Import Question JSON

Current Question (ID: 19629)

Question:

$\text{Three rods of identical cross-sections and lengths are made of three different materials of thermal conductivity } K_1, K_2, K_3, \text{ respectively. They are joined together at their ends to make a long rod (see figure). One end of the long rod is maintained at } 100^\circ C \text{ and the other at } 0^\circ C \text{ (see figure). If the joints of the rod are at } 70^\circ C \text{ and } 20^\circ C \text{ in a steady state and there is no loss of energy from the surface of the rod, the correct relationship between } K_1, K_2, \text{ and } K_3 \text{ is:}$

Options:

1. $K_1 : K_3 = 2 : 3; K_2 : K_3 = 2 : 5$
2. $K_1 < K_2 < K_3$
3. $K_1 : K_2 = 5 : 2; K_1 : K_3 = 3 : 5$
4. $K_1 > K_2 > K_3$

Solution:

$\text{Hint: } \frac{dQ}{dt} = \frac{kA \Delta T}{l}$ $\text{Rods are identical have same length (} l \text{) and area of cross-section (} A \text{)}$ $\text{Combination are in series, so heat current is same for all Rods}$ $\frac{\left( \frac{\Delta Q}{\Delta t} \right)_{AB}}{(100 - 70)K_1A} = \frac{\left( \frac{\Delta Q}{\Delta t} \right)_{BC}}{(70 - 20)K_2A} = \frac{\left( \frac{\Delta Q}{\Delta t} \right)_{CD}}{(20 - 0)K_3A} = \text{Heat}$ $\Rightarrow 30K_1 = 50K_2 = 20K_3$ $3K_1 = 2K_3$ $\frac{K_1}{K_3} = \frac{2}{3} = 2 : 3$ $5K_2 = 2K_3$ $\frac{K_2}{K_3} = \frac{2}{5} = 2 : 5$

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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
}