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Current Question (ID: 19413)

Question:

$\text{A thin circular plate of mass } M \text{ and radius } R \text{ has its density varying as } \rho(r) = \rho_0 r \text{ with } \rho_0 \text{ as constant and } r \text{ is the distance from its center.}$ $\text{The moment of inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is } I = \alpha MR^2. \text{ The value of the coefficient } \alpha \text{ is:}$

Options:

1. $\frac{3}{2}$
2. $\frac{1}{2}$
3. $\frac{8}{5}$
4. $\frac{3}{5}$

Solution:

$\text{Hint: } I = \int r^2 dm$ $M = \int_0^R (\rho_0 r) (2 \pi r dr) = 2 \pi \rho_0 \frac{R^3}{3}$ $I = MR^2 + \int_0^R 2 \pi \rho_0 r^4 dr = MR^2 + \frac{2 \pi \rho_0}{5} R^5$ $I = MR^2 + \frac{3}{5} MR^2 = \frac{8}{5} MR^2$

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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
}