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SolutionDetailed 14-02-2021 | AFTERNOON SESSION

ME

GENERAL APTITUDE1. A box contains 15 blue balls and 45 black balls.

If 2 balls are selected randomly, withoutreplacement, the probability of an outcome inwhich the first selected is a blue ball and thesecond selected is a black ball, is_________

(a) 34 (b) 3

16

(c) 45236 (d) 1

4Ans. (c)

Sol.Total number of balls in the bag

= 15 + 45 = 60Here, 15 balls are blue& 45 balls are blackIf, the two balls are drawn without replace-ment, the required probability of getting firstball blue & second ball back will be

P = 15 45×60 59 = 45

23645P =

236

2. The world is going through the worst pandemicin the past hundred years. The air travelindustry is facing a crisis, as the resultingquarantine requirement for travelers led to weakdemand.In relation to the first sentence above, whatdoes the second sentence do?(a) Restates an idea from the first sentence.(b) Second sentence entirely contradicts the first

sentence.(c) The two statements are unrelated.(d) States an effect of the first sentence.

Ans. (d)

3. The front door of Mr. X’s house faces East. Mr.X leaves the house, walking 50 m straight fromthe back door that is situated directly oppositeto the front door. He then turns to his right,walks for another 50m and stops. The directionof the point Mr. X is located at with respect tothe starting point is ______

(a) North-East (b) South-East(c) West (d) North-West

Ans. (d)

Sol.

50m

End point

50m

back door House

Starting point

Front doorLeft

Right

W E

NN-W

S

W

So, the direction from the starting point will beNorth-West.

4.

The ratio of the area of the inscribed circle tothe area of the circumscribed circle of anequilateral triangle is ______(a) 1/4 (b) 1/6(c) 1/8 (d) 1/2

Ans. (a)

Sol.

aa

B

A

60°

30°a C

ROuter circle ofradius (R)

Equilateraltriangle (side, a)

Inner circle of radius (r)

60°

Or

D

We have to find,

Area of inscribed circle inner circleArea of circumference circle (outer circle)

= 2

2rR




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[ from triangle, ODC, sin30° = r r 1=R R 2 ]

22i

2o

A r 1 1= = =A 2 4R

5. If = 2; = 3; + = 5; × = 10 ,

Then the value of 2– , is:

(a) 4 (b) 0(c) 16 (d) 1

Ans. (d)

Sol.

(i) = 2 = 2

(ii) = 3 = 3

(iii) + = 5 + = 52 3

= 6 and 6= = 23

(iv) × = 10 2 × = 10 = 5

So, 2 2– = 5 – 6 = 1

6. Five persons P, Q, R, S and T are to be seatedin a row, all facing the same direction, but notnecessarily in the same order. P and T cannotbe seated at either end of the row. P should notbe seated adjacent to S.R is to be seated at thesecond position from the left end of the row.The number of distinct seating arrangementspossible is:(a) 4 (b) 3(c) 2 (d) 5

Ans. (b)

Sol.Number of friends = 5 (P, Q, R, S, T)According to the given conditions,Let’s draw the line diagram as follows:

Q PR T S

Fixed

1Possibility

st

Direction

S PR T Q2Possibility

nd

S TR P Q3Possibility

rd

So, there are three possible combination forseating arrangement according to the givencondition.

7. Given below are two statements 1 and 2, andtwo conclusions I and II.Statement 1 : All entrepreneurs are wealthy.Statement 2: All wealthy are risk seekers.Conclusion I: All risk seekers are wealthy.Conclusion II: Only some entrepreneurs are riskseekers.Based on the above statements and conclusions,which one of the following options is CORRECT?(a) Neither conclusion I nor II is correct(b) Only conclusion II is correct(c) Only conclusion I is correct(d) Both conclusions I and II are correct

Ans. (a)

Sol.Statement-I: All entrepreneurs are wealthy.

W

E

Statement-II: All wealthy are risk seekers.

W

R

E

Conclusion-1: All risk seekers are wealthy.That is incorrect because only few risk seekersare wealthy.Conclusion-2: Some entrepreneurs are riskseekers.

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That is also incorrect because all the entrepre-neurs are risk seekers.So, neither 1 nor 2 is correct.

8. A digital watch X beeps every 30 seconds whilewatch Y beeps every 32 seconds. They beepedtogether at 10 AM.The immediate next time that they will beeptogether is _______(a) 10.00 PM (b) 10.08 AM(c) 11.00 AM (d) 10.42 AM

Ans. (b)

Sol.Clock x beeps after = 30 secClock y beeps after = 32 secSo, the LCM of both time will be

= 25 × 3 × 5 = 480 sec.

In 480 sec, there is 48060 = 8 min

So, after 10 AM, both clock will beep simulta-neously at 10:08 AM.

9. Consider a square sheet of side 1 unit. Thesheet is first folded along the main diagonal.This is followed by a fold along its line ofsymmetry. The resulting folded shape is againfolded along its line of symmetry. The area ofeach face of the final folded shape, in squareunits, equal to _______

(a) 18 (b) 1

16

(c) 14 (d) 1

32Ans. (a)

Sol.

diagon

alx2 x

x xx

1 x2

x2

Line of symmetry

Line of symmetry

x2

x2

Area ofsquare, A = x2=1 =1 unit2 2

x2

So, the area of final shape

a =

1 x x× ×2 2 2 =

2x 1 1 1= × A = ×1 =8 8 8 8

square

units.10. Consider the following sentences:

(i) The number of candidates who appear forthe GATE examination is staggering.

(ii) A number of candidates from my class areappearing for the GATE examination.

(iii) The number of candidates who appear forthe GATE examination are staggering.

(iv) A number of candidates from my class isappearing for the GATE examination.

Which of the above sentences are grammaticallyCORRECT?(a) (i) and (ii) (b) (ii) and (iii)(c) (i) and (iii) (d) (ii) and (iv)

Ans. (a)

Sol.(i) The first sentence informs us about the num-

ber which is singular.(ii) ‘A number of’ means several, some. The

second sentence takes about the studentswhich is plural word.

MECHANICAL ENGINEERING1. A high velocity water jet of cross section area =

0.01 m2 and velocity = 35 m/s enters a pipefilled with stagnant water. The diameter of thepipe is 0.32 m. This high velocity water jetentrains additional water from the pipe and thetotal water leaves the pipe with a velocity 6 m/s as shown in the figure.

