ankit kumar singh

of 0
Two mark
1. A rigid vessel of volume 10m
2
is filled with hydrogen at 35C and 500 kPa. Due to
leakage, some gas has escaped from the vessel until the pressure in the vessel drops
down to 200 kPa, and the corresponding temperature of the gas inside the vessel is found
to be 25C. The amount gas leaked (in kg) from the vessel is___
Ans: 2.291 (range 2.2 to 2.5)
Sol: P
1
=500 kPa
V
1
=10m
3
T
1
= 273+35=308 K
157.4
2
314.8
M
R
R
2
2
H
H
11
1
1
500 10
3.905
4.157 308
PV
m kg
RT
P
2
= 200 kPa (final state)
V
2
= 10m
3
(rigid vessel)
T
2
= 273+25=298K
22
2
2
200 10
1.614
4.157 298
PV
m kg
RT
Amount gas leaked = m
1
m
2
= 3.9051.614=2.291kg
2. An air standard diesel engine has a compression ratio of 18 (the ratio of the volume at
the beginning of the compression process to that the end of the compression process),
and a cut off ratio of 3 (the ratio of the volume at the end of the heat addition process to
that at the beginning of the heat addition process). The thermal efficiency (in%) of the
engine is______
Ans: 59 (range 57 to 61)
Sol: r
k
= 18
r
c
= 3
1r
1r
r
1
1
2
y
c
1y
k
th
0.4
1 4.65 1
1
1.4 18 3 1





