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Current Question (ID: 8887)

Question:
$\text{The acceleration } a \text{ (in ms}^{-2}\text{) of a body, starting from rest varies with time } t \text{ (in s) as per the equation } a = 3t + 4. \text{ The velocity of the body at time } t = 2 \text{s will be:}$
Options:
  • 1. $10 \text{ ms}^{-1}$
  • 2. $18 \text{ ms}^{-1}$
  • 3. $14 \text{ ms}^{-1}$
  • 4. $26 \text{ ms}^{-1}$
Solution:
$\text{Hint: } a = \frac{dv}{dt}$ $\text{Step: Find the velocity of the body at the time } t = 2 \text{s.}$ $a = 3t + 4$ $\int_0^v dv = \int_0^2 (3t + 4)dt$ $v = 3\left[\frac{t^2}{2}\right]_0^2 + 4[t]_0^2$ $= \frac{3}{2} \times 4 + 4 \times 2 = 14 \text{ m/s}$ $\text{Hence, option (3) is the correct answer.}$

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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
}