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1.

( ) ( )

( )

3/2 3/22 2 2 2 2 2

3/22 2 2

xi yjF

x y z x y z

zk

x y z

= +

+ + + +

+

+ +

F·ds over a sphere given by R1 = 1 and R2 = 4 ?

Sol. div F =

( )

( )

12 2 2 2 2 2 2 2 2 2

3 2 2 22 2 2

x y z –2x y z x –2y z

x y –2zx y z

0

+ + + + + + + + + +

=

F·ds = div F dv = 0

2. Cost-time slope for crashing is

A. Crash cost/Normal time

B. (Crash cost-Normal cost)/(Normal time – crash time)

C. (Crash cost – Normal Cost)/(Crash time – Normal time)

D. (Crash cost – normal cost)/(crash time)

Ans. C

Sol. CrashCost Normalcos t

cos t timeslopeNormal time Crashtime

−=

−

3. Which function is not Analytic ?

A. Logz B. sinz

C. ez D. z2

Ans. A

4. A strip of thickness 40 mm is to be rolled to a thickness of 20 mm using a two high mill

having rolls of diameter 200 mm. Coefficient of friction and arc length in mm respectively

are :

A. 0.39 mm and 44.72 mm

B. 0.45 mm & 44.72 mm

C. 0.39 mm & 38.84 mm

D. 0.45 mm & 38.84 mm

Ans. ?

Sol. Δh = μ2R

40 – 20 = μ2 × 100

2 200.45

100 =

μ = 0.45

h 20

1 cos cos 1D 200

− = = −

α = 0.451

Arc length = Rα

= 100 × 0.451

Arc length 45.1 mm

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5. lc = 100 mm (Chip length)

l = 250 mm (uncut chip length)

α = 20°, calculate shear angle in degree.

Sol. cl 100r 0.4

l 250= = =

Shear angle, r cos

tan1–r sin

=

( )( )

0.4cos 20

1– 0.4sin 20

=

ϕ = 23.53

6. Crystal structure of Austenite

Sol. FCC

7. Tolerance of hole 0.002 mm

Tolerance of shaft 0.001 mm

Allowance = 0.003 mm

Basic size = 50 mm

Find the maximum size of hole (shaft basis system) ?

Sol. Maximum size of hole = Allowance + Basic size + Tolerance of hole = 50 + 0.003 +

0.002

8. Resultant shear load on B bolt, all dimension in mm

Sol.

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2 2Br 50 50 50 2= + =

'B

P 10P 2.5kN

No. of bolt 4= = =

( )" e BB 2 2 2 2

a b c d

P rP

r r r r=

+ + +

( )

( )

3 –3

2–3

10 10 0.4 50 2 10

4 50 2 10

=

= 14.14 kN

( ) ( )2 2

' " ' "B B B BResul tant P P 2P P cos45= + +

= 16 kN

9. In a biaxial equal tensile leading of 10 MPa and 1 unit is equal to 1 MPa.

A. Mohr’s circle will be point circle at a distance of 10 on x-axis.

B. Mohr’s circle will be a point circle at a distance of 1 on x-axis

C. Mohr’s circle will be a circle of diameter 10 at x-axis

D. Mohr’s circle will be circle of diameter 10 at y-axis

Ans. A

Sol.

10. I = 12 + 0.2t

Where t is in minutes

Surface area of workpiece = 125 cm2

Cathode efficiency = 0.85

Metal diffusion rate = 3250 mm

A – S

What will be the thickness of coating after 20 min ?

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Sol. So, 1 = 12 + 0.2t

= 12 + 0.2 × 20

I = 16 Amp

= =− cathode

250 A thicknesssM.D. rate

A S I time

= =−

−

3cathode

M.D. rate 250 A thicknesss

A S I timemm

Amp sec

125 1000 thickness

25016 20 60 0.35

=

Thickness = 326.4 mm

11. For an incompressible fluid velocity is given by ( )2 2V 2 x – y i Vj 3k,= + + v has value

A. –4xy – 4xz B. 4xy + 4xz

C. 4xy – 4xz D. 4xz – 4xy

Ans. A

Sol. Velocity for incompressible fluid flow,

( )2 2V 2 x – y i Vj 3k= + +

From above velocity relation

u = 2(x2 – y2)

V = V

ω = 3

If the flow is incompressible continuity equation has to be satisfied,

u V

0x y z

+ + =

( )( ) ( )2 2 V2 x – y 3 0

x y z

+ + =

V

4x 0 0y

+ + =

V

–4xy

=

V –4xy C = +

12. Calculate Magnitude of reaction force at point C is _____ kN

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Sol. Whenever we have internal hinge point, separate that portion

Moment at B

40 × 2 = RC × 4

RC = 20 kN

13. The structural member with loading is shown in figure below.

Find the value of P at which buckling occurs ?

