general aptitude - gateforum

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Q. No. 1 - 5 Carry One Mark Each

1. There are five levels {P, Q, R, S, T) in a linear supply chain before a product reaches customers, as

shown in the figure.

At each of the five levels, the price of the product is increased by 25%. If the product is produced at

level P at the cost of Rs. 120 per unit, what is the paid (in rupees) by the customers?

(A) 234.38 (B) 292.96 (C) 366.21 (D) 187.50

Key: (C)

Sol: The price paid by the customers

5 5 5 5 5120

4 4 4 4 4

15 3125366.21

128

=

=

2. While I agree ________ his proposal this time, I do not often agree ______ him.

(A) to, with (B) with, with (C) to, to (D) with, to

Key: (A)

3. In one of the greatest innings ever seen in 142 years of Test history. Ben Stokes upped the tempo in a

five-and-a-half hour long stay of 219 balls including 11 fours and 8 sixes that saw him finish on a 135

not out as England squared the five-match series.

Based on their connotations in the given passage, which one of the following meanings DOES NOT

match?

(A) tempo = enthusiasm (B) upped = increased

(C) saw = resulted in (D) squared = lost

Key: (D)

GENERAL APTITUDE

P Q R S T Customers

125 125 125 125 125120

100 100 100 100 100

=

P Q R S T customers

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4. Select the word that fits the analogy:

White : Whitening : : Light : ________.

(A) Enlightening (B) Lighting (C) Lightening (D) Lightning

Key: (C)

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

(A) decrease (B) speed (C) increase (D) equalize

Key: (C)

Q. No. 6 - 10 Carry Two Marks Each

6. Climate change and resilience deal with two aspects - reduction of sources of nonrenewable energy

resources 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) Adaptation deals with causes of climate change.

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

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

Key: (C)

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7. 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 __________.

(A) 2 : 1 (B) 22 : 9 (C) 9 : 22 (D) 11 : 9

Key: (B)

Sol: Total number of students enrolled in arts 20

5000 1000100

= = [From 1st pie chart]

Number of girls enrolled in arts 30

1500 450100

= = [From 2nd pie chart]

Number of boys enrolled in arts = 1000 – 450 = 550 ….(1)

Number of girls enrolled in management 15

1500 225100

= = …(2) [From 2nd pie chart]

From (1) and (2), the required ratio = 550 : 225 = 22 : 9.

8. 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) 3 days (B) 13 days (C) 10 days (D) 5 days

Key: (A)

Sol: Given that; initially 52 men can finish a work in 10 days.

Percentage of students enrolled in different

streams in a University

Arts

20%

Engineering

20%

Management

15%

Science

20%Commerce

15%

Arts

30%

Engineering

10%

Management

15%

Science

25%Commerce

20%

Percentage of girls enrolled in different

streams

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Men Days Hours Efficiency Men Days Hours Efficiency

work work

=

( )52 12 x52 10

;1 1

− = Where x→ Number of days required to complete a strip.

52 10

x 1340

= = → 10 days (original estimate) + 3days

Required number of days, more than the original estimate = 3 days.

9. Find the missing element in the following figure:

(A) w (B) y (C) e (D) d

Key: (D)

10. An engineer measures THREE quantities X, Y and Z in an experiment. She finds that they follow a

relationship that is represented in the figure below: (the product of X and Y linearly varies with Z)

Then, which of the following statements is FALSE?

(A) For fixed X; Z is proportional to Y (B) For fixed Y; X is proportional to Z

(C) XY/Z is constant (D) For fixed Z; X is proportional to Y

Key: (D)

( )X,Y

O Z

t

5

n

h

x?

9

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Sol: Given that, XY linearly various with Z.

i.e., XY KZ=

For fixed X, 1Y K Z Y Z Z is proportional to Y.=

For fixed Y, 2X K Z X Z X is proportional to Z=

options (A), (B) are true.

If XY XY

XY KZ K is constantZ Z

= =

For fixed ‘Z’; 3XY K X is not proportional to Y.=

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Q. No. 1 to 25 Carry One Mark Each

1. 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 (mm);

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

(A) I-Q, II-U, III-R, IV-T, V-S, VI-P (B) I-R, II-P, III-U, IV-S, V-T, VI-Q

(C) I-U, II-S, III-Q, IV-T, V-R, VI-P (D) I-R, II-Q, III-P, IV-S, V-U, VI-T

Key: (D)

Sol: Surface roughness terminology will be as follows

I-minimum roughness height (mm)-R

II-maximum roughness height (mm)-Q

III-maximum waviness height-P

IV-Maximum waviness width-S

V-Roughness width cutoff-U

VI-Maximum roughness width-T.

So answer is option ‘D’.

MECHANICAL ENGINEERING

I

VI

V

III IV−

II

Lay

I

VI

V

III IV−

II

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2. 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) 3-4-2-1 (B) 4-3-2-1 (C) 3-2-4-1 (D) 1-4-2-3

Key: (C)

Sol: Laser beam welding uses highly focused beam for welding, Hence heat affected zone will be less

compared to all other type of welding.

In submerged arc welding weld pool is completely covered by fluxes hence in this welding also heat

affected zone is less, but more than TIG and MIG, since flux spread over more area.

In metal inertial gas welding also will be using inert gases to protect the welding zone from atmospheric

contamination. Since it uses electrode with inert gas for shielding, heat affected zone is less than

Submerged arc welding.

In arc welding we don’t use any protection to weld pool hence it has more heat affected zone compared

to all other weldings. Hence the answer is option ‘C’.

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

(A) positive, but more than one atmosphere (B) zero

(C) positive, but less than one atmosphere (D) negative

Key: (D)

Sol: In the space above the mercury column in a barometer tube is nothing but zero absolute pressure (or)

Negative gauge pressure. So answer is option (D).

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

4 4 2

S 4 3 7/2

2 7/2 2

−

=

,

0 2 3

V 2 0 7/2

3 -7/2 0

−

= −

, then matrix P is

(A)

4 2 5

6 3 7

1 0 2

− −

(B)

2 9/2 1

1 81/4 11

2 45/2 73/4

− − −

−

(C)

4 6 1

2 3 0

5 7 2

− − − − − −

(D)

4 6 1

2 3 0

5 7 2

− −

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Key: (A)

Sol: We can express any square matrix as the sum of symmetric and skew-symmetric matrices.

