Import Question JSON

Current Question (ID: 10751)

Question:

$\text{Two conducting slabs of heat conductivity } K_1 \text{ and } K_2 \text{ are joined as shown in figure. If the temperature at the ends of the slabs are } \theta_1 \text{ and } \theta_2 \text{ (}\theta_1 > \theta_2\text{), then the final temperature (}\theta\text{)}_{m} \text{ of the junction will be:}$

Options:

1. $\frac{K_1\theta_1 + K_2\theta_2}{K_1 + K_2}$ (Correct)
2. $\frac{K_1\theta_2 + K_2\theta_1}{K_1 + K_2}$
3. $\frac{K_1\theta_2 + K_2\theta_1}{K_1 - K_2}$
4. $\text{None}$

Solution:

\text{At steady state, the rate of heat flow through both slabs must be equal.} \text{For slab 1: } Q = \frac{K_1 A (\theta_1 - \theta) t}{d} \text{For slab 2: } Q = \frac{K_2 A (\theta - \theta_2) t}{d} \text{Setting these equal:} \frac{K_1 A (\theta_1 - \theta) t}{d} = \frac{K_2 A (\theta - \theta_2) t}{d} \text{Simplifying (A, t, and d cancel out):} K_1(\theta_1 - \theta) = K_2(\theta - \theta_2) K_1\theta_1 - K_1\theta = K_2\theta - K_2\theta_2 K_1\theta_1 + K_2\theta_2 = K_1\theta + K_2\theta K_1\theta_1 + K_2\theta_2 = \theta(K_1 + K_2) \text{Therefore: } \theta = \frac{K_1\theta_1 + K_2\theta_2}{K_1 + K_2}

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}