Question 121 of 192

Surface tension of a soap bubble is $2.0\times {10}^{-2}N{m}^{-1}$ . Work done to increase the radius of soap bubble from $3.5cm$ to $7cm$ will be : [Take $\pi =\dfrac{22}{7}$ ]

A$0.72\times {10}^{-4}J$B$5.76\times {10}^{-4}J$C$18.48\times {10}^{-4}J$D$9.24\times {10}^{-4}J$

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Answer: (C) $18.48\times {10}^{-4}J$

As we know,
Work done $=$ change in surface energy of soap bubble
Therefore,
Work done $=2T\left(4\pi {r}_{2}^{2}-4\pi {r}_{1}^{2}\right)$
$=8\pi T\left({r}_{2}^{2}-{r}_{1}^{2}\right)$
$=8\times \dfrac{22}{7}\times 0.02\left({7}^{2}-3.{5}^{2}\right)\times {10}^{-4}J$
$=18.48\times {10}^{-4}J$