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1

(PAPER – II)

MECHANICAL ENGINEERING

1. The time taken to face a workpiece of 80 mm diameter for the spindle speed of 90 rpm and cross feed 0.3

mm/rev will be

(A) 4.12 min

(B) 3.24 min

(D) 2.36 min

(D) 1.48 min

Answer: (D)

Sol: Diameter of work piece (d) = 80 mm

Speed of spindle (N) = 90 rpm

Time taken for facing operation ( )m

Lt

fN=

L = Distance to be travelled by tool = 40 mm

f = feed = 0.3 mm/rev

( )m

40t 1.48 min

0.3 90 = =

Answer is option (D).

2. A feed f for the lathe operation is

(A) m

Nmm rev

L T

(B) m

Lmm rev

N T

(C) mTmm rev

N L

(D) mT Lmm rev

N

Answer: (B)

d 80=



2

Sol: The taken for any lathe machining process is m

LT

fN=

Where, L = Length of cut in mm

f = feed in mm/rev

N = Speed in rpm

Tm = Machining in min

m

Lf

T N =

Answer is option (B).

3. The main advantage of the radial drilling machine is that

(A) It is very compatible and handy for machining

(B) It is accurate, economical, portable and least time consuming while machining

(C) Heavy workpieces can be machined in any position without moving them

(D) Small workpieces can be machined and it can be used for mass production as well

Answer: (C)

Sol: Advantages of radial drilling machine are

1. Larger and heavier work may be machined.

2. Drilling head raised (or) lowered.

3. Drilling head moves rapidly to any desired location while work piece remains clamped.

Answer is option (C).

4. For the purpose of sampling inspection, the maximum percent defective that can be considered

satisfactory as a process average is

(A) Rejectable Quality Level (RQL)

(B) Acceptable Quality Level (AQL)

(C) Average Outgoing Quality Limit (AOQL)

(D) Lot Tolerance Percent Defective (LTPD)

Answer: (B)

Sol: Acceptable quality level is the maximum percent defective (or) the maximum number of defects per

hundred units, that for purposes of sampling inspection, can be considered satisfactory as a process

average.

So correct option is ‘B’.



3

5. Hard automation is also called

(A) Selective automation

(B) Total automation

(C) Group technology

(D) Fixed position automation

Answer: (D)

Sol: Automation of production systems can be classified into three basic types:

1. Fixed automation (Hard automation).

2. Programmable automation (soft automation).

3. Flexible automation.

Fixed automation: It is used in high volume production with dedicated equipment, which has a fixed set

of operation and designed to be efficient for this set. Continuous flow and discrete mass production

systems, use this automation. Ex: Distillation process, conveyors, paint shops, transfer lines etc.

A process using mechanized machinery to perform fixed and repetitive operations in order to produce a

high volume of similar parts. Hence this is a fixed position automation. So, answer is option (D).

6. The method of CNC programming which enables the programmer to describe part geometry using

variables is

(A) Computer assisted part programming

(B) Computer aided drafting programming

(C) Conversational programming

(D) Parametric programming

Answer: (D)

Sol: Parametric programing possesses both computer-related features as well as CNC specific features.

Computer related features like variables, arithmetic, logic statements, and looping are available in

parametric-programming. Like other computer programming languages, parametric programming is also

available in several versions like Fanuc controls. The variables are used to describe part geometry. Hence

answer is option (D).

7. Revolving joint of the robot is referred to as

(A) L joint

(B) O joint

(C) T joint

(D) V joint



4

Answer: (D)

8. Repairing of a machine consists of 5 steps that must be performed sequentially. Time taken to perform

each of the 5 steps is found to have an exponential distribution with a mean of 5 minutes and is

independent of other steps. If these machines break down in Poisson fashion at an average rate of 2/hour

and it there is only one repairman, the average idle time for each machine that has broken down will be

(A) 120 minutes

(B) 110 minutes

(C) 100 minutes

(D) 90 minutes

Answer: (A)

Sol: Given that,

Time taken to complete each step in repairing machine = 5 minutes

Time taken to complete ‘5’ steps = 5 × 5 = 25 minutes

i.e., service time for repairing each machine = 25 minutes

1

units min ute25

=

Machines are break down in Poisson’s ratio with an average rate of 2/hour

i.e., 2units 2

units minhour 60

= =

Average idle time for each machine which has break down =Waiting time of machine in the queue with

single server.

i.e., ( )q

2

60W 125 minutes

1 1 2

25 25 60

= = = −

−

Hence nearest answer is option (A).

9. A problem of the total float within which an activity can be delayed for start without affecting the floats of

preceding activities is called

(A) Safety float

(B) Free float

(C) Independent float

(D) Interfering float

Answer: (C)



5

Sol: Independent float is that portion of the total float within which an activity can be delayed for start without

affecting the float of the preceding activities. It is calculated for an activity by subtracting the tail event

slack from its total float. Hence the answer option is (C).

10. An oil engine manufacturer purchases lubricant cans at the rate of Rs. 42 per piece from a vendor. The

requirement of these lubricant cans is 1800 per year. If the cost per placement of an order is Rs. 16 and

inventory carrying charges per rupee per year is 20 paise, the order quantity per order will be

(A) 91 cans

(B) 83 cans

(C) 75 cans

(D) 67 cans

Answer: (B)

Sol: Given that, annual demand of cans is ( )R 1800 per year=

Ordering cost for placing the order is ( )0C 16 Rs=

Cost of lubricant can = 42 Rs per can

Inventory carrying cost is 20 paise per rupee of can cost (0.2 rupee)

( )CPer year C 0.2 42rupees can year=

Economic order quantity is ( ) 0

C

2C R 2 16 1800EOQ 82.80 83 cans

C 0.2 42

= = =

11. Consider the following data regarding the acceptance sampling process:

N = 10,000, n = 89, c = 2, P = 0.01, and aP 0.9397=

The average total inspection (ATI) will be

(A) 795

(B) 687

(C) 595

(D) 487

Answer: (B)

Sol: Given that,

N = lot size = 10,000, n = sample size = 89

C = Accepted numbers for the given lot (surplus data)

Quality of lot (P) = 0.01

Probability of acceptance ( )aP 0.9397=



6

Average Total Inspection (ATI) will be find as follows,

( )( )

( ) ( )aATI n 1 P N n

89 1 0.9397 10,000 89

686.63 687

= + − −

= + − −

=

So, answer is option ‘B’.

12. The Non-Destructive Inspection (NDI) technique employed during inspection for castings of tubes and

pipes to check the overall strength of a casting in resistance to bursting under hydraulic pressure is

(A) Radiographic inspection

(B) Magnetic particle inspection

(C) Fluorescent penetrant

(D) Pressure testing

Answer: (D)

Sol: Pressure testing is a non-destructive test performed to ensure the integrity of the pressure shell on new

pressure equipment (or) on previously installed pressure and piping equipment that has undergone an

alteration (or) repair to its boundary (s). Pressure testing is required by most piping codes to verify that a

new, modified (or) repaired piping system is capable of safely withstanding its rated pressure and is leak

tight. Hence the answer is option ‘D’.

13. Consider the situation where a microprocessor gives an output of an 8-bit word. This is fed through an 8-

bit digital-to-analogue converter to a control valve. The Control valve requires 6.0 V being fully open. If

the fully open state is indicated by 11111111, the output to the valve for a change of 1-bit will be

(A) 0.061 V

(B) 0.042 V

(C) 0.023 V

(D) 0.014 V

Answer: (C)

Sol: Output voltage for fully open = 6.0V

Number of internals 82=

Change in output voltage 8

60.023V

2= =



7

14. Which of the following factors are to be considered while selecting a microcontroller?

(1) Memory requirements

(2) Processing speed required

(3) Number of input/output pins

(A) 1 and 2 only

(B) 1 and 3 only

(C) 2 and 3 only

(D) 1, 2 and 3

Answer: (D)

15. Which of the following statements regarding interface circuit are correct?

1. Electrical buffering is needed when the peripheral operates at a different voltage or current to that on

the microprocessor bus system or there are different ground references.

2. Timing control is needed when the data transfer rates of the peripheral and the microprocessor are

different.

3. Changing the number of lines is needed when the codes used by the peripherals differ from those

used by the microprocessor.

(A) 1 and 2 only

(B) 1 and 3 only

(C) 2 and 3 only

(D) 1, 2 and 3

Answer: (A)

16. Alternative paths provided by vertical paths from the main rung of a ladder diagram, that is, paths in

parallel, represent

(A) Logical AND operations

(B) Logical OR operations

(C) Logical NOT operations

(D) Logical NOR operations

Answer: (B)

17. The resolution of an encoder with 10 tracks will be nearly

(A) 0.15°

(B) 0.25°

(C) 0.35°

(D) 0.45°



8

Answer: (C)

Sol: Resolution n 10

360 3600.35

2 2= = =

18. Which of the following features is/are relevant to variable reluctance stepper motors?

1. Smaller rotor mass; more responsive

2. Step size is small

3. More sluggish

(A) 1 only

(B) 2 only

(C) 3 only

(D) 1, 2 and 3

Answer: (A)

Sol: Soft iron is used for manufacturing rotor in case of variable reluctance motor, having small rotor mass and

quick response.

19. Which of the following statements regarding hydraulic pumps are correct?

1. The gear pump consists of two close-meshing gear wheels which rotate in opposite directions.

2. In vane pump, as the rotor rotates, the vanes follow the contours of the casing.

3. The leakage is more in vane pump compared to gear pump.

(A) 1, 2 and 3

(B) 1 and 2 only

(C) 1 and 3 only

(D) 2 and 3 only

Answer: (B)

Sol: Gear pump is positive displacement pump. It has two meshed gears revolving about their respective axes.

These pumps have a disadvantage of small leakage due to a gap between teeth and the pump housing.

Vane pumps generate a pumping action by tracking of vanes along the casing wall. In vane pump vanes

are located on the slotted rotor. The rotor is eccentrically placed inside the casing. When the prime mover

rotates the rotor, the vanes are thrown outward due to centrifugal force. In vane pumps leakage is less and

its volumetric efficiency is around 95%. Hence answer is option ‘B’.

20. The selection of the right controller for the application depends on

1. The degree of control required by the application.

2. The individual characteristics of the plants.



9

3. The desirable performance level including required response, steady-state deviation and stability.

Which of the above statements are correct?

(A) 1 and 2 only

(B) 1 and 3 only

(C) 2 and 3 only

(D) 1, 2 and 3

Answer: (D)

21. Consider a system described by

x Ax Bu

y Cx Du

= +

= +

The system is completely output controllable if and only if

(A) The matrix 2 n 1CB CBA CB A CB A D− is of rank n

(B) The matrix 2 n 1CB CBA CA B CA B D− is of rank m

(C) The matrix 2 n 1BC BAC BA C BA C D− is of rank m

(D) The matrix 2 n 1BC ABC CA B CB A D− is rank of n

Where:

x = State vector (n-vector)

u = Control vector (r-vector)

y = output vector (m-vector)

A= n × n matrix

B = n × r matrix

C = m × n matrix

D = m × r matrix

Answer: (B)

22. Which one of the following symbols is used as the notation for designating arm and body of a robot with

joined arm configuration?

