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Current Question (ID: 10323)

Question:
$\text{A solid body rotates about a stationary axis according to the equation } \theta = 6t - 2t^3. \text{ What is the average angular velocity over the time interval between } t = 0 \text{ and the time when the body comes to rest? } (\theta: \text{ angular displacements, } t: \text{ time})$
Options:
  • 1. $1 \text{ rad/s}$
  • 2. $2 \text{ rad/s}$
  • 3. $3 \text{ rad/s}$
  • 4. $4 \text{ rad/s}$
Solution:
$\text{Given, } \theta = 6t - 2t^3 \ldots(i)$ $\therefore \frac{d\theta}{dt} = \omega = 6 - 6t^2$ $\text{As, } \omega = 0 \text{ (given)}$ $6 - 6t^2 = 0$ $t^2 = \pm 1$ $\text{As, } t = -1 \text{ s is not possible, hence } t = 1 \text{ s.}$ $\text{Substituting } t = 1 \text{ s in Eq. (i), we get}$ $\theta_1 = 6 - 2 = 4 \text{ rad}$ $\text{and } \theta_0 = 6 \times 0 - 2 \times 0^2 \text{ } (t = 0) = 0$ $\therefore \text{ Mean value of } \omega = \frac{\theta_1 + \theta_0}{t_1 + t_0} = \frac{4 + 0}{1 + 0} = 4 \text{ rad s}^{-1}$

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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
}