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1. Ans. C. ‘Lesser than’ is apt because the sentence should be in comparative degree.

2. Ans. A. ‘hanged’ means death by hanging ‘hung’ is used only with things and not with people 3. Ans. D. ‘lying prone’ means lie down flat. ‘Prone to’ means vulnerable to

4. Ans. C. Statements i and ii are not logically possible based on the given fact. 5. Ans. B.

2

3 2 6

5 6

6 / 5

6 6 36

5 5 25

3 3 6 6 3 36.

4 4 5 5 4 25

3 361 2.06

4 25

x x

x

x

Areaof square

Areaof squaretriangle

6. Ans. C.

7. Ans. B. 8. Ans. C

9. Ans. A.

10. Ans. C. 11. Ans. A. By the properties of eigen values & eigen vectors, if all

the principal minors of ‘A’ are +Ve then all the eigen values of ‘A’ are also +Ve.

2 2

10

2

1

2

A for k

sok

12. Ans. C.

The function 2

2

3 4

3 4

x xf x

x x

is not continuous at

4&1;x since f(x) does not exists at x=-4 &1.

13. Ans. A. By the L.T of standard functions 14. Ans. B.

From C-R equation; we have

&u v u v

x y y dx

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2 2

, 2

2 2

,

2 ; 2

2 2

1

u x y xy

u uky kx

x yv x y x y

u uv v

x yx yx y

ky y

k

15. Ans. A. f(x) is a linear function

16. Ans. D. Let w is the velocity after collision

17. Ans. B.

max

1 1

max

1

where, is largest principal stress

1

xy

xy

18. Ans. D.

where, p is internal pressure d is internal diameter t is thickness

19. Ans. A.

2

1

2 1

2

exp

Torque=

1 exp

31 1 1 exp 0.25 2.248

2

F

F

F F R

F R

Nm

20. Ans. B. We know that

2

2 2

n

q

m

kq m km

m

21. Ans. C.

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At location Y

At circumference

At location X

At circumference

22. Ans. A.

FBD of Lever taking moment about hinge for clockwise rotation of wheel

0

If b> c self energizing

Nb F Nc

b cFN F N

b c

So for clockwise rotation of the drum, the brake is self energizing.

23. Ans. D. Volume flow rate per unit depth between two streamlines

is given by 1 2

24. Ans. B. the variation in atmospheric pressure with height calculated from fluid statics is exponential

25. Ans. A.

For a hollow cylinder

1 2

2

1

1 2 1 2

2

1

2

ln

ln

2

th

kL T TQ

r

r

T T T T

Rr

r

kL

26. Ans. D.

1 1

1 2

12 2 1 2 1 2 2 1

2 0 0

2 2 2 2

0

0

1

2

2

1 1

i i

i

F

F

R L RAA F A F F

A R L R

RF F

R

27. Ans. D. According to Joule’s law, Internal energy of an ideal gas is a function of temperature only.

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28. Ans. B.

2 7.2. . 4

1.8R

QCOP

W

29. Ans. B. Brayton cycle:-

1

1.4 1

1.4

6

1.4

11

11

6

0.4006

p

brayton

p

r

r

Gas Turbine cycle with perfect regeneration:-

5 2 3 2

5 3

1

4

4 3

6

1.4

1

0.3

Heat supplied =

p

p p

p

r

C T T C T T

T T

T

T

C T T

4 5 2 1

4 3 2 1

T c

p p

p p

Work done W W

C T T C T T

C T T C T T

4 3 2 1

4 3

2 21

12 1 1 1

54 3 434

44

1

1

1 1

4 41

Work done

Heat supplied

1 1

1 1 1

11

11 1

11

1 0.3 6

regenerative

p p

p

p

p

p

C T T C T T

C T T

T TTTT T T T

TT T TTT

TT

rT Tr

T T

r

1.4 1

1.4 0.4994

0.40060.8021

0.4994

brayton

regenerative

30. Ans. B.

1212 / , 24 / 0.5

24hr hr

Let, expected time that a customer spend in queue is qW

0.50.5

1 0.5 6012

0.560 2.5mins

12

q sq

L LW

31. Ans. C.

For aluminium alloy solution hardening process will be used to increase strength and hardness. In this process

component will be heated to 550C above temperature so that solute particles can penetrate into the lattice easily

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32. Ans. B. Submerged arc welding uses a blanket of fusible granular flux

33. Ans. B. Final length = L/2 Initial length = L

0

1ln ln ln 0.69

2 2

f

r

L L

L L

34. Ans. A.

900

200

0.25 /

300 / min

?

