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1

General Aptitude Questions

Q.No-1-5 Carry One Mark Each

1. Choose the word most similar in meaning to the given word: Educe

(A) Exert (B) Educate (C) Extract (D) Extend

Answer: (C)

2. If logx (5/7) = -1/3, then the value of x is

(A) 343/125 (B) 12/343 (C) -25/49 (D) -49/25

Answer: (A)

Exp:

3

1 3 1 35 7 7x x x 2.74

7 5 5

− = ⇒ = ⇒ ⇒ =

3. Operators a b a b

, and aredefined by : a b ;a b ;a b ab.a b a b

− +◊ → = ◊ = → =

+ −� � Find the value

( ) ( )66 6 , 66 6◊� .

(A) -2 (B) -1 (C) 1 (D) 2

Answer: (C)

Exp: 66 6 60 5

66 666 6 72 6

−= = =

+�

6666 6 72 6

666 6 60 5

+= = =

−

(66� 6)→(66 6) 5 6

16 5

= × =

4. Choose the most appropriate word from the options given below to complete the following

sentence.

The principal presented the chief guest with a ____________, as token of appreciation.

(A) momento (B) memento (C) momentum (D) moment

Answer: (B)



2

5. Choose the appropriate word/phrase, out of the four options given below, to complete the

following sentence:

Frogs ____________________.

(A) Croak (B) Roar (C) Hiss (D) Patter

Answer: (A)

Exp: Frogs make ‘croak’ sound.

6. A cube of side 3 units is formed using a set of smaller cubes of side 1 unit. Find the proportion of

the number of faces of the smaller cubes visible to those which are NOT visible.

(A) 1 : 4 (B) 1 : 3 (C) 1 : 2 (D) 2 : 3

Answer: (C)

Exp:

Number of faces per cube = 6

Total number of cubes = 9×3 = 27

∴ Total number of faces = 27×6 = 162

∴ Total number of non visible faces = 162-54 = 108

∴ Number of visible faces 54 1

Number of non visible faces 108 2= =

7. Fill in the missing value

6 5 4

7 4 7 2 1

1 9 2 8 1 2 1

4 1 5 2 3

33

1 1 11

1

1

1 1 1

111

1 1 1



3

Answer: (3)

Exp: Middle number is the average of the numbers on both sides.

Average of 6 and 4 is 5

Average of (7+4) and (2+1) is 7

Average of (1+9+2) and (1+2+1) is 8

Average of (4+1) and (2+3) is 5

Therefore, Average of (3) and (3) is 3

8. Humpty Dumpty sits on a wall every day while having lunch. The wall sometimes breaks. A

person sitting on the wall falls if the wall breaks.

Which one of the statements below is logically valid and can be inferred from the above

sentences?

(A) Humpty Dumpty always falls having lunch

(B) Humpty Dumpty does not fall sometimes while having lunch

(C) Humpty Dumpty never falls during dinner

(D) When Humpty Dumpty does not sit on the wall, the wall does not break

Answer: (B)

9. The following question presents a sentence, part of which is underlined. Beneath the sentence you

find four ways of phrasing the underlined part. Following the requirements of the standard written

English, select the answer that produces the most effective sentence.

Tuberculosis, together with its effects, ranks one of the leading causes of death in India.

(A) ranks as one of the leading causes of death

(B) rank as one of the leading causes of death

(C) has the rank of one of the leading causes of death

(D) are one of the leading causes of death

Answer: (A)

10. Read the following paragraph and choose the correct statement.

Climate change has reduced human security and threatened human well being. An ignored reality

of human progress is that human security largely depends upon environmental security. But on

the contrary, human progress seems contradictory to environmental security. To keep up both at

the required level is a challenge to be addressed by one and all. One of the ways to curb the

climate change may be suitable scientific innovations, while the other may be the Gandhian

perspective on small scale progress with focus on sustainability.

(A) Human progress and security are positively associated with environmental security.

(B) Human progress is contradictory to environmental security.

(C) Human security is contradictory to environmental security.

(D) Human progress depends upon environmental security.

Answer: (D)



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Section Name: Chemical Engineering

Q.No-1-25 Carry One Mark Each

1. Benzene is removed from air by absorbing it in a non-volatile wash-oil at 100kPa in a counter-

current gas absorber. Gas flow rate is 100 mol/min, which includes 2 mol/min of benzene. The

flow rate of wash-oil is 50 mol/min.Vapour pressure of benzene at the column conditions is 50

kPa. Benzene forms an ideal solution with the wash-oil and the column is operating at steady

state. Gas phase can be assumed to follow ideal gas law. Neglect the change in molar flow rates

of liquid and gas phases inside the column.’

For this process, the value of the absorption factor (upto two decimal places) is _________.

Answer: 1.02

Exp: Absorption factor

s

S

LA

mG=

Given Raoult’s law is applicable

for gas phase

satyP xp

y 100 x 50

y 0.5x

So, m 0.5

=⇒ × = ×⇒ =

=

50

A 1.020.5 98

= =×

2. The following set of three vectors

1 x 3

2 , 6 and 4

1 x 2

is linearly dependent when x is equal to

(A) 0 (B) 1 (C) 2 (D) 3

Answer: (D)

Exp: For linearly dependent vectors A 0=

1 x 3

2 6 4 0

1 x 2

=

( ) ( ) ( )12 4x x 4 4 3 2x 6 0

2x 6

x 3

⇒ − − − + − =

⇒ =⇒ =

( )S

liquid

L

mol50min

( )S

Gas 100mol min

G 98mol min

Benzene 2mol min

==



5

3. For which reaction order, the half-life of the reactant is half of the full lifetime (time for 100%

conversion) of the reactant?

(A) Zero order (B) Half order (C) First order (D) Second order

Answer: (A)

Exp: For zero order reaction

OAA

dCk C k

dt− = =

Where k = rate constant

AO AC C kt− =

For, full life time CA = 0

AOt C K=

and for half life

AOA AO 1/2

CC C 2, So t

2k= =

1/2So, t 2t=

4. Match the output signals as obtained from four measuring devices in response to a unit step

change in the input signal.

Output signal time

P: Gas chromatograph, with a long capillary tube

Q : Venturi tube

R : Thermocouple with first order dynamics

S : Pressure transducer with second order dynamics

(A) P-IV, Q-III, R-II, S-I (B) P-III, Q-I, R-II, S-IV

(C) P-IV, Q-I, R-II, S-III (D) P-II, Q-IV, R-III, S-I

Answer: (C)

Exp: If y(t) = output (response)

Gas chromatograph – IV

Ventury tube - I

Thermocouple – II

Pressure transducer with second order dynamics - III

I

II

III

IV

( )I Gas chromatograph

( )IV Venturytube

( )II 1st order

( )III 2nd order

time

( )y t



6

5. Match the chemicals written on the left with the raw materials required to produce them

mentioned on the right.

