---
title: "True length of 1650 m line"
subject: "Surveying I"
chapter: 2
key: "semester_3/surveying_i:tape-1650m-30m-pull-100n"
source: https://www.bctengineer.com/q/semester_3/surveying_i/tape-1650m-30m-pull-100n
site: BCT Engineering
---

# True length of 1650 m line

Surveying I · Chapter 2: Distance Measurements

- Asked in: 2063 Baisakh, 10 marks
- Video: [True Length Calculation of chain in surveying when chain short & long | Numerical Solution](https://www.youtube.com/watch?v=sDa_gGmN_HM) — Learn Civil Tech & Tricks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/tape-1650m-30m-pull-100n
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-2

## Question

A line was measured with a steel tape which was exactly 30 metres at 20°C at a pull of 100 N, the measured length being 1650.00 metres. The temperature during measurement was 30°C and the pull applied was 150 N. Find the length of the line, if the cross-sectional area of the tape was 0.025 sq.cm. The coefficient of expansion of the material of tape per 1°C = $3.5\times10^{-6}$ and the modulus of elasticity of the material of the tape = $2.1\times10^{5}$ N/mm².

## Short answer

Only temperature and pull corrections apply: temperature +0.0578 m, pull +0.1571 m. True length of the line = 1650.215 m.

## Answer

**Assumption.** No weight or support condition is given, so the tape is taken as fully supported and no sag correction is applied.

#### Data
- $L = 1650.00$ m; $T_0 = 20^\circ$C, $T = 30^\circ$C; $P_0 = 100$ N, $P = 150$ N
- $A = 0.025$ cm$^2 = 2.5$ mm$^2$; $E = 2.1\times 10^{5}$ N/mm$^2$, so $AE = 2.5\times 2.1\times 10^{5} = 525{,}000$ N
- $\alpha = 3.5\times 10^{-6}$/$^\circ$C

#### 1. Temperature
$$
C_t = 3.5\times 10^{-6}\times(30 - 20)\times 1650 = +0.0578\ \text{m}
$$

#### 2. Pull
$$
C_p = \frac{(150 - 100)\times 1650}{525{,}000} = +0.1571\ \text{m}
$$

#### 3. True length
$$
L_{true} = 1650.00 + 0.0578 + 0.1571 = 1650.2149\ \text{m}
$$

**Answer: length of the line $= 1650.215$ m.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is no sag correction applied?
   A. No weight or support condition given; taken fully supported
   B. Pull equals
   C. Tape is steel
   D. Sag too big
   Answer: A. Stated assumption. (Key point: Sag)

2. A = 0.025 cm^2 equals
   A. 0.25 mm^2
   B. 2.5 mm^2
   C. 25 mm^2
   D. 0.025 mm^2
   Answer: B. 1 cm^2 = 100 mm^2. (Key point: Area conversion)

3. C_p = 
   A. -0.1571
   B. 0.0578
   C. (150-100) x 1650/525,000 = +0.1571 m
   D. 1650/525,000
   Answer: C. Positive. (Key point: Pull correction)

4. C_t = 
   A. -0.0578
   B. 0.578
   C. 0
   D. 3.5e-6 x 10 x 1650 = +0.0578 m
   Answer: D. Hotter. (Key point: Temperature)

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