|CH-GATE-2023|
www.gateforumonline.com
GENERAL APTITUDE
Q. No. 1-5 Carry One Mark Each
1.
“You are delaying the completion of the task. Send _________ contributions at the earliest.”
(A) you are
(C) you‟re
(B) your
(D) yore
Key:
(B)
Sol:
“Your are delaying the completion of task, send your contribution at the earliest.”
2.
References : : : Guidelines : Implement
(By word meaning)
(A) Sight
(B) Site
(C) Cite
(D) Plagiarise
Key:
(C)
Sol:
References: Cite :: Guidelines : Implement
3.
In the given figure, PQRS is a parallelogram with PS = 7 cm, PT = 4 cm and PV = 5 cm. What is the
length of RS in cm? (The diagram is representative.)
7
P
S
5
4
90
90
Q
(A)
20
7
V
T
(B)
R
28
5
(C)
Key:
(B)
Sol:
1
1
Area of parallelogram   PT  QR   RS  PV
2
2
1
1
 4  7  RS  5
4
2
28
RS 
5
28
RS  PQ 
5
9
2
(D)
35
4
7
P
S
5
4
90
90
Q
T
V
R
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
1
|CH-GATE-2023|
4.
www.gateforumonline.com
In 2022, June Huh was awarded the Fields medal, which is the highest prize in Mathematics.
When he was younger, he was also a poet. He did not win any medals in the International Mathematics
Olympiads. He dropped out of college.
Based only on the above information, which one of the following statements can be logically inferred
with certainty?
(A) Every Fields medalist has won a medal in an International Mathematics Olympiad.
(B) Everyone who has dropped out of college has won the Fields medal.
(C) All Fields medalists are part-time poets.
(D) Some Fields medalists have dropped out of college.
Key:
(D)
5.
A line of symmetry is defined as a line that divides a figure into two parts in a way such that each part is
a mirror image of the other part about that line.
The given figure consists of 16 unit squares arranged as shown. In addition to the three black squares,
what is the minimum number of squares that must be coloured black, such that both PQ and MN form
lines of symmetry? (The figure is representative)
P
N
M
(A) 3
Key:
(B) 4
Q
(C) 5
(D) 6
(C)
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
2
|CH-GATE-2023|
www.gateforumonline.com
Sol:
A
A
5  squares need to be colored
A
Q. No. 6-10 Carry Two Marks Each
6.
Human beings are one among many creatures that inhabit an imagined world. In this imagined world,
some creatures are cruel. If in this imagined world, it is given that the statement “Some human beings
are not cruel creatures” is FALSE, then which of the following set of statement(s) can be logically
inferred with certainty?
(i)
All human beings are cruel creatures.
(ii) Some human beings are cruel creatures.
(iii) Some creatures that are cruel are human beings.
(iv) No human beings are cruel creatures.
(A) only (i)
(B) only (iii) and (iv)
(C) only (i) and (ii)
(D) (i), (ii) and (iii)
Key:
(D)
7.
To construct a wall, sand and cement are mixed in the ratio of 3:1. The cost of sand and that of cement
are in the ratio of 1:2.
If the total cost of sand and cement to construct the wall is 1000 rupees, then what is the cost (in rupees)
of cement used?
(A) 400
Key:
(B) 600
(C) 800
(D) 200
(A)
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
3
|CH-GATE-2023|
Sol:
Concentrations
Costs
www.gateforumonline.com
sand
3
 x
cement 1
sand
1
 x
cement 2
Total cost = 1000
For x units taken
Total cost = 3x + 2x = 1000
x = 200
Cost of cement = 200 × 2 = 400
8.
The World Bank has declared that it does not plan to offer new financing to Sri Lanka, which is battling
its worst economic crisis in decades, until the country has an adequate macroeconomic policy
framework in place. In a statement, the World Bank said Sri Lanka needed to adopt structural reforms
that focus on economic stabilisation and tackle the root causes of its crisis. The latter has starved it of
foreign exchange and led to shortages of food, fuel, and medicines. The bank is repurposing resources
under existing loans to help alleviate shortages of essential items such as medicine, cooking gas,
fertiliser, meals for children, and cash for vulnerable households.
Based only on the above passage, which one of the following statements can be inferred with certainty?
(A) According to the World Bank, the root cause of Sri Lanka‟s economic crisis is that it does not have
enough foreign exchange.
(B) The World Bank has stated that it will advise the Sri Lankan government about how to tackle the
root causes of its economic crisis.
(C) According to the World Bank, Sri Lanka does not yet have an adequate macroeconomic policy
framework.
(D) The World Bank has stated that it will provide Sri Lanka with additional funds for essentials such
as food, fuel, and medicines.
Key:
(C)
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
4
|CH-GATE-2023|
9.
www.gateforumonline.com
The coefficient of x 4 in the polynomial  x  1  x  2  is equal to __________.
3
(B) –3
(A) 33
Key:
(A)
Sol:
 x  1  x  2 
3
3
(C) 30
(C) 21
3
  x 3  1  3x 2  3x  x 3  8  6x 2  12x 
  x 9  8x 3  6x 5  12x 4   x 3  8  6x 2  12x  3x 5  24x 2  18x 4  36x 3  3x 4
Counting all power of x 4  12  3  18  33
10.
Key:
Which one of the following shapes can be used to tile (completely cover by repeating) a flat plane,
extending to infinity in all directions, without leaving any empty spaces in between them? The copies of
the shape used to tile are identical and are not allowed to overlap.
(A) circle
(B) regular octagon
(C) regular pentagon
(D) rhombus
(D)
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
5
|CH-GATE-2023|
www.gateforumonline.com
CHEMICAL ENGINEERING
Q. No. 11-35 Carry One Mark Each
11.
Which one of the following is the CORRECT value of y, as defined by the expression given below?
2x
x 0 e  1
y  lim
x
(A) 1
Key:
(B)
Sol:
y  lim
(B) 2
(C) 0
(D) 
2x
x 0 e  1
x
1
1
 2  2
x 0  e x  1 
1


 x 
 2lim

 ex  1  

lim
 x 0 
  1
 x  

Option (B).
12.


ˆ
The vector v is defined as v  zxiˆ  2xyjˆ  3yzk.

Which one of the following is the CORRECT value of divergence of v, evaluated at the point
 x, y,z   3,2,1 ?
(A) 0
(B) 3
Key:
(D)
Sol:





.V  d.V V   zx    2xy    3yz 
x
y
z
(C) 14
(D) 13
 
 z  2x  3y  3,2,1
 1  2  3  3  2   13

13.
Option (B)
Given that,
F
z1  z 2
z1  z 2
,
Where z1  2  3i and z2  2  3i with i  1, which one of the following options is CORRECT?
(A) F < 0
(B) F < 1
(C) F > 1
(D) F = 1
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
6
|CH-GATE-2023|
Key:
Sol:
www.gateforumonline.com
(B)
z