The length of an iron bar is 100cm at 20°C. At what temperature will its length increase by 0.01%? [Linear expansivity of iron = 1.2 x 10\(^{-5}\)°C]
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Correct Answer: Option C
Explanation:
GIVEN DATA:
where ΔL is the increase in length = 0.01
L1 is the initial length = 100
T1 = 20
T2 = ?
α is the linear expansivity of iron = 1.2 x 10\(^{-5}\)°C
Formula
ΔL = L1αΔT
ΔT = \(\frac{ΔL}{L1α}\)
ΔT is the change in temperature = T2 - T1
T2 = \(\frac{ΔL}{L1α}\) + T1
T2 = \(\frac{0.01}{100 \times 1.2 x 10^{-5}}\) + 20
T2 = 8.3 + 20
T2 = 28.3
GIVEN DATA:
where ΔL is the increase in length = 0.01
L1 is the initial length = 100
T1 = 20
T2 = ?
α is the linear expansivity of iron = 1.2 x 10\(^{-5}\)°C
Formula
ΔL = L1αΔT
ΔT = \(\frac{ΔL}{L1α}\)
ΔT is the change in temperature = T2 - T1
T2 = \(\frac{ΔL}{L1α}\) + T1
T2 = \(\frac{0.01}{100 \times 1.2 x 10^{-5}}\) + 20
T2 = 8.3 + 20
T2 = 28.3