In case there is any kind of discrepancy in questions please write to us at or call us at : ies_master@yahoo.co.in Ph : 011-41013406, 011-

26522064, for rectification/deletion/updation of the same.

IES

MAS

TER

Q.1– Q.25 carry one mark each.

Q.1: In simple exponential smoothing forecasting to give higher weightage to recent demand information, thesmoothing constant must be close to(a) – 1 (b) zero (c) 0.5 (d) 1.0

Sol: (b)

Smaller values of smoothing coefficient ( ) gives more weightage to the recent demand information ascompared to higher values of ( ). Thus, should be near to zero.Thus, option (b) is correct.

Q.2: A steel bar 200 mm in diameter is turned at a feed of 0.25 mm/rev with a depth of cut of 4 mm. Therotational speed of the workpiece is 160 rpm. The material removal rate in mm3/s is(a) 160 (b) 167.6 (c) 1600 (d) 1675.5

Sol: (d)

d = 4 mm

200 mm

Metal removal rate = f × d × v

Where

f = feed in mm/rev.

d = depth of cut.

v = avgD .N

Davg = 0 fD D2

= 200 1962 = 198 mm.

v = 0.198 160 = 99.526 m/min.

IES

MAS

TER

M.R.R = 30.25 4 99.526 10

60 = 1659 mm3/sec.

Therefore, option d is correct.

Q.3: A cube shaped casting solidifies in 5 min. The solidifications time in min for a cube of the samematerial, which is 8 times heavier than the original casting, will be(a) 10 (b) 20 (c) 24 (d) 40

Sol: (b)

Since the new cube is 8 times heavier, its dimensions will change w.r.t. first cube. Let a1 be dimensionof first cube and a2 be of second cube.

a1

a1

a1

a2

a1

a2

Volume of first cube = 31a

Volume of second cube = 32a

But V2 = 8V1

So, 32a = 3

18 a or

2 1a 2a

Solidification time is given by chvorinov's rule

ts = 2

s

VkA

where k mould constantV = volume of casting ; As is surface area of casting.For first cube

IES

MAS

TER

ts1 =5 minutes (given)

5=23

121

ak6a

= 2

1ak6

...(i)

For second cube

ts2 =23

222

ak6a

k is same for both cubes as both cubes are of same material.

ts2 =23

121

8ak24a

= 2

1ak3

...(ii)

Dividing equation (i) by equation (2).

s1

s2

tt =

21

21

ak6ak3

ts2 =

21

21

a53

a6

= 21

21

5 a 36a 9

s2t 20minutes

Hence, option b is correct.

Q.4: For a ductile material, toughness is a measure of(a) resistance to scratching (b) ability to absorb energy up to fracture(c) ability to absorb energy till elastic limit (d) resistance to indentation

Sol: Toughness is the ability to absorb energy upto fracture therefore, option (B) is correct.

IES

MAS

TER

IES Master announces following programsFor

ESE - 2013 & GATE - 2014Civil Engineering -

- 14th Feb, 2013 Mechanical Engineering7th Jan, 2013

1. New Batch for

2. Conventional Ques. Practice Program for

3. Interview Guidance Program for

4. Postal Program for

Civil EngineeringMechanical Engineering

- 30th March, 2013 - 30th March, 2013

Civil Engineering

Electrical EngineeringMechanical Engineering

5. Non-tech class (G.S. & English)Starting from 2nd Feb, 2013.

6. Gate SolutionIES Master takes this opportunity to announce the launch of .GATE Solution (CE) from 2003 to 2012

Civil EngineeringMechanical Engineering

IES

MAS

TER

Q.5: In order to have maximum power from a Pelton turbine, the bucket speed must be(a) equal to the jet speed (b) equal to half of the jet speed(c) equal to twice the jet speed (d) independent of the jet speed

Sol: (b)u = v

2

Q.6: Consider one-dimensional steady state heat conduction along x-axis 0 x L , through a plane wallwith the boundary surfaces (x = 0 and x = L) maintained at temperatures of 0°C and 100°C. Heat isgenerated uniformly throughout the wall. Choose the CORRECT statement.(a) the direction of heat transfer will be from the surface at 100°C to the surface at 0°C(b) the maximum temperature inside the wall must be greater than 100°C(c) the temperature distribution is linear within the wall(d) the temperature distribution is symmetric about the mid-plane of the wall

Sol: (b)Heat conduction is one dimensional, steady state the conduction equation in this situation is

0°C 100C

x = o x x = L

t(x)Uniformheat generation

qgen

2gen

2

qd T 0k1dx

This equation shows that temperature distribution in the wall will not be linear but parabolic. So the curveshows that maximum temperature will be more than 100°C close to x = L and heat will flow to both sidesof this maximum temperature plane. So the correct statement is B

IES

MAS

TER

Q.7: A cylinder contains 5 m3 of an ideal gas at a pressure of 1 bar. This gas is compressed in a reversibleisothermal process till its pressure increases to 5 bar. The work in kJ required for this process is(a) 804.7 (b) 953.2 (c) 981.7 (d) 1012.2

Sol: (a)V1 = 5m3, P1 = 1 bar,Isothermal process i.e. PV = constantP2 = 5barWork done during isothermal compression,

