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Current Question (ID: 11247)

Question:

$\text{The displacement between the maximum potential energy position and maximum kinetic energy position for a particle executing simple harmonic motion is:}$

Options:

1. $\pm \frac{a}{2}$
2. $+a$
3. $\pm a$ (Correct)
4. $-1$

Solution:

$\text{Hint: } P.E = \frac{1}{2}Kx^2$ $\text{Step: Find the displacement between the maximum potential energy position and maximum kinetic energy position.}$ $\text{The variation of the potential and kinetic energy at different position are shown below;}$ $\text{Extreme position: } x = -a, \text{ P.E. (max)}$ $\text{Mean position: } x = 0, \text{ K.E. (Max.)}$ $\text{Extreme position: } x = +a, \text{ (P.E.) Max.}$ $\text{Therefore, the maximum displacement is } \pm a.$ $\text{Hence, option (3) is the correct answer.}$

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
}