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Current Question (ID: 19357)

Question:
$\text{A cord is wound around the circumference of the wheel of radius } r. \text{ The axis of the wheel is horizontal and the moment of inertia about it is } I. \text{ A weight } mg \text{ is attached to the cord at the end. The weight falls from rest. After falling through a distance } h', \text{ the square of the angular velocity of the wheel will be:}$
Options:
  • 1. $\frac{2mgh}{I + 2mr^2}$
  • 2. $\frac{2mgh}{I + mr^2}$
  • 3. $2gh$
  • 4. $\frac{2gh}{I + mr^2}$
Solution:
$\text{Using the conservation of energy we get;} \ mgh = \frac{1}{2} I \omega^2 + \frac{1}{2} mv^2 \ldots (1)$ $\text{We know that;} \ v = \omega \times r$ $\text{Substituting the values in the equation (1) we get;} \ mgh = \frac{1}{2} I \omega^2 + \frac{1}{2} m \omega^2 r^2$ $\Rightarrow \omega^2 = \frac{2mgh}{(I+mr^2)}$ $\text{Hence, option (2) is the correct answer.}$

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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
}