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Current Question (ID: 9661)

Question:
$\text{A particle moves with constant speed } v \text{ along a circular path of radius } r \text{ and completes the circle in time } T. \text{ The acceleration of the particle is:}$
Options:
  • 1. $\frac{2\pi v}{T}$
  • 2. $\frac{2\pi r}{T}$
  • 3. $\frac{2\pi r^2}{T}$
  • 4. $\frac{2\pi v^2}{T}$
Solution:
\text{Hint: } a_c = \frac{v^2}{r} \text{Step: Find the acceleration of the particle.} \text{The acceleration in circular motion is given by: } a_c = \frac{v^2}{r} \text{Since speed is defined as the total distance divided by the time taken, the speed } v \text{ is: } v = \frac{2\pi r}{T} \text{Rearranging for } r\text{: } r = \frac{vT}{2\pi} \text{Therefore, } a_c = \frac{v^2}{\frac{vT}{2\pi}} = \frac{2\pi v}{T} \text{Hence, option (1) is the correct answer.}

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}