NEET Physics Question - Solid and Fluids

A wind with speed 40 m/s blows parallel to the roof of a house. The area of the roof is 250 m2. Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be: \[(\rho_{air}=1.2 kg/m^{3})\]

\[4.8\times 10^{5} N, upwards\]

\[2.4\times 10^{5} N, upwards\]

\[2.4\times 10^{5} N, downwards\]

\[4.8\times 10^{5} N, downwards\]

Solution:

Here is the short, step-by-step solution for your WordPress post:

Solution:

According to Bernoulli's principle, the high-speed wind blowing outside creates a low-pressure region ( $P_{out}$ ) above the roof compared to the atmospheric pressure ( $P_{in}$ ) inside.

\Delta P = P_{in} - P_{out} = \frac{1}{2}\rho v^2

Substituting the given values, the upward lifting force exerted on the roof is:

F = \Delta P \times A = \left(\frac{1}{2} \times 1.2 \times 40^2\right) \times 250 = 2.4 \times 10^5\text{ N (upwards)}

Correct Option: Option 2 ( $2.4\times 10^{5}\text{ N, upwards}$ )

Rankers Physics

Solution:

Solution:

Leave a Reply Cancel reply