Import Question JSON

Current Question (ID: 8880)

Question:

$\text{A ball is dropped from a high-rise platform at } t = 0 \text{ starting from rest. After 6 seconds, another ball is thrown downwards from the same platform with speed } v. \text{ The two balls meet after 18 seconds. What is the value of } v?$

Options:

1. $75 \text{ ms}^{-1}$
2. $55 \text{ ms}^{-1}$
3. $40 \text{ ms}^{-1}$
4. $60 \text{ ms}^{-1}$

Solution:

$\text{Hint: The heights covered by the two balls are the same.}$ $\text{Step 1: Find the height covered by the balls.}$ $\text{For the first ball:}$ $\text{Using } h = ut + \frac{1}{2}gt^2$ $\therefore h_1 = \frac{1}{2}gt^2 = \frac{1}{2} \times 10 \times (18)^2 \quad \ldots (i)$ $\text{For the second ball,}$ $u = v, t = 12\text{s}, g = 10 \text{ m/s}^2$ $h_2 = 12v + \frac{1}{2} \times 10 \times 12^2 \quad \ldots (ii)$ $\text{Step 2: Put } h_1 = h_2.$ $\text{By using equation (1) and equation (2), we get,}$ $\frac{1}{2} \times 10 \times (18)^2 = 12v + \frac{1}{2} \times 10 \times (12)^2$ $\text{Or } 12v = \frac{1}{2} \times 10 \times [(18)^2 - (12)^2]$ $= \frac{1}{2} \times 10 \times [(18 + 12)(18 - 12)]$ $12v = \frac{1}{2} \times 10 \times 30 \times 6$ $\text{Or } v = \frac{\frac{1}{2} \times 10 \times 30 \times 6}{2 \times 12} = 75 \text{ m/s}$ $\text{Hence, option (1) is the correct answer.}$

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
}