Import Question JSON

Current Question (ID: 10117)

Question:

$\text{A man 'A', mass 60 kg, and another man 'B', mass 70 kg, are sitting at the two extremes of a 2 m long boat, of mass 70 kg, standing still in the water as shown. They come to the middle of the boat. (Neglect friction). How far does the boat move on the water during the process?}$

Options:

1. $5 \text{ cm leftward}$
2. $5 \text{ cm rightward}$
3. $7 \text{ cm leftward}$
4. $7 \text{ cm rightward}$

Solution:

$\text{Hint: There is no external force on the system so the position of the centre of mass will remain the same.}$ $\text{Step 1: Draw the diagram and set up coordinate system}$ $\text{Let the initial position of the boat's left end be at origin (0,0)}$ $\text{Initial positions: Man A at left end, Man B at right end, boat center at 1 m}$ $\text{Step 2: Apply conservation of center of mass}$ $\text{Since } F_{ext} = 0 \text{, the center of mass remains fixed:}$ $\sum \Delta m_i \vec{x}_i = 0$ $\text{Let the boat move distance } x \text{ to the right when both men move to center}$ $\text{Final positions: Both men at boat center, boat moved by } x$ $70x + 60(x + 1) + 70(x - 1) = 0$ $70x + 60x + 60 + 70x - 70 = 0$ $200x - 10 = 0$ $x = 5 \text{ cm}$ $\text{Since } x \text{ is positive, the boat moves 5 cm to the right (rightward).}$

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
}