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GATE 2019Mechanical Engineering

Forenoon Session - 02.02.2019

Bhilai | Raipur | Bhopal www.gateacademy.co.in 97131-13156, 95898-94176

GATE 2019 [Forenoon Session]Mechanical Engineering

Since 2004

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© Copyrightwww.gateacademy.co.inH Oead ffice : A/114-115, Smriti Nagar, Bhilai (C.G.), Contact : 9713113156, 9589894176

B Oranch ffice : Raipur, : 79743-90037, Bhopal, : 89591-87052

.Aptitude Part.

Q.1 to Q.5 Carry One Mark Each

Question 1

The sum and product of two integers are 26 and 165 respectively. The difference between these two integers is

(A) 4 (B) 6 (C) 3 (D) 2

Ans. (A)

Sol. Given, 26x y …(i)

165x y

165

yx

Putting the value of y in equation (i),

165

26xx

2 165 26x x

2 26 165 0x x

2 15 11 165 0x x x

15, 11x

Taking positive value 15x .

165

1115

y

So, 4x y

The difference between x and y is 4.

Hence, the correct option is (A).

Question 2

The minister avoided any mention of the issue of women’s reservation in private sector. He was accused of _______ the issue.

(A) Skirting (B) Collaring (C) Belting (D) Tying

Ans. (A)

Sol. Skirting : Avoid or try to avoid fulfilling, answering, or performing.

Collaring : Seize by the size or collar.

Belting : Strike someone by belt.

Tying : The act of tying or binding things together.



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Question 3

John Thomas, an _______ writer, passed away in 2018

(A) Prominent (B) Eminent (C) Imminent (D) Dominant

Ans. (B)

Sol. Prominent : Having a quality that thrusts itself into attention.

Eminent : Standing above others in quality or position that is well qualified.

Imminent : About to occur.

Dominant : Most frequent or common.

Eminent is the suitable word because it defines the qualities that writer should have.

Hence, the correct option is (B).

Question 4

A worker noticed that hour hand on the factory clock hand moved by 225 degree during her stay at the factory. For how long did she stay in the factory?

(A) 3.75 hours (B) 4 hours and 15 minutes

(C) 7.5 hours (D) 8.5 hours

Ans. (C)

Sol. Given : The angle by which hour hand turned 0( ) 225

Since, the angle turned by hour hand in hour 0

036030

12

So, the working time of worker 0

0 0

2257.5

30 30

hour.

Hence, the correct option is (C).

Question 5

________ I permitted him to leave, I wouldn’t have had any problem with him being absent ______ I?

(A) Had, would (B) Have, would (C) Had, wouldn’t (D) Have, wouldn’t

Ans. (A)

Q.6 to Q.10 Carry Two Marks Each

Question 6

Under a certain legal system, prisoners are allowed to make one statement. If their statement turns out to be true then they are hanged. If the statement turns out to be false then they are shot. One prisoner made a statement and the judge had no option but to set him free. Which one of the following could be that statement?

(A) You committed the crime (B) I committed the crime

(C) I will be shot (D) I didn't commit the crime

Ans. (C)


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Question 7

M and N had four children (P, Q, R and S) of them, only P and R are married. They had children X and Y respectively. If Y is the legitimate child of W, which one of the following statement is necessarily FALSE?

(A) W is the wife of P (B) W is the wife of R

(C) M is the grandmother of Y (D) R is the father of Y

Ans. (A)

Sol.

Since, only P and R married and the both have child X and Y respectively whereas Y is also child of W. So, there is no relation between P and W that is W can’t be wife of P.


Question 8

A person divided an amount of Rs. 100,000 into two parts and invested in two different schemes. In one he got 10% profit and in the other he got 12%. If the profit percentages are interchanged with these investments he would have got Rs. 120 less. Find the ratio between his investments in the two schemes.

(A) 11 : 14 (B) 47 : 53 (C) 37 : 63 (D) 9 : 16

Ans. (B)

Sol. Let, the first amount Rs. x and second amount Rs. (100000 )x

According to first given condition of question,

(100000 ) 12% 10% 12000 0.02x x x …(i)

According to second given condition of question,

(100000 ) 10% 12% 10000 0.02x x x …(ii)

Now, as per the question the difference of equation (i) and (ii) will give the profit of Rs. 120.

So, (12000 0.02 ) (10000 0.02) 120x

2000 0.04 120x

2000 120

Rs.470000.04

x

So, the required ratio 47000 47

100000 100000 47000 53

x

x


R W

P

Q

S

M N

X

Y


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Question 9

A firm hires employees at five different skill levels P, Q, R, S, T. The shares of employment at these skill levels of total employment in 2010 is given in the pie chart as shown. There were a total of 600 employees in 2010 and the total employment increased by 15% from 2010 to 2016. The total employment at skill levels P, Q and R remained unchanged during this period. If the employment at skill level S increased by 40% from 2010 to 2016 how many employees were there at skill level T in 2016?

(A) 30 (B) 72 (C) 60 (D) 35 Ans. (C)

Sol. Total employees in 2010 is 600 Employees increased by 15% in 2016

Total employees in 2016 15

600 600 690100

…(i)

Employees in S category in 2010 25

600 150100

Now, S category increased by 40% in 2016

Employees in S category in 2016 40

150 150 210100

…(ii)

Since, there is no change in P, Q, and R categories of employees so total number of employees of all three categories is given by,

Number of employees 20 25 25

600 600 600 420100 100 100

So, total number employees in T category 690 (420 210) 60

Hence, the correct option is (C). Question 10

Congo was named by Europeans Congo’s dictator Mobuto later changed the name of the country and the river to Zaire with the objective of Africanising names of person and space. However the name Zaire was a Portuguese alteration of Nzadi O Nzere, a local African term meaning. River that swallows Rivers. Zaire was the Portuguese name for the Congo river in the 16th and 17th centuries.

Which one of the following statements can be inferred from the paragraph above? (A) Mobuto was not entirely successful in Africanising the name of his country (B) Mobuto’s desire to Africanise names was prevented by the Portuguese (C) As a dictator Mobuto ordered the Portuguese to alter the name of the river to Zaire (D) The team Nzadi O Nzere was of Portuguese origin Ans. (A)

S25

Q25

R25

T5

P20


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.Technical Part.

Q.1 to Q.25 Carry One Mark Each

Question 1

The table presents the demand of a product. By simple three-months moving average method, the demand-forecast of the product for the month of September is

Month December January 450 February 440 March 460 April 510 May 520 June 495

July 475 August 560

(A) 536.67 (B) 530 (C) 490 (D) 510 Ans. (D)

Sol. In three months moving average method we used most recent data for calculating forecast for upcoming period.

