---
title: "Borrow pit to embankment, water added"
subject: "Soil Mechanics"
chapter: 2
key: "semester_4/soil_mechanics:borrow-pit-embankment-water-added"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/borrow-pit-embankment-water-added
site: BCT Engineering
---

# Borrow pit to embankment, water added

Soil Mechanics · Chapter 2: Solids- Water –Air Relations and Index properties of soils

- Asked in: 2077 Chaitra, 6 marks
- Video: [A soil in the borrow pit is at a dry density of 17 kN/m3 with a moisture content of 10%. The soil i](https://www.youtube.com/watch?v=gzBbGjrJXCA) — Unique Series - Civil Mantraa
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/borrow-pit-embankment-water-added
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-2

## Question

A soil in the borrow pit is at a dry density of 17 kN/m³ with a moisture content of 10%. The soil is excavated from this pit and compacted in an embankment to a dry density of 18 kN/m³ with a moisture content of 15%. Compute the quantity of soil to be excavated from the borrow pit and the amount of water to be added for 100 m³ of compacted soil in the embankment.

## Short answer

Excavate 105.9 m³ from the borrow pit; add 90 kN (approx. 9.17 m³ or 9170 litres) of water.

## Answer

#### Given
Borrow pit: $\gamma_d = 17$ kN/m³, $w = 10\%$. Embankment: $\gamma_d = 18$ kN/m³, $w = 15\%$, $V = 100$ m³. $\gamma_w = 9.81$ kN/m³.

#### Dry weight of soil needed
$$
W_s = \gamma_{d,emb} \times V_{emb} = 18 \times 100 = 1800\ \text{kN}
$$

#### Volume to be excavated
The same solids weight must come from the pit:
$$
V_{pit} = \frac{W_s}{\gamma_{d,pit}} = \frac{1800}{17} = 105.9\ \text{m}^3
$$

#### Water to be added
$$
\begin{aligned}
W_{w,\ pit} &= 0.10 \times 1800 = 180\ \text{kN} \\
W_{w,\ required} &= 0.15 \times 1800 = 270\ \text{kN} \\
W_{w,\ added} &= 270 - 180 = 90\ \text{kN}
\end{aligned}
$$
In volume: $\dfrac{90}{9.81} = 9.17$ m³ $\approx 9170$ litres.

**Answer: excavate 105.9 m³ from the borrow pit; add 90 kN (about 9.17 m³ or 9170 litres) of water.**

## Check yourself

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

1. Dry weight of soil needed for 100 m3 embankment:
   A. 1700 kN
   B. 1800 kN
   C. 2070 kN
   D. 1530 kN
   Answer: B. 18 x 100 = 1800 kN. (Key point: Dry weight needed)

2. Volume to excavate from the pit is:
   A. 110 m3
   B. 100 m3
   C. 94.4 m3
   D. 105.9 m3
   Answer: D. 1800/17 = 105.9 m3. (Key point: Pit volume)

3. Why can the same Ws be used for pit and embankment?
   A. Water content is same
   B. Solids weight is conserved when moved
   C. Volume is conserved
   D. Density is same
   Answer: B. Only water and void volume change. (Key point: Principle)

4. Weight of water in the soil at the pit is:
   A. 153 kN
   B. 270 kN
   C. 90 kN
   D. 180 kN
   Answer: D. 0.10 x 1800 = 180 kN. (Key point: Water in pit)

5. Water required in embankment is:
   A. 180 kN
   B. 270 kN
   C. 90 kN
   D. 300 kN
   Answer: B. 0.15 x 1800 = 270 kN. (Key point: Water required)

6. Water to be added is about:
   A. 9.17 kN
   B. 270 kN
   C. 180 kN, 18.3 m3
   D. 90 kN, 9.17 m3
   Answer: D. 270 - 180 = 90 kN; 90/9.81 = 9.17 m3. (Key point: Water added)

## Related in Soil Mechanics

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