Mine Surveying - Numerical Problems from DGMS and PSU Exams

A 20 m chain was found to be 10 cm too long after chaining a distance of 2000 m. It was found to be 18 cm too long at the end of the day’s work after chaining a total distance of 4000 m. What is the true distance if the chain was correct before the commencement of the day’s work?

(a) 3962 m

(b) 4019 m*

(c) 3981 m

(d) 4038 m (DGMS FCMC/METAL/R 2021)

Solution: For first 2000 m, average error = (0 + 10)/2 = 0.05 m

Incorrect length = L’ = 20 + 0.05 = 20.05 m

Measured length l’ = 2000 m

True length = (L’/L) x l’ = (20.05/20) x 2000 = 2005 m

For next 2000 m, average error = (10 +18)/2 = 14 cm = 0.14 m

True length will come out as 2014 m

Hence total true length = 2005 + 2014 = 4019 m

These questions are taken from study material for mining engineering exams (DGMS and PSU) by amiestudycircle.com. WhatsApp channel: Miners Adda

If the net length covered by each photograph is 1.2 km and length of strip is 20 km, the number of air photographs required are

(a) 32

(b) 16

(c) 24

(d) 17

(e) 18* (DGMS FCMC/METAL/R 2021)

Hint: N = (Length of strip/photograph length) + 1

= (20/1.2) + 1

= 17.7 say 18

The focal length of an aerial camera is 12.5 cm and the exposure station was 2500 m above the datum. The scale of the photograph is

(a) none of these

(b) 1 is to 2000

(c) 1 cm = 400 m

(d) 1 is to 1000

(e) 1 cm = 200 m* (DGMS FCMC/METAL/R 2021)

Hint: Scale = f/H = 12.5/(200 x 100)

= 1/20000

i.e. 1 cm = 20000 cm = 200 m

If the focal length of the object glass is 25 cm and the distance from object glass to the trunnion axis is 15 cm, the additive constant is

(a) 0.1

(b) 1.33

(c) 0.6

(d) 0.4*

(e) None of the options (DGMS SCMC/METAL/UR 2021)

Hint: Additive constant = f + d

These questions are taken from study material for mining engineering exams (DGMS and PSU) by amiestudycircle.com. WhatsApp channel: Miners Adda

A standard steel tape of length 30 m and cross-section 15 × 1.0 mm was standardised at 25°C and at 30 kg pull. While measuring a base line at the same temperature, the pull applied was 40 kg. If the modulus of elasticity of steel tape is 2.2 × 10⁶ kg/cm² the correction to be applied is

(a) + 0.0909 m

(b) None of the options

(c) 0.000909 m

(d) All the options

(e) - 0.000909 m* (DGMS SCMC/METAL/UR 2021)

Solution:

Correction = (P – P₀)L/AE

P = 30 kg, P₀ = 30 kg

A = 1.5 x 0.1 cm²

E = 2.2 × 10⁶ kg/cm²

L = 30 m

These questions are taken from study material for mining engineering exams (DGMS and PSU) by amiestudycircle.com. WhatsApp channel: Miners Adda

The offsets taken at 3 m intervals from a survey line are 3.82 m, 4.37 m, 6.82 m, 5.26 m,7.59 m, 8.90 m, 9.52 m, 8.42 m and 6.43 m. Calculate the area of irregular boundary by Simpson's Rule.

