general aptitude - gate e-learning

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1. A box contains 15 blue balls and 45 black balls. If 2 balls are selected randomly, without replacement,

the probability of an outcome in which the first selected is a blue ball and the second selected is a black

ball, is_________

(A) 45

236 (B)

3

4 (C)

1

4 (D)

3

16

Key: (A)

Sol:

15 45 45

Required probability .60 59 236

2. A digital watch X beeps every 30 seconds while watch Y beeps every 32 seconds. They beeped together

at 10 AM. The immediate next time that they will beep together is _______

(A) 10.08 PM (B) 11.00 AM (C) 10.00 PM (D) 10.42 AM

Key: (A)

Sol: Given,

Watch x beeps every 30 sec

Watch y beeps every 32 sec

L.C.M of 30, 32 = 480 seconds = 8 minutes

They beeps together every 8 min

i.e., 10 AM, 10.08 AM; 10.16 AM, 10.24 AM, ……

Hence option (A).

3. Consider the following sentences:

(i) The number of candidates who appear for the GATE examination is staggering.

(ii) A number of candidates from my class are appearing for the GATE examination.

(iii) The number of candidates who appear for the GATE examination are staggering.

(iv) A number of candidates from my class is appearing for the GATE examination.

Which of the above sentences are grammaticallyCORRECT?

(A) (i) and (ii) (B) (ii) and (iv) (C) (i) and (iii) (D) (ii) and (iii)

Key: (A)

GENERAL APTITUDE

15

Blue

45

Black

Box

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4. The world is going through the worst pandemic in the past hundred years. The air travel industry is

facing a crisis, as the resulting quarantine requirement for travelers led to weak demand.

In relation to the first sentence above, what does the second sentence do?

(A) Restates an idea from the first sentence

(B) Second sentence entirely contradicts the first sentence

(C) The two statements are unrelated

(D) States an effect of the first sentence

Key: (D)

5. Given below are two statements 1 and 2, and two conclusions I and II.

Statement 1: All entrepreneurs are wealthy.

Statement 2: All wealthy are risk seekers.

Conclusion I: All risk seekers are wealthy.

Conclusion II: Only some entrepreneurs are risk seekers.

Based on the above statements and conclusions, which one of the following options is CORRECT?

(A) Both conclusion I and II are correct (B) Only conclusion II is correct

(C) Only conclusion I is correct (D) Neither conclusion I nor II is correct

Key: (D)

Sol: EEntrepreneurs

WWealthy

R Risk sectors

All risk sectors are not wealthy Conclusion I is incorrect.

All entrepreneurs are risk seekersConclusion II is incorrect

Hence, neither conclusion I nor II is correct.

6. Five persons P, Q, R, S and T are to be seated in a row, all facing the same direction, but not necessarily

in the same order. P and T cannot be seated at either end of the row. P should not be seated adjacent to

S.R is to be seated at the second position from the left end of the row.

The number of distinct seating arrangements possible is:

(A) 4 (B) 3 (C) 2 (D) 5

Key: (B)

Sol: Positive arrangements are

S R P T Q (OR) Q R P T Q (OR) S R T P Q

Total number of combinations possible is 3.

F

W

E

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7. If 2; 3; 5; 10,

Then, the value of , is :

(A) 0 (B) 16 (C) 1 (D) 4

Key: (C)

8. The front door of Mr. X’s house faces East. Mr. X leaves the house, walking 50 m straight from the back

door that is situated directly opposite to the front door. He then turns to his right, walks for another 50m

and stops. The direction of the point Mr. X is located at with respect to the starting point is ______

(A) North-East (B) South-East (C) West (D) North-West

Key: (D)

Sol:

From the figure, it is clear that, the direction of X is North West

9.

The ratio of the area of the inscribed circle to the area of the circumscribed circle of an equilateral

triangle is _________.

(A) 1

8 (B)

1

2 (C)

1

6 (D)

1

4

Key: (D)

Sol: From the figure,

r 1 r

sin30 R 2rR 2 R

Area of inscribed circle 2r

Area of circumscribed circle 2 2R 4r

Required ratio 2 2 1

r : 4 r 1: 44

N

EW

S

X

Starting point50m

50m

R

r

r radius of inscribed circle

R radius of circumscribed circle

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10. Consider a square sheet of side 1 unit. T sheet is first folded along the main diagonal. This is followed

by a fold along its line symmetry. The resulting folded shape is again folded along its line of symmetry.

The area each face of the final folded shape, in square units, equal to ________.

(A) 1

16 (B)

1

32 (C)

1

8 (D)

1

4

Key: (C)

Sol: Area of each face of the final folded shape.

(Right angle triangle) 1 1 1 1 1

Box weight square units2 2 2 2 8

2

1

1

1

1

1

1

1

2

1

2

1

2

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1. The size distribution of the powder particles used in Powder Metallurgy process can be determined by

(A) Laser absorption (B) Laser scattering

(C) Laser reflection (D) Laser penetration

Key: (B)

Sol: Laser diffraction (or) laser scattering used in size distribution of powder particles used in powder

metallurgy process. So answer option ‘B’.

2. An adiabatic vortex tube, shown in the figure given below is supplied with 5 kg/s of air (inlet 1) at

500kPa and 300K. Two separate streams of air are leaving the device from outlets 2 and 3. Hot air

leaves the device at a rate of 3 kg/s from outlet 2 at 100 kPa and 340 K, while 2 kg/s of cold air stream is

leaving the device from outlet 3 at 100 kPa and 240 K.

Consider constant specific heat of air is 1005 J/kg.K and gas constant is 287 J/kg.K. There is no work

transfer across the boundary of this device. The rate of entropy generation is _______ kW/K (round off

to one decimal places).