Inlet jet35 m/s

Entrainedwater

Entrainedwater

Totalwater out

flow atuniformvelocity6 m/s

The flow rate of entrained water is ________litres/s (round off to two decimal places)

Ans. (132.55)

Sol.Let the flow rate of entrained water is Q.Using conservation of mass,

Q + AiVi = A0V0




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Q =2o

o i id × V – A V4

Q = 2× 0.32 × 6 – 0.01 ×354

= 0.13255 m3/sQ = 132.55 ltr/s

2. A plane frame PQR (fixed at P and free at R)is shown in the figure. Both members (PQ andQR) have length L, and flexural rigidity, EI.Neglecting the effect of axial stress andtransverse shear, the horizontal deflection atfree end, R, is

P

L

L

F

Q

R

(a)34FL

3EI(b)

35FL3EI

(c)32FL

3EI(d)

3FL3EI

Ans. (a)

Sol.

P

L

L EI = Flexural rigidity

F

Q

R

We have to find the deflection at point R. Wecan solve it by strain-energy method:For member QR,

U1 = 2L0

F × xdx

2EI

=2 3F L

6EIand for member PQ:

LP Q

F×LF (We have to neglect

axial strain energy)

U2 = 2L0

F × Ldx

2EI

=2 3F L

2EISo, total strain energy, U = U1 + U2

=2 3 2 3F L F L+

6EI 2EIDeflection at point R,

= UF

=3 31 FL 2 FL+

3 EI 2 EI34 FL=

3 EI

3. The allowance provided in between a holeand a shaft is calculated from the differencebetween(a) upper limit of the shaft and the lower limit

of the hole(b) lower limit of the shaft and the lower limit

of the hole(c) lower limit of the shaft and the upper limit

of the hole(d) upper limit of the shaft and the upper limit

of the holeAns. (a)

Sol.Allowance: It is the difference between thebasic dimensions of the mating part. It can bepositive or negative.

Hole

Shaft

Upper limit

Upper limit

Lower limit

Lower limit

Allowance

In this case, allowance is the difference betweenthe lower limit of hole and the upper limitof shaft.

4. Consider the following differential equation

dy1 + y = ydx

The solution of the equation that satisfies thecondition y(1) = 1 is(a) yey = ex (b) (1 + y)ey = 2ex

(c) 2yey = ex + e (d) y2ey = ex

Ans. (a)

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Sol.

Given: dy1 + ydx = y

1 + y dy

y = dx

1 dy +1dyy = dx

Integrating both sides,

1 dy + 1dyy = dx + C

Here, C is constant ny + y = x + C at x = 1, y = 1 n1+1 = 1 + C C = 0So, ny + y = x lny = (x – y)

y = e(x–y) = x

yee

y xy e = e

5. A two dimensional flow has velocities in x andy directions given by u = 2xyt and v = –y2t,where t denotes time. The equation forstreamline passing through x = 1, y = 1 is

(a) 2x =1y (b) xy2 = 1

(c) x2y = 1 (d) x2y2 = 1Ans. (b)

Sol.u = 2xyt = –y2t

Equation of stream line,dxu = dy

dx

2xyt = 2dy

–y tIntegrating both sides

dx2x =

dy– + Cy

1 2nx = – ny +C

1 2n x y = C

at x = 1 & y = 1, C = 0

2xy = 1

6. A cast product of a particular materials hasdimensions 75 mm × 125 mm × 20 mm. Thetotal solidification time for the cast product isfound to be 2.0 minutes as calculated usingChvorinov’s rule having the index, n = 2. Ifunder the identical casting conditions, the castproduct shape is changed to a cylinder havingdiameter = 50 mm and height = 50 mm, thetotal solidification time will be _______ minutes(round off to two decimal places).

Ans. (2.83)

Sol.Cuboid casting:

125 mm 20mm

75mm

VolumeSurface area = V

A

= 125 ×75 × 20

2 125 ×75 + 75 × 20 + 20 ×125

= 7 mmFor cylindrical casting: D = H = 50 mm

H

D

VA =

2

2

D ×H4

D2 + DH4

=

3

22

D4

D2 + D4

= D6 = 50

6 mm [D = H]

From chovorinov’s rule, t = 2Vk

A

For same material, value of k will be same.

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So, c

cy

tt =

2 2

C cy

V A×A V

cy

2t = 72 ×

2650

Solidification time for cylindrical casting,

cyt = 2.83min

7. Find the positive real root of x3 – x – 3 = 0using Newton Raphson method. If the startingguess (x0) is 2, the numerical value of the rootafter two iterations (x2) is _______ (round off totwo decimal places).

Ans. (1.67)

Sol.f(x) = x3 – x – 3

f x = 3x2 – 1

Initial setting value, x0 = 2Using Newton Raphson method,1st iteration,

x1 =

oo

o

f xx –f x

=3

2

2 – 2 – 32 –

3 ×2 –1

= 32 –11

x1 = 1.727Second iteration,

x2 =

3

2

1.727 –1.727 – 31.727 –

3 × 1.727 –1

This is the required value after second itera-

tion, 2x =1.67

8. The mean and variance, respectively, of abinomial distribution for n independent trialswith the probability of success as p, are(a) np, np(1 – p) (b) np, np

(c) np, np 1 – 2p (d) np, np 1 – p

Ans. (a)

Sol.For binomial distribution

mean = np , where p is the possibility of occur-rence.

& variance = npq = np 1 – p i.e p + q = 1

9. The wheels and axle system lying on a roughsurface is shown in the figure.

g

0.2m

10N0.8m

Axle

Wheel

Each wheel has diameter 0.8 m and mass 1kg. Assume that the mass of the wheel isconcentrated at rim and neglect the mass of thespokes. The diameter of axle is 0.2 m and itsmass is 1.5 kg. Neglect the moment of inertiaof the axle and assume g = 9.8 m/s2. An effortof 10 N is applied on the axle in the horizontaldirection shown at mid span of the axle. Assumethat the wheels move on a horizontal surfacewithout slip. The acceleration of the wheel axlesystem in horizontal direction is ________ m/s2

(round off to one decimal place).Ans. (1.4)

Sol.

0.8m 0.2m 0.1m

0.3m

Centre of Mass

mwheel m =1.5 kgaxle

=1 kg

Rough Surfacef

10N

Due to rough surface, friction force (f) will acton both the wheels.Net force along horizontal direction of motion ofthe whole system.