(I) Single Superphosphate (SSP) (P) Rock phosphate + Sulfuric Acid +

Ammonia

(II) Triple Superphosphate (TSP) (Q) Brine

(III) Diammonium phosphate (DAP) (R) Rock Phosphate + Sulfuric Acid

(IV) Caustic soda (S) Rock phosphate + phosphoric Acid

(A) I Q, II R, III S, IV P− − − − (B) I S, II P, III Q, IV R− − − −

(C) I R, II S, III P, IV Q− − − − (D) I S, II R, III P, IV Qa− − − −

Answer: (C)

Exp: Single super phosphate (SSP) → Phosphate rock + 2 4H SO

Triple super phosphate ( ) 3 4TSP phosphate rock H PO→ +

Diammonium phosphate (DAD) 2 4 3Rock phosphate H SO NH→ + +

Caustic soda → Brine

6. For the matrix 4 3 1

, if3 4 1

is an eigenvector, the corresponding eigenvalue is _______.

Answer: 7

Exp: Let A = 4 3 1

and x3 4 1

=

Let eigen value is λ

So Ax x= λ

[ ][ ] 1 0or A I x 0, where I

0 1

− λ = =

So 0

Ix0

λ = λ

4 3 10

3 4 1

4 30

3 4

7

− λ ⇒ = − λ

− λ + ⇒ = + − λ

⇒ λ =



7

7. The transfer function for the disturbance response in an open-loop process is given by ( )open

dG s .

The corresponding transfer function for the disturbance response in closed-loop feedback control

system with proportional controller is given by ( )closed

dG s . Select the option that is Always correct

( ) ( ){ }O G S represents order of transfer function G s :

(A) ( ) ( )open closed

d dO G s O G s= (B) ( ) ( )open closed

d dO G s O G s≠

(C) ( ) ( )open closed

d dO G s O G s≥ (D) ( ) ( )open closed

d dO G s O G s≤

Answer: (A)

Exp: Open loop transfer function

( )open

OL dG GH G s= =

and closed loop transfer function

( )closed

CL d

GG G s

1 GH= =

+

Order of transfer function is order of denominator

Order of GH and 1 + GH are same

( ) ( )open closed

d dso, O G s O G s =

8. If v,u, s and g represent respectively the molar volume, molar internal energy, molar entropy and

molar Gibbs free energy, then match the entries in the left and right columns below and choose

the correct option.

( ) ( )s

P u v− ∂ ∂ ( )I Temperature

( ) ( )T

Q g P∂ ∂ ( )II Pr essure

( ) ( )P

R g T− ∂ ∂ ( )III V

( ) ( )V

S u s∂ ∂ ( )IV S

(A) P II, Q III, R IV, S I− − − − (B) P II, Q IV, R III, S I− − − −

(C) P I, Q IV, R II, S III− − − − (D) P III, Q II, R IV, S I− − − −

Answer: (A)

Exp: We have

du Tds pdv= −

and dg = vdp sdT−

So, S S

u uP P

v v

∂ ∂= − ⇒ − =

∂ ∂

V

PT

uT

s

g gV and S

p T

∂=

∂

∂ ∂= − =

∂ ∂

x yG

H



8

9. Two different liquids are flowing through different pipes of the same diameter. In the first pipe,

the flow is laminar with centerline velocity, max.1

V , Whereas in the second pipe, the flow is

turbulent. For turbulent flow, the average velocity is 0.82 times the centerline velocity, max.2

V .

For equal volumetric flow rates in both the pipes, the ratio max.1 max.2

V V (up to two decimal

places) is ______.

Answer: 1.64

Exp: For laminar flow

avg.1 max.1V 0.5V=

For Turbulent flow

( )avg.2 max .2V 0.82 V given=

Since volumetric flow rate is some

1 2

Q Q=

1 avg.1 2 avg.2A V A V=

Given 1 2

A A=

So, avg.1 avg.2V V=

max.1 max.2

max.1

max.2

0.5 V 0.82V

V1.64

V

⇒ =

⇒ =

10. Consider linear ordinary differential equation ( ) ( ) ( ) ( )dyp x y r x . Functions p x and r x are

dx+ =

defined and have a continuous first derivative. The integrating factor of this equation is non-zero.

Multiplying this equation by its integrating factor converts this into a:

(A) Homogeneous differential equation (B) Non-linear differential equation

(C) Second order differential equation (D) Exact differential equation

Answer: (D)

Exp: Linear differential equation

( ) ( )1y p x y r x+ =

Multiplying above equation by integrating factor ( )p x dx

e ∫ makes the equation exact

11. A spherical naphthalene ball of 2mm diameter is subliming very slowly in stagnant air at 25OC.

The change in the size of the ball during the sublimation can be neglected. The diffusivity of

naphthalene in air at 25o C is 1.1×10

-6 m

2/s.

The value of mass transfer coefficient is B × 10-3

m/s, where B (up to one decimal place) is

_______.

Answer: 1.1

Exp: For spherical ball

Dimensionless sherwood number = 2

C

AB

k L2

D=



9

Where L = characteristic length = diameter of

Spherical ball = d = 2 mm = 32 10 m−×

6

ABC 3

3

C

2 D 2 1.1 10K

d 2 10

K 1.1 10 m s

So, B 1.1

−

−

−

× × ×= =

×= ×

=

12. An irreversible, homogeneous reaction A products,→ has the rate expression:

2

A AA

A

2C 0.1CRate , where C is the concentration of A.

1 50C

+=

+

A

C varies in the range 0.5 – 50 mol/m3.

