1. Answer: Bgs-blog-images.s3-ap-southeast-1.amazonaws.com/gate-exam/wp... · According to given...

1. Answer: B

The word perquisite (noun) means: prerogative, perk, a special right

The word Privilege (noun) means: a special right, advantage, or immunity granted or available only to a particular

person or group.

Hence, the words Perquisite and Privilege are synonyms.

2. Answer: D

The first sentence is ‘Today, the earth has many satellites besides the moon.’ Now, the second sentence must focus on

this sentence, means in this sentence, elements of the first sentence should be described.

Therefore Only ‘R’ seems fit to be added after sentence 1. In this sentence, qualities of the satellites have been told.

Similarly, sentence R should be described in the next sentence. Hence sentence Q will be added after it. Again sentence

S is in continuation to sentence Q and the left out sentence i.e. P appear at last. Hence, the correct sequence will be

RQSP.

3. Answer: A

Make hay while the sun shines.

It means: to make good use of opportunities while they last.

4. Answer: A

Transient means a person who is staying or working in a place for a short time only. Fleeting has the same meaning

lasting for a very short time. Hence option (A) is correct.

5. Answer: C

The pattern is:

1 + 4 = 5

5 + 7 (=4+3) = 12

12 + 12 (=7+5(3+2)) = 24

24 + 19 (= 12 +7(5+2)) = 43

43 + 28 (=19+9(7+2)) = 71

6. Answer: C

Let age of Simar be x

Age of Ravi = 4 + x

4 years from now

Age of simar = x + 4

Age of Ravi = x+8

According to given x+8 : x+4 = 9 : 8

x = 28

Simar’s age = 28

Ravi’s age = 32

15 years ago Ravi’s age was 32 – 1 5 = 17 years

7. Answer: C

A and B together can finish a piece of work in 20 days

Hence work of (A+B) in one day = 1/20 ………. (1)

B and C together can finish a piece of work in 10 days

Hence work of (B+C) in one day = 1/10 ………. (2)

A and C together can finish a piece of work in 12 days

Hence work of (A+C) in one day = 1/12 ……….. (3)

Now adding equation (1), (2) and (3)

1 1 12(A B C)

20 10 12

5 14 73 6

60 60 30

Or

7(A B C)

60

Therefore A, B, C jointly can finish the same work in 60 48 days.

7 7

8. Answer: B

Production of maize in 1989 = 234 million tonnes


Hence decrease in production of maize as against the previous year = 234 - 228 = 6 MT

% decrease = 6×100/234 = 2.5641%

9. Answer: C

Production of wheat in 1990 = 560 million tonnes

Production of wheat in 1991 = 680 million tonnes

Hence increase in production of wheat = 680 - 560 = 120 MT

Production of rice in 1990 = 240 million tonnes

Production of rice in 1991 = 300 million tonnes

Hence increase in production of rice = 300 - 240= 60 MT



Hence increase in production of maize = 380 - 228 = 152 MT

Production of Other Cereals in 1990 = 420 million tonnes


Hence increase in production of Other Cereals = 460 - 420 = 40 MT

Therefore In 1991, the increase in production over the previous year was maximum for maize.

10. Answer: B

Production of other cereals in 1988 = 350 million tonnes


Hence increase in production of other cereals = 400 - 350 = 50 MT






Hence increase in production of Other Cereals = 460 - 420 = 40 MT




Therefore the increase in the production of other cereals (over the previous year) was minimum during the year 1990:

(20 MT)

11. Answer: C

The pressure at any depth in a liquid is the sum of the pressure at the surface of the liquid and the pressure due to the

liquid at a given depth in the liquid. Since both are case of fresh water so the weight of water above measuring point

will remain same.

12. Answer: B

u = λxy3 - x2 ; v = xy2 – 3/4 y4

Now we know the continuity equation for incompressible flow is ∂u/∂x + ∂v/∂y = 0

∂u/∂x = λy3 - 2xy

∂v/∂y = 2xy – 3y3

Put the values in continuity equation: ∂u/∂x + ∂v/∂y = 0

(λy3 - 2xy) + (2xy – 3y3) = 0

By solving this equation, we get λ = 3.

