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 and (b) on a winter say when the temperature is 6°C? Coefficient of linear expansion of steel = 11 × 10 –6 °C –1 .

Question

Philoid · Accepted 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 = T_s = 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 = T_w = 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

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 and (b) on a winter say when the temperature is 6°C? Coefficient of linear expansion of steel = 11 × 10–6 °C–1.

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 and (b) on a winter say when the temperature is 6°C? Coefficient of linear expansion of steel = 11 × 10^–6 °C^–1.