Import Question JSON

Current Question (ID: 8913)

Question:

$\text{Among the four graphs shown in the figure, there is only one graph for which average velocity over the time interval } (0, T) \text{ can vanish for a suitably chosen } T. \text{ Select the graph.}$

Options:

1. $\text{Graph 1: Exponentially decreasing curve}$
2. $\text{Graph 2: Curve that rises then falls back down}$
3. $\text{Graph 3: Asymptotically increasing curve}$
4. $\text{Graph 4: Curve that dips then rises and levels off}$

Solution:

\text{Hint: For zero velocity, the displacement of the body should be equal to zero.} \text{Step 1: Find the graph in which the displacement is zero for a time interval.} \text{In graph (b) for one value of displacement, there are two different points of time. Hence, for one time, the average velocity is positive and for other times, the average velocity is negative.} \text{As there are opposite velocities in the interval 0 to T hence average velocity can vanish in (b). This can be seen in the figure given below;} \text{Here, } OA = BT \text{ (same displacement) for two different points of time.} \text{Since average velocity } = \frac{\text{total displacement}}{\text{total time}} = \frac{x(T) - x(0)}{T - 0} \text{For average velocity to be zero, we need } x(T) = x(0)\text{, which means the particle returns to its starting position.} \text{Only graph 2 shows this behavior where the particle can return to the same position at a later time.} \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
}