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Current Question (ID: 19937)

Question:
$\text{A particle is performing SHM having position } x = A \cos(30^\circ), \text{ and } A = 40 \text{ cm.}$ $\text{If its kinetic energy at this position is } 200 \text{ J, then the value of force constant is:}$
Options:
  • 1. $10 \text{ kN/m}$
  • 2. $20 \text{ kN/m}$
  • 3. $10000 \text{ kN/m}$
  • 4. $20000 \text{ kN/m}$
Solution:
$\text{Hint: } \omega^2 = \frac{K}{m}$ $\text{Step: Find the value of the force constant.}$ $K = \frac{1}{2} k(A^2 - x^2)$ $\Rightarrow k = \frac{2K}{(A^2 - x^2)} = \frac{2 \times 200}{A^2 - (A \cos 30^\circ)^2}$ $\Rightarrow k = \frac{2 \times 200}{A^2(1 - 3/4)}$ $\Rightarrow k = \frac{2 \times 200}{0.4^2(1 - 3/4)}$ $\Rightarrow k = 10000 \text{ N/m}$ $\Rightarrow k = 10 \text{ kN/m}$ $\text{Hence, option (1) is the correct answer.}$

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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
}