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Current Question (ID: 19425)

Question:

$\text{For a uniform disc, the moment of inertia about diameter is } \frac{MR^2}{4},$ $\text{where } M \text{ is mass and } R \text{ is radius of the disc.}$ $\text{The moment of inertia about tangent parallel to diameter is:}$

Options:

1. $\frac{3}{4}MR^2$
2. $\frac{5}{4}MR^2$
3. $\frac{3}{2}MR^2$
4. $\frac{5}{2}MR^2$

Solution:

$\text{Hint: Use the parallel axis theorem.}$ $\text{Step: Find the moment of inertia about the given axis.}$ $\text{To calculate the moment of inertia of the disc about a tangent}$ $\text{parallel to the diameter, we use the parallel axis theorem, which}$ $\text{states:}$ $I_{\text{tangent}} = I_{\text{diameter}} + Md^2$ $\Rightarrow I_{\text{tangent}} = \frac{MR^2}{4} + MR^2$ $\Rightarrow I_{\text{tangent}} = \frac{5MR^2}{4}$ $\text{Hence, option (2) is the correct answer.}$

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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
}