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Questions with Detailed Solutions

GATE-2020

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MECHANICAL ENGINEERING(Afternoon Session)

ACE Engineering Publications Hyderabad Delhi Pune Bhubaneswar Lucknow Patna Bengaluru Chennai Vijayawada Vizag Tirupati Kolkata Ahmedabad

SUBJECT-WISE WEIGHTAGE

S. No. NAME OF THE SUBJECT 1 MARK QUESTIONS

2 MARKS QUESTIONS

01 Engineering Mechanics 1 1

02 Fluid Mechanics & Turbomachinery 2 3

03 Heat Transfer 1 2

04 Theory of Machines 3 2

05 Thermal Engineering 4 4

06 Strength of materials 1 3

07 Machine Design 1 2

08 IM & OR 2 2

09 Production Technology 4 6

10 Material Science 1 1

11 Engineering Mathematics 5 4

12 Verbal Ability 2 3

13 Numerical Ability 3 2

Total No. of Questions 30 35

GATE - 2020MECHANICAL ENGINEERING

Question with Detailed Solutions

01/02/20

AfternoonSession

2 GATE - 2020 _Questions with Solutions


Section : General Aptitude

01. An engineer measures THREE quantities X, Y andZ in an experiment. She finds that they follow arelationship that is represented in the figure below:(the product of X and Y linearly varies with Z)

ZO

(X,Y)

Then, which of the following statements is FALSE?(a) For fixed Z; X is proportional to Y(b) XY/Z is constant(c) For fixed X; Z is proportional to Y(d) For fixed Y; X is proportional to Z

01. Ans:(a)Sol: (X,Y)

O Z

From the given figureXY∝ ZXY = mZ, where m is constant

Option (a):If Z is fixed, X

Y

1? , X is inversily proportional to

yOption (b):

m = constant XY

2= it is true

Option (c):If X is fixed, then Z is proportional to Y, it is true.Option (d):If X is fixed, then X is proportional to ZHence option (a) is the answer

02. While I agree _______ his proposal this time, I donot often agree ______ him.(a) to, to(b) with, with(c) with, to(d) to, with

02. Ans: (d)Sol: Agree to a proposal and agree with a person.

03. In one of the greatest innings ever seen in 142 yearsof Test history, Ben Stokes upped the tempo in afive-and-a-half hour long stay of 219 balls including11 fours and 8 sixes that saw him finish on a 135 notout as England squared the five-match series.Based on their connotations in the given passage,which one of the following meanings DOES NOTmatch?(a) upped = increased(b) saw = resulted in(c) tempo = enthusiasm(d) squared = lost

03. Ans: (d)Sol: Squared means equalled and so option four does not

match.

04. Find the missing element in the following figure.

5

h

x

9

n

t

?

(a) w (b) y(c) e (d) d

3 MECHANICAL ENGINEERING _ SET - 2


05. It was estimated that 52 men can complete a strip in a newly constructed highway connecting cities P and Q in 10 days. Due to an emergency, 12 men were sent to another project. How many number of days, more than the original estimate, will be required to complete the strip?

(a) 10 days (b) 13 days (c) 5 days (d) 3 days

05. Ans: (d)Sol: For given data, we have the chain rule

W

N D R E

W

N D R E

1

1 1 1 1

2

2 2 2 2# # # # # #=

N1×D1 = N2×D2

N1 = 52 men, D1 = 10 days, N2 = (52 −12) = 40 men, then D2 = ? 52 ×10 = 40 × D2

D2 = 13 days, ∴ More no. of days are required then the estimated = 13 −10 = 3 days

06. The recent measures to improve the output would ______ the level of production to our satisfaction.

(a) equalise (b) increase (c) speed (d) decrease06. Ans: (b)Sol: “to improve” indicates increase.

07. There are five levels {P, Q, R, S, T} in a linear supply chain before a product reaches customers, as shown in the figure:

P CustomersTSRQ

At each of the five levels, the price of the product is increase by 25%. If the product is produced at level P at the cost of Rs. 120 per unit, what is the price paid (in rupees) by the customers?

(a) 292.96 (b) 366.21 (c) 234.38 (d) 187.50

07. Ans: (a) Sol: From the given data , the product will reaches to

the customers is as follows

P Q R S T 25% ↑ 25% ↑ 25% ↑ 25% ↑

Rs.120 per unit

Customers

The price paid (in rupees) by the

.Customers 120 1100

25292 96

4

= + =c m

08. Select the word that fits the analogy: White: Whitening: : Light : ________ (a) Lightning (b) Lightening (c) Lighting (d) Enlightening08. Ans: (2)Sol: Noun and verb analogy.

09. Climate change and resilience deal with two aspects - reduction of sources of non-renewable energy sources and reducing vulnerability of climate change aspects. The terms 'mitigation' and 'adaptation' are used to refer to these aspects, respectively.

Which of the following assertions is best supported by the above information?

(a) Mitigation deals with consequences of climate change.

(b) Mitigation deals with actions taken to reduce the use of fossil fuels

(c) Adaptation deals with actions taken to combat green-house gas emissions.

(d) Adaptation deals with causes of climate change.

09. Ans: (b)Sol: Reduction of sources...suggests option two.



10. The two pie-charts given below show the data of total students and only girls registered in different streams in a university. If the total number of students registered in the university is 5000, and the total number of the registered girls is 1500; then, the ratio of boys enrolled in Arts to the girls enrolled in Management is ________.

Percentage of studentsenrolled in different

streams in a University

Management15%

Arts20%

Engineering30%

Science20%

Commerce15%

Percentage of girlsenrolled in

different streams

Management15%

Arts30%

Engineering10%

Science25%

Commerce20%

(a) 22 : 9 (b) 2 : 1 (c) 9 : 22 (d) 11 : 9

10. Ans:(a)Sol: Total number of students registered in the university = 5000 Total number of the registered girls = 1500 The boys enrolled in arts = 20% of 5000 − 30% of 1500 = 550 The girls enrolled in management = 15 % of 1500 = 225 ∴The ratio of boys enrolled in arts to the girls

enrolled in management :225

550

9

2222 9&= =

Hence option (a) is correct.

Section : MECHANICAL ENGINEERING

01. Consider the following network of activities, with each activity named A-L, illustrated in the nodes of the network.

A FINISHSTART ECB

DL

K4 410

78

5

6

2

976

2

JI

F

HG

The number of hours required for each activity is shown alongside the nodes. The slack on the activity L, is _______ hours.

