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  4. A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface energy. (Surface tension of mercury is 0.465 J m- 2 )

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Q. A drop of mercury of radius $2 \, mm$ is split into $8$ identical droplets. Find the increase in surface energy. (Surface tension of mercury is $0.465 \, J \, m^{- 2}$ )

NTA AbhyasNTA Abhyas 2022

Solution:

Increase in surface energy or work done in splitting a big drop $= 4 \pi R^{2} T \left(\right. n^{1 / 3} - 1 \left.\right)$ $\Rightarrow W=4\pi \times \left(\right. 2 \times \left(10\right)^{- 3} \left.\right)^{2}\times 0.465\left(\right.8^{1 / 3}-1\left.\right)=23.4μJ$