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Current Question (ID: 19327)

Question:
$\text{A thin smooth rod of length } L \text{ and mass } M \text{ is rotating freely with angular speed } \omega_0 \text{ about an axis perpendicular to the rod and passing through its center.}$ $\text{Two beads of mass } m \text{ and negligible size are at the center of the rod initially.}$ $\text{The beads are free to slide along the rod.}$ $\text{The angular speed of the system, when the beads reach the opposite ends of the rod, will be:}$
Options:
  • 1. $\frac{M \omega_0}{M + 3m}$
  • 2. $\frac{M \omega_0}{M + 2m}$
  • 3. $\frac{M \omega_0}{M + m}$
  • 4. $\frac{M \omega_0}{M + 6m}$
Solution:
$\text{Hint: Apply conservation of angular momentum}$ $\text{Step 1: Find initial angular momentum}$ $L_i = \left( \frac{ML^2}{12} \right) \omega_0$ $\text{Step 2: Calculate the final angular momentum}$ $L_f = \left[ \frac{ML^2}{12} + 2m \left( \frac{L}{2} \right)^2 \right] \omega$ $\text{Step 3: Conserve the angular momentum}$ $L_i = L_f$ $\frac{ML^2}{12} \omega_0 = \left( \frac{ML^2}{12} + \frac{mL^2}{2} \right) \omega$ $\omega = \frac{M \omega_0}{M + 6m}$

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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
}