I85PREP
⌘ K
Mining Engineering study materials
Mining Engineering study resources

Mining Engineering practice questions

One question at a time. Build your Mining Engineering understanding.

First 50 of 200 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Mining and Mineral Processing (official domain list not available from source)

In bench blasting, what is the primary purpose of a stemming column placed above the explosive charge in a blast hole?

  1. To couple the detonator shock front to the main charge
  2. To increase the linear charge concentration along the hole
  3. To confine detonation gases and reduce premature venting through the collar Correct answer.
  4. To lower the detonation velocity of the column explosive

Explanation

Stemming is inert material packed above the charge to delay the escape of high-pressure detonation gases. Better gas confinement lengthens the effective heave and reduces airblast and flyrock from collar venting. Stemming does not add energy to the charge or change its detonation velocity.

Question 2Mining and Mineral Processing (official domain list not available from source)

Which ventilation arrangement is normally required by law for a blind development heading such as a dead-end access drive?

  1. Natural draft through the open portal alone
  2. Mechanical forced or exhaust ventilation with a controlled air quantity Correct answer.
  3. Recirculation of return air without treatment
  4. Reliance on diesel exhaust to sweep the face

Explanation

A dead-end heading has no through-flow path, so air must be supplied mechanically, typically by fan and duct, in a quantity sufficient to dilute blasting fumes, dust and diesel emissions. Statutory ventilation plans specify the minimum flow at the last open crosscut. Natural draft and recirculation cannot reliably achieve this.

Question 3Mining and Mineral Processing (official domain list not available from source)

A run-of-mine ore contains 1.2 percent copper by mass. What mass of copper is contained in 5,000 tonnes of this ore?

  1. 60 tonnes Correct answer.
  2. 6 tonnes
  3. 0.6 tonnes
  4. 600 tonnes

Explanation

Contained metal equals ore mass multiplied by grade expressed as a fraction: 5,000 t times 0.012 equals 60 t of copper. Grade percentages are converted by dividing by 100 before multiplying. This contained-metal figure is the basis for comparing against mill recovery and concentrate payability.

Question 4Mining and Mineral Processing (official domain list not available from source)

In a mineral processing flowsheet, what is the function of a hydrocyclone in closed circuit with a ball mill?

  1. To leach soluble values from the ground pulp
  2. To classify particles by size and return coarse material to the mill Correct answer.
  3. To dewater the final concentrate by thermal drying
  4. To magnetically separate ferromagnetic gangue

Explanation

A hydrocyclone separates particles using centrifugal force in a swirling pulp, sending fines to overflow and coarse particles to underflow. In closed circuit the underflow is recycled to the mill so that product size is controlled. It performs size classification rather than leaching, drying or magnetic separation.

Question 5Mining and Mineral Processing (official domain list not available from source)

Which parameter is the standard measure of the strength of an explosive as determined by the ballistic mortar or equivalent test?

  1. Fume class rating
  2. Cartridge diameter tolerance
  3. Weight strength relative to a reference explosive Correct answer.
  4. Shelf life of the detonator

Explanation

Weight strength expresses the energy or work output of an explosive relative to a standard reference explosive such as ANFO or blasting gelatin. It is measured by ballistic mortar, Trauzl lead block or calculated from ideal detonation energy. Diameter tolerance, fume class and detonator shelf life describe other properties.

Question 6Mining and Mineral Processing (official domain list not available from source)

In underground mining, what does the term dilution refer to?

  1. The loss of fine particles during screening
  2. The addition of waste rock or low-grade material to ore during mining Correct answer.
  3. The mixing of two different explosives in a blast hole
  4. The reduction of ore grade caused by oxidation in the stockpile

Explanation

Dilution is the unintended incorporation of waste or sub-grade material into ore, which lowers the grade delivered to the mill. It arises from overbreak, unstable hangingwall, poor ore-waste contacts and blasting beyond the ore boundary. Controlling dilution is a central economic objective in stope design.

Question 7Mining and Mineral Processing (official domain list not available from source)

Which ground support element is a fully grouted rock bolt intended primarily to work?

  1. By draining water from the rock mass
  2. By applying a large pretension through a face plate only
  3. By bonding along its length so that load transfers to the rock mass Correct answer.
  4. By hanging mesh independently of the rock

Explanation

A fully grouted bolt develops load transfer along its entire bonded length through the grout annulus, mobilising the surrounding rock mass. It acts as a reinforcing element rather than merely a surface clamp. Pretension and mesh retention are secondary functions, and drainage is the role of weep holes or drains.

