JEE MAIN - Physics (2023 - 1st February Morning Shift - No. 19)
A mercury drop of radius $$10^{-3}~\mathrm{m}$$ is broken into 125 equal size droplets. Surface tension of mercury is $$0.45~\mathrm{Nm}^{-1}$$. The gain in surface energy is :
$$28\times10^{-5}~\mathrm{J}$$
$$17.5\times10^{-5}~\mathrm{J}$$
$$5\times10^{-5}~\mathrm{J}$$
$$2.26\times10^{-5}~\mathrm{J}$$
Explanation
Initial surface energy $=0.45 \times 4 \pi\left(10^{-3}\right)^2$
$$ \begin{aligned} & \frac{4}{3} \pi\left(10^{-3}\right)^3=125 \times \frac{4 \pi}{3} R_{\text {new }}^3 \\\\ \therefore & 10^{-3}=5 R_{\text {new }} \\\\ \therefore & R_{\text {new }}=\frac{10^{-3}}{5} \mathrm{~m} \end{aligned} $$
So, final surface energy $=0.45 \times 125 \times 4 \pi\left(\frac{10^{-3}}{5}\right)^2$
Increase in energy $=0.45 \times 4 \pi \times\left(10^{-3}\right)^2\left[\frac{125}{25}-1\right]$
$$ \begin{aligned} & =4 \times 0.45 \times 4 \pi \times 10^{-6} \\\\ & =2.26 \times 10^{-5} \mathrm{~J} \end{aligned} $$
$$ \begin{aligned} & \frac{4}{3} \pi\left(10^{-3}\right)^3=125 \times \frac{4 \pi}{3} R_{\text {new }}^3 \\\\ \therefore & 10^{-3}=5 R_{\text {new }} \\\\ \therefore & R_{\text {new }}=\frac{10^{-3}}{5} \mathrm{~m} \end{aligned} $$
So, final surface energy $=0.45 \times 125 \times 4 \pi\left(\frac{10^{-3}}{5}\right)^2$
Increase in energy $=0.45 \times 4 \pi \times\left(10^{-3}\right)^2\left[\frac{125}{25}-1\right]$
$$ \begin{aligned} & =4 \times 0.45 \times 4 \pi \times 10^{-6} \\\\ & =2.26 \times 10^{-5} \mathrm{~J} \end{aligned} $$
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