…(1)

Given that,

4 4 2

7S 4 3 is symmetric

2

72 2

2

− =

And

0 2 3

7V 2 0 is skew symmetric

2

73 0

2

− = − − −

From (i);

4 4 2 0 2 3

7 7S V 4 3 2 0

2 2

7 72 2 3 0

2 2

− − + = +

4 2 5

P 6 3 7

1 0 2

−

= −

4 2 5

P S V; where P 6 3 7

1 0 2

−

= + = −

5. The equation of motion of a spring-mass-damper system is given by

( )2

2

d x dx3 9x 10sin 5t

dtdt+ + =

The damping factor for the system is

(A) 2 (B) 0.25 (C) 0.5 (D) 3

T TA A A Ai.e., A

2 2

+ −= +

symmetric skew-symmetric

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Key: (C)

Sol: ( )2

2

d x 3dx9x 10sin 5t

dt dt+ + =

Comparing this with standard equation

( )mx cx kx F t

m 1kg, c 3 N-sec m, K 9N m.

+ = + =

= = =

Damping factor,

c

C C 3 10.5

C 22 mk 2 1 9 = = = = =

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

(A) Solid solubility limits are depicted by it

(B) It indicates the temperature at which different phases start to melt

(C) It gives information on transformation rates

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

Key: (C)

Sol: Phase diagram will give the following information.

(i) It shows phases present at different compositions and temperatures under slow cooling

(equilibrium) conditions.

(ii) It indicates equilibrium solid solubility of one element/compound in other.

(iii) It indicates the temperature at which an alloy starts to solidity and the range of solidification.

But phase diagrams do not indicate the dynamics (transformation rates) when one phase is transforms in

to other. Hence the answer is Option (C).

7. A closed vessel contains pure water, in thermal equilibrium with its vapour at 25°C (Stage #1), as

shown.

Water vapour

Water

Valve A

Not to scale

25 CAmbient air

at 25 C, 1 atm

Isothermal oven at 80 C

Valve A

Air at 80 C, 1 atm

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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 (Stage #2). If the Valve A is now opened inside the oven, what will happen immediately

after opening the valve?

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

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

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

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

Key: (B)

Sol: In stage ‘1’, water vapour is present means pressure is not atmosphere pressure, it is definitely less than

atmospheric pressure. In stage ‘2’, we kept this vessel in 80°C air oven and vessel receives heat from it

hence the water vapour in the vessel will become super heated and this heating done at constant volume,

hence pressure use will also increases. The new pressure will be find as follows.

Since V = C

1 2

1 2

P P

T V=

1P = Pressure water vapour + water mixture in stage I

= saturated vapour pressure corresponding to 25°C 0.03 bar

1T = Initial temperature of water vapour and water mixture = 25° = 25 + 273 = 298°K

2P = Pressure of water vapour and water mixture in stage II after reacting thermal equilibrium with hot

air, which is at 80°C.

2T = Temperature of water vapour and water after reaching thermal equilibrium with hot air = 80°C

= 80 + 273= 353°K

2P0.03

298 353 =

( )2P 0.0355 bar 1bar Atmospheric pressure=

Since the pressure inside the vessel (0.0355 bar) is less than outside air (1 bar). Hence hot air will go

inside the vessel when the valve is opened. So answer is option ‘B’.

8. The solution of

2

2

d yy 1,

dt− =

which additionally satisfies t 0

t 0

dyy 0

dt==

= = in the Laplace s-domain is

(A) ( )( )

1

s s 1 s 1+ − (B)

( )1

s s 1− (C)

( )1

s s 1+ (D)

1

s 1−

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Key: (A)

Sol: Given D.E is ( ) ( )2

2

d yy 1; y 0 0; y ' 0 0 y '' y 1

dt− = = = − =

Apply Laplace transform on both sides; then

L y'' L y L 1− =

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( )( ) ( )( )

2

2

2

1s L y t sy 0 y ' 0 L y t

s

1s 1 L y t y 0 0 and y ' 0 0

s

1 1L y t

s s 1 s 1s s 1

− − − =

− = = =

= = + −−

9. 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) Base Stock Level (B) Lot-for-Lot

(C) Economic Order Quantity (D) Fixed Period Quantity, for 2 periods

Key: (B)

Sol: Lot for lot approach will be used when setup cost (or) ordering cost is negligible level compared to

inventory cost. So, answer is option (B).

10. Let 21 x

2

x 0 y 0I xy dydx

= == . Then, I may also be expressed as

(A) 1 1

2

y 0 x yyx dxdy

= = (B) 1 1

2

y 0 x yxy dxdy

= =

(C) 1 y

2

y 0 x 0xy dxdy

= = (D) 1 y

2

y 0 x 0yx dxdy

= =

Key: (B)

Sol: Given, double integral is

21 x

2

x 0 y 0

I xy dydx= =

=

Given limits of integration:

2y 0 y x and x 0 x 1= → = = → =

Take a strip parallel to x-axis,

then the limits of integration of x will be: x y x 1and y 0 y 1= → = = → =

By change of order of integration, the given integral I can also be expressed as 1 1

2

y 0 x y

I xy dxdy= =

=

( )0,0

( )1,1

x 1=

( )1,0

2y x=

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11. 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.

Key: (18)

Sol:

In general to avoid buckling of bar, the maximum unsupported length can be only 3 times the diameter.

In that case the maximum increase in cross section, obtained in single stroke is 1.5 times the diameter of

the rod (d) and sometimes it will be less than it.

In the question, it is given that d = 12 mm and L=40 mm > 3d, hence the maximum increase in diameter

will be 1.5 times the diameter of rod (d). Hence the die diameter will be 1.5 times ‘d’.

Die diameter = 1.5 × 12 = 18 mm

12. In a furnace, the inner and outer sides of the brick wall (k1 = 2.5 W/mK) are maintained at 1100°C and

700°C respectively as shown in figure.

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).

1100 C 700 C200 C

2Heat flux 2500 W m=

2Insulation, k

1L2L

1

Brick wall

k 2.5

W m.K

=

d

Grip die

Heating tool

L 3d=

Horizontal forging

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Key: (0.5)

Sol:

Applying steady state head conduction with resistance concept to brick wall,

( )

( )

1 2

1

1

1 2

1

1

1

1

T TQ

L

k A

T TQ

A L

k

1100 7002500 L 0.4m

L

2.5

−=

−=

−= =

It is given that thickness of insulation ( ) 12

LL

4=

2

0.4L 0.1m

4 = =

Apply steady state heat conduction with resistance concept insulation

( )

2 3 2 3

2 2

2 2

2

2

T T T TQQ

AL L

k A k

700 2002500 k 0.5 W m K

0.1

k

− −= =

−= =

1T 1100 C= 2T 700 C= 3200 C T =

2Heat flux 2500 W m=

2Insulation, k

1L2L

1

Brick wall

k 2.5W m.K=

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13. 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.