1W2 = 2

1

V

V

P.dV

Volume after compression,

P2

1

V

P1V1 = P2V2

1 × 5 = 5 × V2

V2 = 1m3

1W2 = 1

5

PdV

= 1

5

dVCV PV = C

= C(1n 1/5) C = P1V1= P2V2

1W2 = P1V1 ln 5= 100 × 5 × 1.6094= 804.719 kJ

Q.8: A long thin walled cylindrical shell, closed at both the ends, is subjected to an internal pressure. Theratio of the hoop stress (circumferential stress) to longitudinal stress developed in the shell is(a) 0.5 (b) 1.0 (c) 2.0 (d) 4.0

Sol: (c)

h = iPd2t

l = iP d4t

l

h = 2

IES

MAS

TER

Q.9: If two nodes are observed at a frequency of 1800 rpm during whirling of a simply supported long slenderrotating shaft, the first critical speed of the shaft in rpm is(a) 200 (b) 450 (c) 600 (d) 900

Sol: (a)Two nodes are observed when a simply supported shaft rotates at 1800 rpm. So the shaft will like.

N1 N2

There are three lobs during the rotation at this speed of rotation. The rotational speed and natural fre-quency relationship is, = n2.n, where n = number of lobs.In this case n = 3 1800 = 32 n

n = 1800 200rpm9

Q.10: A planar closed kinematic chain is formed with rigid links PQ = 2.0 m, QR = 3.0 m, RS = 2.5 m andSP = 2.7 m with all revolute joints. The link to be fixed to obtain a double rocker (rocker-rocker)mechanism is(a) PQ (b) QR (c) RS (d) SP

Sol: (c)When link opposite to shortest link is fixed in inversion of 4-bar mechanism, the double rocker or oscillat-ing-oscillating mechanism is obtained.

2m

3m3m

2.5m

P S

RQ

2.7m

Since in this problem opposite to shortest (PQ) link is RS. So fix link RS to get double rocker mechanism.

IES

MAS

TER

Q.11: Let X be a normal random variable with mean 1 and variance 4. The probability P(X < 0) is(a) 0.5 (b) greater than zero and less than 0.5(c) greater than 0.5 and less than 1.0 (d) 1.0

Sol: (b)

Mean ( ) 1

Variance ( ) = 4From the figure we can observe that

0 1

O < P (x < 0) < 12 ( p(x 1) = 1

2 )

Q.12: Choose the CORRECT set of functions, which are linearly dependent.(a) sin x, sin2x and cos2x (b) cos x, sin x and tan x(c) cos 2x, sin2x and cos2x (d) cos 2x, sin x and cos x

Sol: (c)

sin2 x = 1 cos2x2

cos2 x = 1 cos2x2

We know that,cos 2x = cos2x – sin2xTherefore, cos 2x, sin2x, cos2x are linear dependent

Q.13: The eigenvalues of a symmetric matrix are all(a) complex with non-zero positive imaginary part(b) complex with non-zero negative imaginary part

IES

MAS

TER

(c) real(d) pure imaginary

Sol: (c)

Let A = a cc b

is any symmetric matrix of order 2 × 2.

|A I| = a c 0

c b

2 2(a b) ab c = 0

Discriminent (D) = (a + b)2 – 4ab + 4c2

= (a – b)2 + 4c2 > 0 eigen values of a symmetric matrix are always real.

Q.14: The partial differential equation 2

2u u uut x x

is a

(a) linear equation of order 2 (b) non-linear equation of order 1(c) linear equation of order 1 (d) non-linear equation of order 2

Sol: (d)

Q.15: Match the CORRECT pairs.

Numerical Integration Scheme Order of Fitting Polynomial

P. Simpson’s 3/8 Rule 1. FirstQ. Trapezoidal Rule 2. SecondR. Simpson’s 1/3 Rule 3. Third

(a) P-2, Q-1, R-3 (b) P-3,Q-2,R-1 (c) P-1,Q-2,R-3 (d) P-3,Q-1,R-2

Sol: (d)

IES

MAS

TER

Q.16: A rod of length L having uniform cross-sectional area A is subjected to a tensile force P as shown inthe figure below. If the Young’s modulus of the material varies linearly form E1 to E2 along the lengthof the rod, the normal stress developed at the section-SS is

(a) PA (b)

1 2

1 2

P E EA E E

(c) 2

1

PEAE (d) 1

2

PEAE

Sol: (a)

Normal stress at every point will be P/A.

Q.17: Two threaded bolts A and B of same material and length are subjected to identical tensile load. If theelastic strain energy stored in bolt A is 4 times that of bolt B and the mean diameter of bolt A is 12mm, the mean diameter of bolt B in mm is(a) 16 (b) 24 (c) 36 (d) 48

Sol: (b)

u = 1 p.2

= 1 pp2 AE

l

Since material is same, load is also same, and length is same.

So, u 1A

u 21d

1

2

uu =

2221

dd

4 = 2

2

1

dd

IES

MAS

TER

2

1

dd

= 2

d2 = 2d1 = 2(dA) = 24 mm Ans.