=0.589=59%
3. In a process industry two different streams of water (to be considered incompressible)
are available at 10C and 80C as shown in the figure. Mass flow rates of both the
streams are 2 kg/s. Rather than wasting these resourses, it is desired to connect a
reversible Carnot engine that will continuously extract heat from the hot stream and
supply part of it to the cold stream such that the exit temperature of both the streams T
f
is identical. Heat capacity of water is 4.18 kJ/kgK. The value of T
f
is
(a) 30C (b) 43.06C (c) 47.5C (d) 50C
Ans: (c)
Sol: T
H
= 90C , T
C
= 10C
s/kg1mm
CH
C
p
= 4.18 kJ/kgK
Since the engine is reversible
(dS)
net
= (dS)
1
+ (dS)
2
= 0
0
TT
T
nmC
CH
2
1
P
CHf
CH
2
f
TTT1
TT
T
80 273 10 273
f
T
= 316 K = 43.06C
4. An ideal regenerative Rankine cycle employs one open feed water heater (FWH) which
operate at 1.2 MPa. The extracted steam leaves the turbine at 210C. The condensate
enters the FWH at an enthalpy of 175 kJ/kg
Water property data : h = 2662 kJ/kg at 1.2 MPa and 210C. Saturated liquid enthalpy at
1.2 MPa = 750 kJ/kg.
Neglecting pump work input, the mass flow fraction extracted from the turbine is
(a) 0.127 (b) 0.227 (c) 0.327 (d) 0.427
Ans: (b)
mh
1
+ (1m)h
2
= h
3
T
h
=80C
m
h
=2kg/s
T
h
=10C
m
c
=2kg/s
q
C
q
H
T
f
T
f
W
mh
1
+ h
2
mh
2
= h
3
m(h
1
h
2
) = h
3
h
2
21
21
hh
hh
m
750 175
2662 175
Mass flow fraction m = 0.231
05. A steel component 21mm 61mm is made of 2mm thick sheet. The below figure
shows layout of scrap strip. Then percentage of stock used is____
(a)52.69 (b)82.59 (c)25.29 (d)99.69
Ans: (b)
Sol: The various dimensions will be as follows
Thickness of material (t) = 2 mm
Length of component (L) =21mm
Width of component (H) = 61 mm
B= 1.25t=1.252=2.5 mm
D= L+B=L+B=21+2.5=23.5mm
W= H+B+B=61+2.5+2.5=66mm
Area of strip used to produce a single blanked part
(A)= DW=23.566=1551 mm
2
Area of the component is (a) = LH
2161=1281 mm
2
Percentage of stock used
%59.82
1551
1281
100
A
a
06. A 400 mm diameter bar is turned at 45 rev/min with depth of cut of 2 mm and feed of
0.2 mm/rev the forces measured at the cutting tool point are cutting force=1800N,
feed force = 400N. Then power consumption (kW) and material removal rate
(mm
3
/min) are___
(a) 1.69, 4.498 (b) 2.969, 4.498
(c)1.696, 22.62 10
3
(d) 2.696, 22.6210
3
h
2
= 195
(1-m)
h
1
= 800
h
1
= 2862
m
FEED
L=21
B
B
D
W
t
H=6
1
Ans: (c)
Sol: Given data
Bar dia (D)= 40 mm
Speed (N) =45 rev/min
Feed (f) = 0.2 mm/rev
Cutting force (F
c
)=1800N
Feed force (F
f
) = 400N
Cutting velocity
(i)
sec/m9424.0
100060
45400
100060
DN
V
Cutting power
)kW
1000
VF
P
c
c
kW696.1
1000
9424.01800
Now feed velocity
s/m105.1
100060
452.0
4
Feed power = 4001.510
4
=0.06 W (negligible)
(ii) Metal removal rate = A
c
V
= btV=dfV
=20.240045=22.6210
3
mm
3
/min
07. In electro chemical machining of an iron surface that is 20mm 20 mm in cross
section using Nacl in water as electrolyte. The gap between the tool and the work
piece is 0.2 mm. The supply voltage is 13 V
dc
. The specific resistance of the
electrolyte is 2cm. Then material removal rate is ___ (g/sec). [Hint: Valency (Z) =
2, Atomic weight (A)= 55.85, Density (
a
)= 7860 kg/m
3
coulombs]
(a) 0.00376 (b) 0.124 (c) 0.376 (d) 0.0124
Ans: (c)
Sol: Given data
Cross sectional area of the gap (A
gap
) = 20 20=400mm
2
Gap between the tool and work piece (H)= 0.2 mm
Voltage (V) =13v
= 2cm = 210mm
The gap resistance (R) is given by
01.0
400
2.0102
A
h
R
gap
Current I=
A1300
01.0
13
R
V
I
The material removal rate in ECM by considering 100%
Current efficiency is
193080
72605
965402
130085.55
ZF
AI
MRR
= 0.376 g/sec
08. The casting shown in below figure is to be made in plain carbon steel using a wooden
pattern. Assume the shrinkage allowance is for this steel is 23.0 mm/m only. Then
dimensions (including allowance) of the pattern as P,Q,R,S respectively ______
(a) 93.07, 111.53, 216.83, 164.86 (b) 92.07, 111.53, 214.83, 163.68
(c) 92.07, 112.53, 214.83, 163.68 (d) 93.07, 112.53, 216.83, 163.68
Ans: (c)
Sol: Shrinkage allowance is 23 mm/m
For dimension P=90, allowance is
mm07.2
1000
23
90
mm53.2
1000
23
110isallowacne,110Q
mm83.4
1000
23
210isallowance,210R
mm68.3
1000
23
160isallowance,160S
The dimensions of wooden pattern as follows
P= 92.07 mm \
Q= 112.53 mm
R= 214.83 mm
S= 163.68mm
09. A circular rod of diameter 50 mm is subjected to force as shown in figure. The stress at
the bottom-most support point A is in MPa
l=3d
10kN
d=50mm
A
R = 210
Q = 110
S=160
P= 90
(a) 25.46 (b) -5.09
(c) 20.37 (d) -15.28
Ans: (D)
Sol:
Direct stress
d
=
MPa09.5
50
4
F
2
Bending stress
b
=
2
Fd
M,
d
M32
3
23
d
F16
d2
dF32
MPa37.20
50
10001016
2
Resultant stress at A =
d
+
b
=-5.09 MPa + 20.37 MPa
=-15.28 MPa
10. A 10mm thick steel rectangular plate of size 100mm x 200mm is subjected to biaxial
stresses of
x
= 150 MPa,
y
= 200 MPa as shown below. The Young’s modulus and
Poisson’s ratio are 200 GPa and 0.3 respectively then the change in the thickness of the
plate is
(a) 2.39m (b) 5.25 m
(c) 7.12 m (d) 9.16 m
Ans: (B)
Sol:
EEE
y
xz
z
)0(
Ez
z
zyx
z =
yx
E
z
200150
10200
3.010
3
z = 5.25 m
y
= 200MPa
x
=150MPa
100mm
200mm
11. A bar of circular cross section is clamped at its left end, free of the right and loaded by a
twisting moment ‘t’ per unit length is uniformly distributed along the middle third of
the bar as shown in figure/ The angle of twist at the free end of the bar is
GJ9
tL
)a(
2
(b)
2
tL
6GJ
GJ9
tL2
)c(
2
(d)
GJ3
tL2
2
Ans: (b)
Sol:
Angle of twist at D with respect to A
=
AB
+
BC
+
CD
AB BC CD
Tl Tl Tl
GJ GJ GJ