For very high concentration of A, the reaction order tends to:

(A) 0 (B) 1 (C) 1.5 (D) 2

Answer: (B)

Exp: Rate = 2

A A

A

2C 0.1C

1 50C

++

( )3

A0.5 C 50 mol m< <

For very high value of CA ( )3say 50 mol m

2

A A

A

2

AA

A

0.1C 2C

and 50C 1

2C 1So, rate C

50C 25

So reaction order is one

<<>>

= =

13. A scalar function in the xy-plane is given by ( ) 2 2 ˆ ˆx, y x y . if i amd jφ = + are unit vectors in the x

and y directions, the direction of maximum increase in the value of φat (1,1) is along:

(A) ˆ ˆ2i 2j− + (B) ˆ ˆ2i 2j+ (C) ˆ ˆ2i 2j− − (D) ˆ ˆ2i 2j−

Answer: (B)

Exp: Direction of maximum increase

( )

( ) ( )

( )

at 1.1

ˆ ˆ2x i 2yj at 1.1

ˆ ˆ2i 2 j

= ∇φ

= +

= +



10

14. For a gas phase cracking reaction A B C→ + at 300OC, the Gibbs free energy of the reaction at

this temperature is OG 2750J mol.∆ = − The pressure is 1 bar and the gas phase can be assumed

to be ideal. The universal gas constant R = 8.314J/mol. K. The fractional molar conversion of A

at equilibrium is:

(A) 0.44 (B) 0.50 (C) 0.64 (D) 0.80

Answer: (D)

Exp: A B C→ +

at t 0, 0 0

at t t 1 where, fraction converted

= ⊥= − α α α α =

Equilibrium constant

B C

P

A

2 2

2 2

Y YK .P

Y

1 1.P

1 1 1

1 x

=

α α α α+ α + α = = =

− α − α − α +

Also

O

O

2

2

ln k G RT

G 2750K exp : exp

RT 8.314 573

K 1.78

so, 1.781

0.80

= −∆

−∆ = ×

=

α=

− αα =

15. Two infinitely large parallel plates (I and II) are held at temperatures TI and TII ( )1 IIT T>respectively, and placed at a distance 2d apart in vacuum. An infinitely large flat radiation shield

(III) is placed in parallel in between I and II. The emissivities of all the plates are equal. The ratio

of the steady state radiative heat fluxes with and without the shield is:

(A) 0.5 (B) 0.75 (C) 0.25 (D) 0

1

I

T

III

I1

II

T



11

Answer: (A)

Exp: We have( )4 4

1 212T TQ

1 1A1

σ −=

+ −ε ε

Where ε = emissivity

( )4 4

1 212T TQ

2A1

σ −=

−ε

In presence of shield

( ) ( )

( ) ( )( )

4 4 4 4

1 3 3 213 23

4 44 1 23

4 4 44 41 1 21 313

12

13

12

T T T TQ Q

2 2A A1 1

T Tor T

2

T T T 2T TQSo,

2 2A1 1

Q1

2 A

Q 1So,

Q 2

σ − σ −= = =

− −ε ε+

=

− +σ −= = σ

− −ε ε

=

=

16. A cylindrical packed bed of height 1 m is filled with equal sized spherical particles. The particles

are nonporous and have a density of 1500 kg/m3. The void fraction of the bed is 0.45. The bed is

fluidized using air (density 1kg/m3). If the acceleration due to gravity is 9.8m/s

2, the pressure drop

(in pa) across the bed at incipient fluidization (up to one decimal place ) is _________.

Answer: 8079.61

Exp: For incipient fluidization

( )( )P

Pg 1

L

∆= − ε ρ − ρ

Where 0.45ε =

3 3

P 1500kg m 1kg mρ = ρ =

And L = 1m (height of the bed)

( )( )P 9.8 1 0.45 1500 1 1

8079.61Pa

∆ = × − − ×

=

( )i ( )iii ( )ii

1T

3T

2T



12

17. For a pure liquid, the rate of change of vapour pressure with temperature is 0.1 bar/K in the

temperature range of 300 to 350 K. if the boiling point of the liquid at 2 bar is 320 K, the

temperature (in K) at which it will boil at 1 bar (up to one decimal place) is ___________.

Answer: 310

Exp: Given dP

0.1 dp 0.1dTdT

= ⇒ =

( )P 0.1T C........... 1

Given at T 320K. P 2bar

2 0.1 320 C

C 30

⇒ = +

= == × +

⇒ = −

Equation (1) becomes

P 30 0.1T=− +

Putting P = 1 bar

1 30 0.1T

T 310K

= − +⇒ =

18. For uniform laminar flow over a flat plate, the thickness of the boundary layer, ,δ at a distance x

from the leading edge of the plate follows the relation:

(A) ( ) 1x x−δ α (B) ( )x xδ α (C) ( ) 1 2x xδ α (D) ( ) 1 2x x−δ α

Answer: (C)

Exp: For laminar flow

Boundary layer thickness

( ) 4.99x 4.99x8 x

Re Vx= =

ρµ

Or, ( )1

2x xδ α

19. Match the polymer mentioned on the left with the catalyst used for it manufacture given on the

right.

(I) Low density Polyethylene (P) Ziegler-Natta catalyst

(II) High density Polyethylene (Q) Traces of Oxygen

(III) Polyethylene Terephthalate (R) Butyl Lithium

(IV) Polyvinyl Chloride (S) Antimony

(A) I Q, II R, III S, IV P− − − − (B) I S, II P, III Q, IV R− − − −

(C) I Q, II P, III S, IV R− − − − (D) I S, II R, III P, IV Q− − − −



13

Answer: (C)

Exp:

LDPE Traces of oxygen

HDPE Zeigler – Natta catalyst

PET Antimony

PVC Butyl Lithium

20. Three identical closed systems of a pure gas are taken from an initial temperature and pressure

( )L 1T , P to a final state ( )2 2T , P , each by a different path. Which of the following in Always true

for the three systems? ( ∆ represents the change between the initial and final states: U, S, G, Q

and W are internal energy, entropy, Gibbs free energy, heat added and work done, respectively.)

(A) U, S, Q are same∆ ∆ (B) W, U, G are same∆ ∆

(C) S, W, Q are same∆ (D) G, U, S are same∆ ∆ ∆

Answer: (D)

Exp: U,S and G are state variables (point function)

so, U, S and G∆ ∆ ∆

Remain same between two states 1 and 2.

21. Identify the WRONG statement amongst the following:

(A) Steam distillation is used for mixtures that re immiscible with water.

(B) Vacuum distillation is used for mixtures that are miscible with water.

(C) Steam distillation is used for mixtures that are miscible with water.

(D) Vacuum distillation columns have larger diameters as compared to atmospheric columns for

the same throughout.

Answer: (C)

Exp: Steam distillation is used for the mixture immiscible with water.

Option (C) is the wrong statement

1

( )1 1P . T

( )2 2P . T

2P

T



14

22. For a binary mixture of components A and B, NA and NB denote the total molar fluxes of

components A and B, respectively. JA and JB are the corresponding molar diffusive fluxes. Which

of the following is true for equimolar counter-diffusion in the binary mixture?

(A) A B A B

N N 0 and J J 0+ = + ≠ (B) A B A B

N N 0 and J J 0+ ≠ + =

(C) A B A B

N N 0 and J J 0+ ≠ + ≠ (D) A B A B

N N 0 and J J 0+ = + =

Answer: (A)

23. A complex-valued function, f(z), given below is analytic domain D:

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

Which of the following is NOT correct?

