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Current Question (ID: 19903)

Question:
$\text{When a particle of mass } m \text{ is attached to a vertical spring of spring constant } k \text{ and released, its motion is described by}$ $y(t) = y_0 \sin^2 \omega t, \text{ where 'y' is measured from the lower end of the unstretched spring. Then } \omega \text{ is:}$
Options:
  • 1. $\sqrt{\frac{g}{y_0}}$
  • 2. $\sqrt{\frac{g}{2y_0}}$
  • 3. $\frac{1}{2} \sqrt{\frac{g}{y_0}}$
  • 4. $\sqrt{\frac{2g}{y_0}}$
Solution:
$\text{Hint: } y = y_0 \sin^2 \omega t = \frac{y_0}{2} (1 - \cos 2\omega t).$ $y = y_0 \sin^2 \omega t$ $y = \frac{y_0}{2} (1 - \cos 2\omega t)$ $y = \frac{y_0}{2} - \frac{y_0}{2} \cos 2\omega t$ $\text{Amplitude: } \frac{y_0}{2}$ $\frac{y_0}{2} = \frac{mg}{K}$ $2\omega = \sqrt{\frac{K}{m}} = \sqrt{\frac{2g}{y_0}}$ $\omega = \sqrt{\frac{g}{2y_0}}$ $\text{Therefore Ans is 2}$

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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
}