Solution - IES Master...Mechanical Engineering 1. A flywheel is attached to an engine to keep its...

Solution01-02-2020 | FORENOON SESSIONDetailed

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General Aptitude

1. The bar graph shows the data of the studentswho appeared and passed in an examination forfour schools P, Q, R and S. The average ofsuccess rates (in percentage) of these four schoolsis__________.

800

700

600

500

400

300

200

100

0

500

280330

455

240

400

700

600

School P School Q School SSchool R

Performance of Schools P, Q, R and S

Num

ber o

f stu

dent

s

PassedAppeared

(a) 58.5% (b) 59.0%(c) 58.8% (d) 59.3%

Sol–1: (b)

Success rate =

Number of candidateswho passed the examNumber of appeared

candidatesSo, Average success rate,

=

280 330 455 240500 600 700 400

4

= 0.59 or 59%

Correct option is (b).2. He is known for his unscrupulous ways. He

always sheds__________tears to deceive people.(a) crocodile (b) crocodile’s(c) fox’s (d) fox

Sol–2: (a)Crocodile tears: A false, insincere display ofemotion.

3. P, Q, R and S are to be uniquely coded using and . If P is coded as and Q as ,then R and S, respectively, can be coded as

(a) and (b) and

(c) and (d) and

Sol–3: (c)

P, Q, R, S

4. Select the word that fits the analogy:Build: Building :: Grow : ________(a) Growed (b) Grown(c) Grew (d) Growth

Sol–4: (d)Build : Building (Noun)

so, Grow : Growth (Noun)

5. Jofra Archer, the England fast bowler, is__________than accurate.(a) more faster (b) more fast(c) faster (d) less fast

Sol–5: (b)Jofra Archar, the England fast bowler, ismore fast than accurate.

6. The sum of the first n terms in the sequence 8,88, 888, 8888, ... is _______.

(a) n81 9(10 1) n80 8

(b) n81 9(10 1) n80 8

(c) n80 8(10 1) n81 9

(d) n80 8(10 1) n81 9

Sol–6: (d)S = 8 + 88 + 888 + ...... n

= 8 (9 99 999 ......n)9

= 1 2 n8 (10 1) (10 1) .... (10 1)9

= 1 2 n8 (10 10 ....10 ) n9

= n8 10 (10 1) 8 n

9 10 1 9

= n8 10 8(10 1) n9 9 9




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= n80 8(10 1) n81 9

7. Select the graph that schematically representsBOTH y = xm and y = x1/m properly in theinterval 0 x 1 , for integer values of m, wherem > 1.

(a)

1

0

xm

x1/m

1x

y

(b)

1

0

xm

x1/m

1x

y

(c)

1

0xm

x1/m

1x

y

(d)

1

0

xm

x1/m

1x

y

Sol–7: (c)

Graph of y = xm and y = x1/m for m > 1

1

0xm

x1/m

1x

y

Correct option is (c).

8. I do not think you know the case well enoughto have opinions. Having said that, I agree withyour other point.What does the phrase “having said that” meanin the given text?(a) in addition to what I have said(b) contrary to what I have said(c) despite what I have said(d) as opposed to what I have said

Sol–8: (c)having said that : despite what one just said.

9. Define [x] as the greatest integer less than orequal to x, for each x ( , ) . If y = [x], then

area under y for x [1, 4] is _________.

(a) 6 (b) 1(c) 3 (d) 4

Sol–9: (a)

Graph of y = [x] for x [1, 4]

y

4

3

2

1

01 2 3 4

x

A

C

E F

B

D

The total area, A = (1×1) + (2×1) + (3×1)= 1 + 2 + 3 = 6

10. Crowd funding deals with mobilisation of fundsfor a project from a large number of people, who




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would be willing to invest smaller amountsthrough web-based platforms in the project.Based on the above paragraph, which of thefollowing is correct about crowd funding?(a) Funds raised through large contributions on

web-based platforms.(b) Funds raised through coerced contributions

on web-based platforms.(c) Funds raised through voluntary

contributions on web-based platforms(d) Funds raised through unwilling

contributions on web-based platforms.Sol–10: (c)

In the given paragraph, it is said thatcrowd funding deals with mobilization offunds for a project from a large number ofpeople (It does not mean large contribution.It means, small contribution from a largenumber of people), who would be willing(i.e voluntary or freewill) to invest smalleramounts through web-based platform.so, the correct statement about crowdfunding is to fund raised through voluntarycontributions on web-based platforms.

Mechanical Engineering

1. A flywheel is attached to an engine to keep itsrotational speed between 100 rad/s and 110 rad/s. If the energy fluctuation in the flywheelbetween these two speeds is 1.05 kJ then themoment of inertia of the flywheel is________kg.m2 (round off to 2 decimal places).

Sol–1: (1)

Given: Maximum rotational speed, 1 =110 rad/s

Minimum rotational speed, 2 = 100rad/s

Energy fluctuation, E = 1.05 kJ

E = 2 21 2

1 ( )2

where, I = Moment of inertia of fly wheel

so, 1.05 × 1000 = 2 21 (110 100 )2

21 kg m

2. A single-degree-of-freedom oscillator is subjectedto harmonic excitation F(t) = 0F cos( t) asshown in the figure.

m

F(t)

k c

The non-zero value of , for which the amplitudeof the force transmitted to the ground will beF0, is

(a) km (b) k2

m

(c)k

2m (d) 2km

Sol–2: (d)Transmissibility ratio,

t

o

FF =

2

n22 2

n n

1 2

1 2

Ft = Fo (Given)

so, 1 =

2

n22 2

n n

1 2

1 2

22 2

n n1 2

= 2

n2 1

It can also be written as, 22

n1

= 1

2

n

= 2

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n

= 2

= n2 nkm

so, = k2m

2km

3. Froude number is the ratio of(a) buoyancy forces to viscous forces(b) inertia forces to gravity forces(c) inertia forces to viscous forces(d) buoyancy forces to inertia forces

Sol–3: (b)

Froude number, Fr = i

g

Inertia force (F )Gravity force (F )

4. Multiplication of real valued square matrices ofsame dimension is(a) not always possible to compute(b) commutative(c) always positive definite(d) associative

Sol–4: (d)If A, B and C are real valued squarematrices of same dimension,

(1) AB BA

i.e. multiplication of matrices are notcommutative.

(2) A(BC) = (AB)Ci.e. multiplication of matrices areassociative.so, the correct option is (d).

