Winning and Working - Questions from DGMS and PSU Exams

Question

A mine void of dimension 100.0 m x 2.0 m x 1.2 m is to be filled in 3 hours by hydraulic stowing. The sand to slurry ratio is 0.4. If the hydraulic fill factor is 0.9. the hourly consumption of water for the operation, in m³, is ____ (rounded off to 2 decimal places.

Solution

V_{mine void}​=100.0 × 2.0 × 1.2 = 240.0 m³

V_sand​ = V_void​ × fill factor

= 240.0 × 0.9

= 216.0 m³

V_slurry = V_sand/0.4 = 216/0.4 = 540 m³

V_water = V_slurry  - V_sand

= 540 - 216 = 324 m³

Hourly water consumption

= 324/3 = 108 m³

These questions are taken from study material for mining engineering exams (DGMS and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required. 

Question 

A pit slope has the following information:

Number of benches = 5

Height of each bench = 8 m

Bench slope angle = 70 degree

If width of one bench is 24 m and that of other four benches is 10 m each, the overall pit slope angle, in degree, is(rounded off to 2 decimal places).

Solution

See following figure

x = H/tanθ

Overall pit slope angle (β)

where θ = slope of the bench

B = bench width

H = bench height

Given data:

Bench width = 10 m

Bench height (H) = 8 m

Slope of bench (θ) = 70⁰

Putting these values in the formula

= 28.8⁰

These questions are taken from study material for mining engineering exams (DGMS and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.

Question

Following information are given for three mines of a company receiving explosives from three suppliers.

Considering the initial basic feasible solution of this transportation problem, using North-West comer method, the transportation cost of explosives supplied to the company, in Rs., is

(a) 90000

(b) 109000

(c) 293500*

(d) 385000 

Solution

See following figure which is self explanatory.

Cost = 10 x 1000 + 15 x 1500 + 2 x 500 + 100 x 2000 + 20 x 3000 

= ₹ 293500

Also read: 

Numerical problems from Mine Surveying from DGMS and PSU Exams

Numerical problems from Mining Machinery from DGMS and PSU Exams 

Numerical problems from Mine Ventilation from DGMS and PSU Exams 

Question

In a bord and pillar panel, the following data are obtained.

Number of blasting rounds per shift : 8

Face dimension (m x m) : 4.2 x 2.5

Average pull (m) : 1.2

Specific gravity of coal : 1.4

Manpower per shift : 100

The OMS (output per manshift) of the panel, in tonne, is ____ (rounded off to 2 decimal places).

Solution

Total volume of coal

= Number of shifts x gallery width x gallery height x Average pull x Number of faces x number of rounds

= 1 x 4.2 x 2.5 x 1.2 x 1 x 8 = 100.8 m³

Weight of coal/day

= volume x density

= 100.8 m³ x 1.4 = 141.12 t

OMS = weight per day/total manpower

= 141.12/100 = 1.41ton

These questions are taken from study material for mining engineering exams (DGMS and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.

Question

In a surface mine, blasting is carried out using electronic detonator and cartridge emulsion explosive with following details:

Burden = 3.5 m

Spacing = 4.5 m

Bench height = 10.0 m

Subgrade drilling =1.0 m

Stemming = 4.0 m

Linear charge concentration =16 kg/m

Cost of one detonator = Rs. 800

Cost of explosive = Rs. 30 per kg

The cost of blasting material per cubic meter of blasted rock, in Rs., is ____ (rounded off to 2 decimal places)

Solution

The volume of rock blasted per hole

V = Burden x Spacing x Bench height

= 3.5 m x 4.5 m x 10.0 m = 157.5 m³

Hole length = 10.0 m + 1.0 m = 11.0 m

Explosive column length

= 11.0 m - 4.0 m  = 7.0 m

Explosive weight = 7.0 m x 16 kg/m = 112 kg

Cost per hole = Cost of detonator  + (Explosive weight x Cost per kg of explosive)

Cost per hole = ₹ 800  + (112 kg x ₹ 30/kg)

= ₹ 800 + ₹ 3360 = ₹ 4160

Cost per m³ = Cost per hole/Volume per hole

= 4160/157.5 = ₹ 26.4 /m³ 

These questions are taken from study material for mining engineering exams (DGMS and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.