(A) df v u

idz y y

∂ ∂= +

∂ ∂ (B)

df u vi

dz x x

∂ ∂= +

∂ ∂

(C) df v u

idz y y

∂ ∂= −

∂ ∂ (D)

df v vi

dz y x

∂ ∂= +

∂ ∂

Answer: (A)

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

( ) df u v v u vf ' z i i

dz x x x y y

∂ ∂ ∂ ∂ ∂= = + = = +

∂ ∂ ∂ ∂ ∂ - (i)

For analytic function

u u v

andx y y x

∂ ∂ υ ∂ −∂= =

∂ ∂ ∂ ∂

So, equation ( )i can be written as

df v v u u v

i i idz y x y y x x

∂υ ∂ ∂ ∂ ∂ ∂= + = − = +

∂ ∂ ∂ ∂ ∂ ∂

Only option (A) can not be deduced

24. Which of the following can change if only the catalyst is changed for a reaction system?

(A) Enthalpy of reaction (B) Activation energy

(C) Free energy of the reaction (D) Equilibrium constant

Answer: (B)

Exp: Catalyst changes the activation energy of the reaction



15

25. Match the technologies in Group 1 with the entries in Group 2:

Group – 1 Group 2

(P) Urea manufacture (I) Microencapsulation

(Q) Coal gasification (II) Ultra-low sulphur diesel

(R) Controlled release of chemicals (III) Shale oil

(S) Deep hydrodesulphurization (IV) Prilling tower

(V) Gas hydrates

(VI) Gas – solid non-catalytic

reaction

(A) P I, Q V, R II, S VI− − − −

(B) P IV, Q VI, R I, S II− − − −

(C) P IV, Q I, R III, S II− − − − (D) P V, Q VI, R IV, S II− − − −

Answer: (B)

Exp: P : Urea manufacture IV prilling tower→

Q : coal gasification VI Gas solid non catalytic reaction→ −

R : controlled release of chemical I :micron capsulation.→

S : Deep hydrodesulphurization II ultra low sulphur diesel.→

Q.No-26-55 Carry Two Marks Each

26. Consider a solid block of unit thickness for which the thermal conductivity decreases with an

increase in temperature. The opposite faces of the block are maintained at constant but different

temperatures: T(x = 0) > T(x = 1). Heat transfer is by steady state conduction in x-direction only.

There is no source or sink of heat inside the block. In the figure below, identify the correct

temperature profile in the block.

(A) I (B) II (C) III (D) IV

( )T x 0=

I

II

III

IV( )T x 1=

0 x 1

T



16

Answer: (C)

Exp: Let ( ) 1T x 0 T=

and ( ) 2T x 1 T= =

we have from Fourier’s law

q dT

kA dx

= −

At steady state conduction T1>T2

dT q

k constdx A

− = =

q

dx kdTA

⇒ = − ____(1)

Here K is decreasing with temperature

Let ( )OK K 1 T= − α

⇒ ( )1

x T

o0 T

qdx k 1 T dT

A= − − − α∫ ∫

( ) ( )2 2

O 1 1

qx k T T T T

A 2

α= − + + − ____(11)

Similarly integrating equation (1) between x 0 & x 1= =

( )2

1

x 1 T

Ox 0 T

qdx k 1 T dT

A

=

== − − α∫ ∫

( ) ( ) ( )2 2

o 2 1 2 1

q1 k T T T T (iii)

A 2

α⇒ = − − + − − − −

Dividing equation (ii) and (iii)

( ) ( ) { } ( )2 2 2 2

1 o 1 2 1 0 2 1T T 2k T T x T T 2k T Tα − − − = α − − −

From the above expression for temperature profile it is clear that the temp passes through a

minima



17

27. The schematic diagram of a steady state process is shown below. The fresh feed (F) to the reactor

consists of 96 mol% reactant A and 4 mol% inert I. The stoichiometry of the reaction is A C→ .

A part of the reactor effluent is recycled. The molar flow rate of the recycle stream is 0.3F. The

product stream P contains 50 mol% C. The percentage conversion of A in the reactor based on A

entering the reactor at point 1 in the figure (up to one decimal place) is ________.

Answer: 59.19

Exp: Basis: -

100 mole of feed F

A in feed = 96 mol

Inert in feed = 4mol

Recycle = 0.3F = 30 mole

Product (P) Contains 50%C

At steady state F = P = 100 mole

Product Contains 50 mole C

Product must also contain 4% inert

At point (2) total molar flow rate =100 + 30 = 130mole

Product stream P is

50 mole C

46 mole A

4 mole C

Recycle stream R is

15 mole C

13.8 mole A

1.2 mole I

Conversion of A in the reactor

( )

( )

A converted int o C at point 2

A fed at point A

65100 59.19%

96 13.8

=

= × =+

F

1A C

P

0.3F

0.3F

( )1

A C→( )2

PF



18

28. The impulse response to a tracer pulse experiment for a flow reactor is given below:

In the above figure, c is the exit tracer concentration. The corresponding E or Eθ (normalized E)

curve is correctly represented by which of the following choices? Here, θ is dimensionless time.

(A) (B)

(C) (D)

Answer: (C)

29. Consider a steady state mass transfer process between well-mixed liquid and vapour phases of a

binary mixture comprising of components A and B. The mole fractions of component A in the

bulk liquid (xA) and bulk vapour ( )Ay phases are 0.36 and 0.16, respectively. The mass transfer

coefficients for component A in liquid and vapour phases are 0.1 mol/(m2.s) and 0.05 mol/(m

2.s),

respectively. The vapour-liquid equilibrium can be approximated as *

A Ay 2x= for A

x less than

0.4. The mole fraction of A in the liquid at the interface (up to two decimal places) is

Answer: 0.08

Exp: Given A

x 0.36=

A

y 0.16=

2

L

2

g

k 0.1mol m sec

k 0.05mol m s

=

=

*

A A Ay 2x for x 0.4= <

( )C mol 1

6.0

4.0

2.0

0.00.0 2.0 4.0 6.0

( )t min

( )1

1E min−

0.30

0.30

0.10

0.000.0 2.0 4.0 6.0

( )t min

( )E θ −2.00

1.00

0.000.0 1.0 2.0 3.0

( )θ −

( )1

1E min−

0.60

0.40

0.20

0.000.0 2.0 4.0 6.0

( )t min

( )1

1E min−4.00

2.00

0.000.0 2.0 4.0 6.0

( )t min

6.00



19

So Ai A

y 2x= ……….(i)

At steady state, mass flux is constant

( ) ( )( ) ( )

( ) ( )

L A Ai g Ai A

Ai A

Ai

Ai

k x x k y y

0.1 0.36 x 0.05 2x 0.16

0.11 0.36 x 0.05 2 0.36 0.16

x 0.08

− = −

− = −

− = × −

⇒ =

30. A heated solid copper sphere (of surface area A and volume V) is immersed in a large body of cold

fluid. Assume the resistance to heat transfer inside the sphere to be negligible and heat transfer

coefficient (h), density ( )ρ , heat capacity (C), and thermal conductivity (k) to be constant. Then,

at time t, the temperature difference between the sphere and the fluid is proportional to:

(A) Ahexp t

VC

− ρ

(B) VC

exp th A

ρ−

(C) 4 k

exp tCA

π− ρ

(D) CA

exp t4 k

ρ − π

Answer: (A)

Exp: T = f(time)

Let apply first law of thermodynamics. If at any instant of time T. if T is body temperature then.

