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Current Question (ID: 10767)

Question:

$\text{A ring consisting of two parts, } ADB \text{ and } ACB, \text{ of the same heat conductivity } k, \text{ conducts an amount of heat } H. \text{ The } ADB \text{ part is now replaced with another metal, keeping the temperatures } T_1 \text{ and } T_2 \text{ constant. The amount of heat carried rises to } 2H. \text{ What should be the heat conductivity of the new } ADB \text{ part?}$ $\left(\frac{ACB}{ADB} = 3\right)$

Options:

1. $\frac{7}{3}k$ (Correct)
2. $2k$
3. $\frac{5}{2}k$
4. $3k$

Solution:

$\text{Hint: Net resistance becomes half.}$ $\text{Step 1: Find initial and final thermal resistance}$ $R = \frac{l}{kA}$ $R_1 = \frac{l}{kA}$ $R_2 = \frac{3l}{kA}$ $\frac{1}{R_1'} = \frac{1}{R_1} + \frac{1}{R_2}$ $R_1' = \frac{3l}{4kA}$ $\text{When ADB part is replaced}$ $\frac{1}{R_2'} = \frac{k'A}{l} + \frac{kA}{3l}$ $\text{Step 2: Use the fact that temp. difference is constant}$ $i_T = \frac{\Delta T}{R}$ $i_T R = const.$ $H(R_1') = 2H(R_2')$ $R_2' = \frac{R_1'}{2}$ $\frac{2}{R_1'} = \frac{1}{R_2'}$ $\frac{8kA}{3l} = \frac{k'A}{l} + \frac{kA}{3l}$ $k' = \frac{7k}{3}$

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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
}