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Current Question (ID: 19337)

Question:

$\text{A uniform cylinder of mass } M \text{ and radius } R \text{ is to be pulled over a step of height } a \,(a < R) \text{ by applying a force } F \text{ at its centre } 'O' \text{ perpendicular to the plane through the axes of the cylinder on the edge of the step (see figure). The minimum value of } F \text{ required is:}$

Options:

1. $Mg \sqrt{1 - \frac{a^2}{R^2}}$
2. $Mg \sqrt{\left(\frac{R}{R-a}\right)^2 - 1}$
3. $Mg \frac{a}{R}$
4. $Mg \sqrt{1 - \left(\frac{R-a}{R}\right)^2}$

Solution:

$\text{Hint: } \vec{\tau} = \vec{r} \times \vec{F}$ $\text{(\tau)_P = 0}$ $x = \sqrt{R^2 - (R-a)^2}$ $F = mg \frac{x}{R}$ $F = mg \sqrt{1 - \left(\frac{R-a}{R}\right)^2}$ $= \text{minimum value of force to pull}$

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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
}