Fluid Mechanics

JEE Main level, single correct, medium

87 of 192

Question 87 of 192

Eight mercury drops, each of radius rr, coalesce to form a bigger drop. The surface energy released in this process is _______. (SS is the surface tension of mercury).

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Answer: (B) 16πr2S16\pi r^2 S

Let the radius of the bigger drop be RR. Since the total volume remains conserved during the process:

8×43πr3=43πR38 \times \dfrac{4}{3} \pi r^3 = \dfrac{4}{3} \pi R^3

R3=8r3⇒R=2rR^3 = 8r^3 \Rightarrow R = 2r

Initial surface energy of the 88 drops is:

Ui=8×(4πr2S)=32πr2SU_i = 8 \times (4 \pi r^2 S) = 32 \pi r^2 S

Final surface energy of the bigger drop is:

Uf=4πR2S=4π(2r)2S=16πr2SU_f = 4 \pi R^2 S = 4 \pi (2r)^2 S = 16 \pi r^2 S

The surface energy released in the process is:

ΔU=Ui−Uf=32πr2S−16πr2S=16πr2S\Delta U = U_i - U_f = 32 \pi r^2 S - 16 \pi r^2 S = 16 \pi r^2 S

Answer: 16πr2S16\pi r^2 S

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