|MN-2021|

GENERAL APTITUDE

Q. No: 1-5 Carry One Mark Each

1. The mirror image of the above text about the x-axis is

PHYLAXIS

(A) (B)
(C) (D)
[1-Mark, MCQ]
Key: (A)
Sol: PHYLAXIS
X

(∵ Mirror image of any text about a line is symmetrical about the line).
Hence option (A).

2. Four persons P, Q, R and S are to be seated in a row, R should not be seated at the second position from
the left end of the row. The number of distinct seating arrangement possible is: [1-Mark, MCQ]
(A) 6 (B) 18 (C) 9 (D) 24
Key: (B)
Sol:
( P, Q, S)

  
3choices 3choices 2choices 1choice

The number of distinct seat arrangements = 3  3  21 = 18

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3.  and are two operators on numbers p and q such that

p q = p − q, and p  q = p  q

Then, (9 ( 6  7)) ( 7  ( 6 5)) = [1-Mark, MCQ]

(A) –33 (B) –40 (C) 40 (D) –26

Key: (B)
Sol: ( 9 ( 6  7 ) ) ( 7  ( 6 5) )
= ( 9 − ( 6  9 ) ) − ( 7  ( 6 − 5 ) ) = ( 9 − 42 ) − ( 7  1) = −33 − 7 = −40

4. (i) Arun and Aparna are here.

(ii) Arun and Aparna is here.
(iii) Arun’s families is here.
(iv) Arun’s family is here.
Which of the above sentences are grammatically CORRECT? [1-Mark, MCQ]
(A) (ii) and (iv) (B) (i) and (ii) (C) (i) and (iv) (D) (iii) and (iv)
Key: (C)
F
Sol: F → Football players
H
C → Cricket players C

H → Hockey players
∴ Some football players play hockey

5. Two identical cube shaped dice each with faces numbered 1 to 6 rolled simultaneously. The probability
that an even number is rolled out on each dice is: [1-Mark, MCQ]
1 1 1 1
(A) (B) (C) (D)
12 8 36 4
Key: (D)
Sol: Total number of events in sample space = 6 6 = 62 = 36.
No. of favourable events of event A = 33 = 9
9 1
∴ Required probability = =
36 4

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Q. No: 6-10 Carry Two Marks Each

6. In an equivalent triangle PQR, side PQ is divided into four equal parts, side QR is divided into six equal
parts and side PR is divided into eight equal parts. The length of each subdivided part in cm is an
integer.
The minimum area of the triangle PQR possible, in cm2, is
(A) 144 3 (B) 48 3 (C) 18 (D) 24
[2-Marks, MCQ]
Key: (A)
Sol: Let us assume, the length of side of an equilateral triangle = 24cm. (L.C.M of 4, 6, 8 = 24)
To have minimum area of triangle PQR & length of each sub divided part is an integer).
3 2
∴ Area of an equilateral triangle = a , where a is the length of the side
4

3 3
=  ( 24 ) =  24  24 = 144 3 cm 2
2

4 4

7. In the figure shown above, PQRS is a square. The shaded portion is formed by the intersection of sectors
of circles with radius equal to the side of the square and centers at S and Q.
[2-Marks, MCQ]
P Q

S R
r

The probability that any point picked randomly within the square falls in the shaded area is ______.
  1 
(A) (B) 4 − (C) (D) −1
4 2 2 2
Key: (D)

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Sol: Totalarea = r  r = r 2
P Q
1 1 2 1 2 1 2
Favourable area =  r 2 − r  +  r − r 
4 2  4 2 


Area of Area of
quarter  PSR
circle
r
1 1 
= 1 r 2 − r 2 
4 2  S
r R
1 1 
2  r 2 − r 2 
2  
∴ Required probability = 
4
2
= −1
r 2

P r
Q

r
r
r

S R
1 2 1 2
Shaded portion =  r − r 
4 2 

8. 1. Some football players play cricket.

2. All cricket players play hockey.
Among the options given below, the statement that logically follows from the two statements 1 and 2
above, is
(A) All hockey players play football (B) No football player plays hockey
(C) All football players play hockey (D) Some football players play hockey
[2-Marks, MCQ]
Key: (D)

9. On a planar field, you travelled 3 units East from a point O. Next you travelled 4 units South to arrive at
point P. Then you travelled from P in the North-East direction such that you arrive at a point that is 6
units East of point O. Next, you travelled in the North-West direction, so that you arrive at point Q that
is 8 units North of point P. The distance of point Q to point O, in the same units, should be ______
(A) 3 (B) 6 (C) 4 (D) 5
[2-Marks, MCQ]

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Key: (D) Q
Sol: From right angle triangle ORQ,
2 R
OQ 2 = OR 2 + RQ 2
4
= 32 + 42
= 9 + 16 = 25 O
6
3 8
 OQ = 25 = 5
4

