Answer :

Let T0 be the temperature at which the meter scale measures the correct length, T0 = 16° = 289 K


Let the length of the meter scale be l


Given:


The coefficient of linear expansion of the steel meter scale (α), α = 11× 10-6 C-1


a)


Given:


The temperature at which the scale measures during a summer day = Ts = 46° = 319 K


Therefore, Δ T = 319 – 289 = 30K


The change in length due to linear expansion as a consequence of rise in temperature is given as,


ΔL= lαΔT


So the above equation can be written as,



The percentage error is given as,


% error



b)


Similarly,


The temperature at which the scale measures during a winter day = Tw = 6° = 279 K


Therefore, Δ T = 289 – 279 = 10K


The change in length due to linear expansion as consequence of fall in temperature is given as,


ΔL= lαΔT


So the above equation can also be written as,



The percentage error,


% error



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