Question 8Mining and Mineral Processing (official domain list not available from source)

What is the main purpose of the frother in froth flotation?

  1. To stabilise the air bubbles so a persistent froth can carry particles Correct answer.
  2. To depress the gangue minerals
  3. To make the target mineral surface hydrophobic
  4. To adjust the pulp pH to an alkaline value

Explanation

Frothers are surface-active reagents that adsorb at the air-water interface, lowering surface tension and producing small, stable bubbles. A stable froth is needed to transport hydrophobic particles to the overflow launder. Collectors confer hydrophobicity, depressants suppress gangue, and pH modifiers control pulp chemistry.

Question 9Mining and Mineral Processing (official domain list not available from source)

A stope is designed with a tonnage of 120,000 t at a diluted grade of 2.5 g/t gold. If mill recovery is 90 percent, what mass of gold is recovered, in grams?

  1. 27,000 g
  2. 270,000 g Correct answer.
  3. 300,000 g
  4. 2,700 g

Explanation

Contained gold is 120,000 t times 2.5 g/t, which equals 300,000 g. Applying 90 percent recovery gives 270,000 g of recovered gold. Recovery is applied to contained metal, not to the grade alone.

Question 10Mining and Mineral Processing (official domain list not available from source)

Why is ANFO generally unsuitable for use in wet blastholes without special precautions?

  1. It becomes insensitive to detonators at low temperature
  2. It is highly soluble and loses its explosive performance when water penetrates the charge Correct answer.
  3. It generates excessive nitrogen dioxide in dry holes
  4. It detonates too fast in the presence of water

Explanation

ANFO is a mixture of ammonium nitrate prills and fuel oil, and ammonium nitrate is readily soluble in water. In wet holes the charge can dissolve or desensitise, causing misfires or poor fragmentation. Water-resistant explosives, sleeved charges or dewatering are used instead.

Question 11Mining and Mineral Processing (official domain list not available from source)

In an open pit, which slope failure mode is most directly associated with a persistent bedding plane dipping steeply out of the pit wall?

  1. Planar sliding along the bedding plane Correct answer.
  2. Circular rotational slump in homogeneous soil
  3. Block caving of the pit floor
  4. Toppling of columns perpendicular to the slope

Explanation

A discontinuity set dipping out of the slope at an angle steeper than the friction angle but shallower than the slope face allows a block to slide along that plane. This kinematic condition defines planar wedge sliding. Circular slumps occur in homogeneous material, toppling requires steeply dipping columns into the face, and block caving is an underground mining method.

Question 12Mining and Mineral Processing (official domain list not available from source)

Which condition is necessary for a rockburst to occur in a deep underground mine?

  1. Low in-situ stress with a ductile rock mass
  2. High stress concentration in a stiff, brittle rock mass that can store and release strain energy violently Correct answer.
  3. A high water table above the mining horizon
  4. A thick layer of unconsolidated sand at surface

Explanation

Rockbursts require the accumulation of elastic strain energy in a stiff, strong rock mass under high stress, followed by sudden violent release. Mining-induced stress concentration near the face provides the trigger. Water and weak surface soils do not create the stored energy needed, and ductile behaviour dissipates rather than releases it.

Question 13Mining and Mineral Processing (official domain list not available from source)

In a coal mine, what is the principal purpose of a stone dust barrier in the roadways?

  1. To arrest the propagation of a coal dust explosion by inerting the flame front Correct answer.
  2. To suppress spontaneous combustion in the goaf
  3. To provide additional roof support
  4. To reduce diesel particulate emissions

Explanation

Stone dust barriers consist of shelves or bags of inert dust that are dispersed by the pressure wave of an explosion. The resulting dust cloud dilutes and cools the flame front so that it cannot propagate along the roadway. They are an explosion suppression measure rather than a support or emission control.

Question 14Mining and Mineral Processing (official domain list not available from source)

A flotation circuit treats 200 t/h of solids at 30 percent solids by mass. What is the approximate volumetric pulp flow if the solids density is 2.7 t/m3 and water density is 1.0 t/m3?

  1. About 540 m3/h Correct answer.
  2. About 220 m3/h
  3. About 740 m3/h
  4. About 300 m3/h

Explanation

Solids are 200 t/h, so water is 200 times 70/30, approximately 467 t/h. Solids volume is 200/2.7, about 74 m3/h, and water volume is about 467 m3/h. The total pulp flow is therefore roughly 541 m3/h.

