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Current Question (ID: 16820)

Question:
$\text{A charge } Q \text{ is uniformly distributed on a ring of radius } R \text{ made of an insulating material.}$ $\text{If the ring rotates about the axis passing through its centre and normal to the plane of the ring with constant angular speed } \omega,$ $\text{then what will be the magnitude of the magnetic moment of the ring?}$
Options:
  • 1. $Q \omega R^2$
  • 2. $\frac{1}{2} Q \omega R^2$
  • 3. $Q \omega^2 R$
  • 4. $\frac{1}{2} Q \omega^2 R$
Solution:
$M = iA = i \times \pi R^2$ $\text{also } i = \frac{Q \omega}{2 \pi} \Rightarrow M = \frac{1}{2} Q \omega R^2$

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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
}