Solutions All MOCK Tests_ME

SolutionsME-A

Q.1 TO Q.25 CARRY ONE MARK EACH

MCQ 1.1 A bag contains 8 white and 6 red balls. The probability of drawing two balls of the same color is

(A) 9121 (B) 112

13

(C) 9143 (D) 91

41

SOL 1.1 Correct Option is (C)

Probability CC

CC

14 142

82

2

62= +

9128

9115

9143= + =

MCQ 1.2 Lt xe 1

x

x

0

−"

is

(A) 0 (B) 1

(C) 2 (D) 3

SOL 1.2 Correct Option is (B)

Let f x^ h lim xe 1

x

x

0= −

"

At x 0" , f x^ h is of the form of 00

^ h.

So, applying L-Hospital’s rule

f x^ h lime1x

x

0=

"

11 1=

MCQ 1.3 The value of x in the mean value theorem of ( ) ( ) ( ) ( )f b f a b a f x− = − for ( )f x Ax Bx C2= + + in ,a b^ h is

Page 2 ME-A Chapter 1

(A) b a+ (B) b a−

(C) b a

2+^ h

(D) b a

2−^ h

SOL 1.3 Correct option is (C).We have

f x^ h Ax Bx C2= + + f xl h Ax B2= +Thus from the mean value Theorem, there exist at least one point x c= &

,a b^ h such that

f Cl h b af b f a

= −−^ ^h h

AC B2 + b aAb Bb C Aa Ba C2 2

= −+ + − + +^ ^h h

b aA b a B b a2 2

= −− + −^ ^h h

AC B2 + A b a B= + +^ h

AC2 A b a= +^ h

C b a2= +

or x b a2= +

MCQ 1.4 The function f x x x x6 9 253 2= − + +^ h has (A) a maximum at x 1= and minimum at x 3=

(B) a maximum at x 3= and a minimum at x 1=

(C) no maximum, but a minimum at x 3=

(D) a maximum at x 1= , but no minimum

SOL 1.4 Correct Option is (C)We have

f x^ h x x x6 9 253 2= − + + f xl h x x3 12 92= − + ....(i)

f xm^ h x6 12= − ....(ii)

Substitute f xl h 0= x x3 12 92 − + 0= x x4 32 − + 0= x x x3 32 − − + 0= x x x3 1 3− − −^ ^h h 0= x ,1 3=

Chapter 1 ME-A Page 3

Substitute x 1= in equation (ii)

f xm^ h 6 0<=− (Maxima)

Again substitute x 3= in (ii), we have

f xm^ h 6 0>= (Minima)

So that f x^ h is maximum at x 1= and minimum at x 3=

MCQ 1.5 The value of lim sin tanx x1 1

x−

"3b l; E is

(A) 0 (B) 1

(C) 2 (D) 3

SOL 1.5 Correct Option is (D)

Let F x^ h lim sin tanx1 1

x α= −"3

b l

lim sin sincos

xx1

x α= −"3

b l

lim sincosx

x1x

= −"3

b l

limsin cos

sin2

2x x x

x

2 2

22=

"3

lim tanx2x

="3

b l

This limit does not exist as tanx2b l does not approach any value as x 3−

MCQ 1.6 The resultant of two forces acts along a line perpendicular to one force and

its magnitude is 21 of other force. What will be the angle between the forces

?(A) 54 cθ = (B) 09 cθ =

(C) 150cθ = (D) 1 02 cθ =

SOL 1.6 Correct Option is (C)As per problem statement the forces will be as shown in figure. Here

F1 ?= F2 ?=


R F22=

α 90c= θ ?=In right angle triangle OCB in figure.

cosBOC OCOB

FR

FF

21

2 2

21

2= = = =

BOC+ cos 21 601 c= =−

θ BOC 60 90 150c c c+ α= + = + =

MCQ 1.7 A ball of 2 kg, is dropped from a height of 15 cm on a spring of stiffness 980 /N mk = . What will be the maximum deflection of the spring?

(A) . cm12 5 (B) cm5

(C) 10 cm (D) . cm7 5

SOL 1.7 Correct Option is (C)When the ball falls through a height h on spring as shown in figure, then its potential energy is transferred to compress the spring. If compression in spring is x then total potential energy is due to height h x+ .

Thus mg h x+^ h kx21 2=

Substituting the values


. . x2 9 8 1 15# +^ h x21 980 2# #=

or . x0 15 + x25 2=or . .x x0 04 0 0062 − − 0= x 0.1 m=or x 10 cm=

MCQ 1.8 If Poisson’s ratio of a material is 0.5, then the elastic modulus for the material is(A) three times its shear modulus (B) for times its shear modulus

(C) equal to its shear modulus (D) indeterminate

SOL 1.8 Correct Option is (A)

E G2 1 μ= +^ h

put μ .0 5= E G3=

MCQ 1.9 A beam, built in both ends, carries a uniformly distributed load over its entire span as shown in figure. Which one of the diagram given below represents bending moment distribution along the length of the beam ?



MCQ 1.10 A thick cylinder is subjected to internal pressure of 100 /N mm2. If hoop stress developed at the outer radius of the cylinder is 100 /N mm2, the loop stress developed at the inner radius is(A) 100 /N mm2 (B) 200 /N mm2

(C) 300 /N mm2 (D) 400 /N mm2

SOL 1.10 Correct Option is (B)Hoop stress at inner radius p A2= + (to be found)Hoop stress at outer radius 2 100 /N mmA 2= = where A is lames constant.Therefore stress at inner radius 100 200 200 /N mm2= + =

Other method

hoop stress at outer radius d d

P d202

12

12

=−

100 d d

d2 10002

12

12

# #=−

d3 12 d0

2=Now hoop stress at inner radius

200 /N mmd d

p d dd

d2

100 402

12

1 02

12

12

12

2#=−+

= =^ h

MCQ 1.11 The gear train usually employed in clocks is a(A) reverted gear train (B) simple gear train

(C) sun and planet gear (D) differential gear

SOL 1.11 Correct Option is (A)In reverted gear train and last gear is on the same axis. Such an arrangement has application on speed reducers, clocks (to connect hour hand to minute


hand) and machine tools.

MCQ 1.12 Match List I with List II and select the correct answer using the codes given below the lists :

List-I List-II

a. Quadratic cycle chain 1. Rapson’s slide

b. Single slider crank chain 2. Oscillating cylinder engine mechanism

c. Double slider crank chain 3. Ackermann steering mechanism

d. Crossed slider crank chain 4. Oldham coupling

Codes : a b c d(A) 1 2 4 3(B) 4 3 1 2(C) 3 2 4 1(D) 3 4 2 1

SOL 1.12 Correct Option is (A)4-Bar Mechanism;(1) Quadric cycle chain " Ackermann steering

(2) Single slider crank chain " oscillating cylinder engine mechanism

(3) Double slider crank chain " Oldham coupling

(4) Crossed slider crank chain " Rapson’s slide

MCQ 1.13 A hydraulic power station has the following major items in the hydraulic circuit :1. Draft tube 2. Runner

3. Guide wheel 4. Penstock

The correct sequence of these items in the direction of flow is(A) 4, 2, 3, 1 (B) 4, 3, 2, 1

(C) 1, 2, 3, 4 (D) 1, 3, 2, 4



List-I List-II

a. Pelton wheel (single jet) 1. Medium discharge, low head

b. Francis Turbine 2. High discharge, low head

c. Kaplan Turbine 3. Medium discharge, medium head

4. Low discharge, high head


Codes : a b c(A) 1 2 3(B) 1 3 4(C) 4 1 3(D) 4 3 2

SOL 1.14 Correct Option is (D)Characteristics of Pelton wheel(i) Impulse turbine

(ii) High head turbine 300 2000 m−^ h

(iii) Low specific discharge turbine

(iv) Axial flow turbine

(v) Low specific speed turbine 4 70 rpm−^ h

Characteristic of Francis turbine(i) Reaction turbine

(ii) Medium head turbine m30 500−^ h

(iii) Medium specific discharge

(iv) Radial flow turbine, but modern francis turbine are mixed flow turbine.

(v) Medium specific speed turbine 60 40m rpm0−^ h

Characteristic of Kalpan turbine

(i) Reaction turbine

(ii) Low head turbine 2 m m70−^ h

(iii) High specific discharge

(iv) Axial loading

(v) High specific speed rpm300 1100−^ h

MCQ 1.15 A :1 20 model of a spillway dissipated 0.25 hp. The corresponding prototype horsepower dissipated will, be(A) 0.25 hp (B) 5.00 hp

(C) 447.20 hp (D) 8944.30 hp

SOL 1.15 Correct Option is (D)We haveLinear Scale ratio L 20r = , 0.25 hpPm =Power ratio for prototype and model is given by

PP

m

p L .r3 5=

Pp .20 0 25.3 5#= ^ h


8944.30 hp=

MCQ 1.16 A 1 kW electric resistance heater in a room is turned on and kept on for 50 minutes. The amount of energy transferred to the room by the heater is(A) 1 kJ (B) 50 kJ

(C) kJ3000 (D) kJ3600


W 1 kW= t min sec50 50 60 3000#= = =Thus amount of energy transferred to the room by Leather is

Q W t#= 1 3000 3000 kW#= =

MCQ 1.17 Lumped system analysis of transient heat conduction situations is valid when the Biot number is(A) very small (B) approximately one

(C) very large (D) any real number

SOL 1.17 Correct Option is (A)For the lumped system analysis the biot number is very small.

.Bi 0 1#

When .Bi 0 1< , the variation of temperature with location within the body

is slight and can be approximated as being uniform

MCQ 1.18 A machinist desires to turn a round steel stock of outside diameter 100 mm at 100 rpm. The material has tensile strength of 75 /kg mm2. The depth of cut chosen is 3 mm at a feed rate of 0.3 /mm rev. Which one of the following tool materials will be suitable for machining the component under the specified cutting conditions ?(A) Sintered carbides (B) Ceramic

(C) HSS (D) Diamond

SOL 1.18 Correct option is (B).Ceramic because ceramic tool should be used for low depth of cut and low feed rate but at very high speed, cutting takes place.

MCQ 1.19 In Powder metallurgy, the operation carried out to improve the bearing property of a bush is called(A) Infiltration (B) Impregnation

(C) Plating (D) Heat treatment

SOL 1.19 Correct option is (B).


The bearing property of a bush is improved by providing a film of lubricant. This is done by impregnation process in powder metallurgy. In this process the parts are immersed in a tank of heated oil for a period of time to fill the pores. The porous components impregnated with lubricants in this manner do not need external lubrication during operation.

MCQ 1.20 The purpose of a gas-welding flux is to :(A) Lower the melting point of the metal

(B) Lower the melting point of the oxide

(C) Remove oxides from the surface of the metal

(D) Remove elements from parent metal

SOL 1.20 Correct option is (C).Fluxes are required to increase the fluidity of the fusible iron silicate slag, as well as to aid in the removal of slag. Gas welding fluxes must melt at a lower temperature than the metals being welded so that surface oxides will be dissolved before the metal melts.

MCQ 1.21 Which of the following methods can be used for manufacturing 2 m long seamless metallic tubes ?1. Drawing 2. Extrusion

3. Rolling 4. Spinning

Select the correct answer using the codes given below :Codes :(A) 1 and 2 (B) 2 and 3

(C) 1, 3 and 4 (D) 2, 3 and 4

SOL 1.21 Correct option is (B).Rolling and extrusion are used for manufacturing long seamless metallic tubes.

MCQ 1.22 The recrystallisation temperature for alloys is approximately :(A) . T0 2 m (B) . T0 3 m

(C) . T0 5 m (D) . T0 8 m

SOL 1.22 Correct option is (C).The recrystallisation temperature for alloys is . T0 5 m .

MCQ 1.23 Match List-I with List-II and select the correct answer using the codes given below the lists :

List I (Material/Part) List II (Techniques)

a. Ductile iron 1. Inoculation

b. Malleable iron 2. Chilled


c. Rail steel joints 3. Annealing

d. White cast iron 4. Thermit Welding

5. Isothermal annealingCodes : a b c d(A) 1 3 4 2(B) 2 1 4 5(C) 5 3 2 1(D) 1 4 2 3

SOL 1.23 Correct option is (A).

List I (Material/Part) List II (Techniques)

a. Ductile iron 1. Inoculation

b. Malleable iron 3. Annealing

c. Rail steel joints 4. Thermit Welding

d. White cast iron 2. Chilled

MCQ 1.24 Joule Thomson coefficient is defined as

(A) pT

h22c m (B) p

hT2

2c m

(C) Th

P22b l (D) T

ph2

2c m

SOL 1.24 Correct option is (A).The numerical value of the slope of an isenthalpe on a T-p diagram at any point is called Joule-Thomson coefficient and is denoted by jμ

and jμ pT

h22= c m

MCQ 1.25 Which one of the following methods can be used for forecasting the sales potential of a new product ?(A) Time series analysis

(B) Jury of Executive opinion method

(C) Sales Force composite method

(D) Direct survey method

SOL 1.25 Correct option is (D).Option survey method is relatively simple and practical method for forcasting demands and especially for new products.


Q.26 TO Q.55 CARRY TWO MARK EACH

MCQ 1.26 Eigen values of the matrix A86

2

674

24

3= −

−

−−

R

T

SSSS

V

X

WWWW

are

(A) , ,0 1 15 (B) , ,0 2 15

(C) , ,0 3 15 (D) , ,0 4 15


We have A 86

2

674

24

3= −

−

−−

R

T

SSSS

V

X

WWWW

The characteristic equation for the eigen value is given by

A Iλ− 0=

86

2

674

24

3

100

010

001

λ−−

−− −

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

0=

8

62

67

4

24

3

λλ

λ

−−

−−

−−−

R

T

SSSS

V

X

WWWW

0=

68 8 3 16 6 3 8λ λ λ λ− − − − + − − +^ ^ ^ ^h h h h6 6@ @

2 24 2 7 λ+ − −^ h6 @ 0= 6 28 10 21 16 6 10 2 102λ λ λ λ λ− − + − + − + +^ ^ ^h h h6 @ 0= 8 80 40 10 5 36 60 4 202 3 2λ λ λ λ λ λ λ− + − + − + − + + 0= 18 453 2λ λ λ− + − 0= 18 453 2λ λ λ− + 0= 18 452λ λ λ− +^ h 0=

3 15λ λ λ− −^ ^h h 0= , ,0 3 15λ =

MCQ 1.27 Given vectors 3 2A i j k= − + and 2B i j k= + + . Then projections of A B# parallel to 5i k− is

(A) 261− (B)

261

(C) 262− (D)

262


A 3 2i j k= − + B 2i j k= + −


A B# i j k12

31

21

= −−

R

T

SSSS

V

X

WWWW

i j k3 2 1 4 1 6= − − − − + +^ ^ ^h h h

5 7i j k= + +Then projection of A B# parallel to 5i k− is

A B i k5# $ −^ ^h h 5 7 5i j k i k25 1

#= + ++

−^ ch m

26

5 10 7262= + − = −

MCQ 1.28 A triangle ABC consists of vertex points , , ,A B0 0 1 0^ ^h h and ,C 0 1^ h. The

value of the integral xdxdy2## over the triangle is

(A) 1 (B) 31

(C) 81 (D) 9

1


The equation of line BC is

x y1 1+ 1=

x y1= −Thus

2I x dx dy= ## xdx dy2y

0

1

0

1=

−

##

x dy2

y2

0

1

0

1=

−

: D#

y dy1 2

0

1= −^ h#

y

31 3

0

1

=−−^ h

< F


31

31= − =: D

MCQ 1.29 The temperature field in a body varies according to the equation ,T x y x xy43= +^ h . The direction of fastest variation in temperature at the

point ,1 0^ h is given by(A) 3 8i j+ (B) i

(C) 0.6 0.8i j+ (D) 0.5 0.866i j+


We have T x xy43= +

Td xT

yTi j

22

22= +

3 4x xi k2= +At ,P 1 0^ h Td 3 4i k= +and unit Normal Vector at point P

uP 3 4 0.6 0.8i k i j3 42 2

=+

+ = +^ ^h h

Thus direction of fastest variation in temperature at ,1 0^ h is given by

0.6 0.8i j+

MCQ 1.30 Let ,u x t^ h be the bounded solution of

0tu

xu2

2

22

22− = with ,u x

ee0

11

x

x

2

2

=+−

^ h

then ,lim u t1t"3

^ h equals

(A) /1 2− (B) /1 2

(C) 1− (D) 1


We have tu

22

xu 02

2

22= = ....(1)

Let u X x T t= ^ ^h h

Putting in (1) XTl X T= ll

or XXm T

T P2= =l (say)

Thus X P X 02− =m & T P T 02− =l

A.E. D P12 2− 0= and D P 02

2− = D1 P!= and D P2

2=

Thus X c e c ePx Px1 2= + − and T eP t2

=

Now u c e c e c e tPx Px P1 2 3

2

= + −^ h

,u x 0^ h c e c c e cPx Px1 2 3 3= + −

^ h


Thus ee

11

x

x

2

2

+− c c e c c ePx Px

1 3 2 3= + −

u ee e

11

x

xp t

2

22

=+−

d n

Then, ,limu t1t"3

^ h limee e

1 21

t

P t2

2 12

=+

= −"3

−

MCQ 1.31 Match List I with List II and select the correct answer using the codes given below the lists:

List-I List-II

a. Bell Column refrigeration 1. Compressor

b. Vapour compression refrigeration 2. Generator

c. Absorption refrigeration 3. Flash chamber

d. Jet refrigeration 4. Expansion cylinder(A) a-1 b-4 c-3 d-2 (B) a-4 b-1 c-3 d-2

(C) a-1 b-4 c-2 d-3 (D) a-4 b-1 c-2 d-3

SOL 1.31 Correction option (D)

List-I List-II

a. Bell Column refrigeration 4. Expansion cylinder

b. Vapour compression refrigeration 1. Compressor

c. Absorption refrigeration 2. Generator

d. Jet refrigeration 3. Flash chamber

MCQ 1.32 In an ECM process for machining iron, it is desired to obtain a metal removal rate of 1 /mincm3 . If atomic weight of iron 7. /g cm8 3= , valency at which dissolution occurs 2= , density of iron 7.8 /g cm3= , and Faraday’s constant 1609 minA−= , the amount of current required for the process is(A) Amp560 (B) 448 Amp

(C) Amp224 (D) Amp336


Given atomic weight of iron Aw= 56 g=Valency of iron dissolution v 2= =Density of iron 7.8 /g cm3ρ= =Faraday’s constant 1609F= = A-min

Metal removal rate 1 /mincm3=

FvIAw

ρ=

I . 448 A561609 2 7 8# #= =


MCQ 1.33 The figure shows a sphere resting in a smooth V shaped groove and subjected to a spring force. The spring is compressed to a length of 125 mm from its free length of 150 mm. If the stiffness of spring 10 /N mmk = and weight of sphere is 100 N, the contact reactions at the point A and B are

(A) 303.1 , 175N NR RA B= = (B) . , .N NR R227 3 131 25A B= =

(C) . ,N NR R181 8 105A B= = (D) . , .N NR R378 9 218 75A B= =

SOL 1.33 Correct Option is (A)Deflection of the spring

x 150 125 25 mm= − = FS kx= 25 10 250 N#= =This force will act on sphere vertically downward. Beside this spring force

reaction RA at point A, reaction RB at B and weight of sphere W will be

there as shown in Figure,

F WS + 250 100= + 350 N= vertically downward

Applying Lami’s theorem

sin

W F60 30

S

c c++

^ h

sin sinR R

180 30 180 60B A

c c c c=

−=

−^ ^h h

W FS+ sin sin

R R30 60B A

c c= =

or RB sinW F 30S c= +^ h

350 175 N21

#= = ans.

RA sinW F 60S c= +^ h

3350 2#=

3031 N= ans.

MCQ 1.34 Match List I (Each conditions of columns) with List II (Lowest critical load)


and select the correct answer using the codes given below the lists:

List-I

a. Column with both ends hinged

b. Column with both ends fixed

c. Column with one end fixed and the other end hinged

d. Column with one end fixed and the other end free

List-II

1. /EI L2 2π

2. /EI L2 2 2π

3. /EI L4 2 2π

4. /EI L42 2π

(E is the Young’s modulus of elasticity of column material, L is the length and I is the second moment of area of cross-section of the column.)Codes : a b c d(A) 1 2 3 4(B) 3 2 1 4(C) 1 3 2 4(D) 2 4 3 1


Critical load PE LEIeq2

2π=

Column with both ends pinned L Leq =

PE LEI2

2π=

Column with both ends fixed /2L Leq =

PE LEI42

2π=

Column with one end fixed and other end hinged

Leg L2

=

PE LEI22

2π=

Column with one end fixed and other end free

Leq 2L=

PE LEI

4 2

2π=


MCQ 1.35 A slender bar of 100 mm2 cross-section is subjected to loading as shown in the figure below. If the modulus of elasticity is taken as 200 10 Pa9

# , then the elongation produced in the bar will be

(A) 10 mm (B) 5 mm

(C) 1 mm (D) nil

SOL 1.35 Correct Option is (D)F.B.D

AEPLδ =b l

total elongation

. .AE1 100 0 5 100 1 100 0 5 0# # #= − + =^ h

MCQ 1.36 The given figure (all dimensions are in mm) shows an I-section of the beam.

The shear stress at point P (very close to the bottom of the flange) is 12 MPa. The stress at point Q in the web (very close to the flange) is(A) indeterminable due to incomplete data

(B) 60 MPa (C) 18 MPa

(D) 12 MPa



B

Aττ B

t=

12Bτ 60 MPa100

200B& τ= =

MCQ 1.37 Two links andOA OB are connected by a pin joint at ‘O ’. The link OA turns with angular velocity 1ω radians per second in the clockwise direction and the link OB turns with angular velocity 2ω radians per second in the anti-clockwise direction. If the radius of the pin at ‘O ’ is ‘r ’, then the rubbing velocity at the ping joint ‘O ’ will be(A) r1 2ω ω (B) ( )r1 2ω ω−

(C) ( )r1 2ω ω+ (D) ( ) r21 2ω ω−


Vrubbing r rb1 2b 1 2ω ωw w= − = +^ ^h h

MCQ 1.38 A multiple disc clutch has five plates having four pairs of active friction surfaces. If the intensity of pressure is not to exceed 0.127 /N mm2. The outer and inner radii of friction surfaces are 125 mm and 75 mm respectively. If wear is uniform and coefficient of friction is 0.3, the power transmitted at 500 r.p.m, is(A) . kW14 1 (B) . kW35 6

(C) . kW23 5 (D) 18.8 kW


SOL 1.38 Correct Option is (D)Given : n n 51 2+ = ; n 4= ; 0.127 /N mmp 2= ; 500 . . .r p mN = ; 125 ;mmr1 =

75 mmr2 = ; .0 3μ =We know that for uniform wear, ,p r C= (a constant). Since the intensity of pressure is maximum at the inner radius r2^ h, therefore.

.p r2 C=or C 0.127 75 9.525 /N mm#= =and axial force required to engage the clutch,

W 2 2 9.525 2993 NC r r 125 751 2π π #= − = − =^ ^h h

Mean radius of the friction surface,

R 100 0.1mm mr r2 2

125 751 2= + = + = =

We know that the torque transmitted,

T . . . 4 0.3 2993 0.1 359 Nmn W Rμ # # #= = =

Power transmitted P T N602

60359 2 500# # #π π= =

18800 18.8W kW= =

MCQ 1.39 Given that andT T1 2 are the tensions on the tight and slack side of the belt respectively, the initial tension of the belt taking centrifugal tension TC is equal to

(A) T T T3

C1 2+ + (B) T T T2

2 C1 2+ +

(C) T T T3

3 C1 2+ + (D) T T T4

2 C1 2+ +


Ft1 F F Fi i t2= = − F2 i F Ft t1 2= +

Fi F F F T T T

22

221 c c2 1 2= + + = + +

MCQ 1.40 Water is boiled at 150 Cc in a boiler by hot exhaust gases ( 1.05 / )kJ kg Ccp c= that enter the boiler at 400 Cc at a rate of 0.4 /kg s and leaves at 200 Cc . The surface area of the heat exchanger is 0.64 m2. The overall heat transfer coefficient of this heat exchanger is(A) 940 /W m C2c (B) /W m C1056 2c

(C) 1 /W m C145 2c (D) /W m C1230 2c

SOL 1.40 Correct Option is (B)We have

150 CTw c= , 400 CT ,inh c= , 200 CT ,outh c= , 0.4 / seckgmh =o ,


1.05 /kJ kg Ccph $ c= , 0.64 mAs3=

Heat capacity rate

Ch m Ch ph#= o

. .0 4 1 05#= 0.42 /kW Cc= Cmin 0.42 /kW CCh c= =The maximum heat transfer rate is

Qmaxo C T T,min h in w= −^ h

.0 42 400 150#= −^ h

105 kW=The actual heat transfer in heat exchanger is

Qo C T T, ,in outh h h= −^ h

.0 42 400 200= −^ h

84 kW=Effective of heat exchanger is

ε .QQ

8584 0 8

max

= = =o

o

Also Number of transfer units

NTU ln 1 ε=− −^ h

.ln 1 0 8=− −^ h

.1 61

So that over all heat transfer co-efficient is

U NTUA

Cmin

s

#=

.. .

