Detailed Solution - IES Master · Detailed Solution 03-02-2018 | MORNING SESSION ME ... A four bar mechanism is made up of links of length 100, ... For an Oldham coupling used between

SolutionDetailed 03-02-2018 | MORNING SESSION

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Section: General Aptitude

1. A rectangle becomes a square when itslength and breath are reduced by 10 m and5 m respectively. During this process, therectangle loses 650 m2 of area. What is thearea of the original rectangle in squaremeters?(a) 1125 (b) 2250(c) 2954 (d) 4500

Sol–1: (b)

y

x

(x – 5) = (y – 10)

x = y – 5

(xy) – (x –5)2 = 650

x2 + 5x – (x2 + 25 – 10x) = 650

15 x – 25 = 650

x = 45m

y = 50m

Original Area = xy = 45 × 50 = 2250 m2

2. “Going by the _____ that many hands makelight work the school ______ involved allthe students in the task.”(a) Principle, principal(b) Principal, principle(c) Principle, principle(d) Principal, principal

Sol–2: (a)

Principle an accepted or professed ruleof action or conduct

Principal a chief or head

3. Seven machines take 7 minutes to make 7identical toys. At the same rate, how manyminutes would it take for 100 machines tomake 100 toys?(a) 1 (b) 7(c) 100 (d) 700

Sol–3: (b)

7 machine = 7 toys per 7 minute

1 machine = 1 toy per 7 minute

In every 7 minute = 1 toy is produced permachine.

100 toy per 100 machine is equivalent to 1toy per 1 machine

it will take 7 minutes

4. A number consists of two digits. The sumof the digits is 9. If 45 is subtracted fromthe number, its digits are interchanged.What is the number?(a) 63 (b) 72(c) 81 (d) 90

Sol–4: (b) x + y = 9

(x × 10 + y) – 45 = y × 10 + x

10x + y – 45 = 10y + x

9x – 45 = 9y

9(9 – y) – 45 = 9y

81 – 9y – 45 = 9y

36 = 18y

y = 2

x = 7

Number 10 x + y

70 + 2 = 72

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5. “Her _______ should not be confused withmiserlines; she is ever willing to assist thosein need.”

The word that best fills the blank in theabove sentence is(a) Cleanlines (b) Punctuality(c) Frugality (d) Greatness

Sol–5: (c) Frugality prudent in saving, lackof wastefulness

6. From the time the front of an train entersa platform, it take 25 seconds for the backof the train to leave the platform, whiletravelling at a constant speed of 54 km/h.At the same speed, it take 14 second to passa man running at 9 km/h in the samedirection as the train. What is the length ofthe train and that of the platfrom in meters,respecxtively?(a) 210 and 140 (b) 162.5 and 187.5(c) 245 and 130 (d) 175 and 200

Sol–6: (d) Let length of the train be LT and

Length of the platform be Lp

LT + LP = 1V time

LT + LP = 100025 543600

LT = 2manV V time

= 100054 9 143600

LT = 175m

LP = 200m

7. Consider the following three steatements:

(i) Some roses are red

(ii) All red flowers fade quickly

(iii) Some roses fade quickly

What of the following statements can belogically inferred from the above statements?(a) If (i) is true and (ii) is flase, then (iii) is

false.(b) If (i) is true and (ii) is false, then (iii) is

true.(c) If (i) and (ii) are true, then (iii) is true(d) If (i) and (ii) are false, then (iii) is false.

Sol–7: (c)

Roses Redfadequickely

8. Given that a and b are integers and a +a2b3 is odd, which one of the followingstatements is correct?(a) a and b are both odd(b) a and b are both even(c) a is even and b is odd(d) a is odd and b is even

Sol–8: (d)

a + a2 b3 is odd

from option (1)

if a and b are both odd

a + a2b3 = odd + (odd)2 (odd)3 = odd + odd =even

from option (2)

If a and b both are even

a + a2 b3 = even + (even)2 (even)3 = even +even = even


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from option (3)

if a is even and b is odd

a + a2b3 = even + (even)2 (odd)3 = even +even = even

from option (4)

if a is odd and b is even

a+a2b3 = odd + (odd)2 (even)3 = odd + even= odd

Hence option ‘4’ is correct

9. For integers a, b and c, what would be theminimum and maximum values respectivelyof a + b + c if log |a| + log|b| + log|c|=0?(a) –3 and 3 (b) –1 and 1(c) –1 and 3 (d) 1 and 3

Sol–9: (a)

log |a| + log +|b| + log |c| = 0

i.e. |a| = |b| = |c| = 1

a = b = c = 1

Maximum value of a + b + c = 1+ 1 +1= 3

Minimum value of a + b + c = –1–1–1=–3

10. Which of the following function describe thegraph shown in the below figure?

321

–1–2–3

–3 –2 –1 0 1 2 3x

y

(a) y = ||x| + 1| – 2(b) y = ||x| – 1| – 1

(c) y = ||x| + 1| – 1(d) y = ||x – 1| – 1|

Sol–10: (b)

x 2 1 0 1 2y 0 1 0 1 0

These points are satisfied by

y = ||x|–1|–1

Section: Mechanical Engineering

1. A grinding ratio of 200 implies that the(a) Grinding wheel wears 200 times the

volume of the material removed(b) Grinding wheel wears 0.005 times the

volume of the material removed(c) Aspect ratio of abrasive particles used

in the grinding wheel is 200(d) Ratio of volume of abrasive particle to

that of grinding wheel is 200Sol–1: (b)

Grinding ratio G =

Volume of materialremoved

Volume of wheel wear

If G = 200, it means, volume of materialremoved is 200 times the volume of wheelwear. It means that grinding wheel wear0.005 times the volume of material removed.

2. A four bar mechanism is made up of linksof length 100, 200, 300 and 350 mm. If the350 m link is fixed, the number of linksthat can rotate fully is ______.

Sol–2: (1)

The schematic of 4-bar mechanism

sum of longest and shortest (450 mm) isless than rest two (500 mm).

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Hence Grashof’s law satisfied.

