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

The Correct Option is C

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:

\[ \boxed{2518 \, \text{m}}. \]