(A) TRL

(B) TLL, LTL, LVL

(C) LLL

(D) TRR, VVR

Answer: (D)



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Sol: Robot configuration Symbol

(Arm & Body)

Polar configuration →TRL

Cylindrical configuration → TLL, LTL,

Cartesian configuration →LLL

Joint ed arm configuration →TRR, VVR

23. A complaint motion control of robots can be understood by the problem of controlling of

(A) Position and velocity of joints

(B) Position and acceleration of the end-effector

(C) Manipulator motion and its force interactions with the environment

(D) Joint velocities of given end-effector velocity

Answer: (C)

24. For the vector v = 25i + 10j + 20k, perform a translation by a distance 8 in the x-direction, 5 in the y-

direction and 0 in the z-direction. The translated vector Hv will be

(A)

1

20

33

15

(B)

33

15

20

1

(C)

15

33

1

20

(D)

1

15

20

33

Answer: (B)



11

Sol: Given data, V 25i 10 j 20k= + +

Translation in x direction = 8

Translation in y direction = 5

Translation in z direction = 0

Translation vector

1 0 0 8 25 33

0 1 0 5 10 15

0 0 1 0 20 20

0 0 0 1 1 1

= =

Directions: Each of the next six (06) items consists of two statements, one labelled as ‘Statement (I)’

and the other labelled as ‘Statement (II)’. You are to examine these two statements carefully and select

the answers to these items using the codes given below:

Codes:

(A) Both statement (I) and Statement (II) are individually true and Statement (II) is the correct

explanation of Statement (I).

(B) Both statement (I) and Statement (II) are individually true, but statement (II) is not the correct

explanation of Statement (I).

(C) Statement (I) is true, but Statement (II) is false.

(D) Statement (I) is false, but Statement (II) is true.

25. Statement (I): The greater the chemical affinity of two metals, the more restricted is their solid solubility

and greater is the tendency of formation of compound.

Statement (II): Wider the separation of elements in the periodic table, greater is their chemical affinity.

Answer: (A)

Sol: The greater the chemical affinity of two metals, the more restricted is their solid solubility and the greater

is the tendency towards compound formation. Generally, the farther apart the elements are in the periodic

table, the greater is their chemical affinity.

As we move along the periodic table their atomic number will be increased, hence their difference in

atomic size is more, hence they can’t make solid solution instead they can make chemical compounds due

to their more chemical affinity. Hence statement II is correct explanation for statement-I. Hence answer is

option ‘A’.



12

26. Statement (I): The size of a memory unit is specified in terms of the number of storage locations

available, 1 K is 210 = 1024 locations and thus a 4 K memory has 4096 locations.

Statement (II): Erasable and programmable ROM (EPROM) is a form of memory unit used for ROMs

that can be programmed and their contents altered.

Answer: (B)

27. Statement (I): Microprocessors which have memory and various input/output arrangements, all on the

same chip, are called microcontrollers.

Statement (II): The microcontroller is the integration of a microprocessor with RAM, ROM, EPROM,

EEPROM and I/O interfaces, and other peripherals such as timers, on a single chip.

Answer: (A)

28. Statement (I): Capacitive proximity sensor can only be used for the detection of metal objects and is best

with ferrous metals.

Statement (II): One form of capacitive proximity sensor consists of a single capacitor plate probe with

the other plate being formed by the object, which has to be metallic and earthed.

Answer: (D)

29. Statement (I): SCARA configuration provides substantial rigidity for the robot in the vertical direction,

but compliance in the horizontal plane.

Statement (II): A special version of the Cartesian coordinate robot is the SCARA, which has a very high

lift capacity as it is designed for high rigidity.

Answer: (C)

30. Statement (I): The stopper motor is a device that produces rotation through equal angles, the so-called

steps, for each digital pulse supplied to its input.

Statement (II): Stepper motors can be used to give controlled rotational steps but cannot give continuous

rotation, as a result their applications are limited to step angles only.

Answer: (C)

Sol: Stepper motors are different from all other types of electrical drives in the sense that they operate on

discrete control pulses received and rotating in discrete steps. These motors are widely employed in

industrial control, specifically for CNC machines, where open loop control in discrete steps are

acceptable. These motors can also be adapted for continuous rotation. Hence statement I is correct, and

statement II is wrong. So answer is option ‘C’.



13

31. A stone weighs 400 N in air and when immersed in water it weighs 225 N. If the specific weight of water

is 39810 N m , the relative density of the stone will be nearly

(A) 5.9

(B) 4.7

(C) 3.5

(D) 2.3

Answer: (D)

Sol: Weight of stone in air = 400N

Weight of stone in water = 225N

Buoynace force = Weight of water displaced = 400 – 225 = 175N

Specific weight of water ( )water × Volume of store = 175 N

stone

stone

9810 V 175N

175V

9810

=

=

Weight of stone in air = (Specific weight of stone) × Vstone

stone

stone

175400

9810

9810 400

175

=

=

Relative density of stone stone

water

9810 400

4001752.28 2.3

9810 175

= = = =

Option ‘D’ is correct.

32. A flat plate 20.1m area is pulled at 30 cm/s relative to another plate located at a distance of 0.01 cm from

it, the fluid separating them being water with viscosity of 20.001 Ns m . The power required to maintain

velocity will be

(A) 0.05 W

(B) 0.07 W

(C) 0.09 W

(D) 0.11 W



14

Answer: (C)

Sol: Area of plate ( ) 2A 0.1m=

Velocity of plate ( )30cm 30 m m

V 0.3sec 100 sec sec

= = =

Viscosity of fluid between the plates ( ) 20.001N.sec m =

Shear stress at the moving plate is ( )du

dy =

( ) 2

V 0

y

0.3

0.010.001 3N m100

− =

= =

Shear force acting on the moving plate (F) A 3 0.1 0.3N= = =

Power required to maintain velocity (P) = F × V = 0.3 × 0.3 = 0.09 watts

So, answer is option ‘C’.

33. When the pressure of liquid is increased from 3 MN/m2 to MN/m2, its volume is decreased by 0.1%. The

bulk modulus of elasticity of the liquid will be

(A) 12 23 10 N m

(B) 9 23 10 N m

(C) 8 23 10 N m

(D) 4 23 10 N m

Answer: (B)

Sol: Initial pressure ( ) 2

AP 3MN m=

Final pressure ( ) 2

fP 6MN m=

Charge in pressure ( ) ( ) ( ) 2 6 2

f AP P P 6 3 3MN m 3 10 N m = − = − = =

% Decreased in volume V

100 0.1%V

=

V 0.1

V 100

=

V 30cm sec=

y 0.01cm=



15

Bulk modulus of elasticity ( )( )6

9 23 10P

k 3 10 N mV 0.1

V 100

= = =

So, the answer is option ‘B’.

34. A curve that is everywhere tangent to the instantaneous local velocity vector, is

(A) Streak line

(B) Path line

(C) Normal line

(D) Streamline

Answer: (D)

Sol: Streamline is an imaginary line (curve) drawn in flow field in such a way that tangent drawn at any point

on this line represents the direction of the velocity vector at that instant. Hence option ‘D’ is correct.

35. A 120 mm diameter jet of water is discharging from a nozzle into the air at a velocity of 40 m/s. The

power in the jet with respect to a datum at the jet will be

(A) 380 kW

(B) 360 kW

(C) 340 kW

(D) 320 kW

Answer: (B)

Sol: Diameter of jet (d) = 120 mm = 0.12m

Velocity of Jet ( )V 40m sec=

( ) ( )2 2 2

jeta d 0.12 0.01131m4 4

= = =

Velocity head of jet with respect jet as datum

( )( )

2 2V 40H 81.55 m

2g 2 9.81= = =

Power available at jet,

jet

P QH

9810 a v 81.55

9810 0.01131 40 81.55

361.922 kW

=

=

=

=


Datum

V 40m sec=

d 120 mm=



16

36. Which of the following applications regarding Navier-Stokes equations are correct?

(1) Laminar unidirectional flow between stationary parallel plates.

(2) Laminar unidirectional flow between parallel plates having no relative motion.

(3) Laminar flow in circuit pipes.

(4) Laminar flow between concentric rotating cylinders.

(A) 1, 2 and 3 only

(B) 1, 3 and 4 only

(C) 1, 2 and 4 only

(D) 2, 3 and 4 only

Answer: (A)

Sol: To have laminar flow between two concentric rotating cylinders, the gap should between then should be

very small and also cylinders lengths should be sufficiently large to make this possible. On account of

uniform angular motion of cylinders, the laminar flow must be steady and with these assumption of

laminar flow and steady flow the variation of velocity between the cylinders will be linear whereas the

above three applications i.e., flow between stationary parallel plates, flow between parallel plates having

no relative motion and flow in circular pipes the velocity variation is non linear. Hence, we can say that

laminar flow between concentric case of Navier Stokes equation. This rotating cylinders concept is also

used for measurement of viscosity of fluid between them since the velocity variation is linear and easy to

measure viscosity and hence this application can be taken as limiting case of Navier Stokes equation.

Hence answer is option (A).

37. A crude oil having a specific gravity of 0.9 flows through a pipe of diameter 0.15 m at the rate of 8 lps. If

the value of μ is 0.3 2Ns m , the Reynolds number will be nearly

(A) 295

(B) 235

(C) 205

(D) 165

Answer: (C)

Sol: Specific gravity of oil ( )s 0.9=

Diameter of pipe ( )d 0.15m=

Discharge of oil ( )38 m

Q 8 1ps100 sec

= =

Viscosity of oil ( ) 20.3 N.sec m =



17

Reynold’s number ( )VD

Re

=

3density s 100 0.9 100 900 kg m = = = =

Q

V velocity of oilA

= =

( )2

Qd 900 8 0.15Re 203.72 205

A1000 0.15 0.3

4

= = =

38. Two pipes of lengths 2500 m each and diameters 80 cm and 60c m respectively, are connected in parallel.

The coefficient of friction for each pipe is 0.006 and the total flow is 250 litres/s. The rates of flow in the

pipes are nearly

(A) 3 30.17 m s and 0.1 m s

(B) 3 30.23 m s and 0.1 m s

(C) 3 30.17 m s and 0.4 m s

(D) 3 30.23 m s and 0.4 m s

Answer: (A)

Sol: Coefficient of friction ( )1 2f ' f ' 0.006= =

Friction factor ( )1 2f f 4 0.006 0.024= = =

In parallel flow head loss is same in both pipes

1 2f f

2 2 2 2

1 1 1 2 2 2 1 2

5 5 5 5

1 2 1 2

h h

f L Q f L Q Q

12.1d 12.1d d d

=

= =

( ) ( )

5 22 2

1 2 1

5 5

2

1 2

Q Q Q 0.8

Q 0.60.8 0.6

Q 2.053Q

= =

=

1 2

2 2

3 3

2

3

1

Q Q Q

2502.053Q Q

1000

Q 0.082 m sec 0.1m sec

Q 2.053 0.082 0.168 0.17m sec

= +

= +

=

= =

So, answer is option ‘A’.

Q 250 lines sec=

1L 2500 m=

1d 80 cm=

2d 60 cm=

2L 2500 m=



18

39. A fluid of mass density 31790 kg m and viscosity 22.1 Ns m flows at a velocity of 3 m/s in a 6 cm

diameter pipe. The head loss over a length of 12 m pipe will be nearly

(A) 62.0 m

(B) 54.0 m

(C) 46.5 m

(D) 38.5 m

Answer: (D)

Sol: Density of fluid ( ) 31790 kg m =

Viscosity of fluid ( ) 22.1N.sec m =

Velocity of fluid (V) = 3m/sec

Diameter of pipe ( )d 6cm 0.06m= =

length of pipe ( ) 12m=

Head loss in a pipe ( )fh ?=

Reynolds number ( )e

VD 1790 3 0.06R 153.43 2000

2.1

= = =

Hence, the flow is laminar.

