Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A wind with speed $40 \text{ m/s}$ blows parallel to the roof of a house. The area of the roof is $250 \text{ m}^2$ . 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: $(\rho_{air}=1.2 \text{ kg/m}^3)$

$4.8 \times 10^5 \text{ N}$ , downwards

$4.8 \times 10^5 \text{ N}$ , upwards

$2.4 \times 10^5 \text{ N}$ , upwards

$2.4 \times 10^5 \text{ N}$ , downwards

Step-by-Step Solution

Bernoulli's Principle: Apply Bernoulli's equation for the air inside and outside the roof. Neglecting the difference in height (potential energy), the equation is: $P_{in} + \frac{1}{2}\rho v_{in}^2 = P_{out} + \frac{1}{2}\rho v_{out}^2$
Conditions:

Inside the house, the air is static, so $v_{in} = 0$ . The pressure is atmospheric, $P_{in} = P_{atm}$ .
Outside the house, the wind speed is $v_{out} = 40 \text{ m/s}$ . The pressure is $P_{out}$ .

Pressure Difference: Rearranging the equation: $P_{in} - P_{out} = \frac{1}{2}\rho (v_{out}^2 - v_{in}^2)$ $\Delta P = \frac{1}{2} \times 1.2 \times (40^2 - 0) = 0.6 \times 1600 = 960 \text{ Pa (N/m}^2)$
Calculate Force: The aerodynamic lift (force) is the pressure difference multiplied by the area ( $A = 250 \text{ m}^2$ ). $F = \Delta P \times A = 960 \times 250 = 240,000 \text{ N} = 2.4 \times 10^5 \text{ N}$
Determine Direction: Since the velocity outside ( $40 \text{ m/s}$ ) is greater than inside ( $0 \text{ m/s}$ ), the pressure outside is lower than the pressure inside ( $P_{out} < P_{in}$ ). Therefore, the net force acts from high pressure to low pressure, which is upwards .

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