2
1 tL L
tL L
1 tL
2 3 3
33
GJ GJ 6 GJ







12. A single plate clutch is designed to transmit 10 kW power at 200 rpm. The equivalent
mass and radius of gyration of the input shaft are 20 kg and 75 mm respectively. The
equivalent mass and radius of the output shaft are 35 kg and 125 mm . total energy
dissipated during the clutching operation
(a) 2046.38 J (b) 2146.38 J (c) 2246.38 J (d) 2546.38 J
Ans (A)
Sol:
2
22
11
75
I m k 20 0.1125kg m
1000



2
32
2 2 2
125
I m k 35 0.5468kg m
1000



1
2 N 2 2000
60 60
L/3
t
L/3
L/3
A
B
C
D
t
L/3
L/3
L/3
Two mark
1. A rigid vessel of volume 10m
2
is filled with hydrogen at 35C and 500 kPa. Due to
leakage, some gas has escaped from the vessel until the pressure in the vessel drops
down to 200 kPa, and the corresponding temperature of the gas inside the vessel is found
to be 25C. The amount gas leaked (in kg) from the vessel is___
Ans: 2.291 (range 2.2 to 2.5)
Sol: P
1
=500 kPa
V
1
=10m
3
T
1
= 273+35=308 K
157.4
2
314.8
M
R
R
2
2
H
H
11
1
1
500 10
3.905
4.157 308
PV
m kg
RT
P
2
= 200 kPa (final state)
V
2
= 10m
3
(rigid vessel)
T
2
= 273+25=298K
22
2
2
200 10
1.614
4.157 298
PV
m kg
RT
Amount gas leaked = m
1
m
2
= 3.9051.614=2.291kg
2. An air standard diesel engine has a compression ratio of 18 (the ratio of the volume at
the beginning of the compression process to that the end of the compression process),
and a cut off ratio of 3 (the ratio of the volume at the end of the heat addition process to
that at the beginning of the heat addition process). The thermal efficiency (in%) of the
engine is______
Ans: 59 (range 57 to 61)
Sol: r
k
= 18
r
c
= 3
1r
1r
r
1
1
2
y
c
1y
k
th
0.4
1 4.65 1
1
1.4 18 3 1





=0.589=59%
2
=0
The total energy dissipated during the clutching operation
2
1 2 1 2
12
II
1
E
2 I I
2
209.44 0 0.1125 0.5468
1
E
2 0.1125 0.5468




4092.76
2046.384J
2

13. Pedal arm assembly as shown in the figure with a rider applied force of 1.5 kN at the
pedal, Determine the maximum principal stress in the pedal arm if its cross- section is 20 mm
in diameter. And the pedal attaches to the pedal arm with 12- mm screw thread. What will
be the stress in the pedal screw.
(a) 57.29 N/mm
2
& 520 N/mm
2
(b) 90 N/mm
2
& 297.5 N/mm
2
(c) 297.5 N/mm
2
& 530.5 N/mm
2
(d) 180.2 N/mm
2
& 90 N/mm
2
Ans : (C)
Sol: Maximum Bending moment and torque in the pedal arm
Pedalarm
170
M 1500 225N m
1000
Pedalam
60
T 1500 90N m
1000
2
b
3
3
32M 32 225
286.48N / mm
d
0.02

