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Current Question (ID: 19147)

Question:

$\text{As shown in the figure, a block of mass } \sqrt{3} \text{ kg is kept on a horizontal rough surface of coefficient of friction } \frac{1}{3\sqrt{3}}. \text{ The critical force to be applied on the vertical surface as shown at an angle } 60^\circ \text{ with horizontal such that it does not move, will be } 3x. \text{ The value of } x \text{ will be:}$ $\left( g = 10 \text{ m/s}^2; \sin 60^\circ = \frac{\sqrt{3}}{2}; \cos 60^\circ = \frac{1}{2} \right)$

Options:

1. 1.23
2. 4.5
3. 3.33
4. 9.24

Solution:

$\text{Hint: } f_{\text{max}} = \mu N$ $\text{Step: Find the value of } x. \text{ The free-body diagram of the block is shown in the figure below;}$ $F \cos 60^\circ = \mu N$ $\Rightarrow \frac{F}{2} = \frac{1}{3\sqrt{3}} N \quad \ldots (1)$ $N = F \sin 60^\circ + mg$ $\Rightarrow N = F \frac{\sqrt{3}}{2} + \sqrt{3}g \quad \ldots (2)$ $\text{From the equations (1) and (2) we get;}$ $\frac{F}{2} = \frac{1}{3\sqrt{3}} \left( F \frac{\sqrt{3}}{2} + \sqrt{3}g \right)$ $\Rightarrow \frac{F}{2} = \frac{F}{6} + \frac{g}{3}$ $\Rightarrow \frac{F}{2} - \frac{F}{6} = \frac{g}{3}$ $\Rightarrow \frac{2F}{6} = \frac{g}{3}$ $\Rightarrow F = g = 3x$ $\Rightarrow x = \frac{10}{3} = 3.33$ $\text{Hence, option (3) is the correct answer.}$

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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
}