JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Fluid Mechanics

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A small liquid drop of radius R is divided into 27 identical liquid drops. If the surface tension is T, then the work done in the process will be :

(A) $8 π R^{2} T$ ✓ Correct
(B) $3 π R^{2} T$
(C) $\frac{1}{8} π R^{2} T$
(D) $4 π R^{2} T$

Answer: (A)

Solution

Volume is conserved when the drop of radius

R

breaks into 27 identical drops of radius

r

\frac{4}{3} π R^{3} = 27 \times \frac{4}{3} π r^{3} ⟹ r = \frac{R}{3}

The work done against surface tension equals the increase in surface energy:

W = T Δ A = T (27 \times 4 π r^{2} - 4 π R^{2})

= T (27 \times 4 π \times \frac{R ^{2}}{9} - 4 π R^{2}) = T (12 π R^{2} - 4 π R^{2}) = 8 π R^{2} T

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Single correct option

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