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 $\left[\pi=\frac{22}{7}\right]$

Question

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 $\left[\pi=\frac{22}{7}\right]$

$18 .48 \times 10^{-4} \mathrm{~J}$

$5.76 \times 10^{-4} \mathrm{~J}$

$0.72 \times 10^{-4} \mathrm{~J}$

$9.24 \times 10^{-4} \mathrm{~J}$

Solution

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