10 – 2f = 2 ×1+1.5 ×a a = r ×where r = 0.4 m and = angular acceleration

10 – 2f = 3.5 ×0.4 ×

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ME

f = 5 – 0.7 ..(i)Net torque about centre of mass,

–10 × 0.1 + 2 × f × 0.4 = cmI .

Here, Icm = 22 ×1× 0.4Icm = 0.32 kg-m2

[mass moment of inertia of axle is neglected] f × 0.8 = 0.32 +1

f = 0.4 +1.25 ....(ii)Comparing equation (i) and (ii), we get

5 – 0.7 = 1.25 + 0.4

= 3.751.10

= 3.409 rad/s2

So, the acceleration of the wheel axle system inhorizontal direction,

acm = r.= 0.4 ×3.409 = 1.36 m/s2

acm 1.4 m/s2

10. A PERT network has 9 activities on its criticalpath. The standard deviation of each activity onthe critical path is 3. The standard deviation ofthe critical path is(a) 3 (b) 81(c) 27 (d) 9

Ans. (d)

Sol.Standard deviation of each critical activities, = 3and total number of activity in critical path= 9So, the standard deviation of critical path,

c =2 2 2 2 2

2 2 2 23 + 3 + 3 + 3 + 3+3 + 3 + 3 + 3

= 29 ×3 = 32

c = 9

11. Consider adiabatic flow of air through a duct.At a given point in the duct, velocity of air is300 m/s, temperature is 330 K and pressure is

180 kPa. Assume that the air behaves as aperfect gas with constant cp = 1.005 kJ/kg.K.The stagnation temperature at this point is_____ K (round off to two decimal places).

Ans. (374.78)Sol.

V = 0T = ?

2

021 V = 330m/sT = 330K

1

1

Using SFEE:

21

1Vh +2

= h0

where, ho = stagnation enthalpy

(h0 – h1) =21V2

cp(T0 – T1) =21V2 ph = c T

T0 =21

1p

VT +2c

Putting the perspective values,Stagnation temperature,

T0 = 2300330 +2 ×1005

0T = 374.78 K

12. A surface grinding operation has been performedon a cast iron plate having dimensions 300 mm(length) × 10mm (width) × 50mm (height). Thegrinding was performed using an alumina wheelhaving a wheel diameter of 150 mm and wheelwidth of 12 mm. The grinding velocity used is40 m/s, table speed is 5 m/min, depth of cut perpass is 50 m and the number of grindingpasses is 20. The average tangential and averagenormal forces for each pass are found to be 40N and 60 N respectively. The value of the specificgrinding energy under the aforesaid grindingconditions is _______ J/mm3 (round off to onedecimal place).

Ans. (38.4)

Sol.




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Specific grinding energy, u = c

w

f × vv × w ×d

where, fc = cutting forcev = wheel speedvw = work speedd = depth of cutw = width of cut

u =

–3 3

40 ×40 Nm5 ×10 ×50 ×10 ×1000 mm

60

= 38.4 J/mm3

13. In forced convective heat transfer, Stantonnumber (St), Nusselt number (Nu), Reynoldsnumber (Re) and Prandtl number (Pr) arerelated as

(a) NuPrSt =Re (b) NuReSt =

Pr

(c) NuSt =RePr (d) St = Nu Pr Re

Ans. (c)

Sol.Stanton number,

St = Nusselt Number

Reynolds Number × Prandt Number

NuSt =Re × Pr

14. A spot welding operation performed on two piecesof steel yielded a nugget with a diameter of 5mm and a thickness of 1 mm. The weldingtime was 0.1 s. The melting energy for the steelis 20 J/mm3. Assuming the heat conversionefficiency as 10% the power required forperforming the spot welding operation is _______kW (round off to two decimal places).

Ans. (39.27)

Sol.Heat required to melt the material

Q = 220 × d ×h4

= 220 × 5 ×14

The heat transfer efficiency, = 0.10 it means10% of the input energy is converted (used) tomelt the material.So, Q = 0.10 × (i2 Rt)Here, i2 R = Power input

220 × ×5 ×14 = 20.10 × i × R × 0.1

i2 R = 39269.9 J/s

Power input, inW = 39.27kW

15. In a pure orthogonal turning by a zero rakeangle single point carbide cutting tool, the shearforce has been computed to be 400 N. If thecutting velocity, Vc = 100 m/min, depth of cut,t = 2.0 mm, feed S0 = 0.1 mm/revolution andchip velocity, Vf = 20 m/min then the shearstrength, s of the material will be ________MPa (round off to two decimal places)

Ans. (392.23)

Sol.Rake angle, = 0°Shear force, Fs = 400 N

Cutting velocity, Vc = 100 m/minChip velocity, Vf = 20 m/min

r = c

tt = f

c

vv = 20

100 = 0.2

Shear angle, = –1 r costan1 – rsin

= –1tan r = 0

= –1tan 0.2

= 11.3099°

Shear stress, s = sfbt

sin

where, bt

sin = shear area

For pure orthogonal turning, bt = f.d

So, s = sf ×sinf d

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= 400 × sin 11.3092 ×0.1

s = 392.23 MPa

16. An adiabatic vortex tube, shown in the figuregiven below is supplied with 5 kg/s of air (inlet1) at 500 kPa and 300 K. Two separate streamsof air are leaving the device from outlets 2and 3. Hot air leaves the device at a rateof 3 kg/s from outlet 2 at 100 kPa and340 K, while 2 kg/s of cold air stream is leavingthe device from outlet 3 at 100 kPa and240 K.

2

1

3

Low pressure cold air outlet

Low pressure hotair outlet

High pressure air inlet

Consider constant specific heat of air is 1005 J/kg.K and gas constant is 287 J/kg.K. There isno work transfer across the boundary of thisdevice. The rate of entropy generation is________ kW/K (round off to one decimal places)

Ans. (2.24)

Sol.

