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Current Question (ID: 19343)

Question:

$\text{Consider two uniform discs of the same thickness and different radii } R_1 = R \text{ and } R_2 = \alpha R \text{ made of the same material.}$ $\text{If the ratio of their moments of inertia } I_1 \text{ and } I_2, \text{ respectively, about their axes is } I_1 : I_2 = 1 : 16 \text{ then the value of } \alpha \text{ is:}$

Options:

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

Solution:

$\text{Hint: } I = \frac{MR^2}{2}$ $\frac{I_1}{I_2} = \frac{\frac{MR^2}{2}}{\frac{\pi R^2 t \cdot R^2}{2}} \propto \frac{R_1^4}{R_2^4}$ $\frac{I_1}{I_2} = \frac{R_1^4}{R_2^4} = \frac{1}{16}$ $\therefore \frac{R_1}{R_2} = \frac{1}{2}$

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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
}