Q.18: A link OB is rotating with a constant angular velocity of 2 rad/s in counter clockwise direction and ablock is sliding radially outward on it with an uniform velocity of 0.75 m/s with respect to the rod,shown in the figure below. If OA = 1m, the magnitude of the absolute acceleration of the block atlocation A in m/s2 is

(a) 3 (b) 4 (c) 5 (d) 6

Sol: (c)Centripetal acceleration of slider (A) with respect to O i.e absolute,

ac = slider accn – centripetal accn

= 0 – r2

= 0 – 1 × 4= – 4 m/sec2

Tangential acceleration,at = r. + 2V

= 0 + 2 × 0.75 × 2 = accn of rod= 3m/sec2

Total acceleration of slider A

= 2 2( 4) 3

= 5m/sec2

Q.19: For steady, fully developed flow inside a straight pipe of diameter D, neglecting gravity effects, thepressure drop p over a length L and the wall shear stress w are related by

(a) wpD

4L

(b) 2

w 2pD4L

(c) wpD2L

(d) w4 pL

D

IES

MAS

TER

Sol: (a)

w =2vf

8 . .

=2

e

64 vR 8

=264 v

vD / 8

= v8D

= 21 p8 RD 8 x

=21 P D

D L 4

= ( p) D4 L

Q.20: The pressure, dry bulb temperature and relative humidity of air in a room are 1 bar, 30°C and 70%,respectively. If the saturated steam pressure at 30°C is 4.25 kPa, the specific humidity of the room airin kg water vapour/kg dry air is(a) 0.0083 (b) 0.0101 (c) 0.0191 (d) 0.0232

Sol: (c)

The relative humidity, = v

s

PP

0.7 = vP4.25

Pv = 0.7 × 4.25= 2.975 kPa

Specific humidly, = 0.622 v

0 v

PP P

= 0.622 × 2.975

100 2.975= 0.01907

IES

MAS

TER

Q.21: Consider one-dimensional steady state heat conduction, without heat generation, in a plane wall; withboundary conditions as shown in the figure below. The conductivity of the wall is given by 0k k bT ;where 0k and b are positive constants, and T is temperature

As x increases, the temperature gradient (dT/dx) will(a) remain constant (b) be zero (c) increases (d) decreases

Sol: (d)One dimensional steady state conduction without heat generation,k = k0 + btSince the heat conduction is one dimensional.

Heat flux, q = k dTdx – constant

As x increases, temperature in the wall increases and so the thermal conductivity increases, then value of

dTdx

must decreases to ensure constant heat flux. So ans. is (d)

Q.22: In a rolling process, the state of stress of the material undergoing deformation is(a) pure compression (b) pure shear(c) compression and shear (d) tension and shear

Sol: (a)Rolling is pure compression process. Having Biaxial compression.Hence, option A is correct.

Q.23: Match the CORRECT pairs.

Processes Characteristics / Applications

P. Friction Welding 1. Non-consumable electrodeQ. Gas Metal Arc Welding 2. Joining of thick platesR. Tungsten Inert Gas Welding 3. Consumable electrode wireS. Electroslag Welding 4. Joining of cylindrical dissimilar

materials

IES

MAS

TER

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

Sol: (a)Friction welding – joining of cylindrical dissimilar materials.

Gas metal arc welding – consumable electrode wire

Tungsten inert Gas welding – Non-consumable electrode

Electroslag weld – Joining of thick plate

Thus, P–4, Q–3, R–1, S–2 option (A) is correct.

Q.24: Customers arrive at a ticket counter at a rate of 50 per hr and tickets are issued in the order to theirarrival. The average time taken for issuing a ticket is 1 min. Assuming that customer arrivals forma Poisson process and service times are exponentially distributed, the average waiting time in queuein min is

(a) 3 (b) 4 (c) 5 (d) 6

Sol: (c)

= Arrival rate is 50/hr or 50/60 per minute.

= service rate is 1 minute or 60/hr.

Average waiting time in queue in minutes is

= ( – )

=

56

51 1 –6

= 5 / 61 / 6 = 5 minutes

Q.25: A metric thread of pitch 2 mm and thread angle 60° is inspected for its pitch diameter using 3-wiremethod. The diameter of the best size wire in mm is(a) 0.866 (b) 1.000 (c) 1.154 (d) 2.000

Sol: (c)

The diameter of the best size wire in mm for a 60° thread is given as

= 0.57735 × pitch of thread

= 0.57735 × 2 = 1.154 mm.

Thus, option (C) is correct.

IES

MAS

TER

Q.26–Q.55 carry two marks each.

Q.26: Cylindrical pins of 0.0200.01025

mm diameter are electroplated in a shop. Thickness of the plating is 2.030

micron. Neglecting gage tolerances, the size of the GO gage in mm to inspect the plated componentsis(a) 25.042 (b) 25.052 (c) 25.074 (d) 25.084

Sol: (b)Cylindrical pin may be treated as a shaft, thus GO gauge corresponds to maximum (High limit)dimensions of shaft with plating i.e.

= (25 + 0.020) + (30 + 2) × 10–3 mm.

= 25.052 mm.

Option (B) is correct.

