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Current Question (ID: 11249)

Question:

$\text{The potential energy of a particle oscillating along the } x\text{-axis is given as } U = 20 + (x - 2)^2 \text{ where } U \text{ is in joules and } x \text{ in metres. The total mechanical energy of the particle is 36 J. The maximum kinetic energy of the particle will be:}$

Options:

1. $24 \text{ J}$
2. $36 \text{ J}$
3. $16 \text{ J}$ (Correct)
4. $20 \text{ J}$

Solution:

$\text{Hint: Total mechanical energy is conserved.}$ $\text{Step 1: Find minimum P.E.}$ $U = 20 + (x - 2)^2$ $U_{\text{min}} = 20 \text{ J}$ $\text{Step 2: Use the equation of mechanical energy to find K.E.}$ $M = U + K = 36$ $\Rightarrow 36 = 20 + K$ $\Rightarrow K = 16 \text{ J}$ $\text{Therefore, the kinetic energy of the particle is 16 J.}$ $\text{Hence, option (3) is the correct answer.}$

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
}