GATE Mechanical Engineering 2014 Solved Paper - 3

8/9/2019 GATE Mechanical Engineering 2014 Solved Paper - 3

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GATE 2014: ME03 Solutions

ME: Mechanical Engineering

Duration: 180 minutes Maximum Marks: 100

Read the following instructions carefully.

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3. This question paper consists of 2 sections, General Aptitude (GA) for 15 marks and the subjectspecific GATE paper for 85 marks. Both these sections are compulsory. The GA section consistsof 10 questions. Question numbers 1 to 5 are of 1-mark each, while question numbers 6 to 10

are of 2-mark each. The subject specific GATE paper section consists of 55 questions, out of which question numbers 1 to 25 are of 1-mark each, while question numbers 26 to 55 are of 2-mark each.

4. Depending upon the GATE paper, there may be useful common data that may be required foranswering the questions. If the paper has such useful data, the same can be viewed by clickingon the Useful Common Data button that appears at the top, right hand side of the screen.

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examination. The examination will stop automatically at the end of 180 minutes.7. In each paper a candidate can answer a total of 65 questions carrying 100 marks.

8. The question paper may consist of questions of multiple choice type (MCQ) and numericalanswer type.

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Wiley Acing the GATE (Mechanical Engineering): GATE 2014: ME03Copyright c Wiley India

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GENERAL APTITUDE: GA

Q. 1–Q. 5 carry one mark each.

1. “India is a country for rich heritage and culturaldiversity”Which one of the following facts best supports theclaim made in the above sentence?

(a) India is a union of 28 states and 7 unionterritories

(b) India has a population of over 1.1 billion.

(c) India is home to 22 official languages andthousands of dialects.

(d) The Indian cricket team draws players fromover ten states.

Solution. The diversiy can be seen in the form of different cultures, languages and dialects.

Ans. (c)

2. The value of one U.S. dollar is 65 Indian Rupeestoday, compared to 60 last year. The Indian Rupeehas · · · · · · · · · .

(a) depressed

(b) depreciated

(c) appreciated

(d) stabilized

Solution. Depreciate (verb) means to diminish invalue over a period of time.

Ans. (b)

3. “Advice” is · · · · · · · · · .

(a) a verb

(b) a noun

(c) an adjective

(d) both a verb and a noun

Solution. “Advice” is a noun

Ans. (b)

4. The next term in the series 81, 54, 36, 24, · · · is· · · · · · · · · .Solution. In the series, the difference between81−54 = 27. Now, 27×(2/3) = 18 which isthe difference between 54 and 36. Similarly, thedifference between 36−24 = 12, so, 12×(2/3) = 8should be the difference between 24 and the nextnumber. Hence, the next term in the series is

24−8 = 16.Ans. 16

5. In which of the following options will theexpression P<M be definitively true?

(a) M < R > P > S (b) M > S > P > E

(c) Q < M > F = P (d) P = A > R > M

Solution. The expression P = A > R > Mdescribes that P < M.

Ans. (d)Q. 6–Q. 10 carry two marks each.

6. Find the next term in the sequence: 7G, 11K, 13M,· · ·

(a) 15Q (b) 17Q(c) 15P (d) 17P

Solution. In the series, the numerical value of thealphabet is followed by the alphabet, so 17Q is thecorrect choice.

Ans. (b)

7. The multi-level hierarchical pie chart shows thepopulation of animals in a reserve forest.

Mammal

Reptile

Bird

Insect

E l e p h a n t

TigerBeetle

R e d a nt

H o n e y

b e e

Moth

Hawk

D r o n g

o

B u l b

u l

C r o c o d

i l e

Snake

L e o p a r

d

B u t t e r

fl y

The correct conclusions from this information are:

(i) Butterflies are birds

(ii) There are more tigers in this forest than redants

(iii) All reptiles in this forest are either snakes orcrocodiles

(iv) Elephants are the largest mammals in thisforest

(a) (i) and (ii) only

(b) (i), (ii), (iii) and (iv)(c) (i), (iii) and (iv) only


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(d) (i), (ii) and (iii) only

Solution. The conclusion (iv) cannot be drawnfrom the chart, though as a fact it is true.

Ans. (d)

8. A man can row at 8 km per hour in still water. If it takes him thrice as long to row upstream, as torow downstream, then find the stream velocity inkm per hour.

Solution. Given that man’s speed is 8 kmph. If the speed of the stream is s kmph, then upstreamspeed will be 8−s kmph, and downstream speedwill be 8+ s kmph. Given that for a distance d,

d

8−s = 3× d

8 + s

8 + s = 3×8−3ss = 4 kmph

Ans. 4

9. A firm producing air purifiers sold 200 unitsin 2012. The following pie chart presents theshare of raw material, labour, energy, plantand machinery, and transportation costs in thetotal manufacturing cost of the firm in 2012. Theexpenditure on labour in 2012 is Rs. 4,50,000. In2013, the raw material expenses increased by 30%and all other expenses increased by 20%.

Plant and

30%

T r a n

s p

o r t a t i o n

10%

25%

Energy

Raw material20%

Labour15%

machinery

If the company registered a profit of Rs. 10 lakhsin 2012, at what price (in Rs.) was each air purifiersold?

Solution. Given that the total expenditure inlabor is Rs. 4,50,000. Using the chart, the totalexpenditure (x) will be given by

0.15×x = 4, 50, 000

x = Rs. 30, 00, 000

Given that profit is

p = Rs. 10, 00, 000

So, the total selling price will be

r = x + p

= Rs. 40, 00, 000

Total numbers of units sold is 200, thus, price of each unit is r/200 = Rs. 20, 000.

