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Current Question (ID: 9690)

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:
$\text{Hint: The bullets travel an equal distance in the horizontal direction.}$ $\text{Given:}$ $\text{Distance between buildings: } d = 100 \text{ m}$ $\text{Initial height: } h_0 = 200 \text{ m}$ $\text{Initial velocity of each bullet: } v_0 = 25 \text{ m/s}$ $\text{Acceleration due to gravity: } g = 10 \text{ m/s}^2$ $\text{Step 1: Find the relative velocity and acceleration using relative motion concept}$ $\text{Bullet A moves at 25 m/s to the right, Bullet B moves at 25 m/s to the left}$ $\text{Relative velocity: } v_{A,B} = 25 + 25 = 50 \text{ m/s along x-direction}$ $\text{Relative acceleration: } a_{A,B} = 0 \text{ (both have same downward acceleration)}$ $\text{Step 2: Find the time when bullets meet horizontally}$ $\text{Time to meet: } t = \frac{\text{initial separation}}{\text{relative velocity}} = \frac{100}{50} = 2 \text{ s}$ $\text{Step 3: Find the vertical displacement of the bullets}$ $\text{In 2 s, vertical displacement of each bullet in the ground frame:}$ $y = ut + \frac{1}{2}gt^2$ $\text{Since initial vertical velocity } u = 0:$ $y = 0 + \frac{1}{2} \times 10 \times 2^2 = \frac{1}{2} \times 10 \times 4 = 20 \text{ m}$ $\text{Height from ground: } h = 200 - 20 = 180 \text{ m}$ $\text{Therefore, the bullets collide after 2 s at a height of 180 m.}$

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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
}