A spherical liquid drop of radius ……

Single correct+4 / -1Answer matches NTA’s final key

A spherical liquid drop of radius $R$ acquires the terminal velocity $v_1$ when falls through a gas of viscosity $\eta$ . Now the drop is broken into 64 identical droplets and each droplet acquires terminal velocity $v_2$ falling through the same gas. The ratio of terminal velocities $v_1/v_2$ is ______.

4
0.25
32
16

Solution

Volume is conserved: $64 \cdot \frac43\pi r^3 = \frac43\pi R^3$ , so $r = \frac{R}{4}$ .

From Stokes' law the terminal velocity is $v = \dfrac{2r^2(\rho - \sigma)g}{9\eta} \propto r^2$ , so

$\frac{v_1}{v_2} = \left(\frac{R}{r}\right)^2 = 16.$

Answer: 16.

Answer checked against the official answer key. The solution is our own working.

JEE MAIN 2026 · Properties of Solids and Liquids

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