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Current Question (ID: 10471)

Question:
$\text{The speed of flow past the lower surface of a wing of an airplane is 50 m/s. What speed of flow over the upper surface will give a dynamic lift of 1000 Pa?}$ $\text{(density of air } 1.3 \text{ kg/m}^3\text{)}$
Options:
  • 1. $25.55 \text{ m/s}$
  • 2. $63.55 \text{ m/s}$
  • 3. $13.25 \text{ m/s}$
  • 4. $6.35 \text{ m/s}$
Solution:
$\text{Given,}$ $P_1 - P_2 = 1000 \text{ Pa}$ $\Rightarrow \frac{1}{2}\rho v_2^2 - \frac{1}{2}\rho v_1^2 = P_1 - P_2$ $\Rightarrow v_2^2 = \frac{2(P_1 - P_2)}{\rho} + v_1^2$ $\Rightarrow v_2^2 = \frac{2 \times 1000}{1.3} + (50)^2$ $\Rightarrow v_2 = 63.55 \text{ m/s}$

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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
}