MHTCET 23 Apr 2025 Shift 1 Paper

Concept:Applying Bernoulli's principle to find the pressure difference caused by wind over the roof.Explanation:For horizontal flow, Bernoulli's equation is:$$P + \frac{1}{2}\rho v^{2} = \text{constant}$$Inside the house, air is nearly still, so the pressure is atmospheric, $P_{\text{atm}}$.Outside, the wind speed is $v = 50 \, \mathrm{m/s}$, which reduces the pressure above the roof.The pressure difference is given by:$$\Delta P = \frac{1}{2}\rho v^{2}$$Substitute $\rho = 1.2 \, \mathrm{kg/m^3}$ and $v = 50 \, \mathrm{m/s}$:$$\Delta P = \frac{1}{2}(1.2)(50)^{2} = 0.6 \times 2500 = 1500 \, \mathrm{Pa}$$Force on the roof is:$$F = \Delta P \times A = 1500 \times 300 = 4.5 \times 10^{5} \, \mathrm{N}$$Answer:The magnitude of the force exerted by the wind is $4.5 \times 10^{5} \, \mathrm{N}$, which is Option C.