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Current Question (ID: 19905)

Question:
$\text{In the given figure, a body of mass } M \text{ is held between two massless springs, on a smooth inclined plane.}$ $\text{The free ends of the springs are attached to firm supports.}$ $\text{If each spring has spring constant } k, \text{ the frequency of oscillation of given body is:}$
Options:
  • 1. $\frac{1}{2\pi} \sqrt{\frac{k}{2M}}$
  • 2. $\frac{1}{2\pi} \sqrt{\frac{2k}{Mg \sin \alpha}}$
  • 3. $\frac{1}{2\pi} \sqrt{\frac{2k}{M}}$
  • 4. $\frac{1}{2\pi} \sqrt{\frac{k}{Mg \sin \alpha}}$
Solution:
$\text{Hint: } k_{\text{eq}} = k_1 + k_2$ $\text{Step: Find the frequency of oscillation of given body.}$ $k_{\text{eq}} = k_1 + k_2 = k + k = 2k$ $T = 2\pi \sqrt{\frac{m}{k_{\text{eq}}}} = 2\pi \sqrt{\frac{m}{2k}}$ $f = \frac{1}{T} = \frac{1}{2\pi} \sqrt{\frac{2k}{m}}$ $\text{Hence, option (3) is the correct answer.}$

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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
}