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Current Question (ID: 19339)

Question:

$\text{Moment of inertia of a cylinder of mass } M, \text{ length } L \text{ and radius } R \text{ about an axis passing through its centre and perpendicular to the axis of the cylinder is } I = M \left( \frac{R^2}{4} + \frac{L^2}{12} \right).$ $\text{If such a cylinder is to be made for a given mass of a material, the ratio } \frac{L}{R} \text{ for it to have minimum possible } I \text{ is:}$

Options:

1. $\frac{2}{3}$
2. $\frac{3}{2}$
3. $\sqrt{\frac{2}{3}}$
4. $\sqrt{\frac{3}{2}}$

Solution:

$\text{Hint: For } I \text{ to be minimum, } \frac{dI}{dL} = 0.$ $I = M \left( \frac{R^2}{4} + \frac{L^2}{12} \right) \quad \text{(1)}$ $\text{as mass is constant } = m = pV = \text{constant}$ $V = \text{constant}$ $\pi^2 Rl = \text{constant} = R^2 L = \text{constant}$ $2RL + R^2 \frac{dL}{dR} = 0$ $\text{From equation (1)}$ $\frac{dI}{dR} = M \left( \frac{2R}{4} + \frac{2L}{12} \times \frac{dL}{dR} \right) = 0$ $\frac{R}{2} + \frac{L}{6} \frac{dL}{dR}$ $\text{Substituting value of } \frac{dL}{dR} \text{ from equation}$ $\frac{R}{2} + \frac{L}{6} \left( -\frac{2L}{R} \right) = 0$ $\frac{R}{2} + \frac{L^2}{3R} = \frac{L}{R} = \sqrt{\frac{3}{2}}$

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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
}