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Current Question (ID: 19352)

Question:

$\text{The moments of inertia of four bodies, all having the same mass and radius, are reported as follows:}$ $I_1 \text{ Moment of inertia of thin circular ring about its diameter}$ $I_2 \text{ Moment of inertia of circular disc about an axis perpendicular to the disc and going through the centre}$ $I_3 \text{ Moment of inertia of solid cylinder about its axis}$ $I_4 \text{ Moment of inertia of solid sphere about its diameter}$ $\text{Which of the following relationships is correct?}$

Options:

1. $I_1 + I_3 < I_2 + I_4$
2. $I_1 + I_2 = I_3 + \frac{5}{2} I_4$
3. $I_1 = I_2 = I_3 > I_4$
4. $I_1 = I_2 = I_3 < I_4$

Solution:

$\text{Hint: Recall the perpendicular axis theorem.}$ $\text{Step 1: Find the moment of inertia of each case.}$ $I_1 = \frac{I_z}{2} = \frac{MR^2}{2} \quad [I_z = MR^2]$ $I_2 = \frac{MR^2}{2}$ $I_3 = I_z = \frac{1}{2} MR^2$ $I_4 = \frac{I_z}{2} = \frac{\frac{2}{5} MR^2}{2} = \frac{MR^2}{5}$ $\text{Step 2: Compare the moment of inertia of different cases.}$ $\text{By comparing the order of different moment inertia we get;}$ $I_1 = I_2 = I_3 > I_4$ $\text{Hence, option (3) is the correct answer.}$

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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
}