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# What should be the sum of lengths of an aluminium and steel rod at ${0^ \circ }C$ is, so that at all temperatures their difference in length is $0.25m$ (Take coefficient linear expansion for aluminium and steel at ${0^ \circ }C$ as $22 \times {10^{ - 6}}/ ^\circ C$ and $11 \times {10^{ - 6}}/ ^\circ C$ respectively.

Last updated date: 22nd Jun 2024
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Hint: In this question we are going to find the sum of lengths of two different rods of different metals, hence linear expansion will also be different for each rod. So firstly we find their linear expansions, then applying the conditions of sum and difference we have.

Formula used:
The linear expansion of any linear body (rod/wire) is given by-
$\alpha = \Delta L/L \times \Delta T$
Where $\alpha$ is coefficient of linear expansion, $\Delta L$ is change in length, $L$ is original length, $\Delta T$ is the Temperature difference

Complete step by step solution:
When we increase the temperature of any rod then the length of rod will increase. And the expansion of aluminium is greater than the expansion of steel. But in this question we have only the difference in the lengths, we can algebraically solve this question and find the sum of lengths of the rods. For this suppose the lengths of aluminium and steel rod be ${L_a}$ and ${L_S}$ respectively.
And $\Delta {L_a}$ or $\Delta {L_S}$ are the changes in lengths of both rods. Under the same temperature difference as we are going to find length at ${0^ \circ }C$.
And the linear coefficients of both rods are ${\alpha _a}$ and ${\alpha _s}$ respectively.
Now, according to the question, we have given, ${T_1} = {0^ \circ }C$
$\Delta T = {T_2} - {T_1} \\ \Rightarrow \Delta T = T - 0 \\ \Rightarrow \Delta T = {T^ \circ }C \\$
And the change in length-
${L_s} - {L_a} = 0.25m$ …………(i)
According to the question, we have ${\alpha _a}$=$22 \times {10^{ - 6}}/ ^\circ C$, ${\alpha _s}$=$11 \times {10^{ - 6}}/ ^\circ C$then we have to find ${L_a} + {L_s} =$?
We know that
$\Rightarrow {\alpha _a} = \Delta {L_a}/{L_a} \times \Delta T$
$\Rightarrow \Delta {L_a} = {\alpha _a}{L_a} \times \Delta T$……….(ii)
Similarly, $\Delta {L_s} = {\alpha _s}{L_s} \times \Delta T$…………(iii)
According to the question the change in length at all temperatures is the same.
So $\Rightarrow \Delta {L_s} = \Delta {L_a}$
$\Rightarrow {\alpha _a}{L_a} \times \Delta T$$= {\alpha _s}{L_s} \times \Delta T$
$\Rightarrow {\alpha _a}{L_a}$$= {\alpha _s}{L_s}$
$\Rightarrow 22 \times {10^{ - 6}} \times {L_a} = 11 \times {10^{ - 6}} \times {L_S}$
$\Rightarrow 2{L_a} = {L_S}$
Put this value in equation (i)-
$\Rightarrow {L_s} - {L_a} = 0.25 \\ \Rightarrow 2{L_a} - {L_a} = 0.25 \\ \Rightarrow {L_a} = 0.25m$
Now ${L_s} = 2{L_a}$
$\Rightarrow {L_s} = 2 \times 0.25 \\ \Rightarrow {L_s} = 0.50m$
So
$\Rightarrow {L_a} + {L_s} = 0.25 + 0.50 \\ \Rightarrow {L_a} + {L_s} = 0.75m$

Therefore, the sum of length of both rods at ${0^ \circ }C$ is $0.75m$.

Note: We have to remember that the linear expansion is produced due to thermal strain which we have learnt in the topic elasticity. i.e. $\Delta L/L = \alpha \times \Delta T$ And remember that linear expansion is only applicable on longitudinal problems. We have to keep in mind that the temperature should be changed in kelvin for the calculations.