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Question

A scale which is calibrated at 10C measures the length of a stick to be 60 cm, at temperature 30C . If thermal coefficient of linear expansion of the rod is 2×105/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 B 60.024
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
=60×2×105×(3010)
=0.024 cm
So, correct length =l+Δl=60+0.024
=60.024 cm

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