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1. A box contains 15 blue balls and 45 black balls. If 2 balls are selected randomly, without replacement,
the probability of an outcome in which the first selected is a blue ball and the second selected is a black
ball, is_________
(A) 45
236 (B)
3
4 (C)
1
4 (D)
3
16
Key: (A)
Sol:
15 45 45
Required probability .60 59 236
2. A digital watch X beeps every 30 seconds while watch Y beeps every 32 seconds. They beeped together
at 10 AM. The immediate next time that they will beep together is _______
(A) 10.08 PM (B) 11.00 AM (C) 10.00 PM (D) 10.42 AM
Key: (A)
Sol: Given,
Watch x beeps every 30 sec
Watch y beeps every 32 sec
L.C.M of 30, 32 = 480 seconds = 8 minutes
They beeps together every 8 min
i.e., 10 AM, 10.08 AM; 10.16 AM, 10.24 AM, ……
Hence option (A).
3. Consider the following sentences:
(i) The number of candidates who appear for the GATE examination is staggering.
(ii) A number of candidates from my class are appearing for the GATE examination.
(iii) The number of candidates who appear for the GATE examination are staggering.
(iv) A number of candidates from my class is appearing for the GATE examination.
Which of the above sentences are grammaticallyCORRECT?
(A) (i) and (ii) (B) (ii) and (iv) (C) (i) and (iii) (D) (ii) and (iii)
Key: (A)
GENERAL APTITUDE
15
Blue
45
Black
Box
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4. The world is going through the worst pandemic in the past hundred years. The air travel industry is
facing a crisis, as the resulting quarantine requirement for travelers led to weak demand.
In relation to the first sentence above, what does the second sentence do?
(A) Restates an idea from the first sentence
(B) Second sentence entirely contradicts the first sentence
(C) The two statements are unrelated
(D) States an effect of the first sentence
Key: (D)
5. Given below are two statements 1 and 2, and two conclusions I and II.
Statement 1: All entrepreneurs are wealthy.
Statement 2: All wealthy are risk seekers.
Conclusion I: All risk seekers are wealthy.
Conclusion II: Only some entrepreneurs are risk seekers.
Based on the above statements and conclusions, which one of the following options is CORRECT?
(A) Both conclusion I and II are correct (B) Only conclusion II is correct
(C) Only conclusion I is correct (D) Neither conclusion I nor II is correct
Key: (D)
Sol: EEntrepreneurs
WWealthy
R Risk sectors
All risk sectors are not wealthy Conclusion I is incorrect.
All entrepreneurs are risk seekersConclusion II is incorrect
Hence, neither conclusion I nor II is correct.
6. Five persons P, Q, R, S and T are to be seated in a row, all facing the same direction, but not necessarily
in the same order. P and T cannot be seated at either end of the row. P should not be seated adjacent to
S.R is to be seated at the second position from the left end of the row.
The number of distinct seating arrangements possible is:
(A) 4 (B) 3 (C) 2 (D) 5
Key: (B)
Sol: Positive arrangements are
S R P T Q (OR) Q R P T Q (OR) S R T P Q
Total number of combinations possible is 3.
F
W
E
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7. If 2; 3; 5; 10,
Then, the value of , is :
(A) 0 (B) 16 (C) 1 (D) 4
Key: (C)
8. The front door of Mr. X’s house faces East. Mr. X leaves the house, walking 50 m straight from the back
door that is situated directly opposite to the front door. He then turns to his right, walks for another 50m
and stops. The direction of the point Mr. X is located at with respect to the starting point is ______
(A) North-East (B) South-East (C) West (D) North-West
Key: (D)
Sol:
From the figure, it is clear that, the direction of X is North West
9.
The ratio of the area of the inscribed circle to the area of the circumscribed circle of an equilateral
triangle is _________.
(A) 1
8 (B)
1
2 (C)
1
6 (D)
1
4
Key: (D)
Sol: From the figure,
r 1 r
sin30 R 2rR 2 R
Area of inscribed circle 2r
Area of circumscribed circle 2 2R 4r
Required ratio 2 2 1
r : 4 r 1: 44
N
EW
S
X
Starting point50m
50m
R
r
r radius of inscribed circle
R radius of circumscribed circle
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10. Consider a square sheet of side 1 unit. T sheet is first folded along the main diagonal. This is followed
by a fold along its line symmetry. The resulting folded shape is again folded along its line of symmetry.
The area each face of the final folded shape, in square units, equal to ________.
(A) 1
16 (B)
1
32 (C)
1
8 (D)
1
4
Key: (C)
Sol: Area of each face of the final folded shape.
(Right angle triangle) 1 1 1 1 1
Box weight square units2 2 2 2 8
2
1
1
1
1
1
1
1
2
1
2
1
2
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1. The size distribution of the powder particles used in Powder Metallurgy process can be determined by
(A) Laser absorption (B) Laser scattering
(C) Laser reflection (D) Laser penetration
Key: (B)
Sol: Laser diffraction (or) laser scattering used in size distribution of powder particles used in powder
metallurgy process. So answer option ‘B’.
2. An adiabatic vortex tube, shown in the figure given below is supplied with 5 kg/s of air (inlet 1) at
500kPa and 300K. Two separate streams of air are leaving the device from outlets 2 and 3. Hot air
leaves the device at a rate of 3 kg/s from outlet 2 at 100 kPa and 340 K, while 2 kg/s of cold air stream is
leaving the device from outlet 3 at 100 kPa and 240 K.
Consider constant specific heat of air is 1005 J/kg.K and gas constant is 287 J/kg.K. There is no work
transfer across the boundary of this device. The rate of entropy generation is _______ kW/K (round off
to one decimal places).
