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CGP EDU Academic Team
Published on: September 12, 2026
8 mercury drops coalesce to form one mercury drop, the energy changes by a factor of
Text Solution
Verified by ExpertsThe correct answer is:
C
As volume remains constant therefore $R = n^{1/3} r$
$\frac{\text{Energy of big drop}}{\text{Energy of small drop}} = \frac{4 \pi R^{2} T}{4 \pi r^{2} T} = \frac{R^{2}}{r^{2}} = (8)^{2/3} = 4$
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