The coefficient of static friction between the roller and the ground (including the edge of the step) is μ.

Identify the correct free body diagram (FBD) of the roller when the roller is just about to climb over the

step.

(A) (B)

(C) (D)

Key: (B)

Sol: Since the roller is just about to climb, hence there is no contact between roller and ground, while

climbing, it is just at the tip of the step means it is not rolling. Hence no frictional force at the tip of step.

Hence the answer is option ‘B’.

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

(A) always normally distributed

(B) normally distributed, only if X and Y are independent

(C) normally distributed, only if X and Y have the same mean

(D) normally distributed, only if X and Y have the same standard deviation

F

F

R

W

F

RW

F

R

W

2N

F

2N

1N

2N

W

1N

F

RW

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Key: (A)

Sol: The sum of two normal distributed random variables x and y is always normally distributed.

15. 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 __________.

Key: (1)

Sol: Given that

The eigen value of diagonal matrix are always its diagonal elements.

Eigen values of I are 1, 1, ….. 1. (100 times)

E 1 distinct real eigen value= →

Number of elements in E = 1.

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

the network.

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

L, is __________hours.

START A B C E

2 4 10 4

F

I

J

K

L

HGD

FINISH

7

8

5

6

2

976

100 100

100 100

1 0 0 0

0 1 0 0I diagonal matrix

0 0 0 1

= →

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Key: (2)

Sol:

EFT = Earliest finish time

LFT = Latest finish time

EST = Earliest start time

LST= Latest start time

The critical path for the given project is A B C F I J K− − − − − − and time taken is 42 hours.

EFT and LFT of each activity is shown along the nodes.

Slack (or) Float of any activity = LFT – EFT (or) LST – EST

( )Slack or float of activity 'L' is 42 40 2 hours. = − =

17. A machine member is subjected to fluctuating stress σ = σ0cos(8πt). 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 σ0 is

__________ MPa (round off to 2 decimal places).

Key: (100)

Sol: ( )ocos 8 t =

It is a case of cycle loading so mean a o e

0 and , 350MPa, F.S 3.5 = = = =

START A B C

E

2 4 10

4

F I

J

K

LHG

D

FINISH

78

5

6

29

7

6

EPT LFT 16 16

62

22 24

29 3138 40 40 42

23 23

31 31

20 31 42 42

36 36

|ME-2020|


mean a o

o

y e

1 1100MPa

F.S 350 3.5

+ = = =

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

pairs is

(A) 3 (B) 2 (C) 1 (D) 4

Key: (A)

Sol: When Grashoff’s low is satisfied i.e., s P Q+ + we will get four inversions out of which two

inversions are same. Hence three different inversions (qualitative inversions) we will get and they are

1. Crank- Rocker mechanism when link adjacent to shortest link is fixed.

2. Crank-Crank mechanism when shortest link is fixed.

3. Rocker-rocker mechanism when link opposite to shortest link is fixed.

So, answer is option (A)

19. Which of the following conditions is used lo determine the stable equilibrium of all partially submerged

floating bodies?

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

Key: (D)

Sol: For stable equilibrium of partially submerged bodies (floating bodies), meta centre (M) should be above

centre of gravity (G) and centre of buoyancy (B). So, answer is option ‘D’.

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

is

(A) electrical discharge machining

(B) abrasive jet machining

(C) electrochemical machining

(D) ultrasonic machining

Key: (D)

Sol: The unconventional machining method which uses mechanical energy as input from the given option is

abrasive jet machining (or) ultrasonic machining. Out of which in ultrasonic machining a tapering metal

(ultra sonic horn) bar is commonly used for augmenting the oscillation amplitude provided by an

ultrasonic transducer. Hence the answer is option ‘D’.

|ME-2020|


21. For an air-standard Diesel cycle,

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

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

(C) heat addition is at constant pressure and heat rejection is at constant pressure

(D) heat addition is at constant volume and heat rejection is at constant volume

Key: (A)

Sol: An air standard diesel cycle is shown below.

Where, 1-2 is & isentropic compression

2-3 is constant pressure heat addition

3-4 is & isentropic expansion

4-1 is constant volume heat rejection.

22. 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) 100% (B) zero (C) 75% (D) 50%

Key: (D)

Sol: Given that,

1

h = Enthalpy at inlet of stator

2

h = Enthalpy at inlet of rotor = Enthalpy at outlet of stator

3

h = Enthalpy at outlet of rotor

Degree of reaction ( )

( )

Enthalpy change in rotor moving blades

Total Enthalpy change in stage=

( )( )

Enthalpy change in rotorDOR

EnthalpyEnthalpy change in stator Fixed blades

change in rotor moving blades

=

+

( )

( ) ( )3 2

2 1 3 2

h hDOR

h h h h

− =

− + −

P3

1

2

v

4

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( ) ( )

( )

( )

2 1 3 2

3 2

3 2

Given that, h h h h

h h 1DOR 0.5 50%

2 h h 2

− = −

− = = = =

−

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

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

(A) 3

2r

(B)

r

(C)

2r

(D)

2

3r

Key: (A)

Sol: Let I = instantaneous centre of Disc and ‘I’ should lie on line joining ‘P’ and Q, will pass through centre

of disc (0).

Let =angular velocity of disc, x=distance of ‘I’ from ‘Q’.

( )( ) ( )

( )( ) ( )

( )

( )

( )

( )

( )

p

Q

then 2r x ... 1

and 2 x ... 2

1 2r x

2 2 x

2r x1

2 x

= = −

= =

−= =

− =

( )

x 4r 2x

4r3x 4r x

3

4rand 2

3

3

2r

= −

= =

=

=

Answer is option (A)

P

r

Q

2

P

r

Q

2

o

I

( )2r x−

x

|ME-2020|


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

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

__________ N.

Key: (1500)

Sol:

Q

Let R =Reaction at 'Q' (It has only reaction since it is roller support)

HP

R = Horizontal reaction at hinged support ‘P’

VP

R = Vertical reaction at hinged support ‘P’

Take the moments of all the forces about ‘P’ and equated to zero

i.e., p

M 0 =

( ) ( )( )Q

Q

Q

R 4 1200 5 0

4R 6000

6000R 1500N

4

+ − =

=

= =

5m 5m

1200 N

Q

QR

4m

VPR

HPR P

5m 5m

1200 N

Q

4m

R

P

|ME-2020|


25. 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).

