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Current Question (ID: 10320)

Question:
$\text{For a rigid body rotating about a fixed axis, which of the following quantities is the same at an instant for all the particles of the body?}$
Options:
  • 1. $\text{Angular acceleration}$
  • 2. $\text{Angular velocity}$
  • 3. $\text{Angular displacement in the given time interval}$
  • 4. $\text{All of these}$
Solution:
\text{For a rigid body rotating about a fixed axis, all particles in the body have the same:} \text{Angular acceleration } (\alpha)\text{: Every part of the rigid body speeds up or slows down its rotation at the same rate.} \text{Angular velocity } (\omega)\text{: Every part of the rigid body rotates at the same instantaneous rate.} \text{Angular displacement } (\Delta\theta)\text{: Over any given time interval, every part of the rigid body rotates through the same angle.} \text{Linear quantities like linear speed, linear velocity, linear acceleration, and linear displacement would differ for particles at different distances from the axis of rotation. However, angular quantities are uniform for all particles of a rigid body rotating about a fixed axis.} \text{Therefore, all the listed quantities are the same for all particles of the rigid body.} \text{The final answer is All of these}

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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
}