0 641 61 0 42 103

# #=

1056 /W m C2c=

MCQ 1.41 A surface absorbs %10 of radiation at wavelengths less than 3 mμ and %50 of radiation at wavelengths greater than 3 mμ . If .F 0 8900291 = , the average absorptivity of this surface for radiation emitted by a source at 3000 K is(A) .0 14 (B) 0.22

(C) 0.30 (D) 0.38

SOL 1.41 Correct Option is (A)We have

.0 11α = , .0 52α = , 3000 KT = , 3 mλ μ=So, Tλ 3 3000#=


9000 m Kμ −=The radiation fraction corresponding to T 9000λ = is .F 0 8900291 =Thus average absorptivity is

avgα F F11 1 1 2α α= + −^ h

. . . .0 890029 0 1 1 0 890029 0 5= + −^ ^ ^h h h

.0 14=The radiation fraction correspond to 9000 m KTλ μ −= is .0 890029

MCQ 1.42 What will be the torque and the thrust in drilling from a solid block of steel from the following data :

Diameter of drill D 25 mm= Chisel edge length C 3 mm= Feed 0.2 /mm rev= Specific cutting pressure, p 300 /kg mm2= Friction angle β 35c= Mean true rake angle, α 18c= Point angle 2ρ 120c=

(A) , .kg mm kg3465 262 1− (B) , .kg mm kg5775 436 8−

(C) 4620 , 349.5kg mm kg− (D) , .kg mm kg2310 174 75−

SOL 1.42 Correct Option is (C)Applying formula for drilling torque, we get

Torque kg mmpf D c8

2 2−= −

b l

Substituting 300 /kg mmp 2= , 25 mmD = , 3 mmc = , .f 0 2= we get

Torque 300 0.2 825 3

860 625 9

860 6162 2

##= − =

−=b

^l

h

4620 kg mm−=

Thrust tan sinp D c f p4 2 2# β α= − −^ h: D

4 300. tan sin2

25 32

0 2 35 18 60c#=− −b ^l h; E

. .tan1200 1 1 17 0 866#c= ^ h

349.48 kg=

MCQ 1.43 Resistance spot welding is performed on two plates of 1.5 mm thickness with 6 mm diameter electrode, using 15000 A current for a time duration of 0.25 s. Assuming the interface resistance to be 0.0001 Ω, the heat generated to form the weld is


(A) 5625 secW− (B) 8437 secW−

(C) 22500 secW− (D) 3 secW3750 −

SOL 1.43 Correct option is (A).Given : I 15000 A= , 0.25 sect = , 0.0001R Ω=The heat generated to form the weld is, Q I Rt2= ( ) . .15000 0 0001 0 252

# #= 5625 secW−=

MCQ 1.44 Tds equation can be expressed as

(A) Tds C dT TK dVvβ= + (B) Tds C dT K

T dVv= +

(C) Tds C dT TK dVv β= + (D) Tds C dT KT dpv

β= +

SOL 1.44 Correct option is (A)

Tds c dT TTp dVv

v22= + c m ....(i)

β V Tp1

p22= c m ....(ii)

KT V pV1

T22=− c m ....(iii)

KTβ

TV

TV

Tp

p T v22

22

22

#=− =b b cl l m ....(iv)

From equation (1) and (4)

Tds c dT TK dVvβ= +

MCQ 1.45 The air with enthalpy of 100 /kJ kg is compressed by an air compressor to a pressure and temperature at which its enthalpy becomes 200 /kJ kg. The loss of heat is 40 /kJ kg from the compressor as the air presses through it. Neglecting kinetic and potential energies, the power required for an air mass flow of 0.5 /kJ s is(A) 30 kW (B) 50 kW

(C) 70 kW (D) 90 kW

SOL 1.45 Correct option is (C)First law for steady flow


h Q1 + h W2= + (Steady flow energy equation)

100 40− 200 W= − W 140 /kJ kg= Power required mW= o

P 0.5 140 70 kW#= =

MCQ 1.46 In the network shown below the critical path is along :

(A) 1 2 3 4 8 9− − − − − (B) 1 2 3 5 6 7 8 9− − − − − − −

(C) 1 2 3 4 7 8 9− − − − − − (D) 1 2 5 6 7 8 9− − − − − −


TF LST EST= =

Activities Duration EST LST EFT LFT TotalFload

Remark


1 2− 3 0 0 3 0

2 3− 4 3 3 7 0

3 4− 5 7 10 15 3

2 5− 2 3 5 7 2

4 8− 5 12 15 20 3

5 6− 3 7 7 10 0

6 7− 4 10 10 14 0

7 8− 6 14 14 20 0

8 9− 4 20 20 24 0

4 7− 0 12 14 14 2

3 5− 0 7 7 7 0

Therefore critical path

1 2 3 4 5 6 7 8 9− − − − − − − −and total project duration

3 4 3 4 6 4 24+ + + + + =

MCQ 1.47 An operations consultant for an automatic car wash wishes to plan for enough capacity to handle 60 cars per hour. Each car will have wash time to 2 minutes, but there is to be a %20 allowance for setup, delays and payment transactions. The installation capacity of car wash stalls should be :(A) 3 (B) 4

(C) 5 (D) 6

SOL 1.47 Correct option is (A).Time required to wash one car is . .2 1 2 2 4# =mins No. of cars washed in 1 hr in 1 stall

.2 460 25= =

No. of stalls .2560 2 4 3-= =

Common Data For Q.• 48 and 49.A reversible heat engine operates between two reservoirs at temperatures 700 Cc and 50 Cc . The engine drives a reversible refrigerator which operates between reservoirs at temperatures 50 Cc and 25 Cc− . The heat transfer to the engine is 2500 kJ and the net work output of the combined engine refrigerator plant is 400 kJ.

MCQ 1.48 What will be the heat transfer to the refrigerant and the net heat transfer


to the reservoir at 50 Cc ?(A) 6408.5 kJ (B) 6200 kJ

(C) 6298.6 kJ (D) 3200 kJ

SOL 1.48 Correct option is (C)

Temperature T1 700 273 973 K= + = Temperature T2 50 273 323 K= + = Temperature T3 25 273 248 K=− + =The heat transfer to the heat engine

Qt 2500 kJ=The net work output of the combined engine refrigerator plant

W 400 kJW Wt 2= − =Maximum efficiency of the heat engine cycle is given by

maxη .TT1 1 973

323 0 6681

2= − = − =

Again, QW

1

1 .0 668=

W1 0.668 2500 1670 kJ#= =

COP max^ h .T TT

323 248248 3 306

2 3

3= − = − =

Also COP .WQ 3 306

2

4= =

Since, W W1 2− 400 kJW = W2 W W1= − 1670 400 1270 kJ= − =

Q4 3.306 1270 4198.6 kJ#= = Q3 Q W4 2= + 4198.6 1270 . kJ5468 6= + = Q Q W2 1 1= − 2500 1670 830 kJ= − =


Heat rejection to the 50 Cc reservoir

.Q Q 830 5468 62 3= + = + 6298.6 kJ=

MCQ 1.49 If given that the efficiency of the heat engine and the COP of the refrigerator are each %45 of their maximum possible values, then the net heat transfer to the reservoir at 50 Cc is(A) 2618 kJ (B) 2818 kJ

(C) 2515 kJ (D) 2312 kJ

SOL 1.49 Correct option is (A)Efficiency of actual heat engine cycle

.0 45 maxη η= . . .0 45 0 668 0 3#= = W Q1 1#η= 0.3 2500 750 kJ#= = W2 750 400 350 kJ= − =

COP of the actual refrigerator cycle

COP 0.45 . . .COP 0 45 3 306 1 48max #= = =^ h

Q4 350 1.48 518 kJ#= = Q3 518 350 868 kJ= + = Q2 2500 750 1750 kJ= − =Heat rejection to 50 Cc reservoir

Q Q 1750 8682 3= + = + 2618 kJ=

Common Data For Q.• 50 and 51.The following equation has been obtained when machining AISI 2340 steel with H.S.S. cutting tools having a 8, 22, 6, 6, 15, 6, 0.117 mm tool signature. The work tool system is governed by the following equation :

.26 035 VT f d. . .0 13 0 77 0 37=A min100 tool life was obtained under the following cutting conditions :

Velocity 25 /minm= Feed 0.3125 /mm rev= Diameter of bar 2.50 mm=

MCQ 1.50 What will be the tool life if cutting speed is increased by %20 ?(A) min30 (B) min24

(C) min22 (D) min18

SOL 1.50 Correct option is (B)We are given


Velocity minV 25=^ h

Feed f^ h 0.3125 /mm rev=Dia of bar 2.50 mmd =^ h

For a tool life of min100When V is increased by %20 ; feed and diameter of the bar remain constant.Substituting the values, we have

.26 035 V 100 .0 13= ^ h ...(i)

But .26 035 . V T1 2 .0 13= ^ h ...(ii)

Rewriting Eqs. (i) and (ii), we have

. VV

1 2b l T100 1

.0 13= =b l

or T .0 13 .1 2100 .0 013

= ^ h

T .0 13 ..1 21 82=

T .0 13 .1 51= T min24=

MCQ 1.51 What will be the tool life if cutting speed, feed and depth of cut each is increased by %20 together ?(A) . min4 9 (B) . min6 1

(C) . min3 6 (D) . min2 45


When , ,V f d are all increased by %20

.26 035 V f d100 . . .1 3 0 77 0 37= ^ h

.26 035 . . .V T f d1 2 1 2 1 2. . .0 13 0 77 0 37= ^ ^ ^h h h

Rewriting the above two equations, we have

or T .0 13^ h

. . .1 2 1 2 1 2100

. .

.

0 77 0 37

0 13

# #=

^ ^

^

h h

h

T .0 13^ h . . .

.1 2 1 5 1 07

1 82# #

=

T .0 13^ h .1 23=

T . min4 9=

Common Data For Q.• Linked Answer Q. 52 and 53.A 19 - liter, cylindrical open container with a bottom area of 0.075 m2 is filled with Glycerin ( 1260 / )kg m3ρ = and rests on the floor of an elevator.


MCQ 1.52 When the elevator has an upward acceleration of 0.9 /secm 2, the fluid pressure at the bottom of the container in /N m2 will be(A) 341.4 (B) 34.4

(C) 3414 (D) 3.4


An open container of a liquid that is translating along a straight path with constant acceleration, the relation is given by

dzdp ( )g azρ=− +

dp ( )g a dzzρ=− +Integrating p within limit 0 to pB and h within limit h to 0.

dpp

0

B

# ( )g a dzz

h

0

ρ− + #

pB ( )( ) ( )g a h g a h0z zρ ρ=− + − = + ...(i)

Now v h A#= 19 10 3

#− .h 0 075#=

h 0.253 m=From equation (i), we get

pB ( . . ) .1260 9 81 0 9 0 253# #= + 3414 /N m2-

MCQ 1.53 If the weight of the container is negligible, the resultant force, that container exert on the floor of the elevator ?(A) 2.56 N (B) 256 N

(C) 25.60 N (D) 2560 N

SOL 1.53 Correct option is (B)The FBD of container is shown below.


From FBD.

F p AB= .3414 0 075#= 256.05 N= 256 N-

Common Data For Q.• Linked Answer Q. 54 and 55.Consider the figure shown. The coefficient of friction between the 160 kg block and the ramp are .0 3sμ = and .0 28kμ = .

MCQ 1.54 What tension T0 must the winch exert to start the block moving up the ramp ?(A) . N885 8 (B) . N1166 1

(C) 932.9 N (D) . N839 25

SOL 1.54 Correct Option is (C)We have .u 0 3s = , .u 0 28k = , 160 kgm =The FBD of block is shown below


In equilibrium condition of forces

F 0xΣ = T0 18sinmg Fr= + ....(i)

Friction forces Fr Rs Nμ= 1cosmg 8sμ= ....(ii)

From equation (i) & (ii), we have

T0 sin cosmg mg18 18sμ= + . .sin cos160 9 81 18 0 3 18#= +^ h

932.9 N=

MCQ 1.55 If the tension remains at the value T0 after the block starts sliding, what total work is done on the block as it slides a distance 3 ms = up the ramp?(A) N m112 − (B) 89.6 N m−

(C) . N m66 45 − (D) . N m71 68 −

SOL 1.55 Correct Option is (B)The total work done is given by

W F dsts

s

1

2

Σ= #

sin cosT mg mg ds18 18k00

2μ= − −^ h6 @#

. . sin932 9 160 9 81 180

3

# #= −^ h6#

0.28 160 9.81 18cos ds# #−^ h@

. . ds447 87 417 980

3= −^ h#

28.89 3 89.6 N m# −= =

Q. 56 TO Q. 60 CARRY ONE MARK EACH

MCQ 1.56 Which one of the following is the Antonym of the word PROFESSIONAL ?(A) conservative (B) liberal

(C) amateur (D) legal



MCQ 1.57 Which one of the following is the synonym of the word DISPARAGE ?(A) separate (B) compare

(C) refuse (D) belittle

SOL 1.57 Correct option is (D)

MCQ 1.58 One of the four words given in the four options does not fit the set of words. The odd word from the group is(A) Smog (B) Marsh

(C) Haze (D) Mist

SOL 1.58 Correct option is (B)

MCQ 1.59 A pair of CAPITALIZED words shown below has four pairs of words. The pair of words which best expresses the relationship similar to that expressed in the capitalized pair isATMOSPHERE : STRATOSPHERE(A) Nimbus : Cloud (B) Instrument : Calibration

(C) Aircraft : Jet (D) Climate : Rain


MCQ 1.60 In the following sentence, a part of the sentence is left unfinished. Four different ways of completing the sentence are indicated. The best alternative among the four is

..................., the more they remain the same(A) The more the merrier

(B) The less the dynamism

(C) The more things change

(D) The more pronounced the transformation.



MCQ 1.61 2 2 273 72 71− − is the same as(A) 269 (B) 270

(C) 271 (D) 272


Let X 2 2 273 72 71= − − 4 2 2 2 271 71 71

# #= − −


2 4 2 171= − −^ h

271=

MCQ 1.62 If n x1= + , where x is the product of four consecutive positive integers, then which of the following is/are true?1. n is odd

2. n is prime

3. n is a perfect square

(A) A and C only (B) A and B only

(C) A only (D) None of these

SOL 1.62 Correct option is (A)We have

n x1= +Where x is the product of four consecutive positive integers

Thus

n 1 y y y y1 2 3= + + + +^ ^ ^h h h

Where y is a positive integer

If we substitute y 2= n 1 2 2 1 2 2 2 3= + + + +^ ^ ^h h h

1 2 3 4 5# # #= + 121

Which is a odd number and perfect square of 11

Thus n is odd & perfect square.

DIRECTION FOR Q. 63 TO Q. 65Answer these questions based on the pipeline diagram given :

The following sketch shows the pipelines carrying material from one location to another. Each location has a demand for material. The demand at B is 800, at C is 800, at D is 1400, and at E is 400. Each arrow indicates the direction of material flow through the pipeline. The flow from B to C is 600. The quantity of material flow is such that the demands at all these location are exactly met. The capacity of each pipeline is 2000.


MCQ 1.63 The quantity moved from A to E is :(A) 400 (B) 1600

(C) 1400 (D) 2000

SOL 1.63 Correct option is (D)The capacity of each pipeline is 2000. According to the demand and the flow quantity the block diagram is shown below :We have D 800B = , D 800C = , D 1400D = , D 400E = , f 600BC =

The flow quantity shown in figureThus quantity moved from A to E is 2000

MCQ 1.64 The free capacity available in the A-B pipeline is :(A) 0 (B) 200

(C) 400 (D) 600

SOL 1.64 Correct option is (D)Free capacity available in the pipeline A B− is

2000 1400 600= − =

MCQ 1.65 What is the free capacity available in the E -C pipeline ?(A) 600 (B) 400

(C) 200 (D) 0

SOL 1.65 Correct option is (B)The free capacity available in the pipe line E C− is

2000 1600= − 400=


Mock Test - 1

1. (C) 11. (A) 21. (B) 31. (D) 41. (A) 51. (A) 61. (C)

2. (B) 12. (A) 22. (C) 32. (B) 42. (C) 52. (C) 62. (A)

3. (C) 13. (B) 23. (A) 33. (A) 43. (A) 53. (B) 63. (D)

4. (C) 14. (D) 24. (A) 34. (C) 44. (A) 54. (C) 64. (D)

5. (D) 15. (D) 25. (D) 35. (D) 45. (C) 55. (B) 65. (B)

6. (C) 16. (C) 26. (C) 36. (B) 46. (B) 56. (C)

7. (C) 17. (A) 27. (C) 37. (C) 47. (A) 57. (D)

8. (A) 18. (B) 28. (B) 38. (D) 48. (C) 58. (B)

9. (D) 19. (B) 29. (C) 39. (B) 49. (A) 59. (D)

10. (B) 20. (C) 30. (A) 40. (B) 50. (B) 60. (C)

SolutionsME-B


MCQ 1.1 The directional derivative of , ,f x y z xy yz2 3= +^ h at the point , ,2 1 1−^ h in the direction of vector 2 2i j k+ + is

(A) 1 32− (B) 2 3

2−

(C) 3 32− (D) 4 3

2−


fd y xy z yzi j k2 32 3 2= + + +^ ^h h

3 3 , ,ati j k 2 1 1= − − −^ h

Directional Derivative of f in the direction of 2 3i j k+ +

2 3

3i j ki j k

3 31 2 2 3

22 2 2

$= + −+ +

+ +=^

^h

h

MCQ 1.2 If | |f x x=^ h , then(A) f x^ h is continuous at x 0= (B) f x^ h is not continuous at x 0=

(C) f x^ h is differentiable at x 0= (D) none of the above


f x x=^ h ,,

xx

xx

00<

$= −*

f x^ h is continuous and differentiable for all x except zero. We have to check

at x 0= .

Left hand limit of f x^ h at x 0"

lim f xx 0" −

^ h 0=

Right hand limit of f x^ h at x 0"

Page 2 ME-B Chapter 1

lim f xx 0" +

^ h 0=

Value of f x^ h at x 0= , f x 0x 0 ==^ h

Thus f x^ h is continuous at x 0=Now check the differentiability.

Left hand derivative

L f xl h lim hf h f0 0

h 0= −

− −"

^ ^h h

hh0

1= −− −

=−^ h

Right hand derivative.

R f xl h lim hf h f0 0

h 0=

+ −"

^ ^h h

lim hh 1

h 0= =

"

Since ,L f x R f x!l l^ ^h h therefore f x^ h is not differentiable at x 0= , hence

not differentiable for all real value of x

MCQ 1.3 If logy x x1 2= + +^ h; then 1 xdxd y2

2

2

+^ h xdxdy+ is equal to

(A) 0 (B) 1

(C) 2 (D) none of them


we have y log x x1 2= + +^ h

dxdy

x x xx

11 1

2 12

2 2#=+ +

++; E

x x x

x xx1 1

111

2 2

2

2=

+ + ++ + =

+^ ^h h

dxd y

2

2

x x21 1 2/2 3 2

#= − + −^ h

xx

1 /2 3 2=+−

^ h

MCQ 1.4 Given that u exyz= . The x y z

u3

2222 is equal to

(A) e xyz x y z1 3xyz 3 3 3+ +6 @ (B) e xyz x y z1 3xyz 2 2 2+ +6 @

(C) e xyz x y z1 2xyz 3 3 3+ +6 @ (D) e xyz x y z1 3xyz 2 2 2+ +6 @


We have u exyz=

Chapter 1 ME-B Page 3

or xu

22 e yzxyz

$=

x y

u2

222 e yzxyz

$=

x y

u2

222 ze yze xzxyz xyz

$= +

e z xyzxyz 2= +^ h

x y z

u3

2222 e xyz z xyz e xy1 2xyz xyz2

$ $= + + +^ ^h h

e xyz x y z1 3xyz 2 2 2+ +^ h

MCQ 1.5 If A31

41=

−−> H, then for every positive integer n , An is equal to

(A) n

nn

n1 2 4

1 2+

+> H (B) n

nnn

1 2 41 2

− −+> H

(C) n

nn

n1 2 4

1 2−

+> H (D) n

nnn

1 2 41 2

+ −−> H


A2 31

41

31

41

52

83=

−−

−− =

−−> > >H H H

If we put n 2= in option

MCQ 1.6 For the truss shown in figure, the zero force members are,

(A) andAC DC (B) andCE BC

(C) andAD DE (D) andAD BC

SOL 1.6 Correct option is (A)Consider joint D for the truss shown in figure. We assume force F1 in member DC , F2 in member DA and F3 in member DF . This joint is unloaded and the forces F2 and F3 are collinear so F1 must be zero. Similarly forces in the members EF and HG are zero

MCQ 1.7 If shaft made from ductile material is subjects to combined bending and twisting moments calculations based on which one of the following failure


theories would give the most conservative value ?(A)Maximum principal stress theory

(B)Maximum shear stress theory

(C)Maximum strain energy theory

(D)Maximum distortion energy theory

SOL 1.7 Correct option is (B)Most consecutive value means safest design i.e. largest diameter. For ductile material, maximum shear stresses of theory gives higher value of diameter

MCQ 1.8 Which one of the following Mohr’s circles represents the state of pure shear ?

SOL 1.8 Correct option is (C)For pure shear. Radius is equal to shear stress and centre lies on the point of intersection of axes.

MCQ 1.9 A cube having each side of length a is constrained in all directions and is heated uniformly so that the temperature is raised to CTc . If α is the thermal coefficient of expansion of the cube material and E the modulus of elasticity, the stress developed in the cube is

(A) TE2μ

α (B) ( )

TE1 2μα−

(C) TEμ

α (D) ( )

TE1 2μα+

SOL 1.9 Correct option is (B)Since the cube is constraint is all directionstherefore 0x y z! ! != =If 0x! =


& T E L E L ELα σ μ σ μσ− + + 0=

& T E 1 2α σ μ− −^ h 0=

σ E T1 2μ

α=−^ h

MCQ 1.10 A thin cylindrical shell of diameter ‘d ’, length l and thickness ‘t ’ is subjected to an internal pressure ‘p ’. What is the ratio of longitudinal strain to hoop strain in terms of Poisson’s ratio ( /m1 ) ?

(A) mm2 1

2+

− (B) mm2 1

2−

−

(C) mm

22 1

−− (D) m

m2

2 1−+


Longitudinal stress 1σ^ h tpd

m4 1 2= −b l

hoop stress 2σ^ h tpd

m4 2 1= −b l

m

m2 11 2

2

1σσ =

−

− m

m2 1

2= −−


List-I

a. Interference

b. Dynamic load on tooth

c. Static load

d. Contact ratio

List-II

1. Arc of approach, arc of recess, circular pitch

2. Lewis equation

3. Minimum number of teeth on pinion

4. Inaccuracies in tooth profile

Codes : a b c d(A) 3 4 1 2(B) 1 2 3 4(C) 4 3 2 1


(D) 3 4 2 1

SOL 1.11 Correct option is (D)To avoid interference certain minimum number of teeth must be provided on the gear.

contact ratio Circular pitchArc of contact=

Lewis equation gives load on teeth.

MCQ 1.12 Match List I (Fluid properties) with List II (Relate terms) and select the correct answer using the codes given below the lists :

List-I List-II

a. Capillarity 1. Cavitation

b. Vapour pressure 2. Density of water

c. Viscosity 3. Shear forces

d. Specific gravity 4. Surface tension

Codes : a b c d(A) 1 4 2 3(B) 1 4 3 2(C) 4 1 2 3(D) 4 1 3 2


List-I (Fluid properties) List-II (Related terms)

a. Capillarity 4. Surface tension

b. Vapour pressure 1. Caviation

c. Viscosity 3. Shear forces

d. Specific gravity 2. Density of water

MCQ 1.13 The heat conduction equation in a medium is given in its simplest form as

r rrk

rT e1 0gen2

222 + =ob l

Select the wrong statement below.(A) The medium is of cylindrical shape.

(B) The thermal conductivity of the medium is constant.

(C) Heat transfer through the medium is steady.

(D) There is heat generation within the medium.

SOL 1.13 Correct option is (B)The given conduction equation is


r rrk

rT e1

gen22

22 + ob l 0=

It is the heat conduction equation of cylindrical shape

It is a steady heat conduction equation because in this equation the term

tT22 is zero.

There is heat generation with the medium because in this equation the term

egeno is present

But the thermal conductivity of the medium is not a constant.

MCQ 1.14 The_____number is a significant dimensionless parameter for forced convection and the_____number is a significant dimensionless parameter for natural convection.(A) Reynolds, Grashof (B) Reynolds, Mach

(C) Reynolds, Eckert (D) Grashof, Sherwood

SOL 1.14 Correct option is (A)The Reynolds number is a significant dimensionless parameter for forced convection and the Grashof number is a significant dimensionless parameter of natural convection.

MCQ 1.15 The marking on a grinding wheel is ‘ 51 A 36 L 5 V 93’. The cude ‘36’ represents the(A) structure (B) grade

(C) grain size (D) manufacture’s number

SOL 1.15 Correct option is (C)Nomenclature of grinding wheel.

MCQ 1.16 To reduce consumption of synthetic resins, the ingredient added is(A) accelerator (B) elastomer

(C) modifier (D) filler


SOL 1.16 Correct option is (D)To reduce the consumption of synthetic rerins, the ingredient added is filler. Because of their low cost, fillers are important in reducting the overall cost ofd polymers. Fillers may also improve the strength, hardness, toughness, abrasive resistance, dimensional stability etc.

MCQ 1.17 The compositions of some of the alloy steels are as under :1. 18W 4Cr 1V 2. 12Mo 1W 4Cr 1V

3. 6Mo 6W 4Cr 1V 4. 18W 8Cr 1V

The compositions of commonly used high speed steels would include :(A) 1 and 2 (B) 2 and 3

(C) 1 and 4 (D) 1 an 3

SOL 1.17 Correct option is (C)The composition of commonly used high speed steels include 18 4 1W G V and 18 8 1W G V

MCQ 1.18 Major operations in the manufacture of steel balls used for Ball bearings are given below :1. oil lapping 2. cold heading 3. Annealing 4. Hardening 5. Rough grindingThe correct sequence of these operations is(A) 3, 2, 4, 1, 5 (B) 3, 2, 1, 4, 5

(C) 2, 3, 4, 5, 1 (D) 2, 3, 5, 4, 1

SOL 1.18 Correct option is (C)The correct sequence of these operations is 1. Cold heading 2. Annealing 3. Hardening 4. Rough grinding 5. Oil lapping

MCQ 1.19 Chills are used in moulds to(A) achieve directional solidification

(B) reduce the possibility of blow holes

(C) reduce freezing time

(D) Smoothen metal flow for reducing splatter

SOL 1.19 Correct option is (A)A chill is an object used to promote solidification in a specific portion of a metal casting mould. Normally the metal in the mould cools at a certain rate relative to thickness of casting. When the geometry of the moulding cavity prevents directional solidification from occurring naturally, a chill can be placed to help promote it.

MCQ 1.20 Carburised machine components have higher endurance limit, because carburization(A) raises the yield point of the material


(B) produces a better surface finish

(C) introduces a compressive layer on the surface

(D) suppresses any stress concentration produces in the component.

SOL 1.20 Correct option is (C)Carburizing or Carburization is a heat treatment process by which carbon is introduced into a metal. Carburization can be used to increase the surface hardness of low carbon steal. Because of carburization a compressive layer is produced on the surface which increase the hardness of the metal.