Key: (0.9)

Sol: Given that, 2

T 30 C 273 3 270 K= − = − =

1

T 27 C 273 27 300 K= = + =

Since refrigerator (or) Heat pump operates on Carnot cycle

( ) 2

Ref

1 2

TCOP

T T=

−

( )

( )( )

( )

1

Heat pump

1 2

2709

300 270

TCOP

T T

30010

300 270

= =−

=−

= =−

( )

( )Ref

Heat pump

COP 90.9

COP 10= =

Q. No. 26 to 55 Carry Two Marks Each

26. Moist air at 105 kPa, 30°C and 80% relative humidity flows over a cooling coil in an insulated air-

conditioning duct. Saturated air exits the duct at 100 kPa and 15°C. The saturation pressure of water at

30°C and 15°C 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).

Key: (10)

Sol:

1T 27 C=

2T 3 C=

( )

Refrigirator

or Heatpump

Cooling coil

Insulated duct

Air inlet Air outlet

|ME-2020|


Given that, Total pressure of most air at inlet is ( )1P 105 kPa=

Relative humidity of moist air at inlet is ( ) 0.8 =

Temperature of moist air at inlet is ( )1T 30 C=

Saturation water vapour pressure at inlet is ( )SV 1P 4.24 kPa=

Partial water vapour pressure at inlet ( )V 1P ?=

We know that ( )

( )

( )( )V V1

1 V 1SV 1

P P0.8 P 3.392 kPa

P 4.24 = = =

We know that moist air is a mixture of Dry air and water vapour. Both dry air and water vapour behave

as ideal gases.

Apply ideal gas equation to both dry air and water vapour individually.

( )V 1 V V 11P V m R T= …(1)

( )d 1 d d 11P V m R T= …(2)

( )

( )V v v1

d d d1

P m Requation (1)

equation (2) P m R= =

Specific humidity ( )( )

( )v

d

mass of water vapour m

mass of dry air m =

( )

( )V dv 1

d d V

P Rm

m P R= =

Where dR = Dry air gas constant

( ) ddry air

R R

molecular weight= =

VR =Water vapour gas constant

( ) vwater vapour

R R

molecular weight= =

( )( )

( )V

v

d

P

P=

Given that ( ) ( )d v28.94 g mol, 189 mol = =

Pressure of dry air (Pd) = Total pressure of moist air (P) – Pressure of water vapour (PV)

v v

d V

P

P P

=

−

|ME-2020|


Specific humidity at inlet, ( )

( )Vv 1

1

d 1 V 1

P 18 3.392

P P 28.94 105 3.392

0.02076 kg of water vapour kg of dry air

= = − −

=

At outlet of duct, ( )2 2 SV 2P 100 kPa, T 15 C, P 1.7 kPa= = =

( )2 100% since air is saturated at output of duct =

( )

( )( ) ( )V 2

2 V SV2 2SV 2

PP P 1.7 kPa

P = = =

( )

( )Vv 2

2

d 2 V 2

P 18 1.7

P P 28.94 100 1.7

0.01075 kg water vapour kg of dry air

= = − −

=

Mass of water vapour condensed in the out is ( )1 2= −

( )0.02076 0.01075 0.01001 kg of water vapour kg of dry air

kg of water vapour10

kg of dry air

= − =

27. 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.

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

1D

1D1D2D

1H2H

1 2D D

|ME-2020|


(A) ( ) ( )

( )1 2 1 2

1 2

H H D Dcosec

2 D D

− − − =

− (B)

( )

( )1 2

1 2

H Hsin

D D

− =

−

(C) ( )

( ) ( )1 2

1 2 1 2

D Dcos

2 H H 2 D D

− =

− − − (D)

( )

( ) ( )1 2

1 2 1 2

D Dsin

2 H H D D

− =

− − −

Key: (D)

Sol:

In the triangle ABC,

1 2 1 2

AC AD DC AD BE

D D D DAC

2 2 2

= − = −

− = − =

From the above figure 22

11

DOB H

2

DOA H

2

= −

= −

1 21 2

D DAB OA OB H _ H

2 2

= − = − −

From triangle ABC

( )

( ) ( )

( )

( ) ( )( )

1 2

1 21 2

1 2 1 2

1 1 2 2 1 2 1 2

D D

AC 2sin

D DABH H

2 2

D D D Dsin

2H D 2H D 2 H H D D

−

= =

− − −

− − = =

− − − − − −

Correct answer is option ‘D’

1H

A

B

E

D

2H

O

|ME-2020|


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

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).

Key: (1.52)

Sol:

Density of air ( ) 3

air 1.23 kg m =

Applying continuity equation between section (1) and (2)

( )( ) ( )( )

1 1 2 2

1 1

A v A v

0.2 v 0.02 50 v 5m sec

=

= =

Applying Bernoulli’s equation between section (1) and (2)

1 2

1

1

2

1

p

v

A 0.2m=

stream line2 atm

22

2

p p

v 50 m s

A 0.02 m

=

=

=

1 2

1

1

2

1

p

v

A 0.2m=

stream line2 atm

22

2

p p

v 50 m s

A 0.02 m

=

=

=

|ME-2020|


( )

2 2

1 1 2 21 2

1 2

2 2

1 1 2 2

2 2

1 2

2 atm

1 atm

p v p vz z

g 2g g 2g

since z z

p v p v

2 2

p p5 50

1.23 2 1.23 2

p p

p p 1522.125Pa 1.52 kPa

+ + = + +

=

+ = +

+ = +

=

− = =

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

The bar material has a linear elastic response from O to P followed by a non-linear 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).

Key: (56.9)

Sol: Given data,

1 1

100MPa, 0.002, L 1000mm = = =

2 2

130MPa, 0.005 = =

( )2

2

E 2

e

EIP I AK

L

= =

130

100

( )MPa

P

Q

R

O 0.002 0.005 ( )mm mm

|ME-2020|


2 2 4 2

2E

22e

31

1

2

2

e

2 3

2

2

2

2 3

P EK d d dK , K

A L A 16 4d

100E 50 10 MPa

E 0.002

E

L

K

50 10100

1000

K

100 1000K K 14.2

50 10

d14.2

4

d 56.9mm 57mm

= = = = =

= = =

=

=

= =

=

= =

30. 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, ( ) ( )new oldCR 21 CR 15= = −

= ___________ %. (round off to one decimal place).

