In an experiment to verify Stokes law, a small spherical ball of radius r and density ρ falls under gravity through a distance h in air before entering a tank of water. If the terminal velocity of the

Question

In an experiment to verify Stokes law, a small spherical ball of radius r and density ρ falls under gravity through a distance h in air before entering a tank of water. If the terminal velocity of the ball inside water is same as its velocity just before entering the water surface, then the value of h is proportional to: (ignore viscosity of air)

TestHub · Accepted Answer

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

$v^{2} = u^{2} - 2 g y$

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

$\Rightarrow v = \sqrt{2 g h} . . . (1)$

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

$v_{T} = \frac{2}{9} \frac{r^{2} (ρ - σ) g}{η} . . . (2)$

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

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

$\Rightarrow \sqrt{2 g h} = \frac{2}{9} \frac{r^{2} (ρ - σ) g}{η}$

Squaring the above equation, we get

$\Rightarrow h = \frac{2}{81} \frac{r^{4} g (ρ - σ)^{2}}{η^{2}}$

$\Rightarrow h \propto r^{4}$

Physics - Fluid Question with Solution | TestHub

Options:

Doubts & Discussion