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Current Question (ID: 19414)

Question:

$\text{Two coaxial discs, having moments of inertia } I_1 \text{ and } \frac{I_1}{2} \text{ are rotating}$ $\text{with respective angular velocities } \omega_1 \text{ and } \frac{\omega_1}{2}, \text{ about their common}$ $\text{axis. They are brought in contact with each other and thereafter they}$ $\text{rotate with a common angular velocity. If } E_f \text{ and } E_i \text{ are the final and}$ $\text{initial total energies, then } (E_f - E_i) \text{ is:}$

Options:

1. $\frac{I_1 \omega_1^2}{6}$
2. $\frac{3}{8} I_1 \omega_1^2$
3. $\frac{I_1 \omega_1^2}{12}$
4. $\frac{I_1 \omega_1^2}{24}$

Solution:

$\text{Hint: Apply the conservation of angular momentum}$ $L_i = L_f$ $\text{and kinetic energy } = \frac{1}{2} I \omega^2$ $\text{put the value KE } = -\frac{I_1 \omega_1^2}{24}$

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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
}