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Current Question (ID: 20040)

Question:

$\text{An inclined plane making an angle of } 30^\circ \text{ with the horizontal is placed in a uniform horizontal electric field } 200 \text{ N/C as shown in the figure.}$ $\text{A body of mass } 1 \text{ kg and charge } 5 \text{ mC is allowed to slide down from rest at a height of } 1 \text{ m.}$ $\text{If the coefficient of friction is } 0.2, \text{ find the time taken by the body to reach the bottom.}$ $[g = 9.8 \text{ m/s}^2, \sin 30^\circ = \frac{1}{2}; \cos 30^\circ = \frac{\sqrt{3}}{2}]$

Options:

1. $0.92 \text{ s}$
2. $0.46 \text{ s}$
3. $2.3 \text{ s}$
4. $1.3 \text{ s}$

Solution:

$\text{Hint: } S = ut + \frac{1}{2}at^2$ $N = F_e \sin 30 + F_g \cos 30$ $\text{here } N = 9.8 \cos 30 + 1 \sin 30 \approx 9 \text{ N}$ $ma = F_g \sin 30 - F_e \cos 30 - \mu N$ $\text{so } a = \frac{9.8 \sin 30 - 1 \cos 30 - \mu N}{1}$ $a = 2.233 \text{ m/s}^2$ $S = ut + \frac{1}{2}at^2$ $2h = 0 + \frac{1}{2}at^2$ $2 = \frac{1}{2} (2.33) t^2$ $t^2 = \frac{4}{2.33}$ $t = \sqrt{\frac{4}{2.33}} \Rightarrow t \approx 1.3 \text{ sec}$

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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
}