Q. 155.0( 1 Vote )

# <span lang="EN-US

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  Rate this question :

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