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Current Question (ID: 9885)

Question:
$\text{When an object is shot from the bottom of a long, smooth inclined plane kept at an angle of } 60^\circ \text{ with horizontal, it can travel a distance } x_1 \text{ along the plane. But when the inclination is decreased to } 30^\circ \text{ and the same object is shot with the same velocity, it can travel } x_2 \text{ distance. Then } x_1 : x_2 \text{ will be:}$
Options:
  • 1. $1 : 2\sqrt{3}$
  • 2. $1 : \sqrt{2}$
  • 3. $\sqrt{2} : 1$
  • 4. $1 : \sqrt{3}$
Solution:
$\text{Case 1: Inclined plane at } 60^\circ$ $x_1 \sin 60^\circ = h_1$ $\text{Using energy approach:}$ $\frac{1}{2}mv^2 = mg(x_1 \sin 60^\circ) \text{ ...(1)}$ $\text{Case 2: Inclined plane at } 30^\circ$ $x_2 \sin 30^\circ = h_2$ $\frac{1}{2}mv^2 = mgh_2 = mg(x_2 \sin 30^\circ) \text{ ...(2)}$ $\text{From (1) and (2):}$ $\frac{x_1}{x_2} = \frac{\sin 30^\circ}{\sin 60^\circ} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}$ $\text{Hence, option (4) is the correct answer.}$

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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
}