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Current Question (ID: 19142)

Question:

$\text{The coefficient of static friction between a wooden block of mass } 0.5 \text{ kg and a vertical rough wall is } 0.2. \text{ The magnitude of horizontal force that should be applied on the block to keep it adhere to the wall will be:}$ $\text{(Take } g = 10 \text{ ms}^{-2})$

Options:

1. $10 \text{ N}$
2. $25 \text{ N}$
3. $5 \text{ N}$
4. $30 \text{ N}$

Solution:

$\text{Gravitational force is balanced with frictional force.}$ $\text{Since block is at rest therefore}$ $fr - mg = 0 \quad \cdots \cdots \quad (1)$ $F - N = 0 \quad \cdots \cdots \quad (2)$ $fr \leq \mu N$ $\text{In limiting case}$ $fr = \mu N = \mu F$ $\therefore \mu F = mg$ $\Rightarrow F = \frac{0.5 \times 10}{0.2} = 25 \text{ N}$ $\text{Ans } 25.00$

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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
}