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Current Question (ID: 19384)

Question:

$\text{A uniform solid cylinder of radius } R, \text{ is released from a } 600 \text{ m long}$ $\text{ramp, inclined at } 30^\circ \text{ from the horizontal. The time taken to reach the}$ $\text{bottom of the ramp is:}$ $\text{(Consider sufficient friction for pure rolling.)}$

Options:

1. $60 \text{ s}$
2. $6\sqrt{10} \text{ s}$
3. $3\sqrt{10} \text{ s}$
4. $20 \text{ s}$

Solution:

$mg \sin \theta - f_r = ma$ $\text{Also,}$ $\frac{3}{2} mR^2 a = mg \sin \theta \times R$ $\Rightarrow \frac{3}{2} ma = mg \sin \theta$ $a = \frac{2}{3} g \sin 30^\circ = \frac{g}{3} = \frac{10}{3} \text{ m/s}^2$ $\text{Ramp length, } s = 600 \text{ m}$ $t = \sqrt{\frac{2s}{a}} = \sqrt{\frac{2 \times 600 \times 3}{10}} = 6\sqrt{10} \text{ seconds}$

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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
}