B

C

A D

350 mm

100m

m

300mm

200mm

The line adjoint to shortest (100 mm) is fixed,the mechanism will be crank rocker. Henceone link rotates fully.

3. Which one of the following statements iscorrect for a superheated vapour?(a) Its pressure is less than the saturation

pressure at a given temperature(b) Its temperature is less than the

saturation temperature at a givenpressure

(c) Its volume is less than the volume ofthe saturated vapour at a giventemperature

(d) Its enthalpy is less than the enthalpy ofthe saturated vapour at a given pressure

Sol–3: (a)

P < Psat at given T of superheated vapour

TP=C

T

s

P

Psat

4. A steel column of rectangular secton (15mm × 10 mm) and length 1.5 m is simplysupported at both ends. Assuming modulusof elasticity, E = 200 GPa for steel, thecritical axial load (in kN) is ______ (correctto two decimal places).

Sol–4: (1.1)

For colum hinged at both ends

PC =

2

2EI

=

2 3 3

2 6200 10 15 101.5 10 12

= 1096 N = 1.1 kN

5. According to the Mean value theorem, for acontinuous function f(x) in the interval [a,b], there exists a value in this interval

such that b

af(x)dx

(a) f( )(b a) (b) f(b)( a)

(c) f(a)(b ) (d) 0

Sol–5: (a)

If (a,b) then mean value theorem forintegrals in stated as follows:

b

af(x)dx = b a f

6. A six-faced fair dice is rolled five times. Theprobability (in %) of obtaining “ONE” at leastfour times is(a) 33.3 (b) 3.33(c) 0.33 (d) 0.0033


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Sol–6: (c)

Probability of getting 1 in single throw

P = 16

and not getting 1 is

q =

1 516 6

Now probability of getting 1 atleast 4times in 5 times rolled

Here n = 5

P x 4 = n 44 5 n 5n n4 5C P q C P 5/ 6

5 0

5 15 54 5

1 5P x 4 C 1 / 6 5 / 6 C6 6

=

4 51 5 15 16 6 6

= 3.343 × 10–3

Now % = 3.343 × 10–3 × 100

= 0.33

7. If the wire diameter of a compressive helicalspring is increased by 2%, the change inspring stiffness (in %) is _____ (correct totwo decimal places).

Sol–7: (8.24)

Wire diameter of compressive helical springas increased by 2%.

i.e. d2 = 1.02 d1

Stiffness of spring 2 is given by

K2 =

4

123 3

G 1.02dGd8D N 8D N

=

41

3Gd1.0824

8D N

= 1.0824× K1

Change in spring stiffness (in %) =

2 1

1

K K100

K =

1 1

1

1.0824 K K100

K

= 8.24%

8. For an Oldham coupling used between twoshafts, which among the folloiwngstatements are correct?

(i) Torsional load is transferred along shaftaxis.

(ii) A velocity ratio of 1 : 2 between shaftsis obtained without using gears.

(iii) Bending load is transferred to shaftaxis.

(iv) Rotation is transferred along shaft axis(a) i and iii (b) i and iv(c) ii and iii (d) ii and iv

Sol–8: (b)

The oldham coupling connects parallel shaftswith small offset. The connecting shaftsrotates at same speed. The motion andtorsion load transfer is along axes of shafts.

9. A flat plate of width L = 1 m is pusheddown with a velocity U = 0.01 m/s towardsa wall resulting in the drainage of the fluidbetween the plate and the wall as shown inthe figure. Assume two dimensionalincompresible flow and that the plate remainparallel to the wall. The average velocity,uavg of the fluid (in m/s) draining out at theinstant shown in the figure is _____ (correctto three decimal places).

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U

d = 0.1 muavg uavg

LPlate

Wall

Sol–9: (0.05)

uavg

d=0.1m

L Plate

d=0.1m

uavg

U

Let B = width of plate

By conservation of mass we get

avg2 u d b = U L b

avg2 u 0.1 = 0.01 1

uavg = 0.05 m/s

10. If 1 3and are the algebraically largestand smallest principal stresses respectively,the value of the maximum shear stress is.

(a) 1 3

2 (b) 1 3

2

(c) 1 32 (d) 1 3

2

Sol–10: (b)

Maximum shear stress, 1 2

max 2

11. The time series forecasting method thatgives equal weightage to each of the m mostrecent observations is(a) Moving average method(b) Exponential smoothing with linear trend(c) Triple exponential smoothing(d) Kalman filter

Sol–11: (a)

Simple moving average method gives equalweight to each of the m msot recentobservations. Weighted moving averagemethod gives more weight to the recentvalues.

12. For a pelton wheel with a given water jetvelocity, the maximum output power fromthe pelton wheel is obtained when the ratioof the bucket speed to the water jet speed is______ (correct to two decimal places).

Sol–12: (0.5)

For pelton wheel, maximum output power

When UV = 0.5

U =Bucket speed, V = Jet speed

13. Using the Taylor’s tool life equation withexponent n = 0.5, if the cutting speed isreduced by 50%, the ratio of new tool life tooriginal tool life is(a) 4 (b) 2(c) 1 (d) 0.5

Sol–13: (a)

Taylor’s tool life equation is

VTn = C

Given n = 0.5, V2 = V1/2

n1 1V T = n

2 2V T


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n2

1

TT = 1

2

VV

2

1

TT =

10.5

1/n1

2

V 2 4V

14. The height (in mm) for a 125 mm sine barto measure a taper of 27°32’ on a flat workpiece is ______ (correct to three decimalplaces).

Sol–14: (57.783)

sin =H

sin (27.53) = H125

H = 57.783mm

15. Four red balls, four green balls and fourblue balls are put in box. Three balls arepulled out of the box at random one afteranother without replacement. Theprobability that all the three balls are redis(a) 1/72 (b) 1/55(c) 1/36 (d) 1/27

Sol–15: (b)

(4R, 4G, 4B)

P(RRR) (Without replacement) =

4 3 2

1 1 112 11 10

1 1 1

C C CC C C

= 155

16. For a two dimensional incompressible flowfield given by

ˆ ˆu A(xi yj) , where A > 0,which one of the following statements isfalse?