2

1

3

5 kg/s

3 kg/s 2 kg/s

P = 500 kPaT = 300 K

1

1

P = 100 kPaT = 340 K

2

2

P = 100 kPaT = 240 K

3

3

genS = (m2s2 + m3s3) – m1s1

= 3(s2 – s1) + 2(s3 – s1)

= 2 2p

1 1

3 3p

1 1

T P3 c n – R nT P

T P+2 c n – R nT P

Putting the respective values,

genS = 340 1003 1.005 n – 0.287 n300 500

240 100+2 1.005 n – 0.287 n300 500

genS = 2.24 kW K

17. A factory produces m (i = 1,2,....m) products,each of which requires processing on n(j = 1,2,... n) workstations. Let aij be the amount ofprocessing time that one unit of the ith productrequires on the jth workstation. Let the revenuefrom selling one unit of the ith product. Theplanning horizon consists of T (t = 1, 2, ....T)time periods. The minimum demand that mustbe satisfied in time period t is dit, and thecapacity of the jth workstation in time period tis cjt. Consider the aggregate planningformulation below, with decision variables Sit(amount of product i sold in time period t), Xit(amount of product i manufactured in time periodt) and lit (amount of product i held in inventoryat the end of time period t).

T m

i it i itt=1 i=1

max r s – h l

Subject to

it itS d i,t

< capacity constraint ><inventory balance constraint>

xit, Sit, Iit 0; Ii0 = 0

The capacity constraints and inventory balanceconstraints for this formulation respectively are

(a)m

ij it it it i,t–1 it iti

a X d i,t and I I + S – X i,t

(b)m

ij it jt it i,t–1 it iti

a X c i,t and I I + X – d i,t

(c)m

ij it jt it i,t–1 it iti

a X c j,t and I = I + X – S i,t

(d)m

ij it it it i,t–1 it iti

a X d i,tand I = I + X – S i,t

Ans. (c)

18. Value of 5.2

4lnxdx using Simpson’s one-third

rule with interval size 0.3 is




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(a) 1.06 (b) 1.60(c) 1.83 (d) 1.51

Ans. (c)

Sol.5.2

4nxdx

Step size given, h = 0.3

4 4.3 4.6 4.9 5.2n4 n4.3 n4.6 n4.9 n5.2 y0 y1 y2 y3 y4

Using Simpson’s 1/3rd rule:

I = 0 4 1 3 2h y + y + 4 y + y + 2y3

=

n4 + n5.2 + 40.33 n4.3 + n4.9 + 2 n4.6

I =1.83

19. Ambient air flows over a heated slab havingflat, top surface at y = 0. The local temperature(in Kelvin) profile within the thermal boundarylayer is given by T(y) = 300 + 200 exp(–5y),where y is the distance measured from the slabsurface in meters. If the thermal conductivityof air is 1.0 W/m.K and that of the slab is 100W/m.K, then the magnitude of temperature

gradient dT dy within the slab at y = 0 is_____ K/m (round off the nearest integer).

Ans. (10)

Sol.

airy

x

T(y) = 300 + 200 e–5y

K = 1 W/mKair

Heated slabK = 100 W/mKslab

At just boundary of air and solid medium,

airy=0

dT–K ×dy = slab

y=0

dT–K ×dy

1 × (–1000) =y=0

dT100 ×dy

y=0

dTdy = –10 = 10 K/m

20. Consider the system shown in the figure. Arope goes over a pulley. A mass m, is hangingfrom the rope. A spring of stiffness, k, is attachedat one end of the rope. Assume rope isinextensible, massless and there is no slipbetween pulley and rope.

rk

J

m

+

The pulley radius is r and its mass moment ofinertia is J. Assume that the mass is vibratingharmonically about its static equilibriumposition. The natural frequency of the system is

(a)2

2kr

J – mr(b)

2

2kr

J + mr

(c)2kr

J (d) km

Ans. (b)

Sol.

r

k

x = r

J

m

Here, J = Moment of inertia of pulley

Let the system is given an angular displace-ment .Using energy method:

d KE + PEdt = 0

2 2 2d 1 1 1J + mx + kx

dt 2 2 2

= 0

2 22d 1 1 1J + m r + k rdt 2 2 2

= 0

2 2 2 2 21 d J + mr + kr2 dt

= 0

2 2J + mr × 2 + kr × 2 = 0

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2

2kr+

J + mr

= 0

So, natural frequency of the system,

2

n 2kr=

J + mr

21. An object is moving with a Mach number of 0.6in an ideal gas environment, which is at atemperature of 350 K. The gas constant is 320J/kg.K and ratio of specific heats is 1.3. Thespeed of object is __________ m/s (round off tothe nearest integer)

Ans. (228.9)

Sol.Given data:

Mach No., Ma = 0.6Temperature, T = 350 KGas constant, R = 320 J/kg-K

Specific heat ratio, = 1.3

Ma = VC =

Speed of objectSpeed of sound in that medium

= VRT

0.6 =V

1.3 ×320 ×350

V = 228.95 m s

22. If the Laplace transform of a function f(t) is

given by s + 3

s +1 s + 2 then f(0) is

(a) 1 (b) 3/2(c) 1/2 (d) 0

Ans. (a)

Sol.

L [f(t)] = s + 3

s +1 s + 2Can be written as,

s + 3

s +1 s + 2 = A B+

(s +1) s + 2

(s + 3) = A(s + 2) + B(s + 1)

(After comparing respectively coefficient of bothsides) A + B = 1& 2A + B = 3We get, A = 2

B = –1

So, s + 3

s +1 s + 2 = 2 1–

s +1 s + 2

f(t) =

–1 –12 1L – Ls +1 s + 2

f(t) = 2e–t – e–2t

–at 1e =

s + a

So, f(0) = 2e–0 – e–0

= 2 – 1

f 0 = 1

23. Daily production capacity of a bearingmanufacturing company is 30000 bearings. Thedaily demand of the bearing is 15000. Theholding cost per year of keeping a bearing inthe inventory is Rs. 20. The setup cost for theproduction of a batch is Rs.1800. Assuming 300working days in a year, the economic batchquantity in number of bearings is ______ (ininteger)

Ans. (40249.22)

Sol.Given: P = 30000 unit/day

d = 15000 unit/dayC0 = 1800 Rs./setupCH = 20 Rs/unit/yr

No. of days = 300/yearSo, economic batch quantity,

Q = o

h

2DC P×C P – d

s

=2 ×15000 ×300 ×1800 30000×

20 30000 –15000

Q = 40249.22 unit24. A shell and tube heat exchanger is used as a

steam condenser. Coolant water enters the tubeat 300 K at a rate of 100 kg/s. The overall heattransfer coefficient is 1500 W/m2.K and total

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heat transfer area is 400 m2. Steam condensesat a saturation temperature of 350 K. Assumethat the specific heat of coolant water is 4000 J/kg.K. The temperature of the coolant watercoming out of the condenser is __________K(round off the the nearest integer).