Ans. 20,000

10. A batch of one hundred bulbs is inspected bytesting four randomly chosen bulbs. The batchis rejected even if one of the bulbs is defective.A batch typically has five defective bulbs. Theprobability that the current batch is accepted is· · · · · · · · · .

Solution. The probability for atleast one bulb notto be defective is 95/100 = 0.95. The probabilitythat none of the bulbs is defective will be

P = 0.954

= 0.8145

Ans. 0.8145

END OF THE QUESTION PAPER


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MECHANICAL ENGINEERING: ME

Q. 1–Q. 25 carry one mark each.

1. Consider a 3×3 real symmetric matrix S such thattwo of its eigenvalues are a = 0, b = 0 withrespective eigen vectors

x1

x2

x3

,

y1

y2y3

If a = b, then x1y1+ x2y2 + x3y3 equals

(a) a (b) b

(c) ab (d) 0

Solution. Since the matrix is symmetric withdistinct eigen values, so the eigen vectors of thematrix are orthogonal. So,

x1x2

x3

×

y1

y2

y3

= 0

x1y1+ x2y2+ x3y3 = 0

Ans. (d)

2. If a function is continuous at a point,

(a) the limit of the function may not exist at thepoint

(b) the function must be derivable at the point

(c) the limit of the function at the point tends toinfinity

(d) the limit must exist at the point and the valueof limit should be same as the value of thefunction at that point

Solution. A function f (x) is said to be continuousat x = a if limx→a f (x) exists such that

limx→a f (x) = f (a)

Ans. (d)

3. Divergence of the vector field x2zi + xyˆ j−yz2k at(1,−1, 1) is

(a) 0 (b) 3

(c) 5 (d) 6

Solution. Divergence of the vector is given by

∇·

#»

F = ∂

∂xx2z+

∂

∂y

(xy)+ ∂

∂z−

yz2= 2xy + x−2yz

Value at (1,−1, 1) will be

∇· #»

F = 2+1+2= 5

Ans. (c)

4. A group consists of equal number of men andwomen. Of this group 20% of the men and 50% of the women are unemployed. If a person is selectedat random from this group, the probability of theselected person being employed is · · · · · · · · · .

Solution. Given that

P (M ) = 0.5

P (W ) = 0.5

P (U/M ) = 0.2

P (U/W ) = 0.5

The probability of selecting an unemployed personis

P (U ) = P (M )×P (U/M ) + P (W )×P (U/W )

= 0.5×0.2 + 0.5×0.5

= 0.35

So, the probability of selecting an employed person

isP (E ) = 1−P (U )

= 1−0.35

= 0.65

Ans. 0.65

5. The definite integral 31

(1/x) dx is evaluated usingTrapezoidal rule with a step size of 1. The correctanswer is · · · · · · · · · .

Solution. Given that step size h = 1. The valuesof x and (1/x) are tabulated below:

x 1 2 3

ln x 1 0.5 0.33

Using Trapezoidal rule,

ba

f (x) dx = h

2

f (x0) + f (xn)+ 2

n−1k=1

f (xk)

31

ln(1/x) dx = 1

2[1+0.33+2(0.5)]

= 1.165

Ans. 1.165


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6. A rotating steel shaft is supported at the ends.It is subjected to a point load at the center. Themaximum bending stress developed is 100 MPa.If the yield, ultimate and corrected endurance

strength of the shaft material are 300 MPa, 500MPa and 200 MPa, respectively, then the factor of safety for the shaft is · · · · · · · · · .


σb = 100 MPa

σy = 300 MPa

σu = 500 MPa

σe = 200 MPa

The rotating shaft is subjected to dynamic loadingin which minimum load will be zero. Therefore,

σa = 100+0

2= 50 MPa

σv = 100−0

2= 50 MPa

Factor of safety (N ) can be calculated in thefollowing ways

(a) Soderberg line criteria:

σaσy

+ σvσe

= 1N

50

300 +

50

200 =

1

N N = 2.4

(b) Goodman line criteria:

σaσu

+ σvσe

= 1

N 50

500 +

50

200 =

1

N

N = 2.8

Thus, factor of safety can be taken between 2.4and 2.8.

Ans. 2.4 to 2.8

7. Two solid circular shafts of radii R1 and R2 aresubjected to same torque. The maximum shearstresses developed in the two shafts are τ 1 and τ 2.If R1/R2 = 2, then τ 2/τ 1 is · · · · · · · · · .


R1

R2= 2

Using

τ = 16T

πd3

Ratio of shear stresses for the same value of torqueon both shafts will be

τ 1τ 2

= 23

13

= 8

Ans. 8

8. Consider a single degree-of-freedom system withviscous damping excited by a harmonic force.At resonance, the phase angle (in degree) of thedisplacement with respect to the exciting force is

(a) 0 (b) 45

(c) 90 (d) 135

Solution. At resonance, the displacement vectormakes 90 with respect to the exciting force.

