Import Question JSON

Current Question (ID: 9560)

Question:

$\text{A car turns at a constant speed on a circular track of radius 100 m, taking 62.8 s for every circular lap. The average velocity and average speed for each circular lap, respectively, is:}$

Options:

1. $0, 0$
2. $0, 10 \text{ m/s}$
3. $10 \text{ m/s}, 10 \text{ m/s}$
4. $10 \text{ m/s}, 0$

Solution:

\text{Hint: average speed} = \frac{\text{distance}}{\text{time taken}} \text{Step: Find the average velocity and average speed for each circular lap of the car, respectively.} \text{On a circular path in completing one turn, the distance traveled is } 2\pi r \text{ while displacement is zero.} \text{Hence, average velocity} = \frac{\text{displacement}}{\text{time-interval}} = \frac{0}{t} = 0 \text{Average speed} = \frac{\text{distance}}{\text{time-interval}} = \frac{2\pi r}{t} = \frac{2 \times 3.14 \times 100}{62.8} = 10 \text{ m/s} \text{Note: If a particle moves in a straight line without a change in direction, the magnitude of displacement is equal to the distance traveled otherwise it is always less than it. Thus, } |\text{displacement}| \leq \text{distance} \text{Therefore, the average velocity and average speed for each circular lap of the car, respectively are 0, 10 m/s.} \text{Hence, option (2) is the correct answer.}

Import JSON File

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
}