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Current Question (ID: 19383)

Question:
$\text{Two discs of the same mass, radii } r_1, r_2, \text{ thickness 1 mm and 0.5 mm, have densities in the ratio } 3 : 1. \text{ The ratio of their moment of inertia about their diameter is:}$
Options:
  • 1. $6 : 1$
  • 2. $1 : 6$
  • 3. $3 : 1$
  • 4. $1 : 3$
Solution:
$\text{Mass of both disc is equal:}$ $\text{So, } M_1 = M_2$ $\pi r_1^2 h_1 \rho_1 = \pi r_2^2 h_2 \rho_2$ $r_1^2 \times \frac{h_1}{h_2} \times \frac{\rho_1}{\rho_2} = r_2^2$ $\Rightarrow r_1^2 \times 2 \times \frac{\rho_1}{\rho_2} = r_2^2$ $\Rightarrow \frac{r_1^2}{r_2^2} = \frac{\rho_2}{2 \rho_1} = \frac{1}{6} \left( \therefore \frac{\rho_2}{\rho_1} = \frac{1}{3} \right)$ $\text{Ratio of MOI: } \frac{1}{4} M r_1^2 = \frac{1}{4} M r_2^2$ $\frac{r_1^2}{r_2^2} = \frac{1}{6}$

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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
}