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Current Question (ID: 10172)

Question:

$\text{A solid sphere of mass } M \text{ and the radius } R \text{ is in pure rolling with angular speed } \omega \text{ on a horizontal plane as shown in the figure. The magnitude of the angular momentum of the sphere about the origin } O \text{ is:}$

Options:

1. $\frac{7}{5}MR^2\omega$
2. $\frac{3}{2}MR^2\omega$
3. $\frac{1}{2}MR^2\omega$
4. $\frac{2}{3}MR^2\omega$

Solution:

$\text{Hint: Total angular momentum has two components due to translation of centre of mass and other due to rotation about centre of mass.}$ $\text{Step 1:}$ $\text{Write the formula for angular momentum}$ $\vec{L}_0 = \vec{r}_{CM,O} \times \vec{P} + I\vec{\omega}$ $\text{Step 2:}$ $\text{Calculate angular momentum,}$ $\vec{L}_0 = R(mv)(-\hat{k}) + \frac{2}{5}MR^2\omega(-\hat{k})$ $= \frac{7}{5}MR^2\omega$

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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
}