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Current Question (ID: 10727)

Question:

$\text{If two rods of length L and 2L having coefficients of linear expansion } \alpha \text{ and } 2\alpha \text{ respectively are connected so that their total length becomes 3L, the average coefficient of linear expansion of the composition of rods equals:}$

Options:

1. $\frac{3}{2}\alpha$
2. $\frac{5}{2}\alpha$
3. $\frac{5}{3}\alpha$ (Correct)
4. $\text{none of these}$

Solution:

$\text{Solution:}$ $\text{Let the initial lengths be } l_1 = L \text{ and } l_2 = 2L$ $\text{Let the coefficients of linear expansion be } \alpha_1 = \alpha \text{ and } \alpha_2 = 2\alpha$ $\text{After expansion by temperature change } \Delta t\text{:}$ $l_1' = l_1(1 + \alpha \Delta t) = L(1 + \alpha \Delta t)$ $l_2' = l_2(1 + 2\alpha \Delta t) = 2L(1 + 2\alpha \Delta t)$ $\text{Total length after expansion:}$ $l_1' + l_2' = (l_1 + l_2)(1 + \alpha_{\text{eff}} \Delta t)$ $\text{Substituting the values:}$ $L(1 + \alpha \Delta t) + 2L(1 + 2\alpha \Delta t) = (L + 2L)(1 + \alpha_{\text{eff}} \Delta t)$ $L + L\alpha \Delta t + 2L + 4L\alpha \Delta t = 3L + 3L\alpha_{\text{eff}} \Delta t$ $3L + 5L\alpha \Delta t = 3L + 3L\alpha_{\text{eff}} \Delta t$ $\text{Comparing coefficients:}$ $5L\alpha \Delta t = 3L\alpha_{\text{eff}} \Delta t$ $\text{Therefore:}$ $\alpha_{\text{eff}} = \frac{5\alpha}{3} = \frac{5}{3}\alpha$

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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
}