A water drop of diameter 2 cm is broken into 64 equal droplets. The surface tension of water is 0.075N/m. In this process the gain in surface energy will be:

Option 1 - 2.8 × 10-4 J

Option 2 - 1.5 × 10-3 J

Option 3 - 1.9 × 10-4 J

Option 4 - 9.4 × 10-5 J

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Raj Pandey | Contributor-Level 9

4 months ago

Correct Option - 1

Detailed Solution:

Gain in surface energy, $Δ U = T Δ A$

from volume centenary, $\frac{4}{3} π R^{3} = 64 * \frac{4}{3} π r^{3}$

$\Rightarrow r = \frac{R}{4}$

Initial surface area, Ai = 4pR²

final surface area, $A_{f} = 64 * 4 π r^{2}$

$∴ Δ U = 12 π R^{2} . T = 0.075 * 12 * 3.14 * 1 0^{- 4} = 2.8 * 1 0^{- 4} J$

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$\begin{array}{l} = 75.39 \times 1 0^{- 5} J \\ Δ u = 7.5 \times 1 0^{- 4} J \end{array}$

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