2
xy
3
3
16T 16 90
57.29N / mm
d
0.02

y
=0
2
2
1
286 0 286
57.29
22
 
1
= 297.05 N/mm
2
Stress in Pedal screw:
170 mm
F
60 mm
PedalScrew
60
M 1500 90N m
1000
2
b
3
32 90
530.5N / mm
0.012
14.A solid sphere of diameter ‘D’ and density
s
is completely submerged in a tank filled
with a liquid of density
l
(
s
>
l
). The sphere is held place by a string attached to a hook
above the top of the tank. The tension in the string is
(a)
3
s
4
gD
3
 
(b)
3
s
4
gR
3
(c)
3
s
D
g
3
 
(d)
3
s
D
g
6
 
10. Ans: (d)
Sol: Weight of sphere = W = mg
=
s
gV
(Where V= volume of sphere)
3
s
gD
6
Buoyancy force
3
b
F g D
6

W = T + F
B
T = W F
B
3
s
D
Tg
6
  
15. A jet of water issuing from a stationary nozzle with a uniform velocity 4m/s strikes a
frictionless turning vane mounted on a cart. The vane turns the jet through an angle 60. The
area of jet is
2
1
m
16
. An external mass, m is connected to the cart though the frictional
pulley as shown.
W
T
B
F
sphere
The mass required to hold the cart stationary is _______ kg (Assume g = 10 m/s
2
)
Ans: 50 (Range: 50 to 50)
Sol:
Force exerted by liquid on vane in x-direction,
F
x
= Momentum out-Momentum in
F
x
= T= Q(Vcos V)
T = AV
2
(1cos)
2
1
1000 4 1 cos60
16
T = 500 N
T = mg
500
m 50kg
10

16. A tank has two outlets as shown in figure.
A rounded entry orifice ‘A’ of diameter ‘D’ and.
A pipe ‘B’ with same diameter as or orifice and of length L = 7.5 m. For a head of water
9m in the tank the ratio of discharges from the outlets A and B _______. (Assume friction
losses in the pipe as 0.5 m of water)
16. Ans: 0.75 (Range: 0.75 to 0.75)
Sol: As all other parameters remain constant
H 9m
L 7.5m
B
A
Orifice
Pipe
Mass
Pulley
Controlvolume
60
V
T
x
y
Mass
mg
T
T
x
F
Mass
Pulley
Frictionless
60
Q V
2gh
For orifice,
ori
QH
For pipe,
pipe
Q H L loss
ori
pipe
Q9
Q 9 7.5 0.5

93
0.75
16 4
Putting the value of
u
v
, we get
max
8
1 cos
27
17. A contractor has to supply 10,000 bearings per day to an automobile manufacturer. He
finds that when he starts production run, he can produce 25,000 bearings per day. The
cost of holding a bearing in stock for a year is Rs 200 and the set-up cost of a
production run is Rs 1800. What is the economic batch quantity?
Assume Number of working days in the year are 300 days.
Sol:
0
h
2DC
p
Q*
C p d



units9487
1000025000
25000
200
1800300000,102
18. A new facility has to be designed to do all the welding for 3 products: A, B and C. Unit
welding time for each product is 20 s, 40 s and 50 s respectively. Daily demand forecast
for product A is 450, for B is 360 and for C is 240. A welding line can operate
efficiently for 220 minutes a day. Number of welding lines required is
(a) 5 (b) 4 (c) 3 (d) 2
Ans: (C)
Sol: Time required to weld Product A =
min150450
60
20
Time required to weld product B =
min240360
60
40
Time required to weld product C =
min200240
60
50
Total time required = 150 + 240 + 200 = 590 minutes
H 9m
L 7.5m
B
2
Pipe
 