Key: (2.24)
Sol:
1 1 1
2 2 2
3 3 3
m 5 kg sec, P 500 kPa, T 300K
m 3kg sec,P 100 kPa, T 340K
m 2kg sec, P 100 kPa, T 240K
MECHANICAL ENGINEERING
1
2 3
Highpressureairinlet
LowpressurecoldairoutletLowpressurehotairoutlet
1
2 3
LowpressurecoldairoutletLowpressurehotairoutlet
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pC 1005J kgK ; gas constant R 287J / kgK
p v v pR c c c c R 1005 287 718 J kgK
Since tube is maintained in adiabatic condition
No entropy transfer from the vortex tube
Entropy generation = Entropy change – Entropy transfer
Entropy generation = Entropy change entropy transfer 0
Entropy change of hot air
2 22 p
1 1
T Pm c n R n
T P
340 1003 1005 n 287 n
300 500
1763.1 W K
Entropy change of cold air
3 33 p
1 1
T Pm c n R n
T P
240 1002 1005 n 287 n
300 500
475.30W K
Entropy generation gens Total entropy change 1763.1 475.30 2238.4 W K 2.24 kW K
3. Water flows out from a large tank of cross-sectional area 2
tA 1m through a small rounded orifice of
cross sectional area 2
oA 1cm , located at y = 0. Initially the water level(H), measured from y = 0, is
1 m. Theacceleration due to gravity is 9.8 2m s .
y 0
y H
WATER
OrificeCross-sectional
area
tCorss-sectionalarea, A
0A
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Neglecting any losses, the time taken by waterin the tank to reach a level of y = H/4is____________
seconds (round off to one decimalplace).
Key: (2258)
Sol: Cross sectional area of tank 2
tA 1m
Cross sectional of area orifice 2 4 2
0A 1cm 1 10 m
Initially Height of water, (H) = 1m
2g 9.8 m sec
Time taken to reach a level of water H
y4
will be find out as follows,
in out cu
in
m m m
m 0
0 t
0 t
0 t
dA 2gH A dH
dt
dA 2gH A dH
dt
A 2gHdt d A dH
Integration on both sides
0 t
t
0
0.25 11
t t
4 0.251 0.250 0
A 2gH t 0 A dH
A dHt
A 2g H
1A AdH dHt 2 H
A 2g H HA 29 1 10 2 9.8
t 2258 seconds
4. A plane frame PQR (fixed at P and free at R) is shown in the figure. Both members (PQ and QR) have
length L, and flexural rigidity, EI. Neglecting the effect of axial stress and transverse shear, the
horizontal deflection at free end, R, is
P
L
L
R
Q
F
y 0
y H
WATER
tCorss-sectionalarea, A
0A
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(A) 32FL
3EI (B)
35FL
3EI (C)
34FL
3EI (D)
3FL
3EI
Key: (C)
Sol: Flexural rigidity of members PQ and QR having length ‘L’ are ‘EI’
Bending moment at any distance ‘x’ from ‘R’ is x QRM Fx
Bending moment in ‘PQ’ portion is constant and is equal to PQM FL
Strain energy stored in frame
22L L
x PQQR
0 0
M dx M dxf
2EI 2EI
2 2L L
0 0
L2 3 2 2
L
0
0
2 3 2 2
2 3 2 3
2 3 2 3 2 3
Fx dx FL dxU
2EI 2EI
F x F Ldx
2EI 3 2EI
F L F LL
2EI 3 2EI
F L F L
6EI 2EI
F L 3F L 2 F L
6EI 3 EI
Horizontal deflection at ‘R’ is 3 3
H R
U 2 2FL 4FL
F 3 EI 3EI
So answer is option ‘C’
PL
L
R
Q
F
x
x
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5. The wheels and axle system lying on a rough surface is shown in the figure.
Each wheel has diameter 0.8 m and mass 1 kg. Assume that the mass of the wheel is concentrated at rim
and neglect the mass of the spokes. The diameter of axle is 0.2 m and its mass is 1.5 kg. Neglect the
moment of inertia of the axle and assume g = 9.8 2m s . An effort of 10 N is applied on the axle in the
horizontal direction shown at mid span of the axle. Assume that the wheels move on a horizontal surface
without slip. The acceleration of the wheel axle system in horizontal direction is ________ 2m s (round
off to one decimal place).
Key: (1.25)—(4.9-5.1)
Sol:
Axle
0.8m
Wheel
gg
10N
0.2m
Axle
0.8m
Wheel
g
10N0.2m
Force applied F 10N
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Diameter of wheel (D) = 0.8m, Radius of wheel D
R 0.4m2
Mess of wheel wm 1 kg
Diameter of axle 0.2m,
Radius of axle d 0.2
r 0.1m2 2
Mass of axle am 1.5kg
2g 9.8 m sec
Mass moment inertia of wheels and axle is 2
wI 2m R (Neglecting mass moment of inertia of axle)
2 2I 2 1 0.4 0.32kg.m
fH ma 10 2F ma
where Ff = frictional force between wheel and Road
m = mass of two wheels and axle = 3.5kg
a = linear acceleration of wheel
f
f
f
10 2F 3.5a ... 1
T I
T 2F R I
T 10 0.2 1N.m
1 0.8F 0.32 ... 2
Solving equation (1) and (2)
2
f
2
12.5rad / sec , F 3.75N
a R 0.4 12.5 5m/sec
0.2m 10N
fF
R 0.4m
T 10 0.21N.m
0.2m 10N
fF
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6. The allowance provided in between a hole and a shaft is calculated from the difference between
(A) lower limit of the shaft and the upper limit of the hole
(B) lower limit of the shaft and the lower limit of the hole
(C) upper limit of the shaft and the upper limit of the hole
(D) upper limit of the shaft and the lower limit of the hole
Key: (D)
Sol: Allowance = Lower limit of hole – Upper limit of shaft
So Answer is option ‘D’
7. The value of 0 cos
0 0rsin dr d is
(A) 0 (B) 4
3 (C)
1
6 (D)
Key: (C)
Sol: Given double integral
2 cos
0 r 0
rsin dr d
cos2 222
0 00
r 1sin d sin cos d
2 2
1
2
0
13
0
1t dt Put cosθ= t sin d dt and 0 t 1, t 0
2 2
1 t 1
2 3 6
8. A two dimensional flow has velocities in x and y directions given by u = 2xyt and 2v y t, where t
denotes time. The equation for streamline passing through x = 1, y = 1 is
(A) 2x y 1 (B) 2xy 1 (C) 2 2x y 1 (D) 2x y 1
Key: (B)
Sol: 2u 2xyt, v y t
Equation for steam line is
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2
2
dx dy
u v
dx dy
2xyt y t
dx dy dx dy
2xy y 2x y
Integrating on both sides
1log x log y c
2
at x 1, y 1
1log1 log 1 c
2
at x 1, y 1
1log1 log 1 c
2
c 0
equation of stream line is
x x
2
1log x log y
2
log log y log log y 0
xy 1
xy 1
So answer is option ‘B’.
