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Current Question (ID: 19960)

Question:
$\text{In figure (A), mass '2 m' is fixed on mass 'm' which is attached to two springs of spring constant } k. \text{ In figure (B), mass 'm' is attached to two springs of spring constant } k \text{ and } 2k. \text{ If mass 'm' in (A) and (B) are displaced by distance 'x' horizontally and then released, then time period } T_1 \text{ and } T_2 \text{ corresponding to (A) and (B) respectively follow the relation}$
Options:
  • 1. $\frac{T_1}{T_2} = \frac{3}{\sqrt{2}}$
  • 2. $\frac{T_1}{T_2} = \sqrt{\frac{3}{2}}$
  • 3. $\frac{T_1}{T_2} = \sqrt{\frac{2}{3}}$
  • 4. $\frac{T_1}{T_2} = \frac{\sqrt{2}}{3}$
Solution:
$\text{For figure (A), the effective spring constant is } k_\text{eff} = k + k = 2k. \text{ The time period } T_1 = 2\pi \sqrt{\frac{m}{2k}}.$ $\text{For figure (B), the effective spring constant is } k_\text{eff} = k + 2k = 3k. \text{ The time period } T_2 = 2\pi \sqrt{\frac{m}{3k}}.$ $\text{Therefore, } \frac{T_1}{T_2} = \frac{\sqrt{\frac{m}{2k}}}{\sqrt{\frac{m}{3k}}} = \sqrt{\frac{3}{2}}.$

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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
}