Question 798: Thermodynamics & KTG

Question

Two rods, one of aluminium and the other of steel, having initial lengths ' $\mathrm{L}_1$ ' and ' $\mathrm{L}_2$ ' are connected together to form a single rod of length $\left(L_1+L_2\right)$. The coefficients of linear expansion of aluminium and steel are ' $\alpha_1$ ' and ' $\alpha_2$ ' respectively. If the length of each rod increases by the same amount, when their temperatures are raised by $\mathrm{t}^{\mathrm{L}} \mathrm{C}$, then the ratio $\frac{L_1}{L_1+L_2}$ will be

Accepted Answer

When the two rods, one made of aluminum and the other made of steel, are subjected to the same temperature increase, the length of each rod increases by the same amount. This implies:
For aluminum: $$ \Delta L_{	ext{aluminum}} = L_1 \alpha_1 t $$
For steel: $$ \Delta L_{	ext{steel}} = L_2 \alpha_2 t $$
Given that these increases in lengths are equal, so:
$$ L_1 \alpha_1 t = L_2 \alpha_2 t $$
Cancelling the common factor $ t $ from both sides gives:
$$ L_1 \alpha_1 = L_2 \alpha_2 $$
We need to find the ratio $ \frac{L_1}{L_1 + L_2} $.
From $ L_1 \alpha_1 = L_2 \alpha_2 $, we can express $ L_2 $ in terms of $ L_1 $:
$$ L_2 = \frac{L_1 \alpha_1}{\alpha_2} $$
Substituting this into $ L_1 + L_2 $:
$$ L_1 + L_2 = L_1 + \frac{L_1 \alpha_1}{\alpha_2} = L_1 \left(1 + \frac{\alpha_1}{\alpha_2}ight) $$
Therefore, the desired ratio is:
$$ \frac{L_1}{L_1 + L_2} = \frac{L_1}{L_1 \left(1 + \frac{\alpha_1}{\alpha_2}ight)} $$
Simplifying:
$$ \frac{L_1}{L_1 \left(1 + \frac{\alpha_1}{\alpha_2}ight)} = \frac{1}{1 + \frac{\alpha_1}{\alpha_2}} = \frac{\alpha_2}{\alpha_1 + \alpha_2} $$
So the correct option is:
Option C: $ \frac{\alpha_2}{\alpha_1 + \alpha_2} $
.

Answer

$\frac{\alpha_2}{\alpha_1}$

Answer

$\frac{\alpha_1}{\alpha_2}$

Answer

$\frac{\alpha_2}{\alpha_1+\alpha_2}$

Answer

$\frac{\alpha_1}{\alpha_1-\alpha_2}$

Thermodynamics & KTG

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