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Current Question (ID: 10426)

Question:
\text{Given:} \text{Wing area: } A = 120 \text{ m}^2 \text{Pressure difference: } \Delta P = 2.5 \text{ kPa} = 2500 \text{ Pa} \text{Acceleration due to gravity: } g = 10 \text{ m/s}^2 \text{Solution:} \text{For level flight, lift force equals weight:} F_{\text{lift}} = \Delta P \times A = mg \text{Therefore: } m = \frac{\Delta P \times A}{g} m = \frac{2500 \times 120}{10} = \frac{300000}{10} = 30000 \text{ kg} \text{Mass of airplane} = 30 \text{ tonnes}
Options:
  • 1. $2 \times 10^3 \text{ kg}$
  • 2. $3 \times 10^4 \text{ kg}$
  • 3. $5 \times 10^4 \text{ kg}$
  • 4. $7 \times 10^4 \text{ kg}$
Solution:
Hint: The net vertical force on the airplane is zero. Step 1: Balance the forces. For a level flight, the upward lift force must balance the downward gravitational force (weight). The upward force is created by the pressure difference between the lower and upper surfaces of the wing. Let $P_1$ be the pressure on the lower surface and $P_2$ be the pressure on the upper surface. The pressure difference is $\Delta P = P_1 - P_2$. The total upward force (lift) is $F = \Delta P \times A$, where $A$ is the total wing area. The downward force is the weight of the airplane, which is $W = mg$, where $m$ is the mass of the airplane and $g$ is the acceleration due to gravity. Equating the forces: $(P_1 - P_2)A = mg$ Step 2: Find the mass. Given values: Area of wing, $A = 120 \text{ m}^2$ Pressure difference, $\Delta P = P_1 - P_2 = 2.5 \text{ kPa} = 2.5 \times 10^3 \text{ Pa}$ Acceleration due to gravity, $g = 10 \text{ m/s}^2$ Substitute the values into the equation: $(2.5 \times 10^3 \text{ Pa}) \times (120 \text{ m}^2) = m \times (10 \text{ m/s}^2)$ $m = \frac{2.5 \times 10^3 \times 120}{10}$ $m = 2.5 \times 10^3 \times 12$ $m = 30 \times 10^3 \text{ kg} = 3 \times 10^4 \text{ kg}$ The mass of the airplane is $3 \times 10^4 \text{ kg}$.

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{
  "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
}