A. 2

2

EI

l

B.

2

2

2 EI

l

C. 2

2

2 EI

l

D.

2

2

EI

2l

Ans. B

Sol. In the given diagram, buckling will occur along the diagonal.

Load acting along diagonal = 2P

This load will be equal to buckling load

The small value of P at which buckling occurs,

2

2

EI2P

l

=

2

2

EIP

2 l

=

14. The inlet temperature of a compressor is 310 k. With pressure ratio of 6. If the isentropic

efficiency of compressor = 85%. Find the difference between enthalpy at exit and inlet

of compressor in kJ/kg.

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Sol. Temperature at inlet of compressor (T1) = 310 k

For above (T-S) diagram of Brayton cycle,

Isentropic efficiency (ηisen) = 0.85 = Isentropic work

Actual work

2 1'2 1

h –h0.85

h –h=

( )p 2 1' 2 12 1

C T – Th –hh –h

0.85 0.85= =

Now for (1 – 2) isentropic process =

y–1/y

2 2

1 1

T P

T P

=

⇒ T2 = 517.22k

So actual difference in enthalpy ( )( )p 2 1'

2 1

C T – Th –h

0.85=

( )1.005 517.22 –310

0.85=

= 245 kJ/kg

15. Kaplan Turbine, axial flow, the outlet velocity diagram is given here,

The diameter at this section is 3m. Hub and tip are having diameter of 2m and 4m

respectively. Water volume flow rate = 100m3/sec. Rotational speed of turbine = 300

rpm. Blade outlet angle = ?

Sol.

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From, the above velocity diagram, Blade outlet angle (β) can be found by, f

b

Ctan

C =

where, Cf is flow velocity Cb is blade velocity

meanb

D NBlade velocity (C )

60

=

3 300

60

=

47.12 m/sec.

f

volume flow rateFlow velocity (C )

net change in area=

Net change in area (A)

( )2 24 24

= −

( )16 44

= −

12 34

= =

f

100Flow velocity (C ) 10.61m/sec

3= =

So, blade outlet angle (β),

f

b

C 10.61tan 0.225

C 47.12 = = =

β = tan–1(0.225)

β = 12.69°

16. τmax = MPa

By Max shear stress theory which will fail

Square → 4 × 4 cm2

Diameter → 4 cm

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Sol.

380 10

50 MPa40 40

= =

τmax = 25 MPa

τmax = –25 MPa

3

16T

d =

( )

3

3

16 64 100

4

=

τ = 16 MPa

M

y I

=

My

I =

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( )3

4

4320

2 30 100 30MPaa

12

= = =

17. Sum of first ‘n’ terms with sequence

8, 88, 888, ……

A. ( )n80 810 1 – n

81 9−

B. ( )n81 910 1 n

80 80− +

C. ( )n80 810 1 n

81 9− + −

D. ( )n81 910 1 n

80 8+ +

Ans. A

Sol. Put n = 1

option A gives 8.

18. If x indicates greatest integer function such that [x]; greatest integer less than equal to

x. If y = [x], then area under y for xϵ[1, 4] is

A. 4 B. 1

C. 3 D. 6

Ans. D

Sol. y = [x]

Area under the curve y = [x].

Area = 1 × 1 + 1 × 2 + 1 × 3

= 1 + 2 + 3

= 6

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19. Which is the following graph is correct for xm and x1/m?

Ans. C

Sol. Put m = 2, so y = x2 and y =

1

2x

Satisfy option C.

20. The following bar graph shows the number of students appeared in exam and also the

number of students passed for four schools P, Q R and S.

Average of success rate in percentage for 4 schools is ______ .

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Sol.

