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Current Question (ID: 10762)

Question:

$\text{Two rods (one semi-circular and the other straight) of the same material and of the same cross-sectional area are joined as shown in the figure. The points A and B are maintained at different temperatures. The ratio of the heat transferred through a cross-section of a semi-circular rod to the heat transferred through a cross-section of a straight rod at any given point in time will be:}$

Options:

1. $2 : \pi$ (Correct)
2. $1 : 2$
3. $\pi : 2$
4. $3 : 2$

Solution:

$\text{Using Fourier's law of heat conduction:}$ $\frac{dQ}{dt} = \frac{KA\Delta\theta}{l}$ $\text{For both rods, K, A and } \Delta\theta \text{ are same}$ $\Rightarrow \frac{dQ}{dt} \propto \frac{1}{l} \text{ So}$ $\text{The ratio of heat transfer rates:}$ $\frac{\left(\frac{dQ}{dt}\right)_{\text{semicircular}}}{\left(\frac{dQ}{dt}\right)_{\text{straight}}} = \frac{l_{\text{straight}}}{l_{\text{semicircular}}} = \frac{2r}{\pi r} = \frac{2}{\pi}$

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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
}