ankit kumar singh

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

= 3.905–1.614=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

T

dT

mc

T

dT

mc

f

c

f

H

T

T

T

T

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

+ (1–m)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

=20.240045=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

, faraday constant =96540

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

AD

=

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

209.44rad / sec

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

= 3.905–1.614=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

(1–cos)

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

R + addendum

46.35 mm

Addendum

2.35mm

Minimum addendum = 2.35 mm

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

linear, T(x) is quadratic

Hence the correct temperature distribution is

T

x