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Current Question (ID: 19628)

Question:

$\text{Two different wires having lengths } L_1 \text{ and } L_2, \text{ and respective temperature coefficient of linear expansion } \alpha_1 \text{ and } \alpha_2, \text{ are joined end-to-end. Then the effective temperature coefficient of linear expansion is:}$

Options:

1. $\frac{\alpha_1 \alpha_2}{\alpha_1 + \alpha_2} \frac{L_2 L_1}{(L_2 + L_1)^2}$
2. $2 \sqrt{\alpha_1 \alpha_2}$
3. $\frac{\alpha_1 + \alpha_2}{2}$
4. $\frac{\alpha_1 L_1 + \alpha_2 L_2}{L_1 + L_2}$

Solution:

$\text{Hint: } L = L_0 \alpha \Delta T$ $\text{At } T^\circ C \quad L = L_1 + L_2$ $\text{At } T + \Delta T \quad L'_{eq} = L'_{1} + L'_{2}$ $\text{Where } L'_{1} = L_1 (1 + \alpha_1 \Delta T)$ $L'_{2} = L_2 (1 + \alpha_2 \Delta T)$ $L'_{eq} = (L_1 + L_2) (1 + \alpha_{avg} \Delta T)$ $\Rightarrow (L_1 + L_2) (1 + \alpha_{avg} \Delta T) = L_1 + L_2 + L_1 \alpha_1 \Delta T + L_2 \alpha_2 \Delta T$ $\Rightarrow (L_1 + L_2) \alpha_{avg} \Delta T = L_1 \alpha_1 + L_2 \alpha_2$ $\Rightarrow \alpha_{avg} = \frac{\alpha_1 L_1 + \alpha_2 L_2}{L_1 + L_2}$

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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
}