A scale which is calibrated at 10∘C measures the length of a stick to be 60cm, at temperature 30∘C . If thermal coefficient of linear expansion of the rod is 2×10−5/∘C, the correct length of the stick in cm is:
A
60.015
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B
60.024
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C
60.400
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D
60.600
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Solution
The correct option is B60.024 When measuring at high temperature, the scale will be more than its calibrated length. So, actual length of stick will be more that measured length.
Change in length Δl=lαΔT =60×2×10−5×(30−10) =0.024cm
So, correct length =l+Δl=60+0.024 =60.024cm