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Current Question (ID: 10291)

Question:
$\text{A thin circular ring of mass } M \text{ and radius } R \text{ is rotating in a horizontal plane about an axis vertical to its plane with a constant angular velocity } \omega\text{. If two objects each of mass } m \text{ are attached gently to the opposite ends of the diameter of the ring, the ring will then rotate with an angular velocity:}$
Options:
  • 1. $\frac{\omega(M-2m)}{M+2m}$
  • 2. $\frac{\omega M}{M+2m}$
  • 3. $\frac{\omega(M+2m)}{M}$
  • 4. $\frac{\omega M}{M+m}$
Solution:
$\text{As given in the question that the two objects of equal masses } m \text{ are attached to the opposite ends of the diameter of the ring of mass } M \text{ and the radius } R\text{, then angular momentum will be conserved. According to which we can write:}$ $I_1\omega_1 = I_2\omega_2$ $\Rightarrow MR^2\omega = (MR^2 + 2mR^2)\omega_2$ $\Rightarrow \omega_2 = \frac{MR^2\omega}{(M+2m)R^2} = \frac{M\omega}{M+2m}$

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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
}