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Current Question (ID: 10766)

Question:

$\text{Three rods of identical area of cross-section and made from the same metal form the sides of an isosceles triangle ABC, which is right-angled at B. The points A and B are maintained at temperatures T and } \sqrt{2}T \text{ respectively. In the steady state, the temperature of point C is } T_C. \text{ Assuming that only heat conduction takes place, } \frac{T_C}{T} \text{ is equal to:}$

Options:

1. $\frac{1}{\left(\sqrt{2}+1\right)}$
2. $\frac{3}{\left(\sqrt{2}+1\right)}$ (Correct)
3. $\frac{1}{2\left(\sqrt{2}-1\right)}$
4. $\frac{1}{\sqrt{3}\left(\sqrt{2}-1\right)}$

Solution:

\text{Hint: Since } T_B > T_A \Rightarrow \text{ Heat will flow B to A via two paths (i) B to A (ii) and along BCA as shown.} \text{Rate of flow of heat in path BCA will be same} \text{i.e. } \left(\frac{Q}{t}\right)_{BC} = \left(\frac{Q}{t}\right)_{CA} \text{From the diagram, we can see that the triangle has sides of length } a, a, \text{ and } a\sqrt{2} \text{Using Fourier's law of heat conduction:} \frac{k(\sqrt{2}T - T_C)A}{a} = \frac{k(T_C - T)A}{\sqrt{2}a} \text{Solving for } T_C: \frac{T_C}{T} = \frac{3}{1+\sqrt{2}}

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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
}