3

June July Augsept

F F FF

560 475 495

510units3septF

Hence, the correct option is (D). Question 2

For the equation 27 0dy

x ydx

, if 3

(0)7

y , then the value of y(1) is

(A) 3

73

7e

(B) 7

37

3e

(C) 3

77

3e

(D) 7

33

7e

Ans. (D)

Sol. Given : 27 0dy

x ydx

27dy

x ydx

Using the method of separation of variables,

217dy x dx c

y

3

ln( ) 73

xy c


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Taking antilog both sides,

3 37 7

3 3

x xccy e e e

Let, ce K

37

3x

y Ke

…(i)

Given 3

(0)7

y i.e., for 0x , 3

7y . Substituting in equation (i),

7

03

3

7Ke

3

7K

From equation (i),

37

33

7

xy e

So, the value of (1)y , i.e., value of y at 1x is,

7 7

13 3

3 3(1)

7 7y e e


Evaluation of 4 3

2x dx using a 2 equal segment trapezoidal rule gives a value of _______.

Ans. 63

Sol. Given : 4 3

2I x dx

For two equal segment, 2n

2, 4a b

Step size, 4 2

12

b ah

h

x 2 3 4 3y x 8 27 64

From Trapezoidal rule

0 2 1. ( ) 2( )2

b

a

hy dx y y y

4 3

2

1. (8 64) (2 27)

2x dx

1(72 54)

2

12663

2

Hence, the value of 4 3

2x dx by using trapezoidal rule is 63.


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Question 4

A parabola 2x y with 0 1x as shown in figure. The volume of the solid of rotation obtained by

rotating the shaded area of 0360 around the x-axis is

(A) (B) 4

(C)

2

(D) 2

Ans. (C)

Sol. Given : 2 , 0 1x y x

Volume of revolution about x-axis is,

2b

aV y dx

Here, 2 , 0, 1y x a b .

121

002

xV x dx

[1 0]

2 2


The length of large stocks of titanium rods follow normal distribution with ( ) of 440 mm and standard

deviation (σ) of 1 mm. What is the percentage of rods whose length lie between 438 mm and 441 mm?

(A) 81.85% (B) 68.4% (C) 86.64% (D) 99.75% Ans. (A)

Sol. In normal distribution, distribution of total area in percentage is symmetrical about the mean and is shown in figure.

Here, given 440 mm, 1 mm .

The given range is 438 mm to 441 mm, that can be considered as we have to find the percentage of rods, whose length lie between 2 to .

Required area (34.1 13.6) 34.1 47.7 34.1 81.80%


1x

2x y�

y

z

f z( )

�2� � � �2.1%

13.6%34.1% 34.1%

13.6%2.1%

� 2�438 439 440 441


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Question 6

Consider the matrix

1 1 0

0 1 1

0 0 1

P

. The number of distinct eigenvalues of P is

(A) 0 (B) 3 (C) 1 (D) 2 Ans. (C)

Sol. Given :

1 1 0

0 1 1

0 0 1

A

Given matrix is an upper triangular matrix as the elements below the principle diagonal are zero. The Eigen values of upper triangular or lower triangular matrix are the elements of leading diagonal.

Hence, the Eigen values for the given matrix are, 1 2 31, 1, 1 .

As all the three Eigen values are same. So the number of distinct Eigen values for the given matrix is 1. Hence, the correct option is (C). Question 7

The position vector OP of point P(20, 10) is rotated anticlockwise in XY plane by an angle of 030 such that the point P occurs position Q as shown in the figure. The coordinates (x, y) of Q are

(A) (22.32, 8.26) (B) (12.32, 18.66) (C) (13.40, 22.32) (D) (18.66, 12.32) Ans. (B)

Sol. Given : Point P (20, 10)

From triangle OPD, 10

tan 0.520

026.56

0 0 0 030 30 26.56 56.56 Now, from triangle OQC,

tany

x

tan (56.56) 1.514y

x

Now, we go through option and checking the value which is exactly equal to the above value. Hence, the correct option is (B).

x

y

�

Q

P

O

OX

Y

P(20, 10)

Q x y( , )

20

x

030

DC

10y

�

�


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Question 8

Air of mass 1 kg, initially at 300 K and 10 bar, is allowed to expand isothermally till it reaches a pressure of 1 bar. Assuming air as an ideal gas with gas constant of 0.287 kJ/kg K, the change in entropy of air (in kJ/kg K, round off to two decimal place) is _______.

Ans. 0.6608

Sol. Given : 1 10 bar 1000 kPap , 1 300 KT and 2 1 bar 100 kPap

311

1

1 0.287 3000.0861 m

1000

mRTV

p

1 1 2 2p V p V (for isothermal process)

21000 0.0861 100 V

32 0.861 mV

Change in entropy, 2

1

0.861ln 1 0.287 ln 0.6608 kJ/kg K

0.0861

VS mR

V

Hence, the change in entropy of air is 0.6608 kJ/kg K.

Question 9

Consider an ideal vapor compression refrigeration cycle. If the throttling process is replaced by an isentropic expansion process, keeping all the other processes unchanged, which one of the following statements is true for the modified cycle?

(A) Coefficient of performance is higher than that of the original cycle.

(B) Coefficient of performance is lower than that of the original cycle.

(C) Coefficient of performance is same as that of the original cycle.

(D) Refrigeration effect is lower than that of the original cycle.

Ans. (A)

Sol. If isenthalpic expansion is replaced by an isentropic expansion, then the simple expansion value being using in VCRS has to be replaced by an isentropic expander which will add cost.

But, the isentropic expander will produce some positive work.

Hence, net c eW W W (isentropic expansion)

net cW W (isenthalpic expansion)

Also, from T-S diagram, E

net

QCOP

W

( ) ( )E h c E s cQ Q

( ) ( )net h c net s cW W

Hence, [ ] [ ]s c h cCOP COP


T

S

3

2

4’ 4

B A

Increase inrefrigeratingeffect

S C=S C=

p C=

p C=


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Question 10

During a non-flow thermodynamic process (1-2), executed by a perfect gas, the heat interaction is equal

to the work interaction, 1-2 1-2( )Q W . When the process is

(A) Adiabatic (B) Isothermal (C) Isentropic (D) Polytropic

Ans. (B)

Sol. From first law for process, 1-2 1-2Q U W

For isothermal process, assuming air as ideal gas,

0U

1-2 1-2Q W

Only for isothermal process executed by an ideal gas.