3
1
Number of welding lines required =
368.2
220
590
19. Weekly production requirements of a product are 1000 items. The cycle time of
producing one product on a machine is 10 minutes. The factory works on two shift basis
in which total available time is 16 hours. Out of the available time about 25% is
expected to be wasted on break downs, material unavailability and quality related
problems. The factory works for 5 days in a week. How many machines are required to
fulfill the production requirements?
(a) 2 (b) 3 (c) 4 (d) 6
Ans (B)
Sol: Hours utilized per day = 16 × (1 0.25) = 16 × 0.75 = 12 hours
Hours utilized in a week = 12 × 5 = 60 hours
Number of items produced on a machine =
360
10
6060
Number of machines required =
377.2
360
1000
20. What is the optimal solution for the assignment problem given below.
Time (in minutes)
Worker
Job 1
Job 2
Job 3
A
B
C
4
8
4
2
5
5
7
3
6
(a) 8 min (b) 9 min (c) 10 min (d) 12 min
Ans: (B)
Row Iteration
Column iteration
2
5
0
0
2
1
5
0
2
J
1
J
2
J
3
A
B
C
The optimal assignment is
A 2, B 3, C -1
Total time = 2+ 3 + 4 = 9 min
20. Two identical involute spur gears are in mesh. The module is 4 mm and each gear has
22 teeth. If the operating pressure angle is 20
0
. The minimum value of addendum
needed to ensure continuous transmission of motion is
(a) 7.5 mm (b) 5.5 mm (c) 1.35 mm (d) 2.35 mm
Ans. (d)
Sol. To ensure continuous transmission of motion at least one pair of teeth should be in
contact.
i.e., contact ratio should be
1
Path of contact
Path of contact
1
Circularpitch cos

2
2
a
2 R Rcos Rsin
1
2R
cos
T



Given,
m 4mm
T 22
0
20
mT
R 44
2

a
R 46.35
46.35 mm
2.35mm
21. A flywheel with a man of 5 kN &radius of gyration 1.8m. Find the energy stored in
theflywheel when its speed increases from 315 to 350 rpm.
Sol:
 


2
5
0
0
2
1
5
0
2
J
1
J
2
J
3
A
B
C
0
0
0
 




 


 
  




22. In the mechanism shown in figure can input link act as a crank? If not suggest maximum
possible length (cm) of input link to act crank without changing connecting length and
offset
(A) Yes (B) No, 12 cm (C) No, 15 cm (D) No, 20 cm
Ans. (B)
Sol.
The minimum possible length of
OB 10cm
OA OB AB
(since this position is possible)
This will be the last position after which rotation is not possible.
Input link can’t act as crank.
Input link in order to act as crank without charging offset and connecting rod length.
OA 10 AB
For maximum possible length of crank.
OA + 10 = AB
OA = l2 cm
23. The guide D in figure has an upward velocity 24 cm/s and an acceleration of 48 cm/s
2
.
The velocity of roller A (cm/s) when
= 30
0
and AB = BC = 6 cm is
(A) 8.32 (B) 4.05 (C) 6.93 (D) 5.25
Ans. (C)
Sol. Given V
D
= 24 cm/s
V
D
= od = 24 cm/s
V
B
= ob
vectors directions are taken in such away that V
B
is in vertical and V
A
is in horizontal
direction.
ca
V cos30 24
l cos30 24
24
12 cos30

A
l1
V ab sin30 . .
22
41
6
2
3
12
6.93cm/s
3

24 The string shown in fig is under tension T. Determine the natural frequency of
vibration in a plane of paper and in a direction perpendicular to string assuming
that for small displacement T remains constant.
b
W
m
m
a
l
(a)
1
n
Ta a
w
ml
b)
/
n
w Tl ma l a
(c)
2
n
Tl
w
ma
(d)
2
n
Tl
w
m l a
Sol: (b)
b
W
m
l
m
x
T
T
1
2
a
1
sin sin 0mx T T