Ans. (338.84)

Sol.

Condensing fluid (steam)350K 350K

T = ?co

T = 300Kci

T = T = 350Khi ho

NTU = p min

UAmc

= 1500 × 400100 × 4000 = 1.5

For condenser or evaporator (where phase changeoccurs)Effectiveness, = 1 – e–NTU

& = co ci

hi ci

T – TT – T

So, co ci

hi ci

T – TT – T = 1 – e–NTU

(Putting the respective values)

coT – 300350 – 300 = 1 – e–1.5

So, exit temp of cooling water, Tco = 338.84 K25. A power transmission mechanism consists of a

belt drive and a gear train as shown in thefigure.

250mm

218T

35 6

36T

150 mm

N = 2500 rpm

44T15T

33T16T

+ +

+ + +4 7

Diameters of pulleys of belt drive and number

of teeth (T) on the gears 2 to 7 are indicated inthe figure. The speed and direction of rotationof gear 7, respectively are(a) 255.68 rpm; anticlockwise(b) 575.28 rpm; anticlockwise(c) 255.68 rpm; clockwise(d) 575.28 rpm; clockwise

Ans. (c)

Sol.Let, the CW direction of motion is taken aspositive and counter clockwise as negative.So, speed of pulley 2

2

1

NN = 1

2

DD

2N2500 = 150

250 N2 = 1500 rpm (ccw)Speed of gear 4

3 4

2

N = NN = 2

3

TT

N4 = 18 ×150044

N4 = 613.64 rpm (cw)Speed of gear 5

5

4

NN = 4

5

TT

N5 = 15 × 613.6433

= 278.93 rpm (ccw)Speed of gear 7:

6 7

5

N = NN = 5

6

TT

N7 = 33 × 278.9336

7N = 255.68 rpm cw

26. The cast iron which possess all the carbon inthe combined form as cementite is known as(a) White cast iron(b) Malleable cast iron

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(c) Grey cast iron(d) Spheroidal cast iron

Ans. (a)

Sol.White cast iron is extremely hard but alsovery brittle due to presence of large amount ofcementite phase.

27. A rigid tank of volume 50 m3 contains a puresubstance as a saturated liquid vapour mixtureat 400 kPa. Of the total mass of the mixture,20% mass is liquid and 80% mass is vapour.Properties at 400 kPa are: Saturationtemperature, Tsat = 143.61°C; Specific volumeof saturated liquid, vf = 0.001084 m3/kg; Specificvolume of saturated vapour, vg = 0.46242m3/kg. The total mass of liquid vapour mixturein the tank is__________kg (round off to thenearest integer).

Ans. (135)

Sol.Given data: mass of liquid,

mf = 20% of m = 0.2 mmass of vapor, mg = 80% of m = 0.8 mwhere, m = mf + mg

Vtot = 50 m3

f = 0.001084 m3/kg

g = 0.46242 m3/kg

Vtot = Vf + Vgv=m

50 = f fm + g gm

50 = (0.2m) × 0.001084+ (0.8m) × 0.46242

Total mass, m = 135.07 kg28. Ambient pressure, temperature, and relative

humidity at a location are 101 kPa, 300 K, and60%, respectively. The saturation pressure ofwater at 300 K is 3.6 kPa. The specific humidityof ambient air is__________g/kg of dry air.(a) 35.1 (b) 13.6(c) 21.9 (d) 21.4

Ans. (b)

Sol.Relative humidity, = 0.60

Saturation pressure, Ps = 3.6 kPa

= v

s

PP

Partial pressure, Pv of water vapor = 0.60 ×3.6 = 2.16 kPa

Humidity ratio, = v

v

0.622 × PP – P

= 0.622 × 2.16101 – 2.16

= 0.01359 kg of water va-por/kg of dry air = 13.59 gm of water vapor/kg of dry air.

29. A block of negligible mass rests on a surfacethat is inclined at 30° to the horizontal plane asshown in the figure. When a vertical force of900 N and a horizontal force of 750 N areapplied, the block is just about to side.

900N

750N

30°

The coefficient of static friction between the blockand surface is ______ (round off to two decimalplaces).

Ans. (0.17)

Sol.

30°

750 kN

Mass of the blockin negligible

900N

900N

R

750N

750 cos30°

900 sin30°900 cos30°

750 sin30° fs

30°30°

Fig. FBD of the block

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Let, s = Static coefficient of friction betweenblock & inclined plane.750 cos30° force is greater than the 900 sin30°force, so the block will about to move upwardon the inclined plane.So, friction force will act downward. On theverge of motion (impending motion),

750 cos30° = 900 sin30° + fs

fs = sR

= s 900cos30 +750sin30

Solving the equation, we get

3750 ×

2= s

1900 × +2

3 1900 × + 750 ×

2 2

s = 0.17

30 In a CNC machine tool, the function of aninterpolator is to generate(a) NC code from the part drawing during post

processing(b) reference signal prescribing the shape of the

part to be machined(c) signal for the lubrication pump during

machining(d) error signal for tool radius compensation

during machiningAns. (b)

Sol–30: (b)In contouring systems, the machining path isusually constructed from a combination of lin-ear and circular segments. It is only necessaryto specify the coordinates of the initial & finalpoints of each segment, and the feed rate. Theoperation of producing the required shape basedon this information is termed interpolation &the corresponding unit is the “Interpolator”.So, in CNC machine tool, the function of aninterpolator is to generate reference signal pre-scribing the shape of the part to be machined.

