of a droplet is given by

$U = AT$

$U_{i} = 1000 \times 4 {πr}^{2} \times T$

$U_{f} = 4 {πR}^{2} \times T$

$∴ E = U_{i} - U_{f} = 7.92 \times 10^{- 4} J$

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