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Current Question (ID: 19361)

Question:

$\text{A thin circular ring of mass } M \text{ and radius } R \text{ is rotating with a constant angular velocity of } 2 \text{ rad/sec in a horizontal plane about an axis vertical to its plane and passing through the centre of the ring. If two objects each of mass } m \text{ be attached gently to the opposite ends of a diameter of the ring, the ring will then rotate with an angular velocity of:}$

Options:

1. $\frac{M}{M + m} \text{ rad/s}$
2. $\frac{M + 2m}{2M} \text{ rad/s}$
3. $\frac{2M}{M + 2m} \text{ rad/s}$
4. $\frac{2(M + 2m)}{M} \text{ rad/s}$

Solution:

$\text{Apply conservation of angular momentum.}$ $\text{Find the final angular velocity of the ring.}$ $\text{The rotation of the ring after placing masses is shown in the figure below;} \omega$ $\text{As the net external torque on the system is zero, then the angular momentum of the system is conserved.}$ $\text{The angular velocity of the ring is given by;} L = I \omega$ $\text{By applying the conservation of angular momentum we get;} I_1 \omega_1 = I_2 \omega'$ $2MR^2 = (MR^2 + 2mR^2) \omega'$ $\Rightarrow \omega' = \frac{2M}{M + 2m} \text{ rad/s}$ $\text{Hence, option (3) is the correct answer.}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}