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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

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

Grade:11

2 Answers

Venkat
273 Points
2 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
askIITians Faculty 646 Points
9 months ago
Dear Student,
Please find below the solution to your problem.

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

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