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Current Question (ID: 10148)

Question:
$\text{A uniform rod of mass } 2M \text{ is bent into four adjacent semicircles, each of radius } r\text{, all lying in the same plane. The moment of inertia of the bent rod about an axis through one end A and perpendicular to the plane of the rod is:}$
Options:
  • 1. $22Mr^2$
  • 2. $88Mr^2$
  • 3. $44Mr^2$
  • 4. $66Mr^2$
Solution:
$x = \frac{2r}{\pi}, m = \frac{2M}{4} = \frac{M}{2}$ $\text{Moment of inertia due to first semicircle}$ $I_1 = I_{cm} + mx^2 \text{ But } I_1 = mr^2$ $I_{A1} = I_{cm} + m(x^2 + r^2) = mr^2 - mx^2 + m(x^2 + r^2) = 2mr^2$ $\text{Similarly}$ $I_{A2} = mr^2 + m(3r)^2 = 10mr^2$ $I_{A3} = mr^2 + m(5r)^2 = 26mr^2$ $I_{A4} = mr^2 + m(7r)^2 = 50mr^2$ $I_A = 2mr^2 + 10mr^2 + 26mr^2 + 50mr^2 = 88mr^2 = 44Mr^2$

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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
}