10. The author said, “Musicians rehearse before their concerts. Actors rehearse their roles before the
opening of a new play. On the other hand, I find it strange that many public speakers think they can just
walk on to the stage and start speaking. In my opinion, it is no less important for public speakers to
rehearse their talks”. Based on the above passage, which one of the following is TRUE?
(A) The author is of the opinion that rehearsing is important for musicians, actors and public speakers
(B) The author is of the opinion that rehearsing is more important only for musicians than public
speakers
(C) The author is of the opinion that rehearsal is more important for actors than musicians
(D) The author is of the opinion that rehearsing is less important for public speakers than for musicians
and actors
[2-Marks, MCQ]
Key: (A)

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MINING ENGINEERING

Q. No: 1-25 Carry One Mark Each

1. Tricone roller bit is used with

(A) down-the-hole hammer (B) Jack hammer
(C) rotary-percussive drill (D) rotary drill
Key: (D)

2. Reusing method of mining is practiced for

(A) thick vein deposit (B) massive shallow deposit
(C) narrow vein deposit (D) massive deep-seated deposit
Key: (C)

3. The equipment used for both drop cut and terrace cut in surface mining is
(A) surface miner (B) shovel
(C) dragline (D) bucket wheel excavator
Key: (D)

4. Surface miner does NOT have a

(A) differential gear for turning (B) tensioning arrangement for crawler
(C) scraper plate behind the drum (D) pick cooling system
Key: (A)

5. Induced blasting enhances production in

(A) sublevel stoping (B) block caving
(C) cut and fill mining (D) shrinkage stoping
Key: (B)

6. The measures of dispersion of a dataset are

(A) standard deviation, range and mode (B) standard deviation, range and interquartile range
(C) variance, range and median (D) interquartile range, median and mode
Key: (B)

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7. NONEL is used as down-the-hole initiator to

(A) avoid generation of air overpressure (B) provide precise delay
(C) avoid deflagration of column charge (D) reduce ground vibration
Key: (C)

8. The vector a and b act in a plane as shown below. The magnitude of the vector

( ) (
c = a + b  a − b is )

(A) zero
(B) half to the area bounded by the vectors a and b

(C) equal to the area bounded by the vectors a and b

(D) twice the area bounded by the vectors a and b

Key: (D)
Sol: (
C= a+b  a−b ) ( )
= a a − a b + ba − bb
= 0−2 ab −0( ) ( ba = − a b ( ))
(
= −2 a  b )
 C = 2 a  b , which is twice the are bounded by the vectors a and b

 Option (D)

9. As per MMDR Act 1957, for the allocation of lease of minor minerals
(A) the State Government is authorized to give the permit
(B) the Central Government is authorized to give the permit
(C) the State Government is authorized to give permit but with the consent of Central Government
(D) the Central Government is authorized to give permit but with the consent of State Government
Key: (A)

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10. In photogrammetry, the ‘Tilt of a photograph’ refers to the angle between the
(A) lines joining the opposite fiducial marks of a photograph
(B) normal to the plane of photograph and optical axis
(C) vertical and the axis of the flight
(D) vertical and optical axis of the camera
Key: (D)

11. The hydraulic sand stowing pipeline layout should be such that
(A) the geometric profile must coincide with the hydraulic gradient line
(B) the hydraulic profile should always be below the hydraulic gradient line
(C) the hydraulic profile should always be above the hydraulic gradient line
(D) the geometric profile should always be above the hydraulic gradient line
Key: (B)

12. For a “positive definite” square matrix, the TRUE statement is

(A) the matrix is singular
(B) all the eigen values of the matrix are greater than zero
(C) all the given values of the matrix are zero
(D) some of the eigen values can be less than zero
Key: (B)
Sol: We know that all the eigen values of a “positive definite” square matrix are positive (i.e., greater than
zero).

13. The standard normal distribution is a

(A) non-parametric distribution
(B) single parameter distribution
(C) two-parameter distribution
(D) three-parameter distribution
Key: (C)
Sol: Normal distribution has two parameter  and 

N ( , 2 )

but standard normal distribution is N(0, 1) (i.e., mean = 0, SD = 1)

 It is non-parametric distribution option (C)

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14. Variance of the sum of two statistically independent random variables X and Y, 2X + Y , is

(A) 2X + Y2 (B) 2X + Y2 + 2XY

(C) 2X + Y2 + XY (D) 2X + Y2 − 2XY

Key: (A)
Sol: 2X + Y = Var ( X + Y ) = Var ( X ) + Var ( Y ) , Since X, Y are independent random variables

= 2X + 2Y , option (A).

