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Current Question (ID: 8875)

Question:
$\text{A ball is thrown vertically upwards with a velocity of } 20 \text{ m/s from the top of a multistorey building. The height of the point from where the ball is thrown is } 25.0 \text{ m from the ground. How long will it be before the ball hits the ground?}$ $\text{(Take } g = 10 \text{ ms}^{-2}\text{)}$
Options:
  • 1. $3 \text{ s}$
  • 2. $2 \text{ s}$
  • 3. $5 \text{ s}$
  • 4. $20 \text{ s}$
Solution:
$\text{Hint: Use the second equation of motion.}$ $\text{Step 1: Write the given values.}$ $\text{Let us take the y-axis in the vertically upward direction with zero at the ground, as shown in the figure below.}$ $u = 20 \text{ m/s}$ $a = -10 \text{ m/s}^2$ $y_0 = 25 \text{ m and } y = 0 \text{ m}$ $\text{Step 2: Find the time of flight.}$ $\text{Using the equation:}$ $y = y_0 + ut + \frac{1}{2}at^2$ $\text{We have,}$ $0 = 25 + 20t + \frac{1}{2} \times (-10)t^2$ $\text{Solving this equation, we get, } t = 5 \text{ s}$ $\text{Therefore, the total time taken by the ball before it hits the ground, } t = 5 \text{ s.}$

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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
}