Question

A triaxial test on a sandstone sample is conducted at a confining pressure of 10 MPa. The elastic axial and volumetric strains at axial stress of 50 MPa are recorded to be 4.2 x 10^-3 and 2.0 x 10^-3 respectively. The modulus of elasticity, in GPa. and Poisson’s ratio of the sample, respectively are closest to

(a) 5.11.0.35

(b) 10.22,0.35*

(c) 5.11.0.17

(d) 10.22.0.17 

Solution

Young modulus (E)
=  ∆σ/elastic axial strain

= (50 - 10)/(4.2 x 10^-3)

= 9524 MPa = 9.52 GPa

Poisson ratio

ν = (1/2)[1 - (ε_v/ε_a)]

= (1/2)[1 - (2 x 10^-3/4.2 x 10^-3)]

= 0.26

Also read: 

Numerical problems from Mine Surveying from DGMS and PSU Exams

Numerical problems from Mining Machinery from DGMS and PSU Exams 

Numerical problems from Mine Ventilation from DGMS and PSU Exams 

Question

The relevant information on metal extraction from a copper mine are given below

Selling price of copper (Rs./kg) = 900

Mining cost (Rs./tonne of ore) = 500

Processing cost (Rs./tonne of ore) = 2000

Overall recovery of copper metal (%) = 70

Ignoring all other costs, the breakeven cutoff grade of copper, in %, is ____ (rounded off to 2 decimal places).

Solution

Breakdown cutoff grade

= (mining cost + processing cost)/(selling price of copper  x overall recovery)

= (500 + 2000)/(900,000 x 0.7)

= 0.0040 = 0.40%

These questions are taken from study material for mining engineering exams (DGMS and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.

Question 

Hydraulic fracturing method is used to determine the major principal stress (σ₁) in an underground rock strata having tensile strength of 6 MPa. The minor principal stress (σ₃) in the strata is 8 MPa. If fluid pressure of 10 MPa is required to fracture the vertical borehole in that strata, the magnitude of in MPa. is ____ (rounded off to 2 decimal places). 

Solution 

P_B = 3σ₃ - σ₁ + σ_t

Putting given values

10 = 3 x 8 - σ₃ + 6

∴ σ₃ = 20 MPa

Question

In a development coal face, 12 holes are drilled and charged with explosive. Holes are initiated with electric delay detonators connected in series. The length of a detonator lead wire is 1.5 m. The length of the blasting cable is 120 m.

Data are as given:

Resistance of each detonator: 1.48 Ω

Resistance of lead wire: 0.04 Ω/m

Resistance of one wire of the blasting cable: 0.009 Ω/m

The total resistance of the circuit in Ω, is ____ (round off up to 2 decimals).

Solution

There are two blasting wires, 12 lead wires and 12 detonator.

Resistance of blasting wire 

= 2 x 0.009 x 120

= 2.16 Ω

Resistance of 12 lead wires 

= 12 x 0.04 x 1.5

= 0.72 Ω

Resistance of 12 detonators in series 

= 12 x 1.48 = 17.76 Ω

Total resistance = 2.16 + 0.72 + 17.76

= 20.64 Ω

These questions are taken from study material for mining engineering exams (DGMS and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.

Question 

For a continuous miner (CM) panel, the following data are given.

Data related to CM

Dimension of a working face : 5.0 m (width) x 3.0 m (height)

Web depth, m : 0.6

Time for one web cut up to full height, min : 9

Data related to shuttle car

Bucket capacity of shuttle car, tonne : 10

Fill factor : 0.9

Number of cars : 2

Cycle time of each car including

loading, travel and unloading, min : 6

Assume, unit weight of coal is 1.4 tonne/m³ and its swell factor is 1.2. Consider, 6 working hours per shift.

