Import Question JSON

Current Question (ID: 8903)

Question:

$\text{A ball is thrown vertically upwards. Then the velocity-time (v-t) graph will be:}$

Options:

1. $\text{A straight line starting from positive velocity and decreasing linearly to negative velocity}$
2. $\text{A triangular shape starting at zero, increasing to a peak, then decreasing back to zero}$
3. $\text{A sinusoidal wave pattern}$
4. $\text{An exponential decay curve}$

Solution:

$\text{Hint: } v = u + at$ $\text{Explanation: When a ball is thrown vertically upwards, its velocity changes due to the force of gravity acting against its motion. Initially, the velocity is positive (upward) and decreases until it becomes zero at the highest point (the peak of its trajectory). After reaching the peak, the velocity becomes negative (downward) as the ball starts to descend. The velocity-time (v-t) graph for this motion would look like a curve that starts from a positive value, decreases to zero, and then becomes negative. It would be symmetrical about the point where the velocity becomes zero, representing the highest point of the ball's trajectory.}$ $v = v_0 - gt$ $\text{This equation is similar to } y = mx + c.$ $\text{The velocity-time graph of the ball is shown in the figure below;}$ $\text{Hence, option (1) is the correct answer.}$

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
}