1 1 1
sin tan
x
for small
a
0
T x x
x
m a l a




0
n
Tl Tl
x x w
m a l a ma l a






25 A short semicircular right cylinder of radius r and weight W rests on a horizontal surface and is
pulled at right angles to its geometric axis by a horizontal force P applied at the middle B of
the front edge. Find the angle that the flat face will make with the horizontal plane just
before sliding begins if the coefficient of friction at the of contact A is
1
1 1 1 1
3 3 3 2
(a)sin (b)sin (c)sin (d)sin
7 5 8 5
Ans: (A)
Sol:
4r
P(r rsin ) W sin
3



R = W
W = P
4r
P(r rsin ) sin
3



4sin
(1 rsin )
3



33
sin
4
4 3 7
3


26. Heat is flowing through the wire of radius r
1
and thermal conductivity k
1
W/mK. To
increase the heat transfer rate insulation is provided over the wire. The radius after
insulation of wire be r and thermal conductivity of insulation material be k, it the heat
transfer coefficient be h W/m
2
K between wire and surroundings, then the condition to
increase heat transfer rate is
(a)
rr
r
r
nrr
h
k
1
1
1
1
(b)
1
1
1
rr
r
r
nrr
h
k
(c)
rr
r
r
nrr
h
k
1
1
1
(d)
rr
r
r
nrr
h
k
1
1
1
Ans: (b)
Sol: Without insulation
Let the conductive resistance of wire be R
w
Total resistance without insulation
Lr2h
1
RR
1
w
Where L=length of wire
With insulation
C
A
B
P
Total thermal resistance with insulation
rL2h
1
Lk2
r
r
n
RR
1
wi
To increase heat transfer, R should be greater than R
i
R>R
i
Lt2h
1
Lk2
r
r
n
R
Lr2h
1
R
1
w
1
w
Multiplying with 2l on both side we get
rh
1
k
r
r
n
rh
1
1
1
k
r
r
n
r
1
r
1
h
1
1
1
k
r
r
n
rr
rr
h
1
1
1
1
1
1
1
rr
r
r
nrr
h
k
1
1
1
rr
r
r
nrr
h
k
One mark
1. A pump raises pressure of saturated liquid water at 100 kPa (density =959 kg/m
3
) to 2
MPa. The isentropic efficiency of the pump is 80%. The work done by the pump (in
J/kg) is_____
Ans: 2476.538 (range 2300 to 2600)
Sol:
actual
cs
isen
W
W
92.0
dp1
92.0
vdPW
W
isen
cs
actual
2000 100
1
959 0.8

= 2476.538 J/kg.
2. Volumetric analysis of a hydrocarbon combustion product shows 8% CO
2
, 15% H
2
O
(vapour) 5.5% O
2
and 71.5% N
2
. The combustion product flows steadily through a heat
exchanger at 200 KPa pressure. Assume each component in the mixture to be an ideal
gas. In order to avoid the condensation of H
2
O in the heat exchanger, the minimum
allowable temperature (in C) is___
Saturated H
2
O Table
P/(kPa)
10
20
30
40
50
T
1
(C)
45.83
60.09
69.12
75.82
81.35
Ans: 69.12 C (range 67 to 71)
Sol: Partial pressure of water vapor in exhaust
pressure
mixture%
watter%
200
5.715.5158
15
PP
OH
2
= 30 kPa
Corresponding to 30 kPa from steam tables the temperature is 69.12C
3. The device shown in the figure enables the separation of a mixture of ideal gases into
constituent
Gases A and B. if the pressure and temperature of the gases at the inlet and the outlet of
he device are equal, and W is the work output the device, then
(a) W> 0 (b) W<0
(c) W=0 (d) W>0 or W<0, depending on constituent gases.
Ans: (b)
Sol: isothermal process
dQ = dW
dW= T(ds)
min
A work is being done on the system for splitting
Mixture
Device
W
Gas B
Gas A
04. A typical 2 D transformation operation to convert a point P
1
(x
1
, y
1
) to the point
P
2
(x
2
, y
2
) in a CAD package is represented below
15.15.2
05.00
0075.1
1yx1yx
1122
The transformation operation(s) envisaged around the origin will be
(a) rotation only
(b) scaling only
(c) rotation and translation
(d) scaling and translation
Ans: (c)
05. Match the following List-I (Casting defects) with List-II (meaning)
LIST-I LIST-II
P. shrinkage cavity 1. This is scar covered by thin layers of a metal
Q. Drop 2. Lighter impurities appearing on the top surface of a
casting
R. Blister 3. An irregularly shaped projection on the cope surface of
the casting
S. Dross 4. An improper riser may give rise to a defect
Codes:
P Q R S
(a) 3 1 4 2
(b) 4 3 1 2
(c) 1 4 2 3
(d) 2 1 4 3
Ans: (b)
06. A copper bar 20cm long is fixed by means of a sinking support at its ends which yield
by an amount 0.01 cm. If the temperature of the bar is raised by 20
0
C. The stress
induced in the bar is (Take
c
= 17.5 10
-6
/
0
C, E
c
= 98 GPa)
(a) 34.3 MPa (b) 14.7 MPa
(c) 83.3 MPa (d) zerp
Ans: (D)
Sol:
Free expansion of the bar = /t
= 17.5 10
-6
20 20
= 7 10
-3
cm
= 0.007 cm
Actual extension allowed
= 0.01 cm > 0.007 cm
Induced stress is zero
07. The number of independent reaction components for fixed support are
(a) 0 (b) 1
(c) 2 (d) 3
Ans: (D)
Sol:
08. A Newtonian fluid of viscosity flows between two parallel plates due to the
motion of bottom plate as shown below, which is moved with a velocity V. The top plate is
stationary. Assume the gap between two plates is small.
The velocity profile in the X direction is
(a)
Vy
b
(b)
2
y
V1
b