5. For three vectors ˆ ˆˆ Â 2 j 3k, B 2i k

and

ˆ ˆC 3i j

, where ˆˆ î, j and k are unit vectorsalong the axes of a right-handed rectangular/Cartesian coordinate system, the value of

(A (B C) 6)

is____________

Sol–5: (6)

Given: A

= ˆˆ2 j 3k

B = ˆˆ2i k

and C = ˆ ˆ3i j

(B C) = ˆˆ ˆ ˆ( 2i k) (3i j)

= ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ2i 3i 2i j k 3i k j

ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ[ i i 0, i j k, k i j and k j i]

so, (B C) = ˆ ˆ ˆ0 2k 3 j i

= ˆ ˆ ˆ(2k 3 j i)

Now, A (B C) = ˆ ˆˆ ˆ ˆ(2 j 3k) (2k 3j i)

= ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ4 j k 6 j j 2 j i 6k k 9k j 3k i

ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ[ j k 0, k k 1, j j 1, j i 0, k j 0,

and ˆ ˆk i 0]

So, A (B C) = 0 + 6 + 0 – 6 – 0 – 0

= 0

The value of (A (B C) 6) is 0 + 6 = 6

6. The Laplace transform of a function f(t) is

2 21L(f )

(s )

. Then, f(t) is

(a)1f(t) cos t

(b) 21f(t) (1 cos t)

(c)1f(t) sin t

(d) 21f(t) (1 sin t)

Sol–6: (c)We know that,

2 2L(sin t)s




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Given, L(f) = 2 21

s = 2 2

1s

L(f) =1 L(sin t)

1f(t) sin t

7. Which of the following function f(z), of thecomplex variable z, is NOT analytic at all thepoints of the complex plane?(a) f(z) = ez (b) f(z) = z2

(c) f(z) = log z (d) f(z) = sin zSol–7: (c)

Analytic function: A function f(z) is said tobe analytic in a given region R, if it is continu-ous, single-valued and has definite derivartiveat every point inside the region R.f(z) = log z is differentiable and analytic every-where except at origin.So the correct option is (c)

8. The velocity field of an incompressible flow in aCartesian system is represented by

2 2 ˆˆ ˆV 2(x y ) i v j 3k

Which one of the following expressions for v isvalid?(a) –4xz + 6xy (b) 4xz – 6xy(c) 4xy + 4xz (d) –4xy – 4xz

Sol–8: (d)

Given: V = 2 2 ˆˆ ˆ2(x y )i v j 3k

= ˆˆ û i v j wk

For incompressible flow,

V = 0

(u) (v) (w)x y z

= 0

2 22(x y ) (v) (3)x y z = 0

v4x 0y

= 0

vy = –4x

v = 4 x dy f (x,z) v = 4xy f (x, z)

From the given option, –4xy is present onlyin option (d), so the correct answer is (d).

9. The value of

c 1 x

c 1 xx 1

1 elim1 xe is

(a)c

c 1 (b) c 1c

(c) c (d) c + 1Sol–9: (a)

c(1 x)

c(1 x)x 1

1 eLim1 xe

At x = 1,c(1 x)

c (1 x)1 e

1 xe

=

00

Apply L-Hospital rule,

c (1 x)

c(1 x) c (1 x)x 1

ceLim0 e xe c

At x = 1,

c (1 x)

c (1 x) c (1 x)x 1

ceLim0 e xce

=

c 11 c

=c

1 c

10. The crystal structure of iron (austenite phase)is(a) BCT (b) HCP(c) BCC (d) FCC

Sol–10: (d)

The crystal structure of iron (Austenitephase) is FCC i.e Face Centered Cubic.

11. The base of a brass bracket needs rough grinding.For this purpose, the most suitable grindingwheel grade specification is(a) C30Q12V (b) A30D12V(c) C90J4B (d) A50G8V

Sol–11: (a)For rough grinding of brass material:

(1) Silicon carbide (SiC) is preferred for




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grinding of brass material.(2) For rough cut, open structure is preferred.

Range of structure from dense to openvaries from 0 to 15 i.e. more the number;more the open structure is.So, according to that the suitable grindingwheel is,

C 30 Q 12 V

12. In the Critical Path Method (CPM), the cost-time slope of an activity is given by

(a) Crashcos t Normal costCrash Time

(b)

Normal costCrash Time Normal Time

(c) Crash Cost Normal CostNormal Time – Crash Time

(d) Crash CostCrash Time – Normal Time

Sol–12: (c)In critical path method (CPM), the cost-time slope of an activity,

(C , Ct C)

Cost

t

(N , Nt C)

Time

= Crash cost Normal costNormal time Crash time

= c c

t t

C NN C

13. A sheet metal with a stock hardness of 250HRC has to be sheared using a punch and a diehaving a clearance of 1mm between them. Ifthe stock hardness of the sheet metal increasesto 400 HRC, the clearance between the punchand the die should be ________ mm.

Sol–13: (1.26)Clearance in the punch,

C = 0.0032t )

where, is the shear strength of sheetmetal.

(BHN) (HRC)

We can write,

C HRC

At, (HRC)1 = 250, C1 = 1 mmthen at (HRC)2 = 400,

Clearance, C2 = 21

1

(HRC)C(HRC)

= 4001250

2C 1.26 mm

14. A four bar mechanism is shown below.

400mmP

Q

600mmR

300mm

For the mechanism to be crank-rockermechanism, the length of the link PQ can be(a) 350 mm (b) 300 mm(c) 200 mm (d) 80 mm

Sol–14: (d)According to Grashof’s Law, (for thecontinuous relative motin)

s P Q

Here, = length of largest side

s = length of shortest sideP and Q = length of other two sides.For, s = 80 mm, the above condition issatisfied, i.e., 600 + 80 < 300 + 400To obtain crank-rocker mechanism, theadjacent side of shortest link (s = 80 mm)should be fixed.

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So, for the mechanism to be a crank-rockermechanism, the length of the link PQshould be 80 mm.Let’s check other options too,

(1) For, PQ = 350 mm,

= 600 mm, s = 300 mmIn this case,600 + 300 > 350 + 400i.e. Grashof’s condition is not satisfied.

(2) For, PQ = 300 mm,

= 600 mm, s = 300 mmIn this case,600 + 300 > 300 + 400Again Grashof’s condition is not satisfied.

(3) For, PQ = 200 mm,

= 600 mm, s = 200 mmIn this case,600 + 200 > 300 + 400i.e. Grashof’s condition is also not satisfiedfor this case.

