A rod is measured at 10∘C using a scale calibrated at that temperature. Find the percentage error in length, if thermal coefficients of linear expansion of the rod and scale are 3×10−5/∘C and 2×10−5/∘C respectively, and the rod is measured at 50∘C.
A
0.01%
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B
0.02%
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C
0.03%
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D
0.04%
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Solution
The correct option is D0.04% Let the length of the rod at temperature T0 [actual length] be L0 α0=3×10−5/∘C αs=2×10−5/∘C T0=10∘C T=50∘C Measured length at temperature T L=L0[1+(α0−αs)(T−T0)] Percentage error =(L−L0L0)×100 =[L0[1+(α0−αs)(T−T0)]−L0L0]×100 =[(α0−αs)(T−T0)]×100 =[(10−5)(40)]100 =4×10−2 =0.04%