Import Question JSON

Current Question (ID: 8812)

Question:
$\text{A car moves from } X \text{ to } Y \text{ with a uniform speed } v_u \text{ and returns to } X \text{ with a uniform speed } v_d. \text{ The average speed for this round trip is:}$
Options:
  • 1. $\frac{2v_dv_u}{v_d + v_u}$
  • 2. $\sqrt{v_u v_d}$
  • 3. $\frac{v_dv_u}{v_d + v_u}$
  • 4. $\frac{v_u + v_d}{2}$
Solution:
\text{Hint: Average speed} = \frac{\text{Total distance travelled}}{\text{Total time taken}} \text{Step: Find the average speed of the car.} \text{Let } t_1 \text{ and } t_2 \text{ be times taken by the car to go from } X \text{ to } Y \text{ and then from } Y \text{ to } X \text{ respectively.} \text{Then, } t_1 + t_2 = \frac{XY}{v_u} + \frac{XY}{v_d} = XY \cdot \frac{v_u + v_d}{v_u \cdot v_d} \text{Total distance travelled} = XY + XY = 2XY \text{Therefore, average speed of the car for this round trip is:} \frac{2XY}{XY \cdot \frac{v_u + v_d}{v_u \cdot v_d}} = \frac{2XY \cdot v_u \cdot v_d}{XY \cdot (v_u + v_d)} V_{av} = \frac{2v_u v_d}{v_u + v_d} \text{Hence, option (1) is the correct answer.}

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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
}