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Current Question (ID: 19317)

Question:

$\text{Consider a thin, uniform square sheet made of a rigid material. If its side is } a, \text{ mass } m \text{ and moment of inertia } I \text{ about one of its diagonals, then:}$

Options:

1. $I > \frac{ma^2}{12}$
2. $I = \frac{ma^2}{12}$
3. $\frac{ma^2}{24} < I < \frac{ma^2}{12}$
4. $I = \frac{ma^2}{24}$

Solution:

$\text{Hint: } I = \sum m_i r_i^2$ $\text{Step: Find the moment of inertia } (I) \text{ about one of its diagonals.}$ $\text{Moment of inertia along } z \text{ axis, by the perpendicular axis theorem;} \ I_z = I_x + I_y$ $\Rightarrow I_z = \frac{ma^2}{12} + \frac{ma^2}{12} = \frac{ma^2}{6}$ $\text{Now, again applying the perpendicular axis theorem for diagonals say } AC \text{ and } BD.$ $\text{We get; } I_{AC} + I_{BD} = I_z$ $\text{Also, } I_{AC} = I_{BD} = I$ $\text{Therefore,}$ $\Rightarrow 2I = I_z = \frac{ma^2}{6}$ $\Rightarrow I = \frac{ma^2}{12}$ $\text{Hence, option (2) is the correct answer.}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}