Question

2 roads one of aluminium of an length L1 having coefficient linear expansion Alpha- a and other steel of length L2 having coefficient of linear expansion alpha s are joined end to end together . expansion in both the rod is same on variation of temperature . then the value of L1 upon L1 plus L2 is
with full solution then I'll mark it as brainleast and you'll receive extra 50 points

Answer 1

SOLUTION HINT :

ℓ1αat=ℓ2αst
⇒ℓ2/ℓ1=αa/αs
⇒(ℓ2+ℓ1)/ℓ1=(αa+αs)/αs
⇒ℓ1/(ℓ1+ℓ2)=αs/(αa+αs)
Solution : 
Answer : $\frac{5}{3}$αExplanation :L1=L(1+α△T)L1=L(1+α△T)L2=2L(1+2α△T)L2=2L(1+2α△T)L3=L1+L2L3=L1+L2=3L+5α△T=3L+5α△T=3L(1+53α△T)=3L(1+$\frac{5}{3}$α△T)Therefore , Coefficient of linear expansion of rod =$\frac{5}{3}$α