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Current Question (ID: 9737)

Question:

$\text{A monkey weighing } 20 \text{ kg is holding a vertical rope. The rope will not break when a mass of } 25 \text{ kg is suspended from it but will break if the mass exceeds } 25 \text{ kg. What is the maximum acceleration with which the monkey can climb up the rope? } (g = 10 \text{ m/s}^2)$

Options:

1. $5 \text{ m/s}^2$
2. $10 \text{ m/s}^2$
3. $25 \text{ m/s}^2$
4. $2.5 \text{ m/s}^2$

Solution:

$\text{The maximum tension the rope can withstand is:}$ $T_{\text{max}} = 25g = 25 \times 10 = 250 \text{ N}$ $\text{When the monkey climbs up with acceleration } a\text{, applying Newton's second law:}$ $T_{\text{max}} - mg = ma$ $\text{Where } m = 20 \text{ kg (mass of monkey)}$ $250 - 20 \times 10 = 20 \times a$ $250 - 200 = 20a$ $50 = 20a$ $a = \frac{50}{20} = 2.5 \text{ m/s}^2$

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
}