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Current Question (ID: 10203)

Question:

$\text{A solid body rotates about a stationary axis according to the equation } \theta = 6t - 2t^3\text{. What is the average angular velocity over the time interval between } t = 0 \text{ and the time when the body comes to rest? (}\theta\text{: angular displacement, } t\text{: time)}$

Options:

1. $1 \text{ rad/s}$
2. $2 \text{ rad/s}$
3. $3 \text{ rad/s}$
4. $4 \text{ rad/s}$

Solution:

$\text{First, find the time when the body comes to rest by setting the derivative of the angular displacement (}\theta = 6t - 2t^3\text{) to zero:}$ $\omega = \frac{d\theta}{dt} = 6 - 6t^2 = 0 \longrightarrow t = 1\text{s}$ $\text{Next, calculate the angular displacement at } t = 0 \text{ and } t = 1\text{:}$ $\theta(0) = 6(0) - 2(0)^3 = 0 \text{ rad}$ $\theta(1) = 6(1) - 2(1)^3 = 4 \text{ rad}$ $\text{Finally, calculate the average angular velocity:}$ $\omega_{avg} = \frac{\Delta\theta}{\Delta t} = \frac{\theta(1) - \theta(0)}{1 - 0} = \frac{4 - 0}{1} = 4 \text{ rad/s}$ $\text{The final answer is } 4 \text{ rad/s}$

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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
}