Two meter scales, one of steel and the other of aluminum, agree at `20^@C`. Calculate the ratio aluminum-centimeter/steel-centimeter at (a) `0^@C`, (b) `40^@C` and (c) `100^@C`. `alpha_(steel) = 1.1 xx 10^(-5)C^(-1)` and `alpha_(aluminum)=2.3 xx 10^(-5)C^(-1)`.

Explanation

`L_st=L_al at 20^0C`
`alpha _al= 2.3xx10^-5/^0C`
`alpha _st =1.1xx 10^-5/^0C`
so `L_0st(1-alpha_st xx 20)`
`=L_0al(1-alpha_al xx20)`
(a) `L_0st/L_0al= (1-alpha_al xx 20)/(1-alpha_st xx20 )`
`= (1-2.3xx 10^-5xx 20)/ (1-1.1xx10^-5xx20)`
` 0.99954/0.99978=0.999759`
(b) ` L_40Al/L_40st=L_0Al(1+alpha_al xx 40)/(L_0st(1+alpha_st xx 40)`
`rArr `L_40Al/L_40st=L_0Al/L_0st xx (1+2.3xx 10^-5 xx 40)/(1+1.1xx 10^5 xx 40)`
`=0.99977xx1.00092/1.00044 = 1.0002496`
(c) `L_100Al/L_100st=L_0Al(1+alpha_Al xx 100)/L_0st(1+alpha_st xx 100)`
`=0.99977xx 1.0023/1.0023=1.00096`