/ min1000

300 / min 0.2

300478

0.2

9007.539min

0.25 478

L mm

d mm

f mm rev

v m

t

DNv m

m N

N RPM

Lt

fN

35. Ans. D. In USM,

36. Ans. D.

37. Ans. B.

Since if z=1 lies inside the closed path and z=2 lies

outside of the closed path then by cauchy’s formula.

38. Ans. D. Probability that a packet would have to be replaced i.e.,

1 ?P X [ Let ‘x’ denote the number of defective

screws]

0 5

0

5

1 1 1

1 0

1 5 0.1 0.9

1 0.9 0.40951

P X P X

P X

C

Since by the Binomial distribution when P=probability of defective screw. 39. Ans. D.

0; sin cos

3 3

03

1 1.37 1

b ah f x x x

n

x

y f x

By trapezoidal rule; we have the approximate value of the integral is

0

00

/ 3sin cos 1 1 2 1.37 0.37

2

1.822

Exact value of the integral is

sin cos cos sin 1 1 2

Error Exact value-Approximate value

=2-1.822 0.178

x x dx

x x dx x x

40. Ans. A.

2.02 / sec

1

ii

V R

Vrad

R

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2 2

2

2 2

2 2

1Loss in kinetic energy of Drum

2

1150 2 300

2

1Loss in kinetic energy of block

2

12 0

2

4000

Loss of Potential energy of block = mgh

2000 9.81 0.5

i f

drum

i f

block

J

KE Joule

m v v

KE Joule

P

9810

Total energy loss

300 4000 9810 14110

14.11

block

block block block

E Joule

KE KE PE

Joule

kJ

41. Ans. B. By definition of Torque

42. Ans. D.

Because of torsion angle of twist will be there.

2 2

4

4

2 32 4

.

T P PWhere

G J G d R G

Due to angle of twist, A will reach at A' and B will reach at B' let A'A'' be the vertical displacement

43. Ans. C.

As we know that

and slope at mid-span where moment is applied will be

U

M

(according to Costigliano’s theorem).

44. Ans. D.

F.B.D of point B

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Applying Lame's theorem

sin180 sin135 sin 45

0

Stress in AB 0

BCAB

AB

FF P

F

45. Ans. D.

5

5

200

10 /

2 10

0.2

t C

C

E MPa

mm

We know that, Axial stress gets induced in the rod when

some gap ' ' is provided is

46. Ans. A.

2Given AB

Since Rod AB is rigid, so Axial velocity of A & B should be same

cos60 cos60

2 / sec

is mid point of AB

A B

A B

V V

V V m

C

Alternate Method:

2 2 2

2

2 2cos120

2 2 2

11

2 2 2

32 2 2 3

2

3,so oc willbeperpendicular toab.2

sin 30 12

cc

x

x

x

xy

vv

47. Ans. C. Displace the block “A” & Release

Alternate method:

Energy of system remain conserved,

where,

Static elongation of spring at equilibrium which is

calculated as follows:

mgmg k

k

Differentiating Eqn.(1) w.r.t time, which will be zero because E = constant

0

0 ...(2)

dE

dt

d dvJ mV mgv k y v

dt dt

Since

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there is no slipping between rope & pulley v r

2

2

2

1 10

2

02

150010 / sec

10 5

2

n

v dv dvMr mv mgv k y v

r r dt dt

M d ym kydt

krad

Mm

48. Ans. D. Given

max min m150 , 50 , 100

50 , 200 , 300

400

a e ut

ut

MPa MPa MPa

MPa S MPa S MPa

S MPa

Equation of line 1

1 ...(1)a m

e utS S

Equation of line 2

1tan ...(2)

2

a

m

Solving above two equations to get co-ordinates of point

P ,m aS S

1200 400

2 400 ...(3)

2 ...(4)

a m

a m

m a

S S

S S

S S

from (3) & (4)

100

100. . 2

50

a

a

a

S MPa

Sf o s

49. Ans. C.

3

3

1000 /

700 /

?

0.45

0.55

water

oil

body

oil body

water body

Given kg m

kg m

V V

V V

3

700 0.45 1000 0.55

865 /

oil oil water water body body

body body body body

body

V g V g V g

V V V

kg m

50. Ans. B.

2

2

360

0.05 0.05

0.44 60 / .

