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Current Question (ID: 19367)

Question:

$\sqrt{34} \text{ m long ladder weighing } 10 \text{ kg leans on a frictionless wall. Its feet rest on the floor } 3 \text{ m away from the wall as shown in the figure.}$ $\text{If } F_f \text{ and } F_w \text{ are the reaction forces of the floor and the wall, then ratio of } F_w/F_f \text{ will be: (Take } g = 10 \text{ m/s}^2)$

Options:

1. $\frac{6}{\sqrt{110}}$
2. $\frac{3}{\sqrt{113}}$
3. $\frac{\sqrt{109}}{2}$
4. $\frac{3}{\sqrt{109}}$

Solution:

$\text{Hint: } f = \mu N$ $\text{Step: Find the ratio of } F_w/F_f$ $\text{The forces acting on the ladder are shown in the figure.}$ $\Rightarrow N_2 = F_f \text{ and } N_1 = mg$ $\text{Balancing the torque on the ladder: } mgl \cos \theta = N_2 (2l) \sin \theta$ $\Rightarrow N_2 = \frac{mg \cot \theta}{2} \quad \cdots (1)$ $F_f = \sqrt{N_1^2 + f_s^2} = mg \sqrt{1 + \frac{\cot^2 \theta}{4}} \quad \cdots (2)$ $\text{The required ratio is given by; } F_w/F_f = \frac{\cot \theta}{2 \sqrt{1 + \frac{\cot^2 \theta}{4}}} \quad \left[ \cot \theta = \frac{3}{5} \right]$ $\Rightarrow F_w/F_f = \frac{3}{10 \sqrt{1 + \frac{9}{100}}} = \frac{3}{\sqrt{109}}$ $\text{Hence, option (3) is the correct answer.}$

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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
}