Key: (2.24)

Sol:

1 1 1

2 2 2

3 3 3

m 5 kg sec, P 500 kPa, T 300K

m 3kg sec,P 100 kPa, T 340K

m 2kg sec, P 100 kPa, T 240K

MECHANICAL ENGINEERING

1

2 3

Highpressureairinlet

LowpressurecoldairoutletLowpressurehotairoutlet

1

2 3

LowpressurecoldairoutletLowpressurehotairoutlet

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pC 1005J kgK ; gas constant R 287J / kgK

p v v pR c c c c R 1005 287 718 J kgK

Since tube is maintained in adiabatic condition

No entropy transfer from the vortex tube

Entropy generation = Entropy change – Entropy transfer

Entropy generation = Entropy change entropy transfer 0

Entropy change of hot air

2 22 p

1 1

T Pm c n R n

T P

340 1003 1005 n 287 n

300 500

1763.1 W K

Entropy change of cold air

3 33 p

1 1

T Pm c n R n

T P

240 1002 1005 n 287 n

300 500

475.30W K

Entropy generation gens Total entropy change 1763.1 475.30 2238.4 W K 2.24 kW K

3. Water flows out from a large tank of cross-sectional area 2

tA 1m through a small rounded orifice of

cross sectional area 2

oA 1cm , located at y = 0. Initially the water level(H), measured from y = 0, is

1 m. Theacceleration due to gravity is 9.8 2m s .

y 0

y H

WATER

OrificeCross-sectional

area

tCorss-sectionalarea, A

0A

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Neglecting any losses, the time taken by waterin the tank to reach a level of y = H/4is____________

seconds (round off to one decimalplace).

Key: (2258)

Sol: Cross sectional area of tank 2

tA 1m

Cross sectional of area orifice 2 4 2

0A 1cm 1 10 m

Initially Height of water, (H) = 1m

2g 9.8 m sec

Time taken to reach a level of water H

y4

will be find out as follows,

in out cu

in

m m m

m 0

0 t

0 t

0 t

dA 2gH A dH

dt

dA 2gH A dH

dt

A 2gHdt d A dH

Integration on both sides

0 t

t

0

0.25 11

t t

4 0.251 0.250 0

A 2gH t 0 A dH

A dHt

A 2g H

1A AdH dHt 2 H

A 2g H HA 29 1 10 2 9.8

t 2258 seconds

4. A plane frame PQR (fixed at P and free at R) is shown in the figure. Both members (PQ and QR) have

length L, and flexural rigidity, EI. Neglecting the effect of axial stress and transverse shear, the

horizontal deflection at free end, R, is

P

L

L

R

Q

F

y 0

y H

WATER

tCorss-sectionalarea, A

0A

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(A) 32FL

3EI (B)

35FL

3EI (C)

34FL

3EI (D)

3FL

3EI

Key: (C)

Sol: Flexural rigidity of members PQ and QR having length ‘L’ are ‘EI’

Bending moment at any distance ‘x’ from ‘R’ is x QRM Fx

Bending moment in ‘PQ’ portion is constant and is equal to PQM FL

Strain energy stored in frame

22L L

x PQQR

0 0

M dx M dxf

2EI 2EI

2 2L L

0 0

L2 3 2 2

L

0

0

2 3 2 2

2 3 2 3

2 3 2 3 2 3

Fx dx FL dxU

2EI 2EI

F x F Ldx

2EI 3 2EI

F L F LL

2EI 3 2EI

F L F L

6EI 2EI

F L 3F L 2 F L

6EI 3 EI

Horizontal deflection at ‘R’ is 3 3

H R

U 2 2FL 4FL

F 3 EI 3EI

So answer is option ‘C’

PL

L

R

Q

F

x

x

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5. The wheels and axle system lying on a rough surface is shown in the figure.

Each wheel has diameter 0.8 m and mass 1 kg. Assume that the mass of the wheel is concentrated at rim

and neglect the mass of the spokes. The diameter of axle is 0.2 m and its mass is 1.5 kg. Neglect the

moment of inertia of the axle and assume g = 9.8 2m s . An effort of 10 N is applied on the axle in the

horizontal direction shown at mid span of the axle. Assume that the wheels move on a horizontal surface

without slip. The acceleration of the wheel axle system in horizontal direction is ________ 2m s (round

off to one decimal place).

Key: (1.25)—(4.9-5.1)

Sol:

Axle

0.8m

Wheel

gg

10N

0.2m

Axle

0.8m

Wheel

g

10N0.2m

Force applied F 10N

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Diameter of wheel (D) = 0.8m, Radius of wheel D

R 0.4m2

Mess of wheel wm 1 kg

Diameter of axle 0.2m,

Radius of axle d 0.2

r 0.1m2 2

Mass of axle am 1.5kg

2g 9.8 m sec

Mass moment inertia of wheels and axle is 2

wI 2m R (Neglecting mass moment of inertia of axle)

2 2I 2 1 0.4 0.32kg.m

fH ma 10 2F ma

where Ff = frictional force between wheel and Road

m = mass of two wheels and axle = 3.5kg

a = linear acceleration of wheel

f

f

f

10 2F 3.5a ... 1

T I

T 2F R I

T 10 0.2 1N.m

1 0.8F 0.32 ... 2

Solving equation (1) and (2)

2

f

2

12.5rad / sec , F 3.75N

a R 0.4 12.5 5m/sec

0.2m 10N

fF

R 0.4m

T 10 0.21N.m

0.2m 10N

fF

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6. The allowance provided in between a hole and a shaft is calculated from the difference between

(A) lower limit of the shaft and the upper limit of the hole

(B) lower limit of the shaft and the lower limit of the hole

(C) upper limit of the shaft and the upper limit of the hole

(D) upper limit of the shaft and the lower limit of the hole

Key: (D)

Sol: Allowance = Lower limit of hole – Upper limit of shaft

So Answer is option ‘D’

7. The value of 0 cos

0 0rsin dr d is

(A) 0 (B) 4

3 (C)

1

6 (D)

Key: (C)

Sol: Given double integral

2 cos

0 r 0

rsin dr d

cos2 222

0 00

r 1sin d sin cos d

2 2

1

2

0

13

0

1t dt Put cosθ= t sin d dt and 0 t 1, t 0

2 2

1 t 1

2 3 6

8. A two dimensional flow has velocities in x and y directions given by u = 2xyt and 2v y t, where t

denotes time. The equation for streamline passing through x = 1, y = 1 is

(A) 2x y 1 (B) 2xy 1 (C) 2 2x y 1 (D) 2x y 1

Key: (B)

Sol: 2u 2xyt, v y t

Equation for steam line is

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2

2

dx dy

u v

dx dy

2xyt y t

dx dy dx dy

2xy y 2x y

Integrating on both sides

1log x log y c

2

at x 1, y 1

1log1 log 1 c

2

at x 1, y 1

1log1 log 1 c

2

c 0

equation of stream line is

x x

2

1log x log y

2

log log y log log y 0

xy 1

xy 1

So answer is option ‘B’.

9. The torque provided by an engine is given by T 12000 2500sin 2 N.m, where is the angle

turned by the crank from inner dead center. The mean speed of the engine is 200 rpm and it drives a

machine that provides a constant resisting torque. If variation of the speed from the mean speed is not to

exceed 0.5%, the minimum mass moment of inertiaof the flywheel should be_________kg.m2 (round

off to the nearest integer).

