Import Question JSON

Current Question (ID: 9590)

Question:

$\text{Two bullets are fired horizontally and simultaneously towards each other from the rooftops of two buildings (building being 100 m apart and being of the same height of 200 m) with the same velocity of 25 m/s. When and where will the two bullets collide?}$ $(g = 10 \text{ m/s}^2)$

Options:

1. $\text{After 2 s at a height of 180 m}$
2. $\text{After 2 s at a height of 20 m}$
3. $\text{After 4 s at a height of 120 m}$
4. $\text{They will not collide.}$

Solution:

$\text{Hint: The bullets travel an equal distance in the horizontal direction.}$ $\text{Step 1: Find the relative velocity and acceleration.}$ $\text{Using the relative motion concept, for bullets A and B:}$ $a_{A,B} = 0$ $u_{A,B} = 25 + 25 = 50 \text{ m/s along x-direction}$ $u_{A,B} = 0 \text{ along y-direction}$ $\text{The meeting time is given by:}$ $t = \frac{100}{50} = 2 \text{ s}$ $\text{Step 2: Find the vertical displacements of the bullets.}$ $\text{In 2s, the vertical displacement of each bullet in the ground frame,}$ $y = ut + \frac{1}{2}gt^2$ $\text{Substituting the known values we get,}$ $y = 0 + \frac{1}{2}gt^2$ $\Rightarrow y = \frac{1}{2} \times 10 \times 2^2$ $\Rightarrow y = 20 \text{ m}$ $\text{Thus the height from the ground is given by,}$ $\Rightarrow h = (200 - 20) = 180 \text{ m}$ $\text{Hence, option (1) is the correct answer.}$

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
}