9. The torque provided by an engine is given by T 12000 2500sin 2 N.m, where is the angle
turned by the crank from inner dead center. The mean speed of the engine is 200 rpm and it drives a
machine that provides a constant resisting torque. If variation of the speed from the mean speed is not to
exceed 0.5%, the minimum mass moment of inertiaof the flywheel should be_________kg.m2 (round
off to the nearest integer).
Key: (570.15)
Sol:
mean
T 12000 2500sin 2
N 200 rpm
Coefficient of fluctuation of speed S
0.5C 2
100
SC 0.01
meanmean
2 N 2 200W 20.94 rad sec
60 60
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E maximum fluctuation in energy
Since T is a function of ‘ ’
Hence mean
0 0
1 1T Td 12000 2500sin 2 d
0
1 250012000 0 cos
2
112000 1250 cos360 cos
12000 N.m
At any instant meanT T T
T 12000 2500sin 2 12000
T 2500sin 2
‘ T ’ is zero when 2500sin 2 0
0 , 902
9090 90
00 0
2500cos2E T d 2500sin 2 d 2500 1 1 2500 N.m
2 2
2
mean SE I C
2
2
2500 I 20.94 0.01
I 570.15 kg.m
10. The machining process that involves ablation is
(A) Electrochemical Machining (B) Laser Beam Machining
(C) Abrasive Jet Machining (D) Chemical Machining
Key: (B)
Sol: Ablation means removal of material by melting or evaporation.
From the given options, the only laser beam machining method will remove the material by melting and
vaporization.
So the answer is option ‘B’.
11. The von Mises stress at a point in a body subjected to forces is proportional to the square root of the
(A) total strain energy per unit volume (B) plastic strain energy per unit volume
(C) distortional strain energy per unit volume (D) dilatational strain energy per unit volume
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Key: (C)
Sol: According to distortion energy theorem,
For 2-D dimensional case,
2 2 2
1 2 1 2 yt
2 2
1 2 1 2 yt
s
s
Vonmises stress 2 2
von 1 2 1 2
Where, 2 2
1 2 1 2 represent distortion strain energy per unit volume
Hence vonmises stress is proportional to square root of distortion strain energy per unit volume
So answer is option ‘C’.
12. The figure shows the relationship between fatigue strength (S) and fatigue life (N) of a material.The
fatigue strength of the material for a life of 1000 cycles is 450 MPa, while its fatigue strength for a life
of 106 cycles is 150 MPa.
The life of a cylindrical shaft made of this material subjected to an alternating stress of 200 MPa will
then be ___________ cycles (round off to the nearest integer).
Key: (164058.98)
Sol:
450 150 200 150
10 10 10 10
x
10
log log log log
6 3 6 log
x
10
x
10
5.215
6 log 0.785
log 6 0.785 5.215
x 10 164058.98 cycles
10log S
10log N7
3 x
10log 6 N
10log7
150
10log
s
10log
450
10log
200
10log
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13. A block of negligible mass rests on a surface that is inclined at 30° to the horizontal plane as shown in
the figure. When a vertical force of 900 N and a horizontal force of 750 N are applied, the block is just
about to side.
The coefficient of static friction between the block and surface is ______ (round off to two decimal
places).
Key: (0.17)
Sol: Free body diagram of block is
H 0
Ncos60 750 Ncos30 0
N N 3750 0
2 2
N1 3 750 ...(1)
2
V 0
Nsin 60 900 N sin30 0
N 3 N900 0
2 2
N3 900
2 …(2)
Equation (1)/(2) =
1 3 750
9003
30
900N
750N
30
900N
750N
60N fF
N
30
900N
750N
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51 3 3
6
6 6 3 5 3 5
6 5 3 5 6 3
6 5 30.17
5 6 3
' ’ is negative means our assumed direction of frictional force is not correct.
So, answer is 0.17
14. A shell and tube heat exchanger is used as a steam condenser. Coolant water enters the tube at 300K at a
rate of 100 kg/s. The overall heat transfer coefficient is 1500 W/m2.K, and total heat transfer area is
400m2. Steam condenses at a saturation temperature of 350K. Assume that the specific heat of coolant
water is 4000 J/Kg.K. The temperature of the coolant water coming out of the condenser is
________________ K. (round off to the nearest integer).
Key: (339)
Sol: Mass flow rate of coolant cm 100 kg sec
Inlet temperature of coolant water overall heat transfer ciT 300K
Overall heat transfer Coefficient 2U 1500 W m K
Total heat transfer area 2A 400 m
Saturation temperature of stream s hi hoT 350K T T
Specific heat of coolant p cc 4000 J kg.K
cmin c pC m c because specific heat of steam is
100 4000 400kW K
3min
1500 400UANTU 1.5
C 400 10
For condensing or boiling effectiveness of heat exchanger is
NTU 1.51 e 1 e 0.78
c pc co ci
min hi ci
m C T T
C T T
co
co
T 3000.78 T 339K
350 300
s hi hoT 350K T T
CiT 300K
CoT
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15. Daily production capacity of a bearing manufacturing company is 30000 bearings. The daily demand of
the bearing is 15000. The holding cost per year of keeping a bearing in the inventory is Rs. 20. The
setup costfor the production of a batch is Rs.1800. Assuming 300 working days in a year, the economic
batch quantity in number of bearing is ______. (in integer).
Key: (40250)
Sol: Daily Demand = 15,000 bearings
Annual Demand (R) = 15,000 × 300
3C ordering cost (or) set up cost = 1800 Rs
1C carrying cost = 20 Rs|Bearing|year
K = Production rate = 30, 000, Bearings per day
Economic batch quantity 3
1
2C REBQ
RC 1
K
3
1
2C R k
C k R
2 1800 15,000 300 30,000
20 30,000 15,000
40249.2 units
So answer is 40, 250 units.
16. Let the superscript T represent the transpose operation. Consider the function T T1f x x Qx r x,
2
where x and r are n 1 vectors and Q is a symmetric n n matrix. The stationary point of f(x) is
(A) TQ r (B) r (C) T
r
r r (D) 1Q r
Key: (D)
Sol: Take n= 2 and Let a c
Qc b
be a symmetric matrix, 1 1
2 2
x rx ; r
x r
Then 1 1 2 2
2 1 2 1 2 1 2 1 1 2 2
2 2
x xa c1 1f x x, x r r ax bx 2cx x r x r x
x xc b2 2
Let 2 2
1 21 2 1 2 1 1 2 2
ax bxf x x ,x cx x r x r x
2 2
For stationary points, we have 1 2
0 and 0x x
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1 2 1ax cx r 0 …(1)
And 2 1 2bx cx r 0 …(2)
Equation (1) and (2) in matrix form is
1 1
2 2
x ra c
x rc b
Q.x r
Multiply by 1Q , we get
1x Q r, option (D) is the stationary point.
