Question

A ball of radius \( r \) and density \( \rho \) falls freely under gravity through a distance \( h \) before entering water. The velocity of the ball does not change even on entering water. If the viscosity of water is \( \eta \), the value of \( h \) is given by:
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• A. \( \frac{2}{9} \frac{r^2 (1 - \rho)}{\eta g} \)
• B. \( \frac{2}{81} \frac{r^2 (\rho - 1)}{\eta g} \)
• C. \( \frac{2}{81} \frac{r^4 (\rho - 1)}{\eta^2 g} \)
• D. \( \frac{2}{9} \frac{r^4 (\rho - 1)}{\eta^2 g} \)

Hint:

For motion in a fluid:
- The terminal velocity is given by Stokes’ law.
- If velocity remains constant, net force must be zero.

Answer 1

The Correct Option is C

Given: Radius of the ball: \( r \) Density of the ball: \( \rho \) Distance fallen: \( h \) Viscosity of water: \( \eta \) Acceleration due to gravity: \( g \) We need to find the value of \( h \). Since the velocity of the ball does not change upon entering the water, the drag force and the gravitational force must be in equilibrium when the ball moves through the water at a constant velocity. The drag force \( F_d \) acting on the ball in water is given by Stokes' Law: \[ F_d = 6 \pi \eta r v \] where \( v \) is the velocity of the ball just before entering the water. The gravitational force \( F_g \) acting on the ball is given by: \[ F_g = {Volume} \times {Density} \times g = \frac{4}{3} \pi r^3 \rho g \] For the ball to move with a constant velocity in water, the net force acting on it should be zero: \[ F_g = F_d \] Substituting the expressions for \( F_g \) and \( F_d \): \[ \frac{4}{3} \pi r^3 \rho g = 6 \pi \eta r v \] Simplifying this equation, we get: \[ v = \frac{2}{9} \frac{r^2 \rho g}{\eta} \] The ball falls freely under gravity through a distance \( h \) before entering the water. The velocity \( v \) of the ball just before entering the water is given by: \[ v = \sqrt{2gh} \] Equating the two expressions for \( v \): \[ \sqrt{2gh} = \frac{2}{9} \frac{r^2 \rho g}{\eta} \] Squaring both sides to solve for \( h \): \[ 2gh = \left( \frac{2}{9} \frac{r^2 \rho g}{\eta} \right)^2 \] \[ 2gh = \frac{4}{81} \frac{r^4 \rho^2 g^2}{\eta^2} \] \[ h = \frac{2}{81} \frac{r^4 \rho^2 g}{\eta^2} \] Now, we need to consider the relative density, and thus we substitute \( \rho \) with \( (\rho - 1) \): \[ h = \frac{2}{81} \frac{r^4 (\rho - 1)}{\eta^2 g} \] So, the detailed solution confirms that the correct answer is: \[ h = \frac{2}{81} \frac{r^4 (\rho - 1)}{\eta^2 g} \]