Key: (570.15)

Sol:

mean

T 12000 2500sin 2

N 200 rpm

Coefficient of fluctuation of speed S

0.5C 2

100

SC 0.01

meanmean

2 N 2 200W 20.94 rad sec

60 60

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E maximum fluctuation in energy

Since T is a function of ‘ ’

Hence mean

0 0

1 1T Td 12000 2500sin 2 d

0

1 250012000 0 cos

2

112000 1250 cos360 cos

12000 N.m

At any instant meanT T T

T 12000 2500sin 2 12000

T 2500sin 2

‘ T ’ is zero when 2500sin 2 0

0 , 902

9090 90

00 0

2500cos2E T d 2500sin 2 d 2500 1 1 2500 N.m

2 2

2

mean SE I C

2

2

2500 I 20.94 0.01

I 570.15 kg.m

10. The machining process that involves ablation is

(A) Electrochemical Machining (B) Laser Beam Machining

(C) Abrasive Jet Machining (D) Chemical Machining

Key: (B)

Sol: Ablation means removal of material by melting or evaporation.

From the given options, the only laser beam machining method will remove the material by melting and

vaporization.

So the answer is option ‘B’.

11. The von Mises stress at a point in a body subjected to forces is proportional to the square root of the

(A) total strain energy per unit volume (B) plastic strain energy per unit volume

(C) distortional strain energy per unit volume (D) dilatational strain energy per unit volume

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Key: (C)

Sol: According to distortion energy theorem,

For 2-D dimensional case,

2 2 2

1 2 1 2 yt

2 2

1 2 1 2 yt

s

s

Vonmises stress 2 2

von 1 2 1 2

Where, 2 2

1 2 1 2 represent distortion strain energy per unit volume

Hence vonmises stress is proportional to square root of distortion strain energy per unit volume

So answer is option ‘C’.

12. The figure shows the relationship between fatigue strength (S) and fatigue life (N) of a material.The

fatigue strength of the material for a life of 1000 cycles is 450 MPa, while its fatigue strength for a life

of 106 cycles is 150 MPa.

The life of a cylindrical shaft made of this material subjected to an alternating stress of 200 MPa will

then be ___________ cycles (round off to the nearest integer).

Key: (164058.98)

Sol:

450 150 200 150

10 10 10 10

x

10

log log log log

6 3 6 log

x

10

x

10

5.215

6 log 0.785

log 6 0.785 5.215

x 10 164058.98 cycles

10log S

10log N7

3 x

10log 6 N

10log7

150

10log

s

10log

450

10log

200

10log

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13. A block of negligible mass rests on a surface that is inclined at 30° to the horizontal plane as shown in

the figure. When a vertical force of 900 N and a horizontal force of 750 N are applied, the block is just

about to side.

The coefficient of static friction between the block and surface is ______ (round off to two decimal

places).

Key: (0.17)

Sol: Free body diagram of block is

H 0

Ncos60 750 Ncos30 0

N N 3750 0

2 2

N1 3 750 ...(1)

2

V 0

Nsin 60 900 N sin30 0

N 3 N900 0

2 2

N3 900

2 …(2)

Equation (1)/(2) =

1 3 750

9003

30

900N

750N

30

900N

750N

60N fF

N

30

900N

750N

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51 3 3

6

6 6 3 5 3 5

6 5 3 5 6 3

6 5 30.17

5 6 3

' ’ is negative means our assumed direction of frictional force is not correct.

So, answer is 0.17

14. A shell and tube heat exchanger is used as a steam condenser. Coolant water enters the tube at 300K at a

rate of 100 kg/s. The overall heat transfer coefficient is 1500 W/m2.K, and total heat transfer area is

400m2. Steam condenses at a saturation temperature of 350K. Assume that the specific heat of coolant

water is 4000 J/Kg.K. The temperature of the coolant water coming out of the condenser is

________________ K. (round off to the nearest integer).

Key: (339)

Sol: Mass flow rate of coolant cm 100 kg sec

Inlet temperature of coolant water overall heat transfer ciT 300K

Overall heat transfer Coefficient 2U 1500 W m K

Total heat transfer area 2A 400 m

Saturation temperature of stream s hi hoT 350K T T

Specific heat of coolant p cc 4000 J kg.K

cmin c pC m c because specific heat of steam is

100 4000 400kW K

3min

1500 400UANTU 1.5

C 400 10

For condensing or boiling effectiveness of heat exchanger is

NTU 1.51 e 1 e 0.78

c pc co ci

min hi ci

m C T T

C T T

co

co

T 3000.78 T 339K

350 300

s hi hoT 350K T T

CiT 300K

CoT

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15. Daily production capacity of a bearing manufacturing company is 30000 bearings. The daily demand of

the bearing is 15000. The holding cost per year of keeping a bearing in the inventory is Rs. 20. The

setup costfor the production of a batch is Rs.1800. Assuming 300 working days in a year, the economic

batch quantity in number of bearing is ______. (in integer).

Key: (40250)

Sol: Daily Demand = 15,000 bearings

Annual Demand (R) = 15,000 × 300

3C ordering cost (or) set up cost = 1800 Rs

1C carrying cost = 20 Rs|Bearing|year

K = Production rate = 30, 000, Bearings per day

Economic batch quantity 3

1

2C REBQ

RC 1

K

3

1

2C R k

C k R

2 1800 15,000 300 30,000

20 30,000 15,000

40249.2 units

So answer is 40, 250 units.

16. Let the superscript T represent the transpose operation. Consider the function T T1f x x Qx r x,

2

where x and r are n 1 vectors and Q is a symmetric n n matrix. The stationary point of f(x) is

(A) TQ r (B) r (C) T

r

r r (D) 1Q r

Key: (D)

Sol: Take n= 2 and Let a c

Qc b

be a symmetric matrix, 1 1

2 2

x rx ; r

x r

Then 1 1 2 2

2 1 2 1 2 1 2 1 1 2 2

2 2

x xa c1 1f x x, x r r ax bx 2cx x r x r x

x xc b2 2

Let 2 2

1 21 2 1 2 1 1 2 2

ax bxf x x ,x cx x r x r x

2 2

For stationary points, we have 1 2

0 and 0x x

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1 2 1ax cx r 0 …(1)

And 2 1 2bx cx r 0 …(2)

Equation (1) and (2) in matrix form is

1 1

2 2

x ra c

x rc b

Q.x r

Multiply by 1Q , we get

1x Q r, option (D) is the stationary point.