17. A PERT network has 9 activities on its critical path. The standard deviation of each activity on the
critical path is 3. The standard deviation of the critical path is
(A) 3 (B) 81 (C) 9 (D) 27
Key: (C)
Sol: Number of activities on the critical path = 9
Standard deviation of each activity in the critical path 3
Variance of each activity in the critical path 2 9
Sum of variances of all activities in the critical path 29 9 9 81
Standard deviation of critical path 81 9
So answer option ‘C’.
18. An object is moving with a Mach number of 0.6 in an ideal gas environment, which is at a temperature
of 350K. The gas constant is 320 J/kg.K and ratio of specific heats is 1.3. The speed of object is
_______ m/s. (round off to the nearest integer).
Key: (229)
Sol: Mach number (M) = 0.6
Temperature of air (T) = 350K
Gas constant R 320 J kg.K
Ratio of specific heat p
v
C1.3
C
Speed of object (v) = ?
We know that Mach number
Vvelocity of objectM
sonic velocity C
VM
RT
V M RT 0.6 1.3 320 350 228.9 229 m sec
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19. For a two-dimensional, incompressible flow having velocity components u and v in the x and y
directions, respectively, the expression
2u uv
x y
Can be simplified to
(A) u v
2u ux y
(B)
u u2u v
x y
(C)
u vu u
x y
(D)
u uu v
x y
Key: (D)
Sol: For a two-dimensional in compressible flow, continuity equation is
u v u v v u
0x y x y y x
Given that
2u uv u u v2u v u
x y x y y
u u v2u v u
x y y
u u u u u u u2u v u 2u u v u v
x y x x y x y
So answer is option ‘D’
20. Consider an ideal vapour compression refrigeration cycle working on R-134a refrigerant. The COP of
the cycle is 10 and the refrigeration capacity is 150 kJ/kg. The heat rejected by the refrigerant in the
condenser is ______________ kJ/kg. (round off to the nearest integer).
Key: (165)
COP = 10
Refrigeration capacity = 150 kJ/kg
Work input to refrigeration (W) = ?
2QCOP
W
15010
W
150W 15 kJ kg
10
2Q
1Q
Ref
HT
LT
W
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1
1 2
1
1
Q Heat rejected by refreigerator
W Q Q
15 Q 150
Q 150 15 165 kJ kg
21. A 76.2mm gauge block is used under one end of a 254mm sine bar with roll diameter of 25.4mm. The
height of gauge blocks required at the other end of the sine bar to measure an angle of 30° is
_________mm (round off to two decimal places).
Key: (203.2)
Length of sine bar 254 mm
Diameter of roll d 25.4 mm
Angle measured 30
Height of gauge blocks
On one side 2h 76.2 mm
Height of gauge blocks on another side 1h ?
1 2
1 2 1
1
h r h rsin 30
h h h 76.21
2 254 254
h 203.2 mm
22. Value of5.2
4n x dx using Simpson's one-third rule with interval size 0.3 is
(A) 1.06 (B) 1.60 (C) 1.83 (D) 1.51
Key: (C)
Sol:
0
1
2
3
4
f x nx, h 0.3; a 4, b 5.2
y f 4 n4 1.386
y f a h n 4.3 1.458
y f a 2h n 4.6 1.526
y f a 3h n 4.9 1.589
y f a 4h n 5.2 1.648
By Simpson’s one-third rule,
254 mm
30
1h
2h
Roll
r
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5.2
0 4 1 3 2
4
hnxdx y y 4 y y 2 y
3
0.31.386 1.648 4 1.458 1.589 2 1.526
3
1.827 1.83
23. A cast produce of a particular material has dimensions 75mm × 125mm × 20mm. The total
solidification time for the cast product is found to be 2.0 minutes as calculated usigChvorinov's rule
having the index, n = 2. If under the identical casting conditions, the cast product shape is changed to a
cylinder having diameter = 50mm and height =50mm, the total solidification time will be ______
minutes. (round off to two decimal places).
Key: (2.85)
Sol: Solidification time according to chvarinovs rule is
n
s
Vt k
A
k = moulding constant
n = 2
For cube:
Volume of the cube 3V 75 125 20 18750 mm
Surface are of cube 2A 2 75 125 125 20 20 75 26750 mm
Solidification time st for cube is = 2 minutes
2
1875002 k
26750
k 0.041
For cylinder:
Volume of cylinder 2 2 3V d h 50 50 31250 mm4 4
Surface area of cylinder 22 2A d 2 d 50 50 50 3750 mm
4 2
Solidification of time st for cylindrical is
n 2
s
V 31250t k 0.041 2.85 mins
A 3750
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24. The cast Iron which possesses all the carbon in the combined for as cementite is know as
(A) White Cast Iron (B) Spheroidal Cast Iron
(C) Grey Cast Iron (D) Malleable Cast Iron
Key: (A)
Sol: White cast iron has carbon in the form of cementic hence the answer is option ‘A’
25. Aspot welding operation performed on two pieces of steel yielded a nugget with a diameter of 5mm and
a thickness of 1mm. The welding time was 0.1s. The melting energy for the steel is 20 J/mm3.
Assuming the heat conversion efficiency as 10%, the power required for performing the spot welding
operation is ___________kW. (round off to two decimal places).
Key: (39.27)
Sol: Diameter of Nuggest (d) = 5mm
Thickness of Nugget (t) = 1mm
Volume of Nugget 2 2 3V d t 5 1 6.25 mm4 5
Welding time t 0.1seconds
Melting energy of steel 320 J mm
Heat required for melting
Total heat required RQ 20 6.25 125 J
Heat conversion efficiency 0.10
Heat supplied SQ ?
R
S
S
S
Q
Q
125 1250.10 Q 3926.99 3927 joules
Q 0.1
Heat supplied per second SS
Q 3927Q 39270 watts 39.27 kW
t 0.1
26. Which of the following is responsible for eddy viscosity (or turbulent viscosity) in a turbulent boundary
layer on a flat plate?
