Question 794: Thermodynamics & KTG

Question

Two rods, one of copper ( Cu$)$ and the other of iron ( Fe ) having initial lengths $\mathrm{L}_1$ and $\mathrm{L}_2$ respectively are connected together to form a single rod of length $L_1+L_2$. The coefficient of linear expansion of Cu and Fe are $\alpha_c$ and $\alpha_i$ respectively. If the length of each rod increases by the same amount when their temperatures are raised by $t^{\circ} \mathrm{C}$, then ratio of $\frac{L_1-L_2}{L_1+L_2}$ will be

Accepted Answer

When two rods of different materials are connected and increase by the same amount in length due to temperature change, the expansion can be described as follows:
The change in length for the copper rod is given by:
$$ \Delta L_1 = L_1 \alpha_c t $$
The change in length for the iron rod is:
$$ \Delta L_2 = L_2 \alpha_i t $$
Given that these changes in lengths are equal:
$$ L_1 \alpha_c t = L_2 \alpha_i t $$
By simplifying, we have:
$$ L_1 \alpha_c = L_2 \alpha_i $$
From this, we can write:
$$ \frac{L_1}{L_2} = \frac{\alpha_i}{\alpha_c} $$
We need to find the ratio:
$$ \frac{L_1 - L_2}{L_1 + L_2} $$
Start by expressing $ L_1 $ and $ L_2 $ in terms of a common variable:
Let $ L_2 = x $, then $ L_1 = \frac{\alpha_i}{\alpha_c} x $.
Substitute these into $ \frac{L_1 - L_2}{L_1 + L_2} $:
$$ = \frac{\frac{\alpha_i}{\alpha_c} x - x}{\frac{\alpha_i}{\alpha_c} x + x} $$
Factor out $ x $ from the numerator and the denominator:
$$ = \frac{x \left(\frac{\alpha_i}{\alpha_c} - 1 \right)}{x \left(\frac{\alpha_i}{\alpha_c} + 1 \right)} $$
Cancel out $ x $:
$$ = \frac{\frac{\alpha_i - \alpha_c}{\alpha_c}}{\frac{\alpha_i + \alpha_c}{\alpha_c}} $$
Simplify further:
$$ = \frac{\alpha_i - \alpha_c}{\alpha_i + \alpha_c} $$
Thus, the correct ratio is:
Option C
$$ \frac{\alpha_i - \alpha_c}{\alpha_c + \alpha_i} $$

Answer

$\frac{\alpha_i}{\alpha_c+\alpha_i}$

Answer

$\frac{\alpha_c}{\alpha_c+\alpha_i}$

Answer

$\frac{\alpha_i-\alpha_c}{\alpha_c+\alpha_i}$

Answer

$\frac{\alpha_c-\alpha_i}{\alpha_c+\alpha_i}$

Thermodynamics & KTG

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