31. For a two-dimensional, incompressible flowhaving velocity components u and v in the xand y directions, respectively, the expression

2(u ) (uv)+x y

can be simplified to

(a)u u2u + vx y (b)

u vu + ux y

(c)u uu + vx y (d)

u v2u + ux y

Ans. (c)

Sol.For incompressible flow,

du dv+dx dy = 0

dv du= –dy dx

We can write, 2u + uvx y

=du dv du2 + u + vdx dy dy

=du du du2 – u + vdx dx dy

=du duu + vdx dy

32. Let the superscript T represent thetranspose operation. Consider the function

f(x) = T T1 x Qx – r x2 , where x and r are n × 1

vectors and Q is a symmetric n × n matrix.The stationary point of f(x) is

(a) Tr

r r (b) Q–1r

(c) r (d) QTrAns. (b)

Sol.

f(x) = T T1 x Qx – r x2

Let, Q(2×2) =1 33 2

x(2×1) =1

2

xx

r(2×1) =1

2

rr

f(x) = 11 2

2

x1 31 x xx2 3 2

11 2

2

x– r rx

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= 11 2 1 2

2

x1 x + 3x 3x + 2xx2

1 1 2 2– r x + r x

= 2 21 1 2 1 2 2

1 x +3x x + 3x x + 2x2

1 1 2 2– r x + r x

= 2 21 1 1 1 2 2 2 2

1 x – r x + 6x x + 2x – r x2

For stationary point of f(x)

1

df xdx = 1 1 2

1 2x – r + 6x2 = 0

2x1 + 6x2 = r1 ....(i)

&

2

df xdx = 1 2 2

1 6x + 4x – r2 = 0

6x1 + 4x2 = r2 ...(ii)We can unit these into matrix from

1 2

1 2

x 3x2 + 23x 2x

= 1

2

rr

1

2

x 1 3x 3 2

=

1

2

rr

1

2

x ×Qx

= r

1

2

xx

= Q–1.r

33. The figure shows the relationship between fatiguestrength (S) and fatigue life (N) of a material.The fatigue strength of the material for a life of1000 cycles is 450 MPa, while its fatigue strengthfor a life of 106 cycles is 150 MPa.

7 log N10

log S10

The life of a cylindrical shaft made of thismaterial subjected to an alternating stress of200 MPa will then be ___________ cycles (roundoff to the nearest integer).

Ans. (163840.58)

Sol.

(log 10 , log 450)103

10

(log x, log 200)10 10(log 10 , log 150)10

610

log S10

log N10

Equation of line is given by,

1

2 1

y – yy – y = 1

2 1

x – xx – x

10 10

10 10

log 200 – log 450log 150 – log 450

= 3

10 106 3

10 10

log x – log 10log 10 – log 10

log10 x = 5.2144

x = 163840.58 cylces

34. Consider the mechanism shown in the figure.There is rolling contact without slip betweenthe disc and ground.

1

3R

S 4T

Q2

P

U

+

Select the correct statement about instantaneouscenters in the mechanism.(a) Only points P, Q, R, S, and U are

instantaneous centers of mechanism(b) Only points P, Q and S are instantaneous

centers of mechanism(c) Only points P, Q, S and T are instantaneous

centers of mechanism(d) All points P, Q, R, S, T and U are

instantaneous centers of mechanismAns. (d)




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Sol.

Q2 3

P 1 R

U

S 4T

There are 4 links, so the total number of in-stantaneous centers will be

n n –12

= 4 ×32 = 6

I-centre:2

3

1

4I12 = P

I34 = S

I23 = Q

I14 = T

I12, I23 = I13 = U& I14, I12 or I23, I34, I24 = RSo, All the points P, Q, R, S, T and U areinstantaneous centres of mechanism.

35. A steel cubic block of side 200 mm is subjectedto hydrostatic pressure of 250 N/mm2. Theelastic modulus is 2 × 105 N/mm2 and Poisson’sratio is 0.3 for steel. The side of the block isreduced by ________mm (round off to twodecimal places).

Ans. (0.10)

Sol.

a

aa

P

P

PHere, side of the cubic block, a = 200 mmHydrostatic pressure, P = 250 N/mm2

Elastic modulus, E = 2 × 105 N/mm2

Poisson’s ratio = 0.3Reduce in volume of the block under hydro-static pressure,

V = 3PV 1 – 2E

= 35

3 × 250200 × × 1 – 2 × 0.32 ×10

V = 12000 mm3

Final volume, Vf = Vi – V = (200)3 – 12000= 7988000 mm3

Final side of the block,

af = 3 7988000 = 199.899 mm

So, the side of the block is reduced ky,a = a – af

= 200 – 199.899

a = 0.10mm

36. The size distribution of the powder particles usedin Powder Metallurgy process can be determinedby(a) Laser reflection (b) Laser absorption(c) Laser penetration (d) Laser scattering

Ans. (d)

Sol.In powder metallurgy, there are several meth-ods available for particle size analysis:(i) Sedimentation(ii) Microscopic analysis(iii) Screening(iv) Light scattering from a laser that illu-minates a sample consisting of particles sus-pended in a liquid medium. The particles causethe light to be scattered, and a detector thendigitizes the signals & computes the particlesize distribution.

37. A vertical shaft Francis turbine rotates at 300rpm. The available head at the inlet to theturbine is 200 m. The tip speed of the rotor is40 m/s. Water leaves the runner of the turbinewithout whirl. Velocity at the exit of the drafttube is 3.5 m/s. The head losses in differentcomponents of the turbine are: (i) stator andguide vanes: 5.0 m, (ii) rotor: 10 m, and (iii)draft tube: 2m. Flow rate through the turbineis 20 m3/s. Take g = 9.8 m/s2. The hydraulic

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efficiency of the turbine is __________% (roundoff to one decimal place).

Ans. (91.2)

Sol.

H = Total head at inlet – Hydraulic lossHead inlet

=

22V200 – 5 +10 + 2 +

2g200

=

23.517 +

2 ×9.811 –

200

= 91.18%

38. A plane truss PQRS (PQ = RS, and PQR =90°) is shown in the figure.

S

L

P

QL

R F

The forces in the members PR and RS,respectively, are_________

(a) F 2 (tensile) and F (tensile)

(b) F(tensile) and F 2 (tensile)

(c) F 2 (tensile) and F (compressive)

(d) F(compressive) and F 2 (compressive)

Ans. (c)

Sol.

Q L F

L

90° 45°

45°

P

L

S

R

We have to find the force in member PR andRS and their nature:Free body diagram:

Q F45°

45°P S

R

FP FS

HP

Here, FP + FS = 0 and HP = FAt joint P:

HP = FPR sin45°

F = PR1F ×2

PRF = F × 2

Nature of force is Tensile i.e force is awayfrom the joint.Now, At joint, R:

FPR sin45° = FRS

FRS = 1F × 2 ×2

= F

RSF = F

Nature of force is Compressive i.e force istowards the joint.

