First slide

Surface tension and surface energy

Question

The radius of a soap bubble is increased from $\frac{1}{\sqrt{π}}$ cm to $\frac{2}{\sqrt{π}}$ cm. If the surface tension of water is 30 dynes per cm, then the work done will be

Moderate

A

180 ergs
B

360 ergs
C

720 ergs
D

960 ergs

Solution

$\begin{array}{l} W = 8 πT (r_{2}^{2} - r_{1}^{2}) = 8 πT ({[\frac{2}{\sqrt{π}})}^{2} - {(\frac{1}{\sqrt{π}})}^{2}] \\ ∴ W = 8 \times π \times 30 \times \frac{3}{π} = 720 erg \end{array}$

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