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, GATE and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.
The reading on a 4.0 m staff at appoint is observed as 2.895 m. If the staff was 8 cm out of the plumb line, the correct reading should be
(a) 2.8950 m
(b) 2.8965 m
(c) 2.8961 m
(d) 2.8938 m*
(e) 2.8150 m (DGMS FCMC/METAL/R 2021)
Hint:
AC = √(AD2 – CD2)
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
If the latitude and departure of a line is 85 m and 5 m, then its length is –––– m
(a) 88.88
(b) 100.15
(c) 90.2
(d) 87
(e) 85.15* (DGMS FCMC/METAL/R 2021)
Hint: L = √(L2 + D2)
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
The QB of a line is N 200 W then its whole circle bearing is
(a) 2000
(b) None of the options
(c) 3400*
(d) 1600
(e) 200 (DGMS SCMC/METAL/UR 2021)
Hint: WCB of a line is measured from the true north or magnetic north in clockwise direction.
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, GATE and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.
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, GATE and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.
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 Trapezoidal Rule.
(a) 165.91 Square meter
(b) 160.81 Square meter
(c) 166.91 Square meter
(d) 168.015 Square meter*
(e) 162.015 Square meter (DGMS SCMC/METAL/UR 2021)
Hint: Area = (3 / 2) x [(3.82 + 6.43) + 2{4.37 + 6.82 + 5.26 + 7.59 + 8.90 + 9.52 +
8.42}]
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, GATE and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.
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)
The length of a line is 20 metre and its bearing is N 600 E. The latitude of the line will be metre.
(a) (-) 10.00*
(b) (+) 17.32
(c) (+) 10.00
(d) (-) 17.34
(e) none of these (DGMS SCMC/METAL/UR 2021)
Hint: Latitude = L cos α
Departure = L sin α
α is bearing of line.
These questions are taken from study material for mining engineering exams (DGMS, GATE and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.
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.
Two roadways in a footwall drive are to be connected by a circular curve. The intersection angle is 104024' and the radius of the curve is 75 m. Calculate the length of each chord if the chords are to be 10 equal parts.
(a) 17.45 m
(b) 10.56 m*
(c) 15.76 m
(d) 12.11 m
(e) 13.64 m (DGMS SCMC/METAL/UR 2021)
Hint: L = πR∆/180
R = radius of the circular curve
Δ = Deflection angle = 180° - Intersection angle
∆ = 180 – 104024’
Length of each chord = L/number of chords
These questions are taken from study material for mining engineering exams (DGMS, GATE and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.
In surveying belowground, the bearing of the lines AB and BC are 950 30' and 400 30' respectively. Calculate the included angle ABC.
(a) 780
(b) 1020
(c) 1350
(d) 780
(e) 1250* (DGMS SCMC/METAL/UR 2021)
Hint: (180 – 95030’) + 40030’ = 125 degree
If the included angle is negative, then add 360°.
If the included angle is greater than 360° then deduct 360°.
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, GATE and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.
In a triangulation survey, if an equilateral triangle has side equal to 30 m the area will be equal to –––– m2
(a) 339.71
(b) 389.71*
(c) 359.71
(d) 379.22
(e) 350.34 (DGMS SCMC/METAL/UR 2021)
Hint: (√3/4) x side x side
If a total station has a purported accuracy capability of ± (5 mm + 5 ppm). What error can be expected in measured distance of 800 m?
(a) ± 9 mm
(b) ± 10 mm
(c) ± 6.4 mm*
(d) ± 8 mm
(e) ± 1 mm (DGMS FCMC/METAL/UR/DEC-2021)
Hint: Measured distance = 800 m
Ed = √Ei2 + Er2 + Ec2 + (ppm x D)2
Here Ei = Er = 0
Ed = √ (5)2 + (5 x 10-6 x 800,000)2
= 6.4 mm
The distance of 800 m was converted to mm to obtain unit consistency.
These questions are taken from study material for mining engineering exams (DGMS, GATE and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.
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
In a triangulation survey, if a triangle has three sides equal to 11 m, 12 m and 15 m the area will be equal to –––– m2
(a) 35.13
(b) 45.22
(c) 65.23*
(d) 75.23
(e) 85.23 (DGMS FCMC/METAL/UR/DEC-2021)
Hint: A = √s(s – a)(s – b)(s-c)
where s = semi perimeter = (a + b + c)/2
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, GATE and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.
1460. 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)
= 500[1 - cos(120/2)]
= 250 m
For a dumpy level, the bubble tube has sensitivity of 40" for one division. While taking a staff reading at a distance of 60 m, the bubble is out of centre by 2 divisions. The error in staff reading in mm is______. (GATE 2021)
Solution: Sensitivity, ϕ = 40" per 1 division
Distance between staff and instrument, L = 60 m
Deviation of bubble from center, n= 2
Find, error in staff reading, S =?
See figure.
Sensitivity formula is
φ = 206265 x S/(L x n)
∴ 40 = 206265 x S/(60 x 2)
∴ S = 0.0232 m = 23.2 mm
These questions are taken from study material for mining engineering exams (DGMS, GATE and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.
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)
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, GATE and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.
Two contours of RL 60 m and 70 m are separated by 34 m measured along dip of the seam. The dip of the seam in degree is ________________. (GATE 2018)
Solution: θ = tan-1(70 - 60)/34 = 16.390
The RL of the initial station is 200 m. If Σ BS = 1.54 m and Σ FS = 0.45 m, then the RL of the last station in m is ________________. (GATE 2018)
Solution:
Height of instrument (HI) = BM + BS
HI = 200 + 1.54 = 201.54 m
RL of forward station = HI – FS
R.L = 201.54 – 0.45 = 201.09 m
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, GATE and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.
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 + ... An - 1]
= 15[{3054 + 1992)/2} + 3025 + 2095 + 2008 + 2000]
= 174765 m3
In a triangle ABC. the bearings of the sides AB, BC and CA are 60°, 130°, and 270° respectively. The interior angles A, B, and C in degrees respectively are
(a) 110, 40, 30
(b) 40, 110, 30
(c) 30, 40, 110
(d) 30, 110, 40* (GATE 2015)
See figure.
∠C = 400
∠A= 180°- (∠B + ∠C)
= 180°- (110° + 40°) = 30°
These questions are taken from study material for mining engineering exams (DGMS, GATE and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.
The bearing of side AB of a regular hexagon ABCDEF is S 50010′ E . If the station C is easterly from the station B, the whole circle bearing of the side BC is
(a) 65015′25′′
(b) 69050′25′′
(c) 69015′25′′
(d) 69050′0"* (GATE 2014)
Solution: See figure.
WCB of BC = 1200 - 50010'
= 119060' - 50010'
= 69050'
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
(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, GATE and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.
It is proposed to connect two straights of a road by a simple circular curve. If the maximum speed of the vehicle is 60 km/h and the centrifugal ratio for the road is 1/4, the minimum radius of the curve in m is
(a) 113.26*
(b) 98.18
(c) 25.46
(d) 15.50 (GATE 2011)
Solution: Rmin = V2/127f
= 60 x 60/(127 x 0.25)
= 113.39 m
In an underground correlation survey by the Weisbach triangle (figure below) the following data obtained.
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, GATE and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.
In a closed traverse, ABC, the bearings of two lines AB and BC are given.
(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
Solution:
See figure.
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
These questions are taken from study material for mining engineering exams (DGMS, GATE and PSU) by amiestudycircle.com. With our study material which is prepared by IIT, Roorkee faculty (Retd.), no text book study is required.
Comments