Import Question JSON

Current Question (ID: 9818)

Question:

$\text{A block slides down on a } 45^\circ \text{ rough incline in thrice the time it takes to slide down on a frictionless } 45^\circ \text{ incline of the same length. The coefficient of friction between the block and the rough incline is:}$

Options:

1. $0.6$
2. $0.7$
3. $0.5$
4. $0.9$

Solution:

\text{For a frictionless incline at } 45°\text{:} \text{Acceleration } a_1 = g\sin(45°) = \frac{g}{\sqrt{2}} \text{Using } s = \frac{1}{2}a_1t_1^2\text{, we get } t_1 = \sqrt{\frac{2s}{a_1}} = \sqrt{\frac{2s\sqrt{2}}{g}} \text{For a rough incline at } 45°\text{:} \text{Net force down the incline: } F_{\text{net}} = mg\sin(45°) - \mu mg\cos(45°) \text{Since } \sin(45°) = \cos(45°) = \frac{1}{\sqrt{2}}\text{:} \text{Acceleration } a_2 = g\sin(45°) - \mu g\cos(45°) = \frac{g}{\sqrt{2}}(1 - \mu) \text{Time for rough incline: } t_2 = \sqrt{\frac{2s}{a_2}} = \sqrt{\frac{2s\sqrt{2}}{g(1-\mu)}} \text{Given that } t_2 = 3t_1\text{:} \sqrt{\frac{2s\sqrt{2}}{g(1-\mu)}} = 3\sqrt{\frac{2s\sqrt{2}}{g}} \text{Simplifying: } \frac{1}{\sqrt{1-\mu}} = 3 \text{Squaring both sides: } \frac{1}{1-\mu} = 9 \text{Therefore: } 1-\mu = \frac{1}{9} \text{Solving: } \mu = 1 - \frac{1}{9} = \frac{8}{9} \approx 0.89 \approx 0.9

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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
}