MCQ 1.21 At %100 relative humidity, the wet bulb temperature is(A) more than dew point temperature

(B) same as dew point temperature

(C) less than dew point temperature

(D) equal to ambient temperature

SOL 1.21 Correct option is (B)For fully saturated air

MCQ 1.22 The thermodynamic parameters are :1. Temperature 2. Specific volume

3. Pressure 4. Enthalpy

5. Entropy

The Clapeyron Equation of state provides relationship between :(A) 1 and 2 (B) 2, 3 and 4

(C) 3, 3 and 5 (D) 1, 2, 3 and 4


The Clapeyron equation is dTdp

satc m .T V

hfg

fg=

It enable us to determined the enthalpy of vaporization hfg at a given temperature by simply measuring the slope of saturation curve on p-T diagram and the specific volume of saturated liquid and saturated vapour


at given temperature

MCQ 1.23 Gibb’s free energy ‘G ’ is defined as(A) G H Ts= − (B) G u Ts= −

(C) G u pV= + (D) G H Ts= +

SOL 1.23 Correct option is (A)The Gibbs free energy is the thermodynamic potential that measures the useful work obtainable from a thermodynamics system at a constant temperature and pressure Gibbs Free energy

,G T p^ h H Ts= −

MCQ 1.24 The solution in a transportation model (of dimension m n# ) is said to be degenerate if it has(A) exactly m n 1+ −^ h allocations

(B) fewer than m n 1+ −^ h allocations

(C) more than m n 1+ −^ h allocations

(D) m n#^ h allocations

SOL 1.24 Correct option is (B)In a TP , if the number of non-negative independent allocations is less than m n 1+ − , there exists a degeneracy.

MCQ 1.25 The management is interested to know the percentage of idle time of an equipment. The trial study showed that percentage of idle time would be %20 . The number of random observations necessary for %95 level of confidence and %5+ accuracy is :(A) 6400 (B) 1600

(C) 640 (D) 160


.P S .K pσ=Hence it is given P .0 20= S .0 05= K %2 95= ^ h

& P sk

np P1

=−^ h

& .0 2 .. .

n0 052 0 2 0 8#=

& n 6400=where P is fraction occurrence of events

n = number of observations


s = Precision factor

k = Confidence factor

sk = confidence precision factor


MCQ 1.26 The series ..... ....n1 2 3 3+ + + + + is(A) convergent (B) divergent

(D) both (A) and (B) (D) none of these


Sn .... ......n1 2 3= + + + +

Sn kn n

21

k

n

1

= =+

=

^ h/

Then limSn

n"3

limn n21 1

n3= + =

"3^ h

Thus it is a divergent series

MCQ 1.27 x

e dx1

tan x

2

1

+

−

# , is equal to

(A) tane x x c1 +− (B) tan x c1 +−

(C) e ctan x1

+−

(D) 1x c+


xe dx1

tan x

2

1

+

−

#

Let tan x1− y=Differentiating both sides, we get

1x

dx1 2+

dy=

& x

e dx1

tan x

2

1

+

−

# .e dyy= #

e c e ctany x1

= + = +−

MCQ 1.28 If V is a differentiable vector function and f is a sufficient differentiable scalar function, then fV^ h is equal to :(A) grad curlf fV V# +^ ^ ^h h h (B) 0

(C) curlf V h (D) grad f V#^ ^h h


Let V V V j Vi kx y z= + +


Then curl fV^ h x yk

zf V V Vi j i j kx y z2

222

22= + + + +c _m i8 B

x

fVy

fVz

fV

i j k

x y z

22

22

22=

R

T

SSSSS

V

X

WWWWW

yfV

yfV

xfV

zfVi jz y z x2

222

22

22

= − − −^ ^ ^ ^h h h h; :E D

xfV

yfVk y x2

222

+ −^ ^h h; E

fyV V

yf f

zV V

zf

i zz

zz2

222

22

22

= + − +; E

fxV V

xf f

zV V

zf

j zz

xx2

222

22

22

− + − +; E

fxV

Vxf f

yV V

yf

k yy

xx2

222

22

22

+ + − += G

Separating the equation

Vyf V

zf V

xf V

zf V

xf V

yfi j kz y z x y x2

222

22

22

22

22= − − − + −c c cm m m= G

fyV F

zV f

xV F

zV f

xV

FyVi j kz z z x y x

22

22

22

22

22

22= − − − + −c b em l o> H

grad curlf fV V#= +^ ^ ^h h h

MCQ 1.29 If x y zr i j k= + + is the position vector of point , ,P x y z^ h and | |r r= then r rn

$d is equal to(A) nrn (B) 3n rn+^ h

(C) n r2 n+^ h (D) 0


r rn$d

xxr

yyr

xzrn n n

22

22

22

= + +^ ^ ^h h h where r x y zn n2 2 2 2= + +^ h

r rn$d 2 2x n x y z y n x y z2 2

n n2 2 2 2 2 1 2 2 2 2 2 1= + + + + +− −^ ^h h

2z n x y z r2 3n n2 2 2 2 2 1+ + + +−

^ h

n x y z x y z r3n n2 2 2 2 2 2 2 1= + + + + +−

^ ^h h

nr r n r3 3n n n+ = +^ h

MCQ 1.30 A side and face cutter of 127 mm diameter has 10 teeth. It operates at a cutting speed of 13.5 /minm and table feed of 108 /minmm . The maximum


chip thickness for a cutting depth of 5 mm, is(A) . mm2 64 (B) 1.32 mm

(C) . mm0 66 (D) . mm3 96


Spindle speed . 34 /rev mm12713 5 1000

##

π= =

f1 . 3.30 mm34

13 5 1000101#

#= =

We also know that

tmax f Dd2 t#=

where tmax is the maximum thickness; D is the diameter of cutter; d is the cutting depth

tmax .2 3 3 1275

#=

1.32 mm=

MCQ 1.31 The principal cutting edge is 75c, while the auxiliary (end) cutting edge is 8c. Assume 6c clearance angle, inclination angle as 5c− , and the shank as 30 30 mm# . What will be the side rake and back rake ?(A) ,Side rake Back rake2 13 10 55c c=− =l l

(B) 1 ,Side rake Back rake5 1 13 2c c= =−l l

(C) 1 ,Side rake Back rake3 2 15 1c c=− =l l

(D) 10 55 , 2 13Side rake Back rakec c= =−l l

SOL 1.31 Correct option is (D)We know

tanλ sin tan cos tany xφ α φ α= −where λ = side rake

φ = principal cutting angle 75c= xα = orthogonal angle (back rake) 5c=− yα = orthogonal side rake 10c=Putting the values

tanλ sin tan cos tan75 10 75 5= − −^ h

. . . .0 965 0 176 0 258 0 08# #= − − λ 10 55 18c= l m

We also know that for calculating γ = back rake

tanγ cos tan sin tany xφ α φ α= +


cos tan sin tan75 10 75 5= + −^ h

. . . .0 258 0 176 0 965 0 08# #= + − γ 2 13 33c=− l m

MCQ 1.32 Load W produces a force of magnitude 120 kN in the member AB of the cantilever truss shown in figure. For this loading, forces in members AE and FE of the truss are

(A) . , 180kN kNF F216 34AE FE= = (B) , .kN kNF F180 108 17AE FE= =

(C) 108.17 , 180kN kNF FAE FE= = (D) 108.17 ,kN kNF F 360AE FE= =

SOL 1.32 Correct option is (C)Let a section 1 1− be passed in such a way that it cuts the members AB , AE , FE and divides the truss in two parts and assumed direction of forces is shown in FigureConsider equilibrium of right part of the truss and take moments about the point E ,

W W2 6# #+ F 3 120 31# #= =

or W 45 kN8120 3#= =

To find force F3 in member FE , taking moments about the point A,

W W8 4# #+ F 33 #=or F3 4 4 45 180 kNW #= = = (compressive)

To find force F2 in member AE , resolving forces vertically

sinF2 θ W W= +

or F2 sinW2

θ=

For the geometry of figure we get,

sinθ .AEAF

3 23 0 832

2 2= =

+=

or F2 . 108.17 kN0 8322 45#= = (compressive)

MCQ 1.33 Blocks A and B of masses 40 kg and 60 kg respectively are placed on a


smooth surface as shown in figure and the spring connected between them is stretched a distance 2 m. If they are released from rest, the speeds of both blocks the instant the spring becomes unstreched is (Take 150 /N mk = )

(A) / , 3 /sec secm mV V4A B= = (B) 2 / , /sec secm mV V 6A B= =−

(C) 2 / , /sec secm mV V 3A B= =− (D) / , /sec secm mV V3 2A B=− =

SOL 1.33 Correct option is (C)Before release the total momentum was zero because of rest condition. From the conversation of momentum we can write

0 m v m vA A B B= +or 0 v v40 60A B= +or vA . v1 5 B=− ....(1)

When spring is stretched its has energy and when it is unstretched all its

energy is transferred to both block. Therefore

kz21 2 m v m v2

121

A A B B2 2= +

or 12 150 22# # 40 0v v2

121 6A B

2 2# #= +

or 30 2v v3A B2 2= + ....(2)

Substituting the value of vA from equation (1) we get

30 . v v2 1 5 3B B2 2= − +^ h

or 30 . v7 5 B2=

or vB 2 / secm=and vA 3 / secm=−

MCQ 1.34 A 10 cm long and 5 cm diameter steel rod fits snugly between two rigid walls 10 cm apart at room temperature. Young’s modulus of elasticity and coefficient of linear expansion of steel are 2 10 /kgf cm6 2

# and 12 10 / C6 c# respectively. The stress developed in the rod due to a 100 Cc rise in temperature wall be(A) 6 10 /kgf cm10 2

#− (B) 6 10 /kgf cm9 2

#−

(C) . 10 /kgf cm2 4 3 2# (D) . 10 /kgf cm2 4 4 2

#



l 10 cm= E 2 10 /kgf cm6 2

#= α 12 10 / C6 c#= −

TΔ 100 Cc=a Strain is prevented stress will be induced in steel rod

It is statically indeterminate. So we used one equation on compatibility

L TαΔ AEPL=

σ E TαΔ= 12 10 2 10 1006 6

# # # #= −

2.4 10 /kgf cm3 2#=

MCQ 1.35 A compound train consisting of spur, bevel and spiral gears is shown in the given figure along with the teeth numbers marked against the wheels. Overall speed ratio of the train is

(A) 8 (B) 2

(C) 21 (D) 8

1

SOL 1.35 Correct option is (A)Elements of higher pair like follower in cam is under the action of gravity of spring force.


Train Value speed of the first gearspeed of lost driven or follower=

Train value product of no. of teeth on the driversproduct of no. teeth no the drivers=

Speed ratio speed of the last driven of followerspeed of the first driver=

MCQ 1.36 When a system is taken from state A to state B along the path A-C - ,B 180 kJ of the heat flows into the system and it does 130 kJ of work (as shown in the figure below) :

How much heat will flow into the system along the path A-D -B if the work done by it along the path is 40 kJ ?(A) 40 kJ (B) 60 kJ

(C) 90 kJ (D) 135 kJ

SOL 1.36 Correct option is (C)From First law of Thermodynamics:

u QA + u WB= + u uB A− Q W= − 180 130 50 kJ= − =Since internal energy is property (is independent of path function.)

u uB A− Q W= − 50 Q 40= − Q 90 kJ=

MCQ 1.37 A system of 100 kg mass undergoes a process in which its specific entropy increases from . / . /kJ kgK to kJ kgK0 3 0 4 . At the same time, the entropy of the surroundings decreases from 80 /kJ K to 75 /kJ KThe process is :(A) Reversible and isothermal (B) Irreversible

(C) Reversible (D) Impossible


m 100 kg=


Entropy of system :

s1 0.3 100 30 /kJ K#= = s2 0.4 100 40 /kJ K#= =Entropy change of the system :

s systemΔ^ h s s2 1= − 40 30 10 /kJ K= − =Entropy of surroundings :

s1 80 /kJ K= s2 75 /kJ K=Entropy change of the surroundings

s surroundingΔ^ h s s2 1= − s surroundingΔ^ h 5 /kJ K=−Entropy change of universe:

s universeΔ^ h s ssystem surroundingΔ Δ= +^ ^h h

10 5 5 /kJ K= − =Since s universeΔ^ h 0>hence process is irreversible

MCQ 1.38 A reversible engine operates between temperatures andT T1 2. The energy rejected by this engine is received by a second reversible engine at temperature T2 and rejected to a reservoir at temperature T3. If the efficiencies of the engines are same then the relationship between , andT T T1 2 3 is given by

(A) T T T22

1 3= + (B) T T T2 12

32= +

(C) T T T2 1 3= (D) T T T22

21 3= +



If 1η 2η=

TT1

1

2− TT1

2

3= −

& T22 T T3 1= T T T2 3 1=

MCQ 1.39 The internal energy of certain system is a function of temperature alone and is given by the formula 25 0.25 kJE t= + . If this system executes a process for which the work done by it per degree temperature increase is 0.75 kNm, the heat interaction per degree temperature increase, in kJ, is(A) .1 00− (B) .0 50−

(C) .0 50 (D) .1 00


E . t25 0 25= +

dtdE 0.25 /kJ Cc=

and dtdW 0.75 /kJ Cc=

From the first law of thermodynamics

dQ dE dW= + dQ 0.25 0.75 1.00 /kJ Cc= + =

MCQ 1.40 The given figure shows the Klein’s construction for acceleration of the slider-crank mechanism

Which one of the following quadrilaterals represents the required acceleration diagram ?(A) ORST (B) OPST

(C) ORWT (D) ORPT

SOL 1.40 Correct option is (B)Procedure to get acceleration polygon by Klein’s construction


(1) Firstly, draw the configuration diagram of slider

(2) After getting configuration diagram OCP , now draw a line through ' 'O perpendicular to the line of stroke OP

(3) External the connecting rod length PC to meet this perpendicular, name the intersection joint as M

(4) OCMΔ is a velocity polygon

(5) With C as centre and with CM as radius draw a circle

(6) Draw another circle with diameter as length P

(7) Then KL represents the common chord of these two circle.

(8) Extend KL to meet the line of stoke at N . Also KL is intersecting to PC at Q

(9) Then Prof. Klein suggested that OCQN4 represents, the acceleration polygon of slider crank mechanism.

MCQ 1.41 In the epicyclic gear train shown in the given figure, ‘A’ is fixed. A has 100 teeth and B has 20 teeth. If the arm C makes three revolutions, the number of revolutions made by B will be

(A) 12 (B) 15

(C) 18 (D) 24



N NN N

B C

A C

−− T

TA

B=−

& N 30 3

B −− 100

20=−

15 N 3B= − NB 18=

MCQ 1.42 A single block brake is show in Figure. The diameter of the Drum is 250 mm and the angle of contact is 90c. If the operating force of 700 N is applied at the end of a lever and the coefficient of friction between the drum and the lining is .0 35, determining the torque that may be transmitted by the block brake.

(A) N1225 (B) 83.75 -N m

(C) N350 (D) N525

SOL 1.42 Correct option is (B)Given 250 mmd = or 125 mmr = ; 2 90 /2 radcθ π= = , 700 NP = , .0 35μ =Since the angle of contact is greater than 60c, therefore equivalent coefficient of friction,

μl /

. .sinsin

sinsin

2 24

2 904 0 35 45 0 385# #

cc

θ θμ θ

π= + = + =

Let RN = Normal force pressing the block to the brake drum, and

Ft = Tangential braking force .RNμ= l

Taking moments above the fulcrum O , we have

F700 250 200 50t #+ +^ h 200 200 . 200R F F0 385N

t t

μ# # #= = =l

520 Ft=or F F520 50t t− 700 450#=or Ft 700 450/470 670 N#= =We know that torque transmitted by the block brake

TB 670 125 83750 N mmF rt # # −= = = 83.75 N m−=


MCQ 1.43 Consider a gray and opaque surface at 0 Cc in an environment at 25 Cc . The surface has an emissivity of .0 8. If 300 /W m2 is the radiation incident on the surface, the radiosity of the surface is(A) 60 /W m2 (B) /W m132 2

(C) /W m300 2 (D) /W m312 2

SOL 1.43 Correct option is (D)We have

0 CT c= , 2.5 CT c=3 , .0 8ε = , 300 /W mG 2= , 5.67 10 /W m k8 2 4σ #= −

Radiority J T G273 14# # #ε σ ε= + + −^ ^h h

0.8 5.67 10 55.55 10 .1 0 8 3008 8## # # #= + −− ^ h

.251 97 60= + 312 /W m2=

MCQ 1.44 Consider a horizontal 0.7 m wide and 0.85 m long plate in a room at 30 Cc . Top side of the plate is insulated while the bottom side is maintained at 0 Cc . The rate of heat transfer from the room air to the plate by natural convection is (For air, 0.02476 /W m Ck c= , .Pr 0 7323= , 1.47 10 /m s5 2ν #= − )(A) 36.8 W (B) . W43 7

(C) . W128 5 (D) . W92 7

SOL 1.44 Correct option is (B)We have

0.7 mb = , 0.85 mh = , 30 CT c=3 , 0 CTs c= , 0.02476 /W m Ck c= ,.P 0 7323r = , 1.470 10 /m sv 5 2

#= −

Volume expansion co-efficient

β T1f

= ....(i)

and Tf T T

2s= + 3

288 K2273 303= + =

From (i), β 3.472 10 k2881 3 1

#= = − −

The characteristic length

L . .

. .b hbh

2 2 0 7 0 850 7 0 85#=

+=

+^ ^h h

Rayleigh number

Ra v

T T Lp

2 sr2

3β=

−3^ h


.

9.81 . . .1 470 10

3 472 10 303 273 0 192 0 73235

3 3

#

# ##=−

−

−

^

^ ^ ^ ^

h

h h h h

.24 50 106#=

In natural convection, the Nusselt number is given by

Nus . R0 27 .a0 25=

. .0 27 24 50 10 .6 0 25#= ^ h

19=

Then h NusLk=

..0 192

0 02476 19#=

2.45 /W m K2$=

As b h#= 0.7 0.85 0.595 m2

#= =Thus rate of heat transfer

Qo hA T Ts s= −3^ h

. .2 45 0 595 303 273#= −^ h

43.7 W=

MCQ 1.45 Match List I (Cycle operating between fixed temperature limits) with List II (Characteristics of cycles efficiency η temperature) and select the correct answer using the codes given below the lists :

List-I List-II

a. Otto cycle 1. η depends only upon temperature limits

b. Diesel cycle 2. η depends only on pressure limits

c. Carnot cycle 3. η depends on volume compression ratio

d. Brayton cycle 4. η depends on cut-off ratio and volume compression ratio

Codes : a b c d(A) 3 4 1 2(B) 1 4 3 2(C) 3 2 1 4(D) 1 2 3 4

SOL 1.45 Correct option is (A)Otto Cycle


ottoη r

1 11= − γ−

r = compression ratio VV

2

1=

γ = specific heat ratio CC

v

p=

Diesel Cycle

Diseselη r r

r1 1

11

c

c1 γ= − −

−γ

γ

− ^c

hm

r = compression ratio VV

2

1=

rc = cut off ratio VV

2

3=

Brayton cycle

Braytonη r

1 1/

p1= − γ γ−

where

rp = pressure ratio pp

1

2=

Carnot Cycle

Carnotη TT1

1

2= −

T2 = higher temperature

T1 = lower temperature

MCQ 1.46 Consider the data given in the following table :

Production Plan

Period Demand Regular Prod. Overtime Prod.

Others

1. 500 500 ... ...

2. 650 650 ... ...

3. 800 650 150 ...

4. 900 650 150 ?Given the fact that production in regular and overtime is limited to 650 and 150 respectively, the balance demand of 100 units in the 4th period can be met by :(A) Using overtime in period 2

(B) Using regular production in period 1


(C) Subcontracting

(D) Using any of the steps indicated in (A), (B) and (C).

SOL 1.46 Correct option is (B)In the 4th period since regular production is limited to 650 and overtime is limited to 150, the balance demand of 100 units in the 4th period can be met by using regular production in period .1 a Demand is only 500 but production capacity in regular period can be increased to 600 to meet the demand.

MCQ 1.47 Time estimates of an activity in a PERT network are :Optimistic time t 90 = days; pessimistic time t 21p = days and most likely time t 15m = days. The approximates probability of completion of this activity in 13 days is :(A) %16 (B) %34

(C) %50 (D) %84


texpected t t t

64 m p0= + +

9

64 15 21

15#=+ +

=^ h

. .S D b a

6 621 9 2=

−= − =^ h

Z x x2

13 15 1σ= − = − =−

p 1−^ h .0 1586= .0 16-

Common Data For Q.• 48 and 49.The design specifications of a 2 m long solid circular transmission shaft require that the angle of twist of the shaft not exceed 3° when a torque of 9 kN-m is applied.

MCQ 1.48 Which of the following is the required diameter of the shaft if the shaft is made of a steel with an allowable shearing stress of 90 MPa and a modulus of rigidity of 77 GPa ?(A) 41.06 mm (B) 79.9 mm

(C) 39.9 mm (D) 82.1 mm

SOL 1.48 Correct Option is (D)3 52.360 10 rad3cφ #= = − , 9 10 N mT 3

# −= , 2.0 mL =


φ GJTL

c GTL24π

= =

c4 GTL2

π φ= based on twist angle

τ JTc

cT2

2π= =

c3 T2πτ= based on shearing stress

Steel shaft: 90 10 Pa6τ #= , 77 10 PaG 4#=

Based on twist angle

c4 ( )( . )

( )( )( . )2.842 10 m

77 10 52 360 102 9 10 2 0

4 3

36 4

# #

#

π #= =−−

c 41.06 10 41.06m mm3#= =− , 2 82.1 mmd c= =

Based on shearing stress

c3 ( )

( )63.662 10 m

90 102 9 10

6

36 3

#

#

π #= = −^ h

c 39.93 10 39.93m mm3#= =− , . mmd 79 9=

Required value of d is the larger, so . mmd 82 1=

MCQ 1.49 Which of the following is the required diameter of the shaft if the shaft is made of a brass with an allowable shearing stress of 35 MPaand a modulus of rigidity of 42 GPa?(A) 54.07 mm (B) 109.4 mm

(C) 95.6 mm (D) 47.8 mm

SOL 1.49 Correct Option is (B)Bronze shaft: 35 10 Pa6τ #= , 42 10 PaG 9

#=

Based on twist angle

c4 ( )( . )

( )( )( . )5.2103 10 m

42 10 52 360 102 9 10 2 0

9 3

36 4

# #

#

π #= =−−

c 47.78 10 47.78m mm3#= =− , 2 95.6 mmd c= =

Based on shearing stress

c3 ( )

( )( )163.702 10 m

35 102 9 10

6

36 3

#

#

π #= = −

c 54.70 10 54.70m mm3#= =− , 2 109.4 mmd c= =

Required value of d is the larger 109.4 mmd =

Common Data For Q.• 50 and 51.


The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25 Cc , the pressure gage reads 210 kPa. The volume of the tire is 0.025 m3 and the atmospheric pressure is 100 kPa.

MCQ 1.50 What will be the pressure rise in the tire when the air temperature in the tire rises to 50 Cc ?(A) . kPa19 5 (B) kPa13

(C) 26 kPa (D) kPa52

SOL 1.50 Correct option is (C)Initially, the absolute pressure in the tire is

P1 210 100 310 kPaP Patmg= + = + =Treating air as an ideal gas and assuming the volume of the tire to remain constant, the final pressure in the tire can be determined from

TP V

1

1 1 336 kPaTPV P T

T P 298323 310

2

1 22

1

21= = = =^ h

Thus the pressure rise is

PΔ 336 310 26 kPaP P2 1= − = − =

MCQ 1.51 The amount of air that must be bled off to restore pressure to its original value at this temperature, is(A) 0.0070 kg (B) 0.0 kg140

(C) 0.00 kg35 (D) 0.0 kg0525

SOL 1.51 Correct option is (A)The amount of air that needs to be bled off to restore pressure to its original value is

m1 . ..

0.0906 kgRTP V

0 287 0 298310 0 025

1

1= = =^ ^

^ ^

h h

h h

m2 ..

0.0 kgRTP V

0 287 323310 0 025

8362

1= = =^ ^

^ ^

h h

h h

mΔ 0.0906 0.0836 0.0070 kgm m1 2= − = − =

Common Data For Q.• Linked Answer Q.52 and 53.A uniform rectangular rod having the cross-section of 0.17 0.31m m# is shown in figure


MCQ 1.52 The specific weight of the rectangular rod in /kN m3 will be(A) 3.14 (B) 62.8

(C) 6.28 (D) 0.628

SOL 1.52 Correct option is (C).The FBD of the rod is shown below

Weight of the rectangular rod

W vγ #= ( . . ) .0 17 0 31 10 0 527# # #γ γ= = ...(i)

where γ = specific weight of the rod.The Buoyancy force,

FB vH O submerged2γ #= (0.17 0.31 8) 0.422H O H O2 2γ γ# # #= = ...(ii)Taking the moment about O,

MOΣ 0=

cosW 210

# θ cosF 28

B θ#=

Substitute the values of W & FB from equation (i) & (ii), we get

.0 527 5#γ 0.422 9.8 41# #= γ 6.28 /kN m3-

MCQ 1.53 What will be the tension in string ?(A) 3731 N (B) 826 N

(C) 373.1 N (D) 82.6 N

SOL 1.53 Correct option is (B).From previous part of the question.


In equilibrium condition(vertical direction).

FVΣ 0= W T+ FB= T F WB= − . . . .0 422 9 8 0 527 6 28# #= − 0.826 826kN N= =

Common Data For Q.• Linked Answer Q.54 and 55.A cold rolled steel cup with an inside radius of 30 mm and a thickness of 3 mm is to be drawn from a blank of diameter 40 mm. The shear yield stress and the maximum allowable stress of the material can be taken as 210 /N mm2 and 600 /N mm2 respectively.

