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Current Question (ID: 11296)

Question:

$\text{The radius of the circle, the period of revolution, initial position and direction of revolution are indicated in the figure. The } y \text{-projection of the radius vector of rotating particle } P \text{ will be:}$

Options:

1. $y(t) = 3 \text{cos}\left(\frac{\pi t}{2}\right), \text{ where } y \text{ in m}$ (Correct)
2. $y(t) = -3 \text{cos} \text{ } 2\pi t, \text{ where } y \text{ in m}$
3. $y(t) = 4 \text{sin}\left(\frac{\pi t}{2}\right), \text{ where } y \text{ in m}$
4. $y(t) = 3 \text{cos}\left(\frac{3\pi t}{2}\right), \text{ where } y \text{ in m}$

Solution:

$\text{Hint: Angle made by the radius vector with the } y \text{-axis is given by the angular velocity. Step 1: Find the angular velocity of the particle. The period } T = 4 \text{ sec. The angular velocity is given by } \omega = \frac{2\pi}{T} = \frac{2\pi}{4} = \frac{\pi}{2} \text{ rad/sec. Step 2: Find the projection on the } y \text{-axis. From the figure, the initial position is at the top of the circle, which is the maximum positive } y \text{ displacement. This corresponds to a cosine function with zero phase shift. Thus, the equation for the } y \text{-projection is } y = A \text{cos}\omega t \text{, where } A \text{ is the amplitude (radius). Here, } A = 3 \text{ m. So, } y = 3 \text{cos}\left[\frac{\pi t}{2}\right] \text{. Hence, option (1) is the correct answer.}$

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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
}