17. A PERT network has 9 activities on its critical path. The standard deviation of each activity on the

critical path is 3. The standard deviation of the critical path is

(A) 3 (B) 81 (C) 9 (D) 27

Key: (C)

Sol: Number of activities on the critical path = 9

Standard deviation of each activity in the critical path 3

Variance of each activity in the critical path 2 9

Sum of variances of all activities in the critical path 29 9 9 81

Standard deviation of critical path 81 9

So answer option ‘C’.

18. An object is moving with a Mach number of 0.6 in an ideal gas environment, which is at a temperature

of 350K. The gas constant is 320 J/kg.K and ratio of specific heats is 1.3. The speed of object is

_______ m/s. (round off to the nearest integer).

Key: (229)

Sol: Mach number (M) = 0.6

Temperature of air (T) = 350K

Gas constant R 320 J kg.K

Ratio of specific heat p

v

C1.3

C

Speed of object (v) = ?

We know that Mach number

Vvelocity of objectM

sonic velocity C

VM

RT

V M RT 0.6 1.3 320 350 228.9 229 m sec

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19. For a two-dimensional, incompressible flow having velocity components u and v in the x and y

directions, respectively, the expression

2u uv

x y

Can be simplified to

(A) u v

2u ux y

(B)

u u2u v

x y

(C)

u vu u

x y

(D)

u uu v

x y

Key: (D)

Sol: For a two-dimensional in compressible flow, continuity equation is

u v u v v u

0x y x y y x

Given that

2u uv u u v2u v u

x y x y y

u u v2u v u

x y y

u u u u u u u2u v u 2u u v u v

x y x x y x y

So answer is option ‘D’

20. Consider an ideal vapour compression refrigeration cycle working on R-134a refrigerant. The COP of

the cycle is 10 and the refrigeration capacity is 150 kJ/kg. The heat rejected by the refrigerant in the

condenser is ______________ kJ/kg. (round off to the nearest integer).

Key: (165)

COP = 10

Refrigeration capacity = 150 kJ/kg

Work input to refrigeration (W) = ?

2QCOP

W

15010

W

150W 15 kJ kg

10

2Q

1Q

Ref

HT

LT

W

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1

1 2

1

1

Q Heat rejected by refreigerator

W Q Q

15 Q 150

Q 150 15 165 kJ kg

21. A 76.2mm gauge block is used under one end of a 254mm sine bar with roll diameter of 25.4mm. The

height of gauge blocks required at the other end of the sine bar to measure an angle of 30° is

_________mm (round off to two decimal places).

Key: (203.2)

Length of sine bar 254 mm

Diameter of roll d 25.4 mm

Angle measured 30

Height of gauge blocks

On one side 2h 76.2 mm

Height of gauge blocks on another side 1h ?

1 2

1 2 1

1

h r h rsin 30

h h h 76.21

2 254 254

h 203.2 mm

22. Value of5.2

4n x dx using Simpson's one-third rule with interval size 0.3 is

(A) 1.06 (B) 1.60 (C) 1.83 (D) 1.51

Key: (C)

Sol:

0

1

2

3

4

f x nx, h 0.3; a 4, b 5.2

y f 4 n4 1.386

y f a h n 4.3 1.458

y f a 2h n 4.6 1.526

y f a 3h n 4.9 1.589

y f a 4h n 5.2 1.648

By Simpson’s one-third rule,

254 mm

30

1h

2h

Roll

r

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5.2

0 4 1 3 2

4

hnxdx y y 4 y y 2 y

3

0.31.386 1.648 4 1.458 1.589 2 1.526

3

1.827 1.83

23. A cast produce of a particular material has dimensions 75mm × 125mm × 20mm. The total

solidification time for the cast product is found to be 2.0 minutes as calculated usigChvorinov's rule

having the index, n = 2. If under the identical casting conditions, the cast product shape is changed to a

cylinder having diameter = 50mm and height =50mm, the total solidification time will be ______

minutes. (round off to two decimal places).

Key: (2.85)

Sol: Solidification time according to chvarinovs rule is

n

s

Vt k

A

k = moulding constant

n = 2

For cube:

Volume of the cube 3V 75 125 20 18750 mm

Surface are of cube 2A 2 75 125 125 20 20 75 26750 mm

Solidification time st for cube is = 2 minutes

2

1875002 k

26750

k 0.041

For cylinder:

Volume of cylinder 2 2 3V d h 50 50 31250 mm4 4

Surface area of cylinder 22 2A d 2 d 50 50 50 3750 mm

4 2

Solidification of time st for cylindrical is

n 2

s

V 31250t k 0.041 2.85 mins

A 3750

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24. The cast Iron which possesses all the carbon in the combined for as cementite is know as

(A) White Cast Iron (B) Spheroidal Cast Iron

(C) Grey Cast Iron (D) Malleable Cast Iron

Key: (A)

Sol: White cast iron has carbon in the form of cementic hence the answer is option ‘A’

25. Aspot welding operation performed on two pieces of steel yielded a nugget with a diameter of 5mm and

a thickness of 1mm. The welding time was 0.1s. The melting energy for the steel is 20 J/mm3.

Assuming the heat conversion efficiency as 10%, the power required for performing the spot welding

operation is ___________kW. (round off to two decimal places).

Key: (39.27)

Sol: Diameter of Nuggest (d) = 5mm

Thickness of Nugget (t) = 1mm

Volume of Nugget 2 2 3V d t 5 1 6.25 mm4 5

Welding time t 0.1seconds

Melting energy of steel 320 J mm

Heat required for melting

Total heat required RQ 20 6.25 125 J

Heat conversion efficiency 0.10

Heat supplied SQ ?

R

S

S

S

Q

Q

125 1250.10 Q 3926.99 3927 joules

Q 0.1

Heat supplied per second SS

Q 3927Q 39270 watts 39.27 kW

t 0.1

26. Which of the following is responsible for eddy viscosity (or turbulent viscosity) in a turbulent boundary

layer on a flat plate?

(A) Nikuradse Stresses (B) Prandtl Stresses

(C) Reynolds Stresses (D) Boussinesq Stresses

Key: (C)

|ME-2021-Set-II|


Sol: Since, turbulence is considered as eddying motion and the additional stresses are added to the viscous

stresses due to mean motion in order to explain the complete stress field, it is often said that the apparent

stresses are caused by eddy viscosity. The total stresses given by,

2

xx

xy

uP 2 pu '

x

u vpu 'v '

y x

Terms such as u 'v ' and u ' are called Reynolds stresses or turbulent stresses.

