---
title: "Limestone reserve for cement factory"
subject: "Engineering Geology II"
chapter: 7
key: "semester_4/engineering_geology_ii:reserve-limestone-cement-density"
source: https://www.bctengineer.com/q/semester_4/engineering_geology_ii/reserve-limestone-cement-density
site: BCT Engineering
---

# Limestone reserve for cement factory

Engineering Geology II · Chapter 7: Geology and Construction Materials

- Asked in: 2080 Chaitra, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/engineering_geology_ii/reserve-limestone-cement-density
- Whole chapter: https://www.bctengineer.com/md/semester_4/engineering_geology_ii/chapter-7

## Question

Apparent dip amounts of limestone bed rock along S15°E and S27°W are 1:16 and 1:22 respectively. The vertical thickness of bedrock is 215 m. Calculate the reserve of limestone for cement factory. The extension of the rock having terrain of 3.27 km and 72.05 m. Assume density of limestone is 2.75×10³ kg/m³.

## Short answer

From apparent dips 1:16 S15°E and 1:22 S27°W, true dip 3.58° towards S16.4°E. True thickness = 215 cos3.58° = 214.6 m. Area 3.27 km × 72.05 m yields volume 5.06×10⁷ m³; with density 2.75×10³ kg/m³, reserve = 1.39×10¹¹ kg (139 million tonnes).

## Answer

#### Step 1: true dip from the two apparent dips
Apparent dips: 1:16 along S15°E (bearing 165°) and 1:22 along S27°W (bearing 207°).
$$
\begin{aligned}
g_E\sin 165^\circ + g_N\cos 165^\circ &= \tfrac{1}{16} = 0.06250 \\
g_E\sin 207^\circ + g_N\cos 207^\circ &= \tfrac{1}{22} = 0.04545
\end{aligned}
$$
Solving: $g_E = 0.01761$, $g_N = -0.05999$.
$$
\tan\delta = \sqrt{0.01761^2 + 0.05999^2} = 0.06252 \Rightarrow \delta = 3.58^\circ\ (1:16.0)
$$
Dip direction $= 163.6^\circ$ (S16.4°E).

#### Step 2: true thickness
$$
t = 215\cos 3.58^\circ = 214.58\ \text{m}
$$

#### Step 3: volume
The extent of the rock is read as 3.27 km $\times$ 72.05 m (length $\times$ width in plan):
$$
A = 3270 \times 72.05 = 2.356\times 10^5\ \text{m}^2
$$
$$
V = A \times t = 2.356\times 10^5 \times 214.58 = 5.06 \times 10^7\ \text{m}^3
$$

#### Step 4: mass (reserve in tonnes)
$$
\text{Mass} = V \times \rho = 5.056\times 10^7 \times 2.75\times 10^3 = 1.39 \times 10^{11}\ \text{kg} = 1.39 \times 10^{8}\ \text{tonnes}
$$

**Answer: true dip = 3.58° (1:16) towards S16.4°E; true thickness = 214.6 m; volume = 5.06 × 10⁷ m³; reserve of limestone = 1.39 × 10¹¹ kg (about 139 million tonnes).**

## Check yourself

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

1. What true dip comes from apparent dips 1:16 (S15°E) and 1:22 (S27°W)?
   A. About 7.2° towards S27°W
   B. About 2.6° towards S15°E
   C. About 3.58° (1:16) towards S16.4°E
   D. About 3.58° towards N16.4°W
   Answer: C. Gradient components give tan δ = 0.0625, direction 163.6°. (Key point: True dip)

2. What true thickness is obtained for tv = 215 m?
   A. About 214.6 m
   B. About 13.4 m
   C. About 215 m
   D. About 227 m
   Answer: A. 215 cos 3.58° = 214.58 m. (Key point: True thickness)

3. How is the volume calculated?
   A. 3.27 km x 72.05 km x 214.6 m
   B. 3270 m x 72.05 m = 2.356 x 10^5 m², times 214.6 m = 5.06 x 10^7 m³
   C. 2.356 x 10^5 m² divided by 214.6 m
   D. 3270 m x 72.05 m x 215 m x 2750
   Answer: B. Both lengths are converted to metres, then area x true thickness. (Key point: Area and volume)

4. How is the mass found?
   A. Mass = volume x density = about 1.39 x 10^11 kg (139 million tonnes)
   B. Mass = volume / density
   C. Mass = area x density
   D. Mass = volume x density = 1.39 x 10^8 kg
   Answer: A. 5.06 x 10^7 m³ x 2750 kg/m³ = 1.39 x 10^11 kg; divide by 1000 for tonnes. (Key point: Mass)

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