Import Question JSON

Current Question (ID: 19019)

Question:

$\text{A body is projected vertically upward and then allowed to fall back down under gravity. Which of the following velocity-time } (v-t) \text{ graphs correctly represents its motion throughout the ascent and descent?}$

Options:

1. $\text{Graph 1:}$ $v$ $t$
2. $\text{Graph 2:}$ $v$ $t$
3. $\text{Graph 3:}$ $v$ $t$
4. $\text{Graph 4:}$ $v$ $t$

Solution:

$\text{Hint: The slope of velocity time graph gives the acceleration.}$ $\text{Explanation: When a body is thrown vertically upward, it starts with a positive velocity, which appears as a point above the time axis on a velocity-time graph. As the body ascends, gravity acts in the opposite direction, causing the velocity to decrease linearly. This results in a straight line with a negative slope.}$ $\text{The line eventually intersects the time axis, indicating the moment when the body's velocity becomes zero (the highest point of its motion). At this instant, the body is momentarily at rest before beginning to descend.}$ $\text{As the body falls back down, its velocity becomes negative, reflecting the change in direction. The graph continues below the time axis as a straight line with the same negative slope, showing that the body is accelerating downward at a constant rate due to gravity. Hence, the correct graph is:}$ $\text{Graph 3:}$ $v$ $t$ $\text{Hence, option (3) is the correct answer.}$

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
}