( Var ( ax + by ) = a 2 Var ( x ) + b2 Var ( y ) , if x, y are independent random variables )

15. The difference between depreciation and amortization allowances in tax calculation is that
(A) depreciation is for a tangible asset applicable on its declared life; whereas amortization is for an
intangible asset applicable on a specified period
(B) depreciation is for an intangible asset applicable on its declared life; whereas amortization is for a
tangible asset applicable on a specified period
(C) depreciation is for a tangible asset applicable on a specified period; whereas amortization is for an
intangible asset applicable on its declared life
(D) depreciation is for an intangible asset applicable on a specified period; whereas amortization is for a
tangible asset applicable on its declared life
Key: (A)

16. Owning cost of a machine does NOT include

(A) purchase price (B) insurance (D) interest (D) operating cost
Key: (D)

17. Folds are the structural features resulting from

(A) ductile deformation of earth crust
(B) brittle deformation of earth crust
(C) high impact tectonic stresses of earth crust
(D) fracturing of earth crust
Key: (A)

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18. The CORRECT curve showing relationship between vertical stress on a coal pillar and extraction ratio
of a bord and pillar panel in a horizontal seam is

A
p
Pillar Stress, 
B

C
D

Extraction Ratio, R

(A) Curve A (B) Curve B (C) Curve C (D) Curve D

Key: (C)

19. Given impeller diameter D, speed of a rotation n and air density , for geometrically similar fans, the
fan pressure is proportional to
(A) nD 2 (B) n 2 D 2 (C) n 2 D52 (D) n 3D5
Key: (B)
Sol: Change in pressure is directly proportional to the square as impeller dia D ratio or the square of speed at
ratio or the square of speed at ratio.
H 2 D 22
=
H1 D12
H 2 N 22
=
H1 N12

20. A coal sample having moisture content of 8.0% has unit weight 15.6 kN m3 . The dry unit weight of the
sample, in kN m3 is ___________.
Key: (14.44)
Sol: Given that
m = 80%
 = 15.6 kN m3
Dry unit weight??

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 |MN-2021|

Bulk unit weight 15.6

d = = = 14.44
1+ m 1 + 0.08

The value of the integral I = 

4
21. xdx computed using Simpson’s 1/3 rule with 2 subintervals is
0

________. (round off to 3 decimal places)

Key: (5.1)
b−a
Sol: f ( x ) = x , a = 0, b = 4, n = 2  h = =2
n
y0 = f ( 0 ) = 0 = 0; y1 = f ( a + h ) = 2  1.414

a
y 2 = f ( a + 2h ) = 4 = 2

1
 By Simpson’s rd rule,
3
4
h 2
 xdx = ( y0 + y 2 ) + 4y1  =  2 + 5.656  5.104
0
3 3

22. In the context of sound frequency analysis, the lower and upper frequencies of a 1/1 octave band are
710 Hz and 1420 Hz respectively. The corresponding centre frequency of the band in Hz, is _______.
(round off to the nearest integer).
Key: (1004)
Sol: Central frequency band = f1f 2 = 710  1420 = 1004.09

23. In Battle Environmental Evaluation System (BEES) of Environmental Impact Assessment (EIA), “air
pollution” has a Parameter Important Unit (PIU) value of 52. The Environmental Quality (EQ) score of a
project with respect to air pollution was 0.8 before the project implementation and it becomes 0.6 after
the project implementation. The difference in the “Environmental Impact Unit (EIU)”, before and after
the project implementation is _________. (round off to 2 decimal places)
Key: (10.4)
Sol: Given that, PIU = 52
EQ1 = 0.8
EQ2 = 0.6
Now environmental input unit = Difference of EIU
= 52  0.8 − 52  0.6
= 10.4  Where EIU = PIU − EQ

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24. A system consists of four components connected functionally in a parallel configuration. The reliability
of the individual components is 0.40, 0.60, 0.50 and 0.40. The system reliability, is ________ (round off
to 3 decimal places)
0.4
Key: (0.92)
Sol: System reliability 0.6
= 1 − (1 − R1 )(1 − R 2 )(1 − R 3 )(1 − R 4 )
= 1 − (1 − 0.4 )(1 − 0.6 )(1 − 0.5 )(1 − 0.4 )  0.5
= 0.92

0.4

25. A vehicle is moving at a speed of 12 m/s on a level road. It applies emergency brakes and starts to skid
without rolling in a straight path. The deceleration of the vehicle is constant after braking and it comes to
rest at a distance of 15m. Assuming, g = 10 m s2 , the coefficient of kinetic friction between the tyres
and road is __________ (round off to 2 decimal places)
Key: (0.48)
Sol: Given that,
v = 12 m s
12 m s
g = 10 m s 2 g = 10 m s 
15m Rest
Initial velocity = 12 m/s = U
Final velocity = 0 = u skiding
without collis
From equation motion
2 = u 2 − 2as
2 = 122 − 2  a  15
144
a= = 4.8 m s 2
30
(a will be retardation)
Frictional frame on vehicle = mg = ma
g = a
  10 = 4.8
4.8
= = 0.48
10
Method-II

