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Question Number 74384 by jagannath19 last updated on 23/Nov/19
Commented by jagannath19 last updated on 23/Nov/19
$${please}\:{explain} \\ $$
Answered by Tanmay chaudhury last updated on 23/Nov/19
![moment of inertia of rod =((mL^2 )/(12)) L=length of rod l=moment of inertia T=temparature l=((mL^2 )/(12))⇛ so (dl/dT)=(1/(12))×m×2L×(dL/dT) ((dl/dT)/l)=(((1/(12))×m×2L×(dL/dT))/((mL^2 )/(12)))=((2(dL/dT))/L) (dl/l)=((2dL)/L) [now α=(dL/(LdT))] (dl/l)=2αdT](Q74387.png)
$${moment}\:{of}\:{inertia}\:\:{of}\:{rod}\:=\frac{{mL}^{\mathrm{2}} }{\mathrm{12}} \\ $$$${L}={length}\:{of}\:{rod} \\ $$$${l}={moment}\:{of}\:{inertia} \\ $$$${T}={temparature} \\ $$$${l}=\frac{{mL}^{\mathrm{2}} }{\mathrm{12}}\Rrightarrow\:\:{so}\:\frac{{dl}}{{dT}}=\frac{\mathrm{1}}{\mathrm{12}}×{m}×\mathrm{2}{L}×\frac{{dL}}{{dT}} \\ $$$$\frac{\frac{{dl}}{{dT}}}{{l}}=\frac{\frac{\mathrm{1}}{\mathrm{12}}×{m}×\mathrm{2}{L}×\frac{{dL}}{{dT}}}{\frac{{mL}^{\mathrm{2}} }{\mathrm{12}}}=\frac{\mathrm{2}\frac{{dL}}{{dT}}}{{L}} \\ $$$$\frac{{dl}}{{l}}=\frac{\mathrm{2}{dL}}{{L}}\:\:\:\left[{now}\:\:\alpha=\frac{{dL}}{{LdT}}\right] \\ $$$$\frac{{dl}}{{l}}=\mathrm{2}\alpha{dT} \\ $$$$ \\ $$