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Current Question (ID: 8820)

Question:

$\text{Given below are two statements:}$ $\text{Assertion (A): If the average velocity of a particle is zero in a time interval, it is possible that the instantaneous velocity is never zero in the interval.}$ $\text{Reason (R): If the average velocity of a particle moving on a straight line is zero in a time interval then at least for one moment the instantaneous velocity will also be zero in the interval.}$

Options:

1. $\text{Both (A) and (R) are True and (R) is the correct explanation of (A).}$
2. $\text{Both (A) and (R) are True but (R) is not the correct explanation of (A).}$
3. $\text{(A) is True but (R) is False.}$
4. $\text{Both (A) and (R) are False.}$

Solution:

$\text{Hint: Recall the uniform circular motion.}$ $\text{Explanation: (A) is true. Zero average velocity over a time interval can occur even with non-zero instantaneous velocities, as long as the overall displacement is zero (e.g., uniform circular motion).}$ $\text{(R) is also true. If the particle is confined to a straight line, then zero average velocity does imply at least one moment of zero instantaneous velocity (changing direction). However, (R) is not a general explanation for (A).}$ $\text{The reason (R) is not a perfect explanation because it doesn't consider scenarios beyond straight-line motion. As we saw with the example of circular motion, a particle's path doesn't have to involve changing direction on a straight line for its average velocity to be zero.}$ $\text{Therefore, Both (A) and (R) are true but (R) is not the correct explanation of (A).}$ $\text{Hence, option (2) is the correct answer.}$

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
}