Question

A spherical ball is steadily supported against gravity by an upward air jet as shown in the figure. Take acceleration due to gravity to be \( g = 10 \, \text{m/s}^2 \). The mass flow rate of air, reaching the ball, is 0.01 kg/s and the air reaches the ball at an upward velocity of 3 m/s. Neglecting the buoyancy force, and using the principle of integral momentum balance, the mass (in grams, up to one decimal place) of the ball is \(\underline{\hspace{1cm}}\). \includegraphics[width=0.35\linewidth]{image9.png}

Hint:

For momentum balance problems, the momentum flux of the air jet is equal to the weight of the object being supported.

Answer 1

Correct Answer: 2.9

Using the principle of integral momentum balance, we know that the momentum flux of the air jet is balanced by the weight of the ball. The momentum balance equation is:
\[ \dot{m} v = W \] Where:
- \( \dot{m} = 0.01 \, \text{kg/s} \) is the mass flow rate of the air,
- \( v = 3 \, \text{m/s} \) is the velocity of the air,
- \( W = mg \) is the weight of the ball, with \( m \) being the mass of the ball and \( g = 10 \, \text{m/s}^2 \).
Substituting values into the momentum balance equation, we get:
\[ 0.01 \times 3 = m \times 10 \] Solving for \( m \), we find:
\[ m = \frac{0.03}{10} = 0.003 \, \text{kg}. \] Finally, converting the mass to grams:
\[ m = 3 \, \text{g}. \] Thus, the mass of the ball is approximately \( 3.0 \, \text{g} \).