( ) P

dThA T T C V

dt∞ − = ρ

Where CP = heat capacity of the body

0

T t

PT 0

dT hAdt

T T VC∞

=− ρ∫ ∫

o P

T T hAln t

T T VC

∞

∞

−⇒ = −

− ρ

hAT T e t

VC∞

−⇒ − = ρ

31. An ideal gas is initially at a pressure of 0.1 MPa and a total volume of 2m3. It is first compressed

to 1MPa by a reversible adiabatic process and then cooled at constant pressure to a final volume

of 0.2m3. The total work done (in kJ) on the gas for the entire process (up to one decimal place) is

____________

Data: R = 8.314 J/molK; heat capacity at constant pressure (CP) = 2.5R

Answer: 750

Exp: 1 2 : Re versible adiabatic process→

2 3: Re versible isobaric process→



20

Given 1

P 0.1MPa=

510 Pa=

3

1

6

2 a 3

3

3

V 2m

P 10 P P

V 0.2m

=

= =

=

P V

C 2.5RSo C 1.5R= =

P V

2.5C C 1.667

1.5γ = = =

For process 1 – 2: 1 1 2 2P V P Vγ γ=

1 15 1.667

12 1 6

2

2

2

P 10V .V 2

P 10

V 0.50m

γ = = ×

⇒ =

Work done on gas for process 1 – 2

6 6

1 1 2 21 2

P V P V 0.1 10 2 10 0.5W

1 0.667−

− × × − ×= =

γ −

1 2

W 450kJ−⇒ = −

Work done for process 2 – 3

2 3

W P V− = ∆

( )6

10 0.2 0.5

300kJ

= × −

= −

Net work done = 450 + 350 = 750kJ

32. A centrifugal pump delivers water at the rate of 0.22 m3/s from a reservoir at ground level to

another reservoir at a height H, through a vertical pipe of 0.2m diameter. Both the reservoirs are

open to atmosphere. The power input to the pump is 90 kW and it operates with an efficiency of

75%.

Data:

Fanning friction factor for pipe flow is f = 0.004. Neglect other head losses

Take gravitational acceleration, g = 9.8 m/s2 and density of water is 1000kg/m

3.

The height H, in meters, to which the water can be delivered (up to one decimal place) is _______.

Answer: 36

Exp: Given 2Q 0.22m sec=

( )2

2

0.22V 7m sec

0.24

= =π

×

W 90kW=

390 10

W in head unitQg

×=

ρ

Applying Bernaulli’s equation between (1) and (2)

3 2P

V



21

( )

2

1 11

g

2

2 22

vz hf w

2g

Vz 1

g 2g

ρ+ + + η

ρ

ρ= + + − − −

ρ

2

and Z H=

2 2

f

3

4fHV 4 0.004 H 7h 0.2H m

2gD 2 9.8 0.2

90 10 49(1) 0.2H 0.75 H

0.22 1000 9.8 2 9.8

× × ×= = =

× ×

×⇒ + × = +

× × ×

0.2H 31.308 H 2.5

0.8H 28.808

H 36.01m

So, H 36m

⇒ + = +⇒ =

==

33. A multi-stage, counter-current liquid-liquid extractor is used to separate solute C from a binary

mixture (F) of A and C using solvent B. Pure Solvent B is recovered from the raffinate R by

distillation, as shown in the schematic diagram below:

Locations of different mixtures for this process are indicated on the triangular diagram below. P is

the solvent-free raffinate, E is the extract, F is the feed and ∆ is the difference point from which

the mass balance lines originate. The line PB interects the binodal curve at U and T. The lines P∆

and FB intersect the bimodal at V and W respectively.

The raffinate coming out of extractor is represented in the diagram by the point:

(A) T (B) U (C) V (D) W

Answer: (B)

F

E

R

B

distillationextractorP

B

C

F

E

VP

U

A B

∆



22

34. The solution of the non-linear equation

3x x 0− = is to be obtained using Newton-Raphson method. If the initial guess is x = 0.5, the

method converges to which one of the following values:

(A) -1 (B) 0 (C) 1 (D) 2

Answer: (A)

Exp: ( ) ( )3 2f x x x, f ' x 3x 1= − = −

Initial guess = 0.5 = x0

First iteration

( )( )( )

( )( )( )

0

1 0

0

3

1 2

1

f xx x

f ' x

0.5 0.5 0.375x 0.5 0.5

0.253 0.5 1

x 1

= −

− −⇒ = − = −

−−

⇒ = −

Solution to above equations are 0.1, - 1 but after first iteration it is converging to – 1

35. An isothermal steady state mixed flow reactor (CSTR) of 1m3 volume is used to carry out the

first order liquid-phase reaction A → products. Fresh feed at a volumetric flow rate of Q

containing reactant A at a concentration CA0 mixes with the recycle steam at a volumetric flow

rate RQ as shown in the figure below.

It is observed that when the recycle ratio R = 0.5, the exit conversion Af

X 50%= When the

recycle ratio is increased to R = 2, the new exit conversion (in percent) will be:

(A) 50.0 (B) 54.3 (C) 58.7 (D) 63.2

Answer: (A)

Exp: Mole balance

Input – Output

disappearance = Accum.

At steady state,

( ) ( )A0 A A AQC RQC RQ Q C kC V+ − + =

A0 A A

QC QC kC V⇒ − =

3Q 0.5m min=3

A0C 1mol m=

3Q 0.5m min=

R Q

A A f

A A f

C C

X X

==

A0 3

Q

1molC

m=

RQ

Q

A Af

A Af

C C

x x

==



23

( )

( )

A0 A A

A0 A

A0

A A

A A

C C kC

C Cor

kC 1 x

x xor k

k 1 x 1 x

⇒ − = τ−

τ =−

τ = ⇒τ =− −

Or , A

kx

1 k

τ=

+ τ

So, conversion xA is independent of Recycle ratio R

So final conversion = 50%

36. A typical batch filtration cycle consists of filtration followed by washing, One such filtration unit

operating at constant pressure difference first filters a slurry during which 5 liters of filtrate is

collected in 100 s. This is followed by washing. Which is done for tw seconds and uses 1 liter of

wash water. Assume the following relation to be applicable between the applied pressure drop

P∆ , cake thickness L at time t, and volume of liquid V collected in time t.

