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Current Question (ID: 10311)

Question:
$\text{A string is wrapped along the rim of a wheel of the moment of inertia 0.10 kg-m}^2 \text{ and radius 10 cm. If the string is now pulled by a force of 10 N, then the wheel starts to rotate about its axis from rest. The angular velocity of the wheel after 2 s will be:}$
Options:
  • 1. $40 \text{ rad/s}$
  • 2. $80 \text{ rad/s}$
  • 3. $10 \text{ rad/s}$
  • 4. $20 \text{ rad/s}$
Solution:
$\text{Hint: } \tau = I\alpha$ $\text{Step 1: Find the angular acceleration of the wheel.}$ $\alpha = \frac{\tau}{I}$ $\tau = r \times F = 10 \times 0.1 = 1 \text{ N-m}$ $\alpha = \frac{1}{0.1} = 10 \text{ rad/s}^2$ $\text{Step 2: Find the angular velocity of the wheel.}$ $\omega = \omega_0 + \alpha t$ $= 0 + 10 \times 2$ $= 20 \text{ rad/s}$ $\text{Hence, option (4) is the correct answer.}$

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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
}