(A) Nikuradse Stresses (B) Prandtl Stresses
(C) Reynolds Stresses (D) Boussinesq Stresses
Key: (C)
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Sol: Since, turbulence is considered as eddying motion and the additional stresses are added to the viscous
stresses due to mean motion in order to explain the complete stress field, it is often said that the apparent
stresses are caused by eddy viscosity. The total stresses given by,
2
xx
xy
uP 2 pu '
x
u vpu 'v '
y x
Terms such as u 'v ' and u ' are called Reynolds stresses or turbulent stresses.
Hence, the correct answer is option (C).
27. Consider the following differential equation
dy
1 y ydx
The solution of the equation that satisfies the condition y(1) = 1 is
(A) 2 y xy e e (B) y x1 y e 2e
(C) y x2ye e e (D) y xye e
Key: (D)
Sol: 1 y
DE dy dx, which is VSFy
Integrating, we get
1
1 dy dx Cy
ny y x C …(1)
Using y(1) = 1, (1) gives 0 + 1 = 1+ C C = 0
ny y x
x y y xy e or ye e option (D).
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28. A power transmission mechanism consist of a belt drive and a gear train as shown in the figure.
Diameters of pulleys of belt drive and number of teeth (T) on the gears 2 to 7 are indicated in the figure.
The speed and direction of rotation of gear 7, respectively are:
(A) 575.28 rpm; Clockwise (B) 255.68; anticlockwise
(C) 575.28 rpm; anticlockwise (D) 255.68 rpm; Clockwise
Key: (D)
Sol:
1 2
2
1
d 150 mm, d 250 mm
N 1502500 1500 rpm ACW
N 250
3 2 2
2 3 3
33
N d z
N d z
N 18N 613.64 rpm CW
1500 44
3 4N N , since they are mounted on same shaft
5 4 4
4 5 5
5
5
N d z
N d z
N 15
613.64 33
N 278.93 rpm ACW
18T
250mm
2 N 2500 rpm
150mm
3
5 6
36T
16T
33T44T
15T
47
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6 5 5
5 6 6
6
6
N d z
N d z
N 33
278.93 36
N 255.68 rpm CW
7 6N N , since they are mounted on same shaft hence answer is option ‘D’.
29. The controlling force curves P,Q and R for a spring controlled governor are shown in the figure, where
r1 and r2 are any two radii of rotation.
The characteristic shown by the curves are
(A) P – Stable; Q – Isochronous; R – Unstable
(B) P – Unstable; Q – Isochronous; R – stable
(C) P – Stable; Q – Unstable; R – Isochronous
(D) P – Unstable; Q – Stable; R – Isochronous
Key: (B)
Sol: The controlling force (F) (Vs) radius of rotation (r) in a spring controlled governor is as follows.
F = ar + b
If b < 0 i.e., F = ar – b It is a stable governor
‘R’ is stable governor
If b = 0 i.e., F = ar It is isochronous governor
‘Q’ is isochronous governor
If b > 0 i.e., F = ar + b It is a unstable governor
‘P’ is unstable governor
So answer is option ‘B’.
P
Q
R
2r 1r
Co
ntr
oll
ing
forc
e
Radius of rotation
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30. Find the positive real root of 3x x 3 0 using Newton-Raphson method. If the starting guess 0x is
2, the numerical value of the root after two iterations 2x is ___________ (round off to two decimal
places).
Key: (1.67)
Sol:
3
2
0
f x x x 3
f ' x 3x 1and x 2
By Newton-Raphson method,
0
1 0
0
1
2 1
1
f x 3x x 2 1.73
f ' x 11
f x 0.48x x 1.73 1.67
f ' x 7.98
31. The thickness, width and length of a metal slab are 50mm, 250mm and 3600mm, respectively. A rolling
operation on this slab reduces the thickness by 10% and increases the width by 3%. The length of the
rolled slab is __________mm (Round off to one decimal place).
Key: (3883.5)
Sol: 1
1
1
2
2
2
t 50 mm
W 250 mm
3600 mm
t 0.9 50 45 mm
W 1.03 250 257.5 mm
?
Volume of slab before rolling 1 2V Volume of slab after V Rolling
1 1 1 2 2 2
2
t w t w
50 25 3600 45 257.5
Length of slab after rolling 2 3883.5 mm
32. A surface grinding operations has been performed on a cast Iron plate having dimensions 300mm
(length) × 10mm (width) × 50mm (height). The grinding was performed using an alumina wheel having
a wheel diameter of 150mm and wheel width of 12 mm. The grinding velocity used is 40 m/s, table
speed is 5 m/min, depth of cut per pass is 50μm and the number of grinding passes is 20. The average
tangential and average normal forces for each pass are found to be 40N and 60N respectively. The value
of the specific grinding energy under the aforesaid grinding conditions is _________ J/mm3. (round off
to one decimal places).
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Key: (38.4)
Sol: Given, tF 40N, V 40m/s
tSo, Power F .V 40 40 1600W
Width, W 10mm,
depth of cut d 50 m 0.05mm
Table speed s 5m/min
So,
3
5000MRR W.d.s 10 0.05
60
41.6666mm /s
Hence, specific grinding energy
=
3Power 160038.4J / mm
MRR 41.6666
33. The mean and variance, respectively, of a binomial distribution for n independent trials with the
probability of success as p, are
(A) np, np 1 p (B) np, np
(C) np, np 1 2p (D) np, np 1 p
Key: (D)
Sol: Mean and variance of a binomial distribution are np and npqrespecivley, where q = 1–p i.e.., np and
np 1 p
Option (D).
34. A factory produces m i 1,2,.......m products, each of which requires processing on n j 1,2......,n
workstations. Let ai,j be the amount of processing time that one unit of the ith product requires on the j
th
workstation. Let the revenue from selling one unit of the ith product ri and hi be the holding cost per unit
per time period for the ith product. The planning horizon consists of T t 1,2.....,T time periods. The
minimum demand that must be satisfied in time period t is itd , and the capacity of the jth workstation in
time period t is jtc . Consider the aggregate planning formulation below, with decision variable itS
(amount of product i sold in time period t), itX (amount of product i manufactured in time period t) and
Iit (amount of product i held in inventory at the end of time period t).
