Question

A ball of density $\sigma$ and radius $r$ is dropped on the surface of a liquid of density $\rho$ from certain height. If speed of ball does not change even on entering in liquid and viscosity of liquid is $\eta$, then the height from which ball dropped is

• A. $2 g\left[\frac{(\sigma-\rho) r}{9 \eta}\right]^{2}$
• B. $\frac{2 g(\sigma-\rho)^{2} r^{2}}{9 \eta}$
• C. $\frac{2(\sigma-\rho) g r^{2}}{9 \eta}$
• D. $2 g\left[\frac{(\sigma-\rho) r^{2}}{9 \eta}\right]^{2}$

Answer 1

Answer: (D)

The ball has already reached the magnitude of velocity which is equal to its terminal velocity in fluid.
$v=\sqrt{2 g h}$
$\{$ Velocity of body fallen form height $h=\sqrt{2 g h}\}$
$v_{\text {Terminal }}=\frac{2 r^{2}}{9 \eta}(\sigma-\rho) g$
Equating both
$\sqrt{2 g h} =\frac{2 r^{2}}{9 \eta}(\sigma-\rho) g$
$\Rightarrow h=2 g\left[\frac{(\sigma-\rho) r^{2}}{9 \eta}\right]^{2}$