Q.27: During the electrochemical machining (ECM) of iron (atomic weight = 56, valency = 2) at current of1000 A with 90% current efficiency, the material removal rate was observed to be 0.26 gm/s. IfTitanium (atomic weight = 48, valency = 3) is machined by the ECM process at the current of 2000A with 90% current efficiency, the expected material removal rate in gm/s will be(a) 0.11 (b) 0.23 (c) 0.30 (d) 0.52

Sol: (c)Mass flow rate in ECM process is given as

m = AIZF

A = Atomic weight in grams.

I = Current in amperes.

Z = Valency of cation.

F = Faraday constent (96,500 coulombs).

= Current efficiency.

m = 48 2000 0.93 96500

= 0.2984 gm/sec.

or 0.30 gm/sec.

Thus, option (C) is correct.

IES

MAS

TER

Q.28: A single degree of freedom system having mass 1 kg and stiffness 10 kN/m initially at rest is subjectedto an impulse force of magnitude 5 kN for 10–4 seconds. The amplitude in mm of the resulting freevibration is(a) 0.5 (b) 1.0 (c) 5.0 (d) 10.0

Sol. (c)Mass, m = 1kgStiffness, k = 10 kN/m.Impulsive force, F = 5 kNImpulse time, t = 10–4 sec No elissipative effect in the system Change in momentum at the time of impact

p = F.dt F. t. m (V–0) = F.t

Velocity just after impact,

V = 4F. t 5000 10 0.5m / sec

m 1

From conservation of energy 2 21 1mV kx2 2

x = mVk

X = 0.5 41

10

= 0.5 metre100

= 5 mm

Q.29: A bar is subjected to fluctuating tensile load from 20 kN to 100 kN. The material has yield strengthof 240 MPa and endurance limit in reversed bending is 160 MPa. According to the Soderberg principle,the area of cross-section in mm2 of the bar for a factor of safety of 2 is(a) 400 (b) 600 (c) 750 (d) 1000

Sol. (d)

IES

MAS

TER

Soderberg's creteria of combined loading gives.

1F.O.S = v fm

y e

ka

kf = 1, F.O.S. = 2 (given)

Mean load wm = max minw w2

= 100 202 = 60 kN = 60 × 103N.

Variable load wv = max minw – w2 = 100 – 20

2 = 40 kN = 40 × 103N.

Mean stress ( m ) = meanwArea =

m2

w

d4 = 2

6000 4d

= 2

276394.38 N / mm

d

Variable stress ( v ) = vwA = 2

40000 4d

= 250930

d N/mm2

Substituting all values in soderberg equation

12 = 2 2

76394.38 50930240d 160d

12 = 2 2

318.3 318.3d d

d2 = 636.625 × 2

d 35.68mm

Area = 2d4 = 999.86 mm2 or 1000 mm2

Thus, option (D) is correct)

Q.30: A simply supported beam of length L is subjected to a varying distributed load 1sin(3 / L)Nm ,x wherethe distance x is measured from the left support. The magnitude of the vertical reaction force in N atthe left support is(a) zero (b) L / 3 (c) L / (d) 2L /

Sol: (b)

IES

MAS

TER

L/3

L/3 L/3

Magnitude of load in length = 3L =

/3

0

(3 )sinl x dx

L

=

/3

0

3cos

3

lxL

L

= cos cos03L

= 2L

Here for reaction calculation forces that can be considered as

23L

23L

23L

R

R = 2 23 6L L

R = 26 3

L L

Q.31: Two large diffuse gray parallel plates, separated by a small distance, have surface temperatures of 400K and 300 K. If the emissivities of the surfaces are 0.8 and the Stefan-Boltzmann constant is 5.67 ×10–8 W/m2K4, the net radiation heat exchange rate in kW/m2 between the two plates is(a) 0.66 (b) 0.79 (c) 0.99 (d) 3.96

Sol: (a)

IES

MAS

TER

Net radiation heat exchange between plates

Q12 =4 4

1 2(T T )1 1 11 2

=8 4 45.67 10 (400 300 )

1 1 10.8 0.8

=5.67 175

1.5

= 661.5 k/m2

= 0.6615 watt/m2

Q.32: A hinged gate of length 5 m, inclined at 30° with the horizontal and with water mass on its left, isshown in the figure below. Density of water is 1000 kg/m3. The minimum mass of the gate in kg perunit width (perpendicular to the plane of paper), required to keep it closed is

(a) 5000 (b) 6600 (c) 7546 (d) 9623

Sol: (d)

IES

MAS

TER

30°

B

2.5 m

30°

5wFwater

5 m30°

W = Total wt of gate per unit length inside

A

On the verge of lifting reaction at B = 0

W cos 30° × 2.5 = water2.53

F

W = water 10

33 2.52

F

Fwater = 1 255 5 12 2w w

W = 25 10

32 3 2.52

w

= 94396.769 N

Mass 9622.5 kg/m

Q.33: The pressure, temperature and velocity of air flowing in a pipe are 5 bar, 500 K and 50 m/s, respectively.The specific heats of air at constant pressure and at constant volume are 1.005 kJ/kgK and 0.718 kJ/kgK, respectively. Neglect potential energy. If the pressure and temperature of the surroundings are1 bar and 300 K, respectively, the available energy in kJ/kg of the air stream is

(a) 170 (b) 187 (c) 191 (d) 213

Sol: (b)

IES

MAS

TER

cp =1.005 kJ/kg-K 5bar, 500 Kcv = 0.718 kJ/kg-K 50m/sec

Neglecting potential energy.

