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Current Question (ID: 10761)

Question:

$\text{The thermal conductivities of two identical rods are } k \text{ and } 2k, \text{ respectively. If they are joined in parallel, then the ratio of heat flowing through them per second will be:}$

Options:

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

Solution:

\text{Hint: Heat flow in conduction is analogous to electric current.} \text{Step 1: Find the formula of thermal resistance.} \text{Thermal resistance is given by: } R_T = \frac{l}{kA} \text{Step 2: Find the ratio of heat flowing through the rods.} \text{For rod 1: } R_1 = \frac{l}{kA} \text{For rod 2: } R_2 = \frac{l}{2kA} \text{Since the rods are in parallel, they have the same temperature difference } \Delta T \text{Heat flow rate: } i_H = \frac{\Delta T}{R} \text{Therefore: } \frac{i_1}{i_2} = \frac{R_2}{R_1} = \frac{\frac{l}{2kA}}{\frac{l}{kA}} = \frac{1}{2} \text{Hence, the ratio of heat flowing through the rods is } 1:2

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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
}