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Current Question (ID: 19132)

Question:

$\text{The mass of a hydrogen molecule is } 3.32 \times 10^{-27} \text{ kg. If } 10^{23} \text{ hydrogen molecules strike, per second, a fixed wall of area } 2 \text{ cm}^2 \text{ at an angle of } 45^\circ \text{ to the normal, and rebound elastically with a speed of } 10^3 \text{ m/s, then the pressure on the wall is nearly:}$

Options:

1. $2.35 \times 10^3 \text{ N/m}^2$
2. $4.70 \times 10^3 \text{ N/m}^2$
3. $2.35 \times 10^2 \text{ N/m}^2$
4. $4.70 \times 10^2 \text{ N/m}^2$

Solution:

$\text{Hint: } P = \frac{F}{A} = \frac{n \times 2m \times v \times \cos(\theta)}{A}$ $\text{Step: Find the pressure on the wall.}$ $v_x = v \cos(\theta) = v \times \frac{1}{\sqrt{2}} = \frac{10^3}{\sqrt{2}} \text{ m/s}$ $\Delta p = m \times [v_x - (-v_x)] = m \times (2v_x) = 2mv \cos(\theta)$ $\text{Total change in momentum} = n \times \Delta p = n \times 2mv \cos(\theta) = n \times 2m \times \frac{v}{\sqrt{2}}$ $F = n \times 2m \times v \cos(\theta)$ $P = \frac{F}{A} = \frac{n \times 2m \times v \cos(\theta)}{A}$ $P = \frac{10^{23} \times 2 \times (3.32 \times 10^{-27}) \times (10^3) \times \frac{1}{\sqrt{2}}}{2 \times 10^{-4}}$ $P = 10^{23} \times 3.32 \times 10^{-23} \times \frac{1}{\sqrt{2}} = \frac{3.32}{\sqrt{2}} \times 10^3 \text{ N/m}^2$ $\Rightarrow P \approx 2.35 \times 10^3 \text{ N/m}^2$ $\text{Hence, option (1) is the correct answer.}$

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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
}