Question 15Mining and Mineral Processing (official domain list not available from source)

What is the purpose of a tailings thickener in a mineral processing plant?

  1. To increase the solids concentration of the tailings stream by settling, recovering water for reuse Correct answer.
  2. To leach residual metal from the tailings slurry
  3. To grind the tailings to a finer size before disposal
  4. To filter the tailings to a dry cake for transport

Explanation

A thickener uses flocculant-assisted settling to concentrate solids in the underflow and produce a clarified overflow. The recovered water is returned to the process, and the denser underflow reduces the volume sent to the tailings storage facility. Grinding, leaching and final dewatering are separate unit operations.

Question 16Mining and Mineral Processing (official domain list not available from source)

In mine ventilation network analysis, what does the equivalent orifice of a mine represent?

  1. The area of an ideal sharp-edged opening that would pass the same airflow at the same pressure difference Correct answer.
  2. The total surface area of the ventilation shaft
  3. The leakage area of the stopping doors
  4. The physical cross-section of the main fan drift

Explanation

Equivalent orifice converts the resistance of a whole ventilation network into the area of an ideal opening that would pass the same quantity of air at the same pressure difference. It is a convenient single measure of how easy the mine is to ventilate. It is a derived equivalent, not a physical dimension of any airway.

Question 17Mining and Mineral Processing (official domain list not available from source)

Which mining method is most appropriate for a steeply dipping, competent orebody with strong walls where controlled subsidence of the hangingwall is acceptable and high production rates are required?

  1. Sublevel stoping with delayed backfill Correct answer.
  2. Room and pillar mining
  3. Longwall top coal caving
  4. Hydraulic borehole mining

Explanation

Sublevel stoping suits steeply dipping, competent orebodies with stable walls and allows large-scale mechanised production. Backfill of mined-out stopes controls dilution and regional stability. Room and pillar suits flat-lying deposits, longwall caving is a coal method, and borehole mining is a niche technique for unconsolidated or deep deposits.

Question 18Mining and Mineral Processing (official domain list not available from source)

During flotation, adding lime to the pulp most commonly achieves which combination of effects?

  1. Lowers pH and increases froth stability
  2. Raises pH and acts as a strong collector for galena
  3. Raises pH and depresses pyrite in selective flotation circuits Correct answer.
  4. Lowers pH and activates sphalerite

Explanation

Lime is an alkaline pH modifier that also suppresses pyrite and other iron sulphides in selective flotation. The higher pH reduces the floatability of pyrite while allowing target sulphides to be recovered. It is neither a collector nor an acidifying reagent.

Question 19Mining and Mineral Processing (official domain list not available from source)

What is the principal geotechnical concern when a tailings storage facility is operated using the upstream raising method?

  1. The supernatant pond is eliminated
  2. The raised embankment may be founded on weak, saturated tailings that can liquefy under seismic or static loading Correct answer.
  3. The tailings beach cannot be maintained
  4. The embankment becomes too wide to construct

Explanation

Upstream construction places each raise on top of previously deposited tailings, which may be loose and saturated. Such material is susceptible to liquefaction and flow failure under earthquake or rapid loading. This is why many jurisdictions restrict or prohibit upstream raising for high-risk facilities.

Question 20Mining and Mineral Processing (official domain list not available from source)

A mill processes 1,000 t/d of ore at a head grade of 1.5 percent copper. The concentrate assays 25 percent copper and the tailings assay 0.15 percent copper. What is the approximate copper recovery?

  1. 80 percent
  2. 95 percent
  3. 90 percent Correct answer.
  4. 85 percent

Explanation

Recovery equals concentrate copper divided by feed copper. Feed copper is 1,000 times 0.015, which is 15 t/d. Tailings copper is 1,000 times 0.0015, which is 1.5 t/d, so concentrate copper is 13.5 t/d. The recovery is 13.5/15, or 90 percent.

Question 21Mining and Mineral Processing (official domain list not available from source)

Why is the powder factor expressed in kilograms of explosive per tonne or per cubic metre of rock useful in blast design?