Ans. (c)

9. A mass m1 of 100 kg travelling with a uniformvelocity of 5 m/s along a line collides with astationary mass m2 of 1000 kg. After the collision,both the masses travel together with the samevelocity. The coefficient of restitution is

(a) 0.6 (b) 0.1(c) 0.01 (d) 0


m1 = 100 kg

u1 = 5 m/s

m2 = 1000 kg

u2 = 0

v1 = v2

The coefficient of restitution is calculated as

e = v2−v1

u2−u1

= 0

Ans. (d)

10. Which one of following is not correct?

(a) Intermediate principal stress is ignored whenapplying the maximum principal stress theory

(b) The maximum shear stress theory gives themost accurate results amongst all the failuretheories

(c) As per the maximum strain energy theory,

failure occurs when the strain energy per unitvolume exceeds a critical value


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(d) As per the maximum distortion energy theory,failure occurs when the distortion energy perunit volume exceeds a critical value

Solution. Of all the theories dealing with ductilebehavior, the distortion energy theory agrees bestwith experimental data.

Ans. (b)

11. Gear 2 rotates at 1200 rpm in counter-clockwisedirection and engages with Gear 3. Gear 3 andGear 4 are mounted on the same shaft. Gear 5engages with Gear 4. The numbers of teeth onGears 2, 3, 4 and 5 are 20, 40, 15 and 30,respectively.

5

2

3

4

20 T

30 T40 T15 T

The angular speed of Gear 5 is

(a) 300 rpm counterclockwise(b) 300 rpm clockwise

(c) 4800 rpm counterclockwise

(d) 4800 rpm clockwise


N 2 = 1200 rpm

T 2 = 20

T 3 = 40

T 4 = 15

T 5 = 30

The directions of rotation of the gears are shownin the figure:

5

2

3

4

20 T

30 T40 T15 T

Angular speed of gear 5 will be given by

N 5 = N 2× T 2T 3× T 4

T 5

= 300 rpm (ccw)

Ans. (a)

12. Consider a long cylindrical tube of inner and outerradii, ri and ro, respectively, length, L and thermalconductivity, k. Its inner and outer surfaces aremaintained at T i and T o, respectively (T i >T o). Assuming one-dimensional steady state heatconduction in the radial direction, the thermalresistance in the wall of the tube is

(a) ln (ri/ro)

2πkL (b)

1

2πrikL

(c) ln (ro/ri)

2πkL (d)

ln (ri/ro)

4πkL

Solution. Thermal resistance of the wall of thetube is given by

Rth = ln (ro/ri)

2πκL

Ans. (c)

13. Which one of the following pairs of equationsdescribes an irreversible heat engine?

(a)

δQ > 0 and

δQ

T < 0

(b)

δQ < 0 and

δQ

T < 0

(c)

δQ > 0 and

δQ

T > 0

(d)

δQ < 0 and

δQ

T > 0

Solution. For irreversible engines (using Clausiustheorem),

δQ > 0 δQ

T < 0

Ans (a)

14. Consider the turbulent flow of a fluid through acircular pipe of diameter, D. Identify the correctpair of statements.

I. The fluid is well-mixedII. The fluid is unmixed


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III. ReD < 2300

IV. ReD > 2300

(a) I, III (b) II, IV

(c) II, III (d) I, IV

Solution. In turbulent flow through circular pipe,the fluid get well-mixed and Reynolds number isgreater than 2300.

Ans. (d)

15. For a gas turbine power plant, identify the correctpair of statements.

P. Smaller in size compared to steam power plantfor same power output

Q. Starts quickly compared to steam power plantR. Works on the principle of Rankine cycle

S. Good compatibility with solid fuel

(a) P, Q (b) R, S

(c) Q, R (d) P, S

Solution. Gas turbine power plants are smallerand start quickly compared to the steam powerplants (boiler) for same power output. Gas powerplants work on Brayton cycle while steam powerplants work on Rankine cycle. Gas power plantsuse liquid fuel (kerosene, diesel, petrol) while

steam power plants use solid fuel (coal).

Ans. (a)

16. A source at a temperature of 500 K provides 1000kJ of heat. The temperature of environment is27C. The maximum useful work (in kJ) that canbe obtained from the heat source is · · · · · · · · · .


T 1 = 500 K

Q1 = 1000 kJ

T 2 = 27

C = 300 K

The maximum useful work of the heat source willbe

W = T 1−T 2

T 1×Q1

= 400 kJ

Ans. 400 kJ

17. A sample of moist air at a total pressure of 85 kPahas a dry bulb temperature of 30C (saturation

vapour pressure of water = 4.24 kPa). If theair sample has a relative humidity of 65%, the

absolute humidity (in gram) of water vapour perkg of dry air is · · · · · · · · · .


p = 85 kPaDBT = 30C

ps = 4.24 kPa

φ = 65%

Using

φ = pv ps

pv = 0.65×4.24

= 2.756

Absolute humidity is calculated as

w = 0.622 pv p− pv

= 0.622× 2.756

85−2.756

= 0.0208 kg/kg of dry air

= 20.8 g/kg of dry air

Ans. 20.8 g/kg of dry air

18. The process of utilizing mainly thermal energy forremoving material is

(a) ultrasonic machining

(b) electrochemical machining(c) abrasive jet machining

(d) laser beam machining

Solution. Laser beam machining, a thermoelectricprocess, is accomplished largely by materialevaporation in a controlled manner using highlyfocused and high energy density of laser.

Ans. (d)

19. The actual sales of a product in different monthsof a particular year are given below:

Sep Oct Nov Dec Jan Feb

180 280 250 190 240 ?

The forecast of the sales, using the 4-monthmoving average method, for the month of Februaryis · · · · · · · · · .