MCQ 1.54 What will be drawing force, if the coefficient of friction .0 1μ = and .0 05β = ?(A) kN47 (B) . kN125 4

(C) 62.7 kN (D) . kN31 35

SOL 1.54 Correct option is (C)We first calculate the blank holding force Fh from the given data as

Fh 0.05 40 210 52,778N N2π # #= =Next, we find the value of rσ at r rd= by using Eq. (26.18). Thus

rσ at r rd= . , lnr 40 30 1 52 778 3 210 30

40# #

## #π= = +σ b l

14 80.8 / 94.8 /N mm N mm2 2= + =Now using Equation, we get

rσ 94.8 / 110.9 /N mm N mme . /0 1 2 2 2#= =π

It should be noted that this is much less than the fracture strength though

10 .mmr r t3 33j p− = = i.e., very close to the limit set by the condition of

plastic buckling. From Equation, the drawing force is found to be s

F 2 30 110.9 62,680N Nπ # #= =In this case, 600 /N mmz

2σ = From Eq.

r rrd

σ = / 512.8 /N mm N mme600

.0 052 2= =π

MCQ 1.55 What will be the minimum possible radius of the cup which can be drawn from the given blank without causing a fracture ?(A) . mm12 4 (B) . mm4 65

(C) . mm3 1 (D) 6.2 mm


SOL 1.55 Correct option is (D)Using Equation and taking 52,778 NFh = , we get

ln r 340

p +c m .

,. , . . .420

512 82 21 040 3

0 1 52 778 1 221 0 033 1 888# #

#π= − = − =

or rp 3 6.2 mme40.1 188= − =

Again, it is intersecting to note that 30.8 4mmr r tj p $− = , and this goes

much beyond the limit set by the plastic buckling condition.


MCQ 1.56 Which one of the following is the Antonym of the word SILENCE ?(A) attune (B) babble

(C) achromatic (D) aurora


MCQ 1.57 Which one of the following is the synonym of the word ADMONISH ?(A) warn (B) escape

(C) worship (D) distribute


MCQ 1.58 One of the four words given in the four options does not fit the set of words. The odd word from the group, is(A) Swim (B) Swill

(C) Ablution (D) Bathe


MCQ 1.59 A pair of CAPITALIZED words shown below has four pairs of words. The pair of words which best expresses the relationship similar to that expressed in the capitalized pair, isNUTS : BOLTS(A) Nitty : Gritty (B) Bare : Feet

(C) Naked : Surprise (D) Hard : Soft


MCQ 1.60 In the following sentence, a part of the sentence is left unfinished. Four different ways of completing the sentence are indicated. The best alternative among the four, is

In pursuance of their decision of resist what they saw as anti-labour policies, the company employee’s union launched agitation to...............


(A) Show their virility

(B) reaffirm their commitment of the company

(C) bring down the government

(D) demonstrate their strength.



MCQ 1.61 Let N 1421 1423 1425# #= . What is the remainder when N is divided by 12 ?(A) 0 (B) 9

(C) 3 (D) 6

SOL 1.61 Correct option is (C)When we divide N by 12, the reminder for the expression will be the same as the reminder for 5 7 9# # . Now 35 9# , when divided by 12, we get the reminder 11 9 99# = .Finally divided it by 12, we get the reminder of N as 3.

MCQ 1.62 There are 60 students in a class. These students are divided into three groups A, B , and C of 15, 20 and 25 students each. The groups A and C are combined to form group D . What is the average weight of the students in group D ?(A) more than the average weight of A

(B) more than the average weight of C

(C) less than the average weight of C

(D) cannot be determined

SOL 1.62 Correct option is (D)There is no indication of weights of the students in the question i.e. groups A, B and C . Therefore it is not possible to find the relation between the groups A, B , C and D .

MCQ 1.63 Four cities are connected by a road network as shown in the figure. In how many ways can you start from any city and come back to it without travelling on the same road more than once ?


(A) 8 (B) 12

(C) 16 (D) 20

SOL 1.63 Correct option is (B)Consider a man starts from a city. The condition is that there is no repetition of path during his journey. So two paths are restrict for one pass only. Thus the total number of possible cases

P 1242= =

MCQ 1.64 The roots of the equation ax x3 6 02 + + = will be reciprocal to each other if the value of a is(A) 3 (B) 4

(C) 5 (D) 6


We have 3 6ax x2 + + 0=The roots of this quadratic equation is

x aa

23 9 24!= − −

It is given that the roots are reciprocal to each other. Thus

aa

23 9 24− + −

32

aa9 24

=− − −

or a4 2 9 24a3 2= − − −] ]g g

or a ,0 6=But zero is not a possible root because the reciprocal of zero becomes infinity.

Thus the correct root is 6.

MCQ 1.65 A man starting at a point walks one km east, then two km north, then one km east then one km north, then one km east and then one km north to arrive at the destination. What is the shortest distance from the starting point to the destination ?(A) 2 km2 (B) 7 km

(C) 3 km2 (D) 5 km



Let the man starting from point O and reaches his destination at point A. The shortest distance between O and his destination is

OA 3 4 5 km2 2= + =] ]g g

Mock Test - 2

1. (C) 11. (D) 21. (B) 31. (D) 41. (C) 51. (A) 61. (C)

2. (A) 12. (D) 22. (D) 32. (C) 42. (B) 52. (C) 62. (D)

3. (A) 13. (B) 23. (A) 33. (C) 43. (D) 53. (B) 63. (B)

4. (B) 14. (A) 24. (B) 34. (C) 44. (B) 54. (C) 64. (D)

5. (D) 15. (C) 25. (A) 35. (A) 45. (A) 55. (D) 65. (D)

6. (A) 16. (D) 26. (B) 36. (C) 46. (B) 56. (B)

7. (B) 17. (C) 27. (C) 37. (B) 47. (A) 57. (A)

8. (C) 18. (C) 28. (A) 38. (C) 48. (D) 58. (A)

9. (B) 19. (A) 29. (B) 39. (D) 49. (B) 59. (A)

10. (B) 20. (C) 30. (B) 40. (B) 50. (C) 60. (D)

SolutionsME-C


MCQ 1.1 The minimum point of the function x x33

− is at

(A) x 1= (B) x 1=−

(C) x 0= (D) x3

1=


f x^ h x x33

= −

f xl h x 1 02= − =

& x 1!= f xll h x2=At n 1=− , f x 2 0<=−ll h

So, minimum value at x 1=−

MCQ 1.2 Inverse Laplace Transform of s a

s2 2+

is

(A) sina at1 (B) cosat

(C) sinha at1 (D) cosh at

SOL 1.2 Correct Option is (B)We know

cosL at^ h ,s a

s2 2=+

0s >

Thus

Ls a

s12 2+

−b l cosat=

Page 2 ME-C Chapter 1

MCQ 1.3 A3

141=

− −−> H; then An is

(A) n

nnn

1 2 41 2

+ −−> H (B)

n n31

41

n

n

−−

^

^

h

h> H

(C) nn

nn

1 31

1 41

++

−−> H (D)

nn

nn

1 21

41 2

++

−−> H

SOL 1.3 Correct Option is (C).

We have A 31

41=

−> H

A AA2 = 31

41

31

41=

−−

−−> >H H

9 43 1

12 44 1=

−−

− +− +> H

52

83=

−−> H

If we put n 2= in options, then only D satisfy.

MCQ 1.4 The solution of the differential equation log logxdxdy y y x 1= − +^ ^h h8 B is y =

(A) ecx (B) x exc

(C) x exc2 (D) x ex c2


xdxdy log logy y x 1= − +6 @

dxdy log log logx

y y x hy

hy1 1= − + = +6 9@ C

It is a homogenous differential equation,

Thus substitute

y vx=

dxdy v xdx

dv= +

So that

v xdxdv+ logv v 1= +6 @

xdxdv logv v=

logv vdv x

dx=

Chapter 1 ME-C Page 3

Integrate both the sides

log logv^ h log logx c= + logxc= logv xc= v exc=

xy exc=

or y x exc=

MCQ 1.5 The function f x x 1= +^ h on the interval ,2 0−6 @ is (A) continuous and differentiable

(B) continuous on the integral but not differentiable at all points

(C) neither continuous nor differentiable

(D) differentiable but not continuous

SOL 1.5 Correct Option is (B).

We have

f x^ h x 1= +

f x^ h forforfor

xx

xx

x

11

0

2 11 0

1

# #

# #

#

=− +

+− −−

−

^

^

h

h

Z

[

\

]]

]]

At x 1= Lf x^ h lim f h1

h 0= − −

"^ h

lim h h1 1h 0

= − − + ="

^ h

and Rf x^ h lim f h1h 0

= − +"

^ h

lim h h1 1h 0

= − + + ="

^ h

Thus f x^ h is continuous at x 1=−

Now Lf xl h 1

lim hf h f 1

h 0= −

− − − −"

^ ^h h

lim hh1 1 0

1h 0

= −− − − + −

=−"

^ h

and Rf xl h lim hf h f1 1

h 0=

− + − −"

^ ^h h

lim hh1 1 0

1h 0

=− + + −

="

^ h


Thus it is not differentiable

MCQ 1.6 A bell-crank lever ABC is pivoted at B and carries a weight 14 kN at C as shown in figure. Take 21 cmAB = and 30 cmBC = . The horizontal force F necessary to prevent turning of the lever, is

(A) . kN46 18 (B) 23.09 kN

(C) . kN17 32 (D) . kN28 86

SOL 1.6 Correct option is (B)Consider the figure If lever is in equilibrium then moment about point B must be zero.

AM F# BC 14#=or sinAB F60 #c 30 14#=

or F21 23

# # 30 14#=

or F 2309 kN=

MCQ 1.7 A weight of 10 kN is raised by two pulley system as shown in figure. The force F required to hold the weight in equilibrium, is

(A) kN10 (B) 5 kN

(C) 5 kN1 (D) kN20

SOL 1.7 Correct option is (B)Assume that F goes down through a distance y . From the geometry of Figure it may be easily seen that weight moves upward by y2

1 distance.Using the principal of virtual work


F y W y2# #− 0=

or F W2=

Here W 10 kN=

Thus F 5 kN210= =

MCQ 1.8 In a beam of circular cross-section, the shear stress variation across a cross-section is

.

.

SOL 1.8 Correct option is (A)Shear stress in circular cross-section

τ IF r y3

2 2= −^ h

We observe the following(i) Variation of τ versus y is a parabolic curve

(ii) τ increases as y decreases

(iii) at y r= ; 0τ =

(iv) at y 0= ; τ , is maximum

maxτ d

F rd

F d

3 64 643

24

2

4

2

#

##

π π= =b l


4aread

F F

3

434

2$

π= =

d^

nh

tmax 34

avgτ=

MCQ 1.9 The bending moment diagram shown in figure corresponds to the shear force diagram in

.

SOL 1.9 Correct option is (B)The corresponding beam to the given BMD is given below

The shear at any section is constant. Since the couple is applied at B, the bending moment diagram is discontinuous at midpoint, it is represented by two oblique straight lines and decrease suddenly at mid point by an amount equal to M.NOTE: Shear force does not change at the point of application of a couple and the bending moment change abruptly at the point of application of a couple

MCQ 1.10 A cantilever beam of length l is subjected to a concentrated load P at a distance of /l 3 from the free end. What is the deflection of the free end of the beam ? (EI is the flexural rigidity)

(A) EIPl

812 3

(B) EIPl

813 3


(C) EIPl

8114 3

(D) EIPl

8115 3

SOL 1.10 Correct option is (C)For cantilever

δ EIPL3

3

= and θ EIPI2

2

=

when L /l2 3=

δ ElPl1

8 3

δ=

and slope EIPl

EIPl

184

922 2

= =

Now deflection at free end

δl EIPl l

EIPl

EIPl

EIPl

818

3 92

818

2723 2 3 3

#= + = +

EIpl

8114 3

=

MCQ 1.11 If / 2nω ω = where ω is the frequency of excitation and nω is the natural frequency of vibrations, then the transmissibility of vibrations will be(A) 0.5 (B) 1.0

(C) 1.5 (D) 2.0


For 2nω

ω = transmissibility is 1 s

MCQ 1.12 The total number of instantaneous centres for a mechanism consisting of ‘n ’ links is(A) /n 2 (B) n

(C) n 21− (D)

( )n n2

1−



Total No. of I.C.R. n n

21

=−^ h

MCQ 1.13 Given thatu = velocity in the x -directionu = velocity in the y -directionA two-dimensional flow in x y− plane is irrotational if

(A) xu

yv

22

22= (B)

xu

yu

22

22=

(C) xv

yu

22

22= (D)

xv

yu

22

22=

SOL 1.13 Correct option is (C)Irrotional flow : If the fluid particle don’t rotate about their mass centre while moving in the direction of motion, the flow is called as irrotational flow (Free vortex)

ω xv

yu1 0

22

22

α= − =< F

xv

22

yu

22=

MCQ 1.14 If ‘n ’ variables in a physical phenomenon contained ‘m ’ fundamental dimensions, then the variables can be arranged into(A) n dimensionless terms (B) m dimensionless term

(C) ( )n m− dimensionless term (D) ( )n m+ dimensionless terms

SOL 1.14 Correct option is (C)Buckingham π theorem : The theorem states that if there are ' 'n variable involved contain physical phenomenon and if these variable contain ' 'm primary dimensions (e.g. M , L , T ) then the variable quantities can be expressed in terms of an equation containing n m−^ h dimensionless groups of parameters.

MCQ 1.15 Consider the following statements :1. A draft tube may be fitted to the tail end of a Pelton turbine to increase

the available head.

2. Kaplan turbine is in axial flow reaction turbine with adjustable vanes on the hub.

3. Modern Francis turbine is a mixed flow reaction turbine.

Which of these statements are correct ?(A) 1, 2, and 3 (B) 1 and 2

(C) 2 and 3 (D) 1 and 3

SOL 1.15 Correct option is (C)In a pelton turbine, draft tube can’t help to increase the available head since


the jet from nozzle gets exposed to atmospheric air.

MCQ 1.16 Which one of the followings is the correct expression for one-dimensional, steady-state, constant thermal conductivity heat conduction equation for a cylinder with heat generation?

(A) r rrk

rT e c

tT1

gen22

22

22ρ+ =ob l

(B) r rr

rT

ke

tT1 1gen

22

22

22

α+ =o

b l

(C) r rr

rT

tT1 1

22

22

22

α=b l

(D) r rr

rT

ke1 0gen

22

22 + =

ob l

SOL 1.16 Correct option is (D)The one-dimensional, steady state 0t

T =22_ i, constant thermal conductivity

heat conduction equation for a cylinder with heat generation is

r rr

rT

ke1 gen

22

22

+o

b l 0=

MCQ 1.17 The Biot number can be thought of as the ratio of(A) the conduction thermal resistance to the convective thermal resistance.

(B) the convective thermal resistance to the conduction thermal resistance.

(C) the thermal energy storage capacity to the conduction thermal resistance.

(D) none of the above.

SOL 1.17 Correct option is (A)Biot number is given by

Bi /L k

h

C1=

Convection thermal resistance at the surface of the bodyconduction thermal resistance within the body=

MCQ 1.18 The rake angle in a twist drill(A) varies from minimum near the dead centre to a maximum value at the

periphery.

(B) is maximum at the dead centre and zero at the periphery.

(C) is constant at every point of the cutting edge.

(D) is a function of the size of the chisel edge.

SOL 1.18 Correct option is (A).Normal rake angle α at a distance r from center is given by

α tan sin

tanDr2

1

β

ψ= −

b l> H


where ψ = helix angle

β = halfpoint angle

' 'r being the radius of the point of the cutting edge where the normal rake

is being evaluated.

MCQ 1.19 Match 4 correct pairs

List-I List-II

a. ECM 1. Plastic Shear

b. EDM 2. Erosion/Brittle fracture

c. USM 3. Corrosive reaction

d. LBM 4. Melting and vaporisation

5. Ion displacement

6. Plastic shear and ion displacementCodes : a b c d(A) 1 3 4 2(B) 2 1 4 5(C) 5 1 2 4(D) 1 4 2 3

SOL 1.19 Correct option is (C).

List-I List-II

ECM 5. Ion displacement

EDM 1. Plastic Shear

USM 2. Erosion/Brittle fracture

LBM 4. Melting and vaporisation

MCQ 1.20 Size of a shaper is given by(A) stroke length (B) motor power

(C) weigh of the machine (D) table size

SOL 1.20 Correct option is (A).Shaper size is designated by its longest nominal cutting stroke. On most horizontal shapers, the table can be fed cross-wire a distance at least as large as the stroke.

MCQ 1.21 In blanking operation, the clearance provided is(A) %50 on punch and %50 on die

(B) on die

(C) on punch


(D) on die or punch depending upon designers choice

SOL 1.21 Correct option is (C).The difference in dimensions between the mating members of a die set is called “clearance”. In blanking operation, the clearance provided in on punch.

MCQ 1.22 When a steel is heated in a furnace and then cooled in air at ordinary temperature, the process is one of(A) Annealing (B) Hardening

(C) Normalizing (D) Tempering

SOL 1.22 Correct option is (C).Normalizing is a heat treatment process which is used for heating the metal to a temperature slightly above the critical temperature and then cooling in air. Thus the given process is Normalizing.

MCQ 1.23 Duralumin alloy contains aluminium and copper in the ratio of % Al % Cu

(A) 94 4

(B) 90 8

(C) 98 10

(D) 86 12

SOL 1.23 Correct option is (A).Duralumin alloy contains aluminium and copper in the ratio of :94 4

MCQ 1.24 Which one of the following statements applicable to a perfect gas will also be true for an irreversible process ? (Symbol have the usual meanings)(A) dQ du pdV= + (B) dQ Tds=

(C) Tds du pdV= + (D) None of these

SOL 1.24 Correct option is (C).The first Tds equation or Gibbs equation

Tds du pdV= +is applicable for both the perfect gas an for an irreversible process.

MCQ 1.25 Match List I (Or technique) with List II (Application) and select the correct answer using the codes given below the lists :

List-I List-II

a. Linear programming 1.

b. Transportation 2. Machine allocation decision

c. Assignment 3. Product mix decision


d. Queueing 4. Project management decision

5. Number of servers decisionCodes : a b c d(A) 1 2 3 5(B) 3 1 2 5(C) 1 3 4 5(D) 3 2 1 4


List-I List-II

a. Linear programming 3. Product mix decision

b. Transportation 1. Warehouse location decision

c. Assignment 2. Machine allocation decision

d. Queueing 5. Number of servers decision

Q.26 TO Q.55 CARRY TWO MARKS EACH

MCQ 1.26 A problem in mechanics is given to three students A, B and C whose

chances of solving it are 21 , 3

1 and 41 respectively. The probability that the

problem will be solved, is

(A) 41 (B) 4

2

(C) 43 (D) 5

3


Required Probability P= (at least one of the students

solve the problem)

P1= − (none of the student solves the problem)

1 1 21 1 3

1 1 41= − − − −b b bl l l

1 21

32

43

43

# #= − =

Alternate:

Probability that A will not solve the problem

1 31

32= − =

Probability that B will not solve the problem

1 31

32= − =


Probability that C will not solve the problem

1 41

43= − =

Probability that all three will not solve the problem

21

32

43

41

# #= =

The probability that all the three will be solved the proble

1 41

43= − =

MCQ 1.27 Laplace Transform of sine btat is

(A) s a b

b2 2− +^ h

(B) s a b

s a2 2− +

−^ h

(C) s a b

b2 2− −^ h

(D) s a b

s a2 2− −

−^ h


Let f t^ h e a ib t= +^ h

then L f t^ h" , s a i b s a i b1 1=

− += − −^ h

Multiply and divide by the conjugate, so that

s a ib s a i b

s a i b1#=

− − − +− +

^ ^

^

h h

h

s a bs a i b

2 2=− +− +

^

^

h

h ...(i)

Also e a i b t+^ h cos sine bt i btat= +6 @

cos sine bt i e btat at= +Hence

L f t^ h" , cos sinL e bt i L e btat at= +" ", , ...(ii)

Thus by equation (i) & (ii), we get

sinL e btst^ h s a b

b2 2=

− +^ h

MCQ 1.28 The rank of a 3 3# matrix C AB=^ h, found by multiplying a non-zero column matrix A of size 3 1# and a non-zero row matrix B of size 1 3# , is(A) 0 (B) 1

(C) 2 (D) 3


Let A andabc

B a b c1

1

1

2 2 2= =

R

T

SSSS

6

V

X

WWWW

@


C AB= abc

a b c1

1

1

2 2 2=

R

T

SSSS

6

V

X

WWWW

@

a ab ac a

a bb bc b

a cb cc c

1 2

1 2

1 2

1 2

1 2

1 2

1 2

1 2

1 2

=

R

T

SSSS

V

X

WWWW

It is a 3 3# square matrix. Thus A 3#ρ^ h

c a a b c b c b c b c a b b a c c b a c c1 2 1 1 2 2 1 1 2 2 1 2 1 2 1 2 1 2 1 2= − − −^ ^h h

a c b b a c b b a c1 2 1 2 2 1 1 2 2 1+ −^ h

c 0=So that Aρ^ h 2#

Now consider the minor of order 2 2# .

We see that all the minors of order 2 2# are zero.

Thus Aρ^ h 0=

MCQ 1.29 If the product of matrices

cos

cos sincos sin

sinA2

2

θθ θ

θ θθ= > H and

coscos sin

cos sinsinB

2

2

φφ φ

φ φφ= > H

is a null matrix, then θ and φ differ by(A) an even multiple of /2π (B) an even multiple of π

(C) an odd multiple of /2π (D) an odd multiple of π


AB cos cos coscos sin cos

cos sin cossin sin cos

θ φ θ φφ θ θ φ

θ φ θ φθ φ θ φ=

−−

−−

^

^

^

^

h

h

h

h> H

is a null matrix when cos 0θ φ− =^ h , this happens when θ φ−^ h is an odd

multiple of /2π .

MCQ 1.30 The particular integral for the differential equation dxd y

dxd y

dxdy x6 13

3

2

22− − = +

is given by

(A) x x x21

9253 2− + (B) x x x9

141

12253 2+ −

(C) x x x18 36 108253 2

− − − (D) x x x31

121

36253 2− + −


We have dyd y

dxd y

dxdy63

3

2

2

− − x1 3= +

. .P I D D D

x6

1 13 22=

− −+^ h


D D D

x6

1 122=

− −+

^^h

h

DD D x6

1 1 6 12 1

2= − + − +−

c ^m h

....DD D D D

61 1 6 6

2 2 2

=− − − + − +b l< F x1 2+^ h

.....DD D x6

1 1 6 367 1

22=− − + + +c ^m h

D x x61 1 3 18

72=− + − +b l

x x x x61

3 6 1873 2

=− + − +c m

x x x18 36 108253 2

=− + −

MCQ 1.31 In an ideal refrigeration (reversed Carnot) cycle, the condenser and evaporator temperature are 27 Cc and 13 Cc− respectively. The COP of this cycle would be(A) 6.5 (B) 7.5

(C) 10.5 (D) 15.0


COP .260 300300

40300 7 5= − = =

MCQ 1.32 In a tool life test, doubling the cutting speed reduces the tool life to th81 of

the original. The Taylor’s tool life index is(A) /1 2 (B) /1 3

(C) /1 4 (D) /1 8


V T n1 1 V T T

TVVn

n

2 22

1

1

2= =c m


TT8 n

1

1b l 2= , 2 2 /n3 1=

n 31=

MCQ 1.33 For the manufacture of full depth spur gear by hobbing process, the number of teeth to be cut 30= , module 3 mm= and pressure angle 20c= . The radial depth of cut to be employed should be equal to(A) 3.75 mm (B) 4.50 mm

(C) 6.00 mm (D) 6.75 mm

SOL 1.33 Correct option is (D).

Radial depth of cut = Tooth depth h^ h

Total depth h^ h 2.25 m= 2.25 3 6.75 mm#= =

MCQ 1.34 What will be the percentage change in cutting speed required to give an %80 reduction in tool life (i.e. to reduce tool life to one-fifth of its former

value), when the value of .n 0 12= .(A) . %21 4 (B) %25

(C) . %42 8 (D) %50


V T .1 1

0 12 V T .2 2

0 12=

VV

1

2 TT .

2

10 12

= b l

VV

1

2 /1 51 5

..

0 120 12= =c m

VV

1

2 .1 214=

Increasing in cutting speed . %21 4=

MCQ 1.35 A small tool manufacturer is interested in the development of a single point tool to be used in the machining of small servomotor cylinders, which is required in the hydraulic control system of the loop line of the automatic NC machine tool. The basic requirements established by tool engineer for tool development is the velocity of chip along tool face. To determine above factors, a seem less tubing 40 mm outside diameter is turned on a lathe. The rake angle used on tool is 32c. Cutting speed 15 mpm and feed 0.8 /mm rev is used. The length of continuous chip in one revolution is 60 mm. The cutting force and feed force are 250 kg and 100 kg respectively. What will be the velocity of chip along tool face ?


(A) 7.155 /minm (B) . /minm14 31

(C) . /minm8 94 (D) . /minm5 37

SOL 1.35 Correct option is (A)We are given:

Rake angle α^ h 32c=Outside diameter of tube 40 mm=Cutting speed Vc 12 /minm=Feed 0.8 /mm rev=Ic 60 mm=Feed force 100 kgFt= =Cutting force 250 kgFc= =We know that the coefficient of friction is given by

μ tantan

F FF F

c t

c1

αα= −

+

By substituting the values, we get

μ tantan

250 100 32100 250 32

cc= −

+

..

250 100 0 624100 250 0 624

##= −

+

.250 62 4100 156= −

+

.. .187 6

256 0 1 354= =

Shear plan angle is given by

tanφ sincosr

r1 c

c

αα= −

where rc = chop thickness ratio.

But rc tt

tt

2

1

1

2= =

where t1 = depth of cut (feed in this case)

t2 = thickness of chip

l1 = length of uncut chip

l2 = length of chip 60 mm= Dπ=

rc . .3 14 4060 0 477#

= =

tanφ sincosr

r1 c

c

αα= −


.

.sin

cos1 0 477 320 47 32#

cc= −

. .. .

1 0 477 0 52990 47 0 8480

##= −

..