(a) 168.015 Square meter

(b) 165.91 Square meter*

(c) 160.81 Square meter

(d) 162.015 Square meter

(e) 166.91 Square meter (DGMS FCMC/METAL/UR/DEC-2021)

Solution: By the Simpson’s rule

Area = (3 / 3) x [(3.82 + 6.43) + 4{4.37 + 5.26 + 8.90 + 8.42} + 2{6.82 + 7.59 + 9.52}]

These questions are taken from study material for mining engineering exams (DGMS and PSU) by amiestudycircle.com. WhatsApp channel: Miners Adda

The bearing of a line is N 56⁰ 20′ 45′′ E and declination of the area is 1⁰ 20′ W. The direction of true north is

(a) 57⁰ 40′ 45′′

(b) 55⁰ 00′ 25′′

(c) None of the options

(d) 55⁰ 00′ 45′′*

(e) 57⁰ 20′ 45′′ (DGMS SCMC/METAL/UR 2021)

Hint: TB (True Bearing) = MB (Magnetic bearing) ± Declination

(Use + ve for eastern and – ve for western declination)

These questions are taken from study material for mining engineering exams (DGMS and PSU) by amiestudycircle.com. WhatsApp channel: Miners Adda

The desired sensitivity of a bubble tube with 2 mm divisions is 30". The radius of the bubble tube should be

(a) 13.75 m*

(b) 3.44 m

(c) 1375 m

(d) All the options

(e) None of the options (DGMS SCMC/METAL/UR 2021)

Solution: 30'' = 2 mm,

1' = 4 mm

1degree = 60 x 4 = 240 mm

360 degree = 240 x 360 = 86400 mm

2πr = 86400

∴ r = 13757.9 mm = 13.75 m

At the equator the dip of the needle is

(a) None of the Options*

(b) 270⁰

(c) 45⁰

(d) 180⁰

(e) 90⁰ (DGMS SCMC/METAL/UR 2021)

Hint: What is the dip of a magnetic needle at the equator and pole?

At the magnetic equator, the dip needle rests horizontally at an angle of zero degrees while, at the magnetic poles, the magnetic needle rests vertically, at an angle of 90 degrees. At all the other places, the angle of dip is between 0 and 90 degrees.

A level when set up 25 m from peg A and 50 m from peg B reads 2.847 on a staff held on A and 3.462 on a staff held on B, keeping bubble at its centre while reading. If the reduced levels of A and B are 283.665 m and 284.295 m respectively, the collimation error per 100 m is

(a) 0.045 m

(b) 0.030 m

(c) 0.015 m

(d) 0.060 m*

(e) None of the options (DGMS SCMC/METAL/UR 2021)

Solution:

Actual difference between A and B = 283.295 - 283.665 = 0.630 m

Apparent difference between A and B = 3.462 - 2.847 = 0.615 m

Error in collimation = 0.630 – 0.615 = 0.015

Distance between A and B = 50 - 25 = 25 m

Error in collimation for 100 m = 0.015 x 100/25

= 0.06 m

These questions are taken from study material for mining engineering exams (DGMS and PSU) by amiestudycircle.com. WhatsApp channel: Miners Adda

If reduced level of a floor is 99.995, the staff reading on the floor is 1.405. Inverted staff reading against the roof is 1.895 m, the floor level below the roof is

(a) 3.300 m*

(b) 0.490 m

(c) 2.790 m

(d) 1.895 m

(e) none of these (DGMS FCMC/METAL/UR/DEC-2021)

Hint: 1.405 + 1.895

Keeping the instrument height as 1.5 m, length of staff 4.0 m, the slope of the decline of an underground mine as 1 in 20, the sight of the down slope should be less than

(a) 50 m*

(b) 30 m

(c) 15 m

(d) 25 m

(e) none of these (DGMS FCMC/METAL/UR/DEC-2021)

Hint: tanθ = (staff reading – HI)/horizontal distance

∴ 1/20 = (4 – 1.5)/x

Solving x = 50 m

A star with a declination of 20 degree will have upper culmination at a place of latitude 50 degree when the zenith distance is

(a) 70 degree

(b) 110 degree

(c) 30 degree*

(d) 170 degree

(e) 130 degree (DGMS FCMC/METAL/R/DEC-2021)

Hint: At upper culmination,

Zenith Distance = δ - θ = 50 - 20 =30⁰

These questions are taken from study material for mining engineering exams (DGMS and PSU) by amiestudycircle.com. WhatsApp channel: Miners Adda