26.4 /

top bottom

wall

V Vdu

dy

duJ kg m s

dy

N m

51. Ans. A

Downward force due to water = weight of water above curved surface

22

2

24

2 14

R Lg R L

gR L N

Weight of plug is neglected. 52. Ans. A.

1 2 2 1

2 2

1 /

2 /

(as both are same fluids)

1 80 50 2 10 25

h c

k c

h

c

P P

k P k k c p c c

c c

m kg s

m kg s

C C

m C t t m C t t

t t C

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11

22

1 2 1 2

1 1

2 2

80 25 5570

50 10 4025

ln ln

70 25 55 40

70 55ln ln

25 40

43.705 47.1

. .

p c

c

m m

p mp c m

CC

CC

Q U A Q U A

.

43.7050.928

47.1

cp mp c m

mpc

p mc

A A

A

A

53. Ans. C.

54. Ans. B. 3

1

1

1

3

2

21 1

1

0.4

100

80 353

0.1

Ideal gas & process is isothermal.

ln

0.1100 0.4ln

0.4

55.45

V m

P kPa

T C K

V m

VW PV

V

kJ

55. Ans. C.

Reversible cycle.

3 31 2

1 2 3

33

0

40 370 0

400 370 320

1 32064

320 5 5

Q

T

Q QQ Q

T T T

QQ kJ

56. Ans. B.

Given

1

1

2

3

2 3

236.04 /

0.9322 /

272.05 /

93.42 /

0.05 /

Heat Rejection to environment =m

0.05 272.05 93.42

8.9315

h kJ kg

s kJ kg

h kJ kg

h kJ kg

m kg s

h h

kW

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57. Ans. B.

58. Ans. A.

Ans. A

0.2 0.70.4

0.2 0.7 0.4 0.4

0.2 0.3 ...(1)

Fraction of solid =

0.3

0.2

0.20.40

0.5

L S

L S

L S L S

L S

s

S L

S

S S

59. Ans. A.

60. Ans. B.

1 2

1 1

2 2

100 40 1 & 2

140 250

180 200

10

140 2501 ...( )

180 2001 ...( )

On solving equation a and b, we have

. . 424.6

V mm mm

V V I A

V V I A

V I

CV SCC

aOCV SCC

bOCV SCC

S C C A

61. Ans. C. Cost of metal cutting = Rs 18 C/V Cost of Tooling = Rs 270 C/TV C= Constant ,V = Cutting Speed , T = tool life

1/0.250.25

4

150, 150 150 /

150 /

C VT T V

T V

4

4

3

4

Total cost =18 270

18 270

150

18 270

150

C C

V TV

C CV

V V

C CV

V

On, differentiating total cost

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2

2 4

44

18 270 3

150

18 150

3 270

57.914 / min

C C V

V

V

V m

62. Ans. A.

12 1.5GivenV V V V

1 1 1 1

1

3

33

6

3

10.02 0.02 0.2

10

7860 /

107860 /

10

55.85

7.860 /

k ohm mm ohm cmohm cm

kg m

gm cm

gm

gm cm

5 355.85: : 3.68 10 / sec

7.86 2 96600

AI IMRR Q cm

ZF

4Interelectrodegapgiven=9 m=9 10 cm

4

current destiny J=

0.2 12 1.52333.33

9 10

.

2333.33 . . / 2333.33

k V V

y

I J S A

I S A S A I

Electrode feed rate = MRR/ surface area cm/sec 53.68 10 2333.33

/ sec

0.086 10 60 / min

51.51 / min

Icm

I

mm

mm

63. Ans. D.

2 2

22

2 22

2

, ,

2

2

a R r b D R r C a b

C R r D R r

H R r C

R r R r D R r D R r

H R r D R r D

64. Ans. C.

025000 , 500 / , 20%hD kg C Rs order C of Cu

Qty (kg) Cu (Rs/kg) Ch (Rs/Kg/year)

11 750Q 70 0.2 × 70 = 14

2750 1500Q 65 0.2 × 65 = 13

3 1500Q 60 0.2 × 60 = 12

This problem belongs to inventory model with two price break.

2* o

h

DCQ

C

first checking for least unit price

*

3

2 25000 5001443.37

12Q

Now, 1443.37 < 1500 therefore, the company will not get the item at Rs 60/kg Now, checking for second minimum unit price

*

2

2 25000 5001386.75

13Q

Since, 1386.75 lies between 750 and 1500 Therefore, we need to find

65. Ans. B.

The latest finish time for node 10 is 14 days.

***