1 2

P dvk ; L k V, if L is changing

L dt

∆= =

1 2

k and k can be taken to be constant during filtration and washing. The wash time tw, in seconds

(up to one decimal place) is __________

Answer: 40

Exp: 1 2

P dVGiven k and L k V

L dt

∆= =

1 2

dv 1 Pso

dt k k V

∆=

5 100

0 01 2

1 2

1 2

Pvdv dt

k k

25 P100

2 K K

P 1or

k k 8

∆⇒ =

∆⇒ = ×

∆=

∫ ∫

Also, Final rate of filtration

1 2

dV P 118 5dt k k V 40

∆= = =×

w

volume of wash water usedwashing time t

finalrate of filtration=

( )w

1t 40sec

140

= =



24

37. Which one of the following transfer functions, upon a unit step change in disturbance at t = 0, will

show a stable time domain response with a negative initial slope (i.e., slope at t = 0):

(A) ( ) 1 2G s

s 1 s 4= −

+ + (B) ( ) 1 2

G ss 1 s 4

= ++ +

(C) ( ) 1 2G s

s 1 s 4= +

+ − (D) ( ) 1 2

G ss 1 s 4

= +− −

Answer: (A)

Exp: From option (A) ( ) 1 2G s

s 1 s 4= −

+ +

( ) 1 1 2so y s

s s 1 s 4

= − + +

( ){ } t 4t

z 0

dyd sy s e 2e

dt

dy1

dt

− −

=

= = −

= −

So (A) has –ve initial slope

38. Given that molar residual Gibbs free energy, Rg , and molar residual volume V

R, are related as

R Rp

R O

0

g vdP, find g at T 27 C and P 0.2MPa.

RT RT

= = =

∫ The gas may be assumed to follow the

viral equation of state, Z=1+BP/RT,where B= -10-4

m3/mol at the given conditions

(R = 8.314J/mol.K). The value of gR in J/mol is:

(A) 0.08 (B) 2.4− (C) 20 (D) 20−

Answer: (D)

Exp: Given R R

P

0

g VdP

Rt RT= ∫

We know, ( )R ig ZRT RT RTV V V Z 1

P P P= − = − = −

( )R Z 1V

orRT P

−=

And ( )Z 1BP B

Z 1RT P RT

−= + ⇒ =

R

P

0

g BSo, dP

RT RT= ∫

( )3

R 4 6mg B P 0 10 0.2 10 Pa

mol

−= − = − × ×

Rg 20J mol= −



25

39. A spherical solid particle of 1mm diameter is falling with a downward velocity of 1.7mm/s

through a liquid (viscosity 0.04 Pa.s) at a low Reynolds number (Strokes regime). The liquid is

flowing upward at a velocity of 1 mm/s. All velocities are with respect to a stationary reference

frame. Neglecting the wall effects, the drag force per unit projected area of the particle, in Pa, (up

to two decimal places) is _________

40. In the figure below, the temperature profiles of cold and hot fluids in counter current double pipe

heat exchanges (in different mode of operation) are shown on the left. For each case, match the

heat exchange process for the fluid represented by the bold curve with the options given on the

right.

1.7mm s

1mm s

Temperature

Temperature

Dis tan ce

( )I( )P Heating of sub cooled feed

to super heated vapour

−

Dis tan ce

( )II( )Q Condensation of super heated vapour

Dis tan ce

( )III( )R Boiling of sub cooled liquid−

Dis tan ce

( )IV ( )S Condensation of Saturated vapour

followed by sub cooling−

Temperature

Temperature



26

(A) I P, II Q, III R, IV S− − − − (B) I P, II Q, III S, IV R− − − −

(C) I Q, II P, III S, IV R− − − − (D) I Q, II S, III P, IV R− − − −

Answer: (C)

Exp: (P) Heating of sub cooled feed to super heated vapour

(1) Sub cooled to saturated

(2) Saturated up to boiling point

(3) Saturated to superheat

(Q) Condensation of superheated vapour.

(1) Cooling of superheated vapour to saturated vapour

(2) Condensation of saturated vapour to liquid.

(R) Boiling of sub-cooled liquid

(1) Heating of sub cooled liquid to saturated liquid

(2) Boiling of saturated liquid

(S). Condensation of saturated vapour followed by sub-cooling

(1) Condensation of saturated vapour to saturated liquid

(2) Sub cooling of saturated liquid.

( )1 ( )2

( )3

( )1

( )2

( )1( )2

( )2

( )1

( )1

( )2



27

41. The block diagram for a process with feedback control for output deviation variable h is shown in

the figure below. All transfer functions are given with pre-factor of s in minutes. A unit step

change is made in the set-point at t = 0. The time required for h to reach 50% of its ultimate value,

in minutes (up to two decimal places), is: _____________.

Answer: 0.8664

Exp: We have by making ( )dG s 0= (no load disturbance)

( )( )sp

0.2h s 0.21.5s 1

0.2 1.2 1.5sh s 11.5s 1

+= =++

+

Given ( )sp1h s

s=

( )sp

1 0.2 A BSo, h s .

s 1.5s 1.2 S 1.5S 1.2= = +

+ +

1 1A . and B

6 4

1 1 1So, h(s) .

6s 4 1.5s 1.2

−= =

= −+

Taking inverse laplace transformation

( ) 0.8t1 1h t e

6 6

−= −

Ultimate value

( ) 1h

6= ∞ =

Let t = time required to reach 50% of ultimate value

0.8t 0.8t

0.8t

1 1 1 1 1e e

12 6 6 12 6

0.6931e t 0.8664sec2 0.8

− −

−

− −= − ⇒ =

⇒ =+ ⇒ = =

+( )sph s

( )cG s 0.5s= ( )f

0.2G s

1.5s 1=

+( )p

1G s

0.5s=

( )d

1G s

0.5s=

( )mG s 1=

( )h s



28

42. Adsorption on activated carbon is to be used for reducing phenol concentration in wastewater

from 0.04 mol/1 to 0.008mol/1. The adsorption isotherm at the operating temperature can be

expressed as q = 0.025C1/3

; where q is the phenol concentration in solid (mol/g solid) and C is the

phenol concentration in water (mol/1). The minimum amount of solid (in grams) required per liter

of wastewater (up to one decimal place) is ______________________

43. Consider a control system with the open loop transfer function given by:

( )0.3c

eOL

K eG s

1.5s 1

−

=+

In the above function, pre-factor of s is in minutes and KC is the gain of proportional controller.