280 380 485 280

500 600 700 400100

4

+ + +

=

21. I do not think, you know that cause well enough to have opinion. Having said that I

agree with your other point.

A. Contrary to what I have said

B. Despite what I have said

C. As oppose to what I have said

D. In addition to what I have said

Ans. B

22. Which of the following word fits with respect to analogy?

A. Grown B. Grew

C. Growed D. Growth

Ans. A

23. Fill in the blanks

He is known for his unscrupulous ways. He always sheds _______ tears to deceive

people.

A. Fox B. Crocodile’s

C. Fox’s D. Crocodile

Ans. D

24. There is a flywheel which stress energy 1050J and whose speed varies from 110 rad/s

to 100 rad/s M.I of flywheel = _____?

25. A single degree of freedom oscillation is subjected to harmonic excitation as shown in

figure.

The non-zero value of ω, for which the amplitude of the force transmitted to the ground

will be F0 is

A. k

2m

B. k

m

C. 2k

m D.

k

2m

Ans. C

Sol. t

0

F

F=

Ft = F0 ⇒ ϵ = 1

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n2 =

k

2m

=

2k

m =

26. A four bar mechanism is shown below for the mechanism

A. 200 mm B. 350 mm

C. 80 mm D. 300 mm

Ans. C

Sol. (s + l) ≤ R + Q

600 mm → longest link

P = 300 mm, Q = 400 mm

S + 600 ≤ (300 + 400)

S ≤ 100

For crank rocker, PQ should be shortest

27. X and Y are two random variables with mean 0.5

Z = (X+Y)

And ( ) ( )2 2X 0 Y = =

The correlation coefficient of Z is _____

Sol. ( ) ( ) ( )2 2 2Z X Y 0 = + = Hence, Z is a constant so answer should be 0.

28. pqrs are two coded!

P α α

Q α β

Then what is rs?

A. βα ββ B. αβ ββ

C. ββ αβ D. ββ βα

Ans. A

29. A company hiring 4 person, 3wopmen and 5 men were applied consider equally likely

find probability of atleast 1 women got selected

Sol. probability = 1 – probability that none women is taken (P(0))

Now, (P(0)) = 5C4/8C4 = 5/70

So probability = 1 – 5/70 = 0.92

30.

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Calculate net unbalanced force when mass A is removed

r = 100 mm

ω= 10 rad/s

m = 10 gm

Sol. If mass A is removed, then system becomes unbalanced.

Fresultant = Net unbalanced force

( ) ( )22

x yF F= +

ΣFx = mrw2(cos90° + cos180° + cos270°)

= mrω2(0 – 1 + 0 –1)

Σfx = –1

Σfy = 0

( ) ( )22

r x yF F F= +

Fr = 0.1N = Net unbalanced force.

31. Given that T-S (temperature v/s entropy) at constant pressure & volume are plotted.

The values of CP & CV are given. The ratio of slopes of constant pressure and constant

volume at point of intersection will be

A. CP/CV B. CV/CP

C. P V

P

C C

C

− D. P V

V

C C

C

−

Ans. B

Sol. Tds equation are

Tds = dU + pdv → (1)

Tds = dU + pdV → (2)

From 1st Tds relation

Tds = dU + pdV

At constant volume, dV = 0

Tds = dU = CvdT

VV

dT T

ds C

=

at constant volume

So, at constant volume, slope of (T –S) is T/CV

From 2nd Tds relation, Tds = dH – vdP

At constant pressure, dP = 0

Tds = dH = CpdT

pp c

T T

S C=

=

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So, ratio of slope of constant pressure & volume = v

p v p

CT T/

C C C=

32. Match the column

P. tempering 1. Hardening

Q. quenching 2. Toughening

R. annealing 3. Softening

S. norm aliasing 4. Strengthening

Sol. P. tempering 4. Strengthening

Q. quenching 1. Hardening

R. annealing 3. Softening

S. norm aliasing 2. Toughening

33. Multiplication of real valued sq. matrix is always

A. not possible

B. commutative AB = BA

C. associative (AB) C = A (BC)

D. can’t say

Ans. C

34. ( )

( )

c 1 x

c 1 xx 1

1 elim

1 xe

− −

− −→

−

−

Sol. apply L hospital rule, you get answer as C/C+1.