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 |MN-2021|

1 2
mv = mg cos   dis tan ce
2
1
 12 =   10  cos 0  15
2
 = 0.48

Q. No: 26-55 Carry Two Marks Each

26. In a bord and pillar panel six shuttle cars, each of 10 tonne capacity, are deployed to transport coal
produced by two continuous miners to a belt conveyor. Each shuttle car on an average carries 80% of its
rated capacity and makes 7 round trips in an hour. The belt conveyor has a capacity such that the
effective material cross section area is of 0.09 m 2 and runs at a speed 1.1 m/s. The broken coal has a
bulk density of 1.2 tonne m3 . The ratio between the production and the capacity of the belt conveyor, in
percent is
(A) 65.46 (B) 71.42 (C) 78.56 (D) 82.46
Key: (C)
Sol: Number of shuttle car = 6
Capacity = 10 tonne
Number of condition mixer = 2
Belt A = 0.09 A 2
v = 1.1 m s
b = 1.2 tonne m 3
Pr oduction 6  10  0.8  7
=  100 = 78.56%
Capacity of belt conveyor ( 0.09 1.1 1.2  3600 )

27. With reference to the figure related to rock cutting by point attack tool, match the angle with
corresponding name

Angle Name
P.  1. Cutting angle
Q.  2. Clearance angle
R.  3. Wedge angle
S.  4. Rake angle

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 |MN-2021|


Cutting tool 

Y


(A) P-2, Q-4, R-1, S-3 (B) P-4, Q-2, R-1, S-3
(C) P-2, Q-4, R-3, S-1 (D) P-4, Q-2, R-3, S-1
Key: (A)

28. The pit bottom in a correlation survey is shown in the figure. Points C and D represents two suspended
wires. The bearing of line CD is 2860000 and its length is 4.64m. The angle CED is measured as
000040. The length of line DE is 5.46m. Considering the Weisbach triangle method, the bearing of
the line CE is

D C
E

(A) 2860047 (B) 28559'12.9''

(C) 28600'40'' (D) 28500'47.1''
Key: (B)
Sol: From figure
D

4.64
5.46
N

40
286
E C

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From sine rule

CD ED
=
sin E sin C
4.64 5.46
=
sin 40'' sin C
C = 47.07 ''
Now Bearing of CE = Bearing of CD - < C
= 286 − 47.07''
= 28559'12.93''

29. A dump truck moves up an incline of 50 with constant tractive force of 800 kN. The gross mass of the
truck is 250 tonne and its rolling resistance is 545 kN. The acceleration due to gravity is 10 m 2 s. The
time required, in s, to reach a speed of 3.3 m/s from 1.0 m/s is
(A) 22.0 (B) 15.5 (C) 3.3 (D) 0.2
Key: (B)
Sol: Drawbar pull = Tractive fore – Rolling resistor – sliding force
= 800 − 545 − mgsin 
mgsin 
= 800 – 545 – 250 × 10 × sin 5
5 Weight = 250 tonne
= 37.11 kN
v 2.3
Now F = ma = m = 37.11 = 250   t = 15.5 sec
t t

30. In a longwall panel, face is supported with shield of yield capacity 460 tonne per shield. The distance
from the canopy tip to coal face is 0.15 m when the support is fully advanced. The depth of web is
0.60m. The shields are set skin to skin at the face. Length of the canopy of the shield is 3.25 m and width
1.5 m. Setting capacity is 80% of the yield capacity. The setting resistance at the maximum and
minimum span of the coal face, in tonne m2 , respectively are
(A) 61.33 and 72.15 (B) 63.72 and 75.48
(C) 76.666 and 90.19 (D) 91.99 and 108.22
Key: (A)
Sol: Given that,
Yield capacity of shield = 460 tonnes/shield
Distance (coal to canopy tip) = 0.15 mm
Web = 0.6m
Length canopy = 3.25 m
Width = 1.5m
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 |MN-2021|

Capacity = 80%
80
Setting capacity = 80% of 460 = 460  = 368 tonne shield
100
Maximum area supported by canopy = 4 1.5 = 6m2

canopy

3.25 1.5m

face
0.6 + 0.15

Minimum area supported by canopy Support Resistance = 3.4 1.5 = 5.1 m2

368
Max = = 61.33 tonne m 2
6
368
Min = = 72.15 tonne m 2
5.1

31. A 10 ml sample of wastewater is diluted with water having no BOD, to fill a 300 ml BOD bottle. The
initial DO of the diluted waste water is 9.0 mg/l. If the BODs of the waste water sample is 60 mg/l, the
final DO of the diluted waste water in mg/l, is
(A) 5.0 (B) 6.0 (C) 7.0 (D) 8.0
Key: (C)
DO depletion ( mg m )  Volume diluted
Sol: BOD sample ( mg m )= (m )
Sample volume ( m )
[DO depletion = DOinitial − DOfinal ]

 60 mg m =
( 9 − DOfinal ) mg m
 300 m
10 m
 9 − DOfinal = 2
 DOfinal = 7 mg m

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32. The Mohr circle of stress of a dry porous rock is shown in the figure. If the rock is fully saturated with a
pore pressure p, then the Mohr circle takes the form of

n
3 1

 
(A) (B)

n n
3 − p 1 − p 3 + p 1 + p

(C)  (D) 

n n
3 − p 1 + p 3 + p 1 − p
Key: (A)
Sol: If the lock is fully saturated with a pore pressure P, then P will be subtracted from the principal stress
since, the Mohr diagram shown in under tri-axial condition.