(a) It satisfies continuity equation(b) It is unidirectional when x 0 and y

(c) Its streamlines are given by x = y(d) It is irrotational

Sol–16: (c)

u = ˆA xi yj

ux = Ax, Uy = Ay

For continuity of flow

u Vx y = 0

= A – A = 0

For streamline flow equation is

dxu = dy

v

dxAx =

dyAy

dxx =

dyy

ln x = ln y C

lnx ln y = C

ln xy = C

xy = ec

17. A bar of uniform cross section and weighing100 N is held horizontally using twomassless and inextensible strings S1 andS2 as shown in the figure.

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L/2 L/2

BarS2S1T = ?2T = ?1

Rigid support

The tensions in the strings are

(a) T1 = 100N and T2 = 0 N

(b) T1 = 0 N and T2 = 100 N

(c) T1 = 75 N and T2 = 25 N

(d) T1 = 25 N and T2 = 75 N

Sol–17: (b)

L/2 L/2

BarS2S1T = ?2T = ?1

Rigid support

A

T1 + T2 = 100

Taking moment about A,

2LT2 =

L1002

T2 = 100 N

T1 = 0N

18. The equation of motion for a spring masssystem excited by a harmonic force is

Mx Kx Fcos( t)

Where M is the mass, K is the springstiffness, F is the force amplitude and isthe angular frequency of excitation.Resonance occurs when is equal to

(a) MK (b)

1 K2 M

(c) K2M (d) K

M

Sol–18: (d)

The given equation is

MX + kx F cos wt

It is the equation of frequency under dampedforced vibration

So, Also we know that natural frequency offree vibration is given by

2

n = k/m

n = (k/m)1/2

At resonance, = n therefore

the angular speed at which resonance occuris

= n k / m

19. The rank of the matrix

4 1 11 1 17 3 1

is

(a) 1 (b) 2(c) 3 (d) 4

Sol–19: (b)

A =

2 1R R4 1 1 1 1 11 1 1 4 1 1

7 3 1 7 3 1


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2 2 1

3 3 1

1 1 1R R 4Rthen A ~ 4 1 1

R R 7R 7 3 1

3 3 2R R 2R then

1 1 1A ~ 0 5 3

0 0 0

Hence A = No. of non zero rows = 2

20. Interpolator in a CNC machine(a) Controls spindle speed(b) Coordinates axes movements(c) Operates tool changer(d) Commands canned cycle

Sol–20: (b)

Basically, an interpolater provides twofunction:

It computes individual axis velocity to drivethe tool along the programmed path at agiven feed rate. It generates intermediatecoordinate positions along the programmedpath.

21. In a linearly hardening plastic material thetrue stress beyond initial yielding.(a) Increases linearly with the true strain(b) Decreases linearly with the true strain(c) First increases linearly and then

decreases linearly with the true strain(d) Remains constant

Sol–21: (a)

Plasticity describe non-linear materialbehaviour where material deformpermanently due to application of load.

Plastic

Elastic

Strain

(Without hardening)

Linear hardeningPlastic

Elastic

Since in virtual case, as stress increaseswith strain so, also in real case same willfollow due to hardening mechanism.

22. The type of weld represented by the shadedregion in the figure is

(a) Groove (b) Spot(c) Fillet (d) Plug

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Sol–22: (c)

Fillet weld

23. The number of atoms per unit cell and thenumber of slip systems, respectively, for afacecentered cubic (FCC) crystal are(a) 3, 3 (b) 3, 12(c) 4, 12 (d) 4, 48

Sol–23: (c)

Effective number of atoms

Ne = 1 18 Corners 6 face 48 2

No. of slip system in FCC crystal is = 12

24. F(Z) is a function of the complex variable z= x + iy given by

F(Z) = iz + k Re(Z) + i lm (Z)

For what value of k will F(Z) satisfy theCauchy-Riemann equations?(a) 0 (b) 1(c) –1 (d) y

Sol–24: (b)

F(z) = iz + kRe(z) + i Im(z)

= i(x+iy) + k(x) + i(y)

= (kx y) i(x y)

= u + iv

So u = kx –y and v = x + y

By C-R equation; Vx = Vy and uy = – Vx

(R = 1) = R(–1 = –1)

Hence k = 1

25. An ideal gas undergoes a process from state1 (T1 = 300 K. p1 = 100 kPa) to state 2 (T2= 600 K, p2 = 500 kPa). The specific heats

of the ideal gas are : Cp = 1 kJ/kg-K and Cv= 0.7 kJ/kg-K. The change in specific entropyof the ideal gas from state 1 to state 2(inkJ/kg-K) is _____ (correct to two decimalplaces).

Sol–25: (0.21)

s =

2 2p

1 1

T PC ln RlnT P

=

600 5001 ln 0.3ln300 100

= 0.21 kJ/kgK

26. F(S) is the Laplace transform of the function

f(t) = 2t2 e–t

F(1) is ______ (correct to two decimal places).

Sol–26: (0.5)

f(t) = 2t2 e–t

nL t =

n 1n 1s

L{t2} = 3 32 1 2s s

then L{2t2} = 34s

Hence L{e–t.(2t2)} = 34 F(s)

s 1

so F(1) = 3

4 1 0.521 1

27. An engine working on air standard Ottocycle is supplied with air at 0.1 MPa and35°C. The compression ratio is 8. The heat


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supplied is 500 kj/kg. Property data for air.Cp = 1.005 kJ/kg K, Cy = 0.718 kJ/kg K. R= 0.287 kJ/kg K. The maximumtemperature (in K) of the cycle is ____(correct to one decimal place).

Sol–27: (1404)

P

V

3

2 4

1

V2 = 1V8

Cv(T3 – T2) = 500 kJ/kg

y 12

1

VV = 1

2

TT

1.4 118 =

2

35 273T

T2 = 707.60 K

0.718 × (T3 – 707.60) = 500

T3 = 1403.978 K

28. An explicit forward Euler method is used tonumerically integrate the differential

equation dy ydt using a time step of 0.1.