The non-working time in min, in working hours per shuttle car to dispatch all coal cut by the CM, is ______. (round off up to 2 decimals) 

Solution

Coal miner

Capacity of coal miner to handle loose soil

= (width x height x web depth x swell factor x density )/time

(5 m x 3 m x 0.6 m x 1.2 x 1.4 t/m³)/9

= 1.68 t/min = 100.8 t/hr

Shuttle car

Shuttle car capacity = bucket capacity x fill factor/time

= (10 t x 0.9)/6 min

= 1.5 t/min = 90 t/hr

Capacity of two shuttle cars = 2 x 90 = 180 t/hr

Now, 180 t ® 1 hr

1 t ® 1/180 hr

100.8 t ® (1/180) x 100.8 hr

= 0.56 hr  = 33.6 minutes

∴ Idle time in one hour

= 60 - 33.6 = 26.4 min

Idle time in 6 hour shift = 26.4 x 6 = 158.4 min

Also read:

Numerical problems from Mine Surveying from DGMS and PSU Exams

Numerical problems from Mining Machinery from DGMS and PSU Exams 

Numerical problems from Mine Ventilation from DGMS and PSU Exams  

Question 

A surface coal mine is worked by 10 m³ shovel – 85 tonne dumper combination. If the shovel bucket fill factor is 0.8, the material swell factor is 0.75 and the average in-situ sp.gr. of the material is 3.5 tonne/m³, the number of shovel bucket loads required to fill a dumper will be nearly____.

Solution 

Bucket loose volume  = 10 x 0.8 = 8 m³

This is Volume of material as it sits in the bucket (before swelling).

In-situ weight of material per bucket

= 8 x 3.5 = 28 tonnes

Number of loads = dumper capacity/(swell factor x in-situ weight)

= 85/(0.75 x 28) ≈ 4

The swell factor usually means that the loose volume taken by the shovel is less than the in-situ volume.

These questions are taken from study material for mining engineering exams (DGMS and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.

Question 

What would be the capacity of a flat belt conveyor in ton/ hour, if the bulk density of material is 0.8 t/m³, belt width is 1.0 m, angle of repose is 45° and velocity is 100 m/min? 

Solution  

T = abv

= (1/2)(width x pile height) x b x v

where b is bulk weight and v is velocity.

 

Pile height = 0.5 tan45⁰ = 0.5 m x 1 = 0.5 m

T = (1/2)(1 m x 0.5 m) x 0.8 t/m³ x (100 x 60 t/hr) = 1200 t/hr

Question

In an opencast coal mine, blast vibrations are measured at two locations. A and B simultaneously for a maximum charge per delay (Q) of 1200 kg as given.

Location	Distance from the blast face (m)	PPV (mm/s)
A	100	112.5
B	300	20.3

Assume the relation

PPV = K(D/√Q)^-β

where, 𝐾 and 𝛽 are site constants. The PPV in mm/s, at a distance of 200 m from

the blast face, is ______. (round off up to 2 decimals).

Solution

Taking log on both the sides (as without taking log, it would be difficult to solve equations as it would need a trial and error approach).

log(PPV) = log K -βlog(D/√Q)

Now, log 112.5 = log K - β(100/√1200)

i.e. 2.05 = log K - 0.46β     (1)

Similarly

log 20.3 = log K - β(300/√1200)

i.e. 1.31 = log K - 0.94β     (2)

Solving (1) and (2)

K = 575.4, β = 1.54

Now from given data

log (PPV₂₀₀) = log(575.4) - 1.54log(200/√1200)

∴ PPV₂₀₀ = 38.

These questions are taken from study material for mining engineering exams (DGMS and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required. 

Question

The percentage Fe and corresponding net value for an iron ore mine is given below

Fe (%)	Net value (INR/ton)
58	4000
62	4500

Assuming net value versus grade curve to be a straight line, and mining cost of waste is INR 1000 /m³; the correct representation of stripping ratio, SR (m³/tonne) versus Fe (%) grade curve is

(a) SR   = -3.250 + 0.125 X Fe*

(b) SR  = 3.250 + 0.125 X Fe

(c) SR   = 3250 + 125 x Fe

(d) SR  = -3250 + 125 x Fe

Solution

NV = m(Fe - Fe₁)

= [(4500 - 4000)/(62 - 58)](Fe - 58)

 

∴ NV = 125Fe - 3250

Stripping ratio

SR = NV/mining cost

= (125Fe - 3250)/1000

= 0.125Fe - 3.25

Question

An explosive with a density of 1.2 g/cm³ has a heat of explosion equal to 900 cal/g. If the heat of explosion of ANFO with density of 0.8 g/cm³ is 950 Cal/g, the bulk strength of the explosive relative to ANFO is _____. (round off up to 2 decimals)

Solution

Relative bulk strength

= str₁ x SG₁/str_anfo x SG_anfo

= 900 x 1.2/(950 x 0.8)

= 1.42

These questions are taken from study material for mining engineering exams (DGMS and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required. 

