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Current Question (ID: 19927)

Question:
$\text{For a particle performing simple harmonic motion, the maximum potential energy is } 25 \text{ J.}$ $\text{What is the kinetic energy of the particle when it is at half of its amplitude?}$
Options:
  • 1. $18.75 \text{ kJ}$
  • 2. $18.75 \text{ J}$
  • 3. $9.45 \text{ kJ}$
  • 4. $9.45 \text{ J}$
Solution:
$\text{Hint: Maximum potential energy } = \frac{1}{2} k A^2$ $\text{Step 1: Find the maximum potential energy.}$ $\text{The maximum potential energy of SHM with amplitude } A \text{ is given by;}$ $P. E_{\text{max}} = \frac{1}{2} k A^2 = 25 \text{ J} \quad \ldots (1)$ $\text{Step 2: Find the kinetic energy at the given point.}$ $K. E = \frac{1}{2} k (A^2 - x^2)$ $\text{at } x = \frac{A}{2}$ $K. E = \frac{1}{2} k \left(A^2 - \frac{A^2}{4}\right)$ $\Rightarrow K. E = \frac{3kA^2}{8} \quad \ldots (2)$ $\text{From equation (1) \& (2), we get;}$ $K. E = \frac{75}{4} = 18.75 \text{ J}$ $\text{Hence, option (2) is the correct answer.}$

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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
}