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Current Question (ID: 10150)

Question:

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

Options:

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

Solution:

\text{Hint: MOI of the disc } = \frac{MR^2}{2} \text{Step 1: Find the ratio of masses of the discs.} M = \rho v = \rho \pi r^2 t \Rightarrow M \propto r^2 \frac{m_1}{m_2} = \left(\frac{r_1}{r_2}\right)^2 \text{Step 2: Find the ratio of MOI.} \frac{I_1}{I_2} = \frac{m_1 r_1^2}{m_2 r_2^2} = \left(\frac{r_1}{r_2}\right)^4 \Rightarrow \left(\frac{r_1}{r_2}\right)^4 = \frac{1}{16} \Rightarrow \left(\frac{1}{\alpha}\right)^4 = \frac{1}{16} \Rightarrow \alpha = 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
}