13. Answer: D

In white cast iron, carbon is present in form of iron carbide (Fe3C) which is hard and brittle. Carbon may be present in

form of cementite and graphite in pig iron. The cast iron identified depends on graphite form and part of cementite

that can be white, grey, malleable and ductile iron. Hence the presence of iron carbide increases hardness and makes it

difficult to machine. On the other hand, white cast irons are abrasion resistant. Cementite is a form of carbon present

in white cast iron.

14. Answer: A

The molten metal used during the casting process may trap and contain gases. There are various reasons that gases

are absorbed into the metal melt during manufacture. Turbulent flow of the casting material through the system may

cause it to trap gas from the air. Gases may be trapped from material or the atmosphere in the crucible when the melt

is being prepared. Gases may be trapped from reactions between the molten metal and the mold material.

15. Answer: A

Stress is force per unit area which is the ratio of applied force F and cross section defined force per area. It is seen that

the strain occurs on body and observed a rupture.

16. Answer: B

Shear stress is component of stress coplanar having material cross section that arises from force vector component

which is parallel to cross section. The normal stress arises from force vector component that are perpendicular to

material cross section on which it acts. From the above diagram we see that q = ρ /A. Hence Shear stress τ across

section X-X will be ρ/A

17. Answer: D

The area under the PV curve gives the work done, the maximum work is for the isobaric process indicated by the curve

AB. So the correct option is D.

18. Answer: A

It is Heat Engine where more positive work is done in its environment as it shows outward pressure on expansion. So

the heat engines will have capability of converting heat into more useful work output. On the other hand, refrigerator

will absorb the work and uses energy to move heat from cold to hot reservoirs.

19. Answer: D

In a Turbine Engine, compressor power requirements is very high that ranges from 30% - 80% of power output of

turbine.

20. Answer: D

Solution: In case of rotation the rigid body is rotated about an axis passing through the particular point selected for

translation. Now the rigid body is in a rotation. Since the point selected is not fixed, it is a relative motion.

21. Answer: D

When the same direction of rotation is required for both the driver and the follower, an idler wheel is used.

Transmission or movement ratio can be expressed as μM = ND / NF = NF / ND

where

μM = movement ratio

ND = revolutions of driver (rpm)

NF = revolutions of follower (rpm)

TF = number of teeth on follower

TD = number of teeth on driver

22. Answer: D

A steady-state heat transfer analysis is used to determine the steady-state temperature distribution and heat flow.

This type of analysis can be performed when the temperature at every point within the model, including the surfaces,

is independent of time. For one-dimensional heat conduction the temperature depends on one variable only.

23. Answer: A

It is half plane conduction method.

24. Answer: A

A fin is a rectangular metal strip to the surface of heat transfer which extends from an object to increase rate of heat

transfer to or from environment by increasing convections. They are sometimes known as extended surface.

25. Answer: D

Rubber is stiffer dynamically than in a static mode; and, since the static to dynamic stiffness ratio varies with

individual compounds, it may be advisable to specify the dynamic characteristics of the rubber for such applications.

26. Answer: A

Plastic deformation is not acceptable in most mechanical design situations, because the permanently deformed part

may no longer serve its intended purpose, and from the mechanical design stand point we may say that the part has

failed.

27. Answer: A

One test focuses on the nominal stress required to cause a fatigue failure in some number of cycles. The fatigue

analysis shows relationship between stress amplitude and number of cycles it executes before it fails. It is noted that

stress amplitude increases the number of cycles for failure to decreases.

28. Answer: A

A free-body diagram is a unique type of vector diagram where vector are labelled with different types of force that

represents the object or forces. Tension force F1 is in vertical direction supporting a weight which is equal to force F2

and F3. In the figure, we see that the forces F2 and F3 acting at center balances force acting downward. Hence F2 and

F3 = F1.

29. Answer: B

Truss is designed to support heavy loads in comparison to its weight. The joints in this are of pin type and has tensile

and compression axial forces. The members are free to rotate about the pin. In case of frame, at least one of its

individual member is a multi-force member. In this the members of frames are connected rigidly at joints by means of

welding and bolting. A truss cannot transfer moments and members are subjected to only axial forces.

30. Answer: A

From the above situation, it is noticed that car, tyre and driver will experience similar impulse therefore they all will

have similar change in momentum. But the bug will experience more change in momentum.

31. Answer: D

The simple moving average method can be used if the underlying demand pattern is stationary. This method include

new demand data in the average after discarding some of the earlier demand data.

