JEE MAIN - Physics (2023 - 29th January Morning Shift - No. 4)
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 $$\left[\pi=\frac{22}{7}\right]$$
$$18 .48 \times 10^{-4} \mathrm{~J}$$
$$5.76 \times 10^{-4} \mathrm{~J}$$
$$0.72 \times 10^{-4} \mathrm{~J}$$
$$9.24 \times 10^{-4} \mathrm{~J}$$
Explanation
Surface area of soap bubble $=2 \times 4 \pi \mathrm{R}^{2}$ Work done $=$ change in surface energy $\times \mathrm{T}_{\mathrm{S}}$
$=\mathrm{T}_{\mathrm{S}} \times 8 \pi \times\left(\mathrm{R}_{2}^{2}-\mathrm{R}_{1}^{2}\right)$
$=2 \times 10^{-2} \times 8 \times \frac{22}{7} \times 49 \times \frac{3}{4} \times 10^{-4}$
$=18.48 \times 10^{-4} \mathrm{~J}$
$=\mathrm{T}_{\mathrm{S}} \times 8 \pi \times\left(\mathrm{R}_{2}^{2}-\mathrm{R}_{1}^{2}\right)$
$=2 \times 10^{-2} \times 8 \times \frac{22}{7} \times 49 \times \frac{3}{4} \times 10^{-4}$
$=18.48 \times 10^{-4} \mathrm{~J}$
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