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Current Question (ID: 19128)

Question:

$\text{A block of mass } m \text{ is placed on a surface with a vertical cross section given by } y = \frac{x^3}{6}. \text{ If the coefficient of friction is } 0.5, \text{ then the maximum height above the ground at which the block can be placed without slipping is:}$

Options:

1. $\frac{2}{3} m$
2. $\frac{1}{3} m$
3. $\frac{1}{2} m$
4. $\frac{1}{6} m$

Solution:

$\text{Hint: Balance forces along normal and tangential directions.}$ $N = mg \cos \theta$ $f_{\text{max}} = \mu Mg \cos \theta$ $\text{for limiting equilibrium,}$ $\mu mg \cos \theta = mg \sin \theta$ $\mu = \tan \theta$ $\text{as } y = \frac{x^3}{6} \text{ (given)}$ $\frac{dy}{dx} = \frac{1}{6} (3x^2)$ $\tan \theta = \frac{x^2}{2}$ $\mu = \tan \theta$ $0.5 = \frac{x^2}{2}$ $x = \pm 1$ $\text{So, from equation (1)}$ $y = \frac{1}{6} m$

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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
}