MCQ
A drop of mercury of radius $2\, mm$ is split into $8$ identical droplets. Find the increase in surface energy ....... $\mu J$. (Surface tension of mercury is $0.465\;J/{m^2}$)
- ✓$23.4$
- B$18.5$
- C$26.8$
- D$16.8$
✓
Answer
Correct option: A.
$23.4$
a
(a) Increase in surface energy or work done in splitting a big drop $ = 4\pi {R^2}T({n^{1/3}} - 1)$
(a) Increase in surface energy or work done in splitting a big drop $ = 4\pi {R^2}T({n^{1/3}} - 1)$
$ \Rightarrow W = 4\pi \times {(2 \times {10^{ - 3}})^2} \times 0.465({8^{1/3}} - 1) = 23.4\;\mu \,J$
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