n
3 − p 1 − p

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33. The straight line shown depicts the failure criterion of a rock type. The values of stress at points A and B
are as shown. The safety factor at the points A and B respectively are

250

Major Principal Stress (MPa), 200

150 •
B ( 40,160 )
A (10,90 )
100 •
1 = 43 + 28
50
1

0
−10 0 10 20 30 40 50 60
Minor Principal stress ( MPa ) , 3

(A) 1.175 and 0.755 (B) 1.324 and 0.851

(C) 0.851 and 1.324 (D) 0.755 and 1.175
Key: (D)
Sol: 1 = 43 + 28
for 3 = 10 ( at point A )
1 = 4  10 + 28 = 68
But at A11 = 90
68
So, factor of safety at A = = 0.755
90
At point B,
1 = 160 & 3 = 40
1 = 4  40 + 28 = 188
188
So, factor of safety at B = = 1.175
160

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34. Figure shown relates to the manufacture of roof bolts. With respect to the cost/revenue vs production
level, match the appropriate trend line with corresponding description

A
Line Item
B

Cost/Revenue
P. A 1. Total cost
R. B 2. Indirect operating cost
S. C 3. Revenue
C

Production level

(A) P-1, Q-3, R-2 (B) P-2, Q-1, R-3

(C) P-1, Q-2, R-3 (D) P-3, Q-1, R-2
Key: (D)
Sol: Total cost = fixed cost + variable cost
CT = Cf + PCV

Total revenue = PSP

Where, Cf = fixed cos t
P = Production in tonnes
SP = Selling Price
CV = Variable cost
Note: Indirect costs do not vary substantially within certain production. So are considered to be fixed
cost
A

B A : Revenue
Cost/Revenue

B: Total cost
C : Indirect operating cost
C

Production level

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 |MN-2021|

35. (
The value of lim x x 2 + b2 − x 4 + b4 is
x →
)
b2
(A) 0 (B) (C)  (D) b2
2
Key: (B)

Sol: (
lim x x 2 + b2 − x 4 + b4 →  −  form
x →
)
= lim
(x x 2 + b2 − x 4 + b4 )( x x 2 + b2 + x 4 + b4 )
x →
(x x 2 + b2 + x 4 + b4 )
 x 2 ( x 2 + b2 ) − ( x 4 + b4 ) 
= lim  
x →  4 
 x x 2
+ b 2
+ x 4
+ b 
 b4 
x 2  b2 − 2 
 b2 x  − b4   x 
= lim   = xlim
x →
x x +b + x +b
2 2 4 4

→  b2 b4 
x2  1 + 2 + 1 + 4 
 x x 
b2 − 0 b2
= = , option (B)
1+ 0 + 1+ 0 2

36. In order to check whether iron ore is supplied to the specification of 62% Fe, a steel company has
conducted a hypothesis test with the null hypothesis as H 0 : Fe = 62% and alternative hypothesis Ha:
Fe  62%. A random sample of 5 observation reveal the following grade values of the lot, 58%, 56%,
60%, 64%, 62%. The t-test statistic for the hypothesis is
(A) –3.000 (B) 1.414 (C) –1.414 (D) 3.000
Key: (C)
58 + 56 + 60 + 64 300
Sol: Average grade values, x = = = 60
5 5
Null hypothesis, Ho :  = 62
Alternate hypothesis Ha :   62
n

( x − x)
2
i
= i =1

n −1

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 |MN-2021|

 ( x i − x ) = ( 58 − 60 ) + ( 56 − 60 ) + ( 60 − 60 ) + ( 64 − 60 ) + ( 62 − 60 )
2 2 2 2 2 2

= 4 + 16 + 0 + 16 + 4
= 40

40
= = 10
4
x −  60 − 62 −2
t= = = = − 2 = −1.414
 n 10 5 2

37. Production planning of a small quarry having 3 years of life is shown in the figure. The following
information of revenue and cost data are available.

Ore :3.0 million tonne Ore :3.0 million tonne Ore :3.0 million tonne

3rd year 2nd year 1st year

Waste : 5.0 million m3 Waste : 5.5 million m3 Waste : 5.0 million m3

Selling price of ore = Rs. 1500/tonne

Ore mining cost = Rs. 500/tone
Waste mining cost = Rs. 500/m3
Initial capital = Rs. 1000 million
Discount rate = 10%
By neglecting depreciation, salvage value and corporate tax, the NPV of the mining project, in million
Rs., is _______ (round off to 2 decimal places)
Key: (36.81)
Sol: Initial capital = Rs 1000 million
For 1st year
Ore = 3 million tonne
Waste = 5 million m 3
1500
Selling price = Rs  3 million tonne = Rs 4500 million
tonne
500
Waste mining cost =  5 million m3 = 2500 million
m3