(c)
2
y
V1
b







(d)
y
V1
b






01. Ans: (d)
Sol: Boundary conditions:
At y = 0, u = V
At y =b, u = 0
From similar triangles,
y
V
V
b b y
y
by
VV
b



y
y
V V 1
b




by
y
M
R
H
x
y
fluid
V
b
x
y
V
b
09. Consider the laminar boundary layer developed over a flat plate as shown in the figure
below
The ratio of boundary layer thickness at a point B to that a point C is
3
4
. The ratio of
the drag force on portion AB to that on portion BC
AB
BC
F
F



is ______.
Ans: 2 [Range 2 to 2]
Sol: From von-Karman momentum integral equation,
2
D
F x U b x
Where F
D
(x) = Drag force upto distance x from the leading edge.
(x) = momentum thickness at distance x from the leading edge.
B = width of plate perpendicular to the plane.
AB AB
BC AC AB
FF
F F F
2
B
22
C
CB
B
U b 1
U b U b
1



Now (x) (x)
AB
BC
F1
3
4
F
1
3
10. The rise in pressure of water flowing through a 5 km pipe of 150 mm diameter, and 5 mm
thickness at a velocity of 2.5 m/sec, when the valve is closed in 6 seconds in MPa is
(velocity of pressure wave = 1400 m/sec)
(a) 3.58 (b) 1.64
(c) 2.81 (d) 5.96
Ans: (a)
Sol: = 1000 kg/m
2
,V= 2.5 m/sec, D = 150 mm
The ratio
2L 2 5000
7.14 T
C 1400
Its sudden closure
P = VC = 2.5 1000 1400
= 3588.17 kPa
= 3.58 MPa
11. Vertical intercept of EGL is 9 m and that for HGL is 3 m. The velocity of fluid in the
pipe is
(a) 10 m/sec (b) 16m/sec
(c) 8m/sec (d) 14m/sex
Ans: (a)
Sol: Difference in vertical intercept of EGL and HGL is 6 m which is kinetic head
A
B
C
U
B
C
2
V
6
2g
V 2 9.8 6 10.844m/sec
12.The hemispherical ends of a pressure vessel is fastened to the cylindrical portion of the
pressure vessel with the help of gasket, bolts and lock nuts. The bolts are subjected to
(a) shear stress (b) bearing stress (c) tensile stress (d) none of the
above
Ans: (C)
13. The light from a single car is photographed from a high vantage point with a time
exposure. What is the line that is observed in the photograph?
(a) Streamline (b) Streak line
(c) Pathline (d) time line
Ans: (c)
Sol: A time exposure picture of a light source tracks positions of the light source at
different time. Clearly this gives path followed by the light source.
14. A project comprising A, B and C activities is scheduled for 80 days at a cost of Rs 1500
million. The manager of project decides to reduce the time for completion of project to
75 days. The decision was taken after 40 days.
Activity
A
B
C
Duration (Days)
30
20
30
Crashing cost/day
(Million Rs)
10
40
20
The minimum project cost in million rupees after crashing 5 days is
(a) 1550 (b) 1600 (c) 1700 (d) 1800
Ans: (B)
Sol: The least cost slope activity after 40 days is ‘C’ which has a cost slope of 20
million/day.
Total cost = Project cost + Crash cost
= 1500 + 5 × 20 = 1600
15. If Poisson arrivals are 5 per minute, the probability that exactly 4 arrivals are there in
the next one minute is
(a) 0.156 (b) 0.175 (c) 0.018 (d) 0.326
Ans: (B)
Sol: x = 5
A
B
C
P(n) =
n
e
n!