15. For an ideal gas, the value of the Joule-Thomsoncoefficient is(a) Zero (b) Negative(c) Positive (d) Indeterminate

Sol–15: (a)Joule-Thomson coefficient is defined as,

jh

TP

Also, j =PP

1 VT VC T

For ideal gas,PV = RT

P

VT

= RP

j =P

1 T R VC P

j =P

1 PV VC P

j 0

i.e., for ideal gas, the value of the Joule-Thomson coefficient is zero.

16. The stress state at a point in a material underplane stress condition is equi-biaxial tension witha magnitude of 10 MPa. If one unit on the plane is 1 MPa, the Mohr’s circlerepresentation of the state of stress is given by

(a) a circle with a radius equal to principal stressand its center at the origin of the plane

(b) a point on the axis at a distance of 10units from the origin

(c) a point on the axis at a distance of 10units from the origin

(d) a circle with a radius of 10 units on the plane

Sol–16: (b)

According to the given problem:

10 MPa

10 MPa10 MPa

10 MPaEqui-biaxial tension

Here, x = y = 10 MPa

and xy = yx 0

so, principal stresses,

1, 2 = 2

x y x y 2xy2 2

= 10 10 0 02

1 = 10 MPa

and 2 = 10 MPa




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(MPa)

10(MPa)

Mohr's circle

i.e. Mohr’s circle will be a point on theaxis at a distance of 10 units from the

origin.

17. In a concentric tube counter flow heat exchanger,hot oil enters at 102°C and leaves at 65°C. Coldwater enters at 25 °C and leaves at 42°C. Thelog mean temperature difference (LMTD) is_______ °C (round off to one decimal place)

Sol–17: (49.3)

Given: Temperature of hot oil at the entry,

ihT = 102°C

Temperature of hot oil at the exit,

ehT = 65°C.

Temperature of cold water at entry,

icT = 25°C

and Temperature of cold water at exit,

ecT = 42°C

102°C

42°C

T i

=60°

C

65°C

25°C

Counter-flow heat exchanger

Te=

40°C

LMTD is defined as,

m( T) = i e

i

e

T TTnT

=60 40

60n40

n( T) = 49.3°C

18. A balanced rigid disc mounted on a rigid rotorhas four identical point masses, each of 10grams, attached to four points on the 100mmradius circle shown in the figure.

C

Br

A

D

r = 100mm

The rotor is driven by a motor at uniformangular speed of 10 rad/s. If one of the massesgets detached then the magnitude of theresultant unbalance force on the rotor is ____N (round off to 2 decimal places).

Sol–18: (0.1)Mass, m = 10 grams = 10 × 10–3 kgRadius, r = 100 mm = 100 × 10–3 mAngular speed, = 10 rad/sForce on each point masses,

F = 2mr = 10 × 10–3 × 100 × 10–3 × (10)2

= 0.1 NIf any one of the masses get detached, thetwo masses (opposite to each other) willcancel out or balance each other but onemass will be unbalanced which will producethe unbalance force of magnitude 0.1 N.

19. A helical gear with 20° pressure angle and 30°helix angle mounted at the midspan of a shaftthat is supported between two bearings at theends. The nature of the stresses induced in theshaft is(a) normal stress due to bending in two planes

and axial loading; shear stress due totorsion

(b) normal stress due to bending only(c) normal stress due to bending in two planes;

shear stress due to torsion(d) normal stress due to bending in one plane

and axial loading; shear stress due totorsion




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Sol–19: (a)

o

A

B

D

C

Pa

Pr

P

Pitchcylinder

Centre linesof teeth

Pt give torsionPa give axial loadingPr give bending

20. The compressor of a gas turbine plant, operatingon an ideal intercooled Brayton cycle,accomplishes an overall compression ratio of 6in a two-stage compression process. Intercoolingis used to cool the air coming out from the firststage to the inlet temperature of the first stage,before its entry to the second stage. Air entersthe compressor at 300 K and 100 KPa. If theproperties of gas are constant, the intercoolingpressure for minimum compressor work is_______ kPa (round off to 2 decimal places).

Sol–20: (244.95)

S

T

4

31

2

P2

Pi

P1

Two-stage compressionprocession process

There are two (n = 2) at stages.For minimum work,

2 ii 1 2

i 1

P P P P PP P

Overall pressure ratio, ro = 2

1

PP

6 = 22

P P 600 kPa100

Inter cooling pressure, Pi = 100 600 kPa

= 244.95 kPa21. Match the following:

P : Tempering 1. StrengtheningQ : Quenching 2. TougheningR : Annealing 3. HardeningS : Normalizing 4. Softening

Heat treatment process Effect

(a) P-2, Q-3, R-4, S-1(b) P-4, Q-3, R-2, S-1(c) P-1, Q-1, R-3, S-2(d) P-3, Q-3, R-1, S-3

Sol–21: (a)(i) Tempering: The main effect of tempering

is to impart toughness in the metal.i.e P 2

(ii) Quenching : It increases the hardness ofthe material.i.e Q 3

(iii) Annealing: It is a heat treatment processwhich ‘soften’ a metali.e R 4

(iv) Normalizing: This heat treatment processis used for strengthening and grainrefinement.i.e S 1

22. The members carrying zero force (i.e zero-forcemembers) in the truss shown in the figure, forany load P > 0 with no appreciable deformationof the truss (i.e with no appreciable change inangles between the members) are

1m 1m 1m 1mC

D EBA 45°

F

1m

G H

P

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(a) BF, DH and GC only(b) BF and DH only(c) BF, DH, GC, CD and DE only(d) BF, DH, GC, FG and GH only

Sol–22: (c)To determine zero force members:

(1) If three members are connected at a jointand there is no external force applied tothe joint and two of the members arecolinear.i.e, Joint B, D and G.

(i)A B C

F

here, FBF = 0

(ii)C D E

H

here, FDH = 0

(iii) F

C

HG

here, FGC = 0so, the members carrying zero force in thetruss are BF, DH and GC.

(2)1m 1m

C D E 45°

H

PRn

Force in member CD and DE will be zerobecause there will be no reaction force inhorizontal direction.