With the initial condition y(0) = 1, the valueof y(1) computed by this method is ____(correct to two decimal places).

Sol–28: (2.59)

dydt = f(t, y) y

and y (0) = 1 t0 = 0, y0 = 1 and step size= 0.1 (given)

By forward euler method

Iteration 1 : y1 = y0 + h f (t0, y0)

= 1 + 0.1 [y0]

= 1 + 0.1 [1] = (1.1)

y2 = 1.1 + 0.1 [1.1]

= 1.21

y3 = 1.21 + 0.1 [1.21]

= 1.21 + 0.21 = 1.331

y4 = 1.331 + 0.1 [1.331]

= 1.331 + 0.1331

= 1.4641

y5 = 1.6105

y6 = 1.771

y7 = 1.9487

y8 = 2.14

y9 = 2.3579

y10 = 2.5937

29. Processing times (including setup times) anddue dates for six jobs waiting to be processedat a work centre are given in the table. Theaverage tardiness (in days) using shortestprocessing time rule is ______ (correct totwo decimal places).

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Job Processing time (days) Due date (days)A 3 8B 7 16C 4 4D 9 18E 5 17F 13 19

Sol–29: (6.33)

Jobs arranged according to shortestprocessing time

( )

A 3 8 3 0C 4 4 3 4 7 3E 5 17 7 5 12 0B 7 16 12 7 19 30D 9 18 19 9 28 10F 13 19 28 13 41 22

Job Processing Due Job flow Tardinesstime days Date time

Total 38

Average Tardiness = 38 6.336

30. A point mass is shot vertically up fromground level with a velocity of 4 m/s attime, t = 0. It loses 20% of its impact velocityafter each collision with the ground.Assuming that the acceleration due togravity is 10 m/s2 and that air resistance isneglibible, the mass stops bouncing andcomes to complete rest on the ground aftera total time (in seconds) of(a) 1 (b) 2(c) 4 (d)

Sol–30: (c)

Given, V0 = 4m/s, g = 10m/s2

u

V=0

V = V0 + at

0 = V0 – gt

Time of ascent, t = V0/g

But we know, time of ascent = time ofdescent, = 0.4 s

Total time, t1 = V2g

After rebounding

V1 = 0.8 V0

Similarly t2 = 1V2g

Total time required

t = t1 + t2 + t3 + ......

= 0 1 2V 2V 2V2 ....g g g

= 0 1 22 V V V ....g

= 20 0

2 V 0.8V 0.8 V ...g

=

20

2 V 1 0.8 0.8 ...g

=

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31. The value of the integral r ndS over

the closed surface S bounding a volume V,where

ˆˆ ˆr xi yj zk is the position vectorand n is the normal to the surface S, is

(a) V (b) 2V(c) 3V (d) 4V

Sol–31: (c)

r = ˆˆ ˆxi yj 3k then

div (r) =

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

= 1 + 1 + 1 = 3

Surface is closed with volume V

so by Gauss-Divergence theorem

s ˆ(f n)ds =

vdiv dv

(r n)ds = v 3 dv 3v

32. A tank of volume 0.05 m3 contains a mixtureof saturated water and saturated steam at200°C. The mass of the liquid present is 8kg. The entropy (in kJ/kg K) of the mixtureis _______ (correct to two decimal places).

Property data for saturated steam and waterare :

At 200°C, psat = 1.5538 MPa

vf = 0.001157 m3/kg, vg = 0.12736 m3/kg

sfg = 4.1014 kJ/kg K, sf = 2.3309 kJ/kg K

Sol–32: (2.49)

Volume of liquid = mf × Vf

= 8 × 0.001157 = 9.256× 10–3 m3

= 0.009256 m3

So volume of steam = 0.05 – 0.009256

= 0.040744 m3

Mass of steam = g

Volume of steamv

= 0.0407440.12736

= 0.319912 kg = ms

Total mass of mixture = mf + ms

= 8 + 0.319912

= 8.319912 kg

dryness fraction =

s

f s

mxm m

= 0.3199128.319912

= 0.03845

So, the entropy of the mixture is given by

s = sf + x sfg

= 2.3309 + 0.03845 ×4.1014

= 2.488 kg/kgK

33. A slider crank mechanism is shown in thefigure. At some instant, the crank angle is45° and a force of 40 N is acting towardsthe left on the slider. The length of thecrank is 30 mm and the connecting rod is70 mm. Ignoring the effect of gravity,friction and inertial forces, the magnitudeof the crankshaft torque (in Nm) needed tokeep the mechanism in equilibrium is ____(correct to two decimal places).

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40 N

Sol–33: (1.12)

O

T

30m

m

A

70mm

B

F=40

N=45°

The torque required to hold the mechanismin equilibrium is equal and opposite ofturning moment,

T =

2 2sin2r.F sin

2 n sin

=

22

sin900.03 40 sin45702 sin 4530

=

1 11.22 49 12

9 2

= 1.2 {0.707 + 0.225}

= 1.1184

1.12 N.m

34. As self-aligning ball bearing has a basicdynamic load rating (C10, for 106 revolutions)of 35 kN. If the equivalent radial load on

the bearing is 45 kN, the expected life (in106 revolutions) is(a) Below 0.5 (b) 0.5 to 0.8(c) 0.8 to 1.0 (d) Above 1.0

Sol–34: (a)

C = 35kN

Per = 45kN

Expected life of ball bearing

L90 =

3 3

er

C 35P 45

= 0.47 million revolution

35. A solid block of 2.0 kg mass slides steadilyat a velocity V along a vertical wall as shownin the figure below. A thin oil film ofthickness h = 0.15 mm provides lubricationbetween the block and the wall. The surfacearea of the face of the block in contact withthe oil film is 0.04 m2. The velocitydistribution within the oil film gap is linearas shown in the figure. Take dynamicviscosity of oil as 7 × 103 Pa-s andacceleration due to gravity as 10 m/s2.Neglect weight of the oil. The terminalvelocity V (in m/s) of the block is _____(correct to one decimal place).