According to hagen Poiseuille’s equation,

( )f 2 2

32 VL 32 2.1 3 12h 38.26 38.5m

gd 1790 9.81 0.06

= = =

So, answer is option ‘D’.

40. Which of the following characteristics regarding fluid kinematics is/are correct?

1. Streamline represents an imaginary curve in the flow field so that the tangent to the curve at any

point represents the direction of instantaneous velocity at that point.

2. Path lines, streamlines and streak lines are identical in steady flow.

(A) 1 only

(B) 2 only

(C) Both 1 and 2

(D) Neither 1 nor 2

Answer: (C)

Sol: In steam flow, the streamlines remain fixed with respect to the co-ordinate axes. Hence streamlines in

steady flow also represent the path lines and streak lines.



19

Streamline represents an imaginary curve in the flow field so that the tangent to the curve at any point

represents the direction of instantaneous velocity at that point.

Hence both statements are correct. So, answer is option ‘C’.

41. To maintain 0.08 m3/s flow of petrol with a specific gravity of 0.7, through a steel pipe of 0.3 m

diameter and 800 m length, with coefficient of friction of 0.0025 in the Darcy relation, the power required

will be nearly

(A) 0.6 kW

(B) 1.0 kW

(C) 2.6 kW

(D) 3.0 kW

Answer: (B)

Sol: Flow rate of petrol ( ) 3Q 0.08 m sec=

Specific gravity of petrol (s) = 0.7

Diameter of pipe (d) = 0.3m

Length of pipe ( ) 800m=

Coefficient of friction ( )f ' 0.0025=

Friction factor ( )f 4f ' 4 0.0025 0.01= = =

Head loss according to Darcy Weisbach equation is

2 2 2

f 5 5

fLV fLQ 0.01 800 0.08h 1.741m

2gd 12.1d 12.1 0.3

= = = =

Powered required,

fP Qh 0.7 1000 9.81 0.08 1.741

0.956kW 1.0kW

= =

=

42. The diameter of a nozzle d for maximum transmission of power through it, is

(A)

15 4D

8fL

(B)

15 2D

8fL

(C)

15 48D

fL



20

(D)

15 28D

fL

Where:

D = Diameter of pipe

f = Coefficient of friction

L = Length of pipe

Answer: (A)

Sol:

H = heat available for pipe

fH = head loss in pipe

For maximum power transmission, f

HH

3=

Head available for nozzle f

H 2HH H H

3 3= − = − =

Velocity of jet in nozzle ( )nozzle

2H 4gHV 2g

3 3

= =

Head loss in pipe 2

f 5

fLQH

12.1D=

5 5

fH 12.1D H 12 1DQ

fL 3 fL

= =

Since discharge is same, pipe nozzleQ Q=

52

nozzle

5 5 22 4

H 12.1Dd V

3 fL 4

H 12.1D 4gH H 12.1D 16gHd d

3fL 4 3 3fL 16 3

=

= =

H

D d

Pipe

Nozzle



21

15 4

2

15 4

12.1 D 16d

16 9.81 fL

Dd

8FL

=

=


43. A piston-cylinder device with air at an initial temperature of 30° undergoes and expansion process for

which pressure and volume are related as given below:

P(kPa) 100 37.9 14.4

( )3V m 0.1 0.2 0.4

The work done by the system for n= 1.4 will be

(A) 4.8 kJ

(B) 6.8 kJ

(C) 8.4 kJ

(D) 10.6 kJ

Answer: (D)

Sol: Given device is piston-cylinder, hence it is closed system. In closed system work done by the system is

given by, i i f fP V P VW

n 1

−=

−

It is given that, n = 1.4

And 1

3

1

P 100 kPa

V 0.1m

=

=

2

3

2

P 37.9kPa

V 0.2 m

=

=

3

3

3

P 14.4 kPa

V 0.4 m

=

=

Work done during process changing its state from ‘1’ to ‘3’ is

( ) ( )( )

1 1 3 31 3

3 3

P V P VW

n 1

100 10 0.1 14.4 10 0.410.6 kJ

1.4 1

−

−=

−

− = =

−

44. A domestic food freezer maintains a temperature of –15°C. The ambient air temperature is 30°C. If heat

leaks into the freezer at the continuous rate of 1.75 kJ/s, the least power necessary to pump this heat out

continuously will be nearly

(A) 0.1 kW

(B) 0.2 kW

P

V

1

23



22

(C) 0.3 kW

(D) 0.4 kW

Answer: (C)

Sol: Given that,

2

1

T 15 C 15 273 258K

T 30 C 30 273 303K

= − = − + =

= = + =

It is asked least work required, the work will be minimum

only when refrigerator operates on reversed Carnot cycle.

For reversed Carnot cycle,

( )

( )

( )

2 2

1 2

T QCOP

T T W

258 1.75

303 258 W

W 0.3kW

= =−

=−

=

Hence, correct answer is option ‘C’.

45. An ideal gas is flowing through an insulated pipe at the rate of 3 kg/s. There is a 10% pressure drop from

an inlet to exit of the pipe. The values of R = 0.287 kJ/kg.K and To = 300 K. The rate of energy loss for

the pressure drop due to friction, will be nearly

(A) 34 kW

(B) 30 kW

(C) 26 kW

(D) 22 kW

Answer: (C)

Sol:

Mass flow rat of gas ( )m 3 kg sec=

Atmospheric temperature ( )0T 300K= , R 0.287 kJ kgK=

Pressure at exit of pipe ( )2P 0.9 Pressure= pressure at entrance of pipe ( )1P

2 1P 0.9 P=

Ref

1Q

1T 30 C=

( )2Q 1.75 kJ sec kW=

2T 15 C= −

W ?=

Insulated pipe

1 2



23

According to Gauss-Stodola theorem the rate of loss of available energy (exergy) (or) Irreversibility in a

flow and non-flow process is

( ) ( )( ) ( ) ( ) ( )( )0 0universe system surroundingI T S T S S= = +

Since there is no heat exchange with the surrounding

( )surroundings

S 0 =

( ) ( )( )0 systemI T S =

Applying steady flow energy equation between state ‘1’ and state ‘2’

2 2

1 21 1 2 2

C Ch gz Q h gz w

2000 2000+ + + = + + +

Neglecting change in Kinetic energies and potential energies

1 2 1 2i.e., C C and z z and no work and= =

Het transfers during the process the above equation becomes, 1 2h h=

Since gas is ideal, ph C T=

p p 2C T C T=

( )

( )( )

2 2psystem

1 1

1

1

T PS mc mR n

T P

0.9 P0 3 0.287 n 0.0861kJ K

P

= −

= − = +

( )( )I 300 0.0861 25.83kW 26kW= =

46. A cyclic heat engine operates between a source temperature of 800°C and a sink temperature of 30°C. The

least rate of heat rejection per kW net output of engine will be nearly

(A) 0.2 kW

(B) 0.4 kW

(C) 0.6 kW

(D) 0.8 kW

Answer: (B)

Sol: Least amount of heat rejection occurs when heat engine works on Carnot cycle.

For Carnot engine,

( )1 2 net

H.E

1 1

T T W

T Q

− = =



24

1

2

T 800 273 1073K

T 30 273 303K

= + =

= + =

( )

( ) 1

1

net 1 2

2

2

1073 303 1Q 1.39 kW

1073 Q

W Q Q

1 1.39 Q

Q 1.39 1 0.39 kW 0.4 kW

−= =

= −

= −

= − =

47. A fictitious pressure that, if it acted on the piston during the entire power stroke, would produce the same

amount of net work as that produced during the actual cycle is called

(A) Quasi equivalent pressure

(B) Mean equivalent pressure

(C) Mean effective pressure

(D) Quasi static pressure

Answer: (C)

Sol: Mean effective pressure is a frictions pressure that, if it acted on the piston during the entire power stroke,

would produce the same amount of network as that produced during the actual cycle.

net

m

W mean effective pressure Piston area stroke

P Displacement volume

=

=

48. An ideal cycle based on the concept of combination of two heat transfer processes, one at constant volume

and the other at constant pressure, is called

(A) Otto cycle

(B) Dual cycle

(C) Diesel cycle

H.E

1Q

1T 800 C=

2Q

2T 15 C= −

netW 1kW=

minVmaxV V

netW

P

V

P

minVmaxV

mP



25

(D) Carnot cycle

Answer: (B)

Sol: Dual cycle for an air standard I.C engine is as follows.

In dual cycle heat is supplied during constant volume and constant pressure processes. Hence answer is

option (B).

49. The ideal thermodynamic cycle for the development of gas-turbine engine is

(A) Otto

(B) Stirling

(C) Ericsson

(D) Brayton

Answer: (D)

Sol: An air standard cycle for gas turbine engine is Brayton (or) Joule cycle. So, answer is option ‘D’.

50. If the pressure at exhaust from the turbine is the saturation pressure corresponding to the temperature

desired in the process heater, such a turbine is called

(A) Condensing turbine

(B) Extraction turbine

(C) Pass out turbine

(D) Back pressure turbine

Answer: (D)

Sol: Many industries such as paper, sugar, and textile mills, chemical plants etc., require process heat, in the

form of saturated steam at specified temperatures for heating and drying. Steam is found convenient for

may such applications since isothermal conditions can be maintained during seating on one side while the

steam is condensing on the other side. This steam is obtained by replacing condenser with process heater

in Rankine cycle. The exhaust steam from the turbine severs as process steam. In this arrangement, the

process of the exhaust steam is kept equal to the saturation pressure corresponding to the saturation

V

P

Q

Q

1

52

3 4



26

temperature of process steam desired in the heater. Such a turbine having a specifically desired exhaust

pressure, is referred to as a back-pressure turbine. Hence correct option is ‘D’.

51. The purpose of providing fins on heat transfer surface is to increase

(A) Temperature gradient so as to enhance heat transfer by convection

(B) Effective surface area to promote rate of heat transfer by convection

(C) Turbulence in flow for enhancing heat transfer by convection

(D) Pressure drop of the fluid

Answer: (B)

Sol: Convective heat transfer is given by

( )Q hA T=

From the above we can say that, convective heat transfer can be increased by increasing effective surface

area. This will be done by providing fins on the heat transfer surface. Hence correct answer is option ‘B’.

52. For fully developed laminar pipe flow the average velocity is

(A) One-half of the maximum velocity

(B) One-third of the maximum velocity

(C) One-fourth of the maximum velocity

(D) Two-third of the maximum velocity

Answer: (A)

Sol: In a fully developed laminar flow through circular pipes velocity distribution according to Hagen

Poiseuille’s equation is

( )2 21 dpu R r

4 dx

= − −

For maximum velocity,

du

0dr

=

( )

( )2

max

1 dP2r 0

4 dx

r 0

1 dPu R

4 dx

− − =

=

− =

Discharge through pipe is ( )( )( )2

max2u R1 dP

Q R8 dx 2

− = =

r 0=

r R=

maxu



27

pipe AverageQ A u =

2maxaverage

maxaverage

u RR u

2

uu

2

=

=


53. The overall heat transfer coefficient due to convection and radiation for a steam maintained at 200°C

running in a large room at 30°C is 217.95W m K. If the emissivity of the pipe surface is 0.8; the value of

8 2 45.67 10 W m K ;− = the heat transfer coefficient due to radiation will be nearly

(A) 217 W m K

(B) 214 W m K

(C) 211 W m K

(D) 28W m K

Answer: (C)

Sol: Heat can be transferred from steam to large room by convection and radiation.