01. Ans: 2 Sol:

ECB

DL

K F44 10

7

8

5

6

2

976

JI

F

HG

A2

20 6 16

23

31

36

42 42 42

40 42

382922

20

Slack of activity L = 42 - 40 = 2

02. A circular disk of radius r is confined to roll without slipping at P and Q as shown in the figure.

2v

v

rP

Q



If the plates have velocities as shown, the magnitude of the angular velocity of the disk is

(a) r

v (b) r

v

2

(c) r

v

3

2 (d) r

v

2

3

02. Ans: (d)Sol:

2v

v

Ix

2r - x

B

A

ω

IA IB

2ν ν ω= =

x 2r x

2ν ν= −

2r x 2x x3

2r&− = =

x 2r

3ω ν ν= =

03. A bolt head has to be made at the end of a rod of diameter d = 12 mm by localized forging (upsetting) operation. The length of the unsupported portion of the rod is 40 mm. To avoid buckling of the rod, a closed forging operation has to be performed with a maximum die diameter of _______ mm.

03. Ans: 18 Sol: As upsetting involves longitudinal compression

of the work-piece, so to prevent buckling of the unsupported length, l, to be forged, the following rules are :

• In an open upsetting, l ≤ 3d, where d is the diameter of the work-piece.

• If l > 3d, a closed upsetting operation should be performed with an inner diameter of the die, D ≤ 1.5 d

• If during closed upsetting, l extends beyond the die cavity by an amount l1, then l1 ≤ d

Upsetting, d = 12 mm, l = 40 mm, l > 3d l > 36 D ≤ 1.5d D = 18 mm 04. The equation of motion of a spring-mass-damper

system is given by sin

dt

d x

dt

dxx t3 9 10 5

2

2

+ + = ] g

The damping factor for the system is (a) 0.25 (b) 0.5 (c) 3 (d) 204. Ans: (b)Sol:

sindt

d x

dt

dxx t3 9 10 5

2

2

+ + = ] g

D x 2 Dx x 10sin 5t2

n n2ςω ω+ + =p ^ ]h g

On comparing 9 9n

2n&ω ω= =

And 2 ς wn = 3 ⇒ ς = 0.5

05. Two plates, each of 6 mm thickness, are to be butt-welded. Consider the following processes and select the correct sequence in increasing order of size of the heat affected zone.

1. Arc welding 2. MIG welding 3. Laser beam welding 4. Submerged arc welding



(a) 1-4-2-3 (b) 3-4-2-1 (c) 3-2-4-1 (d) 4-3-2-105. Ans: (c)Sol: Higher the intensity of heat source, lower the weld

pool size, lower the welding temperature level is HAZ. Laser beam and electron beam are very high intensity sources. TIG & MIG have higher intensity and lower weld pool size than SAW.

06. Let I be a 100 dimensional identity matrix and E be the set of its distinct (no value appears more than once in E) real eigenvalues. The number of elements in E is _________________.

06. Ans: 1Sol: Let I be a 100 × 100 matrix It's × all eigen values are 1 Let E = Set of distinct real eigen values ∴ Number of distinct elements in E = 1

07. A matrix P is decomposed into its symmetric part S and skew symmetric part V. If

/

/ ,

/

/S V

4

4

2

4

3

7 2

2

7 2

2

0

2

3

2

0

7 2

3

7 2

0

=−

=−

−

−

J

L

KKKKKKK

J

L

KKKKKKK

N

P

OOOOOOO

N

P

OOOOOOO,

then matrix P is (a)

4

2

5

6

3

7

1

0

2

− −J

L

KKKKKKK

N

P

OOOOOOO (b)

4

2

5

6

3

7

1

0

2

−−

−−− −

J

L

KKKKKKK

N

P

OOOOOOO

(c) /

/

/ /

2

1

2

9 2

81 4

45 2

1

11

73 4

−−−

−J

L

KKKKKKK

N

P

OOOOOOO (d)

4

6

1

2

3

0

5

7

2

−

−

J

L

KKKKKKK

N

P

OOOOOOO

07. Ans: (d)

Sol: P2

1P P

2

1P P

T T= + + −] ]g g P S V= +

Where S

2

1P P

T= +] g & V2

1P P

T= −] g

P

4

4

2

4

3

2

7

2

2

7

2

0

2

3

2

0

2

7

3

2

7

0

=

−

+−

−

−

R

T

SSSSSSSSSSS

R

T

SSSSSSSSSSS

V

X

WWWWWWWWWWW

V

X

WWWWWWWWWWW

P

4

6

1

2

3

0

5

7

2

=−

−

R

T

SSSSSSSS

V

X

WWWWWWWW

Option (d) is correct

08. If a reversed Carnot cycle operates between the temperature limits of 27°C and −3°C, then the ratio of the COP of a refrigerator to that of a heat pump (COP of refrigerator/ COP of heat pump) based on the cycle is ______(round off to 2 decimal places).

08. Ans: 0.9Sol:

COP

COP

30

300

30

270

0.9HP

R = =]]

gg

09. For an air-standard Diesel cycle,(a) heat addition is at constant pressure and heat

rejection is at constant pressure(b) heat addition is at constant volume and heat

rejection is at constant volume(c) heat addition is at constant pressure and heat

rejection is at constant volume.(d) heat rejection is at constant volume and heat

rejection is at constant pressure

09. Ans: (c)Sol: Heat addition is at constant pressure and heat

rejection is at constant volume.



10. Which one of the following statements about a phase diagram is INCORRECT?

(a) It indicates the temperature at which different phases start o melt

(b) Relative amount of different phases can be found under given equilibrium conditions

(c) solid solubility limits are depicted by it(d) it gives information on transformation rates

10. An: (d)Sol: Phase diagram – Useful information: Important information, useful in materials

development and selection, obtainable from a phase diagram: - It shows phases present at different compositions

and temperatures under slow cooling (equilibrium) conditions.

- It indicates equilibrium solid solubility of one element/compound in another.

- It suggests temperature at which an alloy starts to solidify and the range of solidification.

- It signals the temperature at which different phases start to melt.

- Amount of each phase in a two-phase mixture can be obtained.

11. Which of the following conditions is used to determine the stable equilibrium of all partially submerged floating bodies?

(a) Metacentre must be at a higher level than the centre of gravity

(b) Centre of buoyancy must be below the centre of gravity

(c) Centre of buoyancy must be above the centre of gravity

(d) Metacentre must be at a lower level than the centre of gravity

11. Ans: 1Sol: For stability of partially submerged (floating)

objects meta centre must be above centre of gravity.

