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Current Question (ID: 20357)

Question:
$\text{A circular coil has moment of inertia } 0.8 \text{ kgm}^2 \text{ around any diameter}$ $\text{and is carrying current to produce a magnetic moment of } 20 \text{ Am}^2.$ $\text{The coil is kept initially in a vertical position and it can rotate freely}$ $\text{around a horizontal diameter. When a uniform magnetic field of } 4 \text{ T}$ $\text{is applied along the vertical, it starts rotating around its horizontal}$ $\text{diameter. The angular speed the coil acquires after rotating by } 60^\circ$ $\text{will be:}$
Options:
  • 1. $20 \text{ rads}^{-1}$
  • 2. $20\pi \text{ rads}^{-1}$
  • 3. $10\pi \text{ rads}^{-1}$
  • 4. $10 \text{ rads}^{-1}$
Solution:
$\text{Hint: Apply conservation of energy.}$ $I_{\text{dia}} = 0.8 \text{ kg/m}^2$ $M = 20 \text{ Am}^2$ $U_i + K_i = U_f + K_f$ $0 + 0 = -MB \cos 30^\circ + \frac{1}{2} I \omega^2$ $20 \times 4 \times \frac{\sqrt{3}}{2} = \frac{1}{2} (0.8) \omega^2$ $\omega = \sqrt{100\sqrt{3}} = 10(3)^{1/4}$

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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
}