= (h – h0) – T0 (s – s0) + 2

1V2

= cp(T – T0) – T0(s – s0) + 2

1V2

Change in entropy, (s – s0) = cp ln(T/T0) – R ln(P/P0)

= 1.005 ln(500/300) – 0.287 ln(5/1)= 0.5134 – 0.462= 0.0515 kJ/kg-K

Available energy = 1.005 (500 – 300) – 300 × 0.0515 + 250

2000= 201 – 15.45 + 1.25= 186.8 kJ/kg

Q.34: The probability that a student knows the correct answer to a multiple choice question is 2/3. If thestudent does not know the answer, then the student guesses the answer. The probability of the gussedanswer being correct is 1/4. Given that the student has answered the question correctly, the conditionalprobability that the student knows the correct answer is(a) 2/3 (b) 3/4 (c) 5/6 (d) 8/9

Sol: (d)Let K be the event of knowing the correct answer

G be the event of guessing the answerC be the event of answer being correct

Given,P(K) =

23

P(G) = 1 – P(K)

= 213

= 13

P(C/G) = 14

P(C) = P(K) P(C/K) + P(G) P(C/G)

IES

MAS

TER

= 2 1 113 3 4

The required probability

P(K/C) = P(K)P(C/K)

P(C)

=

2 13

2 1 113 3 4

= 2 4 88 1 9

Q.35: The solution to the differential equation 2

2d u duk 0

dxdx where k is a constant, subjected to the boundary

conditions u(0) = 0 and u(L) = U, is

(a) xu UL

(b) kx

kL1 eu U1 e

(c)

kx

kL1 eu U1 e

(d) kx

kL1 eu U1 e

Sol: (b)2

2d y duk

dxdx = 0

k is any constant,u(0) = 0,u(L) = u

(D2 – kD) u = 0 D = 0, k u = c1 + c2 ekx

u(0) = 0 c1 + c2 = 0 c2 = – c1

u(L) = U c1 + c2 ekL = U

c1 = kLU

1 e

u =kx

kL1 eU1 e

IES

MAS

TER

Q.36: The value of the definite integral e

1ln( )dx x x is

(a) 34 2e9 9

(b) 32 4e9 9

(c) 32 4e9 9

(d) 34 2e9 9

Sol: (c)e

e e3/2 3/21 1

l

2 1 2x (x) dx x x x dx3 x 3

ln ln (Integration by parts)

= e3/2 3/2l

2 4e x3 9

= 3/2 3/2 32 4 4 2 4e e e3 9 9 9 9

Q.37: The following surface integral is to be evaluated over a sphere for the given steady velocity vector fieldˆˆ ˆF xi yj zk defined with respect to a Cartesian coordinate system having ˆˆ ˆi, j and k as unit base

vectors.

1 ˆF.n dA4s

where S is the sphere, 2 2 2y z 1x and n is the outward unit normal vector to the sphere. The value ofthe surface integral is(a) (b) 2 (c) 3 / 4 (d) 4

Sol: (a)

F = ˆˆ ˆxi yj zk

F (x) (y) (z)x y z

= 1 + 1 + 1 = 3

F

= 3By Gauss Divergence Theorem

S V

ˆ(F n) dA ( F) dV

s

1 ˆ(F n) dA4

=

v

1 33 dv4 4 (V)

= 33 4 (1)4 3

=

IES

MAS

TER

Q.38: The function f(t) satisfies the differential equation 2

2d f f 0dt

and the auxiliary conditions, f(0) = 0,

df (0) 4dt

. The Laplace transform of f(t) is given by

(a) 2s 1 (b) 4

s 1 (c) 24

s 1 (d) 42

s 1

Sol: (c)Taking Laplce transform of both sides, we gets2f(s) – sf(0) – f(0) + f(s) = 0 …(i) f(0) = 0

f(0) = 4Putting these values in (i), we get(s2 + 1)f(s) – s × 0 – 4 = 0

f(s) = 24

(s 1)

Q.39: Specific enthalpy and velocity of steam at inlet and exit of a steam turbine, running under steady state,are given below :

Specific Enthalpy (kJ/kg) Velocity (m/s)

Inlet Steam Condition 3250 180Exist Steam Condition 2360 5

The rate of heat loss from the turbine per kg of steam flow rate is 5 kW. Neglecting change in potentialenergy of steam, the power developed in kW by the steam turbine per kg of steam flow rate, is(a) 901.2 (b) 911.2 (c) 17072.5 (d) 17082.5

Sol: (a)From steady flow energy equation

h1 + 2 2

1 21 2

V Vq h2 2

IES

MAS

TER

3250 + 2180 525 2360

2000 2000

h = 23602

v = 5m/sec2

5kW/kg

h = 32501

v = 180m/sec1

W = (3250 –2360) + 2 2180 5 5

2000

= 890 + 16.1875 – 5 = 901.2 kW/kg

Q.40: Water is coming out from a tap and falls vertically downwards. At the tap opening, the stream diameteris 20 mm with uniform velocity of 2 m/s. Acceleration due to gravity is 9.81 m/s2. Assuming steadystate, inviscid flow, constant atmospheric pressure everywhere and neglecting curvature and surfacetension effects, the diameter in mm of the stream 0.5 m below the tap is approximately(a) 10 (b) 15 (c) 20 (d) 25

Sol: (b)

22v = 2

1v 2gh

= 4 + 2 × 9.81 × 0.5

v2 = 3.716 m/sNow, A2v2 = A1 v1

22 2d v

4 = 2

1 1d v4

22

1

dd

= 1

2

v 2 0.538v 3.716

d2 = 14.67 mm Ans.