39. Which of the following is responsible for eddyviscosity (or turbulent viscosity) in a turbulentboundary layer on a flat plate?(a) Prandtl stresses(b) Boussinesq stresses(c) Reynolds stresses(d) Nikuradse stresses

Ans. (c)Since, turbulence is considered as eddyingmotion and the additional stresses are added tothe viscous stresses due to mean motion in orderto explain the complete stress field, it is oftensaid that the apparent stresses are caused byeddy viscosity. The total stresses given by,




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xx =

2u–P + 2 – ux

xy =

u v+ – u vy x

Terms such as – u v or – u are calledreynolds stresses or turbulent stresses.

40. Consider an ideal vapour compressionrefrigeration cycle working on R-134arefrigerant. The COP of the cycle is 10 and therefrigeration capacity is 150 kJ/kg. The heatrejected by the refrigerant in the condenser is_________kJ/kg (round off to the nearestinteger).

Ans. (165)

Sol.

14

3

2

T

SFig. T-s diagram of VCRS cycle

h1 – h4 = Refrigeration capacityh2 – h1 = Work input to compressorh2 – h3 = Heat transferred from condenser.

COP = 1 4

2 1

h – hh – h =

in

REW

Given: COP = 10 & RE = 150 kJ//kg

Win = 15010 = 15 kJ/kg

So, h2 – h3 = (h2 – h1) + (h1 – h3) h3 = h4 (Isenthalpic process)

= 15 + 150

2 3h – h = 165 kJ kg

41. The value of /2 cos

0 0r sin dr d

is

(a)16 (b)

(c) 0 (d)43

Ans. (a)

Sol.

I =

2 cos

o orsin drd

=cos22

00

r sin d2

=

2 20

1 cos sin d2

2 2sin + cos = 1

= 2 3

01 sin – sin d2

=

2

01 3 sin3sin – sin + d2 4 4

3sin3 = 3sin – 4sin

= 2

01 sin + sin3 d8

=

2

0

1 cos3– cos –8 3

=

3cos1 cos02– cos – + cos0 +8 2 3 3

=

1 11 +8 3 = 1 4×

8 3

1I =6

42. The machining process that involves ablationis(a) Electrochemical Machining(b) Abrasive Jet Machining(c) Chemical Machining(d) Laser Beam Machining

Ans. (d)

Sol.In LBM (Laser beam machining), the emittedlaser beam is focussed by a lens system and thefocussed beam falls the work surface, removinga small portion of the material by vaporization& high speed ablation.

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43. A cantilever beam with a uniform flexuralrigidity (EI = 200 × 106 N.m2) is loaded with aconcentrated force at its free end. The area ofthe bending moment diagram corresponding tothe full length of the beam is 10000 N.m2. Themagnitude of the slope of the beam at its freeend is__________ micro radian (round off to thenearest integer).

Ans. (50)

Sol.

Area, A = 10000 = 10 N-mBM4

A B

Pl

B.M

Fig. Bending moment diagram

Slope, B A– = 1EI [Area of bending moment

diagram]

=

46

1 × 10200 ×10

= 50 × 10–6 radian

B = 50 radian A = 0

44. Water flows out from a large tank of cross-sectional area At = 1 m2 through a smallrounded orifice of cross sectional area Ao = 1cm2, located at y = 0. Initially the water level(H), measured from y = 0, is 1 m. Theacceleration due to gravity is 9.8 m/s2.

Cross-sectional area, At

Orifice Cross-sectional

area,Ao

y =H

WATER

y =0

Neglecting any losses, the time taken by waterin the tank to reach a level of y = H/4is____________ seconds (round off to one decimalplace).

Ans. (2258.8)

Sol.

y=H=1m aftertime,t

orific

A = 1cm = 1×10 mi2 –4 2H/4

A = 1m02

The height of the water column is decreas-ing with time.

So, odyAdt = i–A 2gy

H 4

Hdy

y =t i0 o

A– × 2g dtA

(Integrating both side)

t = –4

1×2 × 1 – 0.25

2×9.8 ×10

t = 2258.8 sec

45. The demand and forecast of an item for fivemonths are given in the table.

April 225 200May 220 240June 285 300July 290 270

August 250 230

Month Demand Forecast

The Mean Absolute Percent Error (MAPE) in theforecast is ____% (round off to two decimal places)

Ans. (8.07)

Sol.

t t

t t t

Demand Forecast D – FMonth

D F DApril 225 200 0.111May 220 240 0.0909June 285 300 0.0526July 290 270 0.0689Aug 250 230 0.08

So, mean absolute percentage error, (MAPE)

=n

t t

ti=1

D – F1 ×100n D

=0.111 + 0.0909

1 × +0.0526 ×1005 +0.0689 + 0.08

MAPE = 8.068%

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46. Consider an n × n matrix A and a non-zeron × 1 vector p. Their product Ap = 2p , where

and –1, 0, 1 . Based on the giveninformation, the eigen value of A2 is:(a) 4 (b)

(c) 2 (d)

Ans. (a)

Sol.

Given: AP = 2P

2A – I P = 0

A – I = 0 is called characteristic equationof A.where, is the eigen value of A.

Here, 2=

As we know that, if is the eigen value ofmatrix A, then 2 will be the eigen value of A2.

So, the eigen value of matrix A2 will 2 i.e

22 = 4

47. The von Mises stress at a point in a bodysubjected to forces is proportional to the squareroot of the(a) dilatational strain energy per unit volume(b) plastic strain energy per unit volume(c) distortional strain energy per unit volume(d) total strain energy per unit volume

Ans. (c)

Sol.Under simple tension test ie,

1 = Syt & 2 3= = 0

therefore, distortion strain energy per unit vol-ume,

Ud = 2yt

1 + S3E

(Refer V.B Bhandari’s MDbook).So, von-mises stress,

dSyt U

We can say that, the von-mises stress at a point

in a body subjected to forces is proportional tothe square root of the distortion strain energyper unit volume.

48. A 76.2 mm gauge block is used under one endof a 254 mm sine bar with roll diameter of 25.4mm. The height of gauge blocks required at theother end of the sine bar to measure an angleof 30° is___________mm (round off to twodecimal places).

Ans. (203.2)

Sol.

O=30°

B

A

h = Height of slip gauge

1

76.2

mm

l = 254 mm

h2

OAB, sin = 1 2h – h

( Diameter of both rollers are same)

= 1h – 76.2254

1h – 76.2254 = sin30°

1h = 203.2 mm

49. The torque provided by an engine is given byT( ) = 12000 + 2500sin(2 ) N.m, where isthe angle turned by the crank from inner deadcenter. The mean speed of the engine is 200rpm and it drives a machine that provides aconstant resisting torque. If variation of thespeed from the mean speed is not to exceed

0.5% , the minimum mass moment of inertiaof the flywheel should be_________kg.m2 (roundoff to the nearest integer).