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 |MN-2021|

500
Ore mining cost = Rs  3 million tonne = 1500 million
tonne
Total cash flow for 1st year = Rs (4500–2500–1500) million
CF1 = 500 million
For 2nd year
Ore = 3 million tone
Waste = 5.5 million m 3
Selling price = Rs 1500 × 3 = Rs 4500 million
Waste mining cost = Rs 500 × 5.5 = Rs 2750 million
Ore mining cost = Rs 500 ×3 = Rs 1500 million
Total cash flow for 2nd year = Rs (4500-2750–11500) million
CF2 = Rs 250 million
For 3rd year
Ore = 3 million tone
Waste = 5 million m 3
Selling price = Rs 1500 × 3 = Rs 4500 million
Waste mining cost = Rs 500 × 5 = Rs 2500 million
Ore mining cost = Rs 500 × 3 = Rs 1500 million
Total cash flow for 3rd year = Rs (4500 – 2500–1500) = Rs 500 million
 CF CF2 CF3 
NPV = Initial capital −  1
+ + 3
 (1 + i ) (1 + i ) (1 + i ) 
1 2

 500 250 500 

= Rs 1000 −  + + 3
1 + 0.1 (1 + 0.1) (1 + 0.1) 
2

= Rs 1000 − 454.545 + 206.611 + 37.5657

= Rs1000 − 1036.81
= −36.81 million
Since, question is talking about revenue
So, NPV = 36.81 million

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 |MN-2021|

38. A triangular distributed load is applied on top of a beam as shown in the figure. The value of maximum
bending moment in kN-m is _________ (round off to 2 decimal places).

5 kN m

A B

2m

Key: (1.28)
Sol: D

5
x
2

A • B
C

RA RB

Let maximum bending moment occur at C which is x distance from point A

Taking moment along A,
M A = 0
2
 RB  2 − 5  2 = 0
3
20 10
 RB = = kN
3 2 3
Shear force at C = 0
 10  1 5x
  5 −  − + x. = 0
 3 2 2
2
5 5x
 − =0
3 4

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 |MN-2021|

2
 x= m
3
Moment about C
5 2 51 2 
MC =  −   
3 3 3 3 3
5 2  1
=  1 − 
3 3  3
5 2 2 20
=   = = 1.28 kN-m
3 3 3 9 3

39. 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 in mm is _______
(round off to one decimal place).
Key: (23.27)
3
Sol: Sensitivity,  = radian
nD
1 radian = 206265 second
S
=  206265 sec
nD
nod 2  60  40
S= = = 0.02327 m = 23.27 mm
206265 206265

40. 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
9.8cm on the plan. True lease hold mine area, in m2 , is ____________(round off to the nearest integer).
Key: (83506.87)
distance on map 9.8
Sol: Shrinkage factor SF = =
corresponding distance on ground 10
measured area 802
Corrected area = = = 835.0687 cm 2
(S.F) ( 0.98)
2 2

835.0687  106 2
True mine area = 835.0687  (1000 ) cm2 = m = 83506.87 m2
2

104

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 |MN-2021|

41. CO is released from a point source on a level ground at a rate of 25 g/s. The average wind speed is 5m/s.
The dispersion coefficients are 150 m and 200 m in horizontal and vertical directions, respectively, at a
receiver station located on the ground along the downwind direction. Assuming the plume follows
Gaussian dispersion model, the concentration of CO, in g m3 , at the station is _________(round off to
2 decimal places).
Key: (50-55)
Sol: According to Gaussian dispersion mode
 25
C= = = 26.52 g m3
2 y z 2  3.14  5  150  200

42. Assume that COVID-19 growth rate of number of infections per day (c) in a certain population is
represented by the following differential equation.
dc
100 − 7c = 0
dt
Where, t stands for time in days. Time taken for the number of infections per day to double, in days, is
_____________. (round off to the nearest integer).
Key: (10)
dc 7
Sol: DE  − dt is V.S.F
c 100
Integrating, we get
7
nc = t + n (k)
100

constant

 c  7t
 n  =  c = ke7t 100 …(1)
 k  100
Initially (i.e., t = 0) Let c = c0 , (1) gives (where c is number of infections)
c0 = k

When c = 2c0 ( i.e., double ) , t = ?, (1) gives

7t 100
2c0 = k 0e7t 100  = n2  t = n ( 2 ) = 9.902  10 days
100 7

43. Ore is hoisted from 620m depth using a single skip of 7 tonne pay load. The skip winding system has
constant acceleration/deceleration of 1 m s 2 and a constant speed of 10 m/s. The skip loading time and
unloading time are 120 s and 60 s, respectively. Considering the overall utilization of the skip as70%,
the maximum daily capacity of the winding system, in tonne, is __________(round off to the nearest
integer).