Vm=2.0

A = 0.04 m2

Impermetablewall

h = 0.15 mm


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Sol–35: (10.714)

h = 0.15 mm

m =2kg2

V

h = 0.15 mm

A = 0.04m2

= 7 × 10–3 Pa-s

g = 10 m/s2

Terminal velocity is achieved in case of zeroacceleration.

Hence Fnet = 0

Weight – Frictional force = 0

V Ah = mg

3

37 10 V 0.04

0.15 10 = 2 × 10

V = 10.714 m/s

36. A tank open at the top with a water levelof 1 m, as shown in the figure, has a holeat a height of 0.5 m. A free jet leaveshorizontally from the smooth hole. Thedistance X (in m) where the jet strikes thefloor is

Free Jet1 m

0.5 m

x

(a) 0.5 (b) 1.0(c) 2.0 (d) 4.0

Sol–36: (b)

0.5m

Vx

VxFree jet

x

Vy

Vx = 2gh 2 9.8 0.5

= 3.1304 m/s

and, Apply Newton’s law of motion in xdirection

x = x1V t 0 t2

x = 3.1304 t

and y =

2y

1V t gt2

y =

210 9.8 t2

0.5 = 29.8 t2

t2 = 19.8

t = 1 0.319 sec

9.8

Put this value in above equation we get

x = (3.1304) (0.319)

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MASTER


ME



x = 1m

37. Block P of mass 2 kg slides down thesurface and has a speed 20 m/s at the lowestpoint Q, where the local radius of curvatureis 2 mass as shown in the figure. Assumingg = 10 m/s2, the normal force (in N) at Qis _______ (correct to two decimal places).

Q

P

Sol–37: (420 N)

Given, m = 2kg, v = 20m/s, g = 10m/s2,

R = 2m

Q

P

As the body moves on a circular path nearthe end. Hence centrifugal force will alsoact.

Free body diagram

N

V

2mVmg

R

Normal reaction

N =

2 2mv 2 20mg 2 10 420NR 2

38. A machine of mass m = 200 kg is supportedon two mounts, each of stiffness k = 10 kN/m. The machine is subjected to an externalforce (in N) F(t) = 50 cos 5t. Assuming onlyvertical translatory motion, the magnitudeof the dynamic force (in N) transmitted fromeach mount to the ground is ________(correct to two decimal places).

kk

F(t)

m

Sol–38: (33.33)

The force transmissibility,

Ftr =

2

n s22 2 0

n n

1 2FF

1 2

Since damping is zero in the system,

= 0

Ftr =

s2

0

n

F 1F

1

Natural frequency of system,

n =

k 10,000 2m 200


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IES

MASTER

= 100 rad/sec

s

0

FF =

2151

100

=

1 125 1 0.251

100

= 1.33

Force transmitted to foundation,

Fs = 1.33 F0 = 1.33 × 50

= 66.66 N

Force transmitted to ground from each

mounting = sF2 = 33.33 N

39. In a Lagrangian system, the position of afluid particle in a flow is described as x =x0e–kt and y = y0ekt where t is the timewhile x0, y0 and k are constants. The flowis(a) Unsteady and one-dimensional(b) Steady and two-dimensional(c) Steady and one-dimensional(d) Unsteady and two-dimensional

Sol–39: (b)

x = x0 e–kt

y = y0 ekt

1. Here flow is two dimensional because

dxdt 0, dy

dt 0

2. dxdt = u = x0 × (–k) e–kt

= – x0ke–kt

= –kx

dydt = V = kt

0ky e

= ky

For a given point (x, y), velocity is not afunction of time t.

40. A sprinkler shown in the figure rotatesabout its hinge point in a horizontal planedue to water flow discharged through itstwo exit nozzles.

20 cm10 cm

Q/2 Q/2

The toal flow rate Q through the sprinkleris litre/see and the cross-sectional area ofeach exit nozzle is 1 cm2. Assuming equalflow rate through both arms and africtionless hinge, the steady state angularspeed of the rotation (in rad/s) of thesprinkler is ____ (correct to two decimalplaces).

Sol–40: (10)

0.1 0.1

10cm 20cm

V1 V2

Q = 1 litre/sec, A = 1 cm2

First find the velocity of flow

Q2 = A1V1

31 10

2= (1 × 10–4) × V1

V1 = 5m/s

IES

MASTER


ME



If the angular velocity of the shrinkler is then the absolute velocities of flow throughthe nozzle are

V1ab = 5 0.1

V2ab = 5 0.2

Since the moment of momentum of flowentering is zero and there is no friction themomentum leaving the shrinkler must alsobe zero.

Q 5 0.1 0.1 5 0.2 0.22 = 0

1 0.5 0.01 1 0.042 = 0

5.0 1 0.05 = 0

= 10 rad/sec

41. An orthogonal cutting operation is beingcarried out in which uncut thickness is0.010 mm, cutting speed is 130 m/min, rakeangle is 15° and width of cut is 6 mm. Itis observed that the chip thickness is 0.015mm, the cutting force is 60 N and the thrustforce is 25 N. The ratio of friction energy tototal energy is ______ (correct to twodecimal places).

Sol–41: (0.44)

Given t1 = 0.01mm, t2 = 0.015

V = 130 m/min, 15 , b = 6mm

Fc = 60 N, FT = 25 N

Ratio of friction energy to total energy

=

c 1

c 2

F V tsinF V cos t ...(i)

Friction angle,

1 T

C

FtanF

=

1 2515 tan60

= 37.619

Now from (i)

Friction energyTotal energy =

sin37.619 0.01cos 22.619 0.015

= 0.44

42. An epicycle gear train is shown in the figurebelow. The number of teeth on the gears A,B and D are 20, 30 and 20, respectively.Gear C has 80 teeth on the inner surfaceand 100 teeth on the outer surface. If thecarrier arm AB is fixed and the sum gearA rotates at 300 rpm in the clockwisedirection, then the rpm of D in the clockwisedirection is

C

A D

B

(a) 240 (b) –240(c) 375 (d) –375

Sol–42: (c)

Acrtion Arm A B CArm fix 2 0 x 20 2x 30x

30 3 80x rotation to A2x x3 4

Add y y x y 2x xy y3 4


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IES

MASTER

The arm is fixed,

y = 0

The gear A rotates at 100 rpm CW.

x + y = 300

x = 300

Rotation of B,

NB = 2x 2y 0 3003 3

= – 200 rpm (CCW)

Rotation of C.

