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Current Question (ID: 16994)

Question:
$\text{A thin semicircular conducting ring of radius } R \text{ is falling with its plane vertical}$ $\text{in a horizontal magnetic induction } B. \text{ At the position } MNQ, \text{ the speed of the}$ $\text{ring is } v \text{ and the potential difference developed across the ring is:}$
Options:
  • 1. $\text{Zero}$
  • 2. $Bv\pi R^2 / 2 \text{ and } M \text{ is at the higher potential}$
  • 3. $2RBv \text{ and } M \text{ is at the higher potential}$
  • 4. $2RBv \text{ and } Q \text{ is at the higher potential}$
Solution:
$\text{As the ring is falling with speed } V, \text{ distance cover by the ring in a short}$ $\text{time interval } dt \text{ is } Vdt.$ $\text{According to Faraday's law of induction induced emf}$ $e = -\frac{d\phi}{dt} = -B \frac{dA}{dt} = -B \times 2R \times \frac{Vdt}{dt} = B (2RV)$ $\text{The induced current in the ring must generate a magnetic field in the}$ $\text{downward direction. Thus, } Q \text{ is at the higher potential.}$

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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
}