The coefficients of linear expansion of brass | vedclass

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Question

(D) Let the length of the brass rod at temperature $\Delta 	heta$ be ${L_1}$ and the steel rod be ${L_2}$.${L_1} = {l_1}(1 + {\alpha _1}\Delta 	heta

Vedclass · Accepted Answer

${\alpha _1}{l_1} = {\alpha _2}{l_2}$

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(D) Let the length of the brass rod at temperature $\Delta 	heta$ be ${L_1}$ and the steel rod be ${L_2}$.${L_1} = {l_1}(1 + {\alpha _1}\Delta 	heta )$${L_2} = {l_2}(1 + {\alpha _2}\Delta 	heta )$The difference in length at temperature $\Delta 	heta$ is ${L_2} - {L_1} = {l_2}(1 + {\alpha _2}\Delta 	heta ) - {l_1}(1 + {\alpha _1}\Delta 	heta )$.${L_2} - {L_1} = ({l_2} - {l_1}) + \Delta 	heta ({l_2}{\alpha _2} - {l_1}{\alpha _1})$.For the difference in length to remain the same as at $0^{\circ}C$,we require ${L_2} - {L_1} = {l_2} - {l_1}$.This implies $\Delta 	heta ({l_2}{\alpha _2} - {l_1}{\alpha _1}) = 0$.Since $\Delta 	heta 
eq 0$,we must have ${l_2}{\alpha _2} - {l_1}{\alpha _1} = 0$,or ${l_1}{\alpha _1} = {l_2}{\alpha _2}$.

The coefficients of linear expansion of brass and steel are ${\alpha _1}$ and ${\alpha _2}$ respectively. If we take a brass rod of length ${l_1}$ and a steel rod of length ${l_2}$ at $0^{\circ}C$,their difference in length $({l_2} - {l_1})$ will remain the same at any temperature if:

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