  1. It measures the detonation velocity of the explosive
  2. It links the quantity of explosive to the rock volume broken and allows comparisons between blasts Correct answer.
  3. It specifies the legal storage limit for magazines
  4. It determines the required detonator delay sequence

Explanation

Powder factor is the mass of explosive divided by the volume or tonnage of rock broken. It provides a practical design and benchmarking parameter that can be compared across blasts and rock types. It does not describe detonation velocity, delay timing or magazine storage rules.

Question 22Mining and Mineral Processing (official domain list not available from source)

In a paste backfill plant, what is the main reason for adding a hydraulic binder such as cement or slag to the tailings?

  1. To make the fill more permeable to water
  2. To develop strength and stiffness so the fill can support the stope walls Correct answer.
  3. To reduce the pH of the backfill
  4. To increase the settling rate of the tailings

Explanation

Cementitious binder hydrates to bind the tailings particles into a cemented mass with measurable strength and stiffness. This allows the fill to provide wall support and to remain exposed in primary stopes. Binder addition does not primarily aim at settling, pH reduction or permeability.

Question 23Mining and Mineral Processing (official domain list not available from source)

A ventilation network has two parallel airways of equal length and friction factor, but one has twice the cross-sectional area of the other. How does the airflow split between them compare?

  1. The smaller airway carries more airflow because of higher velocity
  2. The larger airway carries twice the airflow of the smaller one
  3. The larger airway carries roughly 5.7 times the airflow of the smaller one Correct answer.
  4. The airways carry equal airflow

Explanation

For parallel airways with equal pressure loss, airflow scales with the square root of conductance, and conductance scales with area to the power 5/2 for turbulent flow. The area ratio is 2, so the flow ratio is 2 to the power 2.5, approximately 5.7. This steep dependence explains why large airways dominate ventilation resistance.

Question 24Mining and Mineral Processing (official domain list not available from source)

In acid mine drainage prediction, why is the net acid-producing potential of waste rock calculated using both the acid potential and the neutralising potential?

  1. Because the calculation determines the blast fragmentation size
  2. Because neutralising potential measures the permeability of the waste
  3. Because acid potential alone always overestimates the sulphide content
  4. Because carbonate minerals in the rock can neutralise part of the acid generated, so the net balance determines the actual risk Correct answer.

Explanation

Acid potential is estimated from sulphide sulphur content, mainly pyrite, while neutralising potential is estimated from carbonate minerals such as calcite and dolomite. The net difference determines whether the waste will generate acid, be neutral or be alkaline. Using only acid potential ignores the buffering capacity of the rock.

Question 25Mining and Mineral Processing (official domain list not available from source)

In a sublevel caving operation, why is the hangingwall allowed to cave progressively as ore is drawn?

  1. Because the method requires the stope to remain open permanently for ventilation
  2. Because ore is recovered from beneath a caving waste cap that follows the retreating draw, and controlled dilution is inherent to the method Correct answer.
  3. Because the hangingwall must be supported by pillars left in place
  4. Because the orebody is flat-lying and needs no wall support

Explanation

Sublevel caving relies on the progressive caving of the hangingwall or ore above the drawpoints. Broken ore is drawn from sublevel drifts while waste follows and dilutes the ore at the draw. This controlled dilution is accepted as part of the method rather than prevented by pillars.

Question 26Mining and Mineral Processing (official domain list not available from source)

In an underground mine ventilation circuit, which instrument is used to measure the volumetric airflow velocity in a return airway?

  1. Clinometer
  2. Piezometer
  3. Scintillation counter
  4. Vane anemometer Correct answer.

Explanation

A vane anemometer measures air velocity by counting the rotation of a vane in the airstream; multiplying velocity by airway cross-sectional area gives volumetric flow. Clinometers measure angles, piezometers measure pressure head, and scintillation counters detect radiation.

Question 27Mining and Mineral Processing (official domain list not available from source)

Which explosive is commonly used as a primer or booster to initiate ANFO in a blasthole because of its high detonation velocity and water resistance?

  1. Detonating cord
  2. Safety fuse
  3. Emulsion explosive Correct answer.
  4. Black powder

Explanation

Emulsion explosives have high detonation velocity, good water resistance, and sufficient critical diameter to reliably initiate ANFO. Black powder and safety fuse are low-velocity initiating materials, and detonating cord transmits a detonation signal rather than serving as the main primer charge.

Question 28Mining and Mineral Processing (official domain list not available from source)

Which type of drilling rig is most commonly used to drill blastholes in an open pit bench operation?

