L' = L + L@(T2-T1)
50.12 = 50 + 50×12×10^(-6)[T2-12]
REARRANGING:
6×10^(-4)[T2-12] = 0.12
T2 -12 = 200
T2 = 212
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L' = L + L@(T2-T1)
50.12 = 50 + 50×12×10^(-6)[T2-12]
REARRANGING:
6×10^(-4)[T2-12] = 0.12
T2 -12 = 200
T2 = 212
yes this is correct for the question
An iron rod has a length of 50.0 cm at 12 degree Celsius at what temperature will its length becomes 50.12 cm coefficient of linear expansion of iron is equal to 12 Into 10 power minus 6 degree Celsius power minus 1
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yes it is correct.....
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