Question 11

Which one of the following welding methods provides the highest heat flux (W/mm2)?

(A) Laser beam welding (B) Oxy-gas acetylene welding

(C) Plasma arc welding (D) Tungsten inert gas welding (TIG)

Ans. (A)

Question 12

A spur gear with 200 full depth teeth is transmitting 20 kW at 200 rad/s. The pitch circle diameter of the gear is 100 mm. The magnitude of the force applied on the gear in the radial direction is

(A) 0.73 kN (B) 1.39 kN (C) 0.36 kN (D) 2.78 kN

Ans. (A)

Sol. Given : 020kW, 20P , 200 rad/s and 3100 mm 100 10 md

Power, .P T

20 200T

0.1 kNmT

3

3

0.1 102000 N

50 102

T

TF

d

Radial force, tanR TF F

02000 tan 20 727.94 N 0.73 kNRF


Question 13

A cylindrical rod of diameter 10 mm and length 1.0 m is fixed at one end. The other end is twisted by an angle of 100 by applying a torque. If the maximum shear strain in the rod is 310 ,p then p is equal to ____

(round off to two decimal places)

Ans. 0.872


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Sol. Given : 010 10 rad180

,

105mm

2R , 310p and 1m 1000mmL

BC

L BC L

Also, angle of twist, BC L

R R

310 1000

10180 5

p

0.872p

Hence, the maximum shear strain in the rod is 310 ,p then p is equal to 0.872.

Question 14

A flat faced follower is driven using a circular eccentric cam rotating at constant angular velocity . At time 0t , vertical position of follower is (0) 0y and the system is in the configuration shown below.

The vertical position of follower face ( )y t is given by

(A) (1 cos 2 )e t (B) (1 cos )e t (C) sine t (D) sin 2e t

Ans. (B)

Sol. A flat face follower,

ee

y t( )

O

Q�

P

eeecos�

x

x

ee

LL

�R

AB

CC

B

CC


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Vertical lift, cosx e e If is constant angular accelerator then,

t

t , cosx e e t (1 cos )e t

Hence, the correct option is (B). Question 15

A block of mass 10 kg rests on a horizontal floor. The acceleration due to gravity is 9.81 m/s2. The coefficient of static friction between the floor and the block is 0.2. A horizontal force of 10 N is applied on the block as shown in the figure. The magnitude of force of friction (in N) on the block is _______.

Ans. 10

Sol. A block of mass,

0yF , 0N W

98.1NN

Now, 0.2 98.1 19.62 NN

But, applied load, 10 19.62 NP

Hence, friction force will be equal to applied load 10 N. Hence, the magnitude of force of friction on the block is 10 N. Question 16

In a casting process a vertical channel through which molten metal flows downward, from pouring basin to runner for reaching the mould cavity is called

(A) Blister (B) Riser (C) Sprue (D) Pinhole Ans. (C)

Question 17

As per common design practice, three types of hydraulic turbines in decreasing order of flow rate are (A) Pelton, Francis, Kaplan (B) Francis, Pelton, Kaplan (C) Kaplan, Francis, Pelton (D) Francis, Kaplan, Pelton Ans. (C)

Sol. Kaplan Francis PeltonQ Q Q


A slender rod of length L and diameter ( )d L d and thermal conductivity 1k is joined with another rod

of identical dimensions, but of thermal conductivity 2k , to form a composite cylindrical rod of length 2L.

The heat transfer in radial direction and contact resistance are negligible. The effective thermal conductivity of the composite rod is

(A) 1 2

1 2

2k k

k k (B) 1 2

1 2

k k

k k (C) 1 2k k (D) 1 2k k

Ans. (A)

10 kg 10 N

10 9.81 98.1 NW � � �

N


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Sol. 1 2th th thR R R

1 2

2

e

L L L

k k k

1 2

1 2 1 2

2 1 1

e

k k

k k k k k

1 2

1 2

2e

k kk

k k

Hence, the correct option is (A). Question 19

The natural frequencies corresponding to the spring-mass systems I and II are 1 and 2 , respectively.

The ratio 1

2

is

System 1 System 2

(A) 1

4 (B) 4 (C)

1

2 (D) 2

Ans. (C)

Sol.

Fig. (i) Fig. (ii)

From figure (i), 1

1 1 1 2

ek k k k

1 2e

kk

From figure (ii), 2

2ek k k k

Natural frequency, 1

1

122

e

kk k

m m m …(i)

2

2

22ek k k

m m m …(ii)

1

2

1/ 2 1

22


mk k

m

k

k

mk k

1m

k

k

k1 k2

L L

T�

dQ

ke

2L

T�

Q


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Question 20

During a high cycle fatigue test a metallic specimen is subjected to cyclic loading with a mean stress of +140 MPa, and minimum stress of – 70 MPa. The R-ratio (minimum stress to maximum stress) for this cyclic loading is _________ (round off to one decimal place)

Ans. – 0.2

Sol. Given : min140, 70mean

max min

2mean

max ( 70)140

2

max280 70

max 280 70 350

Ratio, min

max

70 10.2

350 5

Hence, the R-ratio for this cyclic loading is – 0.2. Question 21

A Solid cube of side 1 m is kept at a temperature of 032 C. The coefficient of linear thermal expansion of

the cube material is 5 01 10 C , and the bulk modulus is 200GPa. If the cube is constrained all around

and heated to 042 C, then the magnitude of volumetric (mean) stress (in MPa) induce due to heating is

________. Ans. 60

Sol. Strain in x direction due to temperature rise, x T … (i)

Strain due to compressive stress offered by the wall,

2x E E

2 1x E

…(ii)

2 1TE

1 2

E T

3 1 2E k

5 3 1 21 10

1 2

k T

5 91 10 3 200 10 42 32

6 260 10 N/m 60 MPa 60 MPa

Hence, the magnitude of volumetric (mean) stress induce due to heating is 60 MPa.


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Question 22

Water flows through pipe with velocity 4 ˆ m/s,v x y jt

where j is the unit vector in y direction,

at ( 0)t is in second and andx y are in metres. The magnitude of total acceleration at the point

( , ) (1,1)x y at 2t seconds is ______ m/s.

