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Current Question (ID: 19370)

Question:

$\text{A solid cylinder and a solid sphere, having same mass } M \text{ and radius } R, \text{ roll down the same inclined plane from top without slipping. They start from rest. The ratio of the velocity of the solid cylinder to that of the solid sphere, with which they reach the ground, will be:}$

Options:

1. $\sqrt{\frac{5}{3}}$
2. $\sqrt{\frac{4}{5}}$
3. $\sqrt{\frac{3}{5}}$
4. $\sqrt{\frac{14}{15}}$

Solution:

$\text{Hint: } v = \sqrt{\frac{2gh}{1+\frac{k^2}{R^2}}}$ $\text{Step 1: Find the acceleration of given bodies.}$ $a = \frac{g \sin \theta}{1+\beta}$ $\beta_s = \frac{2MR^2}{5MR^2} = \frac{2}{5}$ $\beta_c = \frac{MR^2}{2MR^2} = \frac{1}{2}$ $\text{Step 2: Find the ratio of velocities.}$ $v = \sqrt{2as}$ $\frac{v_c}{v_s} = \sqrt{\frac{a_c}{a_s}}$ $\frac{v_c}{v_s} = \sqrt{\frac{1+\beta_s}{1+\beta_c}}$ $= \sqrt{\frac{1+\frac{2}{5}}{1+\frac{1}{2}}}$ $= \sqrt{\frac{14}{15}}$

Import JSON File

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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
}