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Current Question (ID: 10262)

Question:

$\text{Three identical spherical shells, each of mass m and radius r, are placed as shown in the figure. Consider an axis XX', which is touching two shells and passes through the diameter of the third shell. The moment of inertia of the system consisting of these three spherical shells about the XX' axis is:}$

Options:

1. $\frac{11}{5}mr^2$
2. $3mr^2$
3. $\frac{16}{5}mr^2$
4. $4mr^2$

Solution:

$\text{The total moment of inertia of the system is}$ $I = I_1 + I_2 + I_3 \text{ ...(i)}$ $\text{Here, } I_1 = \frac{2}{3}mr^2$ $I_2 = I_3 = \frac{2}{3}mr^2 + mr^2 \text{ [From parallel axis theorem]}$ $= \frac{5}{3}mr^2$ $\text{From Eq. (i)}$ $I = \frac{2}{3}mr^2 + 2 \times \frac{5}{3}mr^2 = mr^2(\frac{2}{3} + \frac{10}{3})$ $I = 4mr^2$

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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
}