Import Question JSON

Current Question (ID: 9911)

Question:

A stone is projected from a horizontal plane. It attains maximum height, }H\text{, and strikes a stationary smooth wall & falls on the ground vertically below the maximum height. Assuming the collision to be elastic, the height of the point on the wall where the ball will strike will be:

Options:

1. $\frac{H}{2}$
2. $\frac{H}{4}$
3. $\frac{3H}{4}$
4. $\text{None of these}$

Solution:

$\text{As the collision is elastic, so the velocity is same after the collision and it takes the same time to reach the ground from the wall as it took to reach the wall from the highest point.}$ $\text{In vertical direction:}$ $\text{From highest point to the wall—}$ $H - \frac{1}{2}gt^2 = h$ $\text{From highest point to the ground—}$ $H = \frac{1}{2}g(2t)^2$ $\text{On solving both equations:}$ $h = \frac{3H}{4}$

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
}