.. .1 0 248

0 4040 75520 404 0 529= − = =

Chip velocity φ .27 5c= Vf V rc c#= 15 0.477 7.155 /m min#= =

MCQ 1.36 When milling a slot 20 mm wide 10 cm# long in a rectangular plate 10 20cm cm# , the cutting speed 60 /minm= , diameter of end drill

20 mm= , number of flutes 8= , feed 0.01 /mm flute= and depth of cut 3 mm= . The cutting time for the operation to be completed in a single

pass, is(A) . min3 0 (B) . min1 5

(C) .5 min4 (D) . min6 0

SOL 1.36 Correct option is (B)Cutting time is given as

Tm ,minFl x y= + +

Where l = length of job 100 mm=

x = Cutter approach 10 mmD2= =

y = Cutter over travel 5 mm= (say)

F = Table feed, mm/min

= f n Nt # #

f1 = feed per tooth, per rev, 0.01 mm= n = number of flutes 8= N = Spindle rev./min

DV 1000

2060 100#

##

π π= =

955 ./minrev=

Tm .0 01 8 955100 10 5

# #= + +

. . min76 4115 1 505= =

MCQ 1.37 A uniform wheel of 60 cm diameter weighting 2 kN weight is to be pulled over a block height of 15 cm by a pull force F as shown in figure. The least


pull through the center of the wheel, required just to overturn the wheel over the block, is

(A) . kN3 46 (B) . kN2 16

(C) . kN1 30 (D) 1.73 kN

SOL 1.37 Correct option is (D)As per problem statement configuration is shown in Figure. If F passes through center O , then it will produce maximum moment about A when F is applied perpendicular to AO .Now from geometry of figure

OA 30 cm= , 30 15 15 cmOB = − =

sinθ .3015 0 5= =

θ 30c=or AB 30 30 25.98cos cmc= =Taking moment about A.

AO F# AB 2#=or F30# .25 98 2#=or F 1.73 kN=In second case F is applied horizontally as shown in Figure

Taking moment about point A.

BD F# AB 2#=or F30 15 #+^ h .25 98 2#=

or F . 1.155 kN4525 98 2#= =

Note that in both case reaction at C will be zero when the wheel is about

to overturn.

MCQ 1.38 A block of mass 20 kg, lying on a smooth plane inclined at 30c to the horizontal is being pulled by a another block of mass 15 kg as shown in figure. The 15 kg block is connected to the first block by a light in extensible string and hangs freely beyond the frictionless pulley. What will be the acceleration, with which the block 15 kg comes down?


(A) 14 / secm 2 (B) / secm21 2

(C) / secm7 2 (D) / secm28 2

SOL 1.38 Correct option is (A)As per problem statement the configuration is shown in figure

Let m1 20 kg= , 15 kgm2 = and 30cα =Let a be the acceleration of mass m2 in downward direction, then a will be

acceleration of m1 in upward direction along the plane. All force acting on

system is shown in figure.

Consider the motion of m2. Applying newton second law we get

m g T2 − m a2= ....(1)

Now consider the motion of m1. Applying newton second law we get

sinT m g1 α− m a1= ....(2)

Adding equation (1) and (2) we get

sinm g m g2 1 α− m a m a2 1= +

or a sing m mm m

2 1

2 1 α= +−

c m

Substituting the all values we get

a 9.8 14 /sin secm20 1515 20 30 2c= +

− =b l

MCQ 1.39 A body of mass 200 kg resting on rough horizontal plane is pulled by a force F applied to the body making an angle of 60c with the horizontal as shown in figure. The body attains a velocity of 80 / secm in 4 second. If coefficient of friction between the body and plane is .0 2, the value of F required is

(A) . N4901 78 (B) . N3267 85


(C) 6535.7 N (D) . N8169 63

SOL 1.39 Correct option is (C)Once the body start moving the force acting on it shown FigureResolving force vertically we get

sinF Rθ + mg=or R sinmg F θ= − . sinF200 9 8 60# c= − . F1960 0 86= −or .R F0 86+ 1960= ...(1)

Resolving force horizontal

cosF RFθ − ma=a can be calculated as

v u at= + 80 a0 4#= +

or a 20 / secm480 2= =

Thus .cosF R60 0 2c − 200 20#=

or .F R2 0 2− 4000=

or . F R2 2 − 20000= ....(2)

Adding equation (1) and (2) we get

. .F F2 5 0 86+ 21960=

or F . 65357 N3 3621960= =

MCQ 1.40 Match List I (State of stress) with List II (Kind of loading) and select the correct answer using the codes give below the lists :


Codes : a b c d(A) 1 2 3 4(B) 2 3 4 1(C) 2 4 3 1(D) 3 4 1 2



MCQ 1.41 The Euler’s crippling load for a 2 m long slender steel rod of uniform cross-section hinged at both the ends is 1 kN. The Euler’s crippling load for a 1 m long steel rod of the same cross-section and hinged at both ends will be(A) 0.25 kN (B) 0.5 kN

(C) 2 kN (D) 4 kN


PE IEI2

2π= for both ends pinged

1000 EI4

2π= ....(i)

.No PE EI1

2π= 1mat I =^ h

PE 4 kN=

MCQ 1.42 A rigid beam of negligible weight is supported in a horizontal position by two rods of steel and aluminium, m2 and 1 m long having values of cross-sectional area 1 cm2 and 2 cm2 and E of 200 GPa an 100 GPa respectively. A load P is applied as shown in the figure below.

If the rigid beam is to remain horizontal then(A) the forces on both sides should be equal

(B) the forces on aluminium rod should be twice the force on steel

(C) the force on the steel rod should be twice the force on aluminium

(D) the force P must be applied at the centre of the beam



For rigid beam is to remain horizontal

l Alδ^ h l .Stδ= ^ h

F2 100

1A1

## F

1 2002.st

##=

F F2A St1 =

MCQ 1.43 Which of the following are example of a kinematic chain ?

Select the correct answer using the codes given below :(A) 1, 3 and 4 (B) 2 and 4

(C) 1, 2 and 3 (D) 1, 2, 3 and 4

SOL 1.43 Correct option is (D)For kinematic chain

l p2 4= −where, l = no. of link

p = no. of pair

here p 4= l 4` =

MCQ 1.44 A counter-flow heat exchanger is used to cool oil ( 2.20 /kJ kg Ccp c= ) from 110 Cc to 85 Cc at a rate of 0.75 /kg s by cold water ( . /kJ kg Cc 4 18p c= ) that enters the heat exchanger at 20 Cc at a rate of 0.6 /kg s. If the overall heat transfer coefficient is 800 /W m2, the heat transfer area of the heat exchanger is(A) 0.745 m2 (B) 0.7 m60 2

(C) 0.7 5 m7 2 (D) 0.7 m90 2



We have110 CT ,inh c= , 85 CT ,outh c= , 0.75 /kg smh =o , 2.20 /kJ kg Ccph c= ,4.18 /kJ kg CCpc c= , 20 CT ,inC c= , 0.6 /kg smC =o , 0.800 /kN m CU 2c=

The rate of heat transfer by the hot fluid

Qo m c T T, ,in outh ph h h= −o ^ h

. .0 75 2 20 110 85= −^ ^ ^h h h

41.25 kW= ....(i)

Heat gain by the cold fluid is

Qo m c T T, ,out inc pc C C= −o ^ h

. . T0 6 4 18 20,outC= −^ ^ ^h h h ....(ii)

Equation (i) and (ii) for energy balance

.41 25 . T2 508 20,outC= −^ h

T ,outC 36.45 Cc=Thus T1Δ T T, ,in outh c= − .110 36 45= − 73.55 Cc=and T2Δ T T, ,out inh c= − 85 20= − 65 Cc=Log-Mean temperature

TlnΔ ln T

TT T

2

1

1 2

ΔΔ

Δ Δ= −

c m

.

. 69.ln

C

6573 55

73 55 65 2c= − =b l

Also Qo UA Tlns Δ=

As U TQ

lnΔ=o

. .. 0.74 m0 800 69 2

41 25 5 2

#= =

MCQ 1.45 Water ( 4180 /J kg Kcp = ) enters a 4 cm diameter tube at 15 Cc at a rate of 0.06 /kg s. The tube is subjected to a uniform heat flux of 2500 /W m2 on the surface. The length of the tube required in order to heat the water to 45 Cc is(A) 6 m (B) m12

(C) m18 (D) m24

SOL 1.45 Correct option is (D)We have


15 CTi c= , 45 CTe c= , 4 0.04cm mD = = , 0.06 /kg sm =o , 2500 /W mq 2=4180 /J kg KCp −=

Heat transfer rate

Qo mc T Tp e i= −o ^ h

.0 06 4180 45 15= −^ ^ ^h h h

7524 W=

Also As qQ=o

3 m25007524 2= =

So that the length of the tube is

L .3

tDA

t 0 04s

#= =

23.87 24 mb=

MCQ 1.46 The crank and slotted lever quick-return motion mechanism is shown in figure. The length of links , andO O O C O A1 2 1 2 are 10 , 20 5cm cm and cm respectively.

The quick return ratio of the mechanism is(A) 3.0 (B) 2.75

(C) 2.5 (D) 2.0



/Cos 2β^ h /1 2= β 120c=

βα 240

120 2= =

MCQ 1.47 A bronze spur pinion rotating at 600 r.p.m drives a cast iron spur gear at a transmission ratio of :4 1. The allowable static stresses for the bronze pinion and cast iron gear are 84 MPa and 105 MPa respectively. The pinion has 16 standard 20c full depth involute teeth of module 8 mm. The face width of both the gears is 90 mm. What will be the power that can be transmitted from the standpoint of strength?(A) . kW15 82 (B) . kW39 55

(C) . kW23 73 (D) 31.64 kW

SOL 1.47 Correct option is (D)Given : 600 . . .r p mNp = ; . . /V R T T 4G P= = ; 84 84 /MPa N mmOP

2σ = = ; 105 105 /MPa N mmOG

2σ = = ; T 16P = ; 8 mmm = ; 90 mmb =We know that pitch circle diameter of the piston,

DP . 8 16 128 0.128mm mm TP #= = = = Pitch line velocity,

v . . 4.02 /m sD N60 60

0 128 600P P # #π π= = =

Since the pitch line velocity v^ h is less than 12.5 /m s, therefore velocity

factor,

Cv . .v33

3 4 023 0 427= + = + =

We know that for 20c full depth involute teeth, tooth form factor for the

pinion

yp 0.154 . 0.154 . 0.097T0 912

160 912

P= − = − =


and tooth form factor for the gear.

yG . . . . .T0 154 0 912 0 154 4 160 912 0 14

G #= − = − =

.... /T T 4G Pa =^ h

yOP P#σ . .84 0 097 8 148#= =and yOG G#σ . .105 0 14 14 7#= =Since yOP Pσ #^ h is less than yOG G#σ^ h, therefore the pinion is weaker. Now

using the Lewis equation for the pinion, we have tangential load on the

tooth (or beam strength of the tooth),

WP . . . . .b m y C b m ywP P OP v P#σ π σ π= = ^ h

.CWP OP Pa σ σ=^ h

84 0.427 90 8 0.097 7870 Nπ# # # # #= = Power that can be transmitted

7870 4.02 31640 31.64W kWW vT # #= = = =

Common Data For Q.• 48 and 49.A vapour compression refrigerator works between the pressure limits of 60 bar and 25 bar. The working fluid is just dry at the end of compression and there is no under cooling of the liquid before the expansion valve.

Pressure(bar)

Satur.Temp.(K)

Enthalpy (kJ/kg) Entropy (kJ/kg K)

liquid kJ/kg

Vapour kJ/kg

Liquid Vapour

60 295 151.96 293.29 0.554 1.0332

25 261 56.32 322.58 0.226 1.2464

MCQ 1.48 What is the COP of cycle?(A) 4.36 / seckg (B) .60 / seckg3

(C) .70 / seckg2 (D) .39 / seckg5

SOL 1.48 Correct option (A)Given : 60 barp p2 3= = , 25 barp p1 4= = , 295 KT T2 3= = , 261 ,KT T1 4= =

151.96 /kJ kgh hf 3 4= = , 56.32 /kJ kghf1 = , 293.29 /kJ kghg2 = ,hg1 =322.58/kJ kg , 0.554 /kJ kg Ksf 2 = , 0.226 /kJ kg Ksf1 = , 1.0332 / ,kJ kg Ks sg2 2= =

1.2464 /kJ kg Ksg1 =The T s- and p h- diagrams are shown in Figure (a) and (b) respectively


Let x1 = Dryness fraction of the vapour refrigerant entering the

compressor at point 1

We know that entropy at point 1,

s1 s x s s x s sf fg f g f1 1 1 1 1 1 1= + = + −^ h s s sg f fg1 1 1a = +^ h

. . . . .x x0 226 1 2464 0 226 0 226 1 02041 1= + − = +^ h ....(i)

and entropy at point 2

s2 1.0332 /kJ kg Ksg2= = ....(ii)

Since the entropy at point 1 is equal to entropy at point 2, therefore equating

equations (i) and (ii).

. . x0 226 1 0204 1+ .1 0332=or x1 .0 791=We know that enthalpy at point 1,

h1 h x h h x h hf fg f g f1 1 1 1 1 1 1= + = + −^ h h h hg f fg1 1 1a = +^ h

56.32 0.791 . . 266.93 /kJ kg322 58 56 32= + − =^ h

COP of the cycle

. .. . 4.36 / seckgh h

h h293 29 266 93266 93 151 96f

2 1

1 3= −− = −

− =

MCQ 1.49 If the fluid flow is at the rate of 5 /minkg then capacity of the refrigerator will be?(A) 1.14 / secl (B) 0.04 / secl

(C) 2.25 / secl (D) 4.68 / secl

SOL 1.49 Correct option(C)

Common Data For Q.• 50 and 51.


Air flows through the device as shown in figure. If the flow rate is large enough, the pressure within the constriction will be low enough to draw the oil up into the tube. Neglect compressibility and viscous effects. (

12 /N mair3γ = , 8.95 /kN moil

3γ = )

MCQ 1.50 What will be the flow rate vo, in /m s3 .(A) .0 335 (B) 3.35

(C) 0.00335 (D) 0.0335


Applying Bernoulli’s equation at section (2) & (3), we have

pg

V z2air

2 22

2γ + + pg

V z2air

3 32

3γ= + +

Here p 03 = (gage pressure), z z2 3= (horizontal pipe)

So, pair

2

γ gV

gV

2 232

22

= − ...(i)

From continuity equation at section (2) & (3),

A V2 2 A V3 3=

D V4 22

2π D V4 3

23

π=

D V22

2 D V32

3=

V2 DD V V V25

50 42

32

3

2

3 3= = =b bl l ...(ii)

From figure p2 hoil #γ=−


. .8 95 10 0 33# #=−

2.68 10 Pa3#=− ...(iii)

Substitute the values of V2 & p2 from equation (ii) and (iii) into equation (i),

.12

2 68 103#−

( )g

Vg

VgV

2 24

2153

23

232

= − = −

V32 . . .12 15

2 68 10 2 9 8 291 823

## # #= =

V3 17.08 /m s=

Hence, flow rate vo (0.050) 17.08 0.0335 /m sA V 43 32 3π#= = =

Note : Same result is obtained from A V2 2

MCQ 1.51 What will be the pressure needed at section (1), to draw the water into section (2).(A) 0 (B) 2 kPa

(C) 0.02 kPa (D) 0.2 kPa


pg

V z21 1

2

1γ + + pg

V z23 3

2

3γ= + +

Here z z1 3= (horizontal pipe) p 03 = (gage pressure) V V1 3= (because A A1 3= )

So, p1

γ 0=

p1 0=

Common Data For Q.• Linked Answer Q.52 and 53.Steam at 5 MPa and 400 Cc enters a nozzle steadily with a velocity of 80 /m s, and it leaves at 2 MPa and 300 Cc . The inlet area of the nozzle is 50 cm2, and heat is being lost at a rate of 120 /kJ s. The inlet and outlet conditions are as given:

Pressure.kPa^ h

Temperature.Cc h

Specific volume/m kg3

^ h

Enthalpy/kJ kg^ h

2000 300 0.12551 3024.2

5000 400 0.057838 3196.7

MCQ 1.52 The mass flow rate of the steam is(A) . kg5 19 (B) . kg13 84

(C) 6.92 kg (D) . kg8 65


SOL 1.52 Correct option is (C).There is only inlet and one exit, and thus m m m1 2= =o o o . The mass flow rate of steam is

mo .v V A10 057838

1 80 50 101

1 14

#= = −^ ^h h

6.92 /kg s=

MCQ 1.53 What will be the exit area of the nozzle ?(A) . 10 m7 71 4 2

#− (B) 15.42 10 m4 2

#−

(C) . 10 m11 57 4 2#

− (D) . 10 m19 27 4 2#

−

SOL 1.53 Correct option is (D).We take nozzle as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

E E

, ,

in out

Rate of Net energy transferby heat work and mass

−o o1 2 344 44

0E, ,

, ,int

systemsteady

Rate of Change in ernal kineticpotential etc energies

0Δ= =3o ^ h

1 2 3444 444

Eino Eout= o

/m h V 21 12+o^ h /Q m h V 2out 2 2

2= + +o o^ h 0since peW , ,Δo_ i

Qout− o m h h V V22 1

22

12

= − + −oc m

Substituting, the exit velocity of the steam is determined to be

120− . 3024 3196.7V

6 916 280

100012

2

= − +−

^^

bchh

lm

It yields V2 562.7 /m s=The exit area of the nozzle is determined from

mo .. .

v V A A Vmv1

562 76 916 0 12551

22 2 2

2

2= = =o ^ ^h h

15.42 10 m4 2#= −

Common Data For Q.• Linked Answer Q.54 and 55.A project consists of eight activities with the following relevant information

Activity Immediate predecessor

Estimated duration (days)

Optimistic Most likely Pessimistic

A - 1 1 7

B - 1 4 7


C - 2 2 8

D A 1 1 1

E B 2 5 14

F C 2 5 8

G ,D E 3 6 15

H ,F G 1 2 3

MCQ 1.54 What will be the expected project completion time ?(A) 24 days (B) 38 days

(C) 19 days (D) 45 days


Activity to tm tp /t t t t4 6e o m p= + +^ h /t t 6p o2−^ h8 B

A 1 1 7 2 1

B 1 4 7 4 1

C 2 2 8 3 1

D 1 1 1 1 0

E 2 5 14 6 4

F 2 5 8 5 1

G 3 6 15 7 4

H 1 2 3 2 /2 6 2^ h

Using the precedence relationship among the activities, the resulting network is shown as Figure

From the above network diagram, we observe : :CPM 1 3 5 6 7" " " " , i.e. B E G H" " "

Expected duration of the project is 19 days.

MCQ 1.55 What duration will have %95 confidence for project completion ?(A) 24 days (B) 45 days


(C) 19 days (D) 38 days


Since, .P z 1 645#^ h . .0 5 0 45= + , i.e. .0 95;

We get T Te

s e

σ− .1 645=

or Ts 19 30.02 1.645 24 days#= + =Hence, 24 days of project completion time will have %95 confidence of

completion in the schedule time.


MCQ 1.56 Which one of the following is the Antonym of the word HOSTILE ?(A) alluvial (B) able

(C) amicable (D) alterable


MCQ 1.57 Which one of the following is the synonym of the word PROTAGONIST ?(A) prophet (B) explorer

(C) talented child (D) leading character


MCQ 1.58 One of the four words given in the four options does not fit the set of words. The odd word from the group, is(A) Coal (B) Humus

(C) Loam (D) Clay


MCQ 1.59 A pair of CAPITALIZED words shown below has four pairs of words. The pair of words which best expresses the relationship similar to that expressed in the capitalized pair, isBRAND : PRODUCT(A) Dalda : Rath (B) Aircraft : Flying Machine

(C) Ram : Boys (D) Sports car : Automobile



The highest reward for a man’s toil is not what he gets for it but what .......


(A) he makes out of it (B) he gets for others

(C) he has overcome (D) he becomes by it



MCQ 1.61 If the LCM of first 100 natural numbers is N , then the LCM of first 105 natural number will be(A) 51 N# (B) 101 103 N# #

(C) 105 /N 103 (D) 4N

SOL 1.61 Correct option is (B).If we look at the numbers 100 105N< # , we see only 101 and 103 do not have their factors in N (because these are primes), So obviously the new LCM will be 101 103 N# # .

MCQ 1.62 What will be the sum of the factors of 3129 ?

(A) 33 1132+^ h (B) 2

3 1129−^ h

(C) 3 12

128+^ h (D) 23 1130−^ h

SOL 1.62 Correct option is (D).If N is a number such that ......N a b cp q r

# #= where , ,a b c are prime number of N and , ,p q r are positive integers.

So, Required sum aa

bb

cc

11

11

11p q r1 1 1

# #= −−

−−

−−+ + +

b b bl l l

MCQ 1.63 If x is the smallest positive integer such that 2880 multiplied by x is the square of an integer, then x must be(A) 3 (B) 5

(C) 8 (D) 10

SOL 1.63 Correct option is (B).Arithmetic properties of numbersTo find the smallest positive integer x such that x2880 is the square of an integer, first find the prime factorization of 2880 by a method similar to the following:

2880 10 288#= (2 5) (2 )144# # #= 2 5 2 (2 )72# # # #= 2 5 2 2 (2 )36# # # # #= 2 5 2 2 2 (2 )18# # # # # #= 2 5 2 2 2 2 (2 )9# # # # # # #=


2 5 2 2 2 2 2 (3 )3# # # # # # # #= 2 3 56 2

# #=To be a perfect square, x2880 must have an even number of each of its prime

factors. At a minimum, x must have one factor of 5 so that x2880 has even

factors of each of the primes 2, 3, and 5. The smallest positive integer value

of x is then 5

MCQ 1.64 31

31

31

31

8 9 9 9+ + + is equal to

(A) 316 (B) 3

28

(C) 3236 (D) 3

48

SOL 1.64 Correct option is (B).Remember that you can only add fraction with the same denominator.Rearrange 3

18 so that it can be added to 3

19 . That is, try and turn 3

18 into a

fraction with 39 in the denominator.Multiply 3

18 by 3

3 to get

31

33

8 # 3 3

333

8 9#

= =

So 31

31

31

31

8 9 9 9+ + + 33

31

31

31

36

9 9 9 9 9= + + + =

Canceling out a factor of 3 gives

369 3 3

3 232

8 8#

#= =

MCQ 1.65 If it is given that m 1 th+^ h , n 1 th+^ h and r 1 th+^ h terms of an AP are in GP and , ,m n r in HP, then the ratio of the first term of the AP to its common difference in terms of will be(A) :n1 (B) : 2n1

(C) :n 2 (D) :n2 3

SOL 1.65 Correct option is (C).Since the m 1 th+^ h , n 1 th+^ h and r 1 th+^ h term of an A.P. are in G.P. so,

a nd 2+^ h a md a rd= + +^ ^h h (i)[assume a and d as the first term and common difference of an AP] Also m, n , ,r are in HP so,

n22 mr

m r= + ....(ii)

By solving the equation you will get

da m r n

n mr2

2

= + −− , put the /m r mr n2+ =

You will get / / :a d n n2 2.=− [–ve sign indicates that either common difference of first terms is –ve]


Mock Test - 3

1. (B) 11. (B) 21. (C) 31. (B) 41. (D) 51. (A) 61. (B)

2. (B) 12. (D) 22. (C) 32. (B) 42. (B) 52. (C) 62. (D)

3. (C) 13. (C) 23. (A) 33. (D) 43. (D) 53. (D) 63. (B)

4. (B) 14. (C) 24. (C) 34. (A) 44. (A) 54. (C) 64. (B)

5. (B) 15. (C) 25. (B) 35. (A) 45. (D) 55. (A) 65. (C)

6. (B) 16. (D) 26. (C) 36. (B) 46. (D) 56. (C)

7. (B) 17. (A) 27. (A) 37. (D) 47. (D) 57. (D)

8. (A) 18. (A) 28. (A) 38. (A) 48. (A) 58. (A)

9. (B) 19. (C) 29. (C) 39. (C) 49. (C) 59. (C)

10. (C) 20. (A) 30. (C) 40. (C) 50. (D) 60. (D)

SolutionsME-D


MCQ 1.1 For the differential equation dtdy y5 0+ = , with y 0 1=^ h , the general solution

is (A) e5 t5 (B) e 5t−

(C) e5t (D) e5 t5−

SOL 1.1 Correct Option is (B).

dtdy y5+ 0=

& ydy dt5=−

Integrating, logy c t5= −At ,t y0 1= =

log1 c 5 0#= −& c 0= logy t5=−& y e t5= −

MCQ 1.2 The product of two matrix A and .adj A is a(A) zero matrix (B) unit matrix

(C) scalar matrix (D) Matrix A itself

SOL 1.2 Correct Option is (C)The product of a matrix A and its adjoint is equal to equal to unit matrix multiplied by the determinant.Consider A be a square matrix, then

Adjoint A A$^ h AdjointA A A I$ $= =^ h

Page 2 ME-D Chapter 1

Let A abc

abc

abc

1

1

1

2

2

2

3

3

3

=

R

T

SSSS

V

X

WWWW

and adj A AAA

BBB

CCC

1

2

3

1

2

3

1

2

3

=

R

T

SSSS

V

X

WWWW

adjA A$ ^ h abc

abc

abc

AAA

BBB

CCC

1

1

1

2

2

2

3

3

3

1

2

3

1

2

3

1

2

3

#=

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

a A a A a Ab A b A b Ac A c A c A

a B a B a Bb B b B b Bc B c B c B

a C a C a Cb C b C b Cc C c C c C

1 1 2 2 3 3

1 1 2 2 3 3

1 1 2 2 3 3

1 1 2 2 3 3

1 1 2 2 3 3

1 1 2 2 3 3

1 1 2 2 3 3

1 1 2 2 3 3

1 1 2 2 3 3

=+ ++ ++ +

+ ++ ++ +

+ ++ ++ +

R

T

SSSS

V

X

WWWW

A

AA

A00

0

0

00

100

010

001

= =

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

A I=Which is the scalar matrix

MCQ 1.3 The unit vector normal to the surface x y z3 2 62 2 2+ + =^ h at , ,P 2 0 1^ h is

(A) i k2

1 +^ h (B) 2

j k1 +^ h

(C) i k+^ h (D) j k+^ h


, ,f x y z^ h x y z3 2 62 2 2= + + −

grad f xf

yf

zfi j k

22

22

22= + +

2 6 4x y zi j k= + +at , ,P 2 0 1^ h

grad f 4 4i k= +

grad f 16 16 32 4 2= + = =The unit vector normal to the surface at point P is

n grad

gradf

f1at P= ^ h

4 4i k i k4 2

12

1= + = +^ ^h h

MCQ 1.4 The function f t xy iy= +^ h is(A) continuous and analytic

(B) discontinuous and is not analytic

(C) discontinuous and analytic

(D) continuous and is not analytic

Chapter 1 ME-D Page 3

SOL 1.4 Correct Option is (D).