Q.41: A steel ball of diameter 60 mm is initially in thermal equilibrium at 1030°C in a furnace. It is suddenlyremoved from the furnance and cooled in ambient air at 30°C, with convective heat transfer coefficient

h = 20 W/m2K. The thermo-physical properties of steel are : density 37800kg / m , conductivity k= 40 W/mK and specific heat c = 600 J/kgK. The time required in seconds to cool the steel ball in airfrom 1030°C to 430°C is(a) 519 (b) 931 (c) 1195 (d) 2144

IES

MAS

TER

Sol: (d)Ti = 1030°C, To = 30°C

h = 20 W/m2K = 7800kg/m2

k = 40 W/mKC = 600 J/kg-KT = 430°C, t = ?

From lump capacity method,

o

i o

T TT T

= – hAe tVC

A/V =

2

3

4 r 3 64 r dr3

–

2430 30 20 6e t

1030 30 7800 6 10 600

4001000 = –4–4.2735 10 te

–44.2735 10 te = 2.5

4.2735 ×10–4 t = 0.9163t = 2144 sec.

Q.42: A flywheel connected to a punching machine has to supply energy of 400 Nm while running at a meanangular speed of 20 rad/s. If the total fluctuation of speed is not to exceed 2% , the mass moment ofinertia of the flywheel in kg-m2 is(a) 25 (b) 50 (c) 100 (d) 125

Sol: (a)

20rad / sec Fluctuation of speed = + 2% Coefficient of fluctuation of speed

= 0.02 – (–0.02)

IES

MAS

TER

K = 0.04Since the flywheel supplies 400 N.m energy at mean rotational speed = 20 rad/secThe energy supplied will be fluctuation of energy ‘e’ e = I 2 K

400 = I(20)2 × 0.04

I = 2400 25kg / m400 0.04

Q.43: A compound gear train with gears P, Q, R and S has number of teeth 20, 40, 15 and 20, respectively.Gears Q and R are mounted on the same shaft as shown in the figure below. The diameter of the gearQ is twice that of the gear R. If the module of the gear R is 2 mm, the centre distance in mm betweengears P and S is

(a) 40 (b) 80 (c) 120 (d) 160

Sol: (b) DQ = 2DR

mR = 2m

AB = RP DD2 2

DR = mR.tR = 2 × 15 = 30 mDR = 2 × 30 = 60 mm

No. of teeth of P are half of teeth of R

DP = QD2 = 60 30m

2

AB = P QD D2

= 30 60

2

= 45 mm

R and S are masking

IES

MAS

TER

m R s

R s

D Dt t

sD3015 20

Ds = 40 mm

BC = R sD D2

= 30 402

= 35 mm AC = AB + BC

= 45 + 35= 80 cm

Q.44: A pin jointed uniform rigid rod of weight W and length L is supported horizontally by an external forceF as shown in the figure below. The force F is suddenly removed. At the instant of force removal, themagnitude of vertical reaction developed at the support is

(a) zero (b) W/4 (c) W/2 (d) W

Sol: (c)Under equilibrium condition

w2

wF2

w

When F is just removed,

Reaction at the support = w2

IES

MAS

TER

Q.45: Two cutting tools are being compared for a machining operation. The tool life equations are :Carbide tool : VT1.6 = 3000HSS tool : VT0.6 = 200

where V is the cutting speed in m/min and T is the tool life in min. The carbide tool will provide higher toollife if the cutting speed in m/min exceeds(a) 15.0 (b) 39.4 (c) 49.3 (d) 60.0

Sol: (a)We can find cutting speed either by finding the speed which gives same tool life for both the tools.For carbide tool

T1 = 1/1.63000

V

For H.S.S. tool

T2 = 1/0.6200

V

If tool life are to be same for both tools T1 = T2, thus1/1.63000

V

= 1/0.6200

V

solving above equation for V

V = 39.47 m/min.

Hence option (B) is correct. Thus, if we increase speed above 39.47 m/min carbide tool will give morelife.

The solution can also be found out by simply substituting various speed values in taylor's equations.i.e. at 39.4 m/min. Both carbide & H.S.S. tool give same tool life of 15 minutes.

Q.46: A linear programming problem is shown below,Maximize 3x + 7y

Subject to 3x 7y 10

4x 6y 8

x,y 0

IES

MAS

TER

It has(a) an unbounded objective function (b) exactly one optimal solution(c) exactly two optimal solutions (d) infinitely many optimal solutions

Sol. (b)Maximize 3x + 7ysubjected to 3x + 7y 10

4x + 6y 8

x, y 0

3x + 7y = 6 at (2,0)28 4at 0,3 3

(2, 0)10 , 03

100,7

40,3

3x + 7y = 0 Exactly one optional solution.