NC = x 300y 04 4

= 75 rpm (CCW)

The gear D meshes with C externally

ND = C out

CD

ZN

Z

= 1000.75 75 520

= 375 rpm (CW)

43. Let X1, X2 be two independent normalrandom variables with means 1 2, and

standard deviations 1 2, , respectively.Consider Y = X1 – X2; 1 2 1 21, 1, 2 . Then

(a) Y is normally distributed with mean 0and variance 1

(b) Y is normally distributed with mean 0and variance 5

(c) Y has mean 0 and variance 5, but isNOT normally distributed

(d) Y has mean 0 and variance 1, but isNOT normally distributed

Sol–43: (b)

E(x1) = 1 =E(x2) and V(x1) = 1

and V(x2) = 4

1 2 1 2E y E X X E X E X 1 1 0

1 2 1 2V y V X X V X V X 1 4 5 1 2Y X X

Hence y is also normally distributed withmean 0 and variance 5

44. A bar is compresed to half of its originallength. The magnitude of true strainproduce in the deformed bar is ________(correct to two decimal places).

Sol–44: (0.69)

True strain, T =

fn

en

2

n 0.5 0.693

Magnitude of T 0.69

45. The true stress ( ) - ture strain ( )diagram of a strian hardening material isshown in figure. first, there is loading up topoint A, i.e., up to stress of 500 MPa andstrain of 0.5. Then from point A, there isunloading up to point B, i.e. to stress of 100MPa. Given that the Young’s modulusE = 200 GPa, the natural strain at pointB B( ) is ______ (correct to three decimalplaces).

IES

MASTER


ME



A

B

0.5

100

(MPa)500

Sol–45: (0.498)

Slope of initial part of stress strain curve

= E =

3E 200 10 MPa

Slope, AB = B

500 100 :0.5 Slope of initial part

of curve

200 × 103 = B

4000.5

B = 0.498

46. A plane slab of thickness L and thermalconductivity k is heated with a fluid on oneside (P) and the other side (Q) is maintainedat a constant temperature, TQ of 25°C, asshown in the figure. The fluid is at 45°Cand the surface heat transfer coefficient, h,is 10 W/m2K. The steady state temperature,TP, (in °C) of the side which is exposed tothe fluid is ______ (correct to two decimalplaces).

h = 10 W/m K2

T = 45°C

k = 2.5 W/mk

L = 20 cm

T = 25°CQ

T P

Sol–46: (33.88)

K=2.5 W/mK

QT 25 C

L=20cm

h=10W/m K2

TP T 45 C

pT T1/h A =

p QT TL/k A

(45 – Tp)×10 = pT 25 2.50.2

45 – Tp = 1.25 TP – 31.25

76.25 = 2.25Tp

Tp = 33.88°C

47. The minimum value of 3x + 5y such that:

3x 5y 154x 9y 813x 2y 2x 0,y 0

is _______.


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IES

MASTER

Sol–47: (0)

Minimise the objective function,

Z = 3x + 5y

Subject to constraints

3x + 5y 15

4x + 9y 8

13x + 2y 2

x 0, y 0

3x + 5y = 15 [(x = 0, y = 3) and

(y = 0, x = 5)]

4x + 9y = 9 [(x = 0, y = 89 ) and

(y = 0, x = 2)]

13x + 2y = 2 [(x = 0, y = 1) and

2y 0, x13

y

(0, 0)0

80, A9

3x+5y 15<

2 96B ,109 109

C

2 ,013

13x+2y 2<

4x+9y 8<x

The value of objective function at points

Z0 = 3 × 0 + 5 × 0 = 0 (min)

ZA = 3 × 0 + 8 5 4.449

ZB = 2 963 5 0.46

109 109

ZC = 23 0 5 4.45

13

Zmin = 0

48. An electrochemcial machining (ECM) is tobe used to cut a through hole into a 12 mmthick aluminum plate. The hole has arectangular cross-section, 10 mm × 30 mm.The ECM operation will be accomplished in2 minutes, with efficiency of 90%. Assumingspecific removal rate for aluminum as 3.44× 102 mm3/(A s), the current (in A) requiredis _____ (correct to two decimal places).

Sol–48: (968.99)

specificM.RR =

L w t 1time I

0.9 × 3.44 × 10–2 =

10 30 12 1

2 60 I

I = 968.99 Ampere

49. The schematic of an external drum rotatingclockwise engaging with a short shoe isshown in the figure. The shoe is mountedat point Y on a rigid lever XYZ hinged atpoint X. A force F = 100 N is applied at thefree end of the lever as shown. Given thatthe coefficient of friction between the shoeand the drum is 0.3 the braking torque (inNm) applied on the drum is _____ (correctto two decimal places).

IES

MASTER


ME



300200

300

X

YF

Z100

(All dimensions are in mm)

Sol–49: (8.18)

300

200

F

FN

R

300

R × 200 = F × 300 + FN = 300

R × 200 = 100 × 300 + R × 300

R(200 – 0.3 × 300) = 100 × 300

R = 272.73 N

Braking torque = FN × Radius of drum

= R r

= 0.3 × 272.73 × 0.1

= 8.18 Nm

50. The maximum reduction in cross-sectionalarea per pass (R) of a cold wire drawingprocess is

R = 1 – e–(n + 1)

where n represent the strain hardeningcoefficient. For the case of a perfectly plasticmaterial, R is.(a) 0.865 (b) 0.826(c) 0.777 (d) 0.632

Sol–50: (d)

The strain hardening exponent (also calledstrain hardening index) ‘n’ is materialconstant which is used in calculation forstress-strain behaviour in work hardeningthis is given by following formula.

= nK

= applied stress on material

= strain, K = strength co-efficient

n lies in betroean 0 to 1.