Key: (4.79)

Sol: Given that,

1 2

1.3, r 15, r 21 = = =

c

Cutt off ratio 2 r= =

We know that efficiency of Air standard diesel( )

c

1

c

r 11cycle is 1

r r 1

−

− = −

−

Efficiency of diesel cycle in the first case is

( ) ( )

1.3

c

1 1 0.3

1 c

r 11 1 2 11 1 0.508 50.8%

r r 1 15 1.3 2 1

−

− − = − = − = =

− −

Efficiency of diesel cycle in the second case is

( ) ( )

1.3

c

2 1 0.3

2 c

r 11 1 2 11 1 0.5487 54.87%

r r 1 21 1.3 2 1

−

− − = − = − = =

− −

Difference in efficiencies i.e., 2 1

54.87 50.08 4.79% − = − =

|ME-2020|


31. 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.

The material of the cylinder has Young's modulus E, Poisson's ratio ν, and coefficient of thermal

expansion α. What is the minimum rise in temperature ΔT 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) 1 pr

T2 tE

= +

(B)

3 prT

2 tE

=

(C) 1 pr

T4 tE

= −

(D)

prT

tE

=

Key: (D)

Sol: Due to uniform heating, change in length of cylinder ( ) T

= =

We know that longitudinal strain of thin cylinder is h

E

= −

Since cylinder is opened at both ends 0 =

h

h

E

T 'E

=

=

We know that, h

p r

t =

p r

TtE

=

To avoid leakage

p r

TtE

=

Rigid

wall

Rigid

wall

r

Pressure vessel

|ME-2020|


32. 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).

Key: (2500)

Sol: Given that,

Initial temperature of black body ( )1T 3000 K=

Wavelength at temperature is maximum and it is ( )max 1 =

Final temperature of black body ( )2T ?=

Wavelength corresponding to 2 2 1

'T ' is 1.2 =

We know that according to Wien’s displacement Law, T constant =

1 1 2 2T T =

( )( ) ( )( )1 1 2 23000 1.2 T T 2500 K. = =

33. 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 ( )–0.4

c 0 1U U t= , 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. c2 c1

c1

F F100

F

−

,

is__________ (round off to one decimal place).

Key: (-5.6)

Sol: Initial tool geometry according to ASA (American standard association) sytem is

e s e

0 9 r r C 30 R− − − − − −

After some the cutting toal geometry becomes

e s e

0 9 r r C 0 R− − − − − −

Given that, Specific energy consumption ( ) ( )0.4

0e V t

−

=

Where t = un cut chip thickness

According ASA system toal signature is

|ME-2020|


Back Rake angle

( )b

Side rake angle

( )s

End relief angle

( )e

Side relief angle

( )s

End cutting edge angel

( )eC

Side cutting edge angle

( )CC

Nose radius

(R)

Specific cutting energy C C C

1 C 1

F V F

A V A= =

From the initial tool geometry

( )b 10 = and from minimum energy consumption according to merchant is ( ) ( ) ( )1 b1 1

2 90 + − =

Thickness of uncut chip thickness, 1

ts

f cos C=

( ) ( ) ( ) ( )11 S 11 1

t f cos C t f cos 30 = =

( ) ( ) ( )

( )

( )

22 s2

0.4

c

0.4

c s

t f cos C f cos 0

given that e V t

e V f cos C

−

−

= =

=

=

( )

( )

( )

( )

( )

( )

2 1 2

1 1

2

1

0.4C

C s

1

0.4

C s

C C C

C C

0.4

s

0.4

s

0.4

0.4

Fe V f cos c

A

F cosC

F F F100 1 100

F F

cosC1 100

cosC

cos01 100

cos30

5.6% 5.6%

−

−

−

−

−

−

= =

− = −

= −

= −

= −

34. 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).

1t

|ME-2020|


Key: (8)

Sol:

Given data, 3 21000 kg m , g 10m sec = =

Since pipe is of uniform cross-section, A B

V V 3.2m sec= =

Applying bernoulli’s equation Between ‘A’ and ‘B’ points

2 2

A A B B

A B L

P V P VZ Z h

g 2g g 2g+ + = + + +

Where L

h loss of head=

( )( )

( )

( )( )

( )( )

( )( )

23 3 2

L

186 10 260 10 3.23.224.5 9.1 h

1000 10 2 10 1000 10 2 10

+ + = + + +

L

h 8 m of water =

35. There are two identical shaping machines S1 and S2. In machine S1, 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 __________ (roundoff to one

decimal place).

Key: (0.8182)

Sol: In shaping machine time taken to machine a job per one pass is

( )

( )Lw 1 m

t minutes1000 Vf

+=

Where L=length of job in mm

W= width of job in mm

m = quick return ratio Return stroke time

cutting stroke time=

V= cutting speed in m/min, f = feed in mm/ double stroke

Given that, In first machine

1 1

w w, f f= =

1Z 24.5m=

AAP 186kPa=

AV 3.2m sec=

BP 260 kPa=

2Z 9.1m=

B

|ME-2020|


In second machine

1 2

w 1.1w, f 0.9f= =

From machining time, we can say that

1 1 2 1

2 1 2 2

1

2

wt

f

t w f t w 0.9f

t f w t f 1.1w

t0.8182

t

= =

=

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

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 _________

kgm2.

Key: (3.58)

Sol: Given data,

mean

T 2000N m= − , S

N 1000rpm, C 2% 0.04, I ?= = = =

Area above the mean Torque line is fluctuation in energy which is

( )

2

S

E 3000 2000 5002

E I C

= − =

=

1000

1500

2000

2500

3000

( )Angle radius

Mom

ent

(Nm

)

|ME-2020|


2

22 1000500 I 0.04 I 3.58 kg-m

60

= =

37. The forecast for the monthly demand of a product is given in the table below.

The forecast is made by using the exponential smoothing method. The exponential smoothing coefficient

used in forecasting the demand is

(A) 0.10 (B) 1.00 (C) 0.40 (D) 0.50

Key: (A)

Sol: We know that, in exponential forecasting method

( )t 1 t t tF F D F+ = + −

Given that for 1st and 2nd month data

1 1 2F 32, D 30 and F 31.80= = =

( )

( )

2 1 1 1F F D F

31.80 32 30 32 0.1

= + −

= + − =

Similarly we can also check with 2nd and 3rd month data

( )

( )

2 2 3

3 2 2 2

F 31.80, D 32 and F 31.82

F F D F

31.82 31.80 32 31.80 0.1

= = =

= + −

= + − =

Answer is option ‘A’