We have F t^ h xy iy= +Compare with F t^ h u iv= + u xy= , v y=From the cauchy-Reimann equation

xu

22 y= and

yv 1

22 =

Thus xu

22

yv

22!−

So that it is not a analytic function.

Since u and v are functions of x and y with non-zero values. Thus ,u v and

hence F z^ h are continuous functions.

MCQ 1.5 If V xy x y y z2 2 2= − + , the value of the .div Vd is(A) 0 (B) z z y z22 2+ +

(C) 2y z yz x2 − −^ h (D) z y y2 2 2− −^ h


Vd xV

yV

zVi j k

22

22

22= + +

z xy yz x x y zi j k2 2 22 2 2= − + − + −^ ^ ^h h h

V$d d h xz xy

yyz x

zx y z2 2 22 2 2

22

22

22

=−

+−

+−^ ^ ^h h h

y z y z y y2 2 2 22 2 2 2=− + − = − −^ h

MCQ 1.6 A block of 50 N weight rests in limiting equilibrium on rough inclined plane whose slope is 30c.The plane is raised to slope 60c. What will be the force F required acting along the plane that will just move the body up?(A) . N86 61 (B) . N72 17

(C) . N43 30 (D) 57.74 N

SOL 1.6 Correct option is (D)In this case

μ tan tan303

1cθ= = =

Figure show the case when slope is 60c and block is just about to move up

and figure shows all the force acting on the block. Here block is just about

to move up along the plane, therefore friction force oppose the motion and

is in down direction along the plane.

For equilibrium

R cosW 60c= ....(1)


F sinW R60c μ= + ....(2)

From equation (1) and (2) we get

Thus F sin cosW W60 60c cμ= + sin cosW W60 60c cμ= +^ h

but μ 3

1= and 50 NW =

Thus F sin cos50 603

1 60c c= +c m

57.74 N=

MCQ 1.7 The given figure shows the shear force diagram for the beam ABCD bending moment in the portion BC of the beam

(A) is a non zero constant

(B) is zero

(C) varies linearly from toB C

(D) varies parabolically from toB C

SOL 1.7 Correct option is (A)Whenever shear force is zero bending moment is constant

MCQ 1.8 Two identical springs labelled as 1 and 2 are arranged in series and subjected to force F as shown in the given figure.

Assume that each spring constant is K . The strain energy stored in spring 1 is(A) /F K22 (B) /F K42

(C) /F K82 (D) /F K162

SOL 1.8 Correct option is (C)Stain energy stored by system (U)

F F kF

kF

21

21

2 2# # #δ δ= = =: D

U kF4

2

=

Strain energy stored by spring....(I)


Uk

F2 8

2

= =

MCQ 1.9 Slenderness ratio of a column is defined as the ratio of its length to its(A) Least radius of gyration (B) Least lateral dimension

(C) Maximum lateral dimension (D) Maximum radius of gyration

SOL 1.9 Correct option is (A)Since buckling of column the crippling load with take place the axis of least resistance (i.e. about the axis about which moment of inertia is least)Hence slenderness ratio

least radius of gyrationlength=

MCQ 1.10 A straight bar is fixed at edges andA B . Its elastic modulus is E and cross-section is A. There is a load 120 NP = acting at C . The reactions at the ends are

(A) 60 N at A, 60 N at B (B) 30 N at A, 90 N at B

(C) 40 N at A, 80 N at B (D) 80 N at A, 40 N at B


RA 120 / 80 /BC AB N mm2#= =^ h

RB 120 / 40 /AC AB N mm2#= =

F.B.D.

R R1 2− 120= (i)

and I I1 2δ δ+^ ^h h 0=

A ER l

A ER l21 2

##

##+ 0=

R1 2R2=− (ii)

From eq. (i) and (ii)

R2 40=− R2 40 N= (opposite direction to our assumption)

and R1 80 N=


MCQ 1.11 The equation of motion for a single degree of freedom system with viscous damping is X X X4 9 16 0+ + =p o , the damping ratio of the system is

(A) 1289 (B) 16

9

(C) 8 2

9 (D) 89


mx cx kx+ +p o 0= m 4= , c 9= , k 16=

ξ km

C2 2 4 16

9169

#= = =

MCQ 1.12 If the energy grade line and hydraulic grade line are drawn for flow through an inclined pipeline the following four quantities can be directly observed :1. Static head 2. Friction head

3. Datum head 4. Velocity head

Starting from the arbitrary datum line, the above types of heads will be in the sequence(A) 3, 2, 1, 4 (B) 3, 4, 2, 1

(C) 3, 4, 1, 2 (D) 3, 1, 4, 2

SOL 1.12 Correct option is (D)For flow through an inclined pipeline the following four quantities will be in the sequence1. Datum head 2. Static head 3. Velocity head 4. Friction head.

MCQ 1.13 Two finned surfaces with long fins are identical, except that the convection heat transfer coefficient for the first finned surface is twice that of the second one. What statement below is accurate for the efficiency and effectiveness of the first finned surface relative to the second one ?(A) Higher efficiency and higher effectiveness

(B) Higher efficiency but lower effectiveness

(C) Lower efficiency and higher effectiveness

(D) Lower efficiency and lower effectiveness

SOL 1.13 Correct option is (D)The efficiency of long fin is given by / /kA hp Lcη = , which is inversely proportional to convection coefficient h . Therefore, efficiency of first finned surface with higher h will be lower. This is also the case for effectiveness since effectiveness is proportional to efficiency, /A Afin baseε η= ^ h.

MCQ 1.14 In turbulent flow, one can estimate the Nusselt number using the analogy between heat and momentum transfer (Colburn analogy). This analogy


relates the Nusselt number to the coefficient of friction, Cf , as(A) 0.5 Re PrNu C /

f1 3= (B) 0.5 Re PrNu C /

f2 3=

(C) Re PrNu C /f

1 3= (D) Re PrNu C /f

2 3=

SOL 1.14 Correct option is (A)In turbulent flow, using the analogy between heat and momentum transfer, the relation between the Nusselt number to the coefficient of friction cf^ h is

Nu 0.5 Re PrC /f

1 3=

MCQ 1.15 Scab is a(A) sand casting defect (B) machining defect

(C) welding defect (D) forging defect

SOL 1.15 Correct option is (A).Scab refers to the rough, thin layer of a metal, protruding above the casting surface, on top of thin layer of sand the layer is held on to the casting by a metal stringer through the sand.

MCQ 1.16 Match List I (A function connected with NC m/c tool) with List II (Associated parameter) and select the correct answer using the codes given below the lists :

List-I List-II

a. Interpolation 1. Tape preparation

b. Parity check 2. Canned cycle

c. Preparatory function 3. Drilling

d. Point to point control 4. Contouring

5. TurningCodes : a b c d(A) 4 1 2 3(B) 4 1 2 5(C) 5 1 3 2(D) 1 4 3 2


rpm to hobrpm of work (gear blank)

N2=

(for double start hub)

& RPM of hobRPM of work (gear blank)

260=

& RPM of work (gear blank)

RPM of hob 602=


& RPM of hob 602= RPM and gear blank

& 1 revolution of hob 602= revolution of gear blank

MCQ 1.17 A grinding wheel of 150 mm diameter is rotating at 3000 rpm. The grinding speed is(A) 7.5 /m sπ (B) 15 /m sπ

(C) 45 /m sπ (D) 450 /m sπ


We have D 150 0.15mm m= = N 3000 rpm=Grinding speed

V DN60

π=

.60

0 15 3000# #π=

7.5 / secmπ=

MCQ 1.18 In gas welding of mild steel using an oxyacetylene flame, the total amount of acetylene used was 10 litre. The oxygen consumption from the cylinder is(A) 5 litre (B) 10 litre

(C) 15 litre (D) 20 litre

SOL 1.18 Correct option is (B).For gas welding of mild steel neutral flame is used and in neutral flame equal amounts of O2 and C H2 2 are mixed and burned.So that the oxygen consumption from the cylinder in 10 litre

MCQ 1.19 Consider the following ingredients used in moulding :1. Dry silica sand 2. Clay

3. Ethyl silicate 4. Phenol formaldehyde

Those used for Lost Wax casting method include :(A) 1, 2 and 4 (B) 2, 3 and 4

(C) 1 and 3 (D) 1, 2, 3 and 4

SOL 1.19 Correct option is (C).The lost wax casting is also known as investment-casting process.In this, the pattern is made of wax or of a plastic (such as polystyrene) by molding The pattern is made by injecting molten wax or plastic into a metal die in the shape as very fine silica & binders, including water, ethyl silicate etc


MCQ 1.20 Match the terms used in connection with heat-treatment of steel with the micro structural/physical characteristics :

Terms Characteristics

a Pearlite 1. Extremely hard and brittle phase

b Martensite 2. Cementite is finely dispersed in ferrite

c Austenite 3. Alternate layers of Cementite and Ferrite

d Eutectoid 4. Can exist only above 723 Cc

5. Pertaining to state of equilibrium between three solid phases

6. Pertaining to state of equilibrium between one liquid and two solid phase

Codes : a b c d(A) 3 1 4 5(B) 2 1 4 6(C) 5 3 2 1(D) 1 4 2 3

SOL 1.20 Correct option is ( A).

Terms Characteristics

a Pearlite 3. Alternate layers of Cementite and Ferrite

b Martensite 1. Extremely hard and brittle phase

c Austenite 4. Can exist only above 723 Cc

d Eutectoid 5. Pertaining to state of equilibrium between three solid phases

MCQ 1.21 The refrigerant used for absorption refrigerators working heat from solar collectors is mixture of water and(A) carbon dioxide (B) Sulphur dioxide

(C) Lithium bromide (D) Freon 12

SOL 1.21 Correct option (C)Absorption refrigerators working by solar collector uses mixture of water and lithium bromide (water-LiBr)

MCQ 1.22 The given diagram shown as isometric cooling process 1-2 of a pure substance.


The ordinate and abscissa are respectively

(A) Pressure and volume (B) Enthalpy and entropy

(C) Temperature and entropy (D) Pressure and enthalpy

SOL 1.22 Correct option is (B).An isometic cooling process of a pure substance on a h s− diagram is given by

MCQ 1.23 The internal energy of a gas obeying Vander Waals equation

( )pva v b RT2+ − =a k depends on its

(A) temperature (B) temperature and pressure

(C) temperature and specific volume (D) pressure and specific volume


Change in internal energy, u = ,f T V= ^ h

du Tu dT

Tu dv

v T22

22= +b bl l

du C dT TVu p dVv

v22= + − −b l; E

MCQ 1.24 Given that, θ = procurement cost per order, D = number of units demanded per year, H = holding cost per unit year, i = rate of interest, p = purchase price per unit. The procurement quantity per order (Q ) is given by

(A) Q H iPD2θ= + (B) Q iH P

D2θ= +

(C) Q H iPD2θ= + (D) Q

D H iP2θ=+^ h



MCQ 1.25 The type of layout suitable for use of the concept, principles and approaches of ‘group technology’ is :(A) Product layout (B) Job-shop layout

(C) Fixed position layout (D) Cellular layout

SOL 1.25 Correct option is (D).Group technology has become an increasingly popular concept in manufacturing that is designed to take advantages of mass production layout and technique in smaller batch production system and cellular layout is used in G.T.


MCQ 1.26 Rank of the matrix 112

246

325

R

T

SSSS

V

X

WWWW

is

(A) 0 (B) 1

(C) 2 (D) 3


Let A 112

246

325

=

R

T

SSSS

V

X

WWWW

It is a square matrix of order 3 3# . Thus

Aρ^ h 3#

A 112

246

325

=

R

T

SSSS

V

X

WWWW

1 20 12 2 5 4 3 6 8= − − − + −^ ^ ^h h h

8 2 6 0= − − =The third order minor is zero. So that

Aρ^ h 2#

Since the second order minor

11

24 4 2 2 0!= − =

Therefore its rank is A 2ρ =^ h

MCQ 1.27 Find the area under the curves 2y x= − 2y x= +(A) 6 unit2 (B) unit9 2


(C) unit5 2 (D) unit8 2


y x 2= −

y x 2=− −^ h y x2& = − ....(i)

y x 2= −^ h y x 2& = − ....(ii)

y x 2= +6 @

y x 2= + & y x 2= + ...(iii)

y x 2= − ...(iv)

MCQ 1.28 For the following set of simultaneous equations : . .x y1 5 0 5 2− = x y z4 2 3 9+ + = x y z7 5 10+ + =(A) the solution is unique

(B) infinitely many solutions exist

(C) the equations are incompatible

(D) finite number of multiple solutions exist.


. .x y1 5 0 5− 2= x y z4 2 3+ + 9= x y z7 5+ + 10=Comparing to x B4 = , we get

A . .

,B1 547

0 521

035

2910

=−

=

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

Augmented matrix is

:A B6 @ . . :

::

1 547

0 521

0 23 95 10

=−

R

T

SSSS

V

X

WWWW

Applying R2 R R42 1" + R3 R R23 1" +

. . :

::

1 51010

0 500

0 23 175 14

=−

R

T

SSSS

V

X

WWWW

Here :A Bρ6 @ 3 & Aρ= 6 @

Thus :A Bρ6 @ Aρ= 6 @, so the system has unique solution


MCQ 1.29 Taking the step size 12π the value of . sin x dx1 0 162

/

2

0

2

−π

# by Simpson’s one third rule is (A) .1 5058 (B) .1 5759

(C) .2 5056 (D) .1 5056


f x^ h . sin x1 0 162 2= −

.cos x

1 0 162 21 2

= −−^ h

. . cos x0 919 0 081 2= +Here Where /h 12π=when x 00 = y0 1=

when /x 121 π= , y1 . . .cos0 919 0 081 30 0 9795c= + =

when /x 62 π= , y2 . . 0.9795cos0 919 0 081 60c= + =

when /x 43 π= , y3 . . 0.cos0 919 0 081 90 9586c= + =when /x 34 π= y4 .0 9372=when /x 5 125 π= , y5 .0 9213=when /x 26 π= y6 .0 9154=

f x dx/

0

2π

^ h# 4 2h y y y y y y y2 0 6 1 3 5 2 4= + + + + + +^ ^ ^h h h6 @

.1 5056=

MCQ 1.30 A heat engine receives 1000 kW of heat at a constant temperature of 285 Cc and rejects 492 kW of heat at 5 Cc . Consider the following thermodynamic cycles in this regard :1. Carnot cycle 2. Reversible cycle

3. Irreversible cycle

Which of these cycles could possibly be executed by the engine ?(A) 1 only (B) 3 only

(C) 1 and 2 (D) None of 1, 2 & 3


carnotη . %TT1 50 179

1

2= = =

engineη . %QQ Q 50 8

1

1 2= − =

None of 1, 2, and 3 are possible

MCQ 1.31 Match List I (The T -s diagram of the thermodynamic cycles) with List II


(names of cycles) and select the correct answer using the codes given below the lists :


Codes : a b c d(A) 1 4 5 2(B) 1 3 4 5(C) 2 4 5 1(D) 2 3 4 1

SOL 1.31 Correct option is (D)(1) Otto Cycle : Two adiabatics and two constant volumes

(2) Ericsson cycle : Two isothermals and two isobars

(3) String cycle : Two isothermals and two constant volumes

(4) Diesel cycle : Two adiabatics and one constant volume and one constant pressure

MCQ 1.32 A 400 mm long shaft has a 100 mm tapered step at the middle with 4c included angle. The tail stock offset required to produce this taper on a lathe would be(A) sin400 4c (B) sin400 2c

(C) sin100 4c (D) sin100 2c


The offset can be calculate as follows :

sinα ABBC=

S sin sinAB Lα α= =If α is very small, then we can approximate

sin 2α

b l tan ID d

2 2α

= = −b l


S L ID d

2=−^ h

This is most general situation where taper is to be limited and as such this method is suitable for small taper over a long length. The disadvantage is that the centres would not be properly bearing in the centre hole and as such there would be nonuniform wearing.

S L ID d

2#= −

sin 2α

b l ID d

2= −

S sin400 2c=

MCQ 1.33 Match the curves in Diagram I (process on p-v plane) with the curves in Diagram II (Process on T -s plane) and select the correct answer :

Codes : a b c d(A) 3 2 4 5(B) 2 3 4 5(C) 2 3 4 1(D) 1 4 2 3


MCQ 1.34 Match List I (Details of process of the cycle) with List II (Name of the cycle)


and select correct answer using the codes given below the lists :

List-I List-II

a. Two isothermal and two adiabatic 1. Otto

b. Two isothermal and two constant volume

2. Joule

c. Two adiabatic and two constant volume

3. Carnot

d. Two adiabatic and two constant pressure

4. Stirling

Codes : a b c d(A) 4 3 1 2(B) 4 3 2 1(C) 3 4 1 2(D) 3 4 2 1

SOL 1.34 Correct option is (C).(1) Otto Cycle : Two adiabatics and two constant volumes

(2) Joule cycle : Two isothermals and two constant pressure

(3) Carnot cycle : Two adiabatics and two isothermal

(4) Stirling cycle: Two isothermal and two constant volume

MCQ 1.35 A slab milling cutter has a radial rake of 7c and a spiral angle of 35c. The effective rake angler is(A) .31 75c (B) .19 05c

(C) 45c (D) .25 4c

SOL 1.35 Correct option is (D)From the geometry of general form of cutting edge, we have (see Figure)

tan pγ tan secrγ σ=where rγ is the radial rake angle; eγ is the effective rake angle; pγ is the primary rake angle; σ is the spiral angle.Substituting the values, we have

tan pγ tan sec7 35c c= . . .0 1228 1 221 0 1499#= = pγ .8 5c= sin eγ .sin cos sin35 35 8 52 c c c= + . . .0 3291 0 671 0 1478#= + ^ h


.0 4283= eγ .25 4c=

MCQ 1.36 In figure shown, the coefficient of friction between block A and the plane is .0 25.What will be the coefficient of friction between the belt and pulley for

the impending motion of block A up the inclined plane as shown?

(A) 0.28 (B) .0 224

(C) 0.448 (D) 0.168

SOL 1.36 Correct option is (B)Figure show all forces acting on the body. Since block A is about to move up so friction force acts downward direction along the planeNow consider the equilibrium of block A.Resolving the forces normal to inclined plane.

cos30 α RA=Resolving forces along the plane.

T1 sin R30 Aα μ= +or T1 sin cos30 30α μ α= +

Here cosα 53= and , .sin 5

4 0 25α μ= =

Thus T1 .30 54 0 25 30 5

3# # #= +

28.5 N=Now consider the pulley shown in Figure

Angle of lap is θ 180 90c c α= − −^ h

Here α 53.1tan 34 31 c= =−

θ 90 53.13 143.13c c c= + =

θ 143.13 180c π#=

25 rad=


Here block B induces impending motion to block A Hence the portion of

string connected the B is subjected to the light tension T2 and that portion

of the string connected to block A is subjected to the slack tension T1 where

.T T>2 1

Thus TT

1

2 e e .2 5= =μθ μ

But from the equilibrium of block B we can say that 50 NT2 =

Thus .50

28 5 e .2 5= μ

or e .2 5μ .1 754=or .2 5μ . .ln 1 754 0 56= =

μ .. .2 5

0 56 0 224= =

MCQ 1.37 A bullet of 30 gm is fired through a gun towards a suspended ball of 10 .kg The string which is attached to the suspended ball is 1 m long. Due to the impact of the bullet, the bullet penetrates into the ball and the ball with bullet swing through an angle of 30c with the vertical. What will be the velocity of the bullet when fired ?(A) . / secm6 67 (B) 5.34 / secm

(C) . / secm9 5 (D) / secm4

SOL 1.37 Correct option is (B)As per problem statement the configuration is shown in figure. Let u be the velocity of bullet impact and v be the velocity of combined ball and bullet. Momentum before impact must be momentum after impact.

Thus mu m M v= +^ h

. u0 03 . .v v0 03 10 10 03= + =^ h ....(1)

After the impact, the combined ball an bullet M m+ is lifted through a

height h and its all kinetic energy is transferred to potential energy lift

height h .

Thus M m v21 2+^ h M m gh= +^ h

or h gv2

2

= ....(2)

From the figure, we can write

cos30c LL h h

11= − = −

or h 1 30 1 0.87 0.13cos mc= − = − =Substituting this value of h is equation (2), we get


.0 13 gv2

2

=

v2 . . .2 9 81 0 13 16 02# #= = v 5.34 / secm=

MCQ 1.38 If two identical helical springs are connected in parallel and to these two, another identical spring is connected in series and the system in loaded by a weight W , as shown in the figure then the resulting deflection will be given by (δ = deflection, S = stiffness, W = load)

(A) SW23δ = (B) S

W2δ =

(C) SW32δ = (D) S

W3δ =


Seq s ss s S

22

32#= + =

δ SW

SW23

eq= =


List-I

a. Bending moment is constant

b. Bending moment is maximum or minimum

c. Bending moment is zero

d. Loading is constant

List-II

1. Point of contraflexure

2. Shear force changes sign

3. Slope of shear force diagram is zero over the portion of the beam


4. Shear force is zero over the portion of the beam

Codes : a b c d(A) 4 1 2 3(B) 3 2 1 4(C) 4 2 1 3(D) 3 1 2 4


List-I

a. Bending moment is constant

b. Bending moment is maximum or minimum

c. Bending moment is zero

d. Loading is constant

List-II

4. Shear force is zero over the portion of the beam

2. Shear force changes sign

1. Point of contraflexure

3. Slope of shear force diagram is zero over the portion of the beam

MCQ 1.40 Circumferential stress in a cylindrical steel boiler shell under internal pressure is 80 MPa. Young’s modulus of elasticity and Poisson’s ratio are respectively 2 10 MPa5

# and 0.28. The magnitude of circumferential strain in the boiler shall be(A) .3 44 10 4

#− (B) .3 84 10 4

#−

(C) 4 10 4#

− (D) .4 56 10 4#

−

SOL 1.40 Correct option is (A)Circumferential stress,

1σ 80 /N mmtpd2

2= =

Circumferential strain Etpd2

1 2μ

= −b l

.2 10

80 1 20 28

5#

= −b l

.3 44 10 4#= −

MCQ 1.41 For the spring mass system shown in the figure 1, the frequency of vibrations is N. What will be the frequency when one more similar spring added in


series, as shown in figure 2 ?

.

(A) N/2 (B) /N 2

(C) /N2 (D) N2


n1 mk N2

11π= =

Equivalent stiffness of fig 2 /k kkk k 2= + =

n2 mk N

21

2 2π= =

MCQ 1.42 Consider a four-bar mechanism shown in the given figure. The driving link DA is rotation uniformly at a speed of 10 . .r p m0 clockwise.

.

The velocity of A will be(A) 300 cm/s (B) 314 cm/s

(C) 325 cm/s (D) 400 cm/s


V rω=

30 314 / seccm602 100#

#= =

MCQ 1.43 Solar radiations is incident on an opaque surface at a rate of 400 /W m2. The emissivity of the surface is .0 65 and the absorptivity to solar radiation is .0 85. The convection coefficient between the surface and the environment at

25 Cc is 6 /W m C2c . If the surface is exposed to atmosphere with an effective


sky temperature of 250 K, the equilibrium temperature of the surface is(A) 281 K (B) 2 8 K9

(C) 3 K03 (D) K317


Gsolar 400 /W m2= ε .0 65= sα .0 85= h 6 /W m C2= Tinfinity 25 273C K= + Tsky 250 K= σ 5.67 10 /W m K8 2 4

# −= −

Ein Eout= Ein G Tsolar skys

4α ε σ# # #= + . . .0 85 400 0 65 5 67 10 2508 4

# # # #= + −

484= Eout T h T Tinfinitysky sky

4ε σ# # #= + −^ h

0.65 5.67 10 250 6 250 2988 4# #= + + −− ^ h

MCQ 1.44 Consider a 0.3 m diameter and 1.8 m long horizontal cylinder in a room at 20 Cc . If the outer surface temperature of the cylinder is 40 Cc , the natural convection heat transfer coefficient is(A) 3.0 /W m C2 c (B) 3. /W m C5 2 c

(C) 3. /W m C9 2 c (D) . /W m C4 6 2 c

SOL 1.44 Correct option is (C)We have

0.3 mD = , 1.8 mL = , 20 CT c=3 , 40 CTs c= , 0.02588 /W m Kk = , .Pr 0 7282= , 1.608 10 / secmV 5 2

#= −

The film temperature

T T T2f

s= + 3 2273 40 273 20

=+ + +^ ^h h

303 K=Volume expansion co-efficient

β 303 KT1 1f

= =

Rayleigh Number

Ra PrV

g T T Ds2

3β=

− 3^ h


.

. ..

1 608 109 81 313 293 0 3

0 72825 23031 3

##=

−−^

^ ^ ^ ^^

h

h h h hh

.4 92 107#=

Nusselt Number

Nu .. /.

PrRa0 6

1 0 5590 387

/ /

/

9 16 8 27

1 6 2

= ++ ^ h

> H$ .

.

..

. .0 6

1 0 72820 559

0 387 4 92 10/ /

/

9 16 8 27

7 1 6 2#= +

+ b

^ ^

l

h hR

T

SSSS

V

X

WWWW' 1

. ..0 6 1 202

7 408 2= +; E

.45 74=So that natural convection heat transfer co-efficient is

h NuDk=

.. 45.740 3

0 02588#=

3.9 /W m K2−=

MCQ 1.45 In which one of the following situations the entropy change will be negative :(A) Air expands isothermally from 6 bars to 3 bars.