Ans. 3

Sol. Given : 4

0, , 0u v x y wt

0x

u u u ua u v w

t x y z

y

v v v va u v w

t x y z

4 4 4

ya x y x y x yt t t y t

2

4 4x y

t t

At, 2sec, 1, 1t x y

22

4 41 1 1 2 2 3 m/s

2 2ya

0z

w w w wa u v w

t x y z

Magnitude of total acceleration 2 2 22 2ya a a 2 2 20 3 0 23 m/s

Hence, the magnitude of total acceleration ( , ) (1,1)x y at 2t seconds is 3 m/s2.

Question 23

For a hydrodynamically and thermally fully developed laminar flow through a circular pipe of constant

cross section, the nusselt number at constant wall heat flux ( )qNu and that at constant wall temperature

( )TNu are related as

(A) q TNu Nu (B) q TNu Nu (C) q TNu Nu (D) 2( )q TNu Nu

Ans. (C)

Sol. For constant heat flux, 4.36qNu

For constant wall temperature,

3.66TNu

1.191q

T

Nu

Nu

q TNu Nu



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Question 24

Consider the stress-strain curve for an ideal elastic- plastic strain hardening metal as shown in the figure. The metal was loaded in uniaxial tension starting from O. Upon loading, the stress-strain curve passes through initial yield point at P, and then strain hardens to point Q, where the loading was stopped. From point Q, the specimen was unloaded to point R, where the stress is zero. If the same specimen is reloaded in tension from point R, the value of stress at which the material yields again is ________ MPa.

Ans. 210

Sol. Initial loading upto Y.P. and then unloading to zero load results in cold working of the material. As a result, Y.S. increases on immediate next reloading. Since it is ideal elastic plastic, material Y.S. on reloading of the specimen remains at 210 MPa.

Hence, the value of stress at which the material yields again is 210 MPa. Question 25

The length, width and thickness of a steel sample are 400 mm, 40 mm and 20 mm, respectively. Its thickness needs to be uniformly reduced by 2 min in a single pass by using horizontal slab milling. The milling cutter (diameter : 100 mm, width : 50 mm) has 20 teeth and rotates at 1200 rpm. The feed per tooth is 0.05 mm. The feed direction is along the length of the sample. It the over travel distance is the same as the approach distance, the approach distance and time taken to complete the required machining task are

(A) 14 mm, 21.4 s (B) 21 mm, 39.4 s (C) 14 mm, 18.4 s (D) 21 mm, 28.9 s Ans. (A)

Sol. Given : Cutter diameter ( ) 100mmD

Cutter width 50mm , Number of teeth ( ) 20z ,

Depth of cut ( ) 2mmd , Rotational speed ( ) 1200rpmN ,

0.05mmf per tooth

Approach distance, 2 2

( )2 2

D DA d d D d

2 (100 2) mm =14 mmA

Time for one pass in slab or slot milling

L A

fzN

(for rough milling)

2L A

fzN

(for finish milling)

210

180

Q

O R Strain

Str

ess

(MP

a)

W = 400 mm

t = 20 mm

L = 400 mm


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As length of approach and over travel are same

400 14 14

min 21.4s0.05 20 1200

L A A

fzN


Q.26 to Q.55 Carry Two Marks Each

Question 26

A project consists of six activities. The immediate predecessor of each activity and the estimated duration is also provided in the table below :

Activity Precedence Activity time (in weeks)

P - 5

Q - 1

R Q 2

S P, R 4

T P 6

U S, T 3

If all activities other than S take the estimated amount of time, the maximum duration (in weeks) of the activity S without delaying the completion of the project is ________.

Sol. 6

Critical path of network :

1st longest path of network, 1 2 5 6 5 6 3 14 weeks

Total project duration is 14 weeks.

2nd longest path of network, 1 2 4 5 6 5 0 4 3 12 weeks

Maximum delay for S will be, 14 12 2 weeks

Hence, maximum duration of S is, 4 2 6 weeks

Hence, the maximum duration of the activity S without delaying the completion of the project is 6 weeks.

51

PU

3

2

4

5 6

3

6

T

Q

1R

2

S

4


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Question 27

Five jobs (J1, J2, J3, J4 and J5) need to be processed in a factory. Each job can be assigned to any of the five different machines (M1, M2, M3, M4 and M5). The time durations taken (in minutes) by the machines for each of the jobs, are given in the table. However, each job is assigned to a specific machine in such a way that the total processing time is minimum. The total processing time is ________ minutes.

M1 M2 M3 M4 M5

J1 40 30 50 50 58

J2 26 38 60 26 38

J3 40 34 28 24 30

J4 28 40 40 32 48

J5 28 32 38 22 44

Ans. 146

Sol.

M1 M2 M3 M4 M5

J1 40 30 50 50 58

J2 26 38 60 26 38

J3 40 34 28 24 30

J4 28 40 40 32 48

J5 28 32 38 22 44

By selecting minimum value of each row and then subtracting it from each value

M1 M2 M3 M4 M5

J1 10 0 20 20 28

J2 0 12 34 0 12

J3 16 10 4 0 6

J4 0 12 12 4 20

J5 6 10 16 0 22

Now applying same as row operation in column by taking value of column in which no value is zero.

M1 M2 M3 M4 M5

J1 10 0 16 20 22

J2 0 12 30 0 6

J3 16 10 0 0 0

J4 0 12 8 4 14

J5 6 10 12 0 16

From the above table it is seen that only 4 assignment is in process which is not applicable.

So, we make a new table for the assignment by taking horizontal and vertical line for covering maximum zero.


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New table is made by taking smallest value from uncovered value and subtracting from uncovered value

and adding on junction value.

Applying row and column operation,

M1 M2 M3 M4 M5

J1 16 0 16 26 22

J2 0 6 24 0 0

J3 22 10 0 6 0

J4 0 6 2 4 8

J5 6 4 6 0 10

Then, the total job assigned are,

M1 M2 M3 M4 M5

J1 30

J2 38

J3 28

J4 28

J5 22

Hence, the total minimum processing time 30 38 28 28 22 146

Hence, the total processing time is 146.

Question 28

The set of equations

1x y z

3 5ax ay z

5 3 6x y az

has infinite solution if a =

(A) 4 (B) 3 (C) – 3 (D) – 4

Ans. (A)

M1 M2 M3 M4 M5

J1

J2

J3

J4

J5


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Sol. Given set of equations in matrix form,

1 1 1 1

3 5

5 3 6

x

a a y

a z

Forming the augmented matrix [ : ]A B ,

1 1 1 : 1

[ : ] 3 : 5

5 3 : 6

A B a a

a

For infinite solutions, ( ) ( : )A A B Number of variables.