38. One kg of air in a closed system undergoes an irreversible process from an initial state of p1 = 1 bar

(absolute) and T1 = 27°C, to a final state of p2 = 3 bar (absolute) and T2 = 127°C. If the gas constant of

air is 287 J/kg.K and the ratio of the specific heats γ = 1.4, then the change in the specific entropy (in

J/kg.K) of the air in the process is

(A) 172.0 (B) 28.4

(C) –26.3 (D) indeterminate, as the process is irreversible

Key: (C)

Sol: Given that,

Month Forecast Actual Sales

1 32.00 30.00

2 31.80 32.00

3 31.82 30.00

|ME-2020|


1 1

2 2

P 1bar; T 27 C 27 273 300 K

P 3bar; T 127 C 127 273 400 K

R 287 J kg K; 1.4

= = = + =

= = = + =

= =

We know that dQ dh vdP= −

( )( )

( )

p

p p

2 2

2 1 P

1 1

P V

p

V

P V

Tds C dT VdP

dT v dT Rds C dP C dP PV RT

T T T P

T PS S C n R n

T P

C C 287

C1.4

C

C 1004.5, C 717.5

= −

= − = − =

− = −

− =

=

= =

( ) ( ) ( )2 1

400 3S S 1004.5 n 287 n

300 1

− = −

26.3 J kg K= −

So, answer is option (C).

39. For the integral ( )/2

08 4cosx dx

+ , the absolute percentage error in numerical evaluation with the

Trapezoidal rule, using only the end points, is ___________. (round off to one decimal place).

Key: (5.2)

Sol: Given integral is ( )2

0

8 4cos x dx

+

Let ( )

0 1

f x 8 4cos x

x 0; x ; h2 2

= +

= = =

Exact value of the integral:

( ) ( )2

2

0

0

8 4cos x dx 8x 4sin x 4 4 1 4 4

+ = + = + = +

Approximate value of the integral:

By trapezoidal rule;

|ME-2020|


( ) ( )

( ) ( )( )

( )

2

0

28 4cos x dx f 0 f

2 2

12 8 f 0 8 4cos 0 8 4 124

20 5 f 1 8 4cos 84 2

+ +

= + = + = + =

= = = + =

Absolute percentage error True value approximate value

100True value

4 4 5100

4 4

4 0.86100 100 5.2%

4 4 16.6

−=

+ − =

+

− += = =

+

40. 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 0.2VT 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).

Key: (26.8)

Sol: Given that, length of bar (L) = 250 mm

Diameter of bar (d) = 25mm, Feed (f) = 0.2 mm/rev

Regrinding cost of tool = 20 Rs.

Tool changing time is ( )ctt 1min=

Tool life equation is 0.2VT 24=

n 0.2, C 24= =

Optimum cutting speed for maximum production rate is

n 0.2

opt

ct

n 1 0.2 1V C 24 18.19 m min

1 n t 1 0.2 1

= = =

− −

If we do machining operation at this speed, the life of tool will be

( )

0.2

opt

0.2

V T 24

18.19 T 24

=

=

T 3.99mins 4 min=

Rotational speed of bar when the cutting speed optV 18.19m min= is

opt

opt

dNV

1000

=

|ME-2020|


optN 231.60rpm 232 rpm =

Time taken for machining ( )

m

L 250t 5.387 min

fN 0.2 232= = =

Number of regrinding required mt 5.3871.34 regrinds

T 4= = =

So optimum tool cost = cost required for 1.34 regrinds = 1.34 × 20 = 26.8 Rs.

41. 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 (roundoff to the nearest integer).

Superheated Steam Table:

Pressure, p

(MPa)

Temperature, T

(°C)

Enthalpy, h

(kJ/kg)

Entropy, s

(kJ/kg.K)

10 500 3373.6 6.5965

1 500 3478.4 7.7621

Saturated Steam Table:

Pressure, p Sat. Temp.

Tsat (°C)

Enthalpy, h

(kJ/kg)

Entropy, s

(kJ/kg.K)

hf hg sf sg

1 MPa 179.91 762.9 2778.1 2.1386 6.5965

20 kPa 60.06 251.38 2609.7 0.8319 7.9085

Key: (1513.73)

Sol:

10 MPa

1MPa

20 kPa

S

T

2

1

3

5

2'

4

6

2"

|ME-2020|


Given that, condenser pressure ( )1 6P P 20 kPa= =

3T 500 C=

3-4 is isentropic expansion in first turbine.

From the given data we can say that it is expanded up to 1 MPa.

Work done (or) Enthalpy change during isentropic expansion process of 3-4 in first turbine,1 3 4w h h= −

3h = Enthalpy of super-heated water vapour at 10 MPa pressure and 500°C temperature= 3373.6 kJ/kg

4h = Enthalpy of saturated water vapour (hg) at 1 MPa and 179.91°C temperature

( )1w 3373.6 2778.1 595.5 kJ kg= − =

Work done (or) enthalpy change during isentropic expansion from state ‘5’ to ‘6’, 2 5 6w h h= −

5h = Enthalpy of super heat water vapour at 1 MPa pressure and 500°C temperature = 3478.4 kJ/kg

Since 5-6 is isentropic process, 5 6s s−

5s = entropy at pressure 1 MPa pressure and 500°C temperature = 7.7621 kJ/kg°K

( )( )5 f g fat 20 kPa

s s x s s= + −

( )( )7.7621 0.8319 x 7.9085 0.8319 x 0.979= + − =

( )( )

( )( )

6 f y fat 20kPa

h h x h h

251.38 0.979 2609.7 251.38 2560.17 kJ kg

= + −

= + − =

( )2 5 6W h h 3478.4 2560.17 918.23 kJ kg= − = − =

Total work done in both the turbines is1 2W W W 595.5 918.23 1513.73 kJ kg= + = + =

42. A cantilever of length , and flexural rigidity EI, stiffened by a spring of stiffness k, is loaded by

transverse force P, as shown.

The transverse deflection under the load is

,EIP

k

|ME-2020|


(A) 3

3

P 3EI

3EI 3EI 2k

+

(B) 3 3P 6EI k

3EI 6EI

−

(C) 3

3

P 3EI

3EI 3EI k

+

(D) 3 3P 3EI k

3EI 3EI

−

Key: (C)

Sol:

( )1 2

3

3 3

P R L R

3EI K

PL RL R

3EI 3EI K

=

−=

− =

( )

3 3

3 3

3

3

PL L 1R

3EI 3EI K

PL KL 3EIR

3EI 3EI.K

PL .KR

KL 3EI

= +

+ =

=+

( )

3 3

2 33

R PL .K PL

K KL 3EIKL 3EI

K

= = =++

43. 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 _________ (round off to 3 decimal

places).