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 |MN-2021|

Key: (1250-1350)
Sol: Depth = 620 mm
Speed, s = 10 m/s
Loading time, LT = 120s
Unloading time UT = 60s
Time taken to hoist 620 m
620
t= = 62sec
10
Total time required to complete one cycle
= 62  2 + 120 + 60
= 304 sec
7 7
Daily capacity =  0.7 =  24  3600  0.7 = 1392.63
304sec 304

44. In an analysis of fragmented blast muck, the mean fragment size is found to be 60 cm with uniformly
index of 1.25. Considering Rosin-Ramler equation, the cumulative mass fraction, in percent, to pass the
grizzly screen size of 100 cm is _________(round off 2 decimal places).
Key: (71-75)
Sol: According to Rosin-Rammler equation
n
 x 
− 
R =e  xc 

R = proportion of material retained on screen

x = screen size
x c = characteristic size
N = index of uniformity
Here, x = 100
x c = 60 cm
n = 1.25
1.25
 100 
− 
R =e  60 
= 0.15
So, % cumulative mass fraction to pass
= (1 − R )  100%
= (1 − 0.15)  100%
= 85%

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 |MN-2021|

45. A single-acting reciprocating ram pump, while running at 120 rpm, delivers water at a rate of 10 litres
per second. Considering the ram diameter is 150 mm and stroke length is 300 mm, the volumetric
efficiency of the pump, in percent is __________. (round off to one decimal place)
Key: (94)
Sol:  = 120 rpm
d = 10 litres sec
dia = 150 mm
L = 300 mm
10
Q= = 0.01 m3 sec
1000
Ad 2 rpm
Capacity = L
4 6
3.14  ( 0.15 )
2
120
=  0.3 
4 60
3.14  0.152
=  0.3
2
= 0.010597
0.01
= = 0.94 = 94%
0.010597

46. In a sand stowing arrangement, the slurry has a sand concentration of 35% by volume. The specific
gravity of sand grain is 2.6. The concentration of sand by weight, in percent, in the slurry is ________
(round off to one decimal place)
Key: (58.83)
Wsand 2.6  0.35
Sol: % sand by weight = = 100 = 58.33%
Wsand + Ww 2.6  0.35 + 0.65

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 |MN-2021|

47. In a surface mine, third bench from the pit bottom is blasted, as shown in the figure. The width, height
and slope angle of each bench are 8m, 6m, and 80°, respectively. A fly rock is projected at an angle of
45° with the horizontal with initial velocity, v. If the acceleration due to gravity is 10 m/s2 then the
minimum velocity (v) in m/s required for the fly rock to reach just beyond toe of the pit slope is
_________. (round off 2 decimal places)
Fly rock V

45

8m 6m

Toe
80
Key: (10)
Sol:
V
45
6
tan 80 =
8m x
6
x=
tan 80
h = 18m

80
Toe
6m
80
x
D

6
D = 8 + 8 + 3x = 16  3  = 19.174m
tan80
Trajectory equation

 g  2
y = ( tan  ) x −  x
 2 ( V cos  ) 
2

Here, y = 18 m
x = 19.174 m

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 |MN-2021|

 
 
−18 = ( tan 45 )19.174 −  10 
(19.174 )
2

  v  2

 2  
  2  
 10 
− (18 + 19.174 ) = −  2  (19.174 )
2

v 
37.174v 2 = 10  (19.174 )
2

10  (19.174 )
2

V =
2
= 98.897
37.174
V = 9.945 m s

48. Injury experience is studied in an underground coal time with a random sample of 132 workers. The
results of the study are tabulated below.
Injured Non-injured Total workers
Roof-bolter operators 13 12 25
Loader operators 35 75 107

Total workers 48 84 132

The odds ratio of experiencing an injury by the roof-bolter operators when compared to the loader
operator is ________ (round off to 2 decimal places).
Key: (2.22)
13  72
Sol: Odal ratio = = 2.22
35  12

49. A random variable X is defined by

 1
−2 probability 3

 1
X =  3 probability
 2
 1
 1 probability
 6

( )
The value of E X2 is ___________(round off to one decimal place)

Key: (6)
Sol: X = −2,1,3 ( discrete RV )

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 |MN-2021|

1 1 1
 P ( X = −2 ) = ; P ( X = 1) = , P ( X = 3) =
3 6 2
1 8 + 1 + 27
 E ( X 2 ) =  x 2 P ( x ) = ( −2 )  + 12  + 32  =
1 1
=6
2

x 3 6 2 6

50. A project network consists of the following activities

Activity Immediate predecessors Duration (days)
A – 3
B – 4
C A, B 5
D B 6
E D 7
F C,E 8
G D 9
H F, G X
If the project completion time is 30 days, then the value of ‘X’, in days, is __________. (in integer)
Key: (5)

51. Rate of fuel consumption f c (litres per hour) of a truck varies with truck speed x, (kmph) as given below

x2
f c = 20 +
50
The fuel price is Rs. 70 per litre. Other costs amount of Rs. 500 per hour. If the truck travels 100 km
from a coal mine to a thermal plant, the speed of the truck, in kmph, that minimizes the total cost is
__________ (round off to one decimal place)
Key: (36.84)
x2
Sol: f c = 20 +
50
100  x2  100 190000
Total cost =  20 +   70 +  500 = + 140x
x  80  20 x

TC −190000
= + 140
x x2
Minimize Tc = 0
−190000
= −140
x2

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 |MN-2021|

 x ' = 1357.1428  x  36.84 km hr

−19000 190000  2 380000
TC'' = 3
 ( −2 ) = =
x x3 x3
TC'' ( at x = +36.84 ) = 7.6
TC'' ( at x = −36.84 ) = −7.6

For minimum T.C''  0

 x = 36.84 km hr
Velocity of vehicle = 36.84 km/hr

52. A cement company has three factories which transport cement to four distribution centres. The daily
production of each factory, the demand at each distribution centre, and the associated transportation cost
per tonne from factory to distribution centre are given in the Table.

