Surface tension of a soap bubble is $2.0 \times 10^{-2} \mathrm{Nm}^{-1}$. Work done to increase the radius of soap bubble from $3.5 \mathrm{~cm}$ to $7 \mathrm{~cm}$ will be:
Take $\left[\pi=\frac{22}{7}\right]$
Solution
Surface area of soap bubble $=2 \times 4 \pi \mathrm{R}^{2}$ Work done $=$ change in surface energy $\times \mathrm{T}_{\mathrm{S}}$
<br/><br/>
$=\mathrm{T}_{\mathrm{S}} \times 8 \pi \times\left(\mathrm{R}_{2}^{2}-\mathrm{R}_{1}^{2}\right)$
<br/><br/>
$=2 \times 10^{-2} \times 8 \times \frac{22}{7} \times 49 \times \frac{3}{4} \times 10^{-4}$
<br/><br/>
$=18.48 \times 10^{-4} \mathrm{~J}$
About this question
Subject: Physics · Chapter: Properties of Solids and Liquids · Topic: Elasticity
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