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Current Question (ID: 19940)

Question:

$\text{A particle performing simple harmonic motion such that its amplitude is } 4 \text{ m and speed of the particle at the mean position is } 10 \text{ m/s.}$ $\text{Find the distance of the particle from the mean position where velocity becomes } 5 \text{ m/s.}$

Options:

1. $\sqrt{3} \text{ m}$
2. $2\sqrt{3} \text{ m}$
3. $\frac{\sqrt{3}}{2} \text{ m}$
4. $\frac{1}{\sqrt{2}} \text{ m}$

Solution:

$\text{Hint: } v = \omega \sqrt{A^2 - x^2}$ $v = \omega \sqrt{A^2 - x^2}$ $\text{In 1st case: at } x = 0, v = 10 \text{ m/s}$ $\text{then } 10 = \omega \sqrt{(4)^2 - 0^2}$ $\omega = \frac{10}{4} = \frac{5}{2} \text{ rad/s}$ $\text{In 2nd case: } 5 = \frac{5}{2} \sqrt{(4)^2 - x^2}$ $x = 2\sqrt{3} \text{ m}$

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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
}