Question

A measuring scale is calibrated to be used at temperature 30°C. The coefficient of linear expansion for the
measuring scale is 5 x 10^-4/°C. Then length measured by this scale at 40°C if actual length is 10 cm is -
(1) 10.05 cm
(2) 9.95 cm
(3) 10 cm
(4) 10.5 cm​

Answer 1

Given that,

Initial temperature = 30° C

Final temperature = 40° C

Actual length = 10 cm

Linear expansion $\alpha=5\times10^{-4}/^{\circ}C$

We need to calculate the length measured by this scale at 40° C

Using formula for length

$l=l_{0}(1+\alpha(\Delta T))$

Where, l₀ = actual length

$\alpha$ = Linear expansion

$\Delta T$ = change in temperature

Put the value into the formula

$l=10(1+5\times10^{-4}\times(40-30))$

$l=10.05\ cm$

Hence, The length measured by this scale at 40° C is 10.05 cm.

(1) is correct option.