Import Question JSON

Current Question (ID: 20034)

Question:

$\text{The electric field in a region is given } \vec{E} = \left( \frac{3}{5} E_0 \hat{i} + \frac{4}{5} E_0 \hat{j} \right) \frac{\text{N}}{\text{C}}.$ $\text{The ratio of the flux of the reported field through the rectangular surface of area } 0.2 \text{ m}^2 \text{ (parallel to } y - z \text{ plane) to that of the surface of area } 0.3 \text{ m}^2 \text{ (parallel to } x - z \text{ plane) is:}$

Options:

1. $1 : 2$
2. $2 : 1$
3. $1 : 3$
4. $3 : 1$

Solution:

$\text{Hint: } \phi = \vec{E} \cdot \vec{A}$ $\vec{E} = \left( \frac{3}{5} E_0 \hat{i} + \frac{4}{5} E_0 \hat{j} \right) \frac{\text{N}}{\text{C}}$ $\vec{A}_1 = 0.2 \text{ m}^2 \text{ [parallel to } y - z \text{ plane]} = \vec{A}_1 = 0.2 \text{ m}^2 \hat{i}$ $\vec{A}_2 = 0.3 \text{ m}^2 \text{ [parallel to } x - z \text{ plane]} = \vec{A}_2 = 0.3 \text{ m} \hat{j}$ $\text{Now}$ $\phi_a = \left[ \frac{3}{5} E_0 \hat{i} + \frac{4}{5} E_0 \hat{j} \right] \cdot [0.2 \hat{i}] = \frac{3 \times 0.2}{5} E_0$ $\phi_b = \left[ \frac{3}{5} E_0 \hat{i} + \frac{4}{5} E_0 \hat{j} \right] \cdot [0.3 \hat{j}] = \frac{4 \times 0.3}{5} E_0$ $\text{Now } \frac{\phi_a}{\phi_b} = \frac{0.6}{1.2} = \frac{1}{2} = \frac{a}{b}$ $\Rightarrow a : b = 1 : 2$ $\Rightarrow a = 1$

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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
}