12. Let I = xy dydxx

2

y 0x 0

12

==

## . Then, I may also be

expressed as (a) y d dx x y

2

x 0

y

0

1

y ==

## (b) x d dy x y2

x y0

1

y

1

==

##

(c) xy d dx y2

x 0

y

0

1

y ==

## (d) x d dy x y2

x y0

1

y

1

==

##

12. Ans: (c)Sol: I xy dy dx

2

y 0

x

x 0

1 2

===

## .......(1)

(1, 1)

y = x2

x = 1 (0, 0)

After changing the order of integration, equation (1) can be expressed as

I xy dydx xy dxdy2 2

x y

1

y 0

1

y 0

x

0

1 2

= ====

####

13. In a furnace, the inner and outer sides of the brick

wall (K1 = 2.5 W/m.K) are maintained at 1100°C and 700°C, respectively as shown in figure.

1100°C 700°C 200°C

Brick wallK1 = 2.5W/m.K

L1 L2

L2 = L1/4

Insulation, K2

Heat flux = 2500 W/m2



The brick wall is covered by an insulating material of thermal conductivity K2. The thickness of the insulation is 1/4th of the thickness of the brick wall. The outer surface of the insulation is at 200°C. The heat flux through the composite walls is 2500 W/m2. The value of K2 is ______ W/m.K (round off to one decimal place).

13. Ans: 0.5 Sol:

T2T1 T3

L1

K1AL2

K2A

K A

L

T T

K A

L

T T

1

1

1 2

2

2

2 3− = −

KL

T TK

L

T T1

1

1 22

2

2 3− = −; ;E E

K4L

T TK

L

T T1

2

1 22

2

2 3− = −; ;E E

2.54

1100 700K 700 2002

− = −: 6D @ K2 = 0.5 W/mK

14. An attempt is made to pull a roller of weight W over a curb (step) by applying a horizontal force F as shown in the figure.

F

θ

The coefficient of static friction between the roller and the ground (including the edge of the step) is m. Identify the correct free body diagram (FBD) of the roller when the roller is just about to climb over the step.

F

N2

µN2µN1

N1

(a)W

θ

F

RW

(d)

θ

F

WN2

µN2

(c)

θ

(b)

F

R

θ

W

14. Ans: (c)Sol: When the roller is about to climb the contact with

floor is lost and hence normal reaction exerted by floor is zero.

15. The values of enthalpies at the stator inlet and rotor outlet of a hydraulic turbomachine stage are h1 and h3 respectively. The enthalpy at the stator outlet (or, rotor inlet) is h2. The condition (h2 − h1) = (h3 − h2) indicates that the degree of reaction of this stage is

(a) zero (b) 100 % (c) 75 % (d) 50 %15. Ans: (d)Sol: P1

h1

h2

h3

h

P2 P31

2

3

s



Degree of reaction is defined as:

Rh

h

h h

hd

stage

MB

SB MB

MB

TT

T TT= =

+^ ]]

]]]h g

gggg

where, (∆h)MB = Enthalpy drop in movable blade

(∆h)SB = Enthalpy drop in stator bladeBut, (∆h)MB = (∆h)SB ⇒ (Given)

Thus, .Rh

h

20 5

MB

MB

TT= =]]gg

16. A closed vessel contains pure water, in thermal

equilibrium with its vapour at 25°C (Stage # 1), as shown.

Stage # 1

Not to scaleValve A

Water vapour

Water

25°CAmbient air

at 25°C, 1 atm

Stage # 2

Valve A

Air at 80°C, 1 atm

Isothermal oven at 80°C

The vessel in this stage is then kept inside an isothermal oven which is having an atmosphere of hot air maintained at 80°C. The vessel exchanges heat with the oven atmosphere and attains a new thermal equilibrium (State # 2). If the Valve A is now opened inside the oven, what will happen immediately after opening the valve?

(a) Hot air will go inside the vessel through Valve A

(b) Water vapor inside the vessel will come out of the Valve A

(c) All the vapor inside the vessel will immediately condense

(d) Nothing will happen-the vessel will continue to remain in equilibrium

16. Ans: (a)Sol: As inside pressure is less than outside pressure, hot

air will flow into the vessel.

17. A beam of negligible mass in hinged at support P and has a roller support Q as shown in the figure.

1200 N

5 m5 m 4 m

RP

Q

A point load of 1200 N is applied at point R. The magnitude of the reaction force at support Q is ______ N.

17. Ans: 1500 Sol:

RPH

RQV

RQH

Q

RPV

5 m 5 mRP

1200 N

Taking moments about point P Σmp = 0 1200 5 R 4 R 10 0Q QH V# # #− − =

As the support at Q is roller support,. It doesn't restrict the motion in vertical direction. R 0QV` =

1200 5 R 4 0QH` # #− =

R 1500NQH` =

The resultant load on Q, R R R 1500NQ Q

2Q2

H V= + =



18. A machine member is subjected to fluctuating stress s = s0 cos (8pt). The endurance limit of the material is 350 MPa. If the factor of safety used in the design is 3.5 then the maximum allowable value of s0 is ______________ MPa (round offf to 2 decimal places).

18. Ans: 100 Sol: Given: s = so . cos (8pt) cos q varies from 1 to −1 ∴s varies from so to − so smax = s0 & smin = −s0

Here, the loading is reversed. For reversed loading design condition is, Amplitude stress

fosa

eσ σ=

Where, se = Corrected endurance strength

3.5

350o`σ =

so = 100 MPa

19. In the space above the mercury column in a barometer tube, the gauge pressure of the vapour is

(a) positive, but less than one atmosphere (b) zero (c) negative (d) positive, but more than one atmosphere

19. Ans: 3Sol:

Patm Patm ∇

Pv of Hg (< Patm)

In space above mercury column in barometer, ideally perfect vaccum is expected. But due to evaporation of mercury , the pressure in that space

is equal to vapour pressure of mercury [nearly 0.2 Pa (abs)] . As absolute pressure in that space is less than atmospheric pressure, gauge pressure is negative.

20. The solution of

dt

d yy 12

2

− = ,

which additionally satisfies ydt

dy0

t 0

t 0 = ==

= in the Laplaces-domain is

(a) s s s1 1

1

+ -^ ^h h (b) s s 1

1

−^ h

(c) s s 1

1

+^ h (d) s 1

1

−

20. Ans (a)Sol: Given:

dt

d yy 12

2

− =

y"(t) - y(t) = 1 Taking laplace transforms both sides L(y"(t) - y(t)) = L(1)

s y s sy 0 y' 0 y ss

12 − − − =] ] ] ]g g g g

s y s y ss

12 − =] ]g g

s 1 y ss

12 − =^ ]h g

y ss s 1

12` =

−] ^g h

s s 1 s 1

1= + −^ ^h h

21. The number of qualitatively distinct kinematic inversions possible for a Grashof chain with four revolute pairs is

(a) 3 (b) 4 (c) 1 (d) 2



21. Ans: (a) Sol: The different kinematic pairs that are possible are,

1 double crank, 2 double rocker, 3 crank rocker.

