Import Question JSON

Current Question (ID: 19042)

Question:

$\text{The position-time graph for a particle is parabolic (as shown in the figure).}$ $\text{Choose the corresponding } (v-t) \text{ graph.}$

Options:

1. $\text{Graph 1}$
2. $\text{Graph 2}$
3. $\text{Graph 3}$
4. $\text{Graph 4}$

Solution:

$\text{Hint: } x \propto t^2$ $\text{Step 1: Analyze the position-time graph (parabolic)}$ $\text{The position-time graph shows a parabolic shape. In a position-time graph, a parabolic curve implies uniformly accelerated motion (constant acceleration).}$ $\text{A parabolic } x-t \text{ graph implies that the position changes at a rate proportional to } t^2, \text{ which is characteristic of uniformly accelerated motion.}$ $\text{Step 2: Find the relation between the position } (x), \text{ velocity } (v), \text{ and acceleration } (a)$ $\text{The velocity is the derivative of the position with respect to time: } v(t) = \frac{dx}{dt}$ $\text{Since the } x-t \text{ graph is parabolic, its derivative (velocity) will be a linear function. This is because the derivative of a quadratic function is linear.}$ $\text{Step 3: Find the nature of the velocity-time } (v-t) \text{ graph.}$ $\text{The velocity-time graph for uniformly accelerated motion should be a straight line as shown in the diagram below;}$ $\text{Hence, option (2) is the correct answer.}$

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
}