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Current Question (ID: 19378)

Question:

$\text{A disk and a solid sphere (dark boundary) of the same radius are rotated as shown in the figure.}$ $\text{If masses of disk and solid sphere are 4 kg and 5 kg respectively then,}$ \frac{I_{\text{disc}}}{I_{\text{solid sphere}}} = $

Options:

1. \frac{7}{5}
2. \frac{25}{28}
3. \frac{5}{7}
4. \frac{28}{25}

Solution:

$\text{Hint: Use the parallel axis theorem.}$ $\text{Step 1: Find the moment of inertia of the disc.}$ $I_{\text{disc}} = \frac{1}{4} m R^2 + m R^2$ $\Rightarrow I_{\text{disc}} = \frac{5}{4} m R^2$ $\Rightarrow I_{\text{disc}} = 5 R^2 \quad [m_{\text{disc}} = 4 \text{ kg}]$ $\text{Step 2: Find the moment of inertia of a solid sphere.}$ $I_{\text{solid sphere}} = \frac{2}{5} m R^2 + m R^2$ $\Rightarrow I_{\text{solid sphere}} = \frac{7}{5} m R^2$ $\Rightarrow I_{\text{solid sphere}} = 7 R^2 \quad [m_{\text{solid sphere}} = 5 \text{ kg}]$ $\text{The required ratio of their moments of inertia is given by:}$ \frac{I_{\text{disc}}}{I_{\text{solid sphere}}} = \frac{5 R^2}{7 R^2} = \frac{5}{7}$ $\text{Hence, option (3) is the correct answer.}$

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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
}