JEE Main 2026 · 2 Apr · Shift 1 · Physics Q45

← JEE Main 2026 · 2 Apr · Shift 1

PhysicsSingle correct+4 / -1Question 45 of 75

A liquid drop of diameter 2 mm breaks into 512 droplets. The change in surface energy is $\alpha\times10^{-6}$ J. The value of $\alpha$ is ________. (Take surface tension of liquid = 0.08 N/m)

Answer and worked solution

Official answer: B

Droplet radius. $512=8^3$ , so $r=\frac R8$ with $R=1$ mm.

Area change. $\Delta A=4\pi\left(512\cdot\frac{R^2}{64}-R^2\right)=28\pi R^2=8.80\times10^{-5}$ m².

Energy. $\Delta E=T\Delta A=0.08\times8.80\times10^{-5}=7.0\times10^{-6}$ J, so $\alpha=7$ .

(B)

Topics:Surface tension

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JEE Main 2026 · 2 Apr · Shift 1, Physics question 45