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Current Question (ID: 19934)

Question:
$\text{For a particle undergoing linear simple harmonic motion (SHM), the graph showing the variation of kinetic energy, } K \text{ with position, } x \text{ of the particle is:}$
Options:
  • 1. $\text{Graph 1}$
  • 2. $\text{Graph 2}$
  • 3. $\text{Graph 3}$
  • 4. $\text{Graph 4}$
Solution:
$\text{Hint: } K = \frac{1}{2} m \omega^2 \left( A^2 - x^2 \right)$ $\text{Step: Identify the correct graph between kinetic energy } (K) \text{ and position } (x).$ $\text{The kinetic energy of a particle undergoing SHM is given by:}$ $K = \frac{1}{2} m \omega^2 \left( A^2 - x^2 \right)$ $K \propto -x^2$ $\text{Therefore, the graph of } K \text{ versus } x \text{ will be a downward parabola.}$ $\text{The kinetic energy is zero at the extreme positions and maximum at the mean position as shown in the figure below;}$ $\text{Hence, option (4) is the correct answer.}$

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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
}