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Current Question (ID: 19345)

Question:
$\text{A wheel is rotating freely with an angular speed } \omega \text{ on a shaft. The moment of inertia of the wheel is } I \text{ and the moment of inertia of the shaft is negligible.}$ $\text{Another wheel of moment of inertia } 3I \text{ initially at rest is suddenly coupled to the same shaft.}$ $\text{The resultant fractional loss in the kinetic energy of the system is:}$
Options:
  • 1. $0$
  • 2. $\frac{1}{4}$
  • 3. $\frac{3}{4}$
  • 4. $\frac{5}{6}$
Solution:
$\text{Hint: Apply conservation of angular momentum.}$ $\omega I + 3I \times 0 = 4I \omega' \Rightarrow \omega' = \frac{\omega}{4}$ $(KE)_i = \frac{1}{2} I \omega^2$ $(KE)_f = \frac{1}{2} \times (4I) \times \left(\frac{\omega}{4}\right)^2 = \frac{I \omega^2}{8}$ $\Delta KE = \frac{3}{8} I \omega^2$ $\text{fraction loss} = \frac{\Delta KE}{KE_1} = \frac{\frac{3}{8} I \omega^2}{\frac{1}{2} I \omega^2} = \frac{3}{4}$

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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
}