For perfectly plastic material

So from question n = 0

R = n 11 e

R = 1 11 e 1e

= 0.632

51. A carpenter glues a pair of cylindricalwooden logs by bonding their end faces atan angle of 30 as shown in the figure.

Log 1

4 M

Pa

Log 2

= 30°

interface

axis

4 M

Pa

The glue used at the interface fails if

Criterion 1 : the maximukm normal stressexceeds 2.5 MPa.


ME



IES

MASTER

Criterian 2 : the maximum shear stressexceeds 1.5 MPa.

Assume tht the interfacxe fails before thelogs fail. When a uniform tensile stress of 4MPa is applied the interface(a) Fails only because of criterion 1(b) Fails only because of criterion 2(c) Fails because of both criterion 1 and 2(d) Does not fail

Sol–51: (c)

Q=30°

=

x y x y2 2

xycos 2 sin 2

= 4 0 4 0

2 2

cos 60 + 0 = 3 MPa

=

4 0 sin60 02

= 1.73 MPa

So interface fails because of both criterion 1and 2

52. The percentage scrap in a sheet metalblanking operation of a continuous strip ofsheet metal as shown in the figure is_______ (correct to two decimal places).

Feed D

D/5

D/5

D/5

D/5

D/5

D/5D/5D/5

D

Sol–52: (53.25)

D/10 D/10 D D/10 D/10

D/5

D/5

%scrap =

Area of hole1 100blank area

=

2D41 100

D D D DD D5 5 10 10

= 53.25%

53. Steam flow through a nozzle at a mass flowrate of m 0.1kg / s with a heat loss of 5kW. The enthalpies at inlet and exit are2500 kJ/kg and 2350 kJ/kg, respectively.Assuming negligible velocity at inlet

1(C 0) , the velocity (C2) of steam (in m/s)

IES

MASTER


ME



at the nozzle exit is ______ (correct to twodecimal places).

h = 2500 kJ/kg1C = 01 C2

h = 2350 kJ/kg2

Q = 5 kW.

m = 0.1 kg/s.

Sol–53: (447.213)

h =2350 kJ/kg2

m 0.1 kg/sh=2

500

kJ/k

g1

Applying steady flow energy equationwe get

21

1 3Cm h

2 10 1z

Q0

0

=

22

2 23Cm h z

2 10

0

22

3Cm Q

2 10 = 1 2m h h

22

3Cm

2 10= m 2500 2350 5

22

3Cm

2 10=

0.1 2500 2350 5

22C =

310 2 100.1

C = 447.213 m/s

54. The state of stress at a point, for a body inplane stress, is shown in the figure below.If the minimum principal stress is 10 kPa,

then the normal stress y (in kPa) is

y

xy = 50 kPa

x = 100 kPa

(a) 9.45 (b) 18.88(c) 37.78 (d) 75.50

Sol–54: (c)

x = 100 kPa

xy = 50 kPa

2 = 10 kPa

Principal stress

2 = 2 2

x y x y xy1 42

10 = 2 2

y y1 100 100 4 502


ME



IES

MASTER

20 = 2 2y y100 100 4 50

y 80 = 2y100 10000

squaring both sides

2y 80 = 2

y100 10000

Squaring both sides,

2y 80 = 2

y100 10000

2 2y y80 2 80 =

2 2y100 y 2 100 10000

y 2 80 2 100 = 1002 + 10000 – 802

y =

2 2100 10000 802 80 2 100

= 37.78 kPa

55. A simply supported beam of width 100 mm,height 200 mm and lenght 4 m is carrying

a uniformly distributed load of intensity 10kN/m. The maximum bending stress (inMPa) in the beam is ______ (correct to onedecimal place).

4 m

10 kN/m

Sol–55: (30)

Maximum Bending moment, Mmax = 2

8

Maximum bending stress, b = max2

6Mbd

=

2

26 1

8 bd

=

3 2

26 10 10 4 1

8 0.1 0.2= 30 MPa

Conventional Test 10:00 A.M. to 1:00 P.M.

Test Type Timing

Conventional Full Length Test Paper-1 10:00 A.M. to 1:00 P.M.Conventional Full Length Test Paper-2 02:00 P.M. to 5:00 P.M.

Note : The timing of the test may change on certain dates. Prior information will be given in this regard.*N.T. : New Topic. *R.T. : Revision Topic

Call us : 8010009955, 011-41013406 or Mail us : [email protected]

ESE-2018 Conventional Test Schedule, Mechanical Engineering

Institute for Engineers (IES/GATE/PSUs)

IES MASTER

Date Topic

N.T.

N.T.

N.T.

N.T.

N.T.

N.T.

N.T.

N.T.

N.T.

N.T.

N.T.

R.T.

R.T.

R.T.

R.T.

R.T.

R.T.

R.T.

R.T.

R.T.

R.T.

R.T.

:

:

:

:

:

:

:

:

:

:

:

TH-1, TH-2, HT-1, RAC-1, MS-1, MS-2

FMM-1, RAC-2, IE-2, RSE-1

MECH-1, MECH-2, HT-2, RE-1

FMM-2, PPE-1,RSE-2

ICE-1,ToM-2, MR-1

ToM-1, MR-2, PROD-1

IE-1, PPE-2, FMM-3,

PPE-3, PROD-2

ToM-3, ICE-2

RE-2, MD-1

:

:

:

:

:

:

:

:

:

:

:

TH-2, MS-1, HT-1

RAC-1, RAC-2, MS-2

HT-1, HT-2, TH-1, FMM-1, IE-2

FMM-2, RSE-1, RSE-2, PPE-1

MS-1, MECH-1, MECH-2,TH-1

PPE-1, MS-2, HT-1, PROD-1,ToM-1, ICE-1

RAC-1, RAC-2, RE-1, IE-1, MR-1, MECH-1

MR-2, RSE-1, RSE-2, HT-1, HT-2, FMM-2

PPE-1, PPE-2, FMM-3, ToM-2, ToM-3

FMM-1, FMM-2, PROD-1, PROD-2, MECH-1, ICE-2, MD-1

Mech-3, MD-2

Full Length-1 (Test Paper-1 + Test Paper-2)



:

11th Mar 2018

25th Mar 2018

01st Apr 2018

08th Apr 2018

15th Apr 2018

22nd Apr 2018

29th Apr 2018

06th May 2018

13th May 2018

20th May 2018

27th May 2018

03rd Jun 2018

10th Jun 2018

17th Jun 2018

Thermodynamic

Heat Transfer

IC Engines

Refrigeration AirConditioning

Fluid Mechanicsand Machinery

Power PlantEngineering

TH-1

HT-1

ICE-1

RAC-1

FMM-1

PPE-1

Thermodynamic systems and processes;

Zeroth, First and Second Laws of Thermodynamics.

properties of pure substance.