23. A company is hiring to fill four managerialvacancies. The candidates are five men and threewomen. If every candidate is equally likely tobe chosen then the probability that at least onewoman will be selected is ________ (round offto 2 decimal places)

Sol–23: (0.93)Number of Men, M = 5Number of Women, N = 3We have to hire four candidates.The required probability (i.e. the Probability that at least one woman will beselected),

P = 5 3 5 35 3

3 1 1 32 28 8 8

4 4 4

C C C CC CC C C

= 3 3 17 7 14 = 13

14 = 0.93

24. Match the following non-dimensional numberswith the corresponding definitions:

Buoyancy forceP : Reynolds number 1. Viscous force

Momentum diffusivityQ : Grashof number 2. Thermal diffusivity

Inertia forceR : Nusselt number 3. Viscous force

S : Pr andtl number 4.

Non - dimensional Definitionnumber

Convective heat transfer Conduction heat transfer

(a) P-3, Q-1, R-4, S-2(b) P-3, Q-1, R-2, S-4(c) P-4, Q-3, R-1, S-2(d) P-1, Q-3, R-2, S-4

Sol–24: (a)

(i) Reynolds number, Re = Inertia forceViscous force

i.e P 3

(ii) Grashof Number, Gr = Buoyancy forceViscous force

i.e Q 1

(iii) Nusselt Number, Nu =

Convective heattransfer

Conduction heattransfer

i.e R 4




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(iv)Prandtl number, Pr =

MomentumdiffusivityThermal

diffusivity

i.e S 2

so, the correct option is (a).25. For an ideal gas, a constant pressure line and

a constant volume line intersect at a point, inthe Temperature (T) versus specific entropy (s)diagram. Cp is the specific heat at constantpressure and Cv is the specific heat at constantvolume. The ratio of the slopes of the constantpressure and constant volume lines at the pointof intersection is

(a)P V

P

C CC (b)

P V

V

C CC

(c) V

P

CC (d) P

V

CC

Sol–25: (c)For ideal gas,Tds = du + Pdv = CvdT + PdvAt constant volume, (dv = 0)Tds = CvdT

v

dTds

=

v

TC ...(i)

Again, for ideal gas.Tds = dh – vdP = CpdT – vdP

For constant pressure,Tds = CpdT

P

dTds

=

P

TC ...(ii)

so, the ratio,

P

v

TdsdTds

= v v

P P

C CTC T C

26. The evaluation of the definite integral

1.4

1x x dx

by using Simpson’s 1/3rd (one third) rule with

step size h = 0.6 yields(a) 0.914 (b) 0.581(c) 1.248 (d) 0.592

Sol–26: (d)1.4

1x x dx

f(x) = –x2 1 x 0

f(x) = x2 0 x 1.4

x 1 0.4 0.2 0.8 1.4f(x) 1 0.16 0.04 0.64 1.96

yo y1 y2 y3 y4

By simpson’s 1/3rd rule

1.4

1x x dx

= o 4 2 1 3h (y y ) 2(y ) 4(y y )3

= 0.6 ( 1 1.96) 2 0.04 4( 0.16 0.64)3

= 0.59227. A vector field is defined as

3 32 2 2 2 2 22 2

32 2 2 2

x yˆ ˆf x,y,z i jx y z x y z

z k̂x y z

where, ˆˆ î, j,k are unit vectors along the axes ofa right handed rectangular Cartesian coordinate

system. The surface integral f ds (where

ds is an elemental surface area vector)evaluated over the inner and outer surfaces ofa spherical shell formed by two concentricspheres with origin as the center, and internaland external radii of 1 and 2, respectively is

(a) 4 (b) 2

(c) 0 (d) 8Sol–27: (c)

f (x,y,z)

= 2 2 2 3/2 2 2 2 3/2x yˆ î j

(x y z ) (x y z )

2 2 2 3/2z k̂

[x y z ]




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By gauss - divergence theorem

sˆFnds

= VF dV

...(i)

So, F

= 32 2 2 2

xx

(x y z )

3 32 2 2 2 2 22 2

y zy z

(x y z ) (x y z )

F

= 3 52 2 2 2 2 22 2

3 x 2x1 2

(x y z ) (x y z )

3 52 2 2 2 2 22 2

3 y 2y1 2

(x y z ) (x y z )

3 52 2 2 2 2 22 2

3 z 2z1 2

(x y z ) (x y z )

=

2 2 2 2 2 2 2 2 2

2 2 2

52 2 2 2

x y z 3x x y z 3y xy z 3z

(x y z )

= 0

So by equation (i),

SˆFndS

= VF dV

= 0

28. The truss shown in the figure has four membersof length l and flexural rigidity EI, and onemember of length 2 and flexural rigidity 4EI.The truss is loaded by a pair of forces ofmagnitude P, as shown in the figure.

P45° l,EI

l,EI l,EI

l,EIP

45°

2,4EIl

The smallest value of P, at which any of thetruss members will buckle is

(a)2

2EI

2

l(b)

2

22 EI

l

(c)2

22 EIl

(d)2

2EI

lSol–28: (b)

45°P

A B

D C 45°

P

Make FBD of joint (A)

P sin 45°

P cos 45°TAB

TAD

TAB = Pcos45° = P2

TAD = Psin45° = P2

For the bar BD,

P2

TBD

TBD

B

D

P2

P2

P2

For the joint B,

TBD

45°

T sin 45°BD

T cos 45°BD

B

P2

P2

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So in verticle direction

BDPT cos452

= 0

TBD = –Pso maximum load acting on the bar BD.so bar BD buckle firstBuckling load,

P =2

24EI

( 2 )

[both ends are hinge]

P =2

22 EI

29. Air (ideal gas) enters a perfectly insulatedcompressor at a temperature of 310 K. Thepressure ratio of the compressor is 6. Specificheat at constant pressure for air is 1005 J/kg.Kand ratio of specific heats at constant pressureand constant volume is 1.4. Assume that specificheats of air are constant. If the isentropicefficiency of the compressor is 85 percent, thedifference in enthalpies of air between the exitand the inlet of the compressor is _______ kJ/kg (round off to nearest integer)

Sol–29: (245)

S

T

2

1

2

2T

T2

T1 = 310 K

2

1

PP = 6

Cp = 1005 J/kgK

c = 85%

= 1.4

Process (1–2) (Isentropic compression)

T2 = 1

21

1

PTP

T2 = 1.4 11.4310 (6)

= 517.24 K

Isentropic efficiency

C = 2 1

2 1

T TT T

0.85 =2

517.24 310T 310

2T = 553.81 K

Increase in enthalpy

h = h2 – h1 = P 2 1C T T

= 1.005 (553.81 – 310)= 245.03 kJ/kg= 245 kJ/kg

30. A rectangular steel bar of length 500 mm, width100 mm and thickness 15 mm is cantileveredto a 200 mm steel channel using 4 bolts asshown. For an external load of 10 kN applied atthe tip of the steel bar, the resultant shear loadon the bolt at B, is _______ kN (round off toone decimal place).

