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Current Question (ID: 9644)

Question:
$\text{The angle turned by a body undergoing circular motion depends on the time as given by the equation, } \theta = \theta_0 + \theta_1 t + \theta_2 t^2. \text{ It can be deduced that the angular acceleration of the body is?}$
Options:
  • 1. $\theta_1$
  • 2. $\theta_2$
  • 3. $2\theta_1$
  • 4. $2\theta_2$
Solution:
\text{Hint: } \omega = \frac{d\theta}{dt} \text{Step: Find the angular acceleration of the body.} \text{The angular displacement of the body is given by:} \theta = \theta_0 + \theta_1 t + \theta_2 t^2 \text{The angular velocity of the body is given by: } \omega = \frac{d\theta}{dt} \omega = \theta_1 + 2\theta_2 t \text{The angular acceleration of the body is given by:} \alpha = \frac{d\omega}{dt} = 2\theta_2 \text{Hence, option (4) 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
}