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Current Question (ID: 8811)

Question:
$\text{If a body travels some distance in a given time interval, then for that time interval, its:}$
Options:
  • 1. $\text{Average speed} \geq |\text{Average velocity}|$
  • 2. $|\text{Average velocity}| \geq \text{Average speed}$
  • 3. $\text{Average speed} < |\text{Average velocity}|$
  • 4. $|\text{Average velocity}| \text{ must be equal to average speed.}$
Solution:
\text{Hint: Distance} \geq \text{Displacement} \text{Step: Find the relationship between average velocity and average speed.} \text{If a body travels some distance, the displacement of the body may be equal to or less than the distance covered by the body. So, the average speed may be equal to or more than the magnitude of the average velocity.} \text{Since:} \text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} \text{Average velocity} = \frac{\text{Total displacement}}{\text{Total time}} \text{And since distance} \geq \text{displacement, we have:} \text{Average speed} \geq |\text{Average velocity}| \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
}