Not to scaleDimensions in mm

Steel barSteel channel

200

F = 10 kN

200

A B

O50

50DC50 50

100 100 300




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IES MASTER

Sol–30: (16)

r1

r4

r2r3

O

P2

P1

45°

A B

D C

P = 10 kNr1 = r2 = r3 = r4

= 2 2(50) (50)

= 70.71 mmEccentricity, e = 400 mm

Primary force, P1 = P 104 4 = 2.5 kN

Secondery force, P2 = 12 2 2 21 2 3 4

P e rr r r r

= 210 400 70.71

4 (70.71)

= 14.14 kN

= 90° – 45° = 45°

Resultant shear force on bolt B,

PB = 2 21 2 1 2P P 2P P cos45

= 2 2(2.5) (14.14) 2 2.5 14.14 cos45

= 16 kN31. Consider two cases as below.

Case 1: A company buys 1000 pieces per yearof a certain part from vendor ‘X’. The changovertime is 2 hours and the price is Rs. 10 perpiece. The holding cost rate per part is 10% peryear.Case 2: For the same part, another vendor ‘Y’offers a design where the changeover time is 6minutes, with a price of Rs. 5 per piece and aholding cost rate per part of 100% per year. Theorder size is 800 pieces per year from ‘X’ and200 pieces per year from ‘Y’.

Assume the cost of downtime as Rs. 200 perhour. The percentage reduction in the annualcost for case 2, as compared to case 1 is_________ (round off to 2 decimal places)

Sol–31: (5.32)Case-I: (For vendor X)Demand, D = 1000Changeover time = 2 hourCost of downtime, C0 = 200 × 2 = Rs. 400One piece cost, C = Rs. 10Holding cost, Ch = 0.1 × 10 = Rs. 1

Annual cost = C × D 0 hD QC CQ 2

= 1000 100010 1000 400 11000 2

(TAC)1 = Rs. 10,900Case-II: (order size is 800 pieces per year fromX and 200 pieces per year from Y)Q = 800 from X, Q = 200 for YChange over time = 6 min.

Cost of downtime, C0 = 6 200 Rs.2060

One piece cost Rs. 5Holding cost, Ch = 1 × 5 = Rs. 5Total annual cost = (TAC)X + (TAC)Y

(TAC)X = h 0Q DC D C C2 Q

= 800 80010 800 1 400 Rs.8,8002 800

(TAC)Y = h 0Q DC D C C2 Q

= 200 2005 200 5 20 Rs.15202 200

(TAC)2 = 8800 + 1520 = Rs. 10,320

% reduction = 10320 10900 10010900

= 5.32%




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32. A strip of thickness 40 mm is to be rolled to athickness of 20 mm using a two high mill havingrolls of diameter 200 mm. Coefficient of frictionand arc length in mm, repectively are(a) 0.39 and 44.72 (b) 0.45 and 44.72(c) 0.39 and 38.84 (d) 0.45 and 38.84

Sol–32: (b)H1 = 40 mmH2 = 20 mmR = 100 mm

H = 40 – 20 = 20 mm

Coefficient of friction ( )

= max( H)R

max[ H ( H) ]

= 20100 = 0.45

Arc length (L) = R( H)

= 100 20 = 44.72 mm

33. A small metal bead (radius 0.5 mm) initially at100°C. When placed in a stream of fluid at20°C attains a temperature of 28°C in 4.35seconds. The density and specific heat of themetal are 8500 kg/m3 and 400 J/kg.K,respectively. If the bead is considered as lumpedsystem, the convective heat transfer coefficient(in W/m2.K) between the metal bead and thefluid stream is(a) 283.3 (b) 299.8(c) 449.7 (d) 149.9

Sol–33: (b)

To = 100°C T = 20°C

T = 28°C = 4.35 s = 8500 kg/m3 C = 400 J/kgK

r = 0.5 mm h = ?Metal bead is assumed as spherical shapeBy lumped system analysis

o

T TT T

=

hAVCe

o

T TT T

=

2

3h 4 r

4V r3e

o

T TT T

=

3hV re

28 20100 20

= 33 h 4.35

8500 400 0.5 10e

h = 299.95 W/m2K34. The magnitude of reaction force at joint C of

the hinge-beam shown in the figure is _______kN (round off to 2 decimal places)

50 kN

A B

10 kN/m

C

4 mHinge

3 m3 m

Sol–34: (20)At hing point moment is zeroFrom bar BC

40 kN

B CRHB

RVB

2m2mRHC

RVC

Take moment about BRVC × 4 = 40 × 2

RVC = 20 kNThere is no force apply in horizontaldirection so

RHB = RHC = 0 NNet force on C,

FC = 2 2HC VC(R ) (R )

= 2 2(0) (20)

FC = 20 kN

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35. A steel part with surface area of 125 cm2 is tobe chrome coated through an electroplatingprocess using chromium acid sulphate as anelectrolyte. An increasing current is applied tothe part according to the following current timerelation:

l 12 0.2t

where, l = current (A) and t = time (minutes).The part is submerged in the plating solutionfor a duration of 20 minutes for plating purpose.Assuming the cathode efficiency of chromiumto be 15% and the plating constant of chromiumacid sulphate to be 2.50 × 102 mm3/A.s theresulting coating thickness on the part surfaceis _______ m (round off to one decimal place)

Sol–35: (5.04)Given: Surface area, A = 125 cm2

Current-time corelation,

= 12 + 0.2t, t is in minutesDuration = 20 minutesPlating constant, C = 2.50 × 10–2 mm3/A.sCathode efficiency, = 0.15

Let, the coating thickness is dx.According to the given question,

A × dx = C dt

125 × 102 × dx = 0.15 × 2.50 × 10–2 ×(12 + 0.2t) dtIntegrating both sides,

x20

125 10 dx = 2(0.15 2.50 10 )

20

0(12 0.2t) 60 dt

125 × 102 × x = (0.15 × 2.50 × 10–2) × 60

× 202

0

0.2t12t2

x = 5.04 × 10–3 mm

= 5.04 m

36. The following data applies to basic shaft system;tolerance for hole = 0.002 mmtolerance for shaft = 0.001 mm

allowance = 0.003 mmbasic size = 50 mmThe maximum hole size is _____ mm (round offto 3 decimal places)