3 3 1 2( )R R R R ,

1 1 1 : 1

[ : ] 3 : 5

5 ( 1) 3 (1 ) (1 3) : 6 (1 5)

A B a a

a a a

For, ( ) ( : ) 3A A B , 3R should be zero.

5 ( 1) 0a

3 (1 ) 0a

(1 3) 0a

All satisfies for 4a .

Hence, for 4a ,

( ) ( : ) 2A A B and the set of equations will have infinite solutions.


Question 29

The variable X takes a value between 0 and 10 with uniform probability distribution. The variable Y takes a value between 0 and 20 with uniform probability distribution. The probability of the sum of variable (X + Y) being greater than 20 is

(A) 0.25 (B) 0.5 (C) 0 (D) 0.33

Ans. (A)

Sol. Given : X uniformly distributed between 0 and 10.

X

P xX ( )

10

1/10


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Y uniformly distributed between 0 and 20.

: Method 1 :

Given 20x y 20y x

10 20

0 20

( 20) ( , )x

P x y P x y dy dx

10 20

0 20

1 1

10 20x

dy dx

[As X and Y are independent ( , ) ( ) ( )X YP x y P x P y ]

10

20

200

1( 20)

200 xP x y y dx

10

0

120 (20 )

200x dx

1010 2

00

1 1( 20)

200 200 2

xP x y x dx

1 1

( 20) [100 0] 0.25400 4

P x y


: Method 2 :

Let Z X Y , then probability distribution function of Z is convolution of PDF’s of X and Y.

Probability of 20Z is area of PDF between 20 and 30,

1 1 1

10 0.252 20 4


Y

P yY ( )

20

1/20

x

f xX ( )

10

1/10

y

f yY ( )

20

1/20

z

f zZ ( )

30

1/20

2010

Requiredarea


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Question 30

A harmonic function is analytic if it satisfies Laplace equation, If 2 2( , ) 2 2 4u x y x y xy is a harmonic

function, then its conjugate harmonic function ( , )v x y is

(A) 24 4 constanty xy (B) 2 24 2 2 constantxy x y

(C) 2 24 2 2 constantxy y x (D) 2 22 2 constantx y xy

Ans. (B)

Sol. Given : 2 2( , ) 2 2 4u x y x y xy ,

We have to time ( , )v x y

As the function is harmonic, so

u v

x y

[It must satisfy Cauchy Riemann equation] …(i)

4 4u

x yx

Checking from options,

(A) 24 4v y xy c

4 4v u

y xy x

(B) 2 24 2 2v xy x y c

4 4v u

x yy x

[Satisfies Cauchy Riemann equation]

(C) 2 24 2 2v xy y x c

4 4v u

x yy x

(D) 2 22 2v x y xy c

4v u

y xy x


Question 31

The value of the following definite integral is ______ (round off to three decimal places).

1

(ln )ex x dx

Ans. 2.097


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Sol. 2 2

1 11

1(ln ) ln

2 2

ee ex xx x dx x dx

x

2 2

1

1 1ln ln 1

2 2 2 2

e

e e

e xe

2 2 2

2

1

1 1(ln ) [ 1]

2 4 2 4 4

e e e ex x dx e

2 12.097

4 4

e

Hence, the value of the definite integral is 2.097.

Question 32

Match the following sand mold casting defects with their respective causes

Defect Cause

P. Blow hole 1. Poor collapsibility

Q. Misrun 2. Mold erosion

R. Hot tearing 3. Poor permeability

S. Wash 4. Insufficient fluidity

(A) P-4, Q-3, R-1, S-2 (B) P-3, Q-4, R-1, S-2

(C) P-2, Q-4, R-1, S-3 (D) P-3, Q-4, R-2, S-1

Ans. (B)

P. Blow hole Poor permeability

Q. Misrun Insufficient fluidity

R. Hot tears Poor collapsibility

S. Wash Mould erosion.


Question 33

A Gas is heated in a duct as it flows over resistance heater. Consider a 101 kW electric heating system. The gas enters the heating section of the duct at 100 kPa and 27 0C with a volume flow rate of 15 m3/s. If heat is lost from the gas in the duct to the surroundings at a rate of 51 kW, the exit temperature of the gas is

(Assume constant pressure, ideal gas negligible change in kinetic and potential energies and constant

specific heat : 1 kJ/kg KpC , 0.5 kJ/kg KR )

(A) 53 0C (B) 37 0C (C) 76 0C (D) 32 0C

Ans. (D)

Sol. Given : 31 15m /sV , 1 100kPap , 1kJ/kgK,pc 1 300K, 0.5kJ/kgKvT c

and 0.5 kJ/kgKp vc c R .


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So, volume of gas at inlet,

11

1

RTv

P 30.5 300

1.5m /kg100

Mass flow rate, 1

1

1510 kg/s

1.5m

v

V

Now, using SFEE,

1 2mh Q mh W (neglecting KE , PE and heat being lost)

1 2mh mh Q W

1 210[ ( )] 50pc T T

21 (300 ) 5T

2 305 KT or 02 305 273 32 CT

Hence, the correct option is (D).

Question 34

A truss is composed of members AB, BC, CD, AD and BD, as shown in the figure. A vertical load of 10 kN is applied at point D. The magnitude of force (in kN) in the member BC is ______.

Ans. 5

Sol.

0,yF 10A CR R

0AM , 2 10 0CR x x

5kNCR

D

CA 450

450

B

10 kN

D

CA 450

450

B

10 kN

450

450

RCRA

x

x

T1 T2

Q = 51 kW

Heating coil= 101 kWW


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Applying sing rule,

0 0

5

sin(45 ) sin(45 )BCF

5kNBCF

Hence, the magnitude of force in the member BC is 5 kN.

Question 35

Consider a prismatic straight beam of length m,L pinned at the two ends as shown in the figure. The

beam has a square cross-section of side 6 mmp . The Young’s modulus 200GPa,E and the

coefficient of thermal expansion 6 13 10 K . The minimum temperature rise required to cause Euler

buckling of beam is ______ K.

Ans. 1

Sol. Given : Square side (p) = 6 mm 36 10 m , 6 03 10 / C and 200 GPaE

Crippling load, 2

max2c

EIP

L

A E A T E A

2

2

EIT E A

4

2

12

pTp

4

12

pT

2 3 2

6

(6 10 )1 K

12 3 10 12

pT

d

Hence, the minimum temperature rise required to cause Euler buckling of beam is 1 K.

