JEE Main 2026 Apr 2 Shift 1, Physics Q20: Properties of Fluids

JEE Main2026Apr 2, Shift 1Physics

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
$7$
C
$8$
D
$11$

Solution

Volume is conserved during fragmentation, so if $R = 1$ mm is the parent radius and $r$ the small droplet radius:

$\frac{4}{3} π R^{3} = 512 \cdot \frac{4}{3} π r^{3} \Rightarrow r = \frac{R}{51 2 ^{1/3}} = \frac{R}{8} = 0.125$ mm

Change in surface energy = $T \times Δ (area)$ :

$Δ S E = T \cdot 4 π (512 r^{2} - R^{2}) = 4 π (0.08) [512 \times (0.125)^{2} - 1^{2}] \times 1 0^{- 6}$ J

$= 4 π (0.08) (8 - 1) \times 1 0^{- 6} \approx 7 \times 1 0^{- 6}$ J

Hence $α = 7$ .

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JEE Main 2026 Apr 2 Shift 1, Physics Q20: Properties of Fluids

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