Import Question JSON

Current Question (ID: 10115)

Question:

$\text{A man hangs from a rope attached to a hot-air balloon. The man's mass is greater than the mass of the balloon and its contents. The system is stationary in still air. If the man now climbs up the balloon using the rope, the centre of mass of the 'man plus balloon' system will:}$

Options:

1. $\text{remain stationary}$
2. $\text{move up}$
3. $\text{move down}$
4. $\text{first moves up and then return to its initial position}$

Solution:

$\text{Hint: Movement of the centre of mass depends only on the external force.}$ $\text{Step 1: Draw the FBD (Free Body Diagram)}$ $\text{The system shows a stationary hot-air balloon with a man hanging from a rope. The forces acting on the system are:}$ $\text{- Weight (W) acting downward on the center of mass}$ $\text{- Buoyant force (F_B) acting upward}$ $\text{Step 2: Calculate net external force}$ $\text{Here the weight of the centre of mass is balanced by the buoyant force.}$ $\text{So, the net force on the system is zero.}$ $\text{Since there is no net external force on the system, the center of mass will remain stationary according to Newton's first law.}$ $\text{When the man climbs up the rope, internal forces within the system change, but the external forces (weight and buoyancy) remain balanced.}$ $\text{Therefore, the centre of mass will remain stationary.}$

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
}