Question 36

Taylor’s tool life equation is given by nVT c , where V is in m/min and T is in min. In a turning operation two tools X and Y are used. For X , 0.3n and 60c and for Y , 0.6n and 90c . Both the tools will have same tool life for cutting speed (in min, round off to two decimal places) of ______.

Ans. 40

L

p

p

L = � m

2

4

12

A p

pI

�

�

2700

2700

5 kNFDC

FBC

450

450


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Sol. Given : For tool X : 1 10.3, 0.60n c

For tool Y : 2 20.6, 90n c

Now, apply Taylor’s tool life equation,

nVT c

For tool ,X 11 1 1

nV T c

1/0.3

11

60T

V

…(i)

For tool Y , 22 2 2

nV T c

1/0.6

22

90T

V

…(ii)

Now, at breakeven point, 1 2T T and 1 2V V V

So, from equation (i) and (ii),

1/0.3 1/0.6

60 90

V V

1 11/0.3

0.3 0.61/0.6

60

90V

5/3 467.843V

3/5467.843 40 m/minV

Hence, the tools have same tool life for cutting speed of 40 m/min. Question 37

A circular shaft having diameter 0.010.0565.00 mm is manufactured by turning process. A 50 m thick

coating TiN is deposited on the shaft. Allowed variation in TiN film thickness is 5 m. The minimum

hole diameter (in mm) just to provide clearance fit as (A) 65.12 (B) 65.01 (C) 65.10 (D) 64.95 Ans. (A)

Sol. Given : Shaft diameter 0.010.05( ) 65.00d mm

Shaft

Hole

0.055 mm

0.055 mm

?D �65 mm


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The minimum hold diameter 65.01 0.055 0.055 65.12mm

Hence, the correct option is (A). Question 38

A single block brake with a short shoe and torque capacity of 250 N-m is shown. The cylindrical brake drum rotates anticlockwise at 100 rpm and the coefficient of friction is 0.25. The value of ,a in mm (round

off to one decimal place), such that the maximum actuating force P is 2000 N, is ______

Ans. 212.5

Sol. Given : Torque ( ) 250 NmT , Radius of drum = a, Speed ( ) 100rpmN ,

Maximum applied force ( ) 2000 NP and Coefficient of friction ( ) 0.25 .

FBD :

T F a

250 F a

250

F Na

250 1000

0.25N

a a

0oM , 1000 0.25 1000

2000 2.5 04

aa a

a a

0.2125 m 212.5 mma

Hence, the value of ,a in mm, such that the maximum actuating force P is 2000 N, is 212.5 mm.

a

a/4

P 1.5a a

a

a/4

P = 2000 N

1.5a a

O

a/4

P 1.5a a

N

�N


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Question 39

A cube of side 100 mm is placed at the bottom of an empty container on one of its faces. The density of

cube material is 3800 kg/m . Liquid of density is 31000 kg/m is now poured into the container. The

minimum height to which the liquid needs to poured into the container for the cube to just lift up is _______ mm.

Ans. 80

Sol. Given : Side of cube = 100 mm.

Let, x is the height of water required to just lift the block,

bF W

L L s sV g V g

1000 (0.1 0.1) 800 (0.1 0.1 0.1)x

800 0.1

0.08m 80 mm1000

x

Hence, the minimum height to which the liquid needs to poured into the container for the cube to just lift up is 80 mm.

Question 40

Consider an elastic straight beam of length 10 m,L with square cross-section of side 5 mm,a and

Young’s modulus 200 GPa.E This straight beam was bent in such a way that the two ends meet, to

form a circle of mean radius R. Assuming that Euler-Bernoulli beam theory is applicable to this bending problem, the maximum tensile bending stress in the bent beam is _____ MPa.

Ans. 100

Sol. Given : 352.5 mm 2.5 10 m

2y , 3200 10 MPaE and 10 mL

Now, circumference of neutral plane,

2 R L

2 10 5mR R

E

yR

3 3200 10 2.5 10

100 MPa5

Hence, the maximum tensile bending stress in the bent beam is 100 MPa.

L

R

Ends if bean

xx

F V gb L L=

W V g= s s

L

R

Ends if bean

a


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Question 41

Two immiscible, incompressible, viscous fluids having same densities but different viscosities are contained between two infinite horizontal parallel plates, 2m apart as shown below. The bottom plate is fixed and the upper plate moves to the right with a constant velocity of 3 m/s. With the assumptions of Newtonian fluid, steady and fully developed laminar flow with zero pressure gradient in all directions, the momentum equations simplify to

2

20

d u

dy

If the dynamic viscosity of the lower fluid 2 is twice that of the upper fluid, 1 , then the velocity at the

interface (round off to two decimal places) is _________ m/s.

Ans. 1

Sol. Velocity profile is laminar in both the fluids,

2

20

d u

dy

1

duc

dy

1 2u c y c

i.e. we can assume linear velocity profile.

If velocity profile is linear, then shear stress will be constant in gap everywhere.

Let u is the velocity at interface.

At interface, 1 2

1 2

1 2

du du

dy dy

1 1

3 02

2 1 1 0

u u

On solving, 1 m/su

Hence, the velocity at the interface is 1 m/s.

Question 42

In ASA system, the side cutting and end cutting edge angles of a sharp turning tool are 045 and 010 respectively. The feed during cylindrical turning is 0.1 mm/rev. The centre line average surface roughness (in m , round off to two decimal place) of the generated surface is _______.

Ans. 3.75

2 m

1 m

�1�1

�2�2

�� 2 1�� 2 1

x

y

2 m

�1�1

�2�2

1 m

1 m

3 m/s

u

Shear stressdistribution

Velocitydistribution

u = 0

�


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Sol. Given : 0 045 , 10S e and 0.1 mm/rev.f

Peak to valley height,

max tan cotS E

fH

max 0 0

0.10.015mm

tan 45 cot10H

The centre line average surface roughness,

3max 0.0153.75 10 mm 3.75 m

4 4a

HR

Hence, the centre line average surface roughness of the generated surface is 21.2 μm .

Question 43

In a four-bar planar mechanism shown in figure, 5 cm, 4 cm and 2 cm.AB AD DC In the

configuration shown, both AB and DC are perpendicular to AD . The bar AB rotates with angular velocity of 10 rad/sec. The magnitude of angular velocity (in rad/s) of bar DC at this instant is

(A) 10 (B) 0 (C) 15 (D) 25

Ans. (D)

Sol. Given : 5 cm, 4 cm, 2 cmAB AD DC and 10 rad/secAB .