Q.47: In a CAD package, mirror image of a 2D point P(5,10) is to be obtained about a line which passesthrough the origin and makes an angle of 45° counterclockwise with the X-axis. The co-ordinates of thetransformed point will be

(a) (7.5, 5) (b) (10, 5) (c) (7.5, –5) (d) (10, –5)

Sol: Option (B) is correct.

Common Data Questions 48-49Water (specific heat, pc 4.18kJ / kgK) enters a pipe at a rate of 0.01 kg/s and a temperature of 20°C. The pipe of

diameter 50 mm and length 3m, is subjected to a wall heat flux qw in 2W / m :

Q.48: If qw = 5000 and the convection heat transfer coefficient at the pipe outlet is 1000 W/m2K, thetemperature in °C at the inner surface of the pipe at the outlet is(a) 71 (b) 76 (c) 79 (d) 81

Sol: (d)

IES

MAS

TER

Now the flux is constant, then the bulk temp.

"wq DL =mcp (Tb – 20)

5 × × 5 × 10–2 × 3 =0.01 × 4.18 (Tb –20)

Tb = 20 + 25 5 3 10

0.01 4.18

Tb = 76.37°CHeat transfer at outlet of pipe

5000 = 1000 (Tw –76.37)Tw = 76.37+5

= 81.37C

Q.49: If qw = 2500x, where x is in m and in the direction of flow (x = 0 at the inlet), the bulk meantemperature of the water leaving the pipe in °C is

(a) 42 (b) 62 (c) 74 (d) 104

Sol: (b)Heat taken by water = Heat due to flux

mcp (T0 – 20) = L "

w0D.dx.q

0.01 × 4.18 (Tb – 20) = D × 22.5 L

2

Tb – 20 = 25 10 2.5 9

2 0.01 4.18

Tb – 20 = 42.28Tb = 20 +42.2Tb = 62.2°C

Common Data Questions 50-51A single riveted lap joint of two similar plates as shown in the figure below has the following geometrical andmaterial details

IES

MAS

TER

width of the plate w = 200 mm, thickness of the plate t = 5 mm, number of rivets n = 3, diameter of the rivet dr =10 mm, diameter of the rivet hole dh = 11 mm, allowable tensile stress of the plate p 200 MPa , allowable shear

stress of the rivet s 100 MPa and allowable bearing stress of the rivet c 150MPa .

Q.50: If the plates are to be designed to avoid tearing failure, the maximum permissible load P in kN is(a) 83 (b) 125 (c) 167 (d) 501

Sol: (c)

For tearing failure

P = pp – d t

= h pw – n d t

=(200 – 3 × 11) × 5 × 200 = 167000N

=167 kN

Here diameter of rivet hole is used in calculating tearing area of plate.

Q.51: If the rivets are to be designed to avoid crushing failure, the maximum permissible load P in kN is(a) 7.50 (b) 15.00 (c) 22.50 (d) 30.00

Sol: (c)Given Data,w = 200 mm; t = 5mm; No. of rivets = 3; dr = 10 mm

Diameter of rivet hole = 11mm; allowable tensile stress p = 200 N/mm2

IES

MAS

TER

Allowable shear stress s = 100 N/mm2; Allowable bearing stress 2c 150N / mm

(i) For crushing failure

permissible crushing load = cn.d.t

where c is bearing stress.

wc = 3 × 10 × 5 × 150 = 22500N = 22.50 kN.

Linked Data Questions 52-53.In a simple Brayton cycle, the pressure ratio is 8 and temperatures at the entrance of compressor and turbine are300 K and 1400 K, respectively. Both compressor and gas turbine have isentropic efficiencies equal to 0.8. For thegas, assume a constant value of cp (specific heat at constant pressure) equal to 1 kJ/kgK and ratio of specific heatsas 1.4. Neglect changes in kinetic and potential energies.

Q.52: The power required by the compressor in kW/kg of gas flow rate is(a) 194.7 (b) 243.4 (c) 304.3 (d) 378.5

Sol: (c)P2/P1 = P3/P4 = 8Temperature T2 at exit at compressor

T2 = T1 (P2/P1) 1

= 300 (8) 1.4 1

1.4

22

1 300K

4 4

31400k

T

S

= 543.43

(T2 – T1) = 2 1

c

T T

= 543.43 300

0.8

= 304.3 KPower required by compressor,= cp (T2–T1)= 1 ×304.3= 304.3 kW/kg

IES

MAS

TER

Q.53: The thermal efficiency of the cycle in percentage (%) is(a) 24.8 (b) 38.6 (c) 44.8 (d) 53.1

Sol: (c)Thermal efficiency at Brayton cycle.

p

11 1r

= 1 – 11.4 18

1.4

= 1 /0 0.552 = 44.8%

Linked Data Questions 54-55.In orthogonal turning of a bar of 100 mm diameter with a feed of 0.25 mm/rev, depth of cut of 4 mm and cuttingvelocity of 90 m/min, it is observed that the main (tangential) cutting force is perpendicular to the friction forceacting at the chip-tool interface. The main (tangential) cutting force is 1500 N.