Key: (0.042)

Sol: Given that, a fair coin is tossed 20 times.

Let ‘A’ be the event that “head” will appear exactly 4 times in the first ten tosses and

‘B’ be the event that “fail” will appear exactly 4 times in the next ten tosses.

The required probability, P A B P A .P B = [ A, B, are independent events]

4 4

4 6 4 6

C C

1 1 1 1P A B 10 .10

2 2 2 2

=

[ using binomial distributing]

10 10

10 10

10 9 8 7 1 10 9 8 7 1

4 3 2 1 2 4 3 2 1 2

210 2100.042

2 2

=

=

P

sP( )1

R

R

2

R

K =

( ) 3

1

P R L

3EI

− =

|ME-2020|


44. 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 of 800 rpm, controlling both the X and Y motion of the

stage, are directly coupled to a pair of lead screw, each with a uniform pitch of 0.5 mm. The time needed

to position the point ‘P’ to the point ‘Q’ is _______ minutes (round off to 2 decimal places).

Key: (1.5)

Sol:

100 200 300 400 500 600 700 800 900 1000

100

200

300

400

500

600

700

Y

X

( )P 200,300

( )Q 800,600

100 200 300

100

200

300

400

500

600Y

X

( )Q 800,600

Y

( )P 200,300

X

400 500 600 700 800

700

|ME-2020|


X 800 200 600mm

Y 600 300 300mm

= − =

= − =

( )N 800rpm, pitch p 0.5mm= =

Time required for ' X' movement ( ) ( )

x

X 600t 1.5 min

N P 800 0.5

= = = =

Time required for ' Y' movement( )

y

Yt

N P

= =

( )300

0.75 min800 0.5

= =

Since two stepper motors are used for both ‘x’ and ‘y’ movements separately.

Hence, time taken for this operation is

( ) ( )x yt max t , t max 1.5, 0.75 1.5 min= = =

45. 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 _________ radians/s (round off to 2

decimal places).

Key: (8.66)

Sol: Since sphere is submerged to half portion in water.

Hence, weight of sphere = weight of water displaced

mass of sphere = Mass of water displaced

( ) ( ) ( )33

water

1 4 1 4m r 1000 0.2 16.75 kg

2 3 2 3= = =

( )Inertia force mx 16.75 x= =

Buoyancy force = weight of water displaced= (volume of water displaced for small displacement of

sphere) × water ×g

Here, volume of water displaced for small displacement of sphere 2r x=

For small displacement radius is almost constant, Hence volume 2r x=

( )( ) ( ) ( )2 2r x 1000 10 0.2 1000 10 x 1256.63x= = =

Equation of motion will be,

Inertia force + Buoyancy force = 0

n

16.75x 1256.63x 0

1256.638.66 rad sec

16.75

= =

= =

xr

xInertia

force

Buoyancy

force

|ME-2020|


46. For a single item inventory system, the demand is continuous, which is 10000 per year. The replacement

is instantaneous and backorders (S units) per cycle are allowed as shown in the figure.

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).

Key: (1137.15)

Sol: Given that, Annual demand ( )R 10,000 units year=

Carrying cost ( )CC Rs. 4/unit/year=

Ordering cost ( )oC Rs 300 / order=

Back order cost (or) shortage cost ( )sC Rs. 25/unit/year=

Economic order quantity when shortage is allowed is

o S C

C S

2C R C C 2 300 10,000 25 4EOQ

C C 4 25

+ + = =

*EOQ Q 1319.09 units= =

Inventory

Time

S

sQ

Q

( )

Invertory

units

Time→

|ME-2020|


Optional shortage level * * C

S

S C

CQ Q

C C

=

+ ( )

41319.09 181.94 units

25 4

= =

+

Maximum amount of inverting level * *

SQ Q 1319.09 181.94 1137.15 units= − = − =

47. A helical spring has spring constant 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) k/2 (B) k (C) 16k (D) 8k

Key: (B)

Sol: 4

3

GdK

8D n=

1 1 1

2 2 2

d d, n n, D D

d 2d, n 2n, D 2D

= = =

= = =

4

3

4 3

1 1 2 2

2 2 1 1

4 3

1

2

1

2

dK

D n

K d D n

K d D n

K d 2D 2n

K 2d D n

K 18 2 1

K 16

=

=

= =

48. 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).

Key: (4.285)

Sol:

Poring basin

Sprue

mould

cavity

2

3

1

sh

ph

th

|ME-2020|


Given that, volume of mould cavity ( ) 3 6 3V 1200cm 1200 10 m−= =

Height of sprue ( )sh 10cm 0.1m= =

Cross-sectional area of Base of sprue ( ) 2

sA 2cm= 4 22 10 m−=

Since height of pouring basin is not given, we can assume that it can be neglected compared to height of

sprue.

Velocity of molten metal at ‘3’ i.e., at end of sprue (or) at the entrance of runner is

3 s

V 2gh 2 9.81 0.1 1.4m sec= = =

Flow rate of molten metal s 3

A V= 4 4 32 10 1.4 Q 2.8 10 m sec− −= =

Time taken to fill the mould cavity ( )( )

( )

Volume Vt

flow rate Q=

6

4

1200 10t 4.285 seconds

2.8 10

−

−

= =

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

teeth.

If the sun (S) and the outer ring (R) gears are rotated in the same direction with angular speed ωS and ωR,

respectively, then the angular speed of the arm AB is

(A) R S

1 3

4 4 + (B) R S

3 1

4 4 + (C) R S

1 1

2 2 − (D) R S

3 1

4 4 −

Key: (B)

Sol: Given that,

Number of teeth on sum gear ( )sz = number of teeth on planet gear (zp)

P B

S A

S

R

R

|ME-2020|


R = Angular velocity of gear ‘R’

s = Angular velocity of gear ‘S’

Radius of gear ‘R’ is RR s p

mzr r 2r

2= = +

psR

R s p

2mzmzmz

2 2 2

z z 2z

= +

= +

( )R p s pz 3z z z= =

Action Arm ‘AB’ Gear ‘S’ Gear ‘P’ Gear ‘R’

Arm is fixed

and give 1

rev anti

clockwise to

gear ‘S’

ABN 0= SN 1rev= − SP

S P

P S

ZN

N Z

N N 1rev

−=

= − = +

( )( )

( )

R P

P R

PP PR

R P

N Z

N Z

1 ZN Z 41N

Z 3 Z 3

=

+= = =

Arm is fixed

and given ‘x’

rev anti

clockwise to

gear ‘S’

ABN 0= –x +x x

3

+

Give ‘y’ revs

to arm

y y–x y+x xy

3+

R

xy

3

= +

…(1)

And ( )s y x = − …(2)

(1)-(2) ( )

( ) ( )

R S

R S S R

xx

3

4x 3x

3 4

− = −

− − − = = −

S

S S R

S R

y x

3 3y

4 4

1 3y

4 4

− =

= − +

= +

Arm speed is AB R S

3 1N y

4 4= = +

So answer is option ‘B’.

