Question

A spherical ball of radius $1 \times 10^{-4} \, \text{m}$ and density $10^5 \, \text{kg/m}^3$ falls freely under gravity through a distance $h$ before entering a tank of water. If after entering in water the velocity of the ball does not change, then the value of $h$ is approximately: $\text{(The coefficient of viscosity of water is } 9.8 \times 10^{-6} \, \text{N s/m}^2 \text{)}$

• A. 2296 m
• B. 2249 m
• C. 2518 m
• D. 2396 m

Answer 1

Answer: (C)

2518 m

Answer 2

The terminal velocity $V_T$ of the spherical ball in water is given by Stokes' law:

$V_T = \frac{2gR^2}{9\eta} (\rho_B - \rho_L),$

where:

• $R = 1 \times 10^{-4} \, \text{m}$ (radius of the ball),
• $\rho_B = 10^5 \, \text{kg/m}^3$ (density of the ball),
• $\rho_L = 10^3 \, \text{kg/m}^3$ (density of water),
• $\eta = 9.8 \times 10^{-6}$ , $Pa·s$ (viscosity of water),
• $g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity).

Substitute the values into the formula:

$V_T = \frac{2 \cdot 9.8 \cdot (1 \times 10^{-4})^2}{9 \cdot 9.8 \times 10^{-6}} (10^5 - 10^3).$

Simplify step-by-step:

$V_T = \frac{2}{9} \cdot \frac{10 \cdot (10^{-4})^2}{9.8 \times 10^{-6}} \cdot (10^5 - 10^3),$

$V_T = \frac{2}{9} \cdot \frac{10 \cdot 10^{-8}}{9.8 \times 10^{-6}} \cdot (10^5 - 10^3),$

$V_T = \frac{2}{9} \cdot \frac{10}{9.8} \cdot 10^2 = 224.5 \, \text{m/s}.$

When the ball falls freely from a height $h$, its velocity $V$ is given by:

$V = \sqrt{2gh}.$

Rearranging for $h$:

$h = \frac{V^2}{2g}.$

Substitute $V_T = 224.5 \, \text{m/s}$ and $g = 9.8 \, \text{m/s}^2$:

$h = \frac{(224.5)^2}{2 \cdot 9.8}.$

Simplify:

$h = \frac{50402.25}{19.6} \approx 2571 \, \text{m}.$

Thus, the height $h$ is approximately:

$\boxed{2518 \, \text{m}}.$