The geometry of a simple planar curve (ADB) is shown below. The value of the; mid-ordinate of the curve in m is ____. (GATE 2020)

Solution: Mid ordinate = R[1 - cos(∆/2)]

= 500[1 - cos(120/2)]

= 250 m

On an old plan of scale 1:1000, leasehold area of a mine is now measured as 802 cm² using a planimeter. The plan is found to have shrunk, such that the original line of 10 cm is now measured as 9.8 cm on the plan. True lease hold mine area, in m², is _______. (GATE 2021)

Solution: SF = 9.8/10 = 0.98

As the plan is shrunk, we will get more plan area.

Plan area = 802 /(0.98)² = 835.07 cm²

On scale of 1/1000, it will be 835.07 x 1000² cm²

= 83507 m²  

Two points A and B are located 150 m apart in the East-West orientation on the bank of a river as shown in the figure. Considering a station C on the north bank the bearings of AC and BC are observed to be 42⁰ and 335⁰ respectively. The width of the river, in m, (rounded off to two decimal places) is __________. (GATE 2019)

Answer: The given is plotted below with various angles calculated from given bearings.

Now, ∆ABC

AC/sin65⁰ = 150/sin67⁰

∴ AC = 147.77 m

Now consider ∆ADC

AC/sin90⁰ = cD/sin48⁰

∴CD = 147.77 x sin48⁰/sin90⁰

= 109.81 m

These questions are taken from study material for mining engineering exams (DGMS and PSU) by amiestudycircle.com. WhatsApp channel: Miners Adda

Two points on the equator have longitudes 55⁰E and 25⁰W. Considering radius of earth as 6400 km, the distance between the two points in km is _______. (GATE 2018)

Solution: Arc length = 2πr(θ⁰/360⁰)

= 2 x 3.14 x 6400 (80/360)

= 8931.5 km

These questions are taken from study material for mining engineering exams (DGMS and PSU) by amiestudycircle.com. WhatsApp channel: Miners Adda

The following readings refer to a reciprocal leveling at staff stations A and B respectively

Inst. Stn	Staff reading		Remarks
	A	B
P	1.725	2.835	RL of A = 125.0 m
Q	0.835	1.545

The reduced level (RL) of staff station B in 1n is ____. (GATE 2017)

Solution: True difference in elevation

h = [(a - b) + (c - d)]/2

= (2.835 - 1.725) + (1.545 - 0.835)]/2

= 1.09

There is a fall from A to B as staff reading is increasing.

∴ RL of B = 125 - 1.09 = 123.9 m

The areas within the contour lines of a proposed site for an overburden dump are as follows

Contour level	Area (m2)
195	3054
180	3025
165	2095
150	2008
135	2000
120	1992

The total volume of overburden in cubic meter that could be dumped within the 120 m and 195 m contour levels is ____. (GATE 2017)

Answer: Use trapezoidal formula

V = d[{A₁ + A₂)/2} + A₂ + A₃ + ... A_{n - 1}]

= 15[{3054 + 1992)/2} + 3025 + 2095 + 2008 + 2000]

= 174765 m³

A 20 m steel tape used in a mine survey is found to be 20 cm short when compared with a standard tape. If the measured volume of a dump using the tape is 4000 m³, its actual volume in m³ is

(a) 3881*

(b) 3902

(c) 3920

(d) 4121 (GATE 2013)

Solution: As the tape is short, it will measure more volume than actual volume.

SF = 19.8/20 = 0.99

As the tape is found short, measured area will be more. It means that actual area should be less.

actual volume = 4000 x (0.99)³ = 3881 m³

The figure shows the values of seven perpendicular offsets and the respective locations along the line XY as observed while carrying out a traverse survey. The area of the plot XABCDEFGY in m² is

 

(a) 26.10

(b) 43.38

(c) 44.92*

(d) 62.50 (GATE 2012)

Solution: If we connect all offset points by straight lines, we get a combination of trapezoids.