Emissivity of pipe ( )1 0.8 =

Heat transfer by radiation from pipe to room is as follows,

( )

( )

( )

4 4

1 2

rad

1 2

1 1 1 12 2 2

12 11

4 4

1 2 1

R ad

1 2

1 12 2 2

T TQ

1 11

A A F A

F 1sin ce F 0 since convex surface

T T AQ

A 11 11

F A

−= − −

+ +

= =

− = −

− + +

Pipe 1T 200 C=

Steam

Large room

230°C=Td



28

Since ‘A2’ is large, 1

2

A0

A→

( )

( )

4 4

R ad 1 1 2 1

8 4 4 2R ad

1

Q T T A

Q0.8 5.67 10 473 303 1888.14 W m

A

−

= −

= − =

To find equivalent convective heat transfer coefficient,

( )

( )

R ad conv

1 1

1 2

2

Q Q

A A

1888.14 h T T

1888.14h 11.1 W m K

170

=

= −

= =

Answer is option ‘C’.

54. Large heat transfer coefficients for vapour condensation can be achieved by promoting

(A) Film condensation

(B) Dropwise condensation

(C) Cloud condensation

(D) Dew condensation

Answer: (B)

Sol: It traces of oil are present during the condensation of steam on a highly polished surface, the film of

condensate is broken into droplets, and the situation is known as dropwise condensation. The present of

condensate acts as a barrier to heat transfer from the vapour to the metal surface, and the drop wise

condensation offers much less resistance to heat flow on the vapour side than the film wise condensation.

So, option ‘B’ is correct.

55. Which one of the following valves is provided for starting the engine manually, during cold weather

conditions?

(A) Starting jet valve

(B) Compensating jet valve

(C) Choke valve

(D) Auxiliary air valve

Answer: (C)

Sol: The choke value is used when starting a cold engine. When doing a cold start, the choke value should be

closed to limit the amount of air going in. This increases the amount of fuel in the cylinder, and helps to



29

keep the engine running, while it is trying to warm up. Once the engine has warmed up, a temperature

sensing spring slowly opens the choke valve to allow the engine to breath fully. Hence the answer is ‘C’.

56. A 4-cylinder, 4-stroke single acting petrol engine consumes 6 kg of fuel per minute at 800 rpm when the

air fuel ratio of the mixture supplied is 9 : 1. The temperature is 650 K and pressure is 12.5 bar at the end

of compression stroke. Take R = 300 Nm/kg.K, diameter of cylinder as 8 cm, stroke of cylinder as 10 cm.

The compression ratio will be nearly

(A) 6.2

(B) 5.7

(C) 5.2

(D) 4.6

Answer: (A)

Sol: Petrol engine will work on Otto cycle, 1-2 is an isentropic

compression process of petrol & air mixture from state

‘1’ to state ‘2’.

( )

C 2

S 1 2

V V clearance volume

V Stroke volume V V

= =

= −

Given that, 2 2P 12.5 bar, T 650K, R 300 N.m kgK= = =

fuel airm 6kg min and m 6 9 54 kg min= = =

Mass flow rate of air and petrol (fuel) supplied per cylinder 54 6 60

15kg min4 4

+ = = =

According to ideal gas equation,

( )2 2 2Air fuel

5

2

3

2 C

P V m R T

1512.5 10 V 300 650

60

V 0.039m sec V

+=

=

= =

Diameter of cylinder (D) = 8cm, Stroke of cylinder (L) = 10cm

Volume flow rate of air in each cylinder is ( )20.08 0.1 800

ALN 4

2 60 2 60

= =

3 3

SV 3.35 10 m sec−=

S CV V this is practically not correct

Hence taking ( )Air Fuel

15kg 15m kg sec

hour 3600+

= =

T

SCV

SV

3

1

2

4



30

5

2

4 3

2 C

1512.5 10 V 300 650

3600

V V 6.5 10 m sec−

=

= =

Compression ratio ( )( )

( )

4 3

C Sc 4

C

6.5 10 3.35 10V Vr 6.15 6.2

V 6.5 10

− −

−

+ += = =

Hence answer is option (A).

57. Ice is formed at 0°C from water at 20°C. The temperature of the brine is –8°C. The refrigeration cycle

used is perfect reversed Carnot cycle. Latent heat of ice = 335 kJ/kg, and pwc 4.18.= The ice formed per

kWh will be nearly

(A) 81.4 kg

(B) 76.4 kg

(C) 71.8 kg

(D) 68.8 kg

Answer: (A)

Sol: Heat to be removed from water ( )( )( )( ) ( )

( )

wP

water water

water ice

m C 20 0 latent of heat

4.18 20 0 335 m 418.6 m

m mass of ice formal m kg

= − +

= − + =

= =

Given that power (work) input 1kWhr 3600 kJ= =

Since the brine is available at -8°C and water should be converted into ice at 0°C from 20°C. Hence

refrigerator should operate between –8°C and 20°C and since it is reversed Carnot engine.

( )( )

( )( )ref

8 273COP 9.464

20 8

− + = =

− −

( )ref

Heat to be removedCOP 9.464

workinput= =

( ) ice418.6 m

9.4643600

=

( )ice

9.464 3600m 81.39kg 81.4 kg

418.6

= =




31

58. A Freon 12 simple saturation cycle operates at temperatures of 35°C and –15°C for the condenser and

evaporator. If the refrigeration effect produced by the cycle is 111.5 kJ/kg and the work required by the

compressor is 27.2 kJ/kg, the value of COP will be nearly

(A) 4.1

(B) 3.6

(C) 3.1

(D) 2.6

Answer: (A)

Sol: 2Q refrigeration effect produced by the cycle

Heat to be removed

111.5kJ kg

=

=

=

Work given to the compressor,

W 27.2kJ kg=

( ) 2

Ref

Q 111.5COP 4.099 4.1

W 27.2 = = =


59. A cold storage is to be maintained at –5°C while the surroundings are at 35°C. The heat leakage from the

surroundings into the cold storage is estimated to be 29 kW. The actual COP of the refrigeration plant

used is one-third of an ideal plant working between the same temperatures. The power required to drive

the plant will be

(A) 10 kW

(B) 11 kW

(C) 12 kW

(D) 13 kW

Answer: (D)

Sol: Heat leakage from the surrounding

= Heat to be removed

2Q 29 kW= =

Theoretical COP = Ideal COP

( )( )

( )

( )( )2

Ideal1 2

5 273TCOP 6.7

T T 35 5

− += = = =

− − −

1T 35 C=

Ref

1Q

2Q

2T 15 C= −

Ref

1Q

1T 35 C=

2Q

2T 5 C= −

W



32

Given that actual ( ) ( )

( )

act Ideal

act

1COP COP

3

1COP 6.7 2.233

3

=

= =

( ) 2

act

QCOP ,

W

here W work required to driverefrigerator

292.233

W

29W 12.98 13kW

2.233

=

=

=

= =

So, answer is option ‘D’.

60. A single acting two-stage air compressor deals with 4 m3/min of air at 1.013 bar and 15°C with a speed of

250 rpm. The delivery pressure is 80 bar. If the inter cooling is complete, the intermittent pressure after

first stage will be

(A) 9 bar

(B) 8 bar

(C) 7 bar

(D) 6 bar

Answer: (A)

Sol: In multistage compression with inter cooling complete (or) Inter cooling is perfect i.e., air is cooled at

constant pressure to its initial pressure. The intermediate pressure will be i 1 2P P P=

1P = Initial pressure (or) Atmospheric pressure = 1.013 bar

2P =Delivery pressure = 80 bar

iP 1.013 80 9 bar = =


61. Which of the following are the advantages of Nano-composite materials?

1. Decreased thermal expansion coefficients

2. Higher residual stress

3. Reduced gas permeability

4. Increased solvent resistance

(A) 1, 2 and 3 only

(B) 1, 3 and 4 only



33

(C) 1, 2 and 4 only

(D) 2, 3 and 4 only

Answer: (B)

62. A rod of copper originally 305 mm long is pulled in tension with a stress of 276 MPa. If the modulus of

elasticity is 110 GPa and the deformation is entirely elastic, the resultant elongation will be nearly

(A) 1.0 mm

(B) 0.8 mm

(C) 0.6 mm

(D) 0.4 mm

Answer: (B)

Sol: Given data,

305 mm, 276 MPa

E 110GPa, ?

= =

= =

3

276L 305 0.76mm 0.8mm

E 110 10

= = =

63. A 1.25 cm diameter steel bar is subjected to a load of 2500 kg. The stress induced in the bar will be

(A) 200 MPa

(B) 210 MPa

(C) 220 MPa

(D) 230 MPa

Answer: (A)

Sol: Given data,

2

2

d 1.25 cm 12.5mm,

P 2500 kg 2500 9.81N

P 4P

A d

4 2500 9.81200MPa

12.5

= =

= =

= =

= =

64. The maximum energy which can be stored in a body up to the elastic limit is called

(A) Proof resilience

(B) Modulus of resilience

(C) Impact toughness

(D) Endurance strength

Answer: (A)



34

65. A cast iron bed plate for a pump has a length of 100 μm If the Young's modulus of cast iron is 210 GN/m2

and the specific surface energy is 10 J/m2, the fracture strength required will be nearly

(A) 8 21.0 10 N m

(B) 8 21.2 10 N m

(C) 8 21.4 10 N m

(D) 8 21.6 10 N m

Answer: (B)

Sol: Given data,

9 2

2 6

E 210 10 N mm

r 10J m , C 100 10 m−

=

= =

Fracture strength

0.52E

C

=

0.59

6

2

8 2

8 2

2 210 10 10

100 10

115,624,457.6 N m

1.15 10 N m

1.2 10 N m

−

=

=

=

=

66. A 13 mm diameter tensile specimen has 50 mm gauge length. If the load corresponding to the 0.2% offset

is 6800 kg, the yield stress will be nearly

(A) 231 kg mm

(B) 243 kg mm

(C) 251 kg mm

(D) 263 kg mm

Answer: (C)

Sol: Given data,

2

yt2

d 13mm, L 50mm

P 6800 kg

P 680051kg mm

A13

4

= =

=

= = =



35

67. The magnitude of the velocity of any point on the kinematic link relative to the other point on the same

kinematic link is the product of

(A) A square root of an angular velocity of the link and the distance between the two points under

consideration

(B) An angular velocity of the link and the square of distance between the two points under

consideration

(C) A square of an angular velocity of the link and the distance between the two points under

consideration

(D) An angular velocity of the link and the distance between the two points under consideration

Answer: (D)

68. In a mechanism, the number of Instantaneous centre (I-centres) N is

(A) ( )n n 1

2

−

(B) ( )n 2n 1

2

−

(C) ( )2n n 1

3

−

(D) ( )n 2n 1

3

−

where : n = Number of links

Answer: (A)

69. In cycloidal motion of cam follower, the maximum acceleration of follower motion maxf at4

= is

(A) 2

2

h

2

(B) 2

2

3h

2

(C) 2

2

2h

(D) 2

2

3h

where:

h = Maximum flower displacement



36

ω = Angular velocity of cam

ϕ = Angle for the maximum follower displacement for cam rotation

Answer: (C)

Sol: Displacement equation,

0 0

1 2x h sin

2

= −

Velocity, 0 0 0

0 0

dx 1 d 2 2 dV h cos

dt dt 2 dt

h 2V 1 cos

= = −

= −

Acceleration,

( )

2 2

2 2

0 0 0 0

0

2 2

2 2

d x h 2 2 d 2 h 2a sin sin

dt dt

,4

2 h 2 ha . 1

= = =

= =

= =

Hence option ‘C’ is correct.