Solution. Forecast of sales for month of February,using 4-month moving average method will begiven by

= 280+250+190+240

4= 240

Ans. 240


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20. A straight turning operation is carried out using asingle point cutting tool on an AISI 1020 steel rod.The feed is 0.2 mm/rev and the depth of cut is 0.5mm. The tool has a side cutting edge angle of 60.

The uncut chip thickness (in mm) is · · · · · · · · · .Solution. Given that

f = 0.2 mm/rev

d = 0.5 mm

ψ = 60

Uncut chip thickness is calculated as

t = f cos ψ

= 0.2×cos 60

= 0.1 mm

Ans. 0.1 mm

21. A minimal spanning tree in network flow modelsinvolves

(a) all the nodes with cycle/loop allowed

(b) all the nodes with cycle/loop not allowed

(c) shortest path between start and end nodes

(d) all the nodes with directed arcs

Solution. Minimum spanning trees have direct

applications in the design of networks. For aconnected but undirected graph, a spanning tree of that graph is a subgraph that is a tree and connectsall the vertices together. A weight can be assignedto each edge that indicates a number representinghow unfavorable it is, such as cost, time, distance.A minimum spanning tree (MST) in an undirectedconnected weighted graph is a spanning tree of minimum weight (among all spanning trees)

Ans. (c)

22. Match the casting defects (Group A) with the

probable causes (Group B).

Group A Group B

P. Hot tears 1. Improper fusion of twostreams of liquid metal

Q. Shrinkage 2. Low permeability of thesand mould

R. Blowholes

3. Volumetric contractionboth in liquid and solidstage

S. Cold shut 4. Differential cooling rate

(a) P-1, Q-3, R-2, S-4(b) P-4, Q-3, R-2, S-1

(c) P-3, Q-4, R-2, S-1

(d) P-1, Q-2, R-4, S-3

Solution. Hot tears are the cracks developed in themetal due to residual stresses in solidified metalcaused by solid shrinkage of the metal. Shrinkagecavity occurs due to volumetric contractions thatcan be avoided using proper riser design. Blowholes are trappings of air and appear as internalvoids because of poor ventilation. Cold shuts arecaused by mating of two cold streams of metalsmoving in opposite directions.

Ans. (b)

23. Cutting tool is much harder than the workpiece.Yet the tool wears out during the tool-workinteraction, because

(a) extra hardness is imparted to the workpiecedue to coolant used

(b) oxide layers on the workpiece surface impartextra hardness to it

(c) extra hardness is imparted to the workpiecedue to severe rate of strain

(d) vibration is induced in the machine tool

Solution. Extra hardness is imparted to workpiecedue to higher rate of deformation.

Ans. (c)

24. The stress–strain curve for mild steel is shown inthe figure given below.

T

P

Q

R

S

Strain ε (%)

S t r e s s σ

( N / m m

2 ) U

Choose the correct option referring to both figureand table.


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Point on thegraph

Description of the point

P 1. Upper yield point

Q 2. Ultimate tensilestrength

R 3. Proportionality limit

S 4. Elastic limit

T 5. Lower yield point

U 6. Failure

(a) P-1, Q-2, R-3, S-4, T-5, U-6

(b) P-3, Q-1, R-4, S-2, T-6, U-5

(c) P-3, Q-4, R-1, S-5, T-2, U-6

(d) P-4, Q-1, R-5, S-2, T-3, U-6

Solution. The sequence of points on the curve isas follows: proportional limit, elastic limit, upperyield point, lower yield point, ultimate tensilestrength, failure.

Ans. (c)

25. The hot tearing in a metal casting is due to

(a) high fluidity

(b) high melt temperature

(c) wide range of solidification temperature

(d) low coefficient of thermal expansion

Solution. Hot tears are the cracks developed in themetal due to residual stresses in solidified metalcaused by solid shrinkage of the metal.

Ans. (c)Q. 26–Q. 55 carry two marks each.

26. An analytic function of a complex variable z =x + iy is expressed as

f (z) = u (x, y) + iv (x, y)

where i = √ −1. If u (x, y) = x2−y2, thenexpression for v (x, y) in terms of x, y and a generalconstant c would be

(a) xy + c (b) x2+ y2

2 + c

(c) 2xy + c (d) (x−y)2

2 + c

Solution. The function z = x + iy is analytic.Hence,

∂u∂y

= −∂v∂x

Given that

u = x2−y2

Therefore,

−∂v

∂x = −2y

v = 2xy + f (y) + c

Thus, out of the given options, (c) is correct.

Ans. (c)

27. Consider two solutions x (t) = x1 (t) and x (t) =x2 (t) of the differential equation

d2x (t)

dt2 + x (t) = 0, t > 0

such that

x1 (0) = 1

dx1 (t)

dt |t=0 = 0

x2 (0) = 0

dx2 (t)

dt |t=0 = 1

The Wronskian function

W (t) =

x1 (t) x2 (t)dx1(t)dt

dx2(t)dt

at t = π/2 is

(a) 1 (b) −1

(c) 0 (d) /π/2

Solution. For the differential equation

d2x (t)

dt2 + x (t) = 0

m2 +1 = 0

m =±

i

The complementary and particular solutions willbe written as

xc = a cos t + b sin t

x p = 0

Thus,

x = a cos t + b sin t

Let

x1 (t) = cos tx2 (t) = sin t


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The Wronskian function for t = π/2 can becalculated as

W (t) = x1 (t) x2 (t)

dx1

(t)dt dx2

(t)dt

=

cos t sin t

− sin t cos t

= cos2 t +sin2 t

= 1

Ans. (a)

