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Current Question (ID: 19316)

Question:

$\text{From a solid sphere of mass } M \text{ and radius } R \text{ a cube of maximum possible volume is cut.}$ $\text{The moment of inertia of a cube about an axis passing through its center and perpendicular to one of its faces is:}$

Options:

1. $\frac{MR^2}{32\sqrt{2\pi}}$
2. $\frac{MR^2}{16\sqrt{2\pi}}$
3. $\frac{4MR^2}{9\sqrt{3\pi}}$
4. $\frac{4MR^2}{3\sqrt{3\pi}}$

Solution:

$\text{Hint: } I = \int dm r^2$ $\therefore \text{ For cube of maximum volume distance from the centre of the sphere to the corner of cube is equal to } R.$ $\text{By figure}$ $\sqrt{3}a = 2R$ $a = \frac{2R}{\sqrt{3}} \text{ (side of cube)}$ $\text{mass of cube}$ $M' = \frac{M}{\frac{4}{3}\pi R^3} \left[ \frac{8R^3}{3\sqrt{3}} \right] \quad (\because M' = \rho a^3)$ $M' = \frac{8M}{4\sqrt{3\pi}}$ $\therefore I = MI \text{ of cube}$ $I = \frac{1}{6} M'a^2$ $= \frac{1}{6} \frac{8M}{4\sqrt{3\pi}} \left( \frac{2r}{\sqrt{3}} \right)^2$ $I = \frac{4MR^2}{9\sqrt{3\pi}}$

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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
}