70. A shaft of span 1m and diameter 25 mm is simply supported at the ends. It carries a 1.5 kN concentrated

load at mid-span. If E is 200 GPa, its fundamental frequency will be nearly

(A) 3.5 Hz

(B) 4.2 Hz

(C) 4.8 Hz

(D) 5.5 Hz

Answer: (D)

Sol: Given data, 9E 200GPa 200 10 Pa, d 25mm 0.025m= = = =

Fundamental natural frequency will be find out by using deflection method i.e., n

1 gf

2=

W 1.5 kN=

d 25 mm=

1m=



37

Where 3 3 3

49

3

W 1.5 10 1

48EI 0.02548 200 10

64

8.1487 10−

= =

=

n 3

1 9.81f 5.52 Hz

2 8.1487 10− = =

So, correct answer is option ‘D’.

71. A vibrating system consists of mass of 50 kg, a spring with stiffness of 30 kN/m and a damper. If damping

is 20% of the critical value, the natural frequency of damped vibrations will be

(A) 16 rad/s

(B) 20 rad/s

(C) 24 rad/s

(D) 28 rad/s

Answer: (C)

Sol: Given data,

m 50 kg, k 30 kN m, 0.2= = =

( )0.5

30.5

2 2

d n

30 101 1 0.2 24 rad sec

50

= − = − =

72. A refrigerator unit having a mass of 35 kg is to be supported on three spring, each having spring stiffness

s. The unit operates at 480 rpm. If only 10% of the shaking force is allowed to transmit to the supporting

structure, the value of stiffness will be nearly

(A) 2.7 N/mm

(B) 3.2 N/mm

(C) 3.7 N/mm

(D) 4.2 N/mm

Answer: (A)

Sol: Given data,

m 35kg, N 480 rpm

0.1, K ?

= =

= =

2

2

2

2

1T.R

1 r

1 10.1 1 r

1 r 0.1

1 r 10

==

−

−= − =

−

− = −



38

( )

n n

n

r 11 3.31

2 480

k603.313.31 3.31 m

= =

= = = =

8071.45

k N m 2690.48 N m 2.7 N mm3

= = =

73. In which one of the following tooth profiles, does the pressure angle remain constant throughout the

engagement of teeth?

(A) Cycloidal

(B) Involute

(C) Conjugate

(D) Epicycloid

Answer: (B)

74. If the axes of the first and last wheels of a compound gear coincide, it is called

(A) Simple gear train

(B) Compound gear train

(C) Epicyclic gear train

(D) Reverted gear grain

Answer: (D)

75. In a reciprocating engine, the force along the connecting rod FQ is

(A) P

2 2

F

n sin−

(B) P

2 2

F

2 n sin−

(C) P

2 2

nF

2 n sin−

(D) p

2 2

nF

n sin−

where:

PF Force on piston

Ln

r

Angle for crank from IDC

=

=

=



39

Answer: (D)

Sol: PQ

FF

nsin sin

sinsin

n

=

=

=

P PQ

2 2

F F .nF

nt sinsin1

n

= =−

−

76. A mass 1m attached to a shaft at radius r1, rotating with angular velocity ω rad/s, can be balanced by

another single mass m2 which is attached to the opposite side of the shaft at radius r2, in the same plane, if

(A) 1 2 2 1m r m r=

(B) 1 2 2 2m r m r=

(C) 1 1 1 2 2 2m r m r =

(D) 1 2 1 2 1 2m r m r =

Answer: (B)

Sol:

If a mass ‘m1’ is placed at distance of ‘ 1r ’ on a rotating shaft with angular velocity ‘ ’, will be subjected

to a centrifugal force ( ) 2

1 1 1 1F m r= in the upward direction. For dynamic balancing of this shaft a mass

‘m2’ is placed at a distance of ‘r2’ on the opposite side of shaft in the same phase then the centrifugal force

produced by this mass ‘m2’ in the vertically downward direction.

i.e., 2

2 2 2F m r= should be equal to 2

1 1 1 1F m r=

2 2

2 2 1 1 1 1 2 2m r m r m r m r = =

Correct answer is option ‘B’.

1F

1m

2m

1r

2r



40

77. For a single cylinder reciprocating engine speed is 500 rpm, stroke is 150 mm, mass of reciprocating parts

is 21 kg; mass of revolving parts is 15 kg at crank radius. If two-thirds of reciprocating masses and all the

revolving masses are balanced, the mass at a radius of 150 mm will be

(A) 7.5 kg

(B) 10.5 kg

(C) 12.5 kg

(D) 14.5 kg

Answer: (D)

Sol: Given data,

2 1

N 500 rpm, 2r 150 mm,

m 21 kg, m 15 kg,

2C ,

3

b 150 mm, B ?

= =

= =

=

= =

( )

( )

1 2Bb m cm r

215 21 75

3B 14.5 kg

150

= +

+

= =

78. If the axes of the rolling of the ship and of the stabilizing rotor are parallel, it will result in

(A) A higher bow and lower stern

(B) A lower bow and higher stern

(C) Turning towards left

(D) No gyroscopic effect

Answer: (D)

Sol: The rolling axes and precision axes are parallel, so gyroscopic couple is zero.

79. Coaxing is a process of

(A) Improving the fatigue properties attained by under-stressing and then raising the stress in small

increments

(B) Decreasing the hardness by full annealing

(C) Increasing the uniaxial tensile strength by heating above recrystallization temperature and

quenching in oil media

(D) Maintaining the ductility of the material by chemical treatment

Answer: (A)



41

Sol: Fatigue limit (or) endurance limit of some metals may be improve by under stressing. A still increase in

the fatigue limit of the same metals will be obtained by doing coaxing process, that is fatigue limit was

said to be coaxed to higher values by a procedure of gradually increasing the applied load in small

increments an allowing a relatively large number of cycles to stress to occur after each new increase in

load. Hence option (A) is correct.

80. According to the distortion-energy theory, the yield strength in shear is

(A) 0.277 times the yield stress

(B) 0.377 times the maximum shear stress

(C) 0.477 times the yield strength in tension

(D) 0.577 times the yield strength in tension

Answer: (D)

Sol: According to Distortion-energy theory,

0.5

2 2

yt 1 2 1 2 = + −

For pure shear, 1 2 = −

yt 13 =

yt

yt 1 yt0.5773

= = =

81. For the prediction of ductile yielding, the theory of failure utilized is

(A) Maximum strain energy theory

(B) Distortion energy theory

(C) Maximum normal strain theory

(D) Mohr theory

Answer: (B)

82. A steel specimen is subjected to the following principal stresses: 120 MPa tensile, 60 MPa tensile and –30

MPa compressive. If the proportionality limit for the steel specimen is 250 MPa; the factor of safety as per

maximum shear stress theory will be nearly

(A) 1.3

(B) 1.7

(C) 2.3

(D) 2.7

Answer: (B)



42

Sol: Given data,

1 2 3

P

120 MPa, 60 MPa and 30 MPa

250 MPa, FOS ?

= = = −

= =

( )

p 2 3 3 1P 1 2safe max , ,

FOS 2 FOS 2 3 2

60 30250 120 60 30 120max , ,

2 FOS 2 2 2

125 12575 FOS 1.66 1.7

FOS 75

− − − = = =

− − − − −=

= = =

83. For which one of the following loading conditions is the standard endurance strength multiplied by a load

factor, eK 0.9?=

(A) Reversed beam bending loads

(B) Reversed axial loads with no bending

(C) Reversed axial loads with intermediate bending

(D) Reversed torsion loads

Answer: (B)

Sol: Since endurance strength is calculated by conducting rotating beam test under completely reversed

bending loads while modifying it so suitable for reversed axial loads we should multiply with load factor.

Load factor 0.85 0.9 for reversed axial loads

Load factor = 0.577 for reversed torsion loads.

Hence answer is option ‘B’.

84. A 120 mm wide uniform plate is to be subjected to a tensile load that has a maximum value of 250 kN and

a minimum value of 100 kN. The properties of the plate material are : endurance limit stress is 225 MPa,

yield point stress is 300 MPa. If the factor of safety based on yield point is 1.5, the thickness of the plate

will be nearly

(A) 12 mm

(B) 14 mm

(C) 16 mm

(D) 18 mm

Answer: (A)

Sol: Given data,

maxb 120 mm ,P 250 kN= =



43

min e

yt

P 100 kN , 225 MPa

300 MPa, FOS 1.5

t ?

= =

= = −

=

mean

V

3 3

mean v

250 100P 175 kN

2

250 100P 75 kN

2

175 10 75 10;

120 t 120 t

+= =

−= =

= =

Soderbergh equation,

mean v

yt e

3 3

1

FOS

175 10 75 10 1

120 t 300 120 t 325 1.5

4.8611 2.777 1

t t 1.5

t 11.45mm 12 mm

+ =

+ =

+ =

=

85. A steel connecting rod having 2 2

ut ytS 1000 MN m , S 900 MN m= = is subjected to a completely reversed

axial load of 50 kN. By neglecting any column action, if the values of e b ak 0.85, k 0.9, k 0.82,= = =

tk 1.5, q 0.6 and N 2,= = = the diameter of the rod will be nearly

(A) 20 mm

(B) 23 mm

(C) 26 mm

(D) 29 mm

Answer: (B)

Sol: Given data,

ut

yt

max,min

S 1000 MPa

S 500MPa

P 50 kN

=

=

=

Corrected endurance strength,

( )a b e ut

e

f

a b e t

k .k .k 0.5

k

k 0.82, k 0.9, k 0.85, q 0.6, k 1.5

=

= = = = =

f

t

k 1q

k 1

−=

−



44

( )fk 0.6 1.5 1 1 1.3= − + =

( )

e

0.82 0.9 0.85 0.5 1000241.269 MPa

1.3

= =

For reverse load, mean v0, amplitude stress = = =

mean v

yt e

3

2

1

FOS

50 10 10

2241.269 d

4

d 22.9mm 23 mm

+ =

+ =

=

86. During crushing or bearing failure of riveted joints

(A) The holes in the plates become oval shaped and joints become loose

(B) There is tearing of the plate at an edge

(C) The plates will crack in radial directions and joints fail

(D) The rivet heads will shear out by applied stress

Answer: (A)

87. The double riveted joint with two cover plates for boiler shell is 1.5 m in diameter subjected to steam

pressure of 1 MPa. If the joint efficiency is 75%, allowable tensile stress in the plate is 83 MN/m2,

compressive stress is 138 MN/m2 and shear stress in the rivet is 55 MN/m2, the diameter of rivet hole will

be nearly

(A) 8 mm

(B) 22 mm

(C) 36 mm

(D) 52 mm

Answer: (B)

Sol: Given data,

t

c

D 1.5 m, p 1MPa

0.75. 83 MPa

138 MPa, 55 MPa

= =

= =

= =

t

pDt C.A

2 .