28. A machine produces 0, 1 or 2 defective pieces ina day with associated probability of 1/6, 2/3 and1/6, respectively. The mean value and the variance

of the number of defective pieces produced by themachine in a day, respectively, are

(a) 1 and 1/3 (b) 1/3 and 1

(c) 1 and 4/3 (d) 1/3 and 4/3

Solution. The data can be tabulated as follows:

x 0 1 2

P (x) 1/6 2/3 1/6

Thus,

x

×P (x) = 1

x2P (x) = 4/3

Variance =

x2P (x)−

x×P (x)2

= 1/3

Ans. (a)

29. The real root of the equation 5x−2cos x−1 = 0(up to two decimal accuracy) is · · · · · · · · · .


f (x) = 5x−2cos x−1f ′ (x) = 5+2 sin x

Using Newton–Raphson’s method,

xn+1 = xn− f (xn)

f ′ (xn)

Let x0 = 1 (rad), hence

x1 = x0− f (x0)

f ′ (x0)

= 1

−

5×1−2×cos(1)−1

5+2sin(1)= 0.5632

Similarly,

x2 = x1− f (x1)

f ′ (x1)

= 0.5632− 5×1−2×cos(0.5632)−15+2sin(0.5632)

= 0.5426

Similarly,

x3 = x2− f (x2)

f ′ (x2)

= 0.5426− 5×1−2×cos(0.5426)−1

5+2sin(0.5426)

= 0.5425

In these calculations, unit of angle is radian.Ans. 0.5425

30. A drum brake is shown in the figure.

Drum

800480

200

100

1000 N

The drum is rotating in anticlockwise direction.The coefficient of friction between the drum andshoe is 0.2. The dimensions shown in the figureare in mm. The braking torque (in N m) for thebrake shoe is · · · · · · · · · .


F = 1000 N

r = 0.2 mµ = 0.2

Normal reaction (Rn) at the brake shoe can bedetermined as

1000×0.8−Rn×0.48−0.2×Rn×0.1 = 0

Rn = 1600 N

Braking torque for the brake shoe is

T = µRn×r

= 64 N m

Ans. 64 Nm


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31. A body of mass (M) 10 kg is initially stationaryon a 45 inclined plane as shown in figure. Thecoefficient of dynamic friction between the bodyand the plane is 0.5. The body slides down the

plane and attains a velocity of 20 m/s.

45

M

The distance travelled (in meter) by the bodyalong the plane is · · · · · · · · · .


M = 10 kg

θ = 45

µ = 0.5

u = 0 m/s

v = 20 m/s

Let a be the acceleration of the block. Therefore,at equilibrium,

ma = mg sin θ−µ×mg cos θ

a = g (sin θ−µ cos θ)

= 3.468 m/s2

Using third equation of motion,

v2−u2 = 2as

s = 57.67 m

Ans. 57.67 m

32. Consider a simply supported beam of length, 50h,with a rectangular cross-section of depth, h, andwidth, 2h. The beam carries a vertical point load,P , at its mid-point. The ratio of the maximum

shear stress to the maximum bending stress in thebeam is

(a) 0.02 (b) 0.10

(c) 0.05 (d) 0.01

Solution. For the given beam configuration,

I = 2h×h3

12

= h4

6

z = h4/6

h/2

= h33

The beam configuration is shown in the figure:

P

P/2 P/2

−P/2

P/2

P L/4

SF

BM

Maximum shear stress is given by

τ max = 3

2τ avg

= 3

2× P/2

2h×h

= 3P

8h2

Maximum bending stress is given by

σmax = M

z

= P/2×50h/2

h3/3

= 75P

2h2

Therefore,

τ max

σmax=

3P ×2h2

75P

×8h2

= 0.01

Ans. (d)

33. The damping ratio of a single degree of freedomspring-mass-damper system with mass of 1 kg,stiffness 100 N/m and viscous damping coefficientof 25 Ns/m is · · · · · · · · · .


m = 1 kg

k = 100 N/mc = 25 Ns/m


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Damping ratio is calculated as

ξ = c

2√

km

= 252√

100×1

= 1.25

Ans. 1.25

34. An annular disc has a mass m, inner radius R andouter radius 2R. The disc rolls on a flat surfacewithout slipping. If the velocity of the center of mass is v, the kinetic energy of the disc is

(a) 9

16mv2 (b)

11

16mv2

(c) 13

16mv2 (d)

15

16mv2

Solution. Moment of inertia of the disc aboutcentral axis is

I c = 1

2×m

(2R)

2−R2

= 3mR2

2

Moment of inertia about the rolling point is

I = 3mR2

2 + m×(2R)2

= 11mR2

2

Angular speed of the disc is

ω = v

2R

Kinetic energy of the disc will be

T = 1

2× 11mR2

2 ×

v

2R

2

=

11mv2

16

Ans. (b)

35. A force P is applied at a distance x from the endof the beam as shown in the figure.

L

L

P

A

x

What would be the value of x so that thedisplacement at A is equal to zero?

(a) 0.5L (b) 0.25L

(c) 0.33L (d) 0.66LSolution. The equivalent configuration of thebeam is shown in the figure:

P

P (L−x)A

Using the principle of superposition, the deflection(downward) at point A is given by

δ = P L3

3EI − P (L−x) L2

2EI

= P L3

3EI − P L3

2EI +

P L2x

2EI

= −P L3

6EI +

P L2x

2EI

For δ = 0,

P L2x

2EI = −P L3

6EI

x = L

3= 0.33L

Ans. (c)

36. Consider a rotating disk cam and a translatingroller follower with zero offset. Which one of thefollowing pitch curves, parametrized by t, lyingin the interval 0 to 2π, is associated with themaximum translation of the follower during onefull rotation of the cam rotating about the centerat (x, y) = (0, 0)?

