A 30 m long tape is standardized at 25 ◦C. It was used to measure the length of a line which came out to be 200 m. The temperature during the measurement was 35 ◦C. The coefficient of expansion of the tape was 11×10⁻⁶ per ^◦C. The cor rection in the measured length (in mm) due to change in temperature is ________(in integer).

The correct answer is 22.

Understanding the Concept:
When a surveying tape is used at a temperature that differs from its standardization temperature, it expands or contracts. This change in the tape's length introduces a systematic error, requiring a temperature correction ($C_t$) to be applied to the measured length.

The formula for temperature correction is given by:

$$C_t=\alpha \times (T_m-T_0)\times L$$

Where:
• $\alpha$ is the coefficient of thermal expansion of the tape.
• $T_m$ is the temperature during the measurement.
• $T_0$ is the temperature at which the tape was standardized.
• $L$ is the measured length of the line.

Given Data:
• Standardized tape length = 30 m (used to measure the line)
• Measured length of the line ($L$) = 200 m
• Temperature during measurement ($T_m$) = 35 °C
• Standard temperature ($T_0$) = 25 °C
• Coefficient of thermal expansion ($\alpha$) = 11 × 10^-6 per °C

Step-by-Step Calculation:

1. Find the temperature difference ($\Delta T$):

$$\Delta T=T_m-T_0$$

$$\Delta T=35-25=10°C$$

2. Substitute the parameters into the correction formula:

$$C_t=(11\times {10}^{-6})\times 10\times 200$$

3. Calculate the value in meters (m):

$$C_t=11\times {10}^{-6}\times 2000$$

$$C_t=22000\times {10}^{-6}=0.022\text{m}$$

4. Convert the result to millimeters (mm):

Since 1 meter equals 1000 millimeters:

$$C_t=0.022\times 1000=22\text{mm}$$

Thus, the temperature correction required for the measured length is 22 mm.