22. In Materials Requirement Planning, if the inventory holding cost is very high and the setup cost is zero, which one of the following lot sizing approaches should be used?

(a) Economic Order Quantity (b) Base Stock Level (c) Lot-for-Lot (d) Fixed Period Quantity, for 2 periods.

22. Ans: (c)Sol: In Lot-for-lot, holding cost is high and ordering

cost is less.

23. The process, the uses a tapered horn to amplify and focus the mechanical energy for machining of glass, is

(a) ultrasonic machining (b) electrical discharge machining (c) electrochemical machining (d) abrasive jet machining 23. Ans: (a)Sol: In Ultrasonic machining, the function of horn (also

called concentrator, it is a tapered metal bar) is to amplify and focus vibration of the transducer to an adequate intensity for driving the tool to fulfill the cutting operation.

24. The figure below shows a symbolic representation of the surface texture in a perpendicular lay orientation with indicative values (I through VI) marking the various specifications whose definitions are listed below.

P: Maximum Waviness Height (mm);

Q: Maximum Roughness Height (mm); R: Minimum Roughness Height (mm); S: Maximum Waviness Width (mm); T: Maximum Roughness Width (mm); U: Roughness Width Cutoff (mm).

II

III - IV

V

VI

I

The correct match between the specifications and the symbols (I to VI) is

(a) I-U, II-S, III-Q, IV-T, V-R, VI-P (b) I-R, II-Q, III-P, IV-S, V-U, VI-T (c) I-R, II-P, III-U, IV-S, V-T, VI-Q (d) I-Q, II-U, III-R, IV-T, V-S, VI-P

24. Ans: (b) Sol: P: Maximum Waviness Height (mm) → III Q: Maximum Roughness Height (mm) → II R: Minimum Roughness Height (mm) → I S: Maximum Waviness Width (mm) → IV T: Maximum Roughness Width (mm) → VI U: Roughness Width Cutoff (mm) → V

25. The sum of two normally distributed random variables X and Y is

(a) always normally distributed. (b) normally distributed, only if X and Y have the

same standard deviation.(b) normally distributed, only if X and Y are

independent.(d) normally distributed, only if X and Y have the

same mean.



25. Ans: (b)Sol: Given: X & Y are normally distributed random variables. If X and Y are independent normal random variables

then its sum X+Y is also a normal random variable.

26. A fair coin is tossed 20 times. The probability that 'head' will appear exactly 4 times in the first ten tosses, and 'tail' will appear exactly 4 times in the next ten tosses is ________ (rounded off to 3 decimal places).

26. Ans: 0.042 Sol: n = 10, p

2

1&q

2

1= =

Let x = Number of heads P(x = 4) = ncx p

xqn-x

10C4 21

214 6

= b bl l

2

10C104= ] g

1024

210= = 0.205

∴Required probability = (0.205)2 = 0.042

27. A rigid block of mass m1 = 10 kg having velocity vo = 2 m/s strikes a stationery block of mass m2 = 30 kg after traveling 1 m along a frictionless horizontal surface as shown in the figure.

Mass-less

m1

1 m 0.25 m

k v0 m2

The two masses stick together and jointly move by a distance of 0.25 m further along the same frictionless surface, before they touch the mass-less buffer that is connected to the rigid vertical wall by means of a linear spring having a spring constant k = 105 N/m. The maximum deflection of the spring is ______ cm (Round off to 2 decimal places).

27. Ans: 1 Sol: Let, V is the velocity with which the two mass

moves after sticking,∴ By using Linear momentum conservation m1V0 = (m1 + m2)r 10 × 2 = (10 + 30)r

V = /m s2

1

At the instant of maximum compression in spring the two sticked mass comes to rest.

∴ By using conservation of mechanical energy.

m m V Kx2

1

2

11 2

2 2+ =^ h

x2

140

2

1

2

110

25 2# # # #=c m

⇒ x = 0.01 m = 1 cm

28. A helical spring of stiffness K. If the wire diameter, spring diameter and the number of coils are all doubled then the spring constant of the new spring becomes

(a) 8 K (b) K (c) K/2 (d) 16K

28. Ans: (b)Sol: Given, K1 = K D2 = 2D1 (D = mean coil diameter) d2 = 2d1 (d = Wire diameter) n2 = 2n1 (n = No. Of Active coils)

Spring constant = 8D .n

G.d3

4

For a given material G = Constant

K

K

d

d

D

D.n

n

1

2

1

24

2

13

2

1` #= b b bl l l

22

1

2

143

# #= b l K2 = K1

K2 = K



29. A point 'P' on a CNC controlled XY-stage is moved to another point 'Q' using the coordinate system shown in the figure below and rapid positioning command (G00).

A pair of stepping motors with maximum speed 800

rpm, controlling both the X and Y motion of the stage, are directly coupled to a pair of lead screws, each with a uniform pitch of 0.5 mm. The time needed to position the point 'P' to the point 'Q' is ______ minutes. (rounded off to 2 decimal places).

29. Ans: 1.5 Sol:

600

300 670.82 mm

Q (800, 600)

P (200, 300)

N = 800 rpm, P = 0.5 mm, Feed rate = 800 × 0.5 = 400 mm/min

Feed rate across ‘X’ & ‘Y’ axis = 400 mm/min Time across ‘X’ axis . min

400

6001 5= =

Time across ‘Y’ axis . min400

3000 75= =

Time = 1.5 min

30. For a single item inventory system, the demand is continuous, which is 10000 per year. The replenishment is instantaneous and backorders (S units) per cycle are allowed as shown in the figure.

Inventory

Time S

Q

As soon as the quantity (Q units) ordered from the supplier is received, the backordered quantity is issued to the customers. The ordering cost is Rs. 300 per order. The carrying cost is Rs. 4 per unit per year. The cost of backordering is Rs. 25 per unit per year. Based on the total cost minimization criteria, the maximum inventory reached in the system is _____ (round off to nearest integer).

30. Ans: 1137.14Sol: Ordering cost, C0 = Rs. 300/order Carrying cost, Ch = Rs. 4/unit/yr Backorder cost, Cs = Rs. 25/unit/yr Annual demand, R = 10,000 units

QC C

RC C C2

S h

s h0= ++^ h

25 4

2 10 300 294

## # #= = 1319.09 units

IC C

C Qm

h s

s#= +

.. units

4 25

25 1319 091137 14

#= + =



31. Keeping all other parameters identical, the compression ratio (CR) of an air standard diesel cycle is increased from 15 to 21. Take ratio of specific heats = 1.3 and cut-off ratio of the cycle rc = 2. The difference between the new and the old efficiency values, in percentage. (ηnew|CR=21) - (ηold|CR=15) = _________% (round off to one decimal place).

