Import Question JSON

Current Question (ID: 8911)

Question:

$\text{The graph between the displacement } x \text{ and time } t \text{ for a particle moving in a straight line is shown in the figure. During the interval OA, AB, BC and CD the acceleration of the particle is:}$

Options:

1. $\text{OA: +, AB: 0, BC: +, CD: +}$
2. $\text{OA: -, AB: 0, BC: +, CD: 0}$
3. $\text{OA: +, AB: 0, BC: -, CD: +}$
4. $\text{OA: -, AB: 0, BC: -, CD: 0}$

Solution:

$\text{Hint: Interpret the rate of change of the slope.}$ $\text{Step: Analyse each option one by one.}$ $\text{Region OA shows that the graph is bending toward the time axis i.e. acceleration is negative.}$ $\text{Region AB shows that the graph is parallel to the time axis i.e. velocity is zero. Hence, the acceleration is zero.}$ $\text{Region BC shows that the graph is bending towards the displacement axis i.e. acceleration is positive.}$ $\text{Region CD shows that the graph is having a constant slope i.e. velocity is constant. Hence, the acceleration is zero.}$ $\text{Analysis summary:}$ $\text{OA: Concave downward curve → decreasing slope → negative acceleration}$ $\text{AB: Horizontal line → zero slope → zero velocity → zero acceleration}$ $\text{BC: Concave upward curve → increasing slope → positive acceleration}$ $\text{CD: Straight line → constant slope → constant velocity → zero acceleration}$ $\text{Hence, option (2) is the correct answer.}$

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
}