P B

S A

Sw

R

RW

|ME-2020|


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

(A) 7

3 (B) ˆ ˆ ˆ3i 3j k− − (C)

7

3− (D) 7

Key: (A)

Sol: Directional derivative of ‘f’ at (–1, 1, 3) in the direction of the vector

( )( )1,1,3

aˆ ˆ â i 2j 2k fa−

= − + = …(1)

( )f f fˆ ˆ ˆf i j k ; where f x,y,z xyzx y z

= + + =

( ) ( ) ( )

( )( )1,1,3

ˆ ˆ ˆf i yz j xz k xy

ˆ ˆ ˆf 3i 3j k

ˆ ˆ â i 2 j 2k 1 4 4 3

−

= + +

= − −

= − + = + + =

From (1);

Directional derivative of f ( )( )ˆ ˆ î 2j 2k 3 6 2 7ˆ ˆ ˆ3i 3j k .

3 3 3

− + + −= − − = =

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

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).

Key: (544)

Sol: Given that, atmospheric pressure ( )atmP 100 kPa=

Initial pressure of air in the cylinder, ( )P 105 kPa=

Stop

8 cm

8 cm

Air

Q

|ME-2020|


Area of piston 2300cm=

Stiffness of spring ( )k 12.5 N mm 12,500 N m= =

Work interaction of air inside the cylinder (dW) =?

Since piston cylinder arrangement is a closed system

Work-done by air (system) (dW) = pdv

( ) ( )dW pdv Assuming pressure is constant=

Work done by air from ‘1’ to ‘2’

( ) ( )( )

( ) ( )

( )

1 2

1 2

3 4

1 2

dW P change in volume from '1' to '2 '

8dW P Area of piston

100

8105 10 300 10

100

dW 252 Nm 252 Joules

−

−

−

−

=

=

=

= =

Work done by air (system) from ‘2’ to ‘3’

( ) ( )( ) ( )

( )

2 3

2

dW P change in volume from '2' to '3' strain energy stored in the spring

8 1P Area of piston kx

100 2

−= +

= +

( )2

3 4

2 3

8 1 8dW 105 10 300 10 12,500

100 2 100

252 40 292 Nm 292 Joules

−

−

= +

= + = =

Work done by air (system) from ‘1’ to ‘3’ is

( ) ( ) ( )1 3 1 2 2 3

dW dW dW 252 292 544 Joules− − −= + = + =

52. 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 it 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

is 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/kg.K. The inner surface temperature at the exit

section of the tube is __________ °C (round off to one decimal place).

Key: (85.7)

Sol:

wi10 C = woT 70 C=

electric heater

L 20m=

8 cm

8 cm

Air

Q

1

2

3

stop

|ME-2020|


Mass flow rate of water water

720 kgm 720 kg hr 0.2 kg sec

3600sec= = =

Diameter of tube (D) 3cm 0.03m= =

Heat flux supplied by heater to tube = Q

Heat flux supplied by heater is used to heat the tube, from the tube heat will be transferred to water

Heat supplied ( )Q = Heat take by water

( ) ( )

( )

water P wo wiwater

ˆ ˆQ m C T T

0.2 4.176 70 10 50.148 kW

= −

= − =

Heat supplied per unit surface are of tube is 2 2

Q Q 50.148 kW kN26.604

A DL 0.03 20 m m

= = =

Heat supplied by heater = convective heat transfer from tube to water at any section of tube

Heat supplied by heater = connective heat transfer from tube to water at exit of tube

( )to woQ h T T= −

Where h = Internal connective heat transfer coefficient 21697 W m K=

toT = Temperature of tube at outlet of tube.

( )3

to to26.604 10 1697 T 70 T 85.67 85.7 C = − =

53. The function f(z) of complex variable z x iy= + , where i 1= − , is given as

( ) ( ) ( )3 2f z x 3xy i x, y= − + . For this function to be analytic, ( )x , y should be

(A) ( )3 2x 3x y constant− + (B) ( )2 33xy y constant− +

(C) ( )2 33x y y constant− + (D) ( )2 2 33x y y constant− +

Key: (C)

Sol: Given,

( ) ( ) ( )3 2f z x 3xy i x,y= − +

Where, ( ) 3 2 2 2u uu x, y x 3xy 3x 3y and 6xy

x y

= − = − = −

From options; consider ( ) 2 3x, y 3x y y constant = − + 2 26xy and 3x 3yx y

= = −

( ) ( )u u

and R equations satified V x,y f z is analyticx y y x

= = − −

|ME-2020|


54. 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 factor

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:t

vK W

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

Key: (10.035)

Sol: Given data,

v

m 1.25, t 20

20 , F 50mm

y 0.322, K 1.26

400MPa, N 1200rpm

= =

= =

= =

= =

t

vK .W

FmY =

( )( )

t

t

3

t

1.26W400

50 1.25 0.322

W 6388.8N

1.25 20 10 1200DNPower W D mT 6388.8 10.035 kW

60 60

−

=

=

= = = =

55. A rigid block of mass m1 = 10 kg having velocity v0 = 2 m/s strikes a stationary block of mass m2 = 30

kg after travelling 1 m along a frictionless horizontal surface as shown in the figure.

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).

Key: (1)

Sol: Given data,

1m 2m

Mass-less

k

1m 0.25m

0V

|ME-2020|


1 0

5

2

m 10kg, V 2m sec

m 30 Kg, K 10 N m

= =

= =

Applying conservation of momentum equation,

( )

( )

1 0 2 1 2m V m 0 m m V

10 2 10 30 V

V 0.5 m sec

+ = +

= +

=

( )( )1 2m m

K.E Strain energy of spring+

=

( )

( )

2 2

1 2

2 5 2

1 1m m V k x

2 2

1 110 30 0.5 10 x

2 2

x 1cm

+ =

+ =

=

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