P(4) =
!4
5e
45
= 0.175
16. Portion of transportation matrix is given. The maximum quantity that can be shifted to
C 3 cell, keeping the demand and supply constraints is
(a) 15 (b) 25 (c) 30 (d) None
Ans: (B)
16. Five jobs are performed, first on machine X and then on machine Y. The time taken, in
hours by each job on each machine is given below:
Job : A B C D E
Machine X : 12 4 20 14 22
Machine Y : 6 14 16 18 10
Determine the optimum sequence of jobs that minimizes the total elapsed time to
complete the jobs?
(a) B-A-E-D-C (b) B-D-C-E-A
(b) B-C-D-E-A (d) B-A-D-E-C
Ans: (B)
Sol:
The optimal sequence according to Johnson’s rule is
The optimal sequence of the jobs is B D C E A
17. The maximum swaying couple is in locomotives:-

  

 
  

 
  

 
  

Ans : (A)
18 The centre of gravity of a trapezium with parallel sides ‘a’ and ‘b’ lies at a distance of y
from the base ‘b’. The value of ‘y’ is
B
D
C
E
A
M/C Y
M/C X
2
3
C
D
30
15
25
a)
2a b
h
ab



b)
h 2a b
2 a b



c)
h 2a b
3 a b



d)
h a b
3 2a b



Ans: (c)
19. Consider a hot, boiled egg in a spacecraft that is filled with air at atmospheric pressure
and temperature at all times. Disregarding any radiation effect, will the egg cool
faster or slower when the spacecraft is in space instead of on the ground.
(a) No difference (b) Faster (c) Slower (d) Insufficient data
Ans: (c)
Sol: From the data it can be observed that the mode of heat transfer from the egg is natural
convection.
Natural convection occurs due to buoyancy effects and buoyancy depends on gravity.
The rate of heat transfer due to natural convectional depends on gravity.
More the gravity, more will be heat transfer rate.
But in space, gravity is very less compared to ground.
Heat transfer rate will be more on ground than in space
20 In a composite material A-B-C (respective thermal conductivities are temperature
independent and related as: k
A
<k
B
; k
B
>k
C
, the steady state heat flux distribution is
shown in the figure below. The corresponding steady state temperature distribution is
given as
Ans: (b)
Sol:
For region A:
,negativeis
dx
dT
Tq
k
Hence,
positiveis
dx
dT
Similarly
dx
dT
is positive for regimes B and C
At A-B interface,
n
B
A
A
dx
dT
k
dx
dT
k
As k
A
<k
B
,
BA
dx
dT
dx
dT
(A)
(B)
(C)
(D)
A
B
C
x
q
x
T
T
T
x
x
x
A
B
C
x
q
x
Similarly at B-c interface,
CB
dx
dT
dx
dT
((k
B
>k
C
)
When
dx
dT
constant, T(x) is linear
When
dx
dT