Sol–36: (50.005)For basic shaft system,

hole

shaft 50mm

0.002

0.003

0.001

Maximum hole size = 50 + 0.003 + 0.002= 50.005 mm

37. One kg of air, initially at a temperature of127°C, expands reversibly at a contant pressureuntil the volume is doubled. If the gas constantof air is 287 J/kg.K, the magnitude of worktransfer is __________ kJ (round off to 2 decimalplaces)

Sol–37: (114.80)m = 1 kgR = 287 J/kgK

T1 = 127°C = 127 + 273= 400 K

P1 = P2 = PV1 = VV2 = 2V

P = C 21

Work done, w = P(V2 – V1)= P (2V – V)= PV = mRT= 1 × 0.287 × 400= 114.80 kJ

38. In a turning process using orthogonal toolgeometry, a chip length of 100mm is obtainedfor an uncut chip length of 250 mm.The cutting conditions are cutting speed = 30m/min, rake angle = 20°The shear plane angle is ______ degrees (roundoff to one decimal place)




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Sol–38: (23.53)

c = 100 mm

= 250 mm

V = 30 m/min. = 20°

Chip thickness ratio (r),

r = c

= 100 0.4250

Shear plane angle ( ) ,

tan =r cos

1 r sin

= 1 0.4 cos20tan1 0.4sin20

= 23.53°

39. An analytic function of a complex variable z =

x + iy i 1 is defined as

2 2f z x y i x,y

where x,y is a real function. The value ofthe imaginary part of f(z) at z = (1 + i) is _______(round off to 2 decimal places)

Sol–39: (2)f(z) = u + iv

u = x2 – y2

v = (x, y) ...(i)

f(z) is an analytic function,

ux

=vy

vx

=uy

so, dv =v vdx dyx y

dv =u udx dyy x

dv = 2y dx + 2x dy

dv = 2 d(xy)v = 2xy ...(ii)

Given, z = 1 + iSo, x = 1

y = 1From equation (1) and (2)

So, v = (1, 1)(x, y) = 2

40. For an assembly line, the production rate was4 pieces per hour and the average processingtime was 60 minutes. The WIP inventory wascalculated. Now, the production rate is kept thesame, and the average processing time isbrought down by 30 percent. As a result of thischange in the processing time, the WIPinventory.(a) increases by 30%(b) decreases by 30%(c) decreases by 25%(d) increases by 25%

Sol–40: (b)WIP means work in processIn first case

(WIP)1 =60460

= 4 pieces

In second case

(WIP)2 =(60 60 0.3)4

60

= 2.8 piece

So, Decrease in (WIP) = 1 2

1

(WIP) (WIP) 100(WIP)

= 4 2.8 1004

= 30%41. For an ideal Rankine cycle operating between

pressures of 30 bar and 0.04 bar, the workoutput from the turbine is 903 kJ/kg and thework input to the feed pump is 3 kJ/kg. Thespecific steam consumption is _________kg/kW.h




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Sol–41: (4)Turbine work, wT = 903 kJ/kgPump work, wP = 3 kJ/kg

wnet = wT – wP

= 903 – 3 = 900 kJ/kg

Specific steam consumption = net

3600 kg / kWhw

=3600900 = 4 kg/kWh

42. Consider two exponentially distributed randomvariables X and Y both having a mean of 0.50.Let Z = X + Y and r be the corrrelationcoefficient between X and Y. If the variance ofZ equals 0, then the value of r is ________(round off to 2 decimal places)

Sol–42: (–1)

Mean of exponential distribution = 1

Standard deviation of exponential

distribution = 1

For x, 1 = 1

0.5 = 2

Standard deviation, x = 0.5

Variance, V(X) = 2x( ) = 0.25

For Y, 2 =1

0.5 = 2

Standard deviation, Y = 0.5

Variance V(Y) = 2Y( ) = 0.25

V(X + Y) = V(X) + V(Y) + 2COV(X, Y)Given: V(X + Y) = 0

0 = 0.25 + 0.25 + 2 COV (X, Y)

COV (X, Y) = –0.25

Corelation factor, r = X Y

COV(X, Y)

= 0.25

0.5 0.5

= –1

43. For a Kaplan (axial flow) turbine, the outletblade velocity diagram at a section is shown infigure.

C : Flow velocityfC : Blade velocityC : Relative velocity

b

r

C fCr

Cb

The diameter at this section is 3m. The huband tip diameters of the blade are 2m and 4m,respectively. The water volume flow rate is 100m3/s. The rotational speed of the turbine is 300rpm. The blade outlet angle is _______ degrees(round off to one decimal place)

Sol–43: (12.69)

Cf

Cr

Cb

Blade velocity (Cb) at the given section

Cb =DN60

=

3 30060

= 47.12 m/sDischarge (Q) = flow area (Af) × flow velocity(Cf)

100 = 2 2t h fD D C

4

100 = 2 2f4 2 C

4

Cf = 10.61 m/s

Now, = 1 f

b

CtanC

= 1 10.61tan47.12

= 12.69°

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44. The barrier shown between two water tanks ofunit width (1m) into the plane of the screen ismodeled as a cantilever.

4 m

1 m

barrier

Taking the density of water as 1000 kg/m3, andthe acceleration due to gravity as 10 m/s2, themaximum absolute bending moment developedin the cantilever is __________ kN-m (roundoff to the nearest integer)

Sol–44: (105)

4m

barrier

A

F1

F2

1Px

2Px

1x =2m

2x =0.5m1m

A

F1F2

1P(4 x )2P(1 x )

F1 = 1wAx = 10 × (4 × 1) × 2 = 80 kN

F2 = 2wAx = 10 × (1 × 1) × 0.50 = 5 kN

1Px = 3

11 4x

12 1 4 2

= 223

= 8 m3

2Px = 3

21 1x

12 1 1 0.5

= 10.56

= 2 m3

The maximum absolute bending momentwill be at fixed support i.e at point A.

(BM)A = 1 21 P 2 PF (4 x ) F (1 x )

= 8 280 4 5 13 3

= 105 kN-m45. A rigid mass-less rod of length L is connected

to a disc (pulley) of mass m and radius r = L/4 through a friction less revolute joint. The otherend of that rod is attached to a wall through afriction less hinge. A spring of stiffness 2k isattached to the rod at its mid span. Aninextensible rope passes over half the discperiphery and is securely tied to a spring ofstiffness k at point C as shown in the figure.There is no slip between the rope and the pulley.The system is in static equilibrium in theconfiguration shown in the figure and the ropeis always taut.