Distribution centre
D1 D2 D3 D4 Supply
(tonnes/day)
F1 20 30 110 70 600
Factory
F2 10 0 60 10 100

F3 50 80 150 90 1000

Demand 700 500 300 200

(tonnes/day)

The initial basis feasible solution using the least-cost rule is ___________(in integer).
Key: (112000)
Sol:
D1 D2 D3 D4 Supply
F1 20 30 110 70 600
F2 10 0 60 10 100
F3 50 80 150 90 1000
Demand 700 500 300 200
From least-cost metro
Solution start from least value

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 |MN-2021|

D1 D2 D3 D4 Supply ( +/day )
F1 20 30 110 70 600
F2 10 0 60 10 100
F3 50 80 150 90 1000
Demand
700 500 300 200
( +/day )
Cost = 20  600 + 0  100 + 50  100 + 80  400 + 150  300 + 90  200 = 112000

53. The grade-tonnage distribution for the ultimate pit of a mine is given below.
Cu grade (%) Cumulative million tonnes below the grade
0.1 0
0.4 15.0
0.5 17.0
0.6 18.0
0.7 19.0
0.9 and above 23.0
The mill cut-off grade is 0.5% Cu. The annual mining capacity (ore + waste) is 4.5 million tonne and
milling capacity is 1.0 million tone. Excavation is planned in such a ways that either of the mine or the
mill runs at full capacity throughout. The planned life of the mine, in years, is __________(round off to
one decimal place)
Key: (6)
Sol: Total cumulative million tonnes (ore + waste) = 23
23
Life of mine run at capacity = = 5.11 years
4.5
As, millo cut of grade = 0.5% cu
Above this grade, excavation is useful
Total excavation for milling = 4+ 1 + 1 = 6 tonnes
6
Lift of mill-run at full capacity = = 6 years
1
Here, planned life of the mine = 6 years.

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 |MN-2021|

54. A coal mine operating in three shifts produces 400 tonnes of coal per day with a face OMS of 1.0 from
panel A, and 200 tonnes of coal with face OMS of 1.0 from panel B. The panel A and panel B are in
parallel with resistance 0.6 Ns−2 m−8 and 0.5 Ns−2 m−8 , respectively. If the panels are supplied with
minimum permissible quantity as per CMR 2018, the requisite regulator resistance to meet the
conditions in Ns−2 m−8 is _________(round off to 2 decimal place)
Key: (1.9)
Sol:
0.6

B
0.5

As per CMR 2018,

Maximum quantity of air per form of daily output is 2.5 m3 min
400  2.5 3
So, for panel A, q A = m s = 16.67 m3 s
60
200  2.5 3
For panel B q B = m s = 8.33 m3 s
60
So, regulator will placed on panel B, having resistance ‘ R r ’
Since, Panel A and Panel B are in parallel, so pressure will be same
P = RQ 2
R A q A2 = ( R B + R r ) q B2
0.6  (16.67 ) = ( 0.5 + R r )(8.33)
2 2

166.733 = ( 0.5 + R r )( 69.3889 )

 2.4028 = 0.5 + R r
 R r = 2.4028 − 0.5 = 1.9

Hence, resistance of regulator = 1.9 Ns−2 m−8

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 |MN-2021|

55. A set of three steel bar of equal cross-sectional area of 0.01 m2 are loaded, as shown in the figure. The
elastic modulus of steel is 200 GPa. The overall change of length of the complete set of bars, in mm, is
____________ (round off to 3 decimal places).

400 kN 500 kN 300 kN 200 kN

250 mm 500 mm 250 mm

Key: (0.05)
Sol:

400 kN 200
400 100 100 200
250 mm 500 mm 250 mm

Area, A = 0.01 m 2
E = 200  109 N m 2
For steel bus A (compression)
P
L P PL
E= A  =  AL =
L L AE AE
L
Compression
−400  250  10−3 400  250  102
L A = m = − m = −0.05 mm
0.01  200  109 200  103
For steel base B (expansion)
100  500  103
L B = mm = 0.025 mm
0.01 200  103
For steel base (compression)
200  103  250
LC = = −0.025 mm
0.01  200  103
Sign convection: Compression (-ve) and expression (+ve)
Overall change in length = = ( −0.05 + 0.025 − 0.025) mm = −0.05 mm

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