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Current Question (ID: 10264)

Question:
$\text{Four identical solid spheres each of mass 'm' and radius 'a' are placed with their centres on the four corners of a square of side 'b'. The moment of inertia of the system about one side of the square where the axis of rotation is parallel to the plane of the square is:}$
Options:
  • 1. $\frac{4}{5}ma^2 + 2mb^2$
  • 2. $\frac{8}{5}ma^2 + mb^2$
  • 3. $\frac{8}{5}ma^2 + 2mb^2$
  • 4. $\frac{4}{5}ma^2$
Solution:
$\text{Hint: Use parallel axis theorem}$ $\text{Step 1: Draw the diagram}$ $\text{Step 2: Use parallel axis theorem}$ $I = 2\left(\frac{2}{5}ma^2\right) + 2\left[\frac{2}{5}ma^2 + mb^2\right]$ $= \frac{8}{5}ma^2 + 2mb^2$

Import JSON File

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
}