As the volume of the mercury will remain same, therefore
34πR3=125(34πr3)⇒R=5r
Now, increase in surface energy
ΔU=125×(S×4πr2)−(S×4πR2)
=0.45×4π×(10−3)2[25125−1]
=4×0.45×4π×10−6
=2.26×10−5J
JEE Main 2023 — Physics Mechanics
A mercury drop of radius 10–3m is broken into 125 equal size droplets. Surface tension of mercury is 0.45Nm–1 . The gain in surface energy is:
Held on 1 Feb 2023 · Verified 6 Jul 2026.
2.26×10–5J
28×10–5J
17.5×10–5J
5×10–5J
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