Import Question JSON

Current Question (ID: 8910)

Question:

$\text{A ball is dropped vertically from a height } h \text{ above the ground. It hits the ground and bounces up vertically to a height of } \frac{h}{2}. \text{ Neglecting subsequent motion and air resistance, its velocity } v \text{ varies with the height } h \text{ as: [Take vertically upwards direction as positive.]}$

Options:

1. $\text{Graph 1: Curved path showing downward then upward motion}$
2. $\text{Graph 2: Semi-circular path}$
3. $\text{Graph 3: Linear relationship}$
4. $\text{Graph 4: Different curved pattern}$

Solution:

$\text{Hint: Use the third equation of motion.}$ $\text{Step 1: Write the equation for the downward motion.}$ $\text{Initial velocity } u = 0, \text{ taking upward as positive direction}$ $\text{For downward motion: } y = H - h$ $v^2 = u^2 + 2gy$ $v^2 = 0^2 + 2g(H - h) = 2g(H - h)$ $\text{when } h = H, v = 0$ $\text{when } h = 0, v = v_{\text{max}} \text{ (negative, since downward)}$ $\text{Step 2: Write the equation for upward motion.}$ $\text{After bouncing, for upward motion:}$ $v^2 = u_1^2 - 2gy$ $\text{At } y = 0 \text{ (ground level), } v = v_{\text{max}} \text{ (positive, since upward)}$ $\text{The velocity decreases as height increases during upward motion}$ $\text{At maximum bounce height } h = \frac{H}{2}, v = 0$ $\text{The graph shows a parabolic relationship between velocity and height for both downward and upward motion phases.}$ $\text{Hence, option (1) is the correct answer.}$

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
}