Q.54: The orthogonal rake angle of the cutting tool in degree is(a) zero (b) 3.58 (c) 5 (d) 7.16

Sol: (a)First make the force diagram according to the data given.

FTF

Fc

Fs

Tools

R

N

Fn

IES

MAS

TER

Now, as we can see it is given that friction force (F) at tool chip interface is perpendicular to cutting(tangential) force. It means that tool face is exactly vertical and rake angle is zero. i.e. 0 . So, option(A) is correct.

Q.55: The normal force acting at the chip-tool interface in N is(a) 1000 (b) 1500 (c) 2000 (d) 2500

Sol: (b)

Furthere, N = Fcos – Fsin

= 1500 cos 0 – Ftsin0°

N = 1500 N

Thus, option (B) is correct.

General Aptitude (GA) QuestionsQ.56–Q.60 carry one mark each

Q.56. Were you a bird, you ___________________ in the sky.(a) would fly (b) shall fly (c) should fly (d) shall have flown

Sol. (a)

Q.57: Choose the grammatically INCORRECT sentence :(a) He is of Asian origin (b) They belonged to Africa

(c) She is an European (d) They migrated from India to Australia

Sol: (c)

Q.58: Complete the sentence :Universalism is to particularism as diffuseness is to ________________ .(a) specificity (b) neutrality (c) generality (d) adaptation

Sol. (a)

IES

MAS

TER

Q.59. What will be the maximum sum of 44, 42, 40 , ...... ?(a) 502 (b) 504 (c) 506 (d) 500

Sol. (c)Sum will be maximum if there are only positive terms Maximum sum = 44 + 42 + 40 + ---- + 2

= 2 (11 + 2 + 3 + ----- + 22)

= 2 × 22 23 5062

Q.60. Which one of the following options is the closest in meaning to the word given below?Nadir

(a) Highest (b) Lowest (c) Medium (d) Integration

Sol. (b)

Q.61–65 carry two marks eachQ.61: A tourist covers half of his journey by train at 60 km/h, half of the remainder by bus at 30 km/h and

the rest by cycle at 10 km/h. The average speed of the tourist in km/h during his entire journey is(a) 36 (b) 30 (c) 24 (d) 18

Sol: (c)Take the total distance as L.C.M. of (60, 30, 10) Total distance = 60 km

D1 = 30 km T1 = 30 1 hr.60 2

D2 = 15 km T2 = 15 1 hr.30 2

D3 = 15 km T3 = 15 3 hr.10 2

Avg. speed = 60 60 241 1 3 5

2 2 2 2

IES

MAS

TER

Q.62: The current erection cost of a structure is Rs. 13,200. If the labour wages per day increase by 1/5 ofthe current wages and the working hours decrease by 1/24 of the current period, then the new cost oferection in Rs. is(a) 16,500 (b) 15,180 (c) 11,000 (d) 10,120

Sol: (b)Let ‘X’ be the labour wages, and ‘R’ be the working hours.Now, total cost is a function of X × RIncrease in wages = 20% Revised wages = 1.2 X

Decrease in labour time = 100 %24

Revised time = 23 R24

Revised total cost = 231.2 X R 1.15 XR24

= 1.15 × 13200 = 15180

Q.63: Out of all the 2-digit integers between 1 and 100, a 2-digit number has to be selected at random. Whatis the probability that the selected number is not divisible by 7?(a) 13/90 (b) 12/90 (c) 78/90 (d) 77/90

Sol: (d)Two digit numbers divisible by 714, 21, ..... 98Tn = a + (n –1)cl98 = 14 + (n – 1) × 7

84 (n 1)7

n = 13There are 13 two digit numbers divisible by 7. Therefore out of 90 two digits nos that are not divisibleby 7 is 90–13 = 77.

P(E) = 7790

IES

MAS

TER

Q.64: After several defeats in wars, Robert Bruce went in exile and wanted to commit suicide. Just beforecommitting suicide, he came across a spider attempting tirelessly to have its net. Time and again, thespider failed but that did not deter it to refrain from making attempts. Such attempts by the spidermade Bruce curious. Thus, Bruce started observing the near impossible goal of the spider to have thenet. Ultimately, the spider succeeded in having its net despite several failures. Such act of the spiderencourage Bruce not to commit suicide. And then, Bruce went back again and won many a battle, andthe rest is history.

Which of the following assertions is best supported by the above information?(a) Failure is the pillar of success

(b) Honesty is the best policy(c) Life begins and ends with adventures(d) No adversity justifies giving up hope

Sol: (d)

Q.65: Find the sum of the expression

1 1 1 1.....1 2 2 3 3 4 80 81

(a) 7 (b) 8 (c) 9 (d) 10

Sol: (b)1 1 1

1 2 2 3 3 4

= 2 2 2 2 2 2 2 22 1 3 2 4 3 81 80

2 1 3 2 4 3 81 80

= ( 2 1) ( 3 2 ) ( 4 3 ) ................ ( 81 80 )

= 81 1

= 9 – 1= 8