A rod of length 1.0 m has a linear expansion coefficient α = 12×10^-6 /°C. If the temperature increases by ΔT = 30°C, by how much does the length change?

A rod of length 1.0 m has a linear expansion coefficient α = 12×10^-6 /°C. If the temperature increases by ΔT = 30°C, by how much does the length change?

• A. 0.00036 m (0.36 mm)
• B. 0.0036 m
• C. 0.000036 m
• D. 0.036 m

Answer: (A)

When a material heats up, its length changes in proportion to its original length: ΔL = α L0 ΔT. Here, L0 is 1.0 m, α is 12×10^-6 /°C, and ΔT is 30°C. Multiply α by ΔT: 12×10^-6 × 30 = 3.6×10^-4. Then multiply by the original length: ΔL = 1.0 m × 3.6×10^-4 = 3.6×10^-4 m, which is 0.00036 m or 0.36 mm. The temperature increase causes the rod to lengthen since α is positive.

When a material heats up, its length changes in proportion to its original length: ΔL = α L0 ΔT. Here, L0 is 1.0 m, α is 12×10^-6 /°C, and ΔT is 30°C. Multiply α by ΔT: 12×10^-6 × 30 = 3.6×10^-4. Then multiply by the original length: ΔL = 1.0 m × 3.6×10^-4 = 3.6×10^-4 m, which is 0.00036 m or 0.36 mm. The temperature increase causes the rod to lengthen since α is positive.