Steady and unsteady heat conduction, Fins,

Radiative heat transfer.

SI and CI Engines, Engine Systems and Components, Fuels.

Vapour compression refrigeration, Refrigerants,

Compressors, Other types of refrigeration systems like Vapour

Absorption, Vapour jet, thermo electric and Vortex tube

refrigeration and Heat pump.

TH-2

HT-2

ICE-2

RAC-2

Entropy, Irreversibility and availability; Real and Ideal gases;

compressibility factor; Gas mixtures.

Free and forced convection, boiling and condensation,

Heat exchanger.

Performance characteristics and testing of IC Engines;

Emissions and Emission Control. Otto, Diesel and Dual Cycles.

Psychometric properties and processes, Comfort chart,

Comfort and industrial air conditioning, Load calculations and

Condensers, Evaporators and Expansion devices.

FMM-2

PPE-2

FMM-3

PPE-3

Basic Concepts and Properties of Fluids,

Manometry, Fluid Statics,

Buoyancy, Equations of Motion such as

velocity potential, Stream Function.

Steam and Gas Turbines, Rankine and

Brayton cycles with regeneration and reheat.

Bernoulli's equation and applications,

Viscous flow of incompressible fluids,

Laminar and Turbulent flows, Flow through

pipes and head losses in pipes.

Reciprocating and Centrifugal pumps,

Hydraulic Turbines and

other hydraulic machines.

Fuels and their properties, Flue gas

analysis, Theory of Jet Propulsion –

Pulse jet and Ram Jet Engines,

Reciprocating and Rotary Compressors.

Boilers, power plant components like

condensers, air ejectors, Electrostatic

precipitators and cooling towers.

RenewableSources of

Energy

RSE-1 RSE-2

Solar Radiation, Solar Thermal Energy collection -

Flat Plate andfocusing collectors their materials

and performance. Solar Thermal Energy Storage, Applications

– heating, cooling and Power Generation.

Solar Photovoltaic Conversion; Harnessing of Wind Energy,

Bio-mass and Tidal Energy – Methods and Applications,

Working principles of Fuel Cells.

EngineeringMaterials

MS-1 MS-2

Basic Crystallography, Alloys and Phase

diagrams, Heat Treatment.

Ferrous and Non Ferrous Metals, Non metallic materials,

Basics of Nano-materials, Mechanical Properties and

Testing, Corrosion prevention and control.

Engineering Mechanics

(SoM)

Mech-1 Mech-2 Mech-3

Analysis of System of Forces, Friction,

Centroid and Centre of Gravity, Dynamics.

Stresses and Strains-Compound Stresses

and Strains, Bending Moment and

Shear Force Diagrams.

Theory of Bending Stresses-Slope and

deflection-Torsion, Thin and thick

Cylinders, Spheres.

Mechanismsand

Machines

ToM-1 ToM-2 ToM-3

Mechanisms, Kinematic Analysis,

Velocity and Acceleration. CAMs with

uniform acceleration, cycloidal motion,

oscillatingfollowers; Effect of

Gyroscopiccouple on automobiles,

ships and aircrafts. Governors.

Vibrations –Free and forced vibration

of undamped and damped SDOF

systems, Transmissibility Ratio, Vibration

Isolation, Critical Speed of Shafts.

Geometry of tooth profiles, Law of

gearing, Interference, Helical, Spiral and

Worm Gears, Gear Trains- Simple,

compound and Epicyclic. Slider crank

mechanisms, Balancing.

Design of Machine Elements

MD-1 MD-2

Design for static and dynamic loading; failure theories;

fatigue strength and the S-N diagram; principles of the design of

machine elements such as riveted, welded and bolted joints.

Shafts, Spur gears, rolling and sliding contact bearings,

Brakes and clutches, flywheels.

Manufacturing,Industrial andMaintenance Engineering

PROD-1

PROD-2

IE-1

IE-2

RE-1

RE-2

Metal casting-Metal forming, Metal Joining,

computer Integrated manufacturing, FMS.

Machining and machine tool

operations, Limits, fits and tolerances,

Metrology and inspection.

Production planning and Control,

Inventory control

Operations research - CPM-PERT

Failure concepts and characteristics-

Reliability, Failure analysis, Machine

Vibration, Data acquisition, Fault

Detection, Vibration Monitoring.

Field Balancing of Rotors, Noise

Monitoring, Wear and Debris Analysis,

Signature Analysis, NDT Techniques in

Condition Monitoring.

MR-1 MR-2

Mechatronics and

Robotics

Microprocessors and Micro controllers: Architecture, programming,

I/O,Computer interfacing, Programmable logic controller. Sensors

and actuators, Piezoelectric accelerometer, Hall effect sensor,

Optical Encoder, Resolver, Inductosyn, Pneumatic and Hydraulic

actuators, stepper motor, Control Systems- Mathematical modeling

of Physicalsystems, control signals, controllability and observability

Robotics, Robot Classification, Robot Specification, notation;

Direct and Inverse Kinematics; Homogeneous Coordinates and

Arm Equation of four Axis SCARA Robot.

Subject Code Details

Documents

Detailed Solution - IES Master · Detailed Solution 03-02-2018 | MORNING SESSION ME ... A four bar mechanism is made up of links of length 100, ... For an Oldham coupling used between