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Current Question (ID: 9760)

Question:

$\text{A uniform chain of length } L \text{ hangs partly from a table which is kept in equilibrium by friction. If the maximum length that can be supported without slipping is } l, \text{ then the coefficient of friction between the table and the chain is:}$

Options:

1. $\frac{l}{L}$
2. $\frac{l}{L+l}$
3. $\frac{l}{L-l}$
4. $\frac{L}{L+l}$

Solution:

\text{Given: A uniform chain of length } L \text{ hangs partly from a table which is kept in equilibrium by friction. The maximum length that can be supported without slipping is } l. \text{For equilibrium, the weight of the hanging part must equal the maximum friction force.} \text{Let the mass per unit length of the chain be } \lambda. \text{Weight of hanging part = } \lambda \cdot l \cdot g \text{Weight of part on table = } \lambda \cdot (L-l) \cdot g \text{Maximum friction force = } \mu \cdot \lambda \cdot (L-l) \cdot g \text{For equilibrium: } \lambda \cdot l \cdot g = \mu \cdot \lambda \cdot (L-l) \cdot g \text{Simplifying: } l = \mu \cdot (L-l) \text{Therefore: } \mu = \frac{l}{L-l}

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
}