The frequency for phase margin of 30O is 40.04rad/min. The value of KC for a gain margin of 1.7

(up to one decimal place) is _______

Answer: 5.019

44. For complex variable Z, the value of the contour integral ( )

2z

c

1 edz

2 i z z 3

−

π −∫ along the clockwise

contour C: ( )z 2 up to two decimal places= is __________

Answer: -0.33

Exp: ( )

2z

c

1 e

2 i z z 3

−

π −∫

C: z 2 2 z 2= − < <

Poles are z 0.3=

Z = 3 is outside the countour C

( ) ( )2 0e 1

R o 0.330 3 3

− ×

= =− = −−

Value of integral = -0.33

45. A proposed chemical plant is estimated to have a fixed capital (FC) of Rs. 24 crores. Assuming

other costs to be small, the total investment may be taken to be same as FC. After commissioning

(at t = 0 years), the annual profit before tax is Rs.10 crores/year (at the end of each year) and the

expected life of the plant is 10 years. The tax rate is 40% per year and a linear depreciation is

allowed at 10% per year. The salvage value is zero. If the annual interest rate is 12% the NPV

(net present value or worth) of the project in crores of rupees (up to one decimal place) is

_______.



29

46. For fanning friction factor f (for flow in pipes) and drag coefficient CD(for flow over immersed

bodies), which of the following statements are true?

P: f accounts only for the skin friction

Q: CD accounts only for the form friction

R: CD accounts for both skin friction and form friction

S: Both f and CD depend on the Reynolds number

T: For laminar flow through a pipe, f doubles on doubling the volumetric flow rate.

(A) R,S,T (B) P,Q,S (C) P,R,S (D) P,Q,S,T

Answer: (C)

Exp: Correct Statements are

P : f accounts only for the skin friction

R : CD accounts for both skin friction and form friction

S : Both f and CD depends on the Reynolds number

47. A binary feed consisting of 25 mol% liquid and 75 mol% vapour is separated in a staged

distillation column. The mole fraction of the more volatile component in the distillate product is

0.95. The molar flow rate of distillate is 50% of the feed flow rate and McCabe-Thiele method

can be used to analyze the column . The q-line intersects the operating line of the enriching

section at (0.35,0.5) on the x-y diagram. The slope of the stripping section operating line (up to

one decimal place) is ________

Answer: 1.4

Exp: Given q = 0.25

D

X 0.95=

Equation of operating line for rectifying section is

Dn 1 n

xRy x

R 1 R 1+ − +

+ +

Since this line passes through (0.35, 0.5).

R 0.95 0.35R 0.950.5 0.35

R 1 R 1 R 1

+= × + =

+ + +

0.5R 0.5 0.35R 0.95

0.15R 0.45

R 3

⇒ + = +⇒ =⇒ =

Reflux ratio = OO

LL 3D

D⇒ =

( )A F D w∞ = +

2D D W⇒ = +

W D and F 2D= =

n.mm

V

OLD

mm 1+

W

F



30

For stripping section

wmm 1 m

m 1 m 1

WxLy x

V V+

+ +

= −

In stripping section Lm = Ln + qF

= 3D + 0.25 * 2D

Lm = 3.5D

and m 1

V Lm W 3D 0.5D 2.5D+ = − = − =

Slope of operating line in stripping

Section = m 1

Lm 3.51.4

V 2.5+

= =

48. A catalyst slab of half-thickness L (the width and length of the slab>> L) is used to conduct the

first order reaction A B→ . At 450 K, the Thiele modulus for this system is 0.5. The activation

energy for the first order rate constant is 100kJ/mol. The effective diffusivity of the reactant in the

slab can be assumed to be independent of temperature, and external mass transfer resistance can

be neglected. If the temperature of the reaction is increased to 470 K, then the effectiveness factor

at 470 K (up to two decimal place) will be _________.

Value of universal gas constant = 8.314 J/mol.K

Answer: 1.875

49. The diameter of sand particles in a sample range from 50 to 150 microns. The number of particles

of diameter x in the sample is proportional to 1

50 x+. The average diameter, in microns, (up to

one decimal place) is ________________

50. Consider two steady isothermal flow configuration shown schematically as Case I and Case II

below. In case I, a CSTR of volume V1 is followed by a PFR of volume V2, while in Case II a

PFR of volume V2 is followed by a CSTR of volume V1. In each case, a volumetric flow rate Q of

liquid reactant is flowing through the two units in series. An irreversible reaction A → products

(order n) takes place in both cases, with a reactant concentration CA0 being fed into the first unit.

Q

1V

A0C

Q

2V

Q

AfCI

Q

2V

Q

1V

Q

A0CfAC II



31

Choose the correct option:

(A) Af

f

I

II

A

C1 for n 1

C> = (B)

Af

f

I

II

A

C1 for n 1

C= =

(C) Af

f

I

II

A

C1 for n 1

C< =

(D) Af

f

I

II

A

C1 for n 1

C= >

Answer: (B)

Exp: We can solve problem by taking 1st order reaction. (n = 1)

For Case I,

Consider 3 1

1 2V V 1m , k 1.sec−= = =

3 3

AOQ 1m sec, and C 1mol m= =

For CSTR, A0 A1

A1

C Ck

C

−τ =

A1

A1

1 C1 1

C

−× =

3

A1C 0.5 mol m⇒ =

For PFR.

Af

A1

CA

CA

dC

kCτ= ∫

3AfAf

A1

C1 ln C 0.184mol m

C⇒ = ⇒ =

Case – II

Q2

VQ

AfCA1C

1V

A0C

Q

1V

Q

2V

Q

A1C

A0C

Q

AfC



32

With same data,

For PFR

A1

A0

Cln 1

Cτ = − =

3

A1C 0.3680 mol m⇒ =

and for CSTR,

A1 Af Af

Af Af

C C 0.3680 CK 1

C C

− −τ = = =

3

AfC 0.1840 mol m .=

Clearly for both the case Af

C is equal

Af

I

II

Af

Ci,e 1 for n 1

C= =

51. Select the wrong statement regarding water gas shift converters from the list below.

(A) Inter-stage cooling is provided between the two stages of shift converters.

(B) Usually high temperature shift (HTS) reactor has a iron-based catalyst and low temperature

shift (LTS) reactor has a copper-based catalyst.

