# A steel scale measures the length of a copper rod as 80cm when both are at 20°c , the caliberation temperature of the scale . what would the scale read for the length of the rod when both are at 40°c , given coefficient of linear expansion of steel =1.1×10^-5°c^-1 and coefficient of linear expansion of copper =1.7×10^-5°c^-1

Venkat
273 Points
4 years ago
A steel scale measures the length of a copper rod as 80cm when both are at 20°c , the caliberation temperature of the scale . what would the scale read for the length of the rod when both are at 40°c , given coefficient of linear expansion of steel =1.1×10^-5°c^-1 and coefficient of linear expansion of copper =1.7×10^-5°c^-1

New length of steel
$\\1.1\times10^{-5}=\frac{L_2-L_1}{L_1\times(T_2-T_1)} \\ \\L_2-L_1=1.1\times10^{-5}\times0.8\times20=0.00176 \\L_2=0.8+0.000176=0.800176=80.0176 cm$

New length of copper
$\\1.7\times10^{-5}=\frac{L_2-L_1}{L_1\times(T_2-T_1)} \\ \\L_2-L_1=1.7\times10^{-5}\times0.8\times20=0.000272 \\L_2=0.8+0.000272=0.800272=80.0272 cm$

Rishi Sharma
2 years ago
Dear Student,

According to the problem the length of the copper wire is measure by the scale at the temperature of 20°C is given,
Now when the temperature increased the both the things will expand,
Let the length of the wire after expanding is L'(c)
L'(c) = L+ (1+α(c) ΔT)
Let the length of the scale after expanding is L'(s)
L'(s) = L+ (1+α(s) ΔT)
Now the apparent length of the wire,
L(ap) = L'(c) -L'(s)
= ( L+ (1+α(c) ΔT)) - ( L+ (1+α(s) ΔT))
= L(α(c) -α(s) ) ΔT
= 80((17×10^(−6)−11×10^(−6))×20
= 80.0096 cm

Thanks and Regards