Let mt = moving average at time t

yt = demand in time t and

n = moving average period

1 1

1t t n

t

y ym

n

32. Answer: D

Production flow analysis (PFA) is a comprehensive method for material analysis, Part family formation, design of

manufacturing cells and facility layout design. These information are taken from the route sheet.

33. Answer: B

We have 2

25 6 0

d y dyy

dx dx

The A.E. is m2 – 5m + 6 = 0

m = 3,2

The CF is yc = C1 e3x + C2 e2x

Since Q = 0, thus y = C1e3x + C2e2x

34. Answer: C

We have X = AX + BU where λ is set of Eigen values

and W = CW+ DU where µ is set of Eigen values

If a liner system is equivalently represented by two sets of state equations, then for both sets, states will be same but

their sets of Eigen values will not be same i.e.

X = W but

35. Answer: C

3

0

I sin d

3

0

(1 cos )sin d

Let cos θ = t

12

1

(1 t )dt

sin d dt, 0,cos0 1 t

,cos 1 t

12

1

(1 t )dt

13

1

tt3

1 11 13 3

2 2 4I3 3 3

36. Answer: C

It's the U-tube manometer, which is a U-shaped glass tube partially filled with liquid needed for measuring large gauge

pressures. In this the pressure at certain level in a liquid is same at all points which is used in manometer. In a

manometer, since the gas in the bulb is pushing more than the atmospheric pressure, so it is clear that Pgas > Patm

37. Answer: C

It is observed that an ideal fluid is non-viscous medium possesses a steady motion which does not experience

turbulence and remains incompressible. It is noted that an incompressible fluid is one in which the fluid density

remains constant immaterial of time and position, hence statement C is correct.

38. Answer: D

We see that average density of fluid:

ρaverage = ρgZaverage = 1000 kg/m3 x 9.81 m/s2 x 1/2 x 10m = 49000Pa

Now R/w = ρaverage x h = 49000 x 10 = 490000 N/m

Now from the figure F + Fh = R

If we take the moment along R, then: 2F = Fh

F/w = 1/10 (R/w) = 490000N/m / 10 = 49 KN/m

39. Answer: C

In non traditional machining process, material removals occur with chip formation as in AJM, chips are of microscopic

size are used. In this, there may not be physical tool present. The tools need not be harder than the work piece

material as copper is used as tool material to machine hardened steels. The non-traditional machining process do not

necessarily use mechanical energy to provide material removal.

40. Answer: A

A self-centering chuck which is also a scroll chuck uses jaws that are interconnected by scroll gear to hold workpiece.

A drill chuck is special self-centering chuck, used to hold drill bits or other rotary tools. A spider typically consists of

ring of metal with screw threads tapped radially into it in which screws serve as independent jaws.

41. Answer: B

The life of a cutting tool is therefore determined by the amount of wear that has occurred on the tool profile and which

reduces the efficiency of cutting to an unacceptable level, or eventually causes tool failure. Tool wear describes the

gradual failure of cutting tools due to regular operation. It is a term often associated with tipped tools, tool bits, or drill

bits that are used with machine tools. It is seen that the rate of tool wear depends on cutting temperature, as any

measures applies to lower the cutting temperature that reduces tool wear.

42. Answer: B

Relationship between Stress and Strain are derived on the basis of the elastic behaviour of material bodies. Hooke’s

law states that stress is proportional to strain upto elastic limit. If I is the stress induced in a material and e the

corresponding strain, then according to Hooke’s law

p / e = E, a constant.

43. Answer: D

The cylindrical shaft is fixed at one end and is twisted at other end. The bar is twisted using the torque T which is the

product of applied force F and distance d. It is noted that torque T = F.d which is applied in plane that is perpendicular

to axis of shaft which is in torsion. So Torque applied will be T = F.d. With this, the option (D) is correct.

44. Answer: C

We see that in an ideal gas:

pV = NkT

so T is maximum when pV is maximum

It shows that the point which is farthest from origin will have maximum temperature of gas. Hence from the graph,

point C is away.

45. Answer: D

The area under the PV curve gives the work done, the maximum work is for the isobaric process indicated by the curve

AB. So the correct option is D.

