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Current Question (ID: 10106)

Question:

$\text{A man of 50 kg mass is standing in a gravity-free space at a height of 10 m above the floor. He throws a stone of 0.5 kg mass downwards with a speed of 2 ms}^{-1}\text{. When the stone reaches the floor, the distance of the man above the floor will be:}$

Options:

1. $\text{9.9 m}$
2. $\text{10.1 m}$
3. $\text{10 m}$
4. $\text{20 m}$

Solution:

$\text{Hint: The position of the centre of mass remains the same.}$ $\text{Step 1: Use the formula of the centre of mass.}$ $\text{As there is no external force (when a particle in gravity-free space), therefore, the centre of mass of the system will also remain unchanged.}$ $\text{Hence, } mr = \text{constant}$ $m_1 r_1 = m_2 r_2$ $\Rightarrow r_2 = \frac{m_1 r_1}{m_2} = \frac{0.5 \times 10}{50} = 0.1 \text{ m}$ $\text{Step 2: Find the distance of the man above the floor.}$ $\text{Thus the distance of the man above the floor (total height) } = 10 + 0.1 = 10.1 \text{ m}$

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
}