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Current Question (ID: 11278)

Question:
$\text{A spring having a spring constant of 1200 N/m is mounted on a horizontal table as shown in the figure. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released. The frequency of oscillations will be:}$
Options:
  • 1. $3.0 \text{ s}^{-1}$
  • 2. $2.7 \text{ s}^{-1}$
  • 3. $1.2 \text{ s}^{-1}$
  • 4. $3.2 \text{ s}^{-1}$ (Correct)
Solution:
$\text{Hint: The amplitude is simply the maximum displacement of the object from the equilibrium position.}$ $\text{Here A=2cm.}$ $\text{Step 1: Angular frequency is given as } \omega = \sqrt{\frac{k}{m}}.$ $\text{Given,}$ $k = 1200 \text{ N/m}$ $\text{and } m = 3 \text{ kg}$ $\text{So,}$ $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{1200}{3}} = 20 \text{ rad/s}$ $\text{Step 2: Use } \omega = 2\pi f$ $\Rightarrow f = \frac{\omega}{2\pi} = \frac{20}{2\pi} \approx 3.2 \text{ s}^{-1}$

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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
}