46. Answer: D

If two systems are in thermal equilibrium with a third system, then they are in thermal equilibrium with each other"

47. Answer: B

Solution: The average temperature at which heat is transferred to steam can be increased without increasing the

boiler pressure by superheating the steam to high temperatures. The effect of superheating on the performance of

vapour power cycle is shown on a T-s diagram the total area under the process curve 3 3− l represents the increase in

the heat input. Thus both the net work and heat input increase as a result of superheating the steam to a higher

temperature. The overall effect is an increase in thermal efficiency, since the average temperature at which heat is

added increases.

48. Answer: D

Slider-crank mechanism is a special case of four bar mechanism which converts rotary motion to reciprocating motion

or vice versa. Its parameters are used to explain angles and link lengths which extends and fold dead centre positions

results when crank and coupler are collinear.

49. Answer: B

In the above figure, P shows the pitch point which divides the line in the centers and its position shows velocity ratio of

two teeth. In this, there are two velocities which is at common point K along N1N2 having similar magnitude and

direction.

So we can write: ω1/ω2 = O2N2/O1N1

As per the intersection of tangents N1N2 and line of center O1O2 shows point P, hence

∆O1N1 P = ∆O2N2 P

So we see that the relationship among angular velocities of driving gear to driven gear or velocity ratio is

ω1/ω2 = O2P/O1P

50. Answer: D

Grash of number Gr, is non-dimensional parameter used in correlation of heat and mass transfer due to thermally

induced natural convection at solid. It is proportional to (buoyancy force) / (viscous force) and is used in heat transfer

in normal and free convection calculations. At higher Grashof numbers, boundary layer is turbulent, while at low Grash

of numbers, boundary layer is laminar.

51. Answer: A

We see that:

λm = ἀ 1/T

λm = b/T

λpeak T = b = Weins constant with 2.89. x 10-3 kelvin

frequency vm = cT/b

We see that λpeak T = 0.5796 x 10-6m

Applying formula for wavelength, we see that wavelength corresponds to quantum energy:

hν = 2.1406

52. Answer: C

We see that:

Fc = Fi – [Kc/Kb + Kc]/Fe

Since clamping force Kc = 300 and Fi = 1500, then:

300 = 1500 – [Kc/Kb + Kc]/Fe

Now since bolt stiffness becomes twice as clamping material stiffness so:

300 = 1500 – [1/1 + 2]/Fe

Solving we get external load Fe as: 900 = 4500 – Fe

Fe = 3600

Now the bolt force will be: Fb = Fi + [Kb/Kb + Kc]/Fe = 1500 + 2/3 * 3600 = 1500 + 2400 = 3900

Bolted joints consist of cap screws or studs and secure it with the mating of screw threads.

53. Answer: A

The notch sensitivity factor q is considered by comparing theoretical stress concentration factor Kt and fatigue notch

factor Kf.

Kt = Smax/S

Kf = Se (unnotched) / Se (notced)

In ideally elastic members, the ratio of these stresses designate the theoretical stress concentration factor KT while

fatigue notch factor Kf relates the unnotched fatigue strength of a member to its notched fatigue strength.

54. Answer: A

We see that the n2=50 cw

Now [n2/A] / [n4/A] = n2 – nA/n4 – nA = 50 – nA/0 – nA

[n2/A] / [n4/A] = [n2/A/n3/A]*[n3/A/n4/A] = - N3/N2(N4/N3)

We see that Ring gear is N2 + 2 N3 = 56

Now:

50 – nA/0 – nA = - N3/N2(N4/N3)

50 – nA/0 – nA = - 21/14(56/21)

50 – nA/0 – nA = - 4

50 – nA = 4nA

nA = 5 cw

55. Answer: A

The stiffness, k, of a body is a measure of the resistance offered by an elastic body to deformation. For an elastic body

with a single degree of freedom, the stiffness is k=F/δ

The basic equation of pvw is - S δl - Z δz = 0

Further we see that:

l =(π/2 –φ)/2 * 2 r sin; δl = =(π/4 –φ/2) δφ

z =Const + 5 r sinφ; δz = 5 r cos φ δ φ

On substitution:

S r cos(π/4 –φ/2) δ φ - Z 5 r cos φ δ φ = 0

Hence

On solving the force in the spring:

S = k (2 r sin 30o - l0)