(a) x (t) = cos t, y (t) = sin t

(b) x (t) = cos t, y (t) = 2 sin t

(c) x (t) = 1

2 + cos t, y (t) = 2 sin t

(d) x (t) = 1

2 + cos t, y (t) = sin t

Solution. All the four options must be checked

to find out the maximum translation during onefull rotation of the cam, that is, by varying the


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parameter t between 0 and 2π, as shown in thefigure:

sin t

cos t

1/2+cos t

2sin t

1

2

0 π2π

It is clear that the pitch curve, shown in option(c), containing 2 sin t would give the maximumtranslational displacement of the follower.

Ans (c)

37. A four-wheel vehicle of mass 1000 kg movesuniformly in a straight line with the wheelsrevolving at 10 rad/s. The wheels are identical,

each with a radius of 0.2 m. Then a constantbraking torque is applied to all the wheels and thevehicle experiences a uniform deceleration. For thevehicle to stop in 10 s, the braking torque (in N m)on each wheel is · · · · · · · · · .


mv = 1000 kg

ωw = 10 rad/s

rw = 0.2 m

t = 10 s

Linear velocity of the vehicle is

v = 10×0.2

= 2 m/s

Acceleration of the vehicle is

a = 2

10

= 0.2 m/s2

Braking force required on the vehicle mass is

F = 1000×0.2= 200 N

Hence, each wheel should exert 200/4 = 50 N.Braking torque on each wheel would be

T = 50×0.2

= 10 N m

Ans. 10 Nm

38. A slider–crank mechanism with crank radius 60mm and connecting rod length 240 mm is shownin figure.

24060

90

The crank is rotating with a uniform angularspeed of 10 rad/s, counter-clockwise. For the givenconfiguration, the speed (in m/s) of the slider is· · · · · · · · · .Solution. Given that

r = 60 mm

= 0.06 m

l = 240 mm

ω = 10 rad/s

Crank angle θ and ratio n are calculated as

sin θ = 240√ 602 +2402

θ = 75.96

n = l

r= 4

Speed of the slider can be calculated as

v p = rω

sin θ +

sin 2θ

2n

= 0.06

×10sin 75.96+

sin (2×75.96)

2×4

= 0.617 m/s

Ans. 0.617 m/s

39. Consider an objective function

z (x1, x2) = 3x1+ 9x2

and the constraints

x1+ x2 ≤ 8

x1+ 2x2 ≤ 4

x1 ≥ 0x2 ≥ 0


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The maximum value of the objective function is· · · · · · · · · .Solution. The coordinates of corner points can bedetermined as

x1+ x2 = 8

x1 + 2x2 = 4

Therefore,

x2 = −4

x1 = 12

The feasible region is shown in the figure:

x1 + x2 = 8

x1+ 2x2 = 4

(12,−4)

(8, 0)

(4, 0)

(0, 8)

(0, 2)

x1

x2

z = 12

z = 24

z = 72

z = 18

Since, x1, x2 ≥ 0, the maximum value of theobjective function is found at (0, 2), z = 18.

Ans. 18

40. A mass-spring-dashpot system with mass m = 10kg, spring constant k = 6250 N/m is excited by aharmonic excitation of 10 cos(25t) N. At the steadystate, the vibration amplitude of the mass is 40mm.

m

k c

F = 10 cos (25t)

The damping coefficient (c, in N s/m) of thedashpot is · · · · · · · · · .


m = 10 kg

k = 6250 N/m

F = 10 cos (25t) N

F 0 = 10 N

ω = 25 rad/sx0 = 0.04 m

Using

x0 = F 0

(k−mω

2

)

2

+ (cω)

2

c = 1

ω

F 0x0

2

−(k−mω2)2

= 10 N s/m

Ans. 10 Ns/m

41. A certain amount of an ideal gas is initiallyat a pressure p1 and temperature T 1. First, itundergoes a constant pressure process 1-2 suchthat T 2 = 3T 1/4. Then, it undergoes a constant

volume process 2-3 such that T 3 = T 1/2. The ratioof the final volume to the initial volume of the idealgas is

(a) 0.25 (b) 0.75

(c) 1 (d) 1.5

Solution. For constant pressure process (1-2),

T 2 = 3T 1

4

Therefore, using ideal gas equation (v ∝ T ),

v2 = v1× T 2T 1

= 3v1

4

For constant volume process (2-3)

v3 = v2

= 3v1

4= 0.75v1

Ans. (b)

42. An amount of 100 kW of heat is transferredthrough a wall in steady state. One side of the wallis maintained at 127C and the other side at 27C.The entropy generated (in W/K) due to the heattransfer through the wall is · · · · · · · · · .

Solution. Entropy generated during the process is

∆S = 100

− 1

127+273 +

1

27+273

= 83.33 W/K

Ans. 83.33 W/K


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43. A siphon is used to drain water from a large tankas shown in the figure below.