Hence, the correct answer is option (C).

27. Consider the following differential equation

dy

1 y ydx

The solution of the equation that satisfies the condition y(1) = 1 is

(A) 2 y xy e e (B) y x1 y e 2e

(C) y x2ye e e (D) y xye e

Key: (D)

Sol: 1 y

DE dy dx, which is VSFy

Integrating, we get

1

1 dy dx Cy

ny y x C …(1)

Using y(1) = 1, (1) gives 0 + 1 = 1+ C C = 0

ny y x

x y y xy e or ye e option (D).

|ME-2021-Set-II|


28. A power transmission mechanism consist of a belt drive and a gear train as shown in the figure.

Diameters of pulleys of belt drive and number of teeth (T) on the gears 2 to 7 are indicated in the figure.

The speed and direction of rotation of gear 7, respectively are:

(A) 575.28 rpm; Clockwise (B) 255.68; anticlockwise

(C) 575.28 rpm; anticlockwise (D) 255.68 rpm; Clockwise

Key: (D)

Sol:

1 2

2

1

d 150 mm, d 250 mm

N 1502500 1500 rpm ACW

N 250

3 2 2

2 3 3

33

N d z

N d z

N 18N 613.64 rpm CW

1500 44

3 4N N , since they are mounted on same shaft

5 4 4

4 5 5

5

5

N d z

N d z

N 15

613.64 33

N 278.93 rpm ACW

18T

250mm

2 N 2500 rpm

150mm

3

5 6

36T

16T

33T44T

15T

47

|ME-2021-Set-II|


6 5 5

5 6 6

6

6

N d z

N d z

N 33

278.93 36

N 255.68 rpm CW

7 6N N , since they are mounted on same shaft hence answer is option ‘D’.

29. The controlling force curves P,Q and R for a spring controlled governor are shown in the figure, where

r1 and r2 are any two radii of rotation.

The characteristic shown by the curves are

(A) P – Stable; Q – Isochronous; R – Unstable

(B) P – Unstable; Q – Isochronous; R – stable

(C) P – Stable; Q – Unstable; R – Isochronous

(D) P – Unstable; Q – Stable; R – Isochronous

Key: (B)

Sol: The controlling force (F) (Vs) radius of rotation (r) in a spring controlled governor is as follows.

F = ar + b

If b < 0 i.e., F = ar – b It is a stable governor

‘R’ is stable governor

If b = 0 i.e., F = ar It is isochronous governor

‘Q’ is isochronous governor

If b > 0 i.e., F = ar + b It is a unstable governor

‘P’ is unstable governor


P

Q

R

2r 1r

Co

ntr

oll

ing

forc

e

Radius of rotation

|ME-2021-Set-II|


30. Find the positive real root of 3x x 3 0 using Newton-Raphson method. If the starting guess 0x is

2, the numerical value of the root after two iterations 2x is ___________ (round off to two decimal

places).

Key: (1.67)

Sol:

3

2

0

f x x x 3

f ' x 3x 1and x 2

By Newton-Raphson method,

0

1 0

0

1

2 1

1

f x 3x x 2 1.73

f ' x 11

f x 0.48x x 1.73 1.67

f ' x 7.98

31. The thickness, width and length of a metal slab are 50mm, 250mm and 3600mm, respectively. A rolling

operation on this slab reduces the thickness by 10% and increases the width by 3%. The length of the

rolled slab is __________mm (Round off to one decimal place).

Key: (3883.5)

Sol: 1

1

1

2

2

2

t 50 mm

W 250 mm

3600 mm

t 0.9 50 45 mm

W 1.03 250 257.5 mm

?

Volume of slab before rolling 1 2V Volume of slab after V Rolling

1 1 1 2 2 2

2

t w t w

50 25 3600 45 257.5

Length of slab after rolling 2 3883.5 mm

32. A surface grinding operations has been performed on a cast Iron plate having dimensions 300mm

(length) × 10mm (width) × 50mm (height). The grinding was performed using an alumina wheel having

a wheel diameter of 150mm and wheel width of 12 mm. The grinding velocity used is 40 m/s, table

speed is 5 m/min, depth of cut per pass is 50μm and the number of grinding passes is 20. The average

tangential and average normal forces for each pass are found to be 40N and 60N respectively. The value

of the specific grinding energy under the aforesaid grinding conditions is _________ J/mm3. (round off

to one decimal places).

|ME-2021-Set-II|


Key: (38.4)

Sol: Given, tF 40N, V 40m/s

tSo, Power F .V 40 40 1600W

Width, W 10mm,

depth of cut d 50 m 0.05mm

Table speed s 5m/min

So,

3

5000MRR W.d.s 10 0.05

60

41.6666mm /s

Hence, specific grinding energy

=

3Power 160038.4J / mm

MRR 41.6666

33. The mean and variance, respectively, of a binomial distribution for n independent trials with the

probability of success as p, are

(A) np, np 1 p (B) np, np

(C) np, np 1 2p (D) np, np 1 p

Key: (D)

Sol: Mean and variance of a binomial distribution are np and npqrespecivley, where q = 1–p i.e.., np and

np 1 p

Option (D).

34. A factory produces m i 1,2,.......m products, each of which requires processing on n j 1,2......,n

workstations. Let ai,j be the amount of processing time that one unit of the ith product requires on the j

th

workstation. Let the revenue from selling one unit of the ith product ri and hi be the holding cost per unit

per time period for the ith product. The planning horizon consists of T t 1,2.....,T time periods. The

minimum demand that must be satisfied in time period t is itd , and the capacity of the jth workstation in

time period t is jtc . Consider the aggregate planning formulation below, with decision variable itS

(amount of product i sold in time period t), itX (amount of product i manufactured in time period t) and

Iit (amount of product i held in inventory at the end of time period t).

