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Current Question (ID: 11272)

Question:

$\text{A particle is attached to a vertical spring and pulled down a distance of 0.01 m below its mean position and released. If its initial acceleration is } 0.16 \text{ m/s}^2\text{, then its time period in seconds will be:}$

Options:

1. $\pi$
2. $\frac{\pi}{2}$ (Correct)
3. $\frac{\pi}{4}$
4. $2\pi$

Solution:

$\text{Hint: The amplitude is the maximum displacement of the object from the equilibrium position.}$ $\text{Step 1: Find the amplitude of the particle.}$ $\text{Here the maximum displacement from mean position = 0.01 m}$ $\text{Amplitude = 0.01 m}$ $\text{Step 2: Find the angular frequency of the particle.}$ $\text{The maximum acceleration of the particle is given as,}$ $a_{\text{max}} = A\omega^2$ $\Rightarrow 0.16 = 0.01\omega^2$ $\Rightarrow \omega = 4 \text{ rad/s}$ $\text{Step 3: Calculate the time period.}$ $\text{Now, } T = \frac{2\pi}{\omega}$ $\Rightarrow T = \frac{\pi}{2} \text{ s}$

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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
}