Fig. Velocity diagram

a d,b c,

VB

VC


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For such situation, B CV V

AB CDAB CD

10 5 2 CD

25 rad/sCD

Hence, the correct option is (D).

Question 44

At a critical point in a component, the state of stress is given as 100MPaxx , 200yy Mpa,

80MPaxy yx and all other stress components are zero. The yield strength of the material is 468

MPa. The factor of safety on the basis of maximum stress theory is _______ (round off to one decimal place).

Ans. 1.8

Sol. Given : 100 MPaxx y , 220MPayy x and 80MPaxy yx xyz .

Major principal stress, and minor principal stress

2

21,2

( )

2 2x y x y

xy

2

1,2

100 220 100 22080

2 2

Taking positive sign,

1 260 MPa and 2 60 MPa

1 2 1 2max max , ,

2 2 2

1max

260130 MPa

2 2

As per maximum shear stress theory,

max 2ytS

FOS

468234 MPa

2 2yt

ys

SS

max

4681.8

2 130 2ytS

FOS

Hence, the factor of safety on basis of maximum stress theory is 1.8.


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Question 45

Three slabs are joined together as shown in the figure. There is no thermal contact resistance at the interfaces. The center slab experiences a non-uniform internal heat generation with an average value equal

to 310000 Wm , while the left and right slabs have no internal heat generation. All slabs have thickness

equal to 1 m and thermal conductivity of each slab is equal to 1 15Wm K . The two extreme faces are

exposed to fluid with heat transfer coefficient 2 1100 Wm K and bulk temperature 030 C as shown.

The heat transfer in the slabs is assumed to be one dimensional and steady, and all properties are constant.

If the left extreme face temperature 1T is measured to be 0100 C, the right extreme face temperature 2T is

_____ 0 C .

Ans. 60

Sol.

10000 ( )gE L A 10000 1 1 10000 W

1 11( ) conv condhA T T q q

1

100 (100 30) condq

1

7000 Wcondq

2 2( )convq hA T T 2100 1 ( 30)T

2 22100( 30)conv condq T q

At steady state, g outE E

1 2g cond condE q q

210000 7000 100( 30)T

1 m 1 m 1 mLeft extremeface = 100 CT1

0

100 W/m K2

30 C0

100 W/m K2

30 C0

T2

T1 = 100 C0

h = 100 W/m K2

T2 = ?

T� = 30

qcond 1 qcond 2

T� = 30

h = 100 W/m K2

qcond 1 qcond 2

g gE q��

�

1 m 1 m 1 m


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33



02

300030 60 C

100T

Hence, the right extreme face temperature 2T is 600C.

Question 46

A steam power cycle with regeneration as shown below on the T-s diagram employs a single open feedwater heater for efficiency improvement. The fluids mix with each other in an open feedwater heater. The turbine is isentropic and the input (bleed) to the feedwater heater from the turbine is at state 2 as shown in the figure. Process 3-4 occurs in the condenser. The pump work is negligible. The input to the boiler is at state 5. The following information is available from the steam tables :

State 1 2 3 4 5 6

Enthalpy (kJ/kg) 3350 2800 2300 175 700 1000

The mass flow rate of steam bled from the turbine as a percentage of the total mass flow rate at the inlet

to the turbine at state 1 is ________.

Ans. 20

Sol. From energy balance equation,

2 4 5(1 ) 1mh m h h

2 4 4 5mh m h h h

2 4 5 4( )m h h h h

5 4

2 4( )

h hm

h h

700 175

100 0.2or 20%(2800 175)

m

Hence, the mass flow rate of steam bled from the turbine as a percentage of the total mass flow rate at the inlet to the turbine at state 1 is 20.

Question 47

A plane-strain compression (forging) of a block is shown in the figure. The strain in the z- direction is

zero. The yield strength ( )yS in uniaxial tension/ compression of the material of the block is 300 MPa and

it follows the Tresca (Maximum shear stress) criterion. Assume that the entire block has started yielding.

At a point where 40MPax (compressive ) and 0xy the stress component y is

T

S

43

2

1

5

6

T

s

43

2

1

5

6

m�

(1 ) kg/sm� �

1

FWH

2

45

1

m�(1 )m� �


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34



(A) 340 MPa (compressive) (B) 260 MPa (compressive)

(C) 340 MPa (Tensile) (D) 260 MPa (Tensile)

Ans. (A)

Sol. Given : 40 MPax , 300 MPaytS .

According to Tresca’s theory,

2 2

x y ytS

FOS

40 300y

340 MPa 340 MPay (compressive)


Question 48

In UTM experiment, a sample of length 100 mm, was loaded in tension until failure. The failure load was 40 kN. The displacement, measured using the cross-head, at failure, was 15 mm. The compliance of the

UTM is constant and is given by 85 10 m/N . The strain at failure in the sample is __________%.

Ans. 2

Sol. In UTM experiment under tension ( ) 100mmL

Failure load ( ) 40kNP

Compliance of UTM is constant 85 10 m/N = ConstantL

AE

We know that, PL

AE

Strain 1L

PL AE L

3 8

23

40 10 5 10Strain 2 10

100 10

22 10 100% 2%

Hence, the strain at failure in the sample is 2%.

Question 49

A gas turbine with air as the working fluid has an isentropic efficiency of 0.70 when operating at a pressure ratio of 3. Now, the pressure ratio of the turbine is increased to 5, while maintaining the same inlet conditions. Assume air as a perfect gas with specific heat ratio 1.4 . If the specific work output remains

the same for both the cases, the isentropic efficiency of the turbine at the pressure ratio of 5 is ______ (round off to two decimal places).

Ans. 0.514

Movingplaten

y

xO

Fixed platen


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Sol. Given : 1 0.7 , 1

1

3

4

3ps

pr

p and

2

2

3

4

5ps

pr

p .

We know that, isentropic efficiency of the turbine is,

act

ideal

W

W

Since, actW or specific work output is same for both the cases.

1 3 4 1 3 4 12

1 2 3 4 2 3 4 2

( ) ( ) ( )

( ) ( ) ( )ideal s s

ideal s s

W h h T T

W h h T T

1

1

2

2

14

32

41

3

11

1

10.7 11

s

p

s

p

T

rTT

Tr

0.4

1.4

2

0.4

1.4

11

3 0.7 0.5141

1

5

Consider it is asked only for turbine portion as for full cycle, data is not sufficient.

Hence, the isentropic efficiency of the turbine at the pressure ratio of 5 is 0.514.