T m

i it i it

t 1 i 1

it it

max rS h I subject to

S d i, t

h 50 mm

300 mm

W 10mm

|ME-2021-Set-II|


it , it , it i0

capacity constraint

inventory balance constraint

X S I 0; I 0

The capacity constraints and inventory balance constraints for this formulation respectively are

(A) m

ij it jt it i, t 1 it it

i

a X c i, t and I I X d i, t

(B) m


i

a X c j, t and I I X S i, t

(C) m

ij it it it i, t 1 it it

i

a X d i, t and I I X S i, t

(D) m


i

a X d i, t and I I S X i, t

Key: (B)

Sol: Capacity constraint is

ij it jta x c

Where, ija processing time that one unit of ith product requires on j

th work station

itx Amount of product ‘i’ manufacture in time period ‘t’.

jtC Capacity work station ‘j’ in time ‘t’

Inventory balance constraint is

it i,t 1 it itI I x s

Where, itI Amount of product ‘i’ held in inventory at the end of time period ‘t’

i,t 1I Amount of product ‘i’ held in inventory at the time period ‘t–1’

itX Amount product ‘i’ manufacture in time period ‘t’

i,t 1I Amount of product ‘i’ held Inventory at the time period ‘t-1’

itX Amount product ‘i’ is manufactured in time period ‘t’

itS Amount of product ‘i’ sold in time period ‘t’.

So answer is option ‘B’

35. A vertical shaft Francis turbine rotates at 300 rpm. The available head at the inlet to the turbine is

200 m. The tip speed of the rotor is 40 m/s. Water leaves the runner of the turbine without whirl.

Velocity at the exit of the draft tube is 3.5 m/s. The head losses in different components of the turbine

are: (i) stator and guide vanes: 5.0 m, (ii) rotor: 10 m, and (iii) draft tube: 2m. Flow rate through the

|ME-2021-Set-II|


turbine is 20 m3/s. Take g = 9.8 m/s

2. The hydraulic efficiency of the turbine is 20m

3/s. Take

2g 9.8m / s . The hydraulic efficiency of the turbine is __________% (round off to one decimal place).

Key: (91.2)

Sol: N 300 rpm

H 200 mm

u 40 m sec

Velocity of water at exit of draft tube 2V 3.5 m sec

Loss in stator and guide vanes sH 5m

Loss in Rotor RH 10m

Los in Draft tube DH 2m

Flow through turbine 3Q 20 m sec

2g 9.8 m sec

Net available head for turbine 2

0net S R D

VH H H H H

2g

2

Net

3.5H 200 5 10 2 182.375m

2 9.8

Hydraulic efficiency netH

H

H 182.375

H 200

91.187%

So answer is 91.2

36. A rigid tank of volume 50 m3 contains a pure substance as a saturated liquid vapour mixture at 400 kPa.

Of the total mass of the mixture, 20% mass is liquid and 80% mass is vapour. Properties at 400 kPa

are: Saturation temperature, satT 143.61 C; Specific volume of saturated liquid, 3

gv 0.001084 m /kg;

Specific volume of saturated vapour, 3

gv 0.46242 m / kg. The total mass of liquid vapour mixture in

the tank is _________kg (round off to the nearest integer).

Key: (135.16)

Sol: Volume of tank 3V 50m

Saturation temperature at 400 KPa, satT 143.61 C

Vapour

Liquid

|ME-2021-Set-II|


3

f

3

g

V 0.001084 m kg

V 0.46242 m kg

Total mass of liquid-vapour-mixture (m) = ?

L Vm m m

Specific volume of vapour 3

VV 0.46242 m kg

Mass of vapour v

g

V 50m 108.13kg

V 0.46242

Given that,

Mass of vapour vm + mass of liquid Lm Total mass (m) of liquid vapour mixture.

L108.13 m m

1081.13 0.2m m

108.13 0.8m

m 135.16 kg

37. Value of 8

1 i , where i 1, is equal to

(A) 16i (B) 4i (C) 16 (D) 4

Key: (C)

Sol:

2 2 2

4 2 2

8 2

1 i 1 i 2i 2i

1 i 2 i 4

1 i 4 16, option (C)

38. The demand and forecast of an item for five months are given in the table.

Month Demand Forecast

April 225 200

May 220 240

June 285 300

July 290 270

August 250 230

The mean absolute percent error (MAPE) in the forecast is _____________%. (round off to two decimal

places)

|ME-2021-Set-II|


Key: (8.07)

Sol:

Month Demand (D) Forecast (F)

Percentage absolute

error

D - F

D

April 225 200 25

11.11%225

May 220 240 20

9.09%220

June 285 300 15

5.26%285

July 290 270 20

6.89%290

August 250 230 20

8%250

Mean absolute percentage

11.11 9.09 5.26 6.89 8

error 8.07%5

39. A cantilever beam with a uniform flexural rigidity 6 2EI 200 10 N.m is loaded with a concentrated

force at its free end. The area of the bending moment diagram corresponding to the full length of the

beam is 10000 N.m2. The magnitude of the slope of the beam at its free end is _______ micro radian

(round off to nearest integer).

Key: (50)

Sol: 6 2

IE 200 10 N.m

Given that, Area of bending moment diagram between

A and B is = 10,000 2N.m

22 2

110,000 W

2

W10,000 W 20,000 N.m

2

Slope at ‘B’ = Area of M

EI

diagram between A and B

A B

W

W

M W

EI EI

|ME-2021-Set-II|


2

5

6

1 W 1 W

2 EI 2 EI

1 20,0005 10 radians 50 micro radians

2 200 10

40. Ambient pressure, temperature, and relative humidity at a location are 101 kPa, 300K, and 60%

respectively. The saturation pressure of water at 300K is 3.6 kPa. The specific humidity of ambient air

is __________ g/kg of dry air.

(A) 21.4 (B) 21.9 (C) 35.1 (D) 13.6

Key: (D)

Sol: Ambient pressure atmP 101kPa P

Temperature of air (T) = 300K

Relative humidity 0.6

Saturation pressure of water at 300K is

satP 3.6 kPa

Specific humidity of air w ?

