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Current Question (ID: 11250)

Question:

$\text{A particle executing simple harmonic motion has a kinetic energy of } K_0 \cos^2(\omega t). \text{ The values of the maximum potential energy and the total energy are, respectively:}$

Options:

1. $0 \text{ and } 2K_0$
2. $\frac{K_0}{2} \text{ and } K_0$
3. $K_0 \text{ and } 2K_0$
4. $K_0 \text{ and } K_0$ (Correct)

Solution:

$\text{Hint: At the extreme position } P.E = K_0$ $\text{Explanation: The kinetic energy of the SHM is given as } K.E = K_0 \cos^2(\omega t) \text{ and the maximum kinetic energy is given by; } K_0$ $\text{When potential energy is maximum at extreme points Kinetic energy is zero. Therefore due to the conservation of energy } P.E = K_0. \text{ Therefore, at the extreme point } P.E = K_0 \text{ and } K.E = 0 \text{ and total energy } = K_0.$ $\text{Hence, option (4) is the correct answer.}$

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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
}