Question

A wind with speed \(50 \mathrm{~m} / \mathrm{s}\) blows parallel to the roof of a house. The area of the roof is \(300 \mathrm{~m}^2\). Assume that the pressure inside the house is atmospheric pressure. Density of air is \(1.2 \mathrm{~kg} / \mathrm{m}^3\). The magnitude of the force exerted by the wind on the roof will be

• A. \(1.5\times 10^{5}\text{ }\mathrm{N}\)
• B. \(3.0\times 10^{5}\text{ }\mathrm{N}\)
• C. \(4.5\times 10^{5}\text{ }\mathrm{N}\)
• D. \(9.0\times 10^{5}\text{ }\mathrm{N}\)

Hint:

The pressure difference across the roof is half rho v squared; multiply by the roof area.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Air flows over the roof, so the pressure above is lower than inside the house. Bernoulli gives \(\Delta P=\dfrac12\rho v^2\).

Step 2: Pressure difference
\(\Delta P=\dfrac12\times1.2\times50^2=1500\ \text{Pa}\).

Step 3: Force
\(F=\Delta P\times A=1500\times300=4.5\times10^5\ \text{N}\).

Final Answer:
The force on the roof is \(4.5\times10^5\ \text{N}\). \[ \boxed{\text{(C) }4.5\times 10^{5}\text{ N}} \]