Import Question JSON

Current Question (ID: 8912)

Question:

$\text{The graph below shows position as a function of time for two trains running on parallel tracks. Which of the following statements is true?}$

Options:

1. $\text{At time } t_B \text{ both the trains have the same velocity}$
2. $\text{Both the trains have the same velocity at some time after } t_B$
3. $\text{Both the trains have the same velocity at some time before } t_B$
4. $\text{Both the trains have the same acceleration}$

Solution:

$\text{Hint: The slope of the position-time graph gives velocity.}$ $\text{Step: Analyse each option one by one.}$ $\text{Train } A \text{ is moving with constant velocity and train } B \text{ is moving with variable velocity. The slopes of the graphs of train } A \text{ and train } B \text{ are the same before } t = t_B. \text{ So, they have the same velocity before } t = t_B.$ $\text{Analysis of the graph:}$ $\text{Train A: Linear position-time graph indicates constant velocity (constant slope)}$ $\text{Train B: Curved position-time graph indicates variable velocity (changing slope)}$ $\text{At some point before } t_B, \text{ the slope of curve B equals the slope of line A}$ $\text{Therefore, both the trains have the same velocity at some time before } t_B.$ $\text{Hence, option (3) is the correct answer.}$

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
}