Inextensible rope

A Mass-less rodC

k

r

BL/2L/2 2kDisc mass m

Lr4

Neglecting the influence of gravity, the naturalfrequency of the system for small amplitudevibration is

(a) 3 k2 m (b) k

m

(c)3 k

m2 (d) k3m

Sol–45: (d)

Inextensible rope

A

Mass-less rodC

k

r

B

2kDisc mass m

Lr4

P 2r

L2

L/2 L/2

rm




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Since the disc is in pure rolling on the string

By rotating the disc with angle results inrotating the rod with angle (let us say) abouthinge point.

Now for same deflectionof point ‘B’

L = r

= rL

Lr4

= L4 L

By energy method

Total energy,

Etotal =

222

p1 1 1 LI k 2r 2k2 2 2 2

Etotal =

222

p1 1 LI k 2r k2 2 2 4

totaldEdt = 0

= 2

2 21 3 1 kLmr 2 k 4r 2 2 02 2 2 64

2 2

23m L 4L 2k kL2 16 16 64

= 0

3m 9k32 32

= 0

3m 9k = 0

3km

= 0

n = k3m

46. The thickness of a steel plate with materialstrength coefficient of 210 MPa has to be reducedfrom 20 mm to 15 mm in a single pass in a two

high rolling mill with a roll radius of 450 mmand rolling velocity of 28 m/min. If the platehas a width of 200 mm and its strain hardeningexponent, n is 0.25, the rolling force requiredfor the operation is _________ kN (round off to2 decimal places).Note: Average flow stress = Material strength

coefficient ×

nTrue strain1 n

Sol–46: (1167.26)(Final volume) = (Initial volume)

f f fb h = i i ib h

f 200 15 = i 200 20

[ width remains same in rolling]

f

i

=

2015

Now,true strain, T = f

in

= 20n15

Average flow stress = Material strength

coefficient ×n

T( )1 n

n = 0.25 (given)

f =

0.2520n15210

1 0.25

= 123.04 MPa

Roll contact length, L = R h

= 450 5

= 47.434 mm

Required force, F = fL b

= 47.43 × 200 × 123.04= 1167255.87 N= 1167.26 kN

47. Bars of square and circular cross-section with0.5 m length are made of a material with shear




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strength of 20 MPa. The square bar cross-sectiondimension is 4 cm × 4 cm and the cylindricalbar cross-section diameter is 4 cm. Thespecimens are loaded as shown in the figure.

Tensile load

80 kN

Compressive load

80 kN

64 NmTorsional load

320 NmBending load

Which specimen(s) will fail due to the appliedload as per maximum shear stress theory?(a) Torsional load specimen(b) Tensile and compressive load specimens(c) None of the specimens(d) Bending load specimen

Sol–47: (b)

4cm4cm

= PA =

380 1040 40

= 50 MPa

According to maximum shear stress theory

max = 1 22

= 50 0

2

= 25 MPa

per max

This component will failSimilarly,The component, loaded with same load incompression, will also fails.

64 Nm

4cm

max = 316T

d

=3

316 64 10

40

= 16 MPa

max per

Component will not fail.320 Nm

max = M y

= 3

3320 10 2040 40

12

= 30 MPaAccording to maximum shear stress theory

max = max min2

= 30 02 = 15 MPa

max per

component will not fail48. Consider steady, viscous, fully developed flow of

a fluid through a circular pipe of internaldiameter D. We know that the velocity profileforms a paraboloid about the pipe centre line,

given by: V =

22 DC r

4 m/s. Where C is a

constant. The rate of kinetic energy (in J/s) atthe control surface A-B, as shown in the figureis proportional to Dn. The value of n is _____

r Velocity

profile

Control

volume

A

B

Diameter= D

Not to scale

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Sol–48: (8)

r Velocityprofile

Controlvolume

A

B

Diameter = D

rdr

Consider a fluid element of dr thickness ata radius of r from the centre.Now, kinetic energy rate of this element.

d(K.E)

= 21 d m v2

= 21 ( dA v) v2

d(K.E) = 31 2 rdr v

2

d(K.E)

= 32D/2 2

0Drdr C r4

K.E

= 32D/23 2

0DC r r dr4

K.E

=6 2 2 2D/23 6 230

D r D DC r r 3 r dr4 44

After integration we get,

K.E

= f(D8)

So, n = 849. The 2 kg block shown in figure (top view) rests

on a smooth horizontal surface and is attachedto a massless elastic cord that has a stiffness5 N/m.

x

v = 1.5 m/sy

Block

Elasticcord 0.

5 m

o

Horizontal Surface

The cord hinged at O is initially unstretched

and always remains elastic. The block is givena velocity v of 1.5 m/s perpendicular to the cord.The magnitude of velocity in m/s of the block atthe instant the cord is stretched by 0.4 m is(a) 1.50 (b) 1.36(c) 0.83 (d) 1.07

Sol–49: (b)Given: Mass, m = 2 kgStiffness of rod, k = 5 N/mInitial velocity when the block isunstretched, Vi = 1.5 m/sNow, the cord is stretched by x = 0.4 m,then velocity of the block is Vf.Using work-energy theorem,

2i

1 m v2 = 2 2

f1 1m v k x2 2

Putting the respective values,

2 × (1.5)2 = 2 2f2 (V ) 5 (0.4)

Vf = 1.36 m/s50. Two business owners Shveta and Ashok run

their business in two different states. Each ofthem, independent of the other, produces twoproducts A and B, sells them at Rs. 2,000 perkg and Rs. 3,000 per kg, respectively, and usesLinear programming to determine the optimalquantity of A and B to maximize their respectivedaily revenue. Their constraints are as follows:(i) For each business owner the productionprocess is such that the daily production of Ahas to be at least as much as B, and the upperlimit for production of B is 10 kg per day, and(ii) the respective state regulations restrictShveta’s production of A to less than 20 kg perday, and Ashok’s production of A to less than15 kg per day. The demand of both A and B inboth the states is very high and everythingproduced is sold.The absolute value of the difference in daily(optimal) revenue of Shveta and Ashok is_______ thousand Rupees (round off to 2 decimalplaces).

