Physics · 2023
JEE Main · 1 February 2023, Shift 1 · Q18
A mercury drop of radius 10^-3 m is broken into 125 equal size droplets. Surface tension of mercury is 0.45 Nm^-1. The gain in surface energy is:
A mercury drop of radius $\displaystyle 10^{-3} \mathrm{~m}$ is broken into $\displaystyle 125$ equal size droplets. Surface tension of mercury is $\displaystyle 0.45 \mathrm{Nm}^{-1}$. The gain in surface energy is:
Official answer
From NTA’s final answer key for this paper.
(4)
$\displaystyle 2.26 \times 10^{-5} \mathrm{~J}$
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JEE Main 2023 Physics question, with the answer from NTA’s final answer key. Where our answers come from.