Introduction to Mining Engineering MCQs Questions & Answers | Mining Engineering

In a Poisson distribution, the occurrence of events (in this case, the occurrence of minerals in a rock) is randomly distributed over a given interval. This distribution is commonly used to model rare events and is often used in mining engineering when studying the distribution of minerals in rocks. Therefore, the correct option is D.

Shaley Roof:
A shaley roof refers to a roof in underground mining that consists of shale, a sedimentary rock composed of fine particles. Shale is known for its layered structure, making it relatively weak and susceptible to damage under specific conditions.

Correct Answer: Option C: It has great affinity to absorb moisture and get departed from immediate roof
Reason:
Shale is a type of sedimentary rock that tends to absorb moisture easily. When a shaley roof is not supported by coal or other stronger materials, it becomes highly unstable because:
The shale absorbs moisture from the surrounding environment.
This absorption leads to the swelling of the shale, causing it to lose its structural integrity.
As a result, the shale starts to separate (or depart) from the immediate roof surface, increasing the risk of roof falls and making it challenging to control during mining operations.

Why Leaving Coal Helps:
Leaving a layer of coal with the shale provides support and reduces direct exposure to moisture. Coal acts as a protective layer, ensuring that the shale does not come into direct contact with environmental factors that might accelerate its degradation.

Other Options:
Option A: It is weak in nature
While shale is generally weaker compared to other rocks, the main challenge in controlling a shaley roof comes from its tendency to absorb moisture and separate, as stated in Option C.

Option B: It has weathering effect
Although weathering might affect shale over time, it is not the primary reason for difficulty in controlling a shaley roof in an underground mining setting.

Option D: It reacts with coaly surface
There is no significant chemical reaction between shale and the coal surface that would contribute to the instability of the roof.

Thus, the correct answer is Option C because moisture absorption and subsequent separation from the immediate roof are the primary challenges with a shaley roof in mining engineering.

For shafts exceeding 130 meters in depth, regulations typically require significant overhead protection when lowering or raising persons using equipment like buckets.

This is to ensure the safety of workers by preventing injuries from falling debris or objects.

The depth of 130 meters is a standard threshold in many mining safety guidelines, beyond which additional precautions, such as overhead covers, become mandatory.

Why other options are incorrect:
- Option B: 190m: This depth exceeds the standard threshold of 130 meters and is not the correct answer.
- Option C: 270m: This depth is also beyond the standard threshold and is not relevant to the question.
- Option D: 500m: This depth is significantly higher than the standard threshold and is not applicable in this context.

Thus, the correct answer is Option A: 130m, as it aligns with standard mining safety regulations for overhead protection in deep shafts.

Definition:
In mining engineering, specific terms such as Access, Haulage, and Mineralisation are crucial for understanding mining operations:
Access: Refers to the means by which miners reach an underground ore body, such as through shafts, declines, or adits.
Haulage: Involves the transportation of mined material from underground to the surface, which can be done using track systems, trackless vehicles, or hoisting equipment.
Mineralisation: Describes the concentration and distribution of valuable minerals in an ore body, which varies based on depth and location.

Explanation:
The correct answer is Option B: P-3-b, Q-2-a, R-1-c.

Matching Details:
P (Shaft): Shafts are vertical access points often associated with hoisting (3) and are typically used for accessing deep-seated (b) mineral deposits.
Q (Decline): Declines are inclined tunnels commonly used for trackless haulage (2) and are ideal for moderate-depth (a) mineral deposits.
R (Adit): Adits are horizontal or near-horizontal tunnels used in areas like hillocks (c) and are associated with track haulage (1).

Why other options are incorrect:
Option A: Incorrect pairing of features and purposes.
Option C: Misalignment of haulage methods and mineralisation depths.
Option D: Does not follow the logical association of methods with corresponding mining conditions.

Thus, the correct matching is P-3-b, Q-2-a, R-1-c, making Option B the correct answer.

Given data:
Number of holes (n) = 9
Hole length (L) = 8m
Hole diameter (D) = 200mm = 0.2m
Spacing (S) = 6m
Burden (B) = 5m
Subgrade drilling length = 1m
Density of in-situ rock (ρ_rock) = 2.43 t/m³
Explosive density (ρ_explosive) = 0.9 t/m³
Stemming length = 2m

First, calculate the volume of a single hole:
Volume of hole (V_hole) = π/4 * D² * L
Volume of hole (V_hole) = π/4 * (0.2m)² * 8m
Volume of hole (V_hole) = 0.0251 m³

Next, calculate the total blasted volume (V_total):
Total blasted volume (V_total) = n * V_hole
Total blasted volume (V_total) = 9 * 0.0251 m³
Total blasted volume (V_total) = 0.2259 m³

Now, calculate the burden-to-spacing ratio (B/S):
Burden-to-spacing ratio (B/S) = B / S
Burden-to-spacing ratio (B/S) = 5m / 6m
Burden-to-spacing ratio (B/S) = 0.833

Calculate the effective volume (V_eff):
Effective volume (V_eff) = V_total * (1 - B/S)
Effective volume (V_eff) = 0.2259 m³ * (1 - 0.833)
Effective volume (V_eff) = 0.0376 m³

Calculate the total volume of explosive (V_explosive):
Total volume of explosive (V_explosive) = V_eff + n * π/4 * D² * subgrade drilling length
Total volume of explosive (V_explosive) = 0.0376 m³ + 9 * π/4 * (0.2m)² * 1m
Total volume of explosive (V_explosive) = 0.1935 m³

Calculate the total weight of explosive (W_explosive):
Total weight of explosive (W_explosive) = V_explosive * ρ_explosive
Total weight of explosive (W_explosive) = 0.1935 m³ * 0.9 t/m³
Total weight of explosive (W_explosive) = 0.1742 t

Calculate the total weight of rock (W_rock):
Total weight of rock (W_rock) = V_total * ρ_rock
Total weight of rock (W_rock) = 0.2259 m³ * 2.43 t/m³
Total weight of rock (W_rock) = 0.5489 t

Finally, calculate the powder factor (P):
Powder factor (P) = W_explosive / W_rock
Powder factor (P) = 0.1742 t / 0.5489 t
Powder factor (P) = 0.3175

The correct powder factor rounded off to two decimal places is approximately 0.32 t/kg, which does not match option D: 3.01. The provided options and calculations still seem to be inconsistent. Please verify the data and calculations to ensure accurate results.