(B) Air is compressed to half the volume at constant pressure

(C) Heat is supplied to air at constant volume till the pressure becomes three folds

(D) Air expands isentropically from 6 bars to 3 bars.


sΔ ln lnC TT R V

Vv

1

2

1

2= −

sΔ ln lnC TT R p

pp

1

2

1

2= −

MCQ 1.46 Arrivals at a telephone booth are considered to be Poisson with an average time of 10 minutes between one arrival and the next. The length of a phone call is assumed to be distributed exponentially with a mean of 3 minutes. The probability that a person arriving at the booth will have to wait, is :(A) .0 043 (B) .0 300

(C) .0 429 (D) .0 700



λ 1060 6= =

μ 360 20= =

ρ .206 0 3μ

λ= = =

Probability that a person arriving at the booth will have to wait is .0 3

MCQ 1.47 A production system has a product type of layout in which there are four machines laid in series. Each machine does a separate operation. Every product needs all the four operations to be carried out. The designed capacity of each of the four machine is 200, 175, 160 and 210 products per day. The system capacity would be(A) 210 products per day (B) 200 products per day

(C) 175 products per day (D) 160 products per day

SOL 1.47 Correct option is (D).The system capacity will be the capacity of shortest station.

Common Data For Q.• 48 and 49.The load on the journal bearing is 150 kN due to turbine shaft of 300 mm diameter running at 1800 r.p.m.

MCQ 1.48 What will be the length of the bearing if the allowable bearing pressure is 1.6 /N mm2 ?(A) . mm468 75 (B) . mm390 62

(C) 312.5 mm (D) . mm234 4

SOL 1.48 Correct option is (C)Given : 150 150 10kN NW 3

#= = ; 300 0.3mm md = = ; 1800 . . .r p mN = ; 1.6 /N mmp 2= ; 0.02 /kg msZ = ; 0.25 mmc =

Let

l = Length of the bearing in mm

We know that projected bearing area,

A 1 300 300 mml d l 2# #= = =

and allowable bearing pressure p^ h,

.1 6 500AW

l l300150 103

#= = =

l 500/1.6 312.5 mm= =

MCQ 1.49 If the bearing temperature is 60 Cc and viscosity of the oil at 60 Cc is


0.02 /kg m s− and the bearing clearance is 0.25 mm, the amount of heat to be removed by the lubricant per second is(A) 46.7 /kJ s (B) . /kJ s93 4

(C) . /kJ s40 9 (D) . /kJ s58 37

SOL 1.49 Correct option is (A)We know that coefficient of friction for the bearing

μ .

..

. .pZ N

cd k

1033

1033

1 60 02 1800

0 25300

0 0028 8#= + = +b b b bl l l l

. . .0 009 0 002 0 011= + =Rubbing velocity,

V . . 28.3 /m sd N60 60

0 3 1800# #π π= = =

Amount of heat to be removed by the lubricant,

Qg . . 0.011 150 10 28.3W N 3μ # # #= = 46695 /J s or W= 46,695 kW= ... 1 /J s W1=^ h

Common Data For Q.• 50 and 51.Air flows steadily in a pipe at 300 kPa, 77 Cc and 25 /m s at a rate of 18 /minkg .

MCQ 1.50 What will be the diameter of the pipe ?(A) 0. m536 (B) . m0 715

(C) 0.0715 m (D) . m0 0536

SOL 1.50 Correct option is (C)The diameter is determined as follows

v 300

.0.3349 /m kgP

RT 0 287 77 273 3= =+

=^

^ ^

h

h h

A / .

0.004018 mVmv

2518 60 0 3349 2= = =o ^ ^h h

D .

0.0715 mA4 4 0 004018π π= = =^ h

The rate of flow energy is determined from

Wflowo / . 30.14 kWmPv 18 60 300 0 3349= = =o ^ ^ ^h h h

MCQ 1.51 What will be the rate of energy transport by mass ?(A) . kW132 4 (B) 105.94 kW

(C) . kW210 5 (D) . kW79 45


SOL 1.51 Correct option is (B)The rate of energy transport by mass is

Emasso m h ke m c T V2

1p

2= + = +o o^ bh l

/ .18 60 1 008 77 273 21 25 1000

12= + +^ ^ ^ ^ bh h h h l; E

105.94 kW=If we neglect kinetic energy in the calculation of energy transport by mass

Emasso / . 105.84 kWmh mc T 18 60 1 005 77 273p= = = + =o o ^ ^ ^h h h

Therefore, the error involved if neglect the kinetic energy is only . %0 09

Common Data For Q.• Statement for linked answer Q.52 and 53.A radial flow centrifugal pump delivers gasoline ( 680 / )kg m3ρ = at 20 Cc. The pump has 18 ,cmd1 = 33 cmd2 = , 10 cmb1 = , 7.5 cmb2 = , 251 cβ = ,

402 cβ = and rotates at 1160 rpm.

MCQ 1.52 The flow rate in /m hour3 is,(A) 17.23 (B) 0.2872

(C) 28.72 (D) 1038

SOL 1.52 Option (D) is correct.

We have N 1160 rpm=

ω 121.5 /rad s602 1160#π= =

u1 121. . 10.9 /m sr 5 20 18 351ω #= = =b l

Vn1 10.9 25tan tanu 351 1 cβ #= = 5. /m s1= vo r b V2 n1 1 1#π= 2 0.09 0.10 5.1π # # #= 0.28 2 /m s8 3= 0.28 2 36008 #= 103 /m hour8 3,

MCQ 1.53 What will be the head ?(A) 48 m (B) 16 m


(C) 32 m (D) 64 m

SOL 1.53 Option (C) is correct.

Vn2 . ..

r bv

2 2 0 165 0 0750 2882

2 2 # #π π= =o 3. /m s71=

u2 121. 0.165r 52ω #= = . /m s20 05= Vt2 40cotu Vn2 2 c= − . 5 3. 40cot20 0 71 c#= − 15.6 /m s0=And Pideal 680 0.28 2 . 5 15.6vu V 8 20 0 0t2 2ρ # # #= =o 6 7 W129=

Finally Head H ( ) . .gv

P680 9 81 0 2882

61297# #ρ= =o 32 m,

Common Data For Q.• Linked Answer Q.54 and 55.The cross-feed on a shaper consists of a lead screw having 0.2 threads per mm. A ratchet and pawl on the end of the lead screw is driven from the shaper crank such that the pawl indexes the ratchet by one tooth during each return stroke of the ram. Ratchet has 20 teeth.

MCQ 1.54 What will be the feed in mm ?(A) 0.25 (B) 0.19

(C) 0.5 (D) 0.31


Pitch of lead screw . 5 mm0 21= =

Now, the pawl indexes rps201=

Cross feed 0.25pitch pawl indexes mm205

#= = =

MCQ 1.55 The ratio of return speed to cutting speed is :2 1, and the length of the stroke is 150 mm. If a plate 100 mm wide has to be machined in 10 minutes, the cutting speed in metres/sec, is (A) 0. 57 (B) 0. 54

(C) 0.15 (D) 0.30


Number of cutting stroke cross feed per strokewidth of job= .0 25

100 400= =

Cutting speed V ( )

minLN K

10001= +

where L - length of ram stroke mm150=


N - Number of full strokes/min 400time400

10 40= = =

K - cutting timereturn time

return speedcutting speed

21= = =

V . 0.15 / secm1000150 40 1 5# #= =


MCQ 1.56 Which one of the following is the Antonym of the word NULLIFY ?(A) void (B) legitimize

(C) repose (D) indomitable


MCQ 1.57 Which one of the following is the synonym of the word REPEAL ?(A) sharp (B) applaud

(C) acceptance (D) abrogation


MCQ 1.58 One of the four words given in the four options does not fit the set of words. The odd word from the group, is(A) Cardigan (B) Pullover

(C) Tuxedo (D) Sweater


MCQ 1.59 A pair of CAPITALIZED words shown below has four pairs of words. The pair of words which best expresses the relationship similar to that expressed in the capitalized pair, isUNITY : DIVERSITY(A) Single : Multiple (B) One : Many

(C) Homogeneous : Heterogeneous (D) Singular : Plural



The interest generated by the soccer World Cup is....................compared to the way cricket.................the nation.(A) milder, fascinates (B) lukewarm, electrifies

(C) tepid, inspires (D) unusual, grips




MCQ 1.61 What is the smallest integer k for which 64 4>k 14 ?(A) 3 (B) 5

(C) 6 (D) 7

SOL 1.61 Correct option is (B).If we don’t want to do a lot of calculations, we’re going to have to manipulate the exponents. The first step is to put the 64 and the 4 in the same terms, so let’s make the 64 43= instead. Now the question is looking for the smallest integer k for which 4 4>k3 14, which is the same as finding the smallest integer k for which k3 14> . The best answer is 5.

MCQ 1.62 The angles of a polygon are in arithmetic progression. If the smallest angle is 120c and the common difference is 10c, then how many sides does the polygon have ?(A) 5 (B) 6

(C) 8 (D) Either (B) or (C)

SOL 1.62 Correct option is (B).Smallest angle 1200= , common difference 10c= . Maximum angle 170c= (a 180c is a straight line).

MCQ 1.63 A certain clock rings two notes at quarter past the hour, four notes at half past, and six notes at three-quarters past. On the hour, it rings eight notes plus an additional number of notes equal to whatever hour it is. How many notes will the clock ring between : 002 P.M. and 5 : 00 P.M., including the rings at : 002 and :5 00 ?(A) 72 (B) 87

(C) 96 (D) 102

SOL 1.63 Correct option is (B).We know that the problem involves only simple arithmetic, because the rings occur in an hourly pattern. You could set up a chart if that helps you see what’s going on, or just take each part at a time by finding the number of rings at each interval of time and then adding up the total rings at each interval. The total rings on the hour 1 8= + +^ h 2 8 3 8 4 8+ + + + +^ ^ ^h h h

9 10 11 12 42= + + + = . The clock rings twice at a quarter past and it does this 3 times, so the total rings at a quarter past 2 3 6= =] g . Likewise, the number of rings at half past 4 13 2= =] g and the number of rings at three-quarters past 6 3 18= =] g . Adding up, 42 6 12 18 78+ + + =


MCQ 1.64 In an increasing sequence of 10 consecutive integers, the sum of the first 5 integers is 660. What is the sum of the last 5 integers in the sequence ?(A) 5 87 (B) 624

(C) 665 (D) 685

SOL 1.64 Correct option is (D).Let the first 5 consecutive integers be represented by x , x 1+, x 2+ , x 3+ and 4.x + Then, since the sum of the integers is 606 , x x x x x1 2 3 4+ + + + + + + +^ ^ ^ ^h h h h 660= . Thus,

x5 10+ 606= x5 506= solve for x

x 130=The first integer in the sequence is 130, so the next integers are 131, 132, 133 and 134. From this, the last 5 integers in the sequence, and thus their sum, can be determined. The sum of the 6th, 7th, 8th, 9th and 10th integers is, 135 136 137 138 139 685+ + + + =This problem can also be solved without algebra: The sum of the last 5 integers exceeds the sum of the first 5 integers by 1 3 5 7 9 25+ + + + = because the 6th

MCQ 1.65 What will be the sum to n terms of the series ....9 99 999+ + + ?(A) 9

10 1n +^ h (B) 10 1n910 −^ h

(C) n910 10 1n

−−^ h (D) n10 1n910 − −^ h6 @

SOL 1.65 Correct option is (C).The series can be written as 10 1nΣ −^ h.

10 1 n10 110 10 1n

n

Σ Σ= − = −−

−^ h


Mock Test - 4

1. (B) 11. (B) 21. (C) 31. (D) 41. (B) 51. (B) 61. (B)

2. (C) 12. (D) 22. (B) 32. (B) 42. (B) 52. (D) 62. (B)

3. (A) 13. (D) 23. (C) 33. (B) 43. (D) 53. (C) 63. (B)

4. (D) 14. (A) 24. (C) 34. (C) 44. (C) 54. (A) 64. (D)

5. (D) 15. (A) 25. (D) 35. (D) 45. (B) 55. (C) 65. (C)

6. (D) 16. (A) 26. (C) 36. (B) 46. (B) 56. (B)

7. (A) 17. (A) 27. (C) 37. (B) 47. (D) 57. (D)

8. (C) 18. (B) 28. (A) 38. (A) 48. (C) 58. (C)

9. (A) 19. (C) 29. (D) 39. (C) 49. (A) 59. (C)

10. (D) 20. (A) 30. (D) 40. (A) 50. (C) 60. (B)

SolutionsME-E


MCQ 1.1 A square matrix A is Skew Hermitian if

(A) A A=c (B) A A=−Τ

(C) A A=−c (D) A AT =

SOL 1.1 Correct Option is (C)A square matrix A aij= ^ h will be called a skew Hermitian matrix if every i jth− element of A is equal to negative conjugate complex of j ith− element of A

Hence aij a ji=−The necessary and sufficient condition for a matrix A to be skew Hermitian

is that

Ac A=−or A l^ h A=−

MCQ 1.2 The Expression curl (grad f), where f is a scalar function is :(A) equal to f2d

(B) equal to div (grad f)

(C) a scalar of zero magnitude

(D) a vector of zero magnitude.

SOL 1.2 Correct Option is (D).

curl grad f^ h f#d d= ^ h

where d x y zi j k

22

22

22= + +c m

Thus

f#d d h x y z xf

yf

zfi j k i j k

22

22

22

22

22

22

#= + + + +c cm m

Page 2 ME-E Chapter 1

x

xf

y

yf

z

zf

i j k

22

22

22

22

22

22

=

R

T

SSSSSS

V

X

WWWWWW

y z

fy z

fx z

fx z

fx y

fx y

fi j k2 2 2 2 2 2

222

222

222

222

222

222= − − − − −e c eo m o

f#d d h 0=Thus expression curl grad f^ h is a seal vector of zero magnitude

MCQ 1.3 The point of maxima of the function sin cosx x2+ over the interval ,x06 @ is

(A) 3π (B) 6

π

(C) 2π (D) 6

5π


f x^ h sin cosx x2= + f xl h cos sinx x2= − ....(i)

f xm^ h sin cosx x2=− − ....(ii)

For maxima and minima, put f x 0=l h

cosx sinx2=

tanx 21=

x .26 56c=At .x 26 56c= f xm^ h 2.236 ve=− −^ h

Thus f x^ h is maximum at .x 26 56c=Hence the appropriate option is (B)

MCQ 1.4 If A111

021

034

=

R

T

SSSS

V

X

WWWW

then Adj (A).A =

(A) 100

123

114

R

T

SSSS

V

X

WWWW

(B) 100

010

001

R

T

SSSS

V

X

WWWW

(C) 413

222

121

R

T

SSSS

V

X

WWWW

(D) 500

050

005

R

T

SSSS

V

X

WWWW

Chapter 1 ME-E Page 3


We have A 111

021

034

=

R

T

SSSS

V

X

WWWW

Adj A^ h 500

143

11

2

511

041

03

2

T

=−

−

−− = −

− −−

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

So that

Adj A A$^ h 511

041

03

2

111

021

034

= −− −

−

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

500

050

005

=

R

T

SSSS

V

X

WWWW

MCQ 1.5 If the Simpson’s rule

.x

dx a b1

1121 3 1 42

0

1

+= + +^ h8 B#

when the interval ,0 16 @ is divided into 4 sub-intervals and a & b are the

values of x1

12+ at two of its division point, then a & b are

(A) .a 1 06251= , .b 1 25

1= (B) .a 1 06251= , .

1b 1 5625=

(C) .a 1 251= , b 1= (D) .a 1 5625

1= , .b 1 251=


a . .y

x11

1 0 251

1 06251

1 2 2= =+

=+

=^ h

b . .y

x11

1 0 751

1 56251

332 2= =

+=

+=

^ h

MCQ 1.6 The state of plane stress in a plate of 100 mm thickness is given as 100 /N mmXX

2σ = , 200 /N mmYY2σ =

Young’s modulus 300 /N mm2= Poisson’s ratio .0 3=The stress developed in the direction of thickness is(A) zero (B) 90 /N mm2

(C) /N mm100 2 (D) 200 /N mm2

SOL 1.6 Correct option is (A)No stress will be developed


MCQ 1.7 A reverted gear train is one in which the output shaft and input shaft(A) rotate in opposite directions (B) are coaxial

(C) are at right angles to each other (D) area at an angle to each other

SOL 1.7 Correct option is (B)Reverted gear train is a special type of compound gear train in which the first and the last gear have the same axis

MCQ 1.8 The amplitude versus time curve of a damped-free vibration is shown in the below figure. Curve labelled ‘A’ is

(A) a logarithmic decrement curve

(B) an exponentially decreasing curve

(C) a hyperbolic curve

(D) a linear curve

SOL 1.8 Correct option is (B)The damping factor such that the vibration motion is allowed, however, the amplitude of motion reduce with time in an exponential manner.

MCQ 1.9 The critical speed of a uniform shaft with a rotor at the centre of the span can be reduced by(A) reducing the shaft length (B) reducing the rotor mass

(C) increasing the rotor mass (D) increasing the shaft diameter


nω mk=

nω m1?

MCQ 1.10 A rotating shaft carries a flywheel which overhangs on the bearing as a cantilever. If this flywheel weight is reduced to half of its original weight, the whirling speed will(A) be double (B) increase by 2 times


(C) decrease by 2 times (D) be half


Whirling sheet 11α

1

2ωω 1

21=

2ω 2 1ω=

MCQ 1.11 A four-bar chain has(A) all turning pairs

(B) one turning pair and the others are sliding pairs

(C) one sliding pair and the other are turning pairs

(D) all sliding pairs

SOL 1.11 Correct option is (A)Here all turning pair

MCQ 1.12 Kinematic similarly between the model and prototype is the similarity of(A) shape (B) discharge

(C) stream (D) forces

SOL 1.12 Correct option is (B)Kinematic similarity : If the ratio of velocity and acceleration at the corresponding points in the model and prototype are same then there exist kinematic similarity

MCQ 1.13 Consider the following assumptions:1. The fluid is compressible

2. The fluid is inviscid.

3. The fluid is incompressible and homogeneous.

4. The fluid is viscous and compressible.

The Euler’s equation of motion requires assumptions indicated in(A) 1 and 2 (B) 2 and 3


(C) 1 and 4 (D) 3 and 4

SOL 1.13 Correct option is (B)Euler’s equation of motion are applicable to compressible or incompressible, non viscous fluids in steady or unsteady state of flow.

MCQ 1.14 The value of friction factor is misjudged by 25%+ in using Darcy-Weisbach equation. The resulting error in the discharge will be(A) %25+ (B) . %18 25−

(C) . %12 5− (D) . %12 5+

SOL 1.14 Correct option is (C)As per Darcy-weisbach equation

h1 gdfI

d244 160

2 2

2

# π=

i.e. Q 11?

If f is misjudged by %25+ , new Q will be proportional to .1 251 i.e. %89

i.e it is reduced %11

MCQ 1.15 During a hydraulic jump in a 10 m wide channel, the flow depth increases from 0.5 m to 4 m. The water flows at a rate of 70 /m s3 . What will be the mechanical power wasted during this jump ?(A) 4.35 kW (B) 43.5 MW

(C) 43.5 kW (D) 4.35 MW

SOL 1.15 Option (D) is correct.The average velocities before and after the jump are

V1 . 14 /m sb yv

10 0 570

1# #= = =o

V2 1.75 /m sb yv

10 470

2# #= = =o

Head loss

hL

( . ) ( ) .( ) ( . )

y y gV V

2 0 5 4 2 9 8114 1 75

1 212

22 2 2

#= − + − = − + −

6.33 m=The mass flow rate of water is

mo 1000 70 7000 /kg sv 0ρ #= = =o

Then the dissipated mechanical power becomes

Pmechanicl mg hL#= o

. .70000 9 81 6 33# #=


4346811 W= 4.35 MW-

MCQ 1.16 Match List I (Ingredients) with List II (Welding functions) and select the correct answer using the codes given below the lists :

List-I List-II

a. Silica 1. Arc stabilizer

b. Potassium silicate 2. Deoxidizer

c. Ferro silicon 3. Fluxing agent

d. Cellulose 4. Gas forming materialCodes : a b c d(A) 3 4 2 1(B) 2 1 3 4(C) 3 1 2 4(D) 2 4 3 1

SOL 1.16 Correct option is (C).Slag which helps to design the contour of the weld bead. The materials used are " cellulose materials (wood pulp, saw dust) asbestos, iron powder etc.Potassium silicates and sodium silicates are the arc stabilizer.Ferromangenese and ferrosilica are used as deoxidizer.

MCQ 1.17 A block of information in N.C. machine program means :(A) one row on tape

(B) a word comprising several rows on tape

(C) one complete instruction

(D) one complete program for a job

SOL 1.17 Correct option is (C).A block of information in N.C. machine program represent one complete instruction. The block containing a number of lines statements / instructions (called NC blocks) in the form of various words, codes and symbols. Some NC blocks written in the word address format punched on a paper tape are shown below : N G Z030 81 85− N G X Y T J050 83 200 200 100 0−

MCQ 1.18 A component requires a hole which must be within two limits of .25 03 and 25.04 mm diameter. Which of the following statements about the reamer size are correct ?1. Reamer size can not be below 25.03 mm


2. Reamer size can not be above 25.04 mm3. Reamer size can be 25.04 mm4. Reamer size can be 25.03 mmSelect the correct answer using the codes given below :(A) 1 and 3 (B) 1 and 2

(C) 3 and 4 (D) 2 and 4

SOL 1.18 Correct option is (D).A reamer is a multiple - cutting edge tool with straight or helically fluted edgesThe Reamers have small margins one reliefSo that Reamer size can be 25.03 mm and its size cannot be above 25.04 mm

MCQ 1.19 Which one of the following is the extensive property of a thermodynamic system ?(A) Volume (B) Pressure

(C) Temperature (D) Density

SOL 1.19 Correct option is (A).Properties are either intensive or extensive. Intensive properties are those that are independent of the size of a system such as temperature, pressure and density. Extensive properties are those whose values depend on the size of system such as mass volume etc.

MCQ 1.20 Number of components (C ), phases (P ) and degree of freedom (F ) are related by Gibbs phase rule as(A) 2C P F− − = (B) 2F C P− − =

(C) C F P 2+ − = (D) P F C 2+ − =

SOL 1.20 Correct option is (D)Gibbs phase rule:

P f+ C 2= + P f C+ − 2=

MCQ 1.21 In a single server queue customers are served at a rate of μ. If W and Wq represent the mean waiting time in the system and mean waiting time in the queue respectively, then W will be equal to(A) Wq μ− (B) Wq μ+

(C) /W 1q μ+ (D) /W 1q μ−


W L 11

S

λ λ ρρ= = −c m


Wq L 1

1q

2

λ λ ρρ= = −d n

a W L L LS q q

λ λρ

λ λρ= = + = +

L L 1q q

#λ μ λλ

λ μ= + = +

W 1q μ= +

MCQ 1.22 Match List I (Methods) with List II (Problems) and select the correct answer using the codes given below the lists :

List-I List-II

a. Moving average 1. Assembly

b. Line balancing 2. Purchase

c. Economic batch size 3. Forecasting

d. Johnson algorithm 4. SequencingCodes : a b c d(A) 1 3 2 4(B) 1 3 4 2(C) 3 1 4 2(D) 3 1 2 4


List-I List-II

a. Moving average 3. Forecasting

b. Line balancing 1. Assembly

c. Economic batch size 2. Purchase

d. Johnson algorithm 4. Sequencing


List-I List-II

a. Control charts for variables 1. Binomial distribution

b. Control chart for number of non-conformities

2. Beta distribution

c. Control chart for fraction rejected

3. Normal distribution


d. Activity time distribution in PERT

4. Poisson distribution

5. Exponential distributionCodes : a b c d(A) 3 4 1 5(B) 5 4 3 1(C) 4 3 1 2(D) 3 4 1 2


List-I List-II

a. Control charts for variables 3. Normal distribution

b. Control chart for number of non-conformities

4. Poisson distribution

c. Control chart for fraction rejected

1. Binomial distribution

d. Activity time distribution in PERT

2. Beta distribution

MCQ 1.24 Which one of the following combinations is valid for product layout ?(A) General purpose machine and skilled labour

(B) General purpose machine and unskilled

(C) Special purpose machine and semi skilled labour

(D) Special purpose machine and skilled labour

SOL 1.24 Correct option is (C).In a product layout, the workstations and equipment are located along the line of flow of work units. The special purpose machine and semi-skilled labour combinations is valid for product layout

MCQ 1.25 Match List I (Charts) with List II (Operations/informations) and select the correct answer using codes given below the lists :

List-I List-II

a. Standard process sheet 1. Operations involving assembly and inspection without machine

b. Multiple activity chart 2. Operations involving the combination of men and machines

c. Right and left hand operation chart

3. Work measurement


d. SIMO chart 4. Basic information for routing

5. TherbligsCodes : a b c d(A) 4 3 1 2(B) 1 2 4 5(C) 1 3 4 2(D) 4 2 1 5


List-I List-II

a. Standard process sheet 4. Basic information for routing

b. Multiple activity chart 2. Operations involving the combination of men and machines

c. Right and left hand operation chart

1. Operations involving assembly and inspection without machine

d. SIMO chart 5. Therbligs

Q.26 TO Q.55 CARRY TWO MARKS EACH

MCQ 1.26 If the probability of a bad reaction from a certain injection is .0 001. The chance that out 2000 individuals, more than two will get a bad reaction is(A) .0 22 (B) .0 32

(C) .0 15 (D) .0 20

SOL 1.26 Correct Option is (B)We haveThe probability of a bad reaction, .p 0 01=Total number of individuals, n 2000= Mean n^ h np= .2000 0 001#= 2=Let x denote the number of individuals, then probability that more that 2

will het a bad reaction

p x p x p x1 2 1 0= − = + = + =^ ^ ^h h h6 @

Using poission’s distribution

p x^ h !xe mm x

=− ^ h

! ! !e e e

1 22

12

022 2 2 1 2 0

= − + +− − −^ ^ ^h h h

' 1


. . .1 0 2706 0 2706 0 1353= − + +^ h

1 0.6766= − . .0 3233 0 32b=

MCQ 1.27 A box contains 5 black balls and 3 red balls. A total of three balls are picked from the box one after another without replacing them back. The probability of getting two black balls and one red ball is

(A) 83 (B) 15

2

(C) 2815 (D) 2

1


No. of black balls 5=No. of red balls 3=No. of total balls 5 3 8= + =The probability of getting two black balls and one red ball is

P cc c

85 3

3

2 1#=

2815=

MCQ 1.28 Consider the following statements :State of stress in two dimensions at a point in a loaded component can be completely specified by indicating the normal and shear stresses on1. a plane containing the point

2. any two planes passing through the point

3. two mutually perpendicular planes passing through the point

Which of these statements are correct ?(A) 1 and 3 (B) 2 only

(C) 1 only (D) 3 only

SOL 1.28 Correct option is (D)Correct option are 2 & 3

MCQ 1.29 The following conditions relate to a surface grinding operation carried out with a 46 grit wheel : 1672 /minmV = , 9.02 /minmv = , .C 294 50= 0.025 mmd = , 17γ = , 203 mmD =What will be the maximum chip thickness ?(A) 81.07 10 mm6

#− (B) 81.07 10 mm9

#−

(C) 81.07 10 mm8#

− (D) 81.07 10 mm7#

−


SOL 1.29 Correct option is (C)We know

t V Cv

Dd4 /1 2

# # γ= d n

Where t is the chip thickness v is the work speed in ft/min; V is the cutting speed in ft/min; C is the number of effective grits per cm2 of grinding wheel surface; γ is the ratio of width of depth of groove cut by a grit; d is the depth of cut in mm D is the wheel diameter in mmNow substituting the values we have

t .. .