Q

P

O

RDatum

Z P Z O

Z R

Z Q

Assume that the level of water is maintainedconstant. Ignore frictional effect due to viscosityand losses at entry and exit. At the exit of thesiphon, the velocity of water is

(a)

2g (Z Q−Z R) (b)

2g (Z P −Z R)

(c)

2g (Z O−Z R) (d)

2gZ Q

Solution. Applying Bernoulli’s equation betweentwo points P and R:

pa− paρg

+ 02−v2R

2g + (Z P −Z R) = 0

vR =

2g (Z P −Z R)

Ans. (b)

44. Heat transfer through a composite wall is shownin figure.

l l

Heat flow

600 K 300 KT i

κ 2κ

Both the sections of the wall have equal thickness(l). The conductivity of one section is κ and thatof the other is 2κ. The left face of the wall is at600 K and the right face is at 300 K. The interfacetemperature T i (in K) of the composite wall is· · · · · · · · · .Solution. Let T i be the intermediate temperature.Thermal resistance of the left side wall is

R1 = lκA

Thermal resistance of the composite wall is

Rc = l

κA +

l

2κA

= 3l2κA

= 3R1

2

Rate of heat transfer ( Q) through the wall will beconstant throughout the composite wall, therefore,

Q = 600−300

Rc=

600−T iR1

300×2

3R1=

600−T iR1

T i = 400 K

Ans. 400 K

45. A fluid of dynamic viscosity 2×10−5 kg/ms anddensity 1 kg/m3 flows with an average velocity of 1m/s through a long duct of rectangular (25 mm ×15 mm) cross-section. Assuming laminar flow, thepressure drop (in Pa) in the fully developed regionper meter length of the duct is · · · · · · · · · .


µ = 2×10−5

kg/m sρ = 1 kg/m3

v = 1 m/s

Equivalent diameter (D) of the duct can becalcualted as

D = 4A

P

= 4×0.025×0.015

2×(0.025+0.015)

= 0.01875 m

Reynolds number of the flow is

Re = ρvD

µ

= 1×1×0.01875

2×10−5

= 937.5

Darcy–Weisbach friction factor (4f ) is calculatedas

4f = 64

Re= 0.0682


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Hence, pressure drop per meter length of the ductis

∆ p

l

= 4f

D ×v2

2g ×ρg

= 4f

D × v2ρ

2

= 0.0682

0.01875× 12×1

2= 1.818 Pa/m

Ans. 1.818 Pa/m

46. At the inlet of an axial impulse turbine rotor, theblade linear speed is 25 m/s, the magnitude of absolute velocity is 100 m/s and the angle between

them is 25

. The relative velocity and the axialcomponent of velocity remain the same betweenthe inlet and outlet of the blades. The blade inletand outlet velocity triangles are shown in thefigure.

58.6 m/s

78 m/s100 m/s

78 m/s

25 m/s

25

Assuming no losses, the specific work (in J/kg) is· · · · · · · · · .Solution. Given that

u = 25 m/s

V 1 = 100 m/s

V 2 = 58.6 m/s

α = 25

V r1 = 78 m/s

V r2 = 78 m/s

V 2

= 5 8

. 6 m / s

V r 2 =

7 8 m / sV

1 = 1 0 0 m / s

V r 1 =

7 8 m / s

u = 25 m/s

25

β 1

β 2

The angles can be determined as

V r1 sin β 1 = V 1 sin α

78sin β 1 = 100 sin 25

β 1 = 32.80

For the given condition that axial components of velocity remain the same,

β 1 = β 2 = 32.80

Specific work shall be given by

w = (V w1+ V w2)×u

= (V r1 cos β 1+ V r2 cos β 2)×u

= 2V r1u cos β 1

= 2×78×25×cos32.80

= 3278.2 J/kg

Ans. 3278.2 J/kg

47. A solid sphere of radius r1 = 20 mm is placed

concentrically inside a hollow sphere of radius r2 =30 mm, as shown in the figure.

r1

r2

21

The view factor F 21 for radiation heat transfer is

(a) 2/3 (b) 4/9

(c) 8/27 (d) 9/4

Solution. For the given surface 1,

F 11 = 0

Using

F 11+ F 12 = 1

F 12 = 1

Using reciprocating law,

F 21×A2 = F 12×A1

F 21 = A1

A2×F 12

= 4πr214πr22

×1

= 202

302

= 4

9

Ans. (b)


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48. A double-pipe counter-flow heat exchanger trans-fers heat between two water streams. Tube sidewater at 19 l/s is heated from 10C to 38C. Shellside water at 25 l/s is entering at 46C. Assume

constant properties of water, density is 1000 kg/m3

and specific heat is 4186 J/kg K. The LMTD (inC) is · · · · · · · · · .


mc = 19 l/s

T ci = 10C

T co = 38C

mh = 25 l/s

T hi = 46C

ρ = 1000 kg/m3

cc = ch = 4186 J/kg K

For energy balance

mccc (T co−T ci) = mhch (T hi−T ho)

19×(38−10) = 25 (46−T ho)

T ho = 24.72C

Hence

∆T 1 = T hi−T co

= 8C

∆T 2 = T ho−T ci= 14.72C

LMTD of the heat exchanger is

LMTD = ∆T 1−∆T 2ln(∆T 1/∆T 2)

= 11.02C

Ans. 11.02C

49. A diesel engine has a compression ratio of 17 andcut-off takes place at 10% of the stroke. Assumingratio of specific heats (γ ) as 1.4, the air-standardefficiency (in percent) is · · · · · · · · · .


r = v1v2

= 17

kf = 10%

γ = 1.4

Hence, cut-off ratio (k = v3/v2) is calculated as

k = 1 + kf (r−1)

