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Physics NCERT Exemplar Solutions Class 12th Chapter Eight Surface Tension Physics Mechanical Properties of Fluids Physics Class 11th

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^-4J

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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Answered by

Raj Pandey | Contributor-Level 9

2 weeks ago

Correct Option - 1

Detailed Solution:

Gain in surface energy, $Δ U = T Δ A$

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

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

Initial surface area, Ai = 4pR²

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

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

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