In an experiment to verify Stokes law, a small spherical ball of...

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

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

$⇒ {v}^{2} = - 2g ( - h )$

$⇒ v = \sqrt{2gh} . . . ( 1 )$

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

${v}_{T} = \dfrac{2}{9}\dfrac{{r}^{2} ( ρ - \sigma ) g}{η} . . . ( 2 )$

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

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

$⇒ \sqrt{2gh} = \dfrac{2}{9}\dfrac{{r}^{2} ( ρ - \sigma ) g}{η}$

Squaring the above equation, we get

$⇒ h = \dfrac{2}{81}\dfrac{{r}^{4}g ( ρ - \sigma{)}^{2}}{{η}^{2}}$

$⇒ h ∝ {r}^{4}$