T m
i it i it
t 1 i 1
it it
max rS h I subject to
S d i, t
h 50 mm
300 mm
W 10mm
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it , it , it i0
capacity constraint
inventory balance constraint
X S I 0; I 0
The capacity constraints and inventory balance constraints for this formulation respectively are
(A) m
ij it jt it i, t 1 it it
i
a X c i, t and I I X d i, t
(B) m
ij it jt it i, t 1 it it
i
a X c j, t and I I X S i, t
(C) m
ij it it it i, t 1 it it
i
a X d i, t and I I X S i, t
(D) m
ij it jt it i, t 1 it it
i
a X d i, t and I I S X i, t
Key: (B)
Sol: Capacity constraint is
ij it jta x c
Where, ija processing time that one unit of ith product requires on j
th work station
itx Amount of product ‘i’ manufacture in time period ‘t’.
jtC Capacity work station ‘j’ in time ‘t’
Inventory balance constraint is
it i,t 1 it itI I x s
Where, itI Amount of product ‘i’ held in inventory at the end of time period ‘t’
i,t 1I Amount of product ‘i’ held in inventory at the time period ‘t–1’
itX Amount product ‘i’ manufacture in time period ‘t’
i,t 1I Amount of product ‘i’ held Inventory at the time period ‘t-1’
itX Amount product ‘i’ is manufactured in time period ‘t’
itS Amount of product ‘i’ sold in time period ‘t’.
So answer is option ‘B’
35. A vertical shaft Francis turbine rotates at 300 rpm. The available head at the inlet to the turbine is
200 m. The tip speed of the rotor is 40 m/s. Water leaves the runner of the turbine without whirl.
Velocity at the exit of the draft tube is 3.5 m/s. The head losses in different components of the turbine
are: (i) stator and guide vanes: 5.0 m, (ii) rotor: 10 m, and (iii) draft tube: 2m. Flow rate through the
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turbine is 20 m3/s. Take g = 9.8 m/s
2. The hydraulic efficiency of the turbine is 20m
3/s. Take
2g 9.8m / s . The hydraulic efficiency of the turbine is __________% (round off to one decimal place).
Key: (91.2)
Sol: N 300 rpm
H 200 mm
u 40 m sec
Velocity of water at exit of draft tube 2V 3.5 m sec
Loss in stator and guide vanes sH 5m
Loss in Rotor RH 10m
Los in Draft tube DH 2m
Flow through turbine 3Q 20 m sec
2g 9.8 m sec
Net available head for turbine 2
0net S R D
VH H H H H
2g
2
Net
3.5H 200 5 10 2 182.375m
2 9.8
Hydraulic efficiency netH
H
H 182.375
H 200
91.187%
So answer is 91.2
36. A rigid tank of volume 50 m3 contains a pure substance as a saturated liquid vapour mixture at 400 kPa.
Of the total mass of the mixture, 20% mass is liquid and 80% mass is vapour. Properties at 400 kPa
are: Saturation temperature, satT 143.61 C; Specific volume of saturated liquid, 3
gv 0.001084 m /kg;
Specific volume of saturated vapour, 3
gv 0.46242 m / kg. The total mass of liquid vapour mixture in
the tank is _________kg (round off to the nearest integer).
Key: (135.16)
Sol: Volume of tank 3V 50m
Saturation temperature at 400 KPa, satT 143.61 C
Vapour
Liquid
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3
f
3
g
V 0.001084 m kg
V 0.46242 m kg
Total mass of liquid-vapour-mixture (m) = ?
L Vm m m
Specific volume of vapour 3
VV 0.46242 m kg
Mass of vapour v
g
V 50m 108.13kg
V 0.46242
Given that,
Mass of vapour vm + mass of liquid Lm Total mass (m) of liquid vapour mixture.
L108.13 m m
1081.13 0.2m m
108.13 0.8m
m 135.16 kg
37. Value of 8
1 i , where i 1, is equal to
(A) 16i (B) 4i (C) 16 (D) 4
Key: (C)
Sol:
2 2 2
4 2 2
8 2
1 i 1 i 2i 2i
1 i 2 i 4
1 i 4 16, option (C)
38. The demand and forecast of an item for five months are given in the table.
Month Demand Forecast
April 225 200
May 220 240
June 285 300
July 290 270
August 250 230
The mean absolute percent error (MAPE) in the forecast is _____________%. (round off to two decimal
places)
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Key: (8.07)
Sol:
Month Demand (D) Forecast (F)
Percentage absolute
error
D - F
D
April 225 200 25
11.11%225
May 220 240 20
9.09%220
June 285 300 15
5.26%285
July 290 270 20
6.89%290
August 250 230 20
8%250
Mean absolute percentage
11.11 9.09 5.26 6.89 8
error 8.07%5
39. A cantilever beam with a uniform flexural rigidity 6 2EI 200 10 N.m is loaded with a concentrated
force at its free end. The area of the bending moment diagram corresponding to the full length of the
beam is 10000 N.m2. The magnitude of the slope of the beam at its free end is _______ micro radian
(round off to nearest integer).
Key: (50)
Sol: 6 2
IE 200 10 N.m
Given that, Area of bending moment diagram between
A and B is = 10,000 2N.m
22 2
110,000 W
2
W10,000 W 20,000 N.m
2
Slope at ‘B’ = Area of M
EI
diagram between A and B
A B
W
W
M W
EI EI
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2
5
6
1 W 1 W
2 EI 2 EI
1 20,0005 10 radians 50 micro radians
2 200 10
40. Ambient pressure, temperature, and relative humidity at a location are 101 kPa, 300K, and 60%
respectively. The saturation pressure of water at 300K is 3.6 kPa. The specific humidity of ambient air
is __________ g/kg of dry air.
(A) 21.4 (B) 21.9 (C) 35.1 (D) 13.6
Key: (D)
Sol: Ambient pressure atmP 101kPa P
Temperature of air (T) = 300K
Relative humidity 0.6
Saturation pressure of water at 300K is
satP 3.6 kPa
Specific humidity of air w ?
V
V
sat
V
PP 0.6 3.6
P
P 2.16 kPa
V
V
0.622 2.160.622Pw
P P 101 2.16
0.01359 kg of water vapour kg of dry air
13.59 gms of water vapour kg of dry air
So answer is option ‘D’
41. In a CNC machine tool, the function of an interpolator is to generate
(A) error signal for tool radius compensation during machining
(B) signal for the lubrication pump during machining
(C) NC code from the part drawing during post processing
(D) reference signal prescribing the shape of the part to be machined
Key: (D)
Sol: In CNC machine tool, the function of an interpolator is to generate reference signal prescribing the shape
of the part to be machined.
