Import Question JSON

Current Question (ID: 19063)

Question:
$\text{A block } A \text{ takes } 2 \text{ s to slide down a frictionless incline of } 30^\circ \text{ and length } l, \text{ kept inside a lift going up with uniform velocity } v. \text{ If the incline is changed to } 45^\circ, \text{ the time taken by the block, to slide down the incline, will be approximately:}$
Options:
  • 1. $2.66 \text{ s}$
  • 2. $0.83 \text{ s}$
  • 3. $1.68 \text{ s}$
  • 4. $0.70 \text{ s}$
Solution:
$\text{For } 30^\circ, \text{ time } t_1 = 2 \text{ s}$ $\text{For } 45^\circ, \text{ time } t_2 = \frac{t_1}{\sqrt{\frac{\sin 45^\circ}{\sin 30^\circ}}} = \frac{2}{\sqrt{\frac{\sqrt{2}/2}{1/2}}} = \frac{2}{\sqrt{\sqrt{2}}} = \frac{2}{\sqrt{2^{3/2}}} = \frac{2}{2^{3/4}} = 1.68 \text{ s}$

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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
}