31. Ans: 4.8Sol: Diesel cycle r1 = 15, r2 = 21, g = 1.3 Cut off ratio (rc) = 2

r r r

r1

1

1

1

rc

c15 11

η = −−−

c

c

= -f _ i p

= .

. %115

1

1 3 2 1

2 1100 50 08.

.

0 3

1 3

#−−−

=d ] g n= G

.

121

1

1 3 2 1

2 1100.

.

r 21 0 3

1 3

2#η = −

−−

= d ] g n= G = 54.874 %

r r21 151 2η η−= = = 4.79% (4.8% with one decimal)

32. A cylindrical bar with 200 mm diameter is being turned with a tool having geometry 0° - 9° - 7° - 8° - 15° - 30° - 0.05 inch (Coordinate system, ASA) resulting in a cutting force FC1. If the tool geometry is changed to 0° - 9° - 7° - 8° - 15° - 0° - 0.05 inch (Coordinate system, ASA) and all other parameters remain unchanged, the cutting force changes to FC2. Specific cutting energy (in J/mm3) is Uc = U0 (t1)

-0.04, where U0 is the specific energy coefficient, and t1 is the uncut thickness in mm. The value of percentage

change in cutting force FC2, i.e., F

F F100

c1

c2 c1 #−c m ,

is ______ (round off to one decimal place).

32. Ans: 5.6 Sol: Given data,

D = 200 mm, 0° - 9° - 7° - 8° - 15° - 30° - 0.05 inch (ASA) FC1 0° - 9° - 7° - 8° - 15° - 0° - 0.05 inch (ASA) FC2 Specific cutting energy, Uc = U0 (t1)

-0.04 J/mm3

where U0 = specific energy coefficient, and t1 = uncut thickness in mm Specific cutting energy =

f d

FU t

c0 1

0.4

#= -] g

F U f cos f dc 00.4 #ψ = -^ h

F cosc0.4

? ψ -^ h

F

F F100

cos

cos1 100

c1

c2 c1

1

20.4

# #ψψ− = −

-

c dm n= G

cos30

cos01 100

0.4

#= −-b l; E

0.866

11 100 5.6

0.4

#= − =−-cb l m

33. In a steam power plant, superheated steam at 10 MPa and 500°C, is expanded isentropically in a turbine until it becomes a saturated vapour. It is then reheated at constant pressure to 500°C. The steam is next expanded isentropically in another turbine until it reaches the condenser pressure of 20 kPa. Relevant properties of steam are given in the following two tables. The work done by both the turbines together is _____ kJ/kg. (round off to the nearest integer).

Superheated Steam Table :Pressure,

p (MPa)

Temperature,

T (°C)

Enthalpy,

h (kJ/kg)

Entropy,

s (kJ/kgK)10 500 3373.6 6.59651 500 3478.4 7.7621

Saturated Steam Table :Pressure,

p

Sat.

Temperature,

Tsat (°C)

Enthalpy,

h (kJ/kg)

Entropy,

s (kJ/kgK)hf hg sf sg

1 MPa 179.91 762.9 2778.1 2.1386 6.596520 kPa 60.06 251.38 2609.7 0.8319 7.9085



33. Ans: 1513Sol: P1 = 10 MPa, T1 = 500°C h1 = 3373.6 kJ/kg At state 2 sat. vap. available. P2 = 1 MPa

At state 3 superheated vap. availableP1 = 10 MPah

P2 = 1 MPa

P3 = 20 kPa1

2

3

4

s

P3 = 20 kPa h2 = hg2 = 2778.1 kJ/kg h3 = 3478.4 kJ/kg For, 3 - 4 = constant s3 = s4 = Sf4 + x4 sfg4

7.7621 = 0.8319 + x4 (7.0766)

.

. .x

7 0766

7 7621 0 83194 = −

= 0.979

h4 = hf4 + x4 hfg4

= 251.38 + 0.979 (2609.7 − 251.38) = 2560.175 kJ/kg Total work = (h1 − h2) + (h3 − h4) = (3373.6 − 2778.1) + (3478.4 − 2560.175) = 1517.724 kJ/kg ≃ 1513 kJ/kg

34. A steel spur pinion has a module (m) of 1.25 mm, 20 teeth and 20° Pressure angle. The pinion rotates at 1200 rpm and transmits power to a 60 teeth gear. The face width (F) is 50 mm, Lewis form factory Y = 0.322 and a dynamic factor Kv = 1.26. The bending stress (σ) induced in a tooth can be calculated by using the Lewis formula given below.

If the maximum bending stress experienced by the pinion is 400 MPa, the power transmitted is _____ kW (round off to one decimal place).

Lewis formula: FmY

K Wvt

σ = , where Wt is the tangential load acting on the pinion.

34. Ans: 10Sol: Given: m = 1.25 mm, z = 20 Diameter of pinion = M.z = 25 mm = 0.025 m Power transmitted by pinion P

60

2 .N.Tπ =

60

2 N. .2

dt pπ ω=

But F.m.y

K .vt

σ ω=

K

.F.m.Yt

Vω σ=

∴Power = 60

2 N.K

.F.m.Y

2

dp

Vπ σb l

60

1.26

2 p 1200 400 50 1.25 0.322

2

0.025# # # # # ##

=

P = 10035.64 W P = 10.0.35 kW P ≃ 10 kW

35. Water (density 1000 kg/m3) flows through an inclined pipe of uniform diameter. The velocity, pressure and elevation at section A are VA = 3.2 m/s, PA = 186 kPa and zA = 24.5 m respectively, and those at section B are VB = 3.2 m/s, PB = 260 kPa and zB = 9.1 M, respectively. If acceleration due to gravity is 10 m/s2 then the head lost due to friction is_______ m. (round off to one decimal place)

35. Ans: 8Sol: Assume flow is taking place from section ‘A’ to ‘B’. By Bernoulli’s equation,

g

P

g

VZ

g

P

g

VZ h

2 2

A AA

B BB f

2 2

ρ ρ+ + = + + +

∴ hg

P PZ Zf

A BA Bρ =

−+ − [ a VA = VB]



. .

1000 10

186 260 1024 5 9 1

3

##

=−

+ −^ ^h h

hf = 8 m As head loss is positive our assumption regarding

flow direction is correct.

36. A thin-walled cylinder of radius r and thickness t is open at both ends, and fits snugly between two rigid walls under ambient conditions, as shown in the figure.

Pressure vessel

Rigid wallr

Rigid wall

The material of the cylinder has Young's modulus E, Poisson's ratio v, and coefficient of thermal expansion a. What is the minimum rise in temperature ∆T of the cylinder (assume uniform cylinder temperature with no buckling of the cylinder) required to prevent gas leakage if the cylinder has to store the gas at an internal pressure of p above the atmosphere ?

