To solve this problem, we will use Bernoulli's principle, which relates the pressure and velocity of a fluid.$\\$According to Bernoulli's equation:$\\ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \\$Where:$\\$• $P_1$ is the atmospheric pressure inside the house.$\\$• $P_2$ is the pressure outside due to the wind.$\\$• $v_1 = 0$ (velocity of air inside the house).$\\$• $v_2 = 40$ ms$^{-1}$ (velocity of wind outside).$\\$• $\rho = 1.2$ kg m$^{-3}$ (density of air).$\\$Since the air inside the house is at rest, $v_1 = 0,$ so the equation becomes:$\\ P_1 = P_2 + \frac{1}{2} \rho v_2^2 \\$Rearranging gives:$\\ P_2 = P_1 - \frac{1}{2} \rho v_2^2 \\$The force exerted by the wind on the roof is due to the pressure difference:$\\ F = (P_1 - P_2) \times A \\$Substitute $P_2: \\ F = \left(P_1 - \left(P_1 - \frac{1}{2} \rho v_2^2\right)\right) \times A \\ F = \frac{1}{2} \rho v_2^2 \times A \\$Substitute the known values:$\\ F = \frac{1}{2} \times 1.2 \times (40)^2 \times 250 \\ F = 0.6 \times 1600 \times 250 \\ F = 240000$ N$\\$The force is exerted upwards because the pressure outside is lower than inside,$\\$causing a net upward force on the roof.$\\$Therefore, the force exerted by the wind on the roof is $2.4 \times 10^5$ N, upwards.$\\$This corresponds to Option 2.