Now stiffness will be:

k =250/20.1 sin 300 - 1

= 5215 N/m

56. Answer: D

It is observed that moving reference were attached to A as:

rB = rA + rB/A

vB = vA + vB/A

aB = aA + aB/A

By comparing:

rB/A = - rA/B

vB/A =- vA/B

aB/A =- aA/B

57. Answer: A

We see that sin θ = b/r = 1.5/1.5 = 1, => θ = 90°

Apply momentum conservation:

mu + 0 = mv' + mv"

u = v cos45 + v cos45 = 2*0.707v = 1.414v = 1

v = 1/1.414 = 0.707 m/s

58. Answer: A

In a cantilever beam, the force equilibrium at point C will be R1 + R2 which is 40 N

Now if take the moments about point A in clockwise direction, then:

40·2 - 20 - 6·R2 = 0

So we have R1 = 30N and R2 = 10N

We see that the bending moment before and after point C will be: 10Nx3m - 20Nx2m = -10Nm

59. Answer: D

Given : Mean cycle time = 10 min

The workers performing at 90% efficiency.

So, Normal time = 90

10 9min100

Allowance = 10%

Standard time = Normal time + Allowance

109 9 9 0.9 9.9 min

100

60. Answer: C

According to 3-2-1 principle, only the minimum locating points should be used to secure location of the work piece in

any one plane.

(A) The workpiece is resting on three pins A, B, C which are inserted in the base of fixed body.

The workpiece cannot rotate about the axis XX and YY and also it cannot move downward. In this case, the five degrees

of freedom have been arrested.

(B) Two more pins D and E are inserted in the fixed body, in a plane perpendicular to the plane containing, the pins A,

B and C. Now workpiece cannot rotate about the Z-axis and also it cannot move towards the left. Hence the addition of

pins D and E restrict three more degrees of freedom.

(C) Another pin F in the second vertical face of the fixed body, arrests degree of freedom 9.

61. Answer: C

4 2

M2 4

4 2

M I2 4

Given eigen vector

4 2 101

02 4 101

(4 )(101) 2(101) 0

4 2 0

6

62. Answer: A

2 2D K 0

D jK

y AcosKx jBsinKx

At x = 0, y = 0

A = 0, y = jB sin Kx

x = a, y = 0

0 = B sin Ka

B 0 else y = 0 always

sin Ka = 0

m xK

a

mm

m xy A sin

a

63. Answer: C

We need absolute maximum of

f(x) = x3 – 9x2 + 24x + 5 in the interval [1,6]

First find local maximum if any by putting

f '(x) 0

2i.e. f '(x) 3x 18x 24 0

2i.e. x 6x 8 0

x 2,4

Now f ''(x) 6x 18

f ''(2) 12 18 6 0

(So x = 2 is a point of local maximum) and

f ''(4) 24 18 6 0

(So x = 4 is a point of local minimum) Now tabulate the values of f at end point of interval and at local maximum point,

to find absolute maximum in given range, as shown below:

Clearly the absolute maxima is at x = 6 and absolute maximum value is 41.

64. Answer: C

P(no. of tosses is odd) = P(no. of tosses is 1, 3, 5, 7 . . .)

P(no. of toss is 1) = P(Head in first toss) 1

2

P(no. of toss is 3) = P(tail in first toss, tail in second toss and head in third toss)

1 1 1 1

2 2 2 8

P(no. of toss is 5) = P(T. T, T, T, H)

51 1

......etc2 32

So P(no. of tosses is odd 1 1 1

...2 8 32

Sum of infinite geometric series with 1

a2

and 1

r4

1 122 2

1 3 314 4

65. Answer: D

Plotting x & f(x):

3

0

( )f x dx is the area under the curve.

So, area of region I (Triangle)

1( ) ( )

2

1(0.09) (0.3)

2

base height

area of region II (Trapezium)

1 2

1( ) ( )

2

1(0.6 0.3) (0.09 0.36)

2

1(0.3)0 (0.09 0.36)

2

height base base

We can observe for each region, the height is the same = (0.3)

Also, observe that each value of f(x) except at x = 3.0 is included in two different regions.

Hence, the solution is as follows:

3

0

1 1 1( ) (0.3)(0.09) (0.3)(0.9 0.36) ... (0.3)(7.29 9.0)

2 2 2f x dx

1 1(0.3)(0.09 0.36 ... 7.29)(2) (0.3)(9.0)

2 2

= 9.045

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