(a) T v4

1

tE

pr

α ∆ = −b l (b) T2 tE

3 pr

αν

∆ =

(c) T v2

1

tE

pr

α ∆ = +b l (d) TtE

pr

αν

∆ =

36. Ans: (d)Sol:

Pressure vessel

Rigid wallr

Rigid wall

Due to pressure inside the cylinder, the diameter increases and hoop stress will be developed along the circumference.

Due to Poisson's effect, length decreases due to increase in diameter.

When length of cylinder decreases than the original length, the gases leaks into the atmosphere.

Therefore, if the length of cylinder is increased to the original length by heating, then leakage of gases can be avoided.

∴To avoid leakage, Decrease in length ≤ Increase in length due to pressure due to heating.

.2tE

p.d. . Tl l# #ν α ∆

T

2tE

.P. 2r$ α

ν∆ ] g

TEt.

.P.r$ α

ν∆

∴ The minimum rise in temperature ∆T to avoid

leakage of gases is .t.E

.P.r

αν

37. A mould cavity of 1200 cm3 volume has to be filled through a sprue of 10 cm length feeding a horizontal runner. Cross-sectional area at the base of the sprue is 2 cm2. Consider acceleration due to gravity as 9.81 m/s2. Neglecting frictional losses due to molten metal flow, the time taken to fill the mould cavity is ______ seconds. (round off to 2 decimal places).

37. Ans: 4.28Sol: Volume = 1200 cm3

he = 10 cm ; Ac = 2 cm2

g = 9.81 m/sec2

t = ?

t

volumeA 2g hc c=

t

12002 2 9.81 10# #=

⇒ t = 4.284 sec



38. Water flows through a tube of 3 cm internal diameter and length 20 m. The outside surface of the tube is heated electrically so that is subjected to uniform heat flux circumferentially and axially. The mean inlet and exit temperatures of the water are 10°C and 70°C, respectively. The mass flow rate of the water 720 kg/h. Disregard the thermal resistance of the tube wall. The internal heat transfer coefficient is 1697 W/m2K. Take specific heat cp of water as 4.179 kJ/kgK. The inner surface temperature at the exit section of the tube is ______°C. (round off to one decimal place).

38. Ans: 85.67Sol: Given: D = 0.03 m , L = 20 m Ti = 10°C, m 720 kg/hr=o , m 0.2 kg/sec=o

h = 1697 W/m2K CP = 4.179 kJ/kgK Assume thermal fully developed flow Q mC T T hPL T TP o i S o= − = −o ^ ^h h mC T T h DL T TP o i S oπ − = −o ^ ^h h 0.2×4.179×103×(70 − 10) =167×p ×0.03×20×(Ts −70) Ts = 85.67°C

39. The directional derivative of f(x,y,z) = xyz at point (-1, 1, 3) in the direction of vector i 2j 2k− +t t t is

(a) 7 (b) 3

7 (c) 3i 3j k− −t t t (d)

3

7−

39. Ans:(b)Sol: Given: F(x, y, z) = xyz f yz i xzj xyK4 = + +

At (−1, 1, 3), f 3i 3j K4 = − −

Let a i 2j 2K= − +

a 1 2 2 32 2 2= + − + =] ] ]g g g

The directional derivative of f(x, y, z) at point (-1,1, 3) in the direction of

a i 2j 2k= − + is given by

f.a

a3i 3j k .

3

i 2j 2k4 = − −

− +_ _i i

3

3 1 3 2 1 2= + − − + −] ] ^ ] ] ]g g h g g g

3

3 6 2

3

7= + − =

40. A hollow spherical ball of radius 20 cm floats in still water, with half of its volume submerged. Taking the density of water as 1000 kg/m3, and the acceleration due to gravity as 10 m/s2, the natural frequency of small oscillations of the ball, normal to the water surface is ______ radian/s (round off to 2 decimal places).

40. Ans: 0Sol: As sphere is symmetric object, for any axis rotation

through centre, the shape of displaced volume does not change after small rotation. Hence, centre of buoyancy remains same. Thus sphere is in neutral equilibrium. For neutral equilibrium time period of oscillation is infinite.

`

W

FB

G

B

∇

Method - II

/GM

IBG

D

D D

3

2

2

64

8

3

23

4

6π

π= − = −c

cm

m

D D16

3

16

30= − =

.f

k

GM g

2

102π

= =



41. The forecast for the monthly demand of a product is given in the table below. Month Forecast Actual sales

1 32.00 30.002 31.80 32.003 31.82 30.00

The forecast is made by using the exponential smoothing method. The exponential smoothing coefficient used in forecasting the demand is

(a) 0.50 (b) 0.10 (c) 1.00 (d) 0.40 41. Ans: (b)Sol:

Month F D1 32 302 31.8 323 31.82 30

F2 = a D1 + (1-a) F1

31.8 = 30 a + (1-a)32 31.8 = 30 a + 32 - 32a 31.8 = 32 - 2 a 0.2 = 2 a a = 0.1 F3 = a D2 + (1-a) F2

31.82 = 32 a + (1-a)31.8 31.82 = 32 a + 31.8 - 31.8a 0.02 = 0.2 a a = 0.1

42. Uniaxial compression test data for a solid metal bar of length 1 m is shown in the figure.

0.005 0.002

s(MPa)

130

100

ε(mm/mm)

R Q

P

The bar material has a linear elastic response from O to P followed by a nonlinear response. The point P represents the yield point of the material. The rod is pinned at both the ends. The minimum diameter of the bar so that it does not buckle under axial loading before reaching the yield point is _____ mm (Round off to one decimal place).

42. Ans: 56.94 Sol: From graph 100MPaytσ =

E0.002

10050 10 MPa

yt 3#εσ= = =

Maximum safe load that can be applied to avoid buckling.

P.E.I

lcr 2

2minπ = ........ (1)

As the loading is axial P Acr yt #σ = ........ (2) From equation (1) & (2) we get,

AEI

lyt 2

2min` #σ π=

100yd

1000

.50 1064d

22

2 3 4

## #π π π

=

d = 56.941 mm

43. The function f(z) of complex variable z = x + iy, where ,i 1= − is given as f(z) = (x3 − 3xy2)+iv(x, y). For this function to be analytic, v(x, y) should be

(a) (3xy2−y3) + constant (b) (x3−3x2y) + constant (c) (3x2y − y3) + constant (d) (3x2y2−y3) + constant43. Ans: (c)Sol: Given: f(z) = (x3 - 3xy2) + iv (x, y) By the definition of total derivative of V dV

x

Vdx

y

Vdy2

222= +

dVy

udx

x

udy2

222= − + (By C-R equations)



dV 6xydx 3x 3y dy2 2= + −_ i

Integrating both sides dV 6xydx 3x 3y dy

2 2

y const= + −_ i### (terms do

not contain x) V = 3x2y − y3 + Constant

44. A cantilever of length ,, and flexural rigidity EI, stiffened by a spring of stiffness k, is loaded by a transverse force P, as shown,

l, EI

k

P

(a) E

P

EI

EI k

3 3

33 3

,, ,−< F (b)

EI

P

EI k

EI

3 3

33

3,

,+< F

(c) EI

P

EI k

EI

3 3 2

33

3,

,+< F (d) EI

P

EI

EI k

3 6

63 3, ,−< F

44. Ans: (b)Sol:

P

P

+

≈

PS

δP

δS

Where Ps is the spring load

Net deflection at free end p sδ δ δ= −

3EI

P.

