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Current Question (ID: 19356)

Question:

$\text{As shown in the figure, four identical solid spheres, each of mass } m \text{ and radius } a, \text{ are placed at the corners of a square of side length } b. \text{ What is the moment of inertia of the system about a vertical axis that lies along one side of the square (i.e., passes through the centres of the two adjacent spheres on that side)?}$

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:

$\text{Hint: Use parallel axis theorem.}$ $\text{Step 1: Sketch the axis of rotation of the system.}$ $\text{Step 2: Find the moment of inertia of the system about one side of the square.}$ $I = 2 \left( \frac{2}{5}ma^2 \right) + 2 \left( \frac{2}{5}ma^2 + mb^2 \right)$ $\Rightarrow I = \frac{8}{5}ma^2 + 2mb^2$ $\text{Hence, option (3) is the correct answer.}$

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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
}