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Current Question (ID: 10245)

Question:
$\text{The angular speed of the wheel of a vehicle is increased from 360 rpm to 1200 rpm in 14 seconds. Its angular acceleration will be:}$
Options:
  • 1. $2\pi \text{ rad/s}^2$
  • 2. $28\pi \text{ rad/s}^2$
  • 3. $120\pi \text{ rad/s}^2$
  • 4. $1 \text{ rad/s}^2$
Solution:
\text{Hint: Use the equation of angular motion.} \text{Step 1: Identify the given information.} \text{Time, } t = 14 \text{ sec} \text{Initial angular velocity, } \omega_0 = 2\pi \times \frac{360}{60} = 12\pi \text{ rad/s} \text{Final angular velocity, } \omega = 2\pi \times \frac{1200}{60} = 40\pi \text{ rad/s} \text{Step 2: Use the first equation of angular motion.} \omega = \omega_0 + \alpha t \alpha = \frac{\omega - \omega_0}{t} = \frac{40\pi - 12\pi}{14} = \frac{28\pi}{14} = 2\pi \text{ rad/s}^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
}