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42. A column with one end fixed and one end free has a critical buckling load of 100 N. For the same
column, if the free end is replaced with a pinned end then the critical buckling load will be __________
N (round off to the nearest integer)
Key: (800)
Case (i)
2
cr 21e
2
2
EIP
L
EI100N
2
2
2
EI400N
Case (ii)
2 2
cr 222e
2
2
EI EIP
L
2
2 EI
2 400 800N
So answer is 800N
43. If the Laplace transform of a function f(t) is given by
s 3,
s 1 s 2
thenf(0) is
(A) 3
2 (B) 0 (C)
1
2 (D) 1
Key: (D)
Sol:
s 3F s L f t
s 1 s 2
s
f 0 limsF s
(Using initial value theorem)
s s
31
s s 3 1slim lim 1option (D)
1 2s s 1 s 2 1 11 1
s s
one end fixed and other end free
Case (i)
one end fixed and other end pinned
Case (ii)
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44. Consider the system shown in the figure. A rope goes over a pulley. A mass, m, is hanging from the
rope. A spring of stiffness, k, is attached at one end of the rope. Assume rope is inextensible, massless
and there is no slip between pulley and rope.
The pulley radius is r and its mass moment of inertia is J. Assume that the mass is vibrating
harmonically about its static equilibrium position. The natural frequency of the system is
(A) 2
2
kr
J mr (B) k m (C)
2
2
kr
J mr (D)
2kr
J
Key: (A)
Sol: J = Mass moment of inertia of pulley about ‘0’
Give a small angular displacement to pulley
sF spring force krsin
2
i 2
dF m rsin
dt
For small angles r sin r 2
2
dmr mr
dt
iT J
0M
i SJ F r cos F r cos 0
J mr r cos krsin r cos 0
For small angels cos 1, sin
k r
J
m
iF
r sin
r sinSF
r cos r r cos
k r
J
m
O
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2 2
2 2
2
n 2
J mr kr 0
J mr kr 0
kr
J mr
So answer is option ‘A’
45. A steel cubic block of side 200 mm is subjected to hydrostatic pressure of 250 N/mm2. The elastic
modulus is 2 × 105 N/mm
2 and Poisson’s ratio is 0.3 for steel. The side of the block is reduced by
________mm (round off to two decimal places)
Key: (0.11)
Sol:
5 2E 2 10 N mm , 0.3
E 3k 1 2
Bulk modulus
5
2
2 10Ek
3 1 2 3 1 2 0.3
k 166666.67 N mm
Bulk modulus
v
v
3 3
v
volumetric stress (or)hydrostatic stressk
volumetric strain e
250166666.67
e
250e 1.499 10 1.5 10
166666.67
v
33 3
v
change in volume Ve
original volume or initial volume
V e a 1.5 10 200 12,000
V Initial volume Final volume
i f
3
f
12,000 V V
V 200 12,000 7988000
Let a ' New size of cube
3
a ' 7988000
a ' 199.89
a a ' 200 199.89 0.11mm
2250 N mm
a 200 mm2250 N mm
a 200 mm
a 200 mm2250 N mm
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46. Consider the open feed water heater (FWH) shown in the figure given below:
Specific enthalpy of steam at location 2 is 2624 kJ/kg, specific enthalpy of water at location 5 is 226.7
kJ/kg and specific enthalpy of saturated water at location 6 is 708.6 kJ/kg. If the mass flow rate of water
entering the open feed water heater (at location 5) is 100 kg/s then the mass flow rate of steam at
location 2 will be _________kg/s (round off to one decimal place).
Key: (25.16)
Sol:
Given that, 2
5
6
5
2
h 2624 kJ kg
h 226.7 kJ kg
h 708.6 kJ kg
m 100 kg sec
m ?
2 2 5 5 6 6
2 6
m h m h m h
m 2624 100 226.7 m 708.6
Let ‘ 5m ’ be the mass of flow rate steam taken at location ‘2’
Open FWH 6 6m h 5 5m h
2 2m h
1
23
Condenser
Open FWH
HP pump
LP pump
4
7
5
6
Ste
am
Gen
erat
or
Turbine
1
23
Condenser
Open FWH
HP pump
LP pump
4
7
5
6
Ste
am
Gen
erat
or
Turbine
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Let 1m be the total mass flow rate of steam entering at location ‘1’
1 2 5 1 6m m m and m m
6 2 5 2
2 2
2 2
2
2
m m m m 100
m 2624 100 226.7 m 100 708.6
2624m 22670 708.6m 70860
1915.4m 48190
m 25.16 kg sec
47. Consider the mechanism shown in the figure.
There is rolling contact without slip between
the disc and ground.
Select the correct statement about instantaneous centers in the mechanism.
(A) All points P, Q, R, S, T and U are instantaneous centers of mechanism
(B) Only points P, Q, S and T are instantaneous centers of mechanism
(C) Only points P, Q and S are instantaneous centers of mechanism
(D) Only points P, Q, R, S, and U are instantaneous centers of mechanism
Key: (A)
Sol:
Q
2
P1
3
R
U
S 4
T
Q
2
P1
3
R
U
S 4
T
12I
24I
13I
34I
14I
23I
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Number of instantaneous centresare
n n 1 4 4 1
2 2
Hence P, Q, R, S, T and U are instantaneous centres of mechanism
Hence answer is option ‘A’
48. Consider adiabatic flow of air through a duct. At a given point in the duct, velocity of air is 300 m/s,
temperature is 330 K and pressure is 180 kPa. Assume that the air behaves as a perfect gas with constant
pc 1.005 kJ kg.K. The stagnation temperature at this point is_____K (round off to two decimal places).
Key: (374.78)
Sol: Velocity of air V 300 m sec
Temperature of air T 330K
Pressure of air P 180 kPa
PC 1.005 kJ kg.K
Stagnation temperature 2
stag
p
VT T
2C
2
stag 3
300T 330 374.78 K
2 1.005 10
49. A machine of mass 100 kg is subjected to an external harmonic force with a frequency of 40 rad/s. The
designer decides to mount the machine on an isolator to reduce the force transmitted to the foundation. The
isolator can be considered as a combination of stiffness (K) and damper (damping factor, ) in parallel.
The designer has the following four isolators:
1. K = 640 kN/m, 0.70 2. K = 640 kN/m, 0.07
3. K = 22.5 kN/m, 0.70 4. K = 22.5 kN/m, 0.07
Arrange the isolators in the ascending order of the force transmitted to the foundation.
