Question 773: 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 $\mathrm{L}_{\mathrm{B}}$ respectively. Their coefficients of linear expansion are $\alpha_A$ and $\alpha_B$ respectively. It is observed that the difference between their lengths remains constant at all temperatures. The ratio $\mathrm{L}_{\mathrm{A}}: \mathrm{L}_{\mathrm{B}}$ is given by

Accepted Answer

To find the ratio $ L_A : L_B $ such that the difference in lengths of rods A and B remains constant at all temperatures, we start by considering their lengths at a temperature $ t $.
The length of rod A at temperature $ t $ is given by:
$ L_{A_t} = L_A + L_A \alpha_A \Delta t $
Similarly, the length of rod B at temperature $ t $ is:
$ L_{B_t} = L_B + L_B \alpha_B \Delta t $
The difference in their lengths at temperature $ t $ is then:
$ \Delta L = L_{A_t} - L_{B_t} = (L_A - L_B) + (L_A \alpha_A - L_B \alpha_B) \Delta t $
For this difference to remain constant regardless of changes in temperature, the coefficient of $\Delta t$ must be zero:
$ L_A \alpha_A - L_B \alpha_B = 0 $
Thus, we have:
$ L_A \alpha_A = L_B \alpha_B $
Rearranging to find the length ratio:
$ \frac{L_A}{L_B} = \frac{\alpha_B}{\alpha_A} $
This relationship tells us that the ratio of lengths $ L_A : L_B $ is $\frac{\alpha_B}{\alpha_A}$.

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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