3EI

P .l l3s3

= −

3EI

P kl3

p sδ δ= − −_ i7 A

3EI

P k.l3 δ = −6 @ [ ∵ d = Compression in spring]

.3EIP K.l3

δ δ= −

3EIK Pl3δ + =; E

3EI K

P.

ll

3

3

δ = +

3EI

p.

3EI K

3EIll

3

3δ = +; E

45. The turning moment diagram of a flywheel fitted to a fictitious engine is shown in the figure.

2p 2

3p

3000

2500

2000 1500

1000

Angle (radians)

Mom

ent (

Nm

)

p 2p

The mean turning moment is 2000 Nm. The average engine speed is 1000 rpm. For fluctuation in the speed to be within ±2% of the average speed, the mass moment of inertia of the flywheel is _____kg.m2.

45. Ans: 3.582 Sol: Given: Mean torque, Tm = 2000 N-m N = 1000 rpm Maximum fluctuation in speed = ± 2 % I = ? ∴Cs = ± 2 % = 4 % = 0.04 N = 1000 rpm

∴ . /rad s60

2 1000104 7ω

π= =] g



Tm = 2000 N-m Turning moment diagram

2p 2

3p

3000

2500

2000 1500

1000

p 2p

The Hatched Area show the excess/ defect area

above and below mean torque line. Convert the diagram will give A

2500 2501 #

π π= =

A 10002

5002 #π π= =

A 10002

5003 #π π= =

A 5002

2504 #π π= =

Assuming energy at point ABE ∴ EA = E EB = E − 250 p EC = E − 250 p + 500p = E+250p ED = E + 250 p + 250p = E Emax = E +250 p Emax = E − 250 p ∴ ∆E = E+250 p − (E − 250p) ∆E = 500 p N−m ∆E = Iw2Cs 500p = I × (104.7)2 × 0.04 ∴ I = 3.582 kg-m2

46. Bars of 250 mm length and 25 mm diameter are to be turned on a lathe with a feed of 0.2 mm/rev. Each regrinding of the tool costs Rs. 20. The time required for each tool change is 1 min. Tool life equation is given as VT0.2 = 24 (where cutting speed V is in m/min and tool life T is in min). The optimum tool cost per piece for maximum production rate is Rs. _____ (Round off to 2 decimal places).

46. Ans: 26.94Sol: l = 250 mm, D = 25 mm, f = 0.2 mm/rev Regrinding cost (Ce) = Rs. 20/- Tc = 1 min, VT0.2 = 24 V

4

2418.18m/min0.2= =] g

Maximum production rate, T

n

11 Topt c= −b l

T0.2

11 1 4minopt = − =b l

Total cost per piece (C3)

C1000 f V

DL

C

Ve

n1

π = b l

201000 0.2 18.18

25 250

24

18.18 0.21

# ## #π = b l

= 26.94

47. Moist air at 105 kPa, 30oC and 80% relative humidity flows over a cooling coil in an insulated air-conditioning duct. Saturated air exits the duct at 100 kPa and 15oC. The saturation pressures of water at 30oC and 15oC are 4.24 kPa and 1.7 kPa respectively. Molecular weight of water is 18 g/mol and that of air is 28.94 g/mol. The mass of water condensing out from the duct is ______ g/kg of dry air (Round off to the nearest integer).

47. Ans: 10Sol:

1 2

CC

Pb 1^ h = 105 kPa (DBT)1 = 30°C RH1 = 0.8



Psat1 = 4.24 kPa

RHP

P

sat

v

11

1=

Pv1 = 0.8 × 4.24 = 3.392 kPa

.P P

P

0 622b v

v

11 1

1ω = −f p

..

.0 622

105 3 392

3 392= −c m

= 0.0207 kg/kg of da = 20.764 g/kg of da Pb 2^ h = 100 kPa tdb2 = 15°C RH2 = 100% (sat.air) Psat2 = 1.7 kPa

Pv2 = Psat2 = 1.7 kPa

.P P

P

0 622b v

v

22 2

2ω = −f p ..

.0 622

100 1 7

1 7= −c m

= 0.0107 kg/kg of da = 10.756 g/kg of da∴ Amount of water vapor condense, D = w1 − w2

= 20.764 − 10.756 = 10.008 g/kg of da ≃ 10 g/kg of da

48. Air is contained in a frictionless piston-cylinder arrangement as shown in the figure.

AirQ

8 cm

stop

8 cm

The atmospheric pressure is 100 kPa and the initial pressure of air in the cylinder is 105 kPa. The area of piston is 300 cm2. Heat is now added and the

piston moves slowly from its initial position until it reaches the stops. The spring constant of the linear spring is 12.5 N/mm. Considering the air inside the cylinder as the system, the work interaction is _____ J (Round off to the nearest integer).

48. Ans: 544Sol: P0 = 100 kPa P1 = 105 kPa AP = 300 cm2

K = 12.5 N/mm (Spring constant) = 12.5 kN/m

AirQ

8 cm

stop

8 cm

From state 1 to 2, the gas expands against 105 kPa and from state 2 to state 3 the gas expands linearly (against the spring force)

dv1 dv2

v

P

P2 = P11 2

3

V2 - V1 = A (x2 - x1) = 300 ×10-4 ×(0.08) = 2.4×10-3 m3

V3 - V2 = A (x3 - x2) = 300 ×10-4 ×(0.08) = 2.4×10-3 m3

P1 = P2 = 105 kPa

P PA

kx3 2= +



105 10

300 10

12.5 10 8 1034

3 2

##

# # #= + -

-

3

415kPa=

Total work done Wtotal = W(1 - 2) + W(2 - 3)

= P V V2

1P P V V1 2 1 2 3 3 2#− + + −^ _ _h i i

= 105 2.4 102

1105

3

4152.4 10

3 3# #+ +- -^ c ^h m h = (252 + 292)×10-3 kJ = 544 J

49. The sun (S) and the planet (P) of an epicyclic gear train shown in the figure have identical number of teeth.

R

wR S

P B

A wS

If the sun (S) and the outer ring (R) gears are rotated in the same direction with angular speed wS and wR, respectively, then the angular speed of the arm AB is

