Import Question JSON

Current Question (ID: 10474)

Question:

$\text{A wind with speed 40 m/s blows parallel to the roof of a house. The area of the roof is 250 m}^2\text{. Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be: } (\rho_{\text{air}} = 1.2 \text{ kg/m}^3)$

Options:

1. $4.8 \times 10^5 \text{ N, downwards}$
2. $4.8 \times 10^5 \text{ N, upwards}$
3. $2.4 \times 10^5 \text{ N, upwards}$
4. $2.4 \times 10^5 \text{ N, downwards}$

Solution:

$\text{From Bernoulli's theorem:}$ $P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$ $\text{where, } P_1 \text{ and } P_2 \text{ are pressure inside and outside the roof and } v_1, v_2 \text{ are velocities of wind inside and outside the roof. Neglect the width of the roof.}$ $\text{Pressure difference is:}$ $P_1 - P_2 = \frac{1}{2}\rho(v_2^2 - v_1^2)$ $\text{This is Bernoulli's equation.}$ $= \frac{1}{2} \times 1.2(40^2 - 0) = 960 \text{ N/m}^2$ $\text{Force acting on the roof is given by:}$ $F = (P_1 - P_2)A = 960 \times 250 = 24 \times 10^4 \text{ N} = 2.4 \times 10^5 \text{ N}$ $\text{As the pressure inside the roof is more than outside it. So the force will act in the upward direction, i.e. } F = 2.4 \times 10^5 \text{ N, upward}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}