  1. Rotary or rotary-percussive blast-hole rig Correct answer.
  2. Raise boring machine
  3. Longwall shearer
  4. Tunnel boring machine

Explanation

Open pit bench blastholes are drilled with rotary or rotary-percussive rigs mounted on trucks or tracks, capable of drilling vertical or inclined holes to bench depth. Raise borers, shearers, and tunnel boring machines are underground or civil excavation machines.

Question 29Mining and Mineral Processing (official domain list not available from source)

In mineral processing, what is the primary purpose of a hydrocyclone in a grinding circuit?

  1. To reduce ore particle size by impact and attrition
  2. To separate minerals by differences in surface wettability
  3. To dewater concentrate by thermal evaporation
  4. To classify particles by size using centrifugal force Correct answer.

Explanation

A hydrocyclone uses centrifugal force in a swirling pulp to separate fine particles (overflow) from coarser particles (underflow), commonly closing a ball mill circuit. Size reduction is done by mills, wettability separation by flotation, and thermal dewatering by dryers.

Question 30Mining and Mineral Processing (official domain list not available from source)

Which test is used to determine the strength of intact rock by applying an axial compressive load to a prepared cylindrical core specimen?

  1. Uniaxial compressive strength test Correct answer.
  2. Slake durability test
  3. Vickers hardness test
  4. Los Angeles abrasion test

Explanation

The uniaxial compressive strength test loads a cylindrical rock core along its axis until failure, giving a fundamental intact rock strength parameter. The Los Angeles test measures aggregate abrasion resistance, Vickers measures hardness, and slake durability assesses wetting-drying breakdown.

Question 31Mining and Mineral Processing (official domain list not available from source)

What is the main purpose of a primary crusher in a mineral processing plant?

  1. To dewater the ore slurry before stockpiling
  2. To grind ore to liberation size for flotation
  3. To reduce run-of-mine ore to a size suitable for conveying and secondary crushing Correct answer.
  4. To separate valuable minerals from gangue by density

Explanation

A primary crusher receives run-of-mine ore and reduces it to a size that can be handled by conveyors and downstream crushers or mills. Fine grinding is done by mills, density separation by gravity concentrators, and dewatering by thickeners or filters.

Question 32Mining and Mineral Processing (official domain list not available from source)

Which term describes the process of sealing mined-out underground voids with waste rock, tailings, or cementitious fill?

  1. Bridging
  2. Blasting
  3. Benching
  4. Backfilling Correct answer.

Explanation

Backfilling places waste material or engineered fill into mined-out stopes to provide ground support, allow pillar recovery, and manage tailings. Benching is a surface excavation step, bridging refers to span formation, and blasting is rock breakage.

Question 33Mining and Mineral Processing (official domain list not available from source)

In a coal preparation plant, which process is most commonly used to clean coarse coal by exploiting differences in density?

  1. Cyanide leaching
  2. Froth flotation
  3. Magnetic separation
  4. Dense medium separation Correct answer.

Explanation

Dense medium separation uses a suspension of fine magnetite to create a bath density between coal and reject, allowing coarse coal to float and mineral matter to sink. Flotation treats fine coal, cyanide leaching is for gold, and magnetic separation is used for ferromagnetic minerals.

Question 34Mining and Mineral Processing (official domain list not available from source)

A stope has a planned extraction of 80,000 t of ore at an in-situ grade of 4.0 g/t gold. If mining dilution adds 15 percent waste at zero grade, what is the diluted grade in grams per tonne?

  1. 4.00
  2. 3.48 Correct answer.
  3. 3.20
  4. 3.80

Explanation

Diluted grade equals in-situ grade multiplied by the ore fraction of total mined tonnage. With 15 percent dilution, ore fraction is 1/1.15 = 0.8696, so diluted grade is 4.0 × 0.8696 = 3.48 g/t.

Question 35Mining and Mineral Processing (official domain list not available from source)

A jaw crusher receives run-of-mine ore at a 95th-percentile feed size of 600 mm and produces a product with a 95th-percentile size of 150 mm. What is the approximate reduction ratio?

  1. 4:1 Correct answer.
  2. 6:1
  3. 9:1
  4. 2:1

Explanation

The reduction ratio is the feed size divided by the product size, so 600 mm / 150 mm = 4. A reduction ratio of 4:1 is typical for a primary jaw crusher.

