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Current Question (ID: 19634)

Question:

$\text{As per the given figure, two plates } A \text{ and } B \text{ of thermal conductivity } K \text{ and } 2K \text{ are joined together to form a compound plate.}$ $\text{The thickness of the plates are } 4.0 \text{ cm and } 2.5 \text{ cm respectively and the area of the cross-section is } 120 \text{ cm}^2 \text{ for each plate.}$ $\text{The equivalent thermal conductivity of the compound plate is } \left(1 + \frac{5}{\alpha}\right) K, \text{ then the value of } \alpha \text{ will be:}$

Options:

1. 21
2. 10
3. 30
4. 54

Solution:

$\text{Hint: The thermal resistance } R = \frac{L}{KA}$ $\text{Step 1: Find the equivalent thermal conductivity } K_{eq}. \text{The thermal resistance is given by; } R = \frac{L}{KA}$ $\text{Where: } L = \text{the thickness of the material, } K \text{ thermal conductivity, } A = \text{the cross-sectional area}$ $\text{Resistance of Plate } A: R_A = \frac{L_A}{K_A A}$ $\text{Resistance of Plate } B: R_B = \frac{L_B}{K_B A}$ $\text{Since the two plates are in series, the total resistance } R_{eq} \text{ is given by; } R_{eq} = R_A + R_B$ $\frac{L_A + L_B}{K_{eq} A} = \frac{L_A}{K_A A} + \frac{L_B}{K_B A}$ $\frac{L_A + L_B}{K_{eq}} = \frac{L_A K_B + L_B K_A}{K_A K_B}$ $K_{eq} = \frac{(K_A K_B)(L_A + L_B)}{L_A K_B + L_B K_A} \Rightarrow \frac{2K^2 \times 6.5}{4 \times 2K + 2.5 \times K}$ $K_{eq} = \frac{26K}{21} \Rightarrow \left(1 + \frac{5}{21}\right) K$ $\text{Therefore compared with the given equation } \left(1 + \frac{5}{\alpha}\right) K, \text{ the value of } \alpha \text{ is 21.}$ $\text{Hence, option (1) is the correct answer.}$

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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
}