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Current Question (ID: 19959)

Question:

$\text{The equation of a particle executing simple harmonic motion is given by } x = \sin \pi \left( t + \frac{1}{3} \right) \text{ m. At } t = 1 \text{ s, the speed of the particle will be:}$ $\text{(Given: } \pi = 3.14)$

Options:

1. $0 \text{ cm s}^{-1}$
2. $157 \text{ cm s}^{-1}$
3. $272 \text{ cm s}^{-1}$
4. $314 \text{ cm s}^{-1}$

Solution:

$V = \frac{dx}{dt} = \pi \cos \pi \left( t + \frac{1}{3} \right)$ $V = \pi \cos \pi \left( 1 + \frac{1}{3} \right)$ $V = \pi \cos \frac{4\pi}{3}$ $V = \pi \times \left( -\frac{1}{2} \right)$ $V = -\frac{\pi}{2}$ $V = -\frac{3.14}{2}$ $V = -1.57 \text{ m/s}$ $V = -157 \text{ cm/s}$

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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
}