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Current Question (ID: 19143)

Question:

$\text{An inclined plane is bent in such a way that the vertical cross-section is given by, } y = \frac{x^2}{4} \text{ where } y \text{ is in vertical and } x \text{ in the horizontal direction. If the upper surface of this curved plane is rough with a coefficient of friction } \mu = 0.5, \text{ the maximum height in(cm) at which a stationary block will not slip downward is:}$

Options:

1. $25$
2. $50$
3. $75$
4. $100$

Solution:

$\text{Hint: } \mu = \tan \theta$ $\therefore \tan \theta = \frac{dy}{dx} = \frac{2x}{4} = \frac{x}{2} = 0.5$ $\Rightarrow x = 1 \text{ and therefore } y = \frac{x^2}{4} = 0.25 \text{ m}$ $= 25 \text{ cm}$ $\therefore \text{ Answer is 25 cm}$ $\text{(Assuming that } x \text{ & } y \text{ in the equation are given in meter)}$

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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
}