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Current Question (ID: 19328)

Question:

$\text{The moment of inertia of a body about a given axis is } 1.5 \text{ kg-m}^2.$ $\text{Initially, the body is at rest. In order to produce rotational kinetic}$ $\text{energy of } 1200 \text{ J, the angular acceleration of } 20 \text{ rad/s}^2 \text{ must be}$ $\text{applied about the axis for a duration of:}$

Options:

1. $5 \text{ s}$
2. $3 \text{ s}$
3. $2.5 \text{ s}$
4. $2 \text{ s}$

Solution:

$\text{Hint: } (KE)_R = \frac{1}{2} I \omega^2$ $\text{Step 1: Find the final angular velocity}$ $\frac{1}{2} I \omega^2 = 1200$ $\omega^2 = 1600$ $\omega = 40 \text{ rad/s}$ $\text{Step 2: Use equation of motion}$ $\omega = \omega_0 + \alpha t$ $40 = 0 + 20t$ $\Rightarrow \ t = 2 \text{ s.}$

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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
}