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Current Question (ID: 11248)

Question:

$\text{The kinetic energy of a particle executing SHM is 16 J when it is in its mean position. If the amplitude of oscillations is 25 cm and the mass of the particle is 5.12 kg, the time period of its oscillation will be:}$

Options:

1. $\frac{\pi}{5} \text{ s}$ (Correct)
2. $2\pi \text{ s}$
3. $20\pi \text{ s}$
4. $5\pi \text{ s}$

Solution:

$\text{Hint: } K.E = \frac{1}{2}m\omega^2(a^2 - x^2)$ $\text{Step: Find the value of the time period of SHM.}$ $\text{At the mean position, the kinetic energy is maximum.}$ $K.E = \frac{1}{2}m\omega^2 a^2 = 16 \text{ J}$ $\Rightarrow \omega^2 = \frac{32}{ma^2} = \frac{32}{5.12 \times 0.25} = 25$ $\Rightarrow \omega = 5 \text{ rad/s}$ $\text{The time period of SHM is given by;}$ $T = \frac{2\pi}{\omega} = \frac{2\pi}{10} = \frac{\pi}{5} \text{ s}$ $\text{Hence, option (1) is the correct answer.}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}