35. In a two stage compression with perfect intercooling the over all pressure ratio (rp) is

6. If initial pressure is 100 kPa. What is the pressure 1st stage of compression?

Sol. Pressure after 1st stage compression (P2) for perfect intercooling.

Overall pressure ratio (rp) overall = 6

( ) 3p overall

1

Pr

P=

For perfect intercooling, intermediate pressure (P2) = 1 3P P

P1 = 100 kPa

P3 = 6P1 = 600 kPa

2P 100 600=

= P2 = 244.9 kPa

36. Value of Joule Thompson coefficient for ideal gas

A. zero B. positive

C. negative D. indeterminate

Ans. A

Sol. Joule Thomson coefficient for real gas,

( )Ph P

T 1 Vµ T V ... 1

P C T

= = −

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For an ideal gas, PV = RT

P

VT V

T

−

So, P

VT P R

T

=

( )P

V R...... 2

T P

=

Putting eqn (2) in eq (1)

P

1 Rµ T V

C P

= −

Now, since, 12T

VP

=

P

1µ V V 0

C= − =

µ = 0 (for ideal gas)

37. Line Balancing

Production rate = 4/hr.

Processing time = 60 minutes.

2nd Case

Production rate = 4/hr

Processing time reduced by 30%

WIP By

A. Increase by 25%

B. Decrease by 30%

C. Increase by 30%

D. Decrease by 25%

Ans. B

Sol. Production = 4 units

2nd case

Production

= 4 × 0.7 = 2.8

2.8 units

% reduction 4 2.8

1004

− =

= 0.3 × 100

= 30%

38. The mechanical efficiency of an Otto engine is 80% Heat input is 10 kW and compression

ratio of 8. Find brake power.

Sol. Thermal efficiency of Otto engine = ( )

1

11

r−

−

Where, r is compression ratio ( )

1.4 1

11

8−

= −

= 0.56 = 56%

I.P

Heat input =

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Indicated Power (I.P) = η × Heat input

= 0.56 × 10 = 5.6 kw

Mechanical efficiency (ηm) = ( )

( )

Brakepowe B.P.

Indicatedpower I.P.=

B.P.

0.85.6

=

B.P. = 4.48 kW

Brake power is 4.48 kW.

39. Aakash & Shweta have two separate business in which two products A & B are made.

Revenue per unit of A is 2000 Rs. & B is Rs. 3000.

Units of A ≥ B

Units of B ≤ 10.

Given units of A < 15 for Aakash’s business and units of A < 20 “Shweta’s”.

Find difference of max. revenue.

Sol. Let units of A = x

Let units of B = y

For Aakash.

X < 15

Given y ≤ 10

But x ≥ y

10 ≤ x ≤ 15

Given above the feasible regions max. revenue will happen at (15, 10) ∴ max revenue

= 14 × 2000 + 10 × 3000 …….(i)

For Shweta

Let units of A = X1

Let units of B = Y2

Given X2 ≤ 10 & X1 < 20 & X1 ≥ X2 ≥ 10.

X2 ≤ 10 & 10 ≤ X1 < 20.

Maxima will occur at (19, 10).

Max revenue = 19 × 2000 + 10 × 3000…….. (ii)

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Difference = (ii) – (i) = 5 × 2000 = 10000 Rs.

40. For 1 kg mass of gas at constant pressure, having initial temperature of 127°C. If the

volume gets doubled. What is the magnitude of work done in KJ ? Take R = 287 J/Kg K.

Sol. For a constant pressure process,

work done (W) = p(V2 – V1)

W = mR(T2 – T1)

[from ideal gas, eqn. pV = mRT)

21

1

TmRT 1

T

= −

2

1

T1 0.287 400 1

T

= −

…(i)

Now, at constant pressure, Ideal gas eqn. becomes

2 2

1 1

V T

V T=

Since, V2 = 2V1

2 1

1 1

T 2V

T V=

2

1

T2

T= …(ii)

Putting eqn. (ii) in eqn. (i) we get

W = 1 × 0.287 × 400[2 – 1]

= 114.8 KJ

41. There is an ideal Rankine cycle operating between 30 bar & 0.04 bar. Work output from

turbine = 903 KJ/Kg. Work input to feed pump = 3 KJ/Kg. Find specific steam

consumption ?

