The coefficient of linear expansion α = 1.2×10^-5 /°C for a metal rod of length L0 = 2.0 m. If the temperature increases by ΔT = 40°C, what is the change in length ΔL?

The coefficient of linear expansion α = 1.2×10^-5 /°C for a metal rod of length L0 = 2.0 m. If the temperature increases by ΔT = 40°C, what is the change in length ΔL?

• A. 9.6×10^-4 m
• B. 4.8×10^-4 m
• C. 1.92×10^-3 m
• D. 9.6×10^-3 m

Answer: (B)

A metal rod expands linearly with temperature, so the change in length is ΔL = α L0 ΔT. Here you first find the fractional expansion by multiplying α by ΔT, then apply that to the original length. Compute α ΔT = (1.2×10^-5 /°C) × 40°C = 4.8×10^-4. Then ΔL = L0 × (α ΔT) = 2.0 m × 4.8×10^-4 = 9.6×10^-4 m, which is 0.96 mm. The key is to multiply by the original length L0; omitting it would give 4.8×10^-4 m, not the actual change for this rod.

A metal rod expands linearly with temperature, so the change in length is ΔL = α L0 ΔT. Here you first find the fractional expansion by multiplying α by ΔT, then apply that to the original length.

Compute α ΔT = (1.2×10^-5 /°C) × 40°C = 4.8×10^-4. Then ΔL = L0 × (α ΔT) = 2.0 m × 4.8×10^-4 = 9.6×10^-4 m, which is 0.96 mm. The key is to multiply by the original length L0; omitting it would give 4.8×10^-4 m, not the actual change for this rod.