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Current Question (ID: 8813)

Question:
$\text{A vehicle travels half the distance } L \text{ with speed } v_1 \text{ and the other half with speed } v_2, \text{ then its average speed is:}$
Options:
  • 1. $\frac{v_1 + v_2}{2}$
  • 2. $\frac{2v_1 + v_2}{v_1 + v_2}$
  • 3. $\frac{2v_1v_2}{v_1 + v_2}$
  • 4. $\frac{L(v_1 + v_2)}{v_1v_2}$
Solution:
\text{Hint: The average speed is defined as the total distance divided by the total time.} \text{Step 1: Find the time taken by the vehicle in two cases.} \text{Time taken to travel the first half distance: } t_1 = \frac{L/2}{v_1} = \frac{L}{2v_1} \text{Time taken to travel second half distance: } t_2 = \frac{L/2}{v_2} = \frac{L}{2v_2} \text{Total time: } t_1 + t_2 = \frac{L}{2v_1} + \frac{L}{2v_2} = \frac{L}{2}\left(\frac{1}{v_1} + \frac{1}{v_2}\right) \text{Step 2: Find the average speed.} \text{We know that:} V_{\text{avg}} = \text{Average velocity} = \frac{\text{total distance}}{\text{total time}} V_{\text{avg}} = \frac{L}{\frac{L}{2}\left(\frac{1}{v_1} + \frac{1}{v_2}\right)} = \frac{2v_1v_2}{v_1 + v_2} \text{Hence, option (3) 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
}