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Current Question (ID: 19923)

Question:

$\text{A particle undergoing SHM, follows the position-time equation given as } x = A \sin(\omega t + \pi/3).$ $\text{If the SHM motion has a time period of } T,$ $\text{for the first time after } t = 0, \text{ the velocity will be maximum at a time,}$

Options:

1. $\frac{T}{2}$
2. $\frac{T}{3}$
3. $\frac{3T}{2}$
4. $\frac{2T}{3}$

Solution:

$x = A \sin \left( \omega t + \frac{\pi}{3} \right)$ $v = A \omega \cos \left( \omega t + \frac{\pi}{3} \right)$ $\cos \left( \omega t + \frac{\pi}{3} \right) = \pm 1$ $\Rightarrow \omega t + \frac{\pi}{3} = \pi$ $\Rightarrow \omega t = \frac{2\pi}{3}$ $\Rightarrow t = \frac{T}{3}$

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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
}