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Current Question (ID: 19320)

Question:

$\text{Seven identical circular planar disks, each of mass } M \text{ and radius } R \text{ are welded symmetrically as shown.}$ $\text{The moment of inertia of the arrangement about the axis normal to the plane and passing through the point } P \text{ is:}$

Options:

1. $\frac{19}{2} MR^2$
2. $\frac{55}{2} MR^2$
3. $\frac{73}{2} MR^2$
4. $\frac{181}{2} MR^2$

Solution:

$\text{Hint: Use parallel axis theorem.}$ $I_{\text{about } O} = \frac{MR^2}{2} + \left[ \frac{MR^2}{2} + M4R^2 \right] \times 6$ $= \frac{MR^2}{2} + \frac{9MR^2}{2} \times 6$ $= \frac{55}{2} MR^2$ $I_{\text{about } P} = \frac{55}{2} MR^2 + 7M \left( 9R^2 \right)$ $= \left( \frac{55 + 63 \times 2}{2} \right) MR^2$ $= \frac{181 MR^2}{2}$

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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
}