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

Answer 1

Answer: (D)

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