1672 294 50 174 9 00

2030 025 /1 2

# ##

#= b l

..

0 492436 10

1001 1096

##=

−

73 10 .100

1 1096# #= −

81.07 10 mm8#= −

MCQ 1.30 What will be the ratio of the solidification times of two steel cylindrical risers of sizes 30 cm in diameter by 60 cm height and 60 cm in diameter by 30 cm in height subjected to identical conditions of cooling ?(A) 1.93 (B) .1 56

(C) . 21 3 (D) 1.72


According to Chvorinov’s rule solidification is AV 2

αb l

Riser 1 AV

dh d

d h

2 4

42

2

#

#

π π

π=

+

30 60 2

1 90041 900 60

225013500 6

# #

# #=

+= =

Riser 2 : AV .

60 30 21 3600

41 3600 30

360027000 7 5

# #

# #=

+= =

Ratio of solidification times, . .67 5 1 5625

2= =b l

MCQ 1.31 The arc length-voltage characteristic of a D.C. arc is given by the equation


24 4V L= + , where V is voltage in volts and L is arc length in mm. The static volt-ampere characteristic of the power source is approximated by a straight line with a no load voltage of V80 and a short circuit current of

A600 . Determine the optimum arc length for maximum power.(A) 3 mm (B) .2 5 mm

(C) 4 mm (D) 6 mm

SOL 1.31 Correct option is (C).24 4LV = +

The static volt-ampere characteristic of power source is given as

V I80 600+ 1=

Now Power, P V I#=

24 4L V1 80 600# #= + −^ bh l

24 4L L80

80 24 4 600# #= + − −^ bh l

6 L L20 56 4 600# #= + −^ h

6 14L L120= + −^ ^h h

84 8L L120 2= + −^ h

For Max P, LP

22 0=

0 120 8 2 1L 20 0##+ − = 4L mm=

MCQ 1.32 In an orthogonal cutting text, the cutting force and thrust force were observed to the 1000 N and 500 N respectively. If the rake angle of tool is zero, the coefficient of friction in chip-tool interface will be(A) /1 2 (B) 2

(C) /1 2 (D) 2


tan β α−^ h FF

c

1=

0α = tanβ 1000500

21= =

μ 21=

MCQ 1.33 A thin cylinder contains fluid at a pressure of 500 /N m2, the internal diameter of the shell is 0.6 m and the tensile stress in the material is to be limited to


9000 /N m2. The shell must have a minimum wall thickness of nearly(A) 9 mm (B) 11 mm

(C) 17 mm (D) 21 mm


1σ tPd2=

9000 .t2

500 0 6##=

t 0.01667 m-

t 17 mm-

MCQ 1.34 The directions of Coriolis component of acceleration, V2ω , of the slider A with respect to the coincident point B is shown in figure 1, 2, 3, and 4. Directions shown by figures.

(A) 2 and 4 are wrong (B) 1 and 2 are wrong

(C) 1 and 3 are wrong (D) 2 and 3 are wrong


f /p Qcor 2 V /pω= θ

2 and 4 are wrong according to above equation. These should be as shown

below



List-I List-II

a. 4 links, 4 turning pairs 1. Compete constraint

b. 3 links, 3 turning pairs 2. Successful constraint

c. 5 links, 5 turning pairs 3. Rigid frame

d. footstep bearing 4. Incomplete constraint

Codes : a b c d(A) 3 1 4 2(B) 1 3 2 4(C) 3 1 2 4(D) 1 3 4 2


If l p2 4 "= −^ h complete constrain

l p2 4> "−^ h Rigid frame

l p2 4< "−^ h incomplete constraints

MCQ 1.36 The permissible stress in a fillet weld is 100 /N mm2. The fillet weld has equal leg lengths of 15 mm each. The allowable shearing load on per cm length of the weld is (A) 22.5 kN (B) 15.0 kN

(C) 10.6 kN (D) 7.5 kN


Shear Stress is given by τ hlp2=

& lp h

2 2100 15#τ= =

Load/cm length is 1060 /N mm= 10.6 /kN cm=

MCQ 1.37 A bronze spur pinion rotating at 600 . . .r p m . drives a cast iron spur gear at a transmission ration of :4 1. The allowable static stresses for the bronze pinion and cast iron gear are 84 MPa and 105 MPa respectively. The pinion has 16 standard 20c full depth involute teeth of module 8 mm. The face width of both the gears is 90 mm. Find the power that can be transmitted from the standpoint of strength (A) 31.64 kW (B) 64.31 kW


(C) 13.46 kW (D) 46.31 kW

SOL 1.37 Correct option is (A)Given : 600 . . .r p mNP = , . /V R T T 4G P= = , 84 84 /MPa N mmOP

2σ = = , 105 105 /MPa N mmOG

2σ = = , T 16P = , 8 mmm = , 90 mmb =We know that pitch diameter of the pinion,

DP . 8 16 128 0.128mm mm TP #= = = = Pitch line velocity,

v . . 4.02 /m sD N60 60

0 128 600P P # #π π= = =

Since the pitch line velocity v^ h is less than 12.5 /m s, therefore velocity

factor,

Cv . .v33

3 4 023 0 427= + = + =

We know that for 20c full depth involume teeth, too the form factor for the

pinion

yP . . . . .T0 154 0 912 0 154 160 912 0 097

P= − = − =

and tooth form factor fort he gear,

yG . . . . .T0 154 0 912 0 154 4 160 912 0 14

G #= − = − =

/T T 4G Pa =^ h

yOP P#σ . .84 0 097 8 148#= =and yOG G#σ . .105 0 14 14 7#= =Since yOP P#σ^ h is less then yOG G#σ^ h, therefore the pinion is weaker. Now using the Lewis equation for the pinion,. we have tangential load on the tooth (or beam strength of the tooth),

WT . . . . .b m y C b m ywP p OP v p#σ π σ π= = ^ h .CWP OP va σ σ=^ h

Power that can be transmitted

7870 4.02 31640 WW vT # #= = = 31.64 kW=

MCQ 1.38 For a 300 mm long sliding lubricated bearing, the viscosity of oil is 0.008 / .kg m s during steady operation at 80 .Cc The average oil film thickness between the shaft and journal is 1.2 mm. If shaft of 80 mm diameter is rotated at 750 rpm, the amount of torque needed to overcome bearing friction would be(A) 0.0063 N m− (B) 0.063 N m−

(C) 0.63 N m− (D) 6.3 N m−


SOL 1.38 Option (B) is correct.Given: 300 0. ,mm mL 3= = 0.008 / . ,kg m sμ = 1.2 0.0012mm mtfilm = = 80 0.08 ,mm mD = = 750 rpmN =Torque is given by

T Areat Rfilm

2μω# #=

T tR RL t

R L2 2film film

2 3

#μω π πμω= = 2A RLs π=

T ( / )

tN R L2 2 60

film

3#πμ π= N

602ω π=

T tNR L

604

film

2 3

#

π μ=

By substituting values

T .. ( . ) .

60 0 00124 0 008 750 0 04 0 32 3

#

# # # # #π=

T 0.063 N m−=

MCQ 1.39 A 20 kW pump is used to raise water to a 25 m height at a rate of 50 /L s. If the irreversible head loss of the piping system is 7 m, the mechanical efficiency of the pump is

(A) 78.5% (B) 75.8%

(C) 7.85% (D) 87.5%

SOL 1.39 Option (A) is correct.The energy equation for a steady incompressible flow through a control volume between these two points is

m p V gz W2 pump1

112

1ρ α+ + +o; E m p V gz W E2 turbine loss2

222

2ρ α= + + + +o; E

Since p p patm1 2= = , V V 01 2= = and 0z1 = (Reference level)

Wpump mgz Eloss2= +o ...(i)Since in the absence of turbine


Eloss E E, ,loss pump loss pipe= + Wpump mgz E E, ,loss pump loss pipe2= + +o

W E ,pump loss pump− mgz mghloss2= +o o E mgh,loss pipe loss= W ,pump useful [ ]mg z hloss2= +o W E W,pump loss pump pump useful− =

[ ]vg z hloss2ρ= +o

W ,pump useful . ( . ) ( )1000 9 81 0 05 25 7# # #= + 50 / 0.050 /L s m s3= 15696 / 15696Nm s W= =

15.696 kW=Then the mechanical Efficiency of the pump

pumpη .WW

2015 696

shaft

useful= =o

o

. . %0 7848 78 5= =

MCQ 1.40 A new facility has to be designed to do all the welding for 3 products : ,A B and C . Per 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

SOL 1.40 Correct option is (C).Time required to weld product (A)

6020 450# 15 mins=

Time required to could product (B)

6040 360# 240 mins=

Time required to could product (C)

6050 240# 200 mins=

Total time period 150 240 200= + +^ h

590 minutes=Number of welding lines required

.220590 2 68 3-= =

MCQ 1.41 For the network shown in the given figure, the earliest expected completion time of the project is


(A) 26 days (B) 27 days

(C) 30 days (D) indeterminable


project duration is 30 days

MCQ 1.42 The water flows through a duct of square cross section as shown in figure with a uniform velocity of 40 /m sV = . Consider the fluid particles lie along line A-B at time t 0= . By using the volume of fluid in the region between lines A-B and Al-Bl, the flow rate in the duct at 0.1 s will be

(A) 10 /m s3 (B) 6 /m s3

(C) 4 /m s3 (D) 1 /m s3

SOL 1.42 Correct option is (A).Since volume is constant in time and space, all particles on line AB moves a distance


l Velocity time#= 40 0.1 4 m#= = from t 0= to 0. .sect 1=Thus the volume of ABA Bl l is, VABA Bl l VelocityA#= (0.5) 4 1 m2 3

#= =

The flow rate is vo . 10 /m stV

0 11ABA B 3

Δ= = =l l

MCQ 1.43 SAE 30 oil at 20 Cc 891 / , 0.29 /kg m kg ms3ρ μ= =^ h flows through a 7 mm diameter glass tube at an average velocity of 5 /m s. The head loss per meter of tube length and pressure drop will be,(A) 1079 m, 9430 kPa (B) 107.9 m, 943 kPa

(C) 1.079 m, 9.43 kPa (D) 10.79 m, 94.3 kPa

SOL 1.43 Option (B) is correct.Glass tube is considered hydraulically “smooth” / 0dε =The Reynold Number

Red .. 108Vd

0 29891 5 0 007# # ,μ

ρ= = (laminar flow)

Thus, f 0.593Re64

10864= = =

Now head loss per meter

hf 0.593 . . 107.9 mf dL

gV2 0 007

12 9 81

52 2

## # # #= = =

and pΔ 891 9.81 107.9ghfρ # #= =

943123 943Pa kPa,=

MCQ 1.44 A parachutist jumps from a plane, using an 8.5 m diameter chute in the standard atmosphere. The total mass of chutist and chute is 135 kg. Assuming a fully open chute in quasi steady motion, the time to fall from 2000 m (

1.0067 /kg mair3ρ = ) to sea level ( 1.225 /kg mair

3ρ = ) will be (take .C 1 2D = )(A) 321 s (B) 168 s

(C) 355 s (D) 337 s

SOL 1.44 Option (D) is correct.If acceleration is negligible,

W FD=

or m g# C U D2 4D2 2ρ π

# #=

or .135 9 81# 1.2 (8.5)U2 42 2ρ π

# # # #=

or U 2 .38 89ρ=


Thus Usea level .. 5.64 /m s1 225

38 89= =

U2000 m ..

1 006738 89=

6.22 /m s=Thus the change in velocity is very small, so we can seasonably estimate the

time-to-fall using the average fall velocity:

tfallΔ Vz

avg

Δ=

[( . . )/ ]5 64 6 22 2

2000 0= +−

tfallΔ 337 s= .

MCQ 1.45 Hot water ( 4.179 /kJ kg Kcp = ) flows through a 200 m long PVC (0.092 /W m Kk = ) pipe whose inner diameter is 2 cm and outer diameter is

2.5 cm at a rate of 1 /kg s, entering at 40 Cc . If the entire interior surface of this pipe is maintained at 35 Cc and the entire exterior surface at 20 Cc , the outlet temperature of water is(A) 39 Cc (B) 38 Cc

(C) 37 Cc (D) 36 Cc

SOL 1.45 Correct option is (B)We have

2.5 cmd0 = , 2 cmdin = , 0.092 /W m Kl −= , 4.179 /kJ kg Kcp = , 35 CT2 c= , 20 CT1 c= , 40 CTin c= , 200 ml = , 1 / seckgm =o

Heat transfer

Q ln

kl T T2

dd

2 1

i

0

π=

−_

^

i

h

2 .

.ln

0 092 80 308 2937771 5.

22 5

π=

−=

^

^ ^ ^

h

h h h

Also Q mc T Toutp in= −o ^ h

Tout TmcQ

inp

= − o

40.

. 38 C1 4 179 10

777 53

#c− =

^ ^h h

MCQ 1.46 Air at state 1 (DPT 10 , 0.0040 /C kg kgw abc = ) mixes with air at state 2 (DPT 18 , 0.0051 /C kg kgw abc = ) in the ratio 1 to 3 by weight. The degree of saturation (%) of the mixture is (the specific humidity of saturated air at 13.6 , 0.01 /C kg kgw abc = )(A) 25 (B) 30


(C) 48 (D) 62

SOL 1.46 Correct option is (C)kg of moisture actually contained mixture

. . .3 10 04 3 0 0051 0 0048#= +

+ =

Kg of moisture is saturated air of fixture

0.01 /kg kg= of air

Degree of saturation ..0 01

0 0048 100#=

%48=

MCQ 1.47 Consider the following linear programming problem :Max. Z A B2 3= + , subject to A B 10#+ , A B4 6 30#+ , A B2 17#+ , A, B 0$

What can one say about the solution ?(A) It may contain alternative optima

(B) The solution will be unbounded

(C) The solution will be degenerate

(D) It cannot be solved by simplex method


Z A B2 3= + A B+ 10#

A B10 10+ 1#

A B4 6+ 30#

.A B7 5 5+ 1#

A B2 + 17#

.A B8 5 17+ 1<


Maximum value of object function 15= and it has alternative optimal

solutions because of same constraint as objective function.

Common Data For Q.• 48 and 49.Six jobs are to be processed through machines A and B in the order AB . The processing time of jobs on different machines are given below :

Job no. 1 2 3 4 5 6

Machine A 8 12 7 10 11 9

Machine B 10 7 10 6 12 8

MCQ 1.48 The optimal sequence of jobs, is(A) 1 2 3 4 5 6− − − − − (B) 2 3 6 5 4 1− − − − −

(C) 3 1 5 6 2 4− − − − − (D) 3 2 1 4 5 6− − − − −

SOL 1.48 Correct option is (C).By Examination the rows, we find the smallest value. It is min6 for job 4 on machine B . Thus we schedule job 4 last as shown below

4

After excluding this set of processing time, the next smaller processing time is 7. There are two minimal values. Processing time 7 for job 3 on Machine A and processing time 7 for job 2 on machine B According to the rules, job 3 is scheduled on the right and job 2 next to job 4 as shown below

3 2 4

After excluding there set, again two minimal values. Processing time 8 for job 1 on machine A and processing time 8 for job 6 on machine B . Thus job 1 scheduled next to job 3 and job 6 next to job 2 as shown below

3 1 6 2 4

Thus complete sequence is obtained to let job 5 in the remaining block

3 1 5 6 2 4

MCQ 1.49 What is the total elapsed time and idle times on both machines ?(A) 63, 6, 15 (B) 60, 15, 6

(C) 63, 6, 9 (D) 60, 6, 15

SOL 1.49 Correct option is (C)Now for calculate the elapsed time and idle time, we make the table according to the sequence of jobs


Job Machine A Machine B Idle Time

In Out In Out Machine A

Machine B

3 0 7 7 18 - 7

1 7 15 18 28 - -

5 15 26 28 40 - -

6 26 35 40 48 - -

2 35 47 48 55 - -

4 47 57 57 63 6 2

Total minimum elapsed time 63= Idle time on machine A 6= Idle time on machine B 9=

Common Data For Q.• 50 and 51.A riveting machine is driven by a constant torque 3 kW motor. The moving parts including the flywheel are equivalent to 150 kg at 0.6 m radius. One riveting operation takes 1 second and absorbs 100 kN m− of energy. Speed of flywheel is 300 r.p.m before riveting.

MCQ 1.50 The speed immediately after riveting will be(A) 280 . .r p m (B) 257.6 . .r p m

(C) 25 . . .r p m3 3 (D) 25 . . .r p m9 7

SOL 1.50 Correct option is (B)Energy supplied by motor,

E1 10000 N m−= EΔ E E1 2= −

,10 000 3000= − 7,000 /N m s−=

Now, 1ω N60

260

2 300#π π= =

31.42 / secrad=

7000 mk21 2

12

22ω ω= −6 @

. .21 150 0 6 31 422 2

22

# # ω= −^ ^h h8 B

. 31.4221 150 0 6 2 2

22

# # ω= −^ ^h h8 B

2ω .26 98=


N2 257.6 rpm=

MCQ 1.51 How many rivet closed per minute ?(A) 16 (B) 18

(C) 20 (D) 22

SOL 1.51 Correct option is (B)Number of rivet that can be closed per minute.

,EE 60 10 000

3000 60 181

2# #= = =

Common Data For Q.• Linked Answer Q.52 and 53.An adiabatic turbine operates with air entering at 550 kPa, 425 K and 150 /m s and leaving at 110 kPa, 325 K and 50 /m s and 25 CT0 c=

MCQ 1.52 The actual work production for this turbine, is(A) . /kJ kg138 87 (B) . /kJ kg83 35

(C) /kJ kg99 (D) 111.1 /kJ kg

SOL 1.52 Correct option is (D).There is only one inlet and one exit, and thus m m m1 2= =o o o . We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

E E

, ,

in out

Rate of Net energy transferby heat work and mass

−o o1 2 344 44

0E, ,

, ,int

systemsteady

Rate of Change in ernal kineticpotential etc energies

0Δ= =3o ^ h

1 2 3444 444

Eino Eout= o

/m h V 21 12+o^ h /W m h V 2out 2 2

2= + +o o^ h

Wouto m h h V V

21 212

22

= − + −o; E

wout c T T V V2p 1 2

12

22

= − + +^ h

Substituting,

wout c T T V V2p 1 2

12

22

= − + −^ h

.1 011 425 325 2150 50

10001

2 2

= − +−

^ ^^ ^

bh hh h

l

111.1 /kJ kg=

MCQ 1.53 What will be the maximum work production ?(A) . /kJ kg151 11 (B) 167.9 /kJ kg

(C) . /kJ kg111 1 (D) . /kJ kg125 92



The entropy change of air is

s s2 1− ln lnc TT R p

pp

1

2

1

2= −

. .ln ln1 011 425325 0 287 550

110= −^ ^h h

0.1907 /kJ kg K$=The maximum (reversible) work is the exergy difference between the inlet

and exit states

w ,rev out c T T V V T s s2p 1 212

22

0 1 2= − + − − −^ ^h h

w T s sout 0 1 2= − −^ h

111.1 .298 0 1907= − −^ ^h h

167.9 /kJ kg=

Common Data For Q.• Linked Answer Q.54 and 55.A Carnot heat engine receives heat at 750 K and rejects the waste heat to the environment at 300 K. The entire work output of the heat engine is used to drive a Carnot refrigerator that removes heat from the cooled space at

15 Cc− at a rate of 400 /minkJ and rejects it to the same environment at 300 K.

MCQ 1.54 What will be the rate of heat supplied to the heat engine ?(A) . /minkJ135 6 (B) 108.5 /minkJ

(C) . /minkJ86 8 (D) . /minkJ81 37

SOL 1.54 Correct option is (B)The coefficient of performance of the Carnot refrigerator is

COP ,R C / 300

.T T 1

1258

1 6 14H L

= − = =^ ^ ^h h h

Then power input to the refrigerator become

W ,net ino . 65.1 /minCOP kJQ

6 14400

,R C

L= = =o

which is equal to the power output of the heat engine W ,net outo

The thermal efficiency of the Carnot heat engine is determined from

,th Cη .TT1 1 750

300 0 60H

L= − = − =

Then the rate of heat input to this heat engine is determined from the definition of thermal efficiency to be


Q ,HEHo .

. 108.5 /minkJW

0 6065 1

,

,

HE

net out

thη= = =o

MCQ 1.55 The total rate of heat rejection to the environment, is(A) 508.5 /minkJ (B) . /minkJ406 8

(C) . /minkJ254 25 (D) . /minkJ559 35


Q ,HELo 108.5 65.1 43.4 /minkJQ Q W, , ,HE HE net outH H= − − = − =o o o

Q ,H Ro 400 65.1 465.1 /minkJQ W, ,R net inL= + = + =o o

and

QAmbiento 43.4 465.1 508.5 /minkJQ Q, ,HE RL H= + = + =o o


MCQ 1.56 Which one of the following is the Antonym of the word FEARLESS ?(A) intrepid (B) craven

(C) vacillate (D) oscillate


MCQ 1.57 Which one of the following is the synonym of the word CONTROVERSIAL ?(A) pulse (B) polemic

(C) record (D) integrity


MCQ 1.58 One of the four words given in the four options does not fit the set of words. The odd word from the group, is(A) Break (B) Hiatus

(C) Pause (D) End


MCQ 1.59 A pair of CAPITALIZED words shown below has four pairs of words. The pair of words which best expresses the relationship similar to that expressed in the capitalized pair, isENTHUSIASTIC : FANATICAL(A) Frugal : Miserly (B) Faithful : Kind

(C) Admonish : Warn (D) Virtuous : Wholesome


MCQ 1.60 In the following sentence, a part of the sentence is left unfinished. Four different ways of completing the sentence are indicated. The best alternative


among the four, is

No doubt, it was our own government but it was being run on borrowed ideas, using................solutions.(A) worn out (B) second hand

(C) impractical (D) appropriate



MCQ 1.61 After striking the floor, a Tennis ball rebounds to 4/5th of the height from which it has fallen. What will be the total distance that it travels before coming to rest if it has been gently dropped from a height of meters90 . ?(A) 675 metres (B) metres810

(C) metres920 (D) 1020 metres

SOL 1.61 Correct option is (B).The first drop is 9 metres0 . After thus the ball will rise by metres72 and fall by metres72 , now this process will be continue in the form of an infinite GP with common ratio .0 8 and first term 72.The required answer will be got by

90 2 .1 0 872= + −b l

MCQ 1.62 The ratio of cost price and marked price of a cell phone is :2 3 and ratio of profit percentage and discount percentage is :3 2.What is the discount percentage ?(A) %8 (B) %12

(C) . %16 66 (D) 19.5%

SOL 1.62 Correct option is (C). CP : MP :x x2 3=& Profit x=%^ h profit : %^ h discount :3 2=

Let CP 200, SP 300= =

But x x1003 200 100

2 300# #+ 100=

& x . %8 33= Discount = x2 . %16 66=

MCQ 1.63 If prices are reduced 2 %0 and sales increase %15 , what is the net effect on gross receipts ?(A) They increase by %8 (B) They decrease by %8


(C) They remain the same (D) They increase by %10

SOL 1.63 Correct option is (B).Let original price p= , and original sales s= . Therefore, original gross receipts ps= . Let new price .80p0= , and new sales 1. s15= . Therefore, new gross receipts .92ps0= . Gross receipts are only 9 %2 of what they were.

MCQ 1.64 Reduction of %20 in the price of sugar enables a housewife to purchase kg5 more for Rs. 250. What is the original price per kg of sugar ?(A) Rs. 10 per kg (B) Rs. .12 5 per kg

(C) Rs. 15 per kg (D) Rs. 21 per kg

SOL 1.64 Correct option is (B).Reduction in price increase in amount

%20 51

.b l % kg41 25 5- =^ h

It means original amount of sugar needed 5 4 2 kg0#= =

Original price of sugar .1 . .Rs per kg20250 2 5= =

MCQ 1.65 A milkman purchases the milk at Rs. x per litre and sells it at Rs. x2 per litre still he mixes 1 litres water with every 5 litres of pure milk. What is the profit percentage ?(A) 11 %5 (B) 1 . %27 5

(C) %132 (D) %140

SOL 1.65 Correct option is (D).Let the cost price of 1 litre pure milk be Re. 1, then

..

litres milklitres water

CP RsCP Rs CP

51

50

"

""

==

^

^

h

h* 4 .Rs 5= only

and 6 litre mixture 2SP 6" " # .1Rs 2=

Profit 100512 5

5700

#= − = %140=


Mock Test - 5

1. (C) 11. (A) 21. (C) 31. (C) 41. (C) 51. (B) 61. (B)

2. (D) 12. (B) 22. (D) 32. (A) 42. (A) 52. (D) 62. (C)

3. (B) 13. (B) 23. (D) 33. (C) 43. (B) 53. (B) 63. (B)

4. (D) 14. (C) 24. (C) 34. (A) 44. (D) 54. (B) 64. (B)

5. (B) 15. (D) 25. (D) 35. (D) 45. (B) 55. (A) 65. (D)

6. (A) 16. (C) 26. (B) 36. (C) 46. (C) 56. (B)

7. (B) 17. (C) 27. (C) 37. (A) 47. (A) 57. (B)

8. (B) 18. (D) 28. (D) 38. (B) 48. (C) 58. (D)

9. (C) 19. (A) 29. (C) 39. (A) 49. (C) 59. (A)

10. (B) 20. (D) 30. (B) 40. (C) 50. (B) 60. (B)

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