V

V

sat

V

PP 0.6 3.6

P

P 2.16 kPa

V

V

0.622 2.160.622Pw

P P 101 2.16

0.01359 kg of water vapour kg of dry air

13.59 gms of water vapour kg of dry air


41. In a CNC machine tool, the function of an interpolator is to generate

(A) error signal for tool radius compensation during machining

(B) signal for the lubrication pump during machining

(C) NC code from the part drawing during post processing

(D) reference signal prescribing the shape of the part to be machined

Key: (D)

Sol: In CNC machine tool, the function of an interpolator is to generate reference signal prescribing the shape

of the part to be machined.

|ME-2021-Set-II|


42. A column with one end fixed and one end free has a critical buckling load of 100 N. For the same

column, if the free end is replaced with a pinned end then the critical buckling load will be __________

N (round off to the nearest integer)

Key: (800)

Case (i)

2

cr 21e

2

2

EIP

L

EI100N

2

2

2

EI400N

Case (ii)

2 2

cr 222e

2

2

EI EIP

L

2

2 EI

2 400 800N

So answer is 800N

43. If the Laplace transform of a function f(t) is given by

s 3,

s 1 s 2

thenf(0) is

(A) 3

2 (B) 0 (C)

1

2 (D) 1

Key: (D)

Sol:

s 3F s L f t

s 1 s 2

s

f 0 limsF s

(Using initial value theorem)

s s

31

s s 3 1slim lim 1option (D)

1 2s s 1 s 2 1 11 1

s s

one end fixed and other end free

Case (i)

one end fixed and other end pinned

Case (ii)

|ME-2021-Set-II|


44. Consider the system shown in the figure. A rope goes over a pulley. A mass, m, is hanging from the

rope. A spring of stiffness, k, is attached at one end of the rope. Assume rope is inextensible, massless

and there is no slip between pulley and rope.

The pulley radius is r and its mass moment of inertia is J. Assume that the mass is vibrating

harmonically about its static equilibrium position. The natural frequency of the system is

(A) 2

2

kr

J mr (B) k m (C)

2

2

kr

J mr (D)

2kr

J

Key: (A)

Sol: J = Mass moment of inertia of pulley about ‘0’

Give a small angular displacement to pulley

sF spring force krsin

2

i 2

dF m rsin

dt

For small angles r sin r 2

2

dmr mr

dt

iT J

0M

i SJ F r cos F r cos 0

J mr r cos krsin r cos 0

For small angels cos 1, sin

k r

J

m

iF

r sin

r sinSF

r cos r r cos

k r

J

m

O

|ME-2021-Set-II|


2 2

2 2

2

n 2

J mr kr 0

J mr kr 0

kr

J mr

So answer is option ‘A’

45. A steel cubic block of side 200 mm is subjected to hydrostatic pressure of 250 N/mm2. The elastic

modulus is 2 × 105 N/mm

2 and Poisson’s ratio is 0.3 for steel. The side of the block is reduced by

________mm (round off to two decimal places)

Key: (0.11)

Sol:

5 2E 2 10 N mm , 0.3

E 3k 1 2

Bulk modulus

5

2

2 10Ek

3 1 2 3 1 2 0.3

k 166666.67 N mm

Bulk modulus

v

v

3 3

v

volumetric stress (or)hydrostatic stressk

volumetric strain e

250166666.67

e

250e 1.499 10 1.5 10

166666.67

v

33 3

v

change in volume Ve

original volume or initial volume

V e a 1.5 10 200 12,000

V Initial volume Final volume

i f

3

f

12,000 V V

V 200 12,000 7988000

Let a ' New size of cube

3

a ' 7988000

a ' 199.89

a a ' 200 199.89 0.11mm

2250 N mm

a 200 mm2250 N mm

a 200 mm

a 200 mm2250 N mm

|ME-2021-Set-II|


46. Consider the open feed water heater (FWH) shown in the figure given below:

Specific enthalpy of steam at location 2 is 2624 kJ/kg, specific enthalpy of water at location 5 is 226.7

kJ/kg and specific enthalpy of saturated water at location 6 is 708.6 kJ/kg. If the mass flow rate of water

entering the open feed water heater (at location 5) is 100 kg/s then the mass flow rate of steam at

location 2 will be _________kg/s (round off to one decimal place).

Key: (25.16)

Sol:

Given that, 2

5

6

5

2

h 2624 kJ kg

h 226.7 kJ kg

h 708.6 kJ kg

m 100 kg sec

m ?

2 2 5 5 6 6

2 6

m h m h m h

m 2624 100 226.7 m 708.6

Let ‘ 5m ’ be the mass of flow rate steam taken at location ‘2’

Open FWH 6 6m h 5 5m h

2 2m h

1

23

Condenser

Open FWH

HP pump

LP pump

4

7

5

6

Ste

am

Gen

erat

or

Turbine

1

23

Condenser

Open FWH

HP pump

LP pump

4

7

5

6

Ste

am

Gen

erat

or

Turbine

|ME-2021-Set-II|


Let 1m be the total mass flow rate of steam entering at location ‘1’

1 2 5 1 6m m m and m m

6 2 5 2

2 2

2 2

2

2

m m m m 100

m 2624 100 226.7 m 100 708.6

2624m 22670 708.6m 70860

1915.4m 48190

m 25.16 kg sec

47. Consider the mechanism shown in the figure.

There is rolling contact without slip between

the disc and ground.

Select the correct statement about instantaneous centers in the mechanism.

(A) All points P, Q, R, S, T and U are instantaneous centers of mechanism

(B) Only points P, Q, S and T are instantaneous centers of mechanism

(C) Only points P, Q and S are instantaneous centers of mechanism

(D) Only points P, Q, R, S, and U are instantaneous centers of mechanism

Key: (A)

Sol:

Q

2

P1

3

R

U

S 4

T

Q

2

P1

3

R

U

S 4

T

12I

24I

13I

34I

14I

23I

|ME-2021-Set-II|


Number of instantaneous centresare

n n 1 4 4 1

2 2

Hence P, Q, R, S, T and U are instantaneous centres of mechanism

Hence answer is option ‘A’

48. Consider adiabatic flow of air through a duct. At a given point in the duct, velocity of air is 300 m/s,

temperature is 330 K and pressure is 180 kPa. Assume that the air behaves as a perfect gas with constant

pc 1.005 kJ kg.K. The stagnation temperature at this point is_____K (round off to two decimal places).

Key: (374.78)

Sol: Velocity of air V 300 m sec

Temperature of air T 330K

Pressure of air P 180 kPa

PC 1.005 kJ kg.K

Stagnation temperature 2

stag

p

VT T

2C

2

stag 3

300T 330 374.78 K

2 1.005 10

49. A machine of mass 100 kg is subjected to an external harmonic force with a frequency of 40 rad/s. The

designer decides to mount the machine on an isolator to reduce the force transmitted to the foundation. The

isolator can be considered as a combination of stiffness (K) and damper (damping factor, ) in parallel.

The designer has the following four isolators:

1. K = 640 kN/m, 0.70 2. K = 640 kN/m, 0.07

3. K = 22.5 kN/m, 0.70 4. K = 22.5 kN/m, 0.07

Arrange the isolators in the ascending order of the force transmitted to the foundation.