Sol–50: (10)Let x1 = daily production of A




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x2 = daily production of BObjective functionMaximise Z = 2000 x1 + 3000 x2

For shveta

Constraints 1 2x x

2x 10

1x 20

x =102 B (10, 10)x =x1 2

(0, 10)

x2

A(0, 0) x1 (20, 0)

C(20, 10)

x =201

DFeasibleregion

At A Z = 0At B Z = 2000 × 10 + 3000 × 10

= 50000At C Z = 2000 × 20 + 3000 × 10

= 70000At D Z = 2000 × 20 + 3000 × 0

= 40000Zmax = 70000

For Ashok,

Constraints, 1 2x x

2x 10

1x 15

B (10, 10)(0, 10)

x2

A(0, 0) x1 (15, 0)

C(15, 10)

DFeasibleregion

x =x1 2

At C,Zmax = 2000 × 15 + 3000 × 10

= 60000

Difference = 70000 – 60000 = Rs. 10000= 10 thousand Rupees.

51. A slot of 25 mm × 25 mm is to be milled in aworkpiece of 300 mm length using a side andface milling cutter of diameter 100 mm, width25 mm and having 20 teeth.For a depth of cut 5 mm, feed per tooth 0.1mm, cutting speed 35 m/min and approach andover travel distance of 5 mm each, the timerequired for milling the slot is _________minutes (round off to one decimal place).

Sol–51: (1.616)AA Lw O

Length of workpiece, Lw = 300 mmDiameter of cutter, Dc = 100 mmNumber of teeth, Z = 20Depth of cut = 5 mmFeed per tooth, fz = 0.10 mmCutting speed, v = 35 m/min.Approach (A) = over travel (O) = 5 mmTo find, Machining time, (for single pass)

tm =r

length of cut (L)feed rate (f )

L = wA A L O

L = 100 5 300 5 360mm

2

fr = fz × Z × N = zc

vf ZD

=335 100.10 20

100

= 222.82 mm/min.

tm = 360 1.616min

222.82

The time required for milling the slot(25 mm × 25 mm) will be 1.616 min




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52. A cam with a translating flat-face follower isdesired have the follower motion

2y 4 2 ,0 2

Contact stress considerations dictate that theradius of curvature of the cam profile shouldnot be less than 40 mm anywhere. Theminimum permissible base circle radius is__________ mm (round off to one decimal place)

Sol–52: (48)

Given: y( ) = 24 2 , 0 2

y ( ) = 8 8

y ( ) = –8

The minimum permissible base circleradius,

o min(r ) = min miny( ) y ( )

Here, min = 40 mm (given in the question)

y( ) y ( ) = 24[2 ] 8 , 0 2

miny( ) y ( ) = 24 2 0 0 8

= –8so, (ro)min = 40 – (–8) = 48 mm

53. The indicated power developed by an engine withcompression ratio of 8, is calculated using anair-standard Otto cycle (constant properties). Therate of heat addition is 10 kW. The ratio ofspecific heats at constant pressure and constantvolume is 1.4. The mechanical efficiency of theengine is 80 percent.The brake power output of the engine is _____kW (round off to one decimal place)

Sol–53: (4.5)Compression ratio, r = 8

Heat supplied, sQ = 10 kW

Mechanical efficiency, m 80%

Efficiency of otto cycle, = 111

r

We also know that, = work output

heat supplied

Here, work output is indicated power, .P.(given in the question)

so,s

.P.Q = 1

11r

.P.10

= 1.4 111

(8)

.P. = 5.65 kW

Now, mechanical efficiency,

m = Brake power

Indicated power

0.8 =B.P.5.65

B.P = 4.5 kW54. In a disc-type axial clutch, the frictional contact

takes place within an annular region with outerand inner diameters 250mm and 50mm,respectively. An axial force F1 is needed totransmit a torque by a new clutch However, totransmit the same torque, one needs an axialforce F2 when the clutch wears out. If contactpressure remains uniform during operation of anew clutch while the wear is assumed to beuniform for an old clutch, and the coefficient offriction does not change, then the ratio F1/F2 is_______ (round off to 2 decimal places)

Sol–54: (0.87)Given:Inner diameter, Di = 50 mm

or Inner radius, Ri = 502 = 25 mm

and outer diameter, Do = 250 mm

or outer radius, Ro = 2502 = 125 mm

(1) For uniform pressure theory (for newclutch)

Torque transmitted, 1 1 1 mF R

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where, mR =3 3o i2 2o i

(R R )23 (R R )

i.e, 1 =3 3o i

1 1 2 2o i

R R2F3 R R

(2) For uniform wear theory (for old clutch),

Torque transmitted, 2 = 2 2 mF R

where, mR = o iR R2

i.e., 2 = o i2 2

(R R )F2

Torque transmitted are same andcoefficient of friction is also same.

so, 3 3o i

1 2 2o i

R R2F3 R R

= o i

2(R R )F

2

1

2

FF

= 2 2o i

o i3 3o 1

R R3 (R R )4 R R

Putting the respective values,

1

2

FF =

2 2

3 33 125 25 (125 25)4 125 25

1

2

F 0.87F

55. Air discharge steadily through a horizontalnozzle and impinges on a stationary verticalplate as shown in figure.

O

p = 0.36 kPainlet

patm

The inlet and outlet areas of the nozzle are 0.1m2 and 0.02 m2, respectively. Take air densityas constant and equal to 1.2 kg/m3. If the inletgauge pressure of air is 0.36 kPa. the gauge

pressure at point O on the plate is _________kPa (round off to two decimal places).

Sol–55: (0.38)

A =0.1m12

A =0.02m22

21Patm

P = 0.36 kPainlet

O

Apply Bernoulli’s equation between point 1and 2,

21 1P Vg 2g

=2

2 2P Vg 2g

1 2( z z )

Here, P1 = 0.36 kPa (guage)Patm = 0 (guage)

We get,

2 22 1V V = 12P

= 2 0.36 1000

1.2

= 600 ...(i)From continuity equation,

A1V1 = A2V2

0.1 × V1 = 0.02 × V2

2

1

VV = 5 ...(ii)

Solving both equations (i) and (ii),We get, V2 = 25 m/s

The pressure between 2 and 0 isatmospheric. So, the only pressure at pointO will be dynamic pressure due to air

which is equal to 21 V2 .

so, Gauge pressure at point O on the plate,

Po = 21 V2 =

21 1.2 (25)2 1000

Po = 0.375 kPa 0.38 kPa(Gauge)

Date post:	14-Mar-2021
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