(a) 4

3

4

1R Sω ω+

(b)

2

1

2

1R Sω ω−

(c) 4

3

4

1R Sω ω−

(d) 4

1

4

3R Sω ω+

49. Ans: (a)Sol: Condition Arm Ts

(Sun)TP (Planet)

TR (Ring)

let arm is fixed and sun gear is rotating with +X rpm

0 +XT

Tx

P

S−T

T

T

Tx

R

P

R

S+ −b l

Let arm is rotating with y rpm

y y+Xy x

T

T

P

S− y xT

T

P

S−

TP = TS (given)

And RR = RS + 2 RP

mR = mS = mP

⇒ TR = TS + 2 TP

⇒ TR = 3 TP

ws = y + X

y xT

TR

R

sω = −

y xT

Ty

3

x

R

S= − = −

y3

xy

3

xR R&ω ω= − − =

y 3y 3S Rω ω= + −

= 4y - 3wR

y4 4

3s Rω ω= +



50. Consider a flow through a nozzle, as shown in the figure below,

p2 = patm

A2 = 0.02 m2

streamline p1 v1

A1 = 0.2 m2

(1) (2)

v2= 50 m/s

The air flow is steady, incompressible and inviscid. The density of air is 1.23 kg/m3. The pressure difference (p1 − patm) is _____ kPa (Round off to 2 decimal places).

50. Ans: 1.52Sol: By continuity equation, A1V1 = A2V2

0.2 V1 = 0.02× 50 V1 = 5 m/s By Bernoulli’s equation,

g

P

g

VZ

g

P

g

VZ

2 2

1 12

12 2

2

2ρ ρ+ + = + +

g

P P

g

V VZ Z

2

1 2 22

12

1 2ρ −

=−

+ −

g

P P

g

V V

20

atm1 22

12

ρ −

=−

+

∴P P V V2

atm1 22

12ρ

− = −_ i

.

2

1 2350 52 2#= −^ h

= 1.52 kPa

51. There are two identical shaping machines S1 and S2. In machine S2, the width of the workpiece is increased by 10% and the feed is decreased by 10%, with respect to that of S1. If all other conditions remain the same then the ratio of total time per pass in S1 and S2 will be _____ (Round off to one decimal place).

51. Ans: 0.81 Sol: T

f N

Lm =

S2 ⇒ L2 = 1.1 L1 ;

f2 = 0.9f2

T

T

f N

L

f N

L

m2

m1

2 2

2

1 1

1

=

f N

4

1.1L

0.9 f N

1 1 1

1 1#= = 0.81

52. The spectral distribution of radiation from a black body at T1 = 3000 K has a maximum at wavelength λmax. The body cools down to a temperature T2. If the wavelength corresponding to the maximum of the spectral distribution at T2 is 1.2 times of the original wavelength λmax, then the temperature T2 is ______ K (Round off to the nearest integer).

52. Ans: 2500 Sol: Given, T1 = 3000 K , λmax = λ1 ; T2 = ? λ2 = 1.2 λ1

Wein's displacement law, λT = constant λ1 T1 = λ2 T2 λ1× 3000 = 1.2 λ2 × T2

T1.2

30002500K2 = =

53. One kg of air in a closed system undergoes an irreversible process from an initial state of p1 = 1 bar (absolute) and T1 = 27oC, to a final state of p2 = 3 bar (absolute) and T2 = 127oC. If the gas constant of air is 287 J/kg.K and the ratio of the specific heats = 1.4, then the change in the specific entropy (in J/kg.K) of the air in the process is

(a) 28.4 (b) indeterminate, as the process is irreversible (c) −26.3 (d) 172.0



53. Ans:(c)Sol: m = 1 kg ; P1 = 1 bar = 100 kPa T1 = 27 + 273 = 300 K; P2 = 3 bar = 300 kPa T2 = 127 + 273 = 400 K; R = 0.287 kJ/kg.K g = 1.4

s s nT

TR n

P

Pc

2 1 P

1

2

1

2

, ,=− −e eo o

1. n3

4. n

1

3005

00

000 287, ,= −c cm m

= 0.2891 − 0.3153 = −0.0262 kJ/kg.K = − 26.2 J/kg.K

54. Two rollers of diameters D1 (in mm) and D2 (in mm)

are used to measure the internal taper angle in the V-groove of a machined component. The heights H1 (in mm) and H2 (in mm) are measured by using a height gauge after inserting the rollers into the same V-groove as shown in the figure.

a a

H1

D1

a a

H2 D2

D1 > D2

Which one of the following is the correct relationship to evaluate the angle a as shown in the figure?

(a) sinH H D D

D D

2 1 2 1 2

1 2α =

− − −

−

__

_ii

i

(b) cosecD D

H H D D

2 1 2

1 2 1 2α =

−

− − −__

_ii

i

(c) sinD D

H H

1 2

1 2α =

−

−__ i

i

(d) cosH H D D

D D

2 21 2 1 2

1 2α =

− − −

−

__

_ii

i

54. Ans: (a)Sol:

H2 – D2/2

h1

D2/2 O2

H2

O1 D1/2 A

a

D2

D1

H1 –

D1/2

H1

O O H2

DH

2

D1 2 1

12

2= − − −b bl l

H H2

D D1 2

1 2= − − −^ ^h h

sin

H H2

D D

D D /2

1 21 2

1 2α =− − −

−

^^

^hh

h

sin2 H H D D

D D

1 2 1 2

1 2α = − − −−

^^

^hh

h



55. For the integral cosx dx8 4

/

0

2

+r

_ i# , the absolute

percentage error in numerical evaluation with the Trapezoidal rule, using only the end point, is ______ (Round off to one decimal place).

55. Ans: 5.132 Sol: 8 4cosx dx 8x 4sin x 0

0

2

2+ = +

r

r^ ^h h#

82

4sin2

π π= +b bl l = 4p + 4 = 16.56 (exact value) Using trapezoidal rule

x 02π

y 12 8

8 4cosx dx2

h y y 2 y y ...... y

4 y y ..... y

o n 2 4 n 2

1 3 n 10

2

+ =+ + + + +

+ + + +

r

-

-

^ __

_h ii

i* 4#

412 8

π = +^ h" ,

204

π = = 5p

= 15.7 (approximate value)

Absolute error = | 16.56 − 15.7 | = 0.85 Absolute percentage error =

16.56

0.85100#

= 5.132

hyderabad | delhi | ahmedabad | pune | …...ia ib νν 2 == ω x 2r x νν= 2 − 2r x2 xx 3 2r −...

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