Question 36Mining and Mineral Processing (official domain list not available from source)

A ventilation shaft delivers 120 m³/s of air to a mine. If the shaft has a circular cross-section of 4.0 m diameter, what is the approximate average air velocity in the shaft?

  1. 2.4 m/s
  2. 4.8 m/s
  3. 9.5 m/s Correct answer.
  4. 19.1 m/s

Explanation

Velocity equals volumetric flow divided by cross-sectional area. Area = π × (2.0 m)² = 12.57 m², so velocity = 120 / 12.57 = 9.55 m/s.

Question 37Mining and Mineral Processing (official domain list not available from source)

In a flotation bank, a rougher concentrate is typically subjected to which downstream process to improve final concentrate grade?

  1. Primary crushing
  2. Tailings thickening
  3. Cleaner flotation Correct answer.
  4. Scavenger flotation

Explanation

Rougher concentrate contains entrained gangue and is re-floated in cleaner stages to raise grade, while scavenger stages treat rougher tailings to recover additional values. Thickening and crushing do not upgrade the rougher concentrate.

Question 38Mining and Mineral Processing (official domain list not available from source)

A tailings slurry at 25 percent solids by mass is thickened to 55 percent solids by mass. For 1,000 t of dry solids, what is the approximate mass of water removed per tonne of dry solids?

  1. 1.82 t
  2. 2.18 t Correct answer.
  3. 1.22 t
  4. 3.00 t

Explanation

Water mass per tonne of solids is (1/C − 1), where C is the solids mass fraction. At 25 percent, water is 3.00 t/t; at 55 percent, water is 0.818 t/t; the difference removed is 3.00 − 0.818 = 2.18 t/t.

Question 39Mining and Mineral Processing (official domain list not available from source)

In open pit mine planning, which pit slope design parameter directly determines the total volume of waste rock that must be removed to access the ore?

  1. Haul road gradient
  2. Bench height
  3. Crest width
  4. Overall slope angle Correct answer.

Explanation

The overall slope angle controls how far the pit walls must be pushed back, and therefore the stripping ratio and waste volume. Bench height and crest width affect local wall geometry, and haul road gradient affects haulage, not the total waste volume directly.

Question 40Mining and Mineral Processing (official domain list not available from source)

In an underground mine, which ventilation control device is used to regulate and direct airflow by opening or closing a passage between airways?

  1. Air crossing
  2. Regulator Correct answer.
  3. Ventilation door
  4. Stoppings

Explanation

A regulator is an adjustable restriction, often a sliding door or louvre, used to control the quantity of air flowing into a split. A ventilation door provides access and separates airways, a stopping permanently seals an opening, and an air crossing carries one airway over another.

Question 41Mining and Mineral Processing (official domain list not available from source)

A gold heap leach pad contains 2.0 Mt of ore at a grade of 1.2 g/t. If the overall recovery is 70 percent and gold price is USD 1,800/oz (31.1 g/oz), what is the approximate gross revenue in USD?

  1. 97 million Correct answer.
  2. 68 million
  3. 139 million
  4. 49 million

Explanation

Contained gold is 2.0 × 10⁶ t × 1.2 g/t = 2.4 × 10⁶ g. Recovered gold is 70 percent, or 1.68 × 10⁶ g, which equals 54,019 oz. At USD 1,800/oz, revenue is about USD 97 million.

Question 42Mining and Mineral Processing (official domain list not available from source)

In underground coal mining, what is the main purpose of drilling and grouting ahead of the face in a roadway development?

  1. To increase coal seam gas content for later capture
  2. To reduce the need for roof bolting
  3. To cool the strata before mining
  4. To pre-drain and consolidate water-bearing strata ahead of the face Correct answer.

Explanation

Advance drilling and grouting is used to intercept and drain or consolidate water-bearing or weak strata ahead of a development face, reducing water inflow and instability. It is not intended to increase gas content, cool rock, or replace roof support.

Question 43Mining and Mineral Processing (official domain list not available from source)

In a comminution circuit, which operating change is most likely to reduce the specific energy consumption (kWh/t) of a ball mill while maintaining product fineness?

  1. Increasing the mill feed rate without adjusting other parameters
  2. Optimising the ball charge and pulp density Correct answer.
  3. Increasing the circulating load without limit
  4. Reducing the mill speed to the minimum

Explanation

Optimising ball charge, pulp density, and mill speed improves grinding efficiency and reduces specific energy for a given product size. Increasing feed rate without control or running at very low speed usually degrades product fineness, and excessive circulating load raises pumping and classification energy.

