JEE Main 2020 — Mechanical Properties of Fluids Question with Solution

The velocity of ball just before striking the surface of water is given by using kinematical equation of motion under the gravity

$v^2=u^2-2gy$

$\Rightarrow v^2=-2g(-h)$

$\Rightarrow v=\sqrt{2gh} ...\left(1\right)$ 

Now, the terminal velocity of spherical ball inside the water tank is given by,

$v_T=\frac{2}{9}\frac{r^2(\rho -\sigma )g}{\eta } ...\left(2\right)$

After falling through $h$ the velocity of ball should be equal to its terminal velocity.

From equation (1) and (2) we get,

$\Rightarrow \sqrt{2gh}=\frac{2}{9}\frac{r^2(\rho -\sigma )g}{\eta }$

Squaring the above equation, we get

$\Rightarrow h=\frac{2}{81}\frac{r^{4}g(\rho -\sigma {)}^2}{{\eta }^2}$

$\Rightarrow h\propto r^4$