Question 6 of 20intermediate🔧 ApplyNumerical3 marks

A steel tape 1m long is correctly calibrated for a temperature of 27.0 °C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0 °C. What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when the temperature is 27.0 °C? Coefficient of linear expansion of steel = 1.20×10−5 K−11.20 \times 10^{-5} \text{ K}^{-1}.

Correct Answer

(i) The actual length of the steel rod at 45.0 °C is 63.01 cm.

(ii) The length of the steel rod at 27.0 °C is 63.0 cm.

Exercise: EXERCISES | Q: 10.6 | (Page 23)
For More Understanding

Explanation

This question tests the understanding of thermal expansion and measurement errors due to expansion of measuring instruments. The steel tape expands at higher temperatures, causing each 'cm' mark to represent a longer actual length. The key insight is that both the tape and the steel rod expand by the same factor since they are made of the same material (steel). When calculating the actual length at 45°C, we must account for the tape's expansion. When finding the length at 27°C, we account for the rod's contraction from 45°C to 27°C.

Solution Steps

  1. Step 1: Calculate the temperature change from calibration temperature: ΔT = 45.0 - 27.0 = 18 K

  2. Step 2: Calculate the fractional expansion of the tape: ΔL/L = α × ΔT = 1.20×101.20 \times 10⁻⁵ × 18 = 2.16×102.16 \times 10⁻⁴

  3. Step 3: Calculate the actual length of the rod at 45°C: Actual length = Measured length × (1 + α × ΔT) = 63.0 × (1 + 2.16×102.16 \times 10⁻⁴) = 63.0×1.00021663.0 \times 1.000216 = 63.01 cm

  4. Step 4: For length at 27°C, calculate the contraction from 45°C: ΔL = L × α × ΔT = 63.01×1.2063.01 \times 1.20 × 10⁻⁵ × (-18) = -0.0136 cm

  5. Step 5: Calculate length at 27°C: L₂₇ = 63.01 + (-0.0136) = 63.0 cm

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