Question 44Mining and Mineral Processing (official domain list not available from source)

In a flotation circuit, which change is most likely to increase the recovery of a valuable sulfide mineral without significantly reducing concentrate grade?

  1. Optimising collector dosage and pulp pH for the target mineral Correct answer.
  2. Adding more frother to create a very stable, deep froth
  3. Reducing the pulp residence time in the rougher cells
  4. Increasing the feed particle size to coarse gravel

Explanation

Collector dosage and pulp pH control the hydrophobicity and selectivity of the target mineral, so optimising them improves recovery while limiting gangue flotation. Excessive frother, short residence time, or very coarse particles reduce recovery or grade.

Question 45Mining and Mineral Processing (official domain list not available from source)

In a room-and-pillar coal mine, which factor most directly controls the pillar strength used in pillar design?

  1. Pillar width-to-height ratio and coal strength Correct answer.
  2. Seam gas content and permeability
  3. Roof bolting pattern and mesh type
  4. Mine ventilation airflow rate

Explanation

Empirical and analytical pillar strength formulas depend primarily on pillar width-to-height ratio and the strength of the coal. Gas content, roof support, and ventilation affect other hazards but are not the primary inputs to pillar strength.

Question 46Mining and Mineral Processing (official domain list not available from source)

In an underground mine, a stope is planned with a 30 m strike length, 20 m width, and 40 m height. If the ore density is 2.8 t/m³ and mining recovery is 85 percent, what is the approximate ore tonnage recovered?

  1. 57,120 t Correct answer.
  2. 48,000 t
  3. 80,640 t
  4. 67,200 t

Explanation

Stope volume is 30 × 20 × 40 = 24,000 m³. At 2.8 t/m³ the in-situ mass is 67,200 t, and applying 85 percent recovery gives 57,120 t.

Question 47Mining and Mineral Processing (official domain list not available from source)

In a sublevel stoping operation, which factor most strongly governs the required span between sublevels?

  1. The number of available drill rigs
  2. The ability of the stope walls and back to remain stable over the exposed span Correct answer.
  3. The ventilation airflow through the stope
  4. The capacity of the underground haulage fleet

Explanation

Sublevel spacing is governed by geotechnical stability of the exposed stope walls and back; the span must be kept within the stable hydraulic radius for the rock mass. Haulage capacity, drill rig availability, and ventilation affect productivity and safety but not the geotechnical span limit.

Question 48Mining and Mineral Processing (official domain list not available from source)

A processing plant treats 500 t/h of ore containing 1.0 percent copper. The concentrate assays 28 percent copper and the tailings assay 0.10 percent copper. What is the approximate copper recovery?

  1. 90.0 percent Correct answer.
  2. 85.0 percent
  3. 96.4 percent
  4. 80.0 percent

Explanation

Recovery = (feed grade − tailings grade) / (feed grade − tailings grade adjusted to concentrate) using the two-product formula: R = C(f − t) / (f(c − t)). With f = 1.0, t = 0.10, c = 28, R = (1.0 − 0.10) / (1.0 − 0.10) × ... Simplifying, R = (f − t)/(f × (1 − t/c)) ≈ 0.90, or 90 percent.

Question 49Mining and Mineral Processing (official domain list not available from source)

In an open pit, a 15 m high bench is blasted with a powder factor of 0.6 kg/m³. If the blast pattern covers 10,000 m³ of rock, what is the approximate total explosive charge?

  1. 9,000 kg
  2. 3,000 kg
  3. 6,000 kg Correct answer.
  4. 1,500 kg

Explanation

Powder factor is explosive mass per unit rock volume, so total charge = 0.6 kg/m³ × 10,000 m³ = 6,000 kg. The bench height is not needed once the rock volume is given.

Question 50Mining and Mineral Processing (official domain list not available from source)

A ventilation network has two parallel airways of equal length and friction factor, but one has twice the cross-sectional area of the other. What is the approximate ratio of airflow in the larger airway to that in the smaller airway?

  1. About 1.4:1
  2. About 4.0:1
  3. About 2.8:1 Correct answer.
  4. About 2.0:1

Explanation

For parallel airways, resistance is inversely proportional to area cubed for equal length and friction, and flow varies as the square root of the inverse of resistance. The larger airway therefore carries about 2^(3/2) = 2.83 times the flow of the smaller one.