= 1+0.1(17−1)= 2.6

Therefore, efficiency of diesel cycle is

η = 1− 1

rγ −1×

kγ −1

γ (k−1)

= 1− 1170.4

×

2.61.4−11.4 (2.6−1)

= 59.6%

Ans. 59.6%

50. Consider the given project network, where num-bers along various activities represent the normaltime.

1 3 6 7

2

4

5

3

2

2

2

355

4

The free float on activity 4-6 and the projectduration, respectively, are

(a) 2, 13 (b) 0, 13

(c) −2, 13 (d) 2, 12

Solution. Calculations of [ES, LS] and critical path(double line arrows) are shown in the figure:

1 3 6 7

2

4

5

3

2

2

2

3

55

4

[0,0]

[2,4]

[3,3]

[2,3]

[5,5]

[8,8]

[13,13]

The network has project duration of 13 days.Free float of an activity is the delay that can bepermitted in an activity so that the succeedingactivities in the path are not affected. Therefore,

Free float = ESjk−EFij

For activity 4−6

Free float = ES6−7−EF4−6

= 8−(2+4)

= 2

Ans. (a)


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51. A manufacturer can produce 12,000 bearingsper day. The manufacturer received an order of 8,000 bearings per day from a customer. Thecost of holding a bearing in stock is Rs. 0.20

per month. Setup cost per production run is Rs.500. Assuming 300 working days in a year, thefrequency of production run should be

(a) 4.5 days (b) 4.5 months

(c) 6.8 days (d) 6.8 months


p = 12000 units/day

d = 8000 units/day

D = 8000×300 bearings/annum

h = Rs. 0.2

×12/bearings per annum

A = Rs. 500/production

Therefore, EOQ shall be

Q∗ =

2AD

h (1−d/p)

=

2×500×8000×300

0.2×12(1−8000/12000)

= 54772.25 per order

Number of production orders per annum is

n = D

Q∗

= 8000×300

54772.25= 43.81

Therefore, frequency of production run will be

t = 300

75.89= 6.84 days

Ans. (c)

52. A cylindrical blind riser with diameter d and heighth, is placed on the top of the mold cavity of a closedtype sand mold, as shown in the figure.

Sprue basin

d

hRiser

Mold cavity

If the riser is of constant volume, then the rate of solidification in the riser is the least when the ratioh : d is

(a) 1:2 (b) 2:1

(c) 1:4 (d) 4:1

Solution. Given that riser volume (V ) i s aconstant, therefore,

V = πd2h

4

h = 4V

πd2

Surface area for heat transfer in the riser will be

A = πd2

4 + πdh

= πd

2

4 + πd× 4V πd2

= πd2

4 +

4V

d

Rate of solidification in the riser will be leastwhen surface area for heat transfer is minimum,for which

dA

dd = 0

−4V

d2 +

πd

2 = 0

V = πd3

8πd2h

4 =

πd3

8h

d =

1

2

Ans. (a)

53. The diameter of a recessed ring was measured byusing two spherical balls of diameter d2 = 60 mmand d1 = 40 mm, as shown in the figure.

H

D

H 2H 1

R

A

B

C

d1

d2

Recessed ring


18 of 19



The distance H 2 = 35.55 mm and H 1 = 20.55mm. The diameter D (in mm) of the ring gauge is· · · · · · · · · .Solution. Let XX denote the top surface fromwhere H 1 and H 2 are measured. Using the givendimensions,

AC = 60

2 +

40

2= 50 mm

BC =

35.55 +

60

2

−

20.55 + 40

2

= 25 mm

R =

AC2−BC2

= 502

−252

= 43.30 mm

Therefore,

D = R + 60

2 +

40

2= 93.30 mm

Ans. 93.30 mm

54. Which pair of the following statements is correctfor orthogonal cutting using a single-point cuttingtool?

P. Reduction in friction angle increases cuttingforce

Q. Reduction in friction angle decreases cuttingforce

R. Reduction in friction angle increases chipthickness

S. Reduction in friction angle decreases chipthickness

(a) P and R (b) P and S

(c) Q and R (d) Q and S

Solution. The effect of friction angle can beexamined as follows:

(a) Cutting force is related as

F c = R cos(λ−α)

Keeping other parameters constant, reductionin friction angle (i.e. friction force) will reducethe cutting force.

(b) According to Ernst-Merchant formula,

φ = π

4 +

α−λ

2

If friction angle (λ) is reduced, shear angle φwill increase. Chip thickness ratio is given by

r = t1t2

= sin φ

cos(φ−α)

Therefore, t2 will reduce with reduction in λ.

Ans. (d)

55. For spot welding of two steel sheets (base metal)each of 3 mm thickness, welding current of 10000A is applied for 0.2 s. The heat dissipated to thebase metal is 1000 J. Assuming that the heatrequired for melting 1 mm3 volume of steel is 20J and interfacial contact resistance between sheetsis 0.0002 Ω, the volume (in mm3) of weld nuggetis · · · · · · · · · .


h = 3 mm

I = 10000 A

t = 0.2 s

E d = 1000 J

U = 20 J/mm3

R = 0.0002 Ω

Let v be the volume of the weld nugget. For heatbalance

Uv = I 2Rt−E d

v = I 2Rt−E d

U

= 100002×0.0002×0.2−100020

= 150 mm2

Ans. 150 mm2

∗∗∗∗∗∗

Wiley Acing the GATE (Mechanical Engineering): GATE 2014: ME03C i h Wil I di

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