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Current Question (ID: 19377)

Question:

$\text{Let } I_{cm} \text{ be the moment of Inertia of the disc passing through centre}$ $\text{and perpendicular to its plane. } I_{AB} \text{ be the moment of inertia about}$ $\text{axis } AB \text{ that is in the plane of the disc and } \frac{2r}{3} \text{ distance from centre.}$ $\text{Then } \frac{I_{cm}}{I_{AB}} =$

Options:

1. $\frac{1}{4}$
2. $\frac{18}{25}$
3. $\frac{17}{9}$
4. $\frac{1}{2}$

Solution:

$\text{Hint: } I = \sum m_i r_i^2$ $\text{Step 1: Find the ratio of the moment of inertia.}$ $\text{The moment of inertia of the disc about its center is given by}$ $I_{cm} = \frac{1}{2}mr^2 \quad \cdots (1)$ $\text{Step 2: The moment of inertia of the disc about perpendicular to its}$ $\text{plane is given by}$ $\text{Using the parallel axis theorem states } I_{AB} = I_{cm} + md^2$ $I_{AB} = \frac{1}{4}mr^2 + m\left(\frac{2r}{3}\right)^2 \Rightarrow \frac{25}{36}mr^2 \quad \cdots (2)$ $\text{Therefore, from equations (1) and (2) we will get}$ $\frac{I_{cm}}{I_{AB}} = \frac{\frac{1}{2}mr^2}{\frac{25}{36}mr^2} \Rightarrow \frac{18}{25}$ $\text{Hence, option (2) is the correct answer.}$

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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
}