1. Speed gained by the sphere while falling through air

Since viscosity of air is ignored, the ball falls freely through height $h$ before entering water.

Starting from rest, $v^2 = 2gh$ So, the speed just before entering water is $v = \sqrt{2gh}$

2. Terminal velocity inside water using Stokes' law

For a small sphere falling in a viscous liquid, terminal velocity is $v_t = \frac{2}{9}\frac{r^2(\rho-\rho_w)g}{\eta}$ where:

• $r$ = radius of sphere
• $\rho$ = density of sphere
• $\rho_w$ = density of water
• $\eta$ = viscosity of water

Thus, $v_t \propto r^2$

3. Given condition

The problem states that the terminal velocity in water is equal to the velocity just before entering water: $\sqrt{2gh} = v_t$

Squaring both sides, $2gh = v_t^2$

Since $v_t \propto r^2$, $v_t^2 \propto r^4$ Therefore, $h \propto r^4$

4. Option check

• A: $r$ ❌
• B: $r^4$ ✅
• C: $r^3$ ❌
• D: $r^2$ ❌

Hence, the correct option is B.