(A) 3-1-2-4 (B) 1-3-2-4 (C) 1-3-4-2 (D) 4-3-1-2
Key: (D)
Sol: Mass of machine (m) = 100 kg
Applied frequency 40 rad sec
Transmissibility
2
T
2 22 0
1 2 r FTR
F1 r 2 r
1 2
4 3
Page 39
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Where, T
0
F Force transmitted
F Applied force
1. K 640 kN m, 0.70
3
n
n
k 640 1080 rad sec
m 100
40r 0.5
80
2
T
2 220
T
0
1 2 0.70 0.5F
F1 0.5 2 0.7 0.5
F 1.221.189
F 1.026
2. K 640 kN m, 0.07
n
2
T
2 220
k80 rad sec, r 0.5
m
1 2 0.07 0.5F 11.32
F 0.7531 0.5 2 0.07 0.5
3. K 22.5 kN m, 0.70
3
n
n
2
T
2 220
k 225 1015 rad sec
m 100
40r 2.67
15
1 2 0.7 2.67F 3.870.54
F 7.181 2.67 2 0.7 2.67
4. K 22.5 kN m, 0.07
n
2
T
2 220
15 rad sec r 2.67
1 2 0.07 2.67F 1.0670.17
F 6.141 2.67 2 0.07 2.67
So ascending order is 4–3–1–2.
So answer is option ‘D’
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50. Ambient air flows over a heated slab having flat, top surface at y = 0. The local temperature (in Kelvin)
profile within the thermal boundary layer is given by T(y) = 300 + 200 exp(–5y), where y is the distance
measured from the slab surface in meters. If the thermal conductivity of air is 1.0 W/m.K and that of the
slab is 100 W/m.K, then the magnitude of temperature gradient dT / dy within the slab at y = 0 is
_____ K/m (round off the nearest integer).
Key: (10)
Sol: Thermal conductivity of air airk 1 W mK
Thermal conductivity of slab slabk 100 W mK
At y = 0
Conduction heat transfer into slab = Conductive heat transfer from air to slab
slab air
y 0 y 0slab Air
5yair
y 0slaby 0 y 0slab air
dT dTk k
dy dy
kdT dT 1 15 200 e 5 200 10 k m
dy k dy 100 100
Magnitude of temperature gradient within the slab at y = 0 is ‘10’
51. A plane truss PQRS (PQ = RS, and PQR = 90°) is shown in the figure.
P
Q
L
L
S
F
R
Air
T y 300 200 exp 5y
slab
y 0
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The forces in the members PR and RS, respectively, are ________.
(A) F tensile and F 2 tensile (B) F 2 tensile and F compressive
(C) F compressive and F 2 compressive (D) F 2 tensile and F tensile
Key: (B)
Sol: PQ RS
PQR 90
At junction ‘R’
PR RS
PR RS
RSPR RS
PRRS
F cos45 F 0
F cos45 F
FF 2F
cos45
FF
2
PR PR
PRRS RS
FF sin 45 F F 2F
sin 45
F 2FF F F F
2 2
The force in ‘RS’ is compressive and it is ‘F’, the force in ‘PR’ is tensile and it is 2 F
So answer is option ‘B’.
52. In forced convective heat transfer, Stanton number (St), Nusselt number (Nu), Reynolds number (Re)
and Prandtl number (Pr) are related as
(A) Nu Pr
StRe
(B) St Nu Pr Re (C) Nu
StRePr
(D) Nu Re
StPr
Key: (C)
Sol: Stanton Number (st) p p
h h T
c c T
p
h T
Tk
Nu NuL
c T Pr.ReLPr
Tk
L
So answer is ‘C’.
P
Q
L
L
S
F
R
45
45
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53. In a pure orthogonal turning by a zero rake angle single point carbide cutting tool, the shear force has
been computed to be 400 N. If the cutting velocity, Vc = 100 m/min, depth of cut, t = 2.0 mm, feed S0 =
0.1 mm/revolution and chip velocity, Vf = 20 m/min then the shear strength,s of the material will be
________ MPa (round off to two decimal places)
Key: (392.23)
Sol: Rake angle 0
Shear force SF 400N
Cutting velocity CV 100 min min
Depth of cut (t) = 2mm
Feed 0S 0.1mm rev
Chip flow velocity fV 20m min
Shear strength of material s ?
Chip thickness ratio (r) 1 f
2 c
t V 200.2
t V 100
Shear angle ‘ ’ will be find out as follows,
1
0.2 cos0rcostan
1 rsin 1 0.2 sin 0
tan 0.2 11.31
1t uncut chip thickness = depth of cut = 2mm
b = width of chip = feed = 0.1 mm
Shear strength (or) shear stress in the work piece is
ss
1
F 400sin sin 11.31 392.23 MPa
t b 2 0.1
54. A high velocity water jet of cross section area = 0.01 2m and velocity = 35 m/s enters a pipe filled with
stagnant water.
Entrained
water
Entrained
water
Inlet Jet
35 m s
Total
water out
flow at
uniform
velocity
6 m s
Pipe
b
1t
CV
Tool
2t
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The diameter of the pipe is 0.32m. This high velocity water jet entrains additional water from the pipe
and the total water leaves the pipe with a velocity 6 m/s as shown in the figure.
The flow rate of entrained water is _______ litres/s (round off to two decimal places)
Key: (130)
Sol:
d = diameter of jet, velocity of jet jetV 35 m sec
a = cross-sectional area of jet 20.01m
D = Diameter of pipe = 0.32m,
Cross-sectional area of pipe
2
2 2
A D4
A 0.32 0.080 m4
0V Total water outlet velocity = 6 m/sec
= Density of water = 1000 3kg m
Applying conservation of mass of principle,
Mass flow rate of jet + Mass flow rate of entrained water (or) stagnant water = total mass flow rate
jet oaV e flow rate stagnent water or entrained AV
jet 0aV Q AV
3
0.01 35 Q 0.080 6
Q 0.13 m sec 130 lit sec
1. 55. Consider an n × n matrix A and non-zero n × 1 vector p. Their product 2Ap p, where
amd 1,0,1 . Based on the given information, the eigen value of 2A is:
(A) 2 (B) (C) 4 (D)
Key: (C)
Entrained
water
Entrained
water
Inlet Jet
35 m s
Total
water out
at 6 m s
Pipe
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Sol: Given, 2Ap p …(1)
Pre-multiply both sides by A, we get
2
2
A AP A p
AA p Ap
. 2 2 2
4
A p p by (1)
p
4 is the eigenvalue of A2