Question 724: Thermodynamics & KTG

Question

At certain temperature, $\operatorname{rod} \mathrm{A}$ and $\operatorname{rod} \mathrm{B}$ of different materials have lengths $\mathrm{L}_{\mathrm{A}}$ and $\mathrm{L}_B$ respectively. Their co-efficients of linear expansion are $\alpha_A$ and $\alpha_B$ respectively. It is observed that the difference between their lengths remain constant at all temperatures. The ratio $L_A / L_B$ is given by

Accepted Answer

Let's analyze the problem step by step:
Assume that at some initial temperature, rod A has length $$L_A$$ and rod B has length $$L_B$$. When the temperature increases by $$\Delta T$$, the new lengths become:
$$	ext{Rod A: } L_A(1 + \alpha_A \Delta T)$$
$$	ext{Rod B: } L_B(1 + \alpha_B \Delta T)$$
The condition given is that the difference between the lengths remains constant at all temperatures. This means:
$$L_B(1 + \alpha_B \Delta T) - L_A(1 + \alpha_A \Delta T) = L_B - L_A$$
Expanding the left-hand side:
$$L_B + L_B\alpha_B\Delta T - L_A - L_A\alpha_A\Delta T = L_B - L_A$$
Notice that the terms $$L_B$$ and $$-L_A$$ on both sides cancel, leaving:
$$L_B\alpha_B\Delta T - L_A\alpha_A\Delta T = 0$$
Since $$\Delta T 
eq 0$$, we can divide both sides by $$\Delta T$$ to obtain:
$$L_B\alpha_B - L_A\alpha_A = 0$$
Rearranging the equation gives:
$$L_A\alpha_A = L_B\alpha_B$$
Finally, dividing both sides by $$L_B\alpha_A$$ yields:
$$\frac{L_A}{L_B} = \frac{\alpha_B}{\alpha_A}$$
Thus, the required ratio is:
$$\frac{L_A}{L_B} = \frac{\alpha_B}{\alpha_A}$$
This corresponds to Option B.

Answer

$\frac{\alpha_A}{\alpha_B}$

Answer

$\frac{\alpha_B}{\alpha_A}$

Answer

$\frac{\alpha_A+\alpha_B}{\alpha_A}$

Answer

$\frac{\alpha_A+\alpha_B}{\alpha_B}$

Thermodynamics & KTG

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