1 1500t 2 14.048 mm

2 83 0.75

= +

= + =

d 6 t 6 14.04 22.4 mm= = =



45

88. A bearing supports a radial load of 7000 N and a thrust load of 2100 N. The desired life of the ball bearing

is 160 × 106 revolutions at 300 rpm. If the load is uniform and steady, service factor is 1, radial factor is

0.65, thrust factor is 3.5, k = 3 and rotational factor is 1, the basic dynamic load rating of a bearing will be

nearly

(A) 65 kN

(B) 75 kN

(C) 85 kN

(D) 95 kN

Answer: (A)

Sol: Given data,

r T

6

10

F 7000 N, F 2100 N

L 160 10 rev, N 300 rpm

X 0.65, Y 3.5

S.F 1, k 3

= =

= =

= =

= =

( )

( ) ( )

e r aF XF YF S.F

0.65 7000 3.5 2100

11,900

= +

= +

=

( )

k

10

1 3

CL

P

C 160 .11900 64.6kN 65 kN

=

= =

89. A solid cast iron disk, 1 m in diameter and 0.2 m thick is used as a flywheel. It is rotating at 350 rpm. It is

brought to rest in 1.5 s by means of a brake. If the mass density of cast iron is 7200 kg/m3, the torque

exerted by the brake will be nearly

(A) 3.5 kN m

(B) 4.5 kN m

(C) 5.3 kN m

(D) 6.3 kNm

Answer: (A)

Sol: Given data,

3

D 1m, t 0.2m

N 350rpm, t 1.5sec

7200 kg m

= =

= =

=

2Mass of disc D t 7200 1 0.2 1130.97 kg

4 4

= = =



46

Mass moment of Inertia 2 2mR 1130.97 0.5

2 2

= =

2I 141.37 kg m= −

22 350

24.43 rad sect 60 1.5

= = =

Torque I 141.37 24.43 3454.32N m 3.5 kN-m= = = − =

90. The torque transmitting capacity of friction clutches can be increased by

(A) Use of friction material with a lower coefficient of friction

(B) Decreasing the mean radius of the friction disk

(C) Increasing the mean radius of the friction disk

(D) Decreasing the plate pressure

Answer: (C)

91. The ideal gas-refrigeration cycle is the same as the

(A) Brayton cycle

(B) Reversed Brayton cycle

(C) Vapour compression refrigeration cycle

(D) Vapour absorption refrigeration cycle

Answer: (B)

Sol: Ideal gas-refrigeration cycle works on reversed Brayton (or) Joule cycle. Hence answer is option ‘B’.

92. If the atmospheric conditions are 20° C, 1.013 bar and specific humidity of 0.0095 kg/kg dry air, the

partial pressure of vapour will be nearly

(A) 0.076 bar

(B) 0.056 bar

(C) 0.036 bar

(D) 0.016 bar

Answer: (D)

Sol: Given that, total pressure of air (P) = 1.013 bar

Specific humidity of air ( ) ( )W 0.0095 kg kg of dry air=

We know that, V

V

0.622P

P P=

−

Where, VP = Partial pressure of water vapour



47

( ) ( )

V

V

3

V V

3

V

V

0.622 P0.0095

1.013 P

9.6235 10 0.0095 P 0.622P

0.6315 P 9.6235 10

P 0.0152 bar

−

−

=

−

− =

=

=

So nearest answer is ‘D’.

93. In air-conditioning systems, air may be cooled and dehumidified by

1. Spraying chilled water to air in the form of fine mist.

2. Circulating chilled water or brine in a tube placed across the air flow.

3. Placing the evaporator coil across the air flow.

Which of the above statements are correct?

(A) 1 and 2 only

(B) 1 and 3 only

(C) 2 and 3 only

(D) 1, 2 and 3

Answer: (C)

Sol: In order to cool and dehumidify air the temperature of cooling medium should be below ‘DPT’ of air. This

will be possible only with help of chilled water (or) brine in a tube (or) by phasing an evaporator coil

across the air flow. But if we spray chilled water the air will cool but water vapour may not condense from

air and also there is a chance of vaporizing of water droplets from chilled water due to heat energy taken

from air. Hence option ‘C’ is correct.

94. A duct of rectangular cross-section 600 mm × 400 mm carries 90 m3/ min of air having density of

1.2kg/m3. When the quantity of air in both cases is same, the equivalent diameter of a circular duct will be

nearly

(A) 0.86 m

(B) 0.76 m

(C) 0.64 m

(D) 0.54 m

Answer: (D)

Sol: Quantity of air is same for both rectangular and circular ducts

i.e., ( ) ( )rectangular circular ductduct

air rectangle air air circle air

m m

A V A V

=

=



48

Assume air'V ' is same in both ducts

( )

Rectangle circle

2

A A

0.6 0.4 d4

d 0.5m, so nearest by answer is option 'D'

=

=

=

95. A room having dimensions of 5 m × 5 m × 3 m contains air at 25° C and 100 kPa at a relative humidity of

75%. The corresponding value of ps is 3.169 kPa. The partial pressure of dry air will be nearly

(A) 106 kPa

(B) 98 kPa

(C) 86 kPa

(D) 78 kPa

Answer: (B)

Sol: Given that,

Total pressure of air ( )P 100kPa=

Temperature of air (T)=25°C

Relative humidity ( ) 75% 0.75 = =

Saturation pressure of water vapour ( )satat 25 C is P 3.169 kPa =

We know that ( ) v

sat

P

P =

vP0.75

3.169=

Partial pressure water vapour, ( )vP 2.376kPa=

Total pressure of air ( ) v dP P P= +

Where dP Partial= pressure of dry air

d

d

100 2.376 P

P 97.62 98 kPa

= +

=

So, answer is option is (B)

96. A measure of feeling warmth or coolness by the human body in response to the air temperature, moisture

content and air motion is called

(A) Dry bulb temperature

(B) Effective temperature



49

(C) Wet bulb temperature

(D) Dew point temperature

Answer: (B)

Sol: The degree of warmth (or) cold felt by a human body depends mainly on following factory

a. Pry bulb temperature

b. Relative humidity

c. Air velocity.

Effective temperature is used to evaluate the combined effect of all these factors. It is defined as that index

which correlates the combined effects of air temperature, air velocity and relative humidity on human

body.

The numerical value of effective temperature is made equal to temperature of still saturated air (i.e., 5 8

m/min air velocity) which produces same sensation of warmth (or) coolness as produced under the given

conditions.

97. While designing a Pelton wheel, the velocity of wheel ‘u’ is

(A) uK gH

(B) u2K gH

(C) uK 2gH

(D) u2K 2gH

where:

uK = Speed ratio

H = Net head on turbine

g = Gravity

Answer: (C)

Sol: In Pelton wheel,

Speed ratio ( )( )( )

( )u

Bucket speed u or velocity of wheelK

Theoretical velocity of Jet V=

V 2gH,= where H=net head on turbine

u

u

uK

2gH

u K 2gH

=

=

So, correct answer is option ‘C’.



50

98. The turbines of the same shape will have the same

(A) Thomas number

(B) Reynolds number

(C) Specific speed

(D) Rotational speed

Answer: (C)

Sol: Specific speed of turbine is defined as the speed of a geometrical similar turbine of a such size as to

generate one unit of power under one unit of head at maximum efficiency. For geometrically similar

turbines (of same shape) specific speed should be same. So, answer is option ‘C’.

99. A centrifugal pump is required to lift 30.0125m s of water from a well with depth 30 m. If rating of the

pump motor is 5 kW, and the density of water is 31000kg m , the efficiency of the pump will be nearly

(A) 82%

(B) 74%

(C) 66%

(D) 58%

Answer: (B)

Sol: Discharge of water from well ( ) 3Q 0.0125 m sec= , depth of well ( )H 30m=

Power required to pump the water to a height of 30 m is

( )

( )

( )

P QH gQH

1000 9.81 0.0125 30

P 3678.75 watts 3.678 kW

= =

=

= =

But pump is operated by a motor of 5 kW

Remaining 1.322 kW accounts for mechanical and other hydraulic losses.

pump

3.678100 73.56% 74%

5 = =


100. An inward flow reaction turbine has an external diameter of 1 m and its breadth at inlet is 250 mm. If the

velocity of flow at inlet is 2 m/s and 10% of the area of flow is blocked by blade thickness, the weight of

water passing through the turbine will be nearly

(A) 10 kN/s

(B) 14 kN/s



51

(C) 18 kN/s

(D) 22 kN/s

Answer: (B)

Sol: Given that, diameter of turbine ( )PD 1m=

Width of turbine blade (B1) = 250mm = 0.25 m at inlet

Velocity flow at inlet ( )1f

V 2m sec=

Because of thickness of blade, flow is blocked by 10%

Flow rate of water at ( ) ( )11 1 finlet Q D B v 0.9=

31 0.25 2 0.9 1.414 m sec= =

Weight of water at inlet = gQ 1000 9.81 1.484=

So, answer is option ‘B’ 13.87 kN sec 14kN sec=

101. The process of abstracting steam at a certain section of the turbine and subsequently using it for heating

feed water supplied to the boiler is called

(A) Reheating

(B) Regeneration

(C) Bleeding

(D) Binary vapour cycle

Answer: (C)

Sol: Process of abstracting steam at a certain section of the turbine is called bleeding, and this bled steam can

used for feed water heating before supplying to the boiler. So, answer is option ‘C’.

102. When blade speed ratio is zero, no work is done because the distance travelled by the blade is zero even if

the torque on the blade

(A) is minimum

(B) is zero

(C) is maximum

(D) remains the same

Answer: (C)

Sol: Blade speed ratio ( )

( )b

Blade velocity V

Steam velocity at inlet V= =

Power developed by Turbine (P) = (Force exerted by steam on blades) (Blade velocity)



52

( )bP Force V

P Force r

P T

=

=

=

Given that, blade speed ratio is zero, that means bV 0 0= =

So, from P T=

Power will be zero even when torque on the blades is maximum, so answer is option ‘C’.

103. In an axial flow turbine, the utilization factor has an absolute maximum value of unity, for any degree of

reaction if the value of nozzle angle α is

(A) 270°

(B) 180°

(C) 90°

(D) 0°

Answer: (C)

Sol: In a reaction axial flow turbine, the maximum utilization factor

2

1max 12

1

cos; where Nozzle angle

1 Rsin

= =

− , R=Degree of reaction

The value of ( )max

will be ‘1’ irrespective of Degree of reaction (R) when nozzle angle ( )1is 0 =

So answer is option ‘D’.

104. Which of the following are essential for a good combustion chamber of turbojet engine?

1. It should allow complete combustion of fuel.

2. It should maintain sufficiently high temperatures in the zone of combustion in addition to proper

atomization of fuel thus leading to continuous combustion.

3. It should not have high rate of combustion.

4. The pressure drop should be as small as possible.

(A) 1, 2 and 4 only

(B) 1, 2 and 3 only

(C) 1, 3 and 4 only

(D) 2, 3 and 4 only

Answer: (A)



53

105. If fm is the mass of fuel supplied per kg of air in one second, then the mass of gases leaving the nozzle of

turbojet will be

(A) ( )f1 m kg s−

(B) ( )f

1kg s

1 m+

(C) ( )f1 m kg s+

(D) ( )f

1kg s

1 m−

Answer: (C)

Sol: Mass flowrate of air to the turbojet = 1 kg/sec

Mass flow rate of fuel supplied = fm kg sec

Mass flow rate of gases leaving the nozzle of turbojet ( )f1 m kg sec= +

So, answer is option ‘C’.

106. Which one of the following may be considered as a single cylinder two stroke reciprocating engine

running at 2400 rpm to 2700 rpm for rapid chain of impulses?

