JEE MAIN - Physics (2024 - 29th January Evening Shift - No. 1)
A small liquid drop of radius $$R$$ is divided into 27 identical liquid drops. If the surface tension is $$T$$, then the work done in the process will be:
$$4 \pi \mathrm{R}^2 \mathrm{~T}$$
$$8 \pi R^2 \mathrm{~T}$$
$$\frac{1}{8} \pi R^2 T$$
$$3 \pi R^2 \mathrm{~T}$$
Explanation
Volume constant
$$\begin{aligned} & \frac{4}{3} \pi R^3=27 \times \frac{4}{3} \times \pi r^3 \\ & R^3=27 r^3 \\ & R=3 r \\ & r=\frac{R}{3} \\ & r^2=\frac{R^2}{9} \end{aligned}$$
$$\begin{aligned} & \text { Work done }=T . \Delta A \\ & =27 T\left(4 \pi r^2\right)-T 4 \pi R^2 \\ & =27 T 4 \pi \frac{R^2}{9}-4 \pi R^2 T \\ & =8 \pi R^2 T \end{aligned}$$
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