Import Question JSON

Current Question (ID: 8896)

Question:

$\text{Which of the following velocity-time graphs shows a realistic situation for a body in motion?}$

Options:

1. $\text{Graph 1: A curved loop that intersects itself, showing multiple velocity values for some time instants}$
2. $\text{Graph 2: A continuous smooth curve that represents a mathematical function with one velocity value for each time instant}$
3. $\text{Graph 3: A curved line that loops back on itself, showing multiple velocity values for some time instants}$
4. $\text{Graph 4: A closed oval loop showing multiple velocity values for the same time instants}$

Solution:

$\text{Hint: Time cannot go back.}$ $\text{Explanation: In the graph of options (1), (3) and (4) we observed that for the same value of time, there are two or more than two values of velocity or these graphs show more than one velocity of the particle at a single instant of time which is not practically possible.}$ $\text{For a realistic velocity-time graph, each time instant must correspond to exactly one velocity value. This means the graph must be a mathematical function where no vertical line intersects the curve at more than one point.}$ $\text{Only option (2) satisfies this condition, showing a continuous curve where each time value has a unique corresponding velocity value.}$ $\text{Hence, option (2) is the correct answer.}$

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
}