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

Key: (D)

Sol: Mass of machine (m) = 100 kg

Applied frequency 40 rad sec

Transmissibility

2

T

2 22 0

1 2 r FTR

F1 r 2 r

1 2

4 3

|ME-2021-Set-II|


Where, T

0

F Force transmitted

F Applied force

1. K 640 kN m, 0.70

3

n

n

k 640 1080 rad sec

m 100

40r 0.5

80

2

T

2 220

T

0

1 2 0.70 0.5F

F1 0.5 2 0.7 0.5

F 1.221.189

F 1.026

2. K 640 kN m, 0.07

n

2

T

2 220

k80 rad sec, r 0.5

m

1 2 0.07 0.5F 11.32

F 0.7531 0.5 2 0.07 0.5

3. K 22.5 kN m, 0.70

3

n

n

2

T

2 220

k 225 1015 rad sec

m 100

40r 2.67

15

1 2 0.7 2.67F 3.870.54

F 7.181 2.67 2 0.7 2.67

4. K 22.5 kN m, 0.07

n

2

T

2 220

15 rad sec r 2.67

1 2 0.07 2.67F 1.0670.17

F 6.141 2.67 2 0.07 2.67

So ascending order is 4–3–1–2.


|ME-2021-Set-II|


50. Ambient air flows over a heated slab having flat, top surface at y = 0. The local temperature (in Kelvin)

profile within the thermal boundary layer is given by T(y) = 300 + 200 exp(–5y), where y is the distance

measured from the slab surface in meters. If the thermal conductivity of air is 1.0 W/m.K and that of the

slab is 100 W/m.K, then the magnitude of temperature gradient dT / dy within the slab at y = 0 is

_____ K/m (round off the nearest integer).

Key: (10)

Sol: Thermal conductivity of air airk 1 W mK

Thermal conductivity of slab slabk 100 W mK

At y = 0

Conduction heat transfer into slab = Conductive heat transfer from air to slab

slab air

y 0 y 0slab Air

5yair

y 0slaby 0 y 0slab air

dT dTk k

dy dy

kdT dT 1 15 200 e 5 200 10 k m

dy k dy 100 100

Magnitude of temperature gradient within the slab at y = 0 is ‘10’

51. A plane truss PQRS (PQ = RS, and PQR = 90°) is shown in the figure.

P

Q

L

L

S

F

R

Air

T y 300 200 exp 5y

slab

y 0

|ME-2021-Set-II|


The forces in the members PR and RS, respectively, are ________.

(A) F tensile and F 2 tensile (B) F 2 tensile and F compressive

(C) F compressive and F 2 compressive (D) F 2 tensile and F tensile

Key: (B)

Sol: PQ RS

PQR 90

At junction ‘R’

PR RS

PR RS

RSPR RS

PRRS

F cos45 F 0

F cos45 F

FF 2F

cos45

FF

2

PR PR

PRRS RS

FF sin 45 F F 2F

sin 45

F 2FF F F F

2 2

The force in ‘RS’ is compressive and it is ‘F’, the force in ‘PR’ is tensile and it is 2 F


52. In forced convective heat transfer, Stanton number (St), Nusselt number (Nu), Reynolds number (Re)

and Prandtl number (Pr) are related as

(A) Nu Pr

StRe

(B) St Nu Pr Re (C) Nu

StRePr

(D) Nu Re

StPr

Key: (C)

Sol: Stanton Number (st) p p

h h T

c c T

p

h T

Tk

Nu NuL

c T Pr.ReLPr

Tk

L

So answer is ‘C’.

P

Q

L

L

S

F

R

45

45

|ME-2021-Set-II|


53. In a pure orthogonal turning by a zero rake angle single point carbide cutting tool, the shear force has

been computed to be 400 N. If the cutting velocity, Vc = 100 m/min, depth of cut, t = 2.0 mm, feed S0 =

0.1 mm/revolution and chip velocity, Vf = 20 m/min then the shear strength,s of the material will be

________ MPa (round off to two decimal places)

Key: (392.23)

Sol: Rake angle 0

Shear force SF 400N

Cutting velocity CV 100 min min

Depth of cut (t) = 2mm

Feed 0S 0.1mm rev

Chip flow velocity fV 20m min

Shear strength of material s ?

Chip thickness ratio (r) 1 f

2 c

t V 200.2

t V 100

Shear angle ‘ ’ will be find out as follows,

1

0.2 cos0rcostan

1 rsin 1 0.2 sin 0

tan 0.2 11.31

1t uncut chip thickness = depth of cut = 2mm

b = width of chip = feed = 0.1 mm

Shear strength (or) shear stress in the work piece is

ss

1

F 400sin sin 11.31 392.23 MPa

t b 2 0.1

54. A high velocity water jet of cross section area = 0.01 2m and velocity = 35 m/s enters a pipe filled with

stagnant water.

Entrained

water

Entrained

water

Inlet Jet

35 m s

Total

water out

flow at

uniform

velocity

6 m s

Pipe

b

1t

CV

Tool

2t

|ME-2021-Set-II|


The diameter of the pipe is 0.32m. This high velocity water jet entrains additional water from the pipe

and the total water leaves the pipe with a velocity 6 m/s as shown in the figure.

The flow rate of entrained water is _______ litres/s (round off to two decimal places)

Key: (130)

Sol:

d = diameter of jet, velocity of jet jetV 35 m sec

a = cross-sectional area of jet 20.01m

D = Diameter of pipe = 0.32m,

Cross-sectional area of pipe

2

2 2

A D4

A 0.32 0.080 m4

0V Total water outlet velocity = 6 m/sec

= Density of water = 1000 3kg m

Applying conservation of mass of principle,

Mass flow rate of jet + Mass flow rate of entrained water (or) stagnant water = total mass flow rate

jet oaV e flow rate stagnent water or entrained AV

jet 0aV Q AV

3

0.01 35 Q 0.080 6

Q 0.13 m sec 130 lit sec

1. 55. Consider an n × n matrix A and non-zero n × 1 vector p. Their product 2Ap p, where

amd 1,0,1 . Based on the given information, the eigen value of 2A is:

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

Key: (C)

Entrained

water

Entrained

water

Inlet Jet

35 m s

Total

water out

at 6 m s

Pipe

|ME-2021-Set-II|


Sol: Given, 2Ap p …(1)

Pre-multiply both sides by A, we get

2

2

A AP A p

AA p Ap

. 2 2 2

4

A p p by (1)

p

4 is the eigenvalue of A2

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