A₁ = (1/2)(7.2 + 11.9) x (0.6 - 0) = 5.73 m²

A₂ = (1/2)(11.9 + 6) x (1.4 - 0.6) = 10.52 m²

Similarly

A₃ = 10.2 m²

A₄ = 1.815 m²

A₅ = 9.3975 m²

A₆ = 7.26 m²

Total area = 44.92 m²

These questions are taken from study material for mining engineering exams (DGMS and PSU) by amiestudycircle.com. WhatsApp channel: Miners Adda

In an underground correlation survey by the Weisbach triangle (figure below) the following data obtained.

AB = 3.481 m

BC = 2.674 m

CA = 6.155 m  

ACD = 179°14'33”

BCD = 179°10'17” and bearing of  AB = 115°23'49''.

The bearing of traverse CD is

(a) 102°27'16”

(b) 114°41'49”*

(c) 115°27'16”

(d) 179°14'16” (GATE 2011)

Solution: In triangle ABC

∠ACB = 179⁰14'33⁰ - 179⁰10'17" = 4'16"

As the sides of very small angles are proportional to the angles themselves, hence

∠BAC = (BC x ∠ACB)/AB

= 2.674 x 4'16"/3.481 = 3'17"

Azimuth of AC = 115⁰23'49" + 0⁰3'17"

= 115⁰27'06"

Azimuth of CA = 115⁰27'06" + 180⁰00'00"

= 295⁰27'06"

Azimuth of CD = Azimuth of CA + ∠ACD

= 295⁰27'06" + 179⁰14'33"

= 474⁰41'39" - 360

= 114⁰41'39"

These questions are taken from study material for mining engineering exams (DGMS and PSU) by amiestudycircle.com. WhatsApp channel: Miners Adda

In a closed traverse, ABC, the bearings of two lines AB and BC are given.

The length, in m and bearing of line CA, in degree, respectively, are

(a) 190.7 and 303°*

(b) 190.7 and 240°

(c) 160.3 and 240°

(d) 160.3 and 303° (GATE 2024)

Solution: Σ Latitude = 0

Σ (Lcosθ) = 0

100 cos90⁰ + 120 cos150⁰ + Lcosθ = 0

Lcosθ = -160⁰     (1)

Σ Departure = 0

Σ (Lsinθ) = 0

100 sin90⁰ + 120 sin150⁰ + Lsinθ = 0

Lsinθ = -160     (2)

Solving θ = -57⁰

As CA is in 4th quadrant,

θ = 360 - 57 = 303⁰

From (2)

L = -160/sin(303⁰) = 190.77 m

In a tacheometry survey, the readings observed are given

The additive and multiplying constants of the instrument are 0 and 100, respectively. The length of the line AB in m, is ______. (round off up to 2 decimals) (GATE 2024)

Solution:

See figure.

S₁ = 2.2 - 1.2 = 1.0 m

K = 100

C = 0

θ₁ = 8⁰

h₁ = 1.7 m

PA = KS₁cosθ₁ + Ccosθ₁ + h₁sinθ₁

= 100 x 1 x cos8⁰ + 0 + 1.7sin8⁰

= 99.24 m

Similarly

S₂ = 1.6 - 0.8 = 0.8

K = 100

C = 0

θ₂ = 3⁰

h₂ = 1.2 m

PB = 100 x 0.8 x cos3⁰ + 0 + 1.2sin3⁰

= 79.95 m

Angle BPA = 205 - 145 = 60⁰

Now, AB² = PA² + PB² - 2(PA)(PB) cosθ

= 99.24² + 79.95² - 2 x 99.24 x 79.95 cos60⁰

= 8306.34

∴ AB = 91.14 m

Subscribe to our newsletter for updates on DGMS and other  mining engineering exam 

Study material for DGMS and other mining engineering exams 

WhatsApp channel: Miners Adda