Question 50

In orthogonal turning of a cylindrical tube of wall thickness 5 mm, the axial and tangential cutting forces were measured as 1259 N and 160 1 N, respectively. The measured chip thickness after machining was

found to be 0.3 mm. The rake angle was 010 and axial feed was 100 mm/min. The rotational speed of the spindle was 1000 rpm. Assuming the material to be perfectly plastic and Merchant’s first solution, the shear strength of the material is closest to

(A) 875 MPa (B) 920 MPa (C) 722 MPa (D) 200 MPa

Ans. (C)

Sol. Given : Orthogonal turning 090

Wall thickness Depth of cut ( ) 5mmd (assumed)

Axial force, 1259 N, xF 1259 Nsin 90

xT x

FF F

Tangential force, 1601Nz CF F

0.3mmct , 010 , 100mm / minf N

1000 rpm or 1000 100or 0.1mm/revf f

s

T

4s

4

3


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36



0sin 0.1sin 90 0.1mmt f

5

5mmsin sin 90

db

0.1

0.330.3c

tr

t

0

00

cos 0.33cos10tan or 19.02

1 sin 1 0.33sin10

r

sin ( cos sin ) sins s C T

ss

F F F F

A bt bt

(1601cos19.02 1259sin19.02) sin19.02

5 0.1s

719.12 MPa 722 MPas


If one mole of 2H gas is occupies a rigid container with a capacity of 1000 litres and the temperature is

raised from 027 C to 037 C , the change in pressure of the contained gas (round off to two decimal places), assuming ideal gas behaviour is ______ Pa.

Ans. 83.14

Sol. Given : Volume of rigid container V , Initial temperature of the gas 1T and

Final temperature of the gas 2T

1 1PV nRT

2 2PV nRT

2 1 2 1( ) ( )P P V nR T T

1 1 8.314 (37 27)P

83.14 PaP

Hence, the change in pressure of the contained gas (round off to two decimal places), assuming ideal gas behaviour is 83.14 Pa.

Question 52

The wall of a constant diameter pipe of length 1 m is heated uniformly with flux q” by wrapping a heater coil around it. The flow at the inlet to the pipe is hydro-dynamically fully developed. The fluid is incompressible and the flow is assumed to be laminar and steady all through the pipe. The bulk

temperature of the fluid is equal to 00 C at the inlet and 050 C at the exit. The wall temperatures are measured at three locations, ,P Q and R as shown in the figure. The flow thermally develops after some

distance from the inlet. The following measurements are made :

Point P Q R

Wall temperature ( 0C ) 50 80 90


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37



Among the locations P, Q and R the flow is thermally developed at (A) P, Q and R (B) Q and R only (C) R only (D) P and Q only Ans. (B)

Sol. In case of uniform heat flux, bulk mean temperature varies linearly. The difference between bulk mean temperature and wall temperature is constant in thermally developed region so bulk temperature,

( )T x A Bx

At 0,x 00 CT

00 CA

At 1x , 050 CT

050 CB ( ) 50T x x

Location Wall temperature ( )wT Bulk temperature ( )T x ( )wT T T x

, 0.4P x 050 C 020 C 030 C

, 0.6Q x 080 C 030 C 050 C

, 0.8R x 090 C 040 C 050 C

Hence, the correct option is (B). Question 53

A car is having weight W is moving in the direction as shown in the figure. The centre of gravity (CG) of the car is located at height h from the ground, midway between the front and rear wheels. The distance between the front and rear wheels, is l . The acceleration of the car is a , and acceleration due to gravity

is g. The reactions on the front wheels ( )fR and rear wheels ( )rR are given by,

(A) ,2 2f r

W W h W W hR a R a

g l g l

(B) 2f r

W W hR R a

g l

(C) 2f r

W W hR R a

g l

(D) ,2 2f r

W W h W W hR a R a

g l g l

0.2 m 0.2 m 0.2 m 0.2 m 0.2 m

P Q R0 C

050 C

0

Constant wall flux

W

CG

h

Rr Rfll

Direction of motiona


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38



Ans. (D)

Sol. Inertial force, iF ma in opposite direction of motion of car.

0VF , r fR R W …(i)

Taking moment about A,

2i f

WlF h R l

2f

Wlmah R l

2f

W W hR a

g l

…(ii)

From equation (i) and (ii),

2r

W W hR a

g l


A uniform thin disk of mass 1 kg and radius 0.1 m is kept on a surface as shown in the figure. A spring of

stiffness 1 400 N/mk is connected to the disk center A and another spring of stiffness 2 100 N/mk is

connected at point B just above point A on the circumference of the disk. Initially, both the springs are unstretched. Assume pure rolling of the disk. For small disturbance from the equilibrium, the natural frequency of vibration of the system is ______ rad/s (round off to one decimal place)

Ans. 23.094

Sol. Given : Disc mass 1 kgm and radius 0.1 mR

Torque equation about point ‘O’,

1 2( ) ( 2 )2 0I k R R k R R

I about 2

2

2

mRO mR

23

2I mR

2 2 21 2

3[ (4 )] 0

2mR k R k R

2 2

1 2

2

40

32

k R k R

mR

B

A

k1

k2

Wh

Rr Rfll

Fi

A B

�

R�

2R�

2 2k R� �

1k R� �

O


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39



2 2

1 2

2

43

2

n

k R k R

mR

2 2

2

400 0.1 4 100 0.123.094 rad/s

1.5 1 0.1n

Hence, the natural frequency of vibration of the system is 23.094 rad/s.

Question 55

The rotor of a turbojet engine of an aircraft has a mass 180 kg and polar moment of inertia 210 kgm about

the rotor axis. The rotor rotates at a constant speed of 1100 rad/s in the clockwise direction when viewed from the front of the aircraft. The aircraft while flying at a speed of 800 km per hour takes a turn with a radius of 1.5 km to the left. The gyroscopic moment exerted by the rotor on the aircraft structure and the direction of motion of the nose when the aircraft turns, are

(A) 162.9 Nm, the nose goes up (B) 1629.6 Nm, the nose goes down

(C) 1629.6 Nm, and the nose goes up (D) 162.9 Nm, and the nose goes down

Ans. (B)

Sol. Given : 180 kgm , 210 kg-mI , 1100 rad/s and 1.5 km 1500 mR .

5

800 km/hr 800 222.222 m/s18

V

0.14814 rad/sp

V

R

Gyroscopic couple pI 10 1100 0.14814 1629.629 Nm

Sign convention : C.W. ve

Front ve

Left ve

Resultant ( ve) ( ve) ( ve) ve

The resultant sign i.e., ve shows the nose goes down.


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