(A) Turbo jet

(B) Pulse jet

(C) Ram jet

(D) Athodyd jet

Answer: (B)

107. In jet propulsion of ships, when the inlet orifices are at right angles to the direction of motion of the ships,

the efficiency of propulsion η is

(A) 22u

V u+

(B) ( )

2

2Vu

V u+

Turbo jet

1kg sec ( )f1 m kg sec+

fm



54

(C) ( )

2

2u

V u+

(D) 2Vu

V 2u+

where:

V = Absolute velocity of the issuing jet

u = Velocity of the moving ship

Answer: (B)

108. 0.8 kg of air flows through a compressor under steady state conditions. The properties of air at entry are:

pressure 1 bar, velocity 10 m/s. specific volume 30.95m kg and internal energy 30 kJ/kg. The

corresponding values at exit are 8 bar, 6 m/s, 30.2m kg and 124 kJ/kg. Neglecting change in potential

energy, the power input will be

(A) 117 kW

(B) 127 kW

(C) 137 kW

(D) 147 kW

Answer: (B)

Sol:

Given that, 3

1 1 1 1v 10m sec, 0.95 m kg, u 30 kJ kg, P 1bar= = = =

( ) ( )( )

( ) ( )

3

2 2 2 2

1 1 1 1

2 2 1 1

v 6m sec, 0.2m kg, u 124 kJ kg, P 8bar

Enthalpy at 1 is h u p 30 100 0.95 125kJ kg

Enthalpy at 2 is h u p 124 800 0.2

= = = =

= + = + =

= + = +

2h 284 kJ kg

mass flow rate of air 0.8 kg sec

=

=

Applying steady flow energy equation

2 2

1 21 1 2 2

v vm h gz Q m h gz

2000 2000w

+ + + = + + +

compressor



55

Since, heat transfer Q 0= and neglecting charge in potential energy 1 2Z Z=

( )

2 210 6125 0.8 284 0.8 w

2000 2000

w 127.17 kW

+ = + +

=


109. In a power plant, the efficiency of the electric generator, turbine, boiler, cycle and the overall plant are

0.97, 0.95, 0.92, 0.92, 0.42 and 0.33 respectively. In the generated electricity, the auxiliaries will consume

nearly

(A) 7.3%

(B) 6.5%

(C) 5.7%

(D) 4.9%

Answer: (A)

Sol: (A)

In a power plant, over efficiency is given by

overall boiler cycle generator auxilaries turbine =

auxilaries

auxilaries

0.33 0.92 0.42 0.97 0.95

0.927

power consumed 1 0.927 0.073 7.3%

=

=

= − = =


110. The higher power requirements for compression in a steam power plant working on Carnot vapour cycle

(A) Increases the plant efficiency as well as work ratio

(B) Reduces the plant efficiency as well as work ratio

(C) Does not affect the plant efficiency as well as work ratio

(D) Increases the plant efficiency and reduces work ratio

Answer: (B)

Sol: Efficiency of cycle ( )

( )netNetwork done w

Heat supplied Q=

cycle

T Ccycle

Turbine work compressor work

Heat supplied

w w

Q

− =

− =



56

Since C'w ' is increased net'w ' will be decreased which will intron reduces cycle efficiency

net

T

wwork ratio

w=

As net'w ' decreases work ratio will also reduce.

So, answer is option ‘B’

111. For the same compression ratio, the Brayton cycle efficiency is

(A) Same as the Diesel cycle efficiency

(B) Equal to the Diesel cycle efficiency

(C) More than the Diesel cycle efficiency

(D) Less than the Otto cycle efficiency

Answer: (B)

Sol: Efficiency of Brayton cycle

( )Brayton 1

p

11

r−

= −

Where pr = pressure ratio 2

1

p

p

=

Efficiency of Otto cycle ( )

Otto 1

11

r−

= −

Where r= compression Ratio 1

2

V

V

=

Since process 1-2 i.e., Isentropic compression is same in both bray ton and Otto cycle 1 1 2 2p v p v =

2 1

1 2

p V

p V

=

( )1

1

2 1

1 2

p v

p v

−−

=

Hence for same compression ratio both bray ton Otto cycle will have same efficiency.

112. An economizer in a steam generator performs the function of preheating the

(A) Combustion air

(B) Feed water

(C) Input fuel

(D) Combustion air as well as input fuel



57

Answer: (B)

Sol: Economizer is used to preheat the feed water before sending it to the boiler by taking the heat form

exhaust gases leaving boiler.

113. Air enters the compressor of a gas turbine plant operating on Brayton cycle at 1 bar and 27°C. The

pressure ratio in the cycle is 6. If the relation between the turbine work WT and compressor work WC is

WT = 3WC and γ = 1.4, the cycle efficiency will be nearly

(A) 40%

(B) 50%

(C) 60%

(D) 70%

Answer: (A)

Sol:

Given that, 1 1T 27 C, P 1bar, r 1.4= = =

Pressure ratio ( ) 2p

1

Pr 6,

P= =

Turbine work ( ) ( )T CW 3 compressor work W=

Brayton cycle efficiency ( )( ) ( )

Brayton 1 0.4

1.4p

1 11 1 40%

6r−

= − = − =

So, correct answer is option (A).

114. A fluidized bed combustion system having an output of 35 MW at 80% efficiency when using a coal of

heating value 26 MJ/kg with a Sulphur content of 3.6% requires a particular limestone to be fed to it at a

calcium-sulphur molar ratio of 3.0 so as to limit emissions of SO2 adequately. The limestone used contains

85% CaCO3. The required flow rate of limestone will be

(A) 2405 kg/h

(B) 2805 kg/h

T 2

1

3

4

S



58

(C) 3205 kg/h

(D) 3605 kg/h

Answer: (A)

Sol: 35

Coal burning rate 1.683kg/s=6057.7kg/h0.80 26

= =

Flow rate of Sulphur6057.7 0.036

6.81kmol/h32

= =

For a Ca: S molar ratio of 3.0, the flow rate of calcium is 6.81×3.0 = 20.44 kmol/h

Mass flow rate of CaCO3 required = 20.44×100 = 2044kg/h ( )3Molecular weight of CaCO 100=

Mass flow of limestone 2044

2404.7kg/h0.85

= = which is 40% of coal burning rate.

115. In Orsat apparatus, when the percentage of carbon dioxide, oxygen and carbon monoxide are known, the

remaining gas is assumed to be

(A) Hydrogen

(B) Sulphur dioxide

(C) Nitrogen

(D) Air

Answer: (C)

116. The partial vacuum created by the fan in the furnace and flues, draws the product of the combustion from

the main flue and allows them to pass up to the chimney. Such a draught is called

(A) Balanced draught

(B) Forced draught

(C) Induced draught

(D) Artificial draught

Answer: (C)

Sol: In induced draught system a fan (or) blower is located at (or) near the base of the chimney. The pressure

over the fuel bed is reduced below that of the atmosphere. By creating a partial vacuum in the furnace and

flues, the products of combustion are drawn from the main flow and they pass up the chimney. Hence the

answer is option (C).



59

117. Which of the following are applied (used) ways of compounding steam turbines?

1. Pressure compounding

2. Temperature compounding

3. Velocity compounding

(A) 1, 2 and 3

(B) 1 and 2 only

(C) 2 and 3 only

(D) 1 and 3 only

Answer: (D)

Sol: A steam turbine can be compounded by the following ways to reduce the exist loss

1. Velocity compounding

2. Pressure compounding

3. Pressure velocity compounding

Hence the answer is option ‘D’ is correct.

118. A steam ejector which removes air and other non-condensable gases from the condenser is known as

(A) Wet air pump

(B) Dry air pump

(C) Centrifugal pump

(D) Circulating pump

Answer: (A)

119. In a heat exchanger, 50 kg of water is heated per minute from 50°C to 110°C by hot gases which enter the

heat exchanger at 250°C. The value of Cp for water is 4.186 kJ/kg.K and for air is 1 kJ/kg.K. If the flow

rate of gases is 100 kg/min, the net change of enthalpy of air will be nearly

(A) 17.6 MJ/min

(B) 15.0 MJ/min

(C) 12.6 MJ/min

(D) 10.0 MJ/min

Answer: (C)

Sol: Mass flow rate of water ( )wm 50 kg min=

Initial temperature of water ( )wiT 50 C=

Final temperature of water ( )woT 110 C=

Mass flow rate of hot gases ( )hm 100 kg min=



60

Initial temperature of hot gas ( )hiT 250 C=

Final temperature of hot gas ( )hoT ?=

Specific heat of water ( )pwC 4.186 kJ K=

Specific heat of hot gas ( )phC 1kJ K=

In heat exchanger,

Heat lost by hot gas (air) = Heat gained by water

Enthalpy charge of hot gas = Heat lost by hot gas

( ) ( )w pw wo wim c T T 50 4.186 110 50 12.56 12.6 MJ min= − = − =

120. The phenomenon that enables cooling towers to cool water to a temperature below the dry bulb

temperature of air is termed as

(A) Chemical dehumidification

(B) Adiabatic evaporative cooling

(C) Cooling and dehumidification

(D) Sensible cooling

Answer: (B)

Sol: In cooling water hot water is sprayed from the top of tower to expose a large surface area and air is

supplied from the bottom of tower. Water evaporates by taking heat from water itself and some of heat

will be transferred to air, hence water will be cooled and air will be heated and humidified and no heat

transfer takes places from (or) to cooling tower, hence it is adiabatic. Hence the total process is adiabatic

evaporative cooling. So, answer is option (B).

121. The angle through which the Earth must turn to bring the meridian of a point directly in line with the

Sun’s rays is called

(A) Altitude angle

(B) Hour angle

(C) Solar azimuth angle

(D) Zenith angle

Answer: (B)

122. In which type of collector is solar radiation focused into the absorber form the top, rather than from the

bottom?

(A) Fresnel lens

(B) Paraboloidal



61

(C) Concentrating

(D) Compound parabolic

Answer: (A)

123. A flat plate collector is 150 cm wide and 180 cm high and is oriented such that it is perpendicular to the

sun rays. Its active are is 90% of the panel size. If it is in a location that receives solar insolation of 1000

W/m2 peak, the peak power delivered to the area of the collector will be

(A) 1.23 kW

(B) 2.43 kW

(C) 4.46 kW

(D) 6.26 kW

Answer: (B)

Sol: Area of collector = 1.5 × 1.8 = 2.7 m2

Insolation is the solar radiation that reaches the earth. Since the collector plate is phased perpendicular to

solar rays all insolation will fall on collector. But its effective area is only 90%. Hence total solar

insolation delivered to the area of collector = 0.9 × 1000 × 2.7 = 2430 watts = 2.43 kW.

124. A surface having high absorptance for short-wave radiation (less than 2.5 μm) and a low emittance of

long-wave radiation (more than 2.5 μm), is called

(A) Absorber

(B) Emitter

(C) Selective

(D) Black

Answer: (C)

Sun rays Collector

180 cm

150 cm



62

125. In a solar tower power system, each mirror is mounted on a system called

(A) Regenerator

(B) Linear Fresnel

(C) Dish

(D) Heliostat

Answer: (D)

126. The ratio of PV cell’s actual maximum power output to tis theoretical power output is called

(A) Quantum factor

(B) Fill factor

(C) Quantum efficiency

(D) PV factor

Answer: (B)

127. With respect to the wind turbine blades, TSR means

(A) Tip Swift Ratio

(B) Tip Sharp Ratio

(C) Tip Speed Ratio

(D) Tip Swing Ratio

Answer: (C)

128. For a wind turbine 10 m long running at 20 rpm is 12.9 kmph wind, the TSR will be nearly

(A) 3.6

(B) 5.8

(C) 7.6

(D) 9.8

Answer: (B)

Sol: Wind speed 5

12.918

=

Blade tip speed ( )10 2 20DN

60 60

= =

blade tip speed 10 2 20 18

TSR 5.8wind speed 60 12.9 5

= = =



63

129. Which one of the following is an enclosure or housing for the generator, gear box and any other parts of

the wind turbine that are on the top of the tower?

