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Current Question (ID: 19415)

Question:

$\text{A solid sphere of mass } M \text{ and radius } R \text{ is divided into two unequal parts.}$ $\text{The first part has a mass of } \frac{7M}{8} \text{ and is converted into a uniform disc of radius } 2R.$ $\text{The second part is converted into a uniform solid sphere.}$ $\text{Let } I_1 \text{ be the moment of inertia of the disc about its axis and } I_2 \text{ be the moment of inertia of the new sphere about its axis.}$ $\text{The ratio } \frac{I_1}{I_2} \text{ is given by:}$

Options:

1. 65
2. 140
3. 185
4. 285

Solution:

$\text{Hint: Recall the moment of inertia of a solid sphere and disc.}$ $I_1 = \left(\frac{7M}{8}\right) \left(2R^2\right) = \frac{7}{4}MR^2$ $I_2 = \frac{2}{5} \left(\frac{M}{8}\right) R_1^2 = \frac{MR_1^2}{80}$ $R = 2R_1$ $\frac{I_1}{I_2} = \frac{7}{4} \times 80 = 140$

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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
}