Question

A sphere of 150 mm diameter is held in equilibrium by a vertical air stream of velocity 15 m/s. If the density of air is \( 1.225 \, \text{kg/m}^3 \) and the coefficient of drag is 0.43, the weight of the sphere is ...........

• A. 0.78 N
• B. 1.04 N
• C. 1.29 N
• D. 1.56 N

Hint:

When an object is held in equilibrium by a fluid stream, the drag force equals its weight. Use the drag equation to solve such problems.

Answer 1

The Correct Option is B

When the sphere is held in equilibrium by the air stream, the drag force balances the weight.
The drag force is given by the formula:
\[ F_D = \frac{1}{2} C_D \, \rho \, A \, V^2 \]
where:
\( C_D = 0.43 \) is the drag coefficient
\( \rho = 1.225 \, \text{kg/m}^3 \) is the density of air
\( V = 15 \, \text{m/s} \) is the velocity of air
\( A \) is the projected area of the sphere = \( \frac{\pi d^2}{4} \)
\( d = 150 \, \text{mm} = 0.15 \, \text{m} \)
\[ A = \frac{\pi (0.15)^2}{4} = \frac{\pi \times 0.0225}{4} \approx 0.01767 \, \text{m}^2 \]
Substitute values into the drag force equation:
\[ F_D = \frac{1}{2} \times 0.43 \times 1.225 \times 0.01767 \times (15)^2 \]
\[ F_D = 0.5 \times 0.43 \times 1.225 \times 0.01767 \times 225 \]
\[ F_D \approx 1.04 \, \text{N} \]
Therefore, the weight of the sphere is \( \boxed{1.04 \, \text{N}} \).