Question

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\(^{-1}\).

Answer 1

Given

• Steel tape length at 27.0 °C: \( L = 1.00 \,\text{m} = 100.0 \,\text{cm} \)
• Reading of steel rod on tape at 45.0 °C: \( l_{\text{read}} = 63.0 \,\text{cm} \)
• Calibration temperature of tape: \( 27.0^\circ\text{C} \)
• Hot day temperature: \( 45.0^\circ\text{C} \)
• Coefficient of linear expansion of steel: \( \alpha = 1.20 \times 10^{-5} \,\text{K}^{-1} \)

1. Length of the tape at 45 °C

Temperature rise:

\( \Delta T = 45.0 - 27.0 = 18.0 \,\text{K} \)

Linear expansion of the 1 m tape:

\( L_{45} = L \big(1 + \alpha \Delta T \big) \) \( = 100.0 \left[1 + (1.20 \times 10^{-5}) \times 18.0 \right] \,\text{cm} \) \( = 100.0 \left[1 + 2.16 \times 10^{-4} \right] \,\text{cm} \) \( = 100.0 \times 1.000216 = 100.0216 \,\text{cm} \)

So at 45 °C, the scale that is marked “100 cm” is actually 100.0216 cm long. Therefore, 1 “cm” marking on the tape at 45 °C corresponds to:

\( 1\,\text{scale-cm} = \dfrac{100.0216}{100.0} = 1.000216 \,\text{true cm} \)

2. Actual length of the rod at 45 °C

Reading is 63.0 scale-cm, so true length:

\( l_{\text{actual, 45}} = 63.0 \times 1.000216 \,\text{cm} = 63.0136 \,\text{cm} \)

Actual length of the steel rod at 45.0 °C ≈ 63.01 cm.

3. Length of the same rod at 27 °C

At 27.0 °C the tape is correctly calibrated; 1 scale-cm = 1 true cm. The rod’s length at 27.0 °C will therefore be the reading on that correctly calibrated tape.

Length of the rod at 27.0 °C = 63.0 cm.