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Current Question (ID: 20503)

Question:
$\text{A circular loop of radius } \frac{10}{\sqrt{\pi}} \text{ cm is placed in a linearly varying}$ $\text{perpendicular magnetic field which has magnitude } 0.5 \text{ T at time } t = 0.$ $\text{The magnetic field reduces to zero at } t = 0.5 \text{ s. The emf}$ $\text{induced in the loop at } t = 0.25 \text{ s is:}$ $\text{(where } r \text{ is the radius of the circular loop)}$
Options:
  • 1. 0.01 \text{ V}
  • 2. 0.005 \text{ V}
  • 3. 0.02 \text{ V}
  • 4. 0.03 \text{ V}
Solution:
$\varepsilon_{\text{ind}} = \frac{\Delta \phi}{\Delta t} = \frac{\Delta (BA)}{\Delta t} = A \frac{\Delta (B)}{\Delta t}$ $= \pi \times \left( \frac{10}{\sqrt{\pi}} \right)^2 \times 10^{-4} \times \left( \frac{0.25}{0.25} \right)$ $= 10^{-2} \times 1 = 0.01 \text{ V}$

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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
}