Question

The void ratio of an unconsolidated soil heap of volume 1000 m³ is 1.0. If the soil heap is consolidated to a volume of 800 m³, the corresponding void ratio is________. (round off up to 2 decimals).

Solution

Void ratio, e = V_v/V_s = (V - V_s)/V_s

Case 1:

1 = (1000 - V_s)/V_s

Solving, V_s = 500 m³

This volume of soil particles (V_s) will remain constant.

Case 2:

e = (800 - 500)/500 = 0.6

Question

A rectangular development heading of dimension 3 m × 2.8 m is to be blasted with holes of 2.4 m in length. If the pull factor is 0.95 and swell factor is 1.20, the volume of blasted rock per round, in m³, is______. (round off up to 2 decimals).

Solution

Volume of blasted rock = H x W x L x pull factor x swell factor

= 3 x 2.8 x 2.4 x 0.95 x 1.2

= 22.98 m³

Also read:

Numerical problems from Mine Surveying from DGMS and PSU Exams

Numerical problems from Mining Machinery from DGMS and PSU Exams 

Numerical problems from Mine Ventilation from DGMS and PSU Exams  

Question

A Bord and Pillar panel is developed at a depth of 250 m in a flat coal seam. The vertical stress gradient is 0.027 MPa in. If the strength of [a square pillar is 12.5 MPa. the extraction ratio of the pillar for a safety factor of 1.5. is ____ (round off up to 2 decimals). 

Solution 

Load on pillar = vertical stress gradient x depth x [(a +b)/a]²

= 0.027 x 250 x [(a + b)/a]² MPa

= 6.75 x [(a + b)/a]² MPa

Strength of pillar = 12.5 MPa

Now, factor of safety = strength of pillar/load on pillar

∴ 1.5 = 12.5/[(a + b)/a]²

Giving [(a + b)/a]²  = 1.23

Now extraction ratio = 1 - [a/(a + b)]²

= 1 - (1/1.23) = 0.18

Question

A Mohr-Coulomb envelop between shear stress, τ and normal stress, 𝜎_𝑛 of a sandstone rock is given as 𝜏 = 7.5 + 0.84 𝜎_𝑛 (unit of stresses is MPa)

A sandstone sample is tested in triaxial mode with confining pressure of 5.0 MPa. The value of the shear stress, 𝜏 in MPa at the failure, is______. (round off up to 2 decimals).

Solution

Comparing given equation with

τ = c + σ_ntanφ

tanφ = 0.84

∴ φ = 40⁰

Now, σ₁ = mσ₃ + σ_c

where m = (1 + sinφ)/(1 - sinφ)

= (1 + sin40⁰)/(1 - sin40⁰)

= 4.6

Now, σ_c = 2c x cosφ/(1 - sinφ)

= 2 x 7.5 x cos40⁰/(1 - sin40⁰)

= 32.184

σ₁ = 4.6 x 5 + 32.184 = 55.184

τ = [(σ₁ - σ₃)/2]sin(90 - φ)

= [(55.184 - 5)/2]sin(90⁰ - 40⁰)

Giving τ = 19.22 MPa

These questions are taken from study material for mining engineering exams (DGMS and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required. 

 Question

A circular tunnel is constructed at a depth of 100 m. The average unit weight of overburden rock is 27.0 kN/m³. If the tangential stress measured at point A located at the horizontal boundary of the tunnel as shown, is 5.0 MPa, the tangential stress at point B in MPa, is______. (round off up to 2 decimals).

Solution

σ_A = P(3 - k) when θ = 90⁰

and σ_B = P(3k - 1) when θ = 90⁰

Given that d = 100 m

σ_A = 5 MPa = 5 x 10⁶ N/m²

P = 27 kN/m³ = 27000 N/m³

Now, σ_A = P(3 - k)

5 x 10⁶ = (27000 x 100)(3 - k)

Giving k = 1.15

Now, σ_B = P(3k - 1)

= (27000 x 100) (3 x 1.15 - 1)

= 6615000 N/m² = 6.615 MPa