(A) Turbine blade

(B) Nacelle

(C) Turbine head

(D) Gear box

Answer: (B)

130. The force required for producing tides in the ocean is

(A) 70% due to Moon and 30% due to Sun

(B) 30% due to Moon and 70% due to Sun

(C) 45% due to Moon and 55% due to Sun

(D) 55% due to Moon and 45% due to Sun

Answer: (A)

131. Which of the following are related to the Proton Exchange Membrane Fuel Cell (PEMFC)?

1. Polymer electrolyte

2. Hydrogen fuel and oxygen

3. Pure water and small amount of electricity

4. Nitrogen gas

(A) 1 and 3 only

(B) 2 and 4 only

(C) 1 and 2 only

(D) 3 and 4 only

Answer: (C)

132. Which of the following are the essential functions of fuel cells?

1. The charging (or electrolyser) function in which the chemical AB is decomposed to

A and B.

2. The storage function in which A and B are held apart.

3. The charge function in which A and B are charged with the simultaneous generation of electricity.

(A) 1 and 3 only

(B) 2 and 3 only

(C) 1 and 2 only

(D) 1, 2 and 3

Answer: (C)



64

133. The position of centroid can be determined by inspection, if an area has

(A) Single axis of symmetry

(B) Two axes of symmetry

(C) An irregular shape

(D) Centre axes of symmetry

Answer: (B)

134. Which of the following statements of D’Alembert’s principle are correct?

1. The net external force F actually acting on the body on the inertia force F1 together keep the body in

a state of fictitious equilibrium.

2. The equation of motion may be written as F + (–ma) = 0 and the fictitious force (–ma) is called an

inertia force.

3. It tends to give solution of a static problem an appearance akin to that of a dynamic problem.

(A) 1 and 3 only

(B) 1 and 2 only

(C) 2 and 3 only

(D) 1, 2 and 3

Answer: (B)

Sol: Net external force ‘F’ actually acting on the body is equal to product of mass (m) and acceleration (a) of

that body. If we want to bring the body to static equilibrium an inertia force (Fi) equal to ‘ma’ should

apply in the opposite direction and therefore their sum i.e., ( )iF F ma+ = will become zero. This is called

of D’Alembert’s principle and a dynamic problem (akin) a static problem. Hence statement ‘1’ and ‘2’ are

correct, but statement ‘3’ is wrong. So, answer is option (B).

135. The linear relationship between stress and strain for a bar in simple tension or compression is expressed

with standard notations by the equation

(A) E =

(B) E v=

(C) G v =

(D) G =

Answer: (A)



65

136. A punch is used for making holes in steel plates with thickness 8 mm. If the punch diameter is 20 mm and

force required for creating a hole is 110 kN, the average shear stress in the plate will be nearly

(A) 139 MPa

(B) 219 MPa

(C) 336 MPa

(D) 416 MPa

Answer: (B)

Sol: Given data,

3

t 8 mm, d 20 mm

F 110 kN

F 110 10218.83 MPa

dt 20 8

= =

=

= = =

137. A rod of length 2 m and diameter 50 mm is elongated by 5 mm when an axial force of 400 kN is applied.

The modulus of elasticity of the material of the rod will be nearly

(A) 66 GPa

(B) 72 GPa

(C) 82 GPa

(D) 96 GPa

Answer: (C)

Sol: Given data,

L 2m, d 50 mm

L 5mm, P 400kN

= =

= =

2 3

PPLAE 81,487.33 N mm 81.4 10 MPa

L A L

L

81.4 GPa 82 GPa

= = = = =

=

138. A beam of span 3 m and width 90 mm is loaded as shown in the figure. If the allowable bending stress is

12 MPa, the minimum depth required for the beam will be

12kN 5kN 12kN

0.6m 0.9m 0.9m 0.6m



66

(A) 218 mm

(B) 246 mm

(C) 318 mm

(D) 346 mm

Answer: (B)

Sol: Given data,

L 3m, b 90 mm

12 MPa

= =

=

Reaction at left is ( ) ( )12 5 12

2.4 0.6 14.5 kN3 2 3

= + + =

Maximum bending moment ( ) ( )4.5 1.5 12 0.9 10.95 kN-m= − =

2

6

2

M 6M

2 bd

6 10.95 1012

90 d

d 246.64 mm

= =

=

=

139. A vertical hollow aluminium tube 2.5 m high fixed at the lower end, must support a lateral load 12 kN at

its upper end. If the wall thickness is 1

8th of the outer diameter and the allowable bending stress is 50

MPa, the inner diameter will be nearly

(A) 186 mm

(B) 176 mm

(C) 166 mm

(D) 156 m

Answer: (D)

Sol: Given data,

i

L 2.5m, P 12 kN

Dt , 50 MPa

8

D ?

= =

= =

=

0

i

i

0

D D

DD D 2t D 2 0.75D

8

Dk 0.75

D

=

= − = − =

= =

12kN

2.5m



67

( )3 4

0

32M

d 1 k =

−

( )

3

3 4

0

0

i

32 12 10 250050

d 1 0.75

d 207.54

d 155.66mm 156mm

= −

=

=

140. A wooden beam AB supporting two concentrated loads P has a rectangular cross-section of width = 100

mm and height = 150 mm. The distance from each end of the beam to the nearest load is 0.5 m. If the

allowable stress in bending is 11 MPs and the beam weight is negligible, the maximum permissible load

will be nearly

(A) 5.8 kN

(B) 6.6 kN

(C) 7.4 kN

(D) 8.2 kN

Answer: (D)

Sol: Given data,

b 100

d 150

11 MPa

=

=

=

max

3

B.M P 0.5 0.5P N m

0.5P 10 N mm

= = −

= −

2

3

2

6M

bd

6 0.5P 1011

100 150

P 8,250 8.25kN

=

=

= =

141. Which of the following statements regarding thin and thick cylinders, subjected to internal pressure only,

is/are correct?

1. A cylinder is considered thin when the ratio of its inner diameter to the wall thickness is less than

15.

2. In thick cylinders, tangential stress has highest magnitude at the inner surface of the cylinder and

gradually decreases towards the outer surface.

0.5m 0.5m

P P



68

(A) 1 only

(B) 2 only

(C) Both 1 and 2

(D) Neither 1 nor 2

Answer: (B)

Sol: In this cylinder the ratio of inner diameter (d) to thickness (t) of cylinder should be greater than (or) equal

to 15. i.e., d

15t

Hence statement ‘1’ is wrong.

In thick cylinders, tangential stress (or) hoop stress ( )h is maximum at inner surface of cylinder and

minimum at outer surface as follows.

Variation of ‘ h ’ with respect to ‘r’ is h 2

ba

r = +

As ‘r’ increases ‘ h ’ decreases, hence statement ‘2’ is correct.

So, option (B) is correct.

142. A cylindrical storage tank has an inner diameter of 600 mm and a wall thickness of 18 mm. The transverse

and longitudinal strains induced are 6255 10− mm/mm and 60×10-6 mm/mm, and if G is 77 GPa, the

gauge pressure inside the tank will be

(A) 2.4 MPa

(B) 2.8 MPa

(C) 3.2 MPa

(D) 3.6 MPa

Answer: (D)

Sol: Diameter of pipe (d) = 600 mm

Thickness of pipe (t) = 18 mm

Shear modulus (G) = 77 GPa 377 10 MPa=

hVariation of ' '

ir

0r



69

Longitudinal strain ( )pd pd pd 1

e4tE 2tE 2tE 2

= − = −

Transverse strain (or) hoop strain ( )h

pd pd pde 1

2tE 4tE 2tE 2

= − = −

( )( )

( )( )

( )

( )

( )( )( )

( )( )

6

6h

3 5 2

6

5

160 10e 2

e 255 101

2

1 2 60

2 255

255 510 120 60

450 135 0.3

E 2G 1

E 2 77 10 1 0.3 2 10 N mm

pd 1R

2tE 2

p 60060 10 0.5 0.3 P 3.6 MPa

2 18 2 10

−

−

−

−

= =

−

− =

−

− = −

= =

= +

= + =

= −

= − =

So, answer is option (D).

143. A compressed air spherical tank having an inner diameter of 450 mm and a wall thickness of 7 mm is

formed by welding. If the allowable shear stress is 40 MPa, the maximum permissible air pressure in the

tank will be nearly

(A) 3 MPa

(B) 5 MPa

(C) 7 MPa

(D) 9 MPa

Answer: (B)

Sol: Given data,

D 450 mm, t 7mm

40 MPa

= =

=

Maximum shear stress,

hmax

pD

2 8t

p 45040 p 4.977 MPa 5MPa

8 7

= =

= =



70

144. A solid bar of circular cross-section having a diameter of 40 mm of 1.3 m is subjected to torque of 340

N.m. If the shear modulus of elasticity is 80 GPa, the angle of twist between the ends will be

(A) 1.26°

(B) 1.32°

(C) 1.38°

(D) 1.44°

Answer: (A)

Sol: Given data,

d 40mm, L 1.3m

T 340N m, G 80 GPa

?

= =

= − =

=

3

3 4

TL 340 10 13000.021rad

GJ80 10 40

32

1800.021 1.259

= = =

= =

145. Which one of the following statements regarding screw dislocation is correct?

(A) It lies parallel to its Burgers vector.

(B) It moves in the direction parallel to its Burgers vector.

(C) It initially requires very less force to move.

(D) It moves very fast as compared to edge dislocation.

Answer: (A)

146. The percentage of pearlite in a slowly cooled melt of 0.5% carbon steel is

(A) 48.5%

(B) 52.5%

(C) 58.5%

(D) 62.5%

Answer: (C)

147. In the study of phase diagrams, the rule which helps to calculate the relative proportions of liquid and

solid material present in the mixture at any given temperature is known as

(A) Hume -Rothery rule

(B) Lever rule

(C) Gibb’s phase rule

(D) Empirical rule

Answer: (B)



71

148. The phenomenon that artificially increasing the dielectric constant of plastics containing fillers is known

as

(A) Gamma polarization

(B) Interfacial polarization

(C) Post-forming drawing

(D) Reinforcement drawing

Answer: (B)

149. The addition of alloying elements nickel to cast iron will primarily improve

(A) Wear resistance

(B) Toughness

(C) Carbide formation

(D) Machinability

Answer: (B)

150. A unification fibre-epoxy composite contains 65% by volume fibre and 35% epoxy resin. If the relative

density of the fibre is 1.48 and of the resin is 1.2, the percentage weight of fibre will be nearly

(A) 70%

(B) 75%

(C) 80%

(D) 85%

Answer: (A)

Sol: Let, V = volume of fibre-epoxy composite

fV volume of fibre 0.65V= =

eV = Volume of epoxy region = 0.35V

Density of fibre ( ) 3

fe 1.48 1000 1480 kg m= =

Density of epoxy region ( ) 3

ee 1.2 1000 1200 kg m= =

Mass of fibre ( ) ( )( ) ( )( )f fmg e v 1480 0.65V 962V= = =

Mass of epoxy region ( ) ( )( ) ( )( )e e em e v 1200 0.35V 420V= = =

% weight (or) % mass of fibre ( ) ( )

f

f e

m 962V100 100 69.61 70%

m m 962V 420V= = =

+ +

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