Properties of Matter — JEE Main practice

1. A falling sphere reaches terminal velocity when the viscous drag balances the net downward force from gravity and buoyancy, so Stokes' Law and the force balance apply. $F_g = \frac{4}{3}\pi r^3 \rho g, \quad F_b = \frac{4}{3}\pi r^3 \sigma g, \quad F_v = 6\pi \eta r v_t$

2. Setting the viscous drag equal to the net weight at terminal velocity gives the terminal velocity formula. $6\pi \eta r v_t = \frac{4}{3}\pi r^3(\rho - \sigma)g$ So, $v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}$

3. Since $\rho$, $\sigma$, $g$, and $\eta$ are the same for the big drop and the droplets, the terminal velocity depends only on the square of the radius. $v \propto r^2$

4. Because the total volume is conserved when the drop breaks into 64 droplets, we can relate the radius of a droplet to $R$. $\frac{4}{3}\pi R^3 = 64 \times \frac{4}{3}\pi r^3$ Therefore, $R = 4r \Rightarrow r = \frac{R}{4}$

5. Using $v \propto r^2$ for the big drop ($v_1$, radius $R$) and small droplet ($v_2$, radius $r$): $\frac{v_1}{v_2} = \frac{R^2}{r^2} = \frac{R^2}{(R/4)^2} = 16$

Hence, the answer is D.