Sol. specific steam consumption (ssc) = net

3600

W

Net work (Wnet) = Turbine work – Pump work

= 903 – 3 = 900 KJ/Kg

specific steam consumption = net

3600

W

3600

900=

= 4 Kg/Kwh

42. Match the following

Reynolds No. Buoyant/viscous

Grashoff Momentum diffusivity /thermal diffusivity

Nusselt Inertia force/Viscous force

Prandtl No. Conv. H.T/cond. H.T.

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Sol.

Reynolds No. Inertia force/Viscous force

Grashoff Buoyant/viscous

Nusselt Conv. H.T/cond. H.T.

Prandtl No. Momentum diffusivity /thermal diffusivity

43. Concentric tube counter flow

Thi = 102°C Tci = 25°C

Tho = 65°C Tco = 42°C

Sol. 1 2

1

2

T TLMTD

Tn

T

− =

∆T1 = Thi – Tco = 60°

∆T2 = Tho – Tci = 40°

60 40

LMTD 49.33 C60

n40

−= =

44.

A. BF, DH, GC only

B. BF, DH, GC, FG, GH only

C. BF, DH only

D. BF, DH, GC, CD, DE

Sol. Moment at A

ER(P 2 ) 4

2 =

E

PR

2 =

Pt. E

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E1

RF sin45 0

2 + =

1 EF R0

2 2+ =

F1 = –RE

1

PF

2= −

E1 2

RF cos45 F 0

2+ + =

2

P PF 0

2 2

−+ + =

2F 0=

Correction option D

BF = 0

DH = 0

GC = 0

45. A spherical body (radius = 0.5 mm) is initially at 100°C is kept in a fluid having temp of

20°C. the temp of body is 28°C at time, t = 1.38 sec. Density of fluid = 850 kg/m3 Cp =

400 J/kg k. Consider lumped body, calculate convective heat transfer coefficient.

A. 292.2 B. 288.4

C. 299.2 D. 312

Sol.

hAt

3VCp

0

T Te

T T

−

−=

−

h 1 1.35400

0.5 3850 103

28 20e

100 20

−

− −=

−

h 0.02388e

80

− =

–2.3 = –h × 0.0238

h = 96.64

46. 6 Ratio of force while new clutch (uniform pressure) F1 and uniform wear F2 while

Diameter are give 50 mm and 250 mm respectively. H is same in both cases. F2/F2 =

_____

Sol. Ratio

1

2

F?

F=

Sol. U.PT

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3 3

1 2 2

P D dT

3 D d

−=

−

( )1

12 2

FP

D d4

=

−

( )2

PT D d

4

= +

D = 250 mm

D = 50 mm

( )

3 31

2 21

22

P D d

3 D dT

PTD d

4

− −

=

+

( )1

3 3 3 32

2 2 2 2

D d 300P 754 4 0.871P 86.111 D d 1 250 50

3 3D d 250 50

+

= = = = − − − −

1

2

P0.871

P=

( )

22

FP

dD d

2

=

−

47. If Lf(t) = 2 2

1

S + . Find f(t).

Sol. 2 2

1Lf(t)

S=

+

1

2 2

1L f(t)

S

− =

+

or

2 2

aL sinat

S a=

+

2 2

L sin tS

=

+

1

2 2

1 sin tL

S

− =

+

48. Froud Number is the ratio of____

Sol. Froud Number is the ratio of inertia force/gravity force.

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49. ρ = 1000 kg/m3, g = 10 m/s

Calculate the Net moment about A

Sol. Hydrostatic force in 1st & 2nd reservoir = gAx

A = h × 1 as width is unity

A = h

x → centroid of centre of gravity

h

2=

2

1 2

h ghF F gA

2 2

= = =

21

1

ghF

2

=

22

2

ghF

2

=

Moment around A = F1 × h* – F2h*

GICentre of pressure (h*) XAX

= +

3 3

G

bh hI

12 12= =

F gAh=

2h gh

g h 12 2

= =

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3hh 2h12h*

h 2 3h 1

2

= + =

Now this centre of pressure is from top

from bottom distance of centre of pressure = h

3

So, Net moment around A

2 21 21 2

h h1 1gh gh

2 3 2 3= −

3 31 2h h1

g2 3 3

= −

3 31000 104 1

2 3

= −

= 105 KNm

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