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Current Question (ID: 10305)

Question:

$\text{A circular platform is mounted on a frictionless vertical axle. Its radius is R = 2m and its moment of inertia about the axle is 200 kg m}^2\text{. Initially, it is at rest. A 50 kg man stands on the edge of the platform and begins to walk along the edge at a speed of 1 m s}^{-1}\text{ relative to the ground. The time taken by the man to complete one revolution is:}$

Options:

1. $\pi \text{ sec}$
2. $\frac{3\pi}{2} \text{ sec}$
3. $2\pi \text{ sec}$
4. $\frac{\pi}{2} \text{ sec}$

Solution:

\text{From conservation of angular momentum: } I\omega = mvr 200 \times \omega = 50 \times 2 \times 1 \omega = \frac{1}{2} \text{ rad/s} v = r\omega = 1 \text{ m/s} \text{Velocity of the man w.r.t. platform } = 1 - (-1) = 2 \text{ m/s} \text{So, } T = \frac{2\pi r}{2} = \frac{2\pi \times 2}{2} = 2\pi \text{ s}

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}