(C) HTS reactor is followed by LTS reactor.

(D) LTS reactor is followed by HTS reactor.

Answer: (D)

Exp: The correct sequence is

Wrong statement is D

52. A vector u = - 2yi 2xj,+ where i and j are unit vector in x and y directions, respectively. Evaluate

the line integral

c

I u.dr= ∫�

Where C is a closed loop formed by connecting points (1,1) , (3,1), (3,2) and (1,2) in the order.

The value of I is ___________.

High Temp

Reactor

Inter

Cooler

Low temp

reactor



33

Answer: 8

Exp: Given ˆ ˆu 2y i 2xj= − +

ˆ ˆdr dxi dy j= +�

So, ( )c c c

u.dx 2y.dx 2x.dy 1= − + −∫ ∫ ∫� � �

Now, for A B→ y 1, dy 0 x changes from 1 to 3= =

( )3

1c

1 u.dr 2 dx 4⇒ =− × = −∫ ∫�

For B C, x 3, dx 0, ychanges from 1 to 2→ = =

( )2

1c

1 u.dr 2 3dy 6⇒ = × =∫ ∫�

For C D, y 2, dy 0, x changes from 3 to 1→ = =

( )1

3C

1 u.dr 2 2dx 8⇒ = − × =∫ ∫�

For D A, x 1, dx 0, ychanges from 2 to 1→ = =

( )C 2

1 u.dr 2x1dy 2⇒ = = −∫ ∫�

For whole loop C

u.dr 4 6 8 2 8= − + + − =∫�

53. A binary mixture of components (1) and (2) forms an azeotrope at 130OC and x1 = 0.3. The liquid

phase non-ideality is described by ln 2

1 2Axγ = and ln 2

2 1Ax ,γ = where 1 2,γ γ are the activity

coefficients, and 1 2

x , x are the liquid phase mole fractions. For both components, the fugacity

coefficients are 0.9 at the azeotropic composition. Saturated vapor pressures at 130OC are

sat

1P 70= bar sat

2P 30= bar.

The total pressure in bars for the above azeotopic system (up to two decimal places) is

Answer: 27.54

Exp: Given ( )az az az az

1 1 1x 0.3 y At a zeotrope x y= = =

sat sat

1 2P 70bar, P 30bar= =

2

1 2ln Axγ =

and 2

n 2 1l Axγ =



34

So ( ) ( )2 212 1 2 1 2 1

2

ln A x x A(x x ) x xγ

= − = − +γ

( )az11

2

ln A 1 2xγ

= −γ

_____(1)

Also

az az sat

1 1 1 1 1y P x Pφ = γ ______(2)

az az sat

2 2 2 2 2y P x Pφ = γ ______(3)

We know az az az az

1 1 2 2x y and y x= =

So, (2)/(3)

sat

1 1 1

sat

2 2 2

P1

s

φ γ⇒ = =

φ γ

sat

1 2

sat

2 1

P 30 3

P 70 7

γ⇒ = = =

γ

From (1)

( )3ln A 1 2 0.3

7= − ×

A 2.1182= −

So ( )22

1 2ln Ax 2.1182 0.7γ = = − ×

1

0.3542γ =

and ( )22

2 1ln Ax 2.1182 0.3γ = = − ×

2

0.8264γ =

Adding equation (2) and (3) ( )( )1 2 sayφ = φ ≠ φ

sat sat

2 1 1 1 2 2p p p p

2 p 0.3542 70 0.8264 30

P 27.54bar

φ + φ = γ + γφ = × + ×=

54. Air is flowing at a velocity of 3m/s perpendicular to a long pipe as shown in the figure below. The

outer diameter of the pipe is d = 6 cm and temperature at the outside surface of the pipe is

maintained at 100o C. The temperature of the air far from the tube is 30

OC.

Data for air Kinematic Viscosity, 6 2V 18 10 m s;−= × Thermal conductivity, k = 0.3 W/(m.K)

Using the Nusselt number correlation : 0.8hdNu 0.024 Re

k= = × , the rate of heat loss per unit



35

Length ( )W m from the pipe to air (up to one decimal place) is ____________________

Answer: 250.945

Exp: Given v 3m s=

2

6 2

d 6 10 m

18 10 m sec

k 0.03w mk

−

−

= ×

υ = ×=

Given

0.8

e

hdNu 0.024 R

k= =

2

6

Vd Vd 3 6 10Re 10,000

18 10

−

−

ρ × ×= = = =

µ υ ×

( )0.8

2

2

hdSo, 0.024 10,000 38.037

k

0.03 38.037h 19.0187 w m k

6 10−

= × =

×⇒ = =

×

Rate of heat loss/length

( )( )( )

S a

2

h. d T T

19.0187 6 10 100 30

250.945w m

−

= π −

= × π × × × −

=

OSurface temperature 100 C6cm

OAir velocity 3m s, Temperature 30 C



36

55. The cost of two independent process variable f1 and f2 affects the total cost CT (in lakhs of

rupees) of the process as per the following function:

2

T 1 2

1 2

1000C 100f 20f 50

f f= + + +

The lowest total cost CT, in lakhs of rupees (up to one decimal place), is ________.

Answer: 572.8

Exp: Given 2

T 1 2

1 2

1000C 100f 20f 50

f f= + +

T

2

1 2 1

C 1000P 100

f f f

∂= = =

∂

T22

2 2 1

C 1000and q 40f

f f f

∂ −= = +

∂

For maxima of minima P = 0, q = 0

22 2

2 1 1 2

1000 1000100 and 40f

f f f f= =

2

1 2f f 10.............⇒ = (1)

3

1 2f f 25............(II)⇒ =

Dividing equation (ii) by equation (1)

( ) ( )ii i

3

1 2

2

1 2

2 2

2 21

1

f f2.5

f f

f f2.5 f

f 2.5

⇒ =

⇒ = ⇒ =

From (1)

( )

( )

4

222

5

2 2

2

1

1 2

ff 10

2.5

f 62.5 f 2.2865

2.2865so, f 2.091

2.5

f 2.091, f 2.2865

⇒ =

⇒ = ⇒ =

= =

= =



37

Now, 2

T

2 3 3

1 2 1 1 2

C 1000 2 2000r

f f f f f

∂ + ×= = =

∂

2

T

2 2

1 2 1 2

2

T

2 3

2 2 1

C 1000S

f f f f

C 2000t 40

f f f

∂= = +

∂ ∂

∂= = +

∂

2rt s 0 and r 0− > >

So, ( )1 2f , f is point of minima

Lowest cost

2

T

1000C 100 2.091 20 2.865 50

2.091 2.2865= × + + × +

×

T

C 572.82lakhs=

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