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Current Question (ID: 19631)

Question:

$\text{The temperature } \theta \text{ at the junction of two insulating sheets, having thermal resistances } R_1 \text{ and } R_2 \text{ as well as top and bottom temperatures } \theta_1 \text{ and } \theta_2 \text{ (as shown in the figure) is given by:}$

Options:

1. $\frac{\theta_2 R_2 - \theta_1 R_1}{R_2 - R_1}$
2. $\frac{\theta_1 R_2 - \theta_2 R_1}{R_2 - R_1}$
3. $\frac{\theta_1 R_2 + \theta_2 R_1}{R_1 + R_2}$
4. $\frac{\theta_1 R_1 + \theta_2 R_2}{R_1 + R_2}$

Solution:

$\text{Hint: The rate of heat flow will be same through both insulating sheets.}$ $\text{Step: Find the temperature } \theta \text{ at the junction of two insulating sheets.}$ $\text{We know that heat rate } \left( \frac{dQ}{dt} \right) \text{ at steady state is same at every point. Hence, we can write;}$ $H = \frac{\Delta T}{R} \Rightarrow H = \frac{\theta_1 - \theta}{R_1} = \frac{\theta - \theta_2}{R_2}$ $\text{The heat flow rate will be same through both we get;}$ $\Rightarrow \frac{\theta_1 - \theta}{R_1} = \frac{\theta - \theta_2}{R_2}$ $\Rightarrow R_2 \theta_1 - R_2 \theta = R_1 \theta - R_1 \theta_2$ $\Rightarrow \theta = \frac{R_2 \theta_1 + R_1 \theta_2}{R_1 + R_2}$ $\text{Hence, option (3) is the correct answer.}$

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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
}