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Current Question (ID: 19170)

Question:

$\text{A block of mass } 10 \text{ kg is subjected to a force } F, \text{ as shown in the figure.}$ $\text{If the coefficient of friction is } 0.25, \text{ what is the minimum value of } F \text{ required to make the block slide?}$

Options:

1. $\frac{200}{4\sqrt{3} + 1} \text{ N}$
2. $\frac{200}{4\sqrt{3} - 1} \text{ N}$
3. $\frac{100}{4\sqrt{3} + 1} \text{ N}$
4. $50 \text{ N}$

Solution:

$\text{Hint: } f_{\text{max}} = \mu N$ $\text{Step: Find the minimum force required to make block slides.}$ $\text{The free-body diagram of the block is shown in the figure below;}$ $N + F \sin 30^\circ = mg$ $\Rightarrow N = mg - F \sin 30^\circ \quad \ldots (1)$ $F \cos 30^\circ \geq \mu (mg - F \sin 30^\circ)$ $\Rightarrow F \cos 30^\circ + \mu F \sin 30^\circ = \mu mg$ $\Rightarrow F (\cos 30^\circ + \mu \sin 30^\circ) = \mu mg$ $\Rightarrow F = \frac{\mu mg}{\cos 30^\circ + \mu \sin 30^\circ}$ $\Rightarrow F = \frac{0.25 \times 10 \times 10}{\frac{\sqrt{3}}{2} + \frac{1}{4} \times \frac{1}{2}}$ $\Rightarrow F = \frac{25}{\frac{4\sqrt{3} + 1}{8}}$ $\Rightarrow F = \frac{200}{4\sqrt{3} + 1} \text{ N}$ $\text{Hence, option (1) is the correct answer.}$

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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
}