Import Question JSON

Current Question (ID: 19038)

Question:

$\text{A bullet is shot vertically downwards with an initial velocity of } 100 \text{ m/s from a certain height.}$ $\text{Within } 10 \text{ s, the bullet reaches the ground and instantaneously comes to rest due to a perfectly inelastic collision.}$ $\text{The velocity-time curve for total time } t = 20 \text{ s is: (take } g = 10 \text{ m/s}^2)$

Options:

1. $\text{Graph with initial velocity } -100 \text{ m/s, decreasing to } -200 \text{ m/s at } 10 \text{ s, then } 0 \text{ m/s till } 20 \text{ s}$
2. $\text{Graph with initial velocity } +100 \text{ m/s, constant till } 20 \text{ s}$
3. $\text{Graph with initial velocity } -100 \text{ m/s, decreasing linearly}$
4. $\text{Graph with initial velocity } -100 \text{ m/s, decreasing to } -200 \text{ m/s at } 10 \text{ s, then constant}$

Solution:

$\text{Hint: } v = u + at$ $\text{Explanation: The bullet is fired vertically downward with an initial velocity of } 100 \text{ m/s, which we take as } -100 \text{ m/s because downward is negative.}$ $\text{Due to gravity } -10 \text{ m/s}^2, \text{ its velocity keeps increasing in the negative direction.}$ $\text{After 10 seconds, its velocity becomes } v = u + at = -100 + (-10)(10) = -200 \text{ m/s.}$ $\text{At that instant, the bullet strikes the ground and, because the collision is perfectly inelastic, it comes to rest immediately.}$ $\text{From that moment onward, its velocity remains zero for the rest of the time.}$ $\text{Therefore, the velocity-time graph starts at } -100 \text{ m/s, decreases linearly to } -200 \text{ m/s at } 10 \text{ s, then suddenly jumps to } 0 \text{ and stays flat at } 0 \text{ till } 20 \text{ s.}$ $\text{Hence, option (1) is the correct answer.}$

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
}