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Current Question (ID: 10143)

Question:

$\text{A thin wire of length } L \text{ and uniform linear mass density } \rho \text{ is bent into a circular loop with the centre at } O \text{ as shown. The moment of inertia of the loop about the axis } XX' \text{ is:}$

Options:

1. $\frac{\rho L^3}{8\pi^2}$
2. $\frac{\rho L^3}{16\pi^2}$
3. $\frac{5\rho L^3}{16\pi^2}$
4. $\frac{3\rho L^3}{8\pi^2}$

Solution:

\text{Mass per unit length of the wire} = \rho \text{Mass of L length, } M = \rho L \text{Since the wire is bent into a circular loop: } 2\pi R = L \text{Therefore: } R = \frac{L}{2\pi} \text{Moment of inertia of loop about given axis} = \frac{3}{2}MR^2 = \frac{3}{2}\rho L \times \left(\frac{L}{2\pi}\right)^2 = \frac{3\rho L^3}{8\pi^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
}