A ball of radius r and density ρ falls freely under gravity through a distance h before entering water. Velocity of ball does not change even on entering water. If viscosity of water is η the value of h is given by

Concept :

When the ball falls freely under gravity, it accelerates due to the gravitational force. However, upon entering the water, the ball will experience a viscous drag force that counteracts its motion. The terminal velocity is reached when the gravitational force is balanced by the buoyant force and the viscous drag force.

The formula for the terminal velocity v_t of a spherical object falling through a viscous fluid is given by:

v_t = (2/9) (r²(ρ - 1)g)/η

Using energy conservation principles, the distance h can be found by equating the kinetic energy gained during free fall to the work done against viscous drag.

Calculation:

We have the terminal velocity:

⇒ v_t = (2/9) (r²(ρ - 1)g)/η ... (1)

The potential energy lost by the ball during free fall is converted into kinetic energy:

⇒ mgh = (1/2) m v_t²

Where m is the mass of the ball, g is the acceleration due to gravity, and h is the distance fallen.

Substituting the value of v_t from equation (1):

⇒ mgh = (1/2) m [(2/9) (r²(ρ - 1)g)/η]²

Solving for h:

⇒ gh = (1/2) [(2/9) (r²(ρ - 1)g)/η]²

⇒ h = (1/2g) [(2/9) (r²(ρ - 1)g)/η]²

⇒ h = (1/2g) [4/81 (r⁴(ρ - 1)²g²)/η²]

⇒ h = (1/2g) (4/81) (r⁴(ρ - 1)²g²)/η²

⇒ h = (2/81) (r⁴(ρ - 1)²g)/η²

∴ The value of h is \(\frac{2}{81} r^{4}\left(\frac{\rho-1}{\eta}\right)^{2} g \).