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NEET PHYSICSMechanical Properties of FluidMedium

Question

A wind with speed 40 m/s40 \text{ m/s} blows parallel to the roof of a house. The area of the roof is 250 m2250 \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: (ρair=1.2 kg/m3)(\rho_{air}=1.2 \text{ kg/m}^3)

A

4.8×105 N4.8 \times 10^5 \text{ N}, downwards

B

4.8×105 N4.8 \times 10^5 \text{ N}, upwards

C

2.4×105 N2.4 \times 10^5 \text{ N}, upwards

D

2.4×105 N2.4 \times 10^5 \text{ N}, downwards

Step-by-Step Solution

  1. 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: Pin+12ρvin2=Pout+12ρvout2P_{in} + \frac{1}{2}\rho v_{in}^2 = P_{out} + \frac{1}{2}\rho v_{out}^2

  2. Conditions:

  • Inside the house, the air is static, so vin=0v_{in} = 0. The pressure is atmospheric, Pin=PatmP_{in} = P_{atm}.
  • Outside the house, the wind speed is vout=40 m/sv_{out} = 40 \text{ m/s}. The pressure is PoutP_{out}.
  1. Pressure Difference: Rearranging the equation: PinPout=12ρ(vout2vin2)P_{in} - P_{out} = \frac{1}{2}\rho (v_{out}^2 - v_{in}^2) ΔP=12×1.2×(4020)=0.6×1600=960 Pa (N/m2)\Delta P = \frac{1}{2} \times 1.2 \times (40^2 - 0) = 0.6 \times 1600 = 960 \text{ Pa (N/m}^2)
  2. Calculate Force: The aerodynamic lift (force) is the pressure difference multiplied by the area (A=250 m2A = 250 \text{ m}^2). F=ΔP×A=960×250=240,000 N=2.4×105 NF = \Delta P \times A = 960 \times 250 = 240,000 \text{ N} = 2.4 \times 10^5 \text{ N}
  3. Determine Direction: Since the velocity outside (40 m/s40 \text{ m/s}) is greater than inside (0 m/s0 \text{ m/s}), the pressure outside is lower than the pressure inside (Pout<PinP_{out} < P_{in}). Therefore, the net force acts from high pressure to low pressure, which is upwards .

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Mechanical Properties of Fluid. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMechanical Properties of Fluidparallelassumingpressureinsideatmospheric

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