Ans. (569.93)

Sol.

T = 12000 + 2500sin 2

Tmean = o1 12000 + 2500sin2 d

= 12000 N-m

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0

E

/2

Tmean

T

E = 2meano

T – T d

= 2

o2500sin2 d

= 2500 J

E = 2I Cs

Here, mean angular speed

= 2 N60

= 2 × 200 rad s60

and coefficient of speedCS = 0.5%

= 2×0.5 = 0.01100

So, 2500 =22 ×200I × ×0.01

60

Moment of inertia of flywheel,I = 569.93 kg-m2

50. A column with one end fixed and one end freehas a critical buckling load of 100 N. For thesame column, if the free end is replaced with apinned end then the critical buckling load willbe __________ N (round off to the nearestinteger).

Ans. (800)

Sol.

FreeP P

Effective length = 2 × l le

lColumn

Fixed

Critical buckling load, Pcr =

2

2e

EI =

2

2EI

2

Pcr =2

2EI

4

Now, 2nd case:

Hinged

Fixed

P

e =2

New, critical buckling load, crP =

2

2e

EI

=

2

2EI

2

=

2

22 EI

=

2

2EI8 ×

4

crP = 8 × Pcr

New critical buckling load, crP = 8 × 100

crP = 800N

51. Value of (1 + i)8, where i = –1 , is equal to

(a) 4 (b) 16(c) 16i (d) 4i

Ans. (b)

Sol.

(1 + i)8 =

8

2 cos + i sin4 4

=

8 84i2 × e

i[ e = cos + isin ]

=

2 i16 × e

= 16 cos2 + isin2

81 + i = 16

cos2 = 1 &sin2 = 0

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52. A machine of mass 100 kg is subjected to anexternal harmonic force with a frequency of 40rad/s. The designer decides to mount themachine on an isolator to reduce the forcetransmitted to the foundation. The isolator canbe considered as a combination of stiffness (K)and damper (damping factor, ) in parallel.The designer has the following four isolators:(1) K = 640 kN/m, = 0.70(2) K = 640 kN/m, = 0.07(3) K = 22.5 kN/m, = 0.70(4) K = 22.5 kN/m, = 0.07

Arrange the isolators in the ascending order ofthe force transmitted to the foundation.(a) 1-3-2-4 (b) 3-1-2-4(c) 4-3-1-2 (d) 1-3-4-2

Ans. (c)

Sol.m = 100 kgw = 40 rad/s

(i) 1st and 2nd isolators

n = km = 640 ×1000

100

n = 80 rad/s

n 1,2

= 4080 = 0.50

(ii) For 3rd & 4th isolators:

n = km = 22.5 ×1000

100= 15

n 3,4

= 4015 = 2.67

=0.07=0.0

7

=0.7 =0.7=0.071

0

n 1,2

n

1 2n 3,4

Here we can see that, at n 1,2

= 0.5, the value

of transmitted force for = 0.07 is higher than

that of = 0.7 but, for n 3,4

= 2.67, the value

of transmitted force for = 0.07 is lower thanthat of = 0.7 . So, the correct ascending orderof the force transmitted to the foundation will be,

4 < 3 < 1 < 2So, option (c) is correct.Note: We can calculate the value of transmit-ted force (or transmissibility ratio) for all thefour isolator but it is time taking from the exampoint view. So, it would be smart idea, if wewould try to answer it from the graph (if youknow about it).

53. The thickness, width and length of a metal slabare 50 mm, 250 mm and 3600 mm, respectively.A rolling operation on this slab reduces thethickness by 10% and increases the width by3%. The length of the rolled slab is_________mm(round off to one decimal place).

Ans. (3883.5)

Sol.Given:Initial slab condition:

Thickness, t1 = 50 mmWidth, wi = 250 mmLength, li = 3600 mm

Final slab condition:tf = 0.90 × 50 mm

(10% decrease in thickness)wf = 1.03 × 250 mm

(3% increase in width)

Final length, f = ?

In rolling operation, initial volume = final vol-ume

i i i× w × t = f f f× w × t

tf = 3600 × 250 × 500.90 ×50 ×1.03 × 250

Final length of the rolled product,

f = 3883.49 mm

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54. Consider the open feed water heater (FWH)shown in the figure given below:

Turbine

Stea

m G

ener

ator

Condenser

Open FWH65

LP pump

4

2

HP pump

3

1

7

Specific enthalpy of steam at location 2 is 2624kJ/kg, specific enthalpy of water at location 5 is226.7 kJ/kg and specific enthalpy of saturatedwater at location 6 is 708.6 kJ/kg. If the massflow rate of water entering the open feed waterheater (at location 5) is 100 kg/s then the massflow rate of steam at location 2 willbe_________kg/s (round off to one decimal place).

Ans. (25.2)

Sol.

Open FWH

xh2

h6 h5(1–x)

Fig. A part of the given schematic diagramHere, x is the fraction of total mass flow rate

tm at location 2.

So, applying energy balance

2 5xh + 1 – x h = h6

Putting the given values,x × 2624 + (1 – x) × 226.7 = 708.6 x = 0.201 (1 – x) × mt = 100

tm = 125.159 k/s

So, the mass flow rate of steam at location 2will be, tx ×m = 0.201 × 125.159

2m = 25.16 kg s

55. The controlling force curves P, Q and R for aspring controlled governor are shown in thefigure, where r1 and r2 are any two radii ofrotation.

P

Q

R

Cont

rolli

ng fo

rce

Radius of rotati nor2 r1

The characteristics shown by the curves are(a) P-Stable; Q-Isochronous; R-Unstable(b) P-Unstable; Q-Stable; R-Isochronous(c) P-Stable; Q-Unstable; R-Isochronous(d) P-Unstable; Q-Isochronous; R-Stable

Ans. (d)

Sol.

N1N2

F

Speed curve R(stabl

e)Q (is

ochron

ous)

P(unstable

)

r1 r2 1r

2r r

1N2N

For stable governor (Line R): As the speeddecreases, the radius of ball decreases and vice-versa. Here, with increase in radius, the slopeof speed increases.Unstable governor (Line P): As radius in-creases, the slope of speed curve decreases itmeans at smaller radius the speed is higherand vice-versa.

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