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Current Question (ID: 11300)

Question:
$\text{Which of the following relationships between the acceleration } a \text{ and the displacement } x \text{ of a particle involves simple harmonic motion?}$
Options:
  • 1. $a = 0.7x$
  • 2. $a = -200x^2$
  • 3. $a = -10x$ (Correct)
  • 4. $a = 100x^3$
Solution:
$\text{Hint: For simple harmonic motion } a = -\omega^2 x$ $\text{Explanation: The acceleration of the particle performing SHM is given by,}$ $a = -\omega^2 x \Rightarrow a \propto x$ $\text{Therefore, the acceleration of the particle is performing SHM is } a = -10x.$ $\text{Hence, option (3) is the correct answer.}$

Import JSON File

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
}