A liquid drop of diameter 2 mm breaks into 512 droplets. The change in surface energy is $\alpha \times 10^{-6} \mathrm{~J}$. The value of $\alpha$ is ____. (Take surface tension of liquid $=0.08 \mathrm{~N} / \mathrm{m}$)

JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Fluid Mechanics

A liquid drop of diameter 2 mm breaks into 512 droplets. The change in surface energy is

α \times 1 0^{- 6} J

. The value of

α

is ____. (Take surface tension of liquid

= 0.08 N / m

)

(A)10
(B)7✓ Correct
(C)8
(D)11

Answer: (B)

Solution

The parent drop has radius

R = 1

mm. Conserving volume,

\frac{4}{3} π R^{3} = 512 \cdot \frac{4}{3} π r^{3}

, so

r = \frac{R}{51 2 ^{1/3}} = \frac{R}{8}

. The energy stored in the newly created surface equals the surface tension times the gain in area:

Δ E = T (512 \cdot 4 π r^{2} - 4 π R^{2}) = 4 π T R^{2} (\frac{512}{64} - 1) = 4 π T R^{2} (8 - 1) = 28 π T R^{2}

. Substituting

T = 0.08

N/m and

R = 1 0^{- 3}

Δ E = 28 π (0.08) (1 0^{- 3})^{2} = 2.24 π \times 1 0^{- 6} \approx 7 \times 1 0^{- 6}

J. The value of

α

7

JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Single correct option

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