A meter scale is made up of steel and measures correct length at `16^@C`. What will be the percentage error if this scale is used (a) on a summer day when the temperature is `46^@C` .

Explanation

(a) Length at `16^0C =L`,
`t_1=16_0C` , t_2=46_0C`
`alpha=1.1xx 10^-5/^0C`
`Delta L= L alpha Delta theta= Lxx1.1xx10_-5xx 30`
% of error ` =(Delta L/L xx 100)%`
`=(L alpha Delta theta/Lxx 100)%`
`=-1.1xx10^-5xx10xx100%`
`=-3.3xx10^-2=-0.033%`
(b) `t_2=6^0C`
`% of error= (Delta L/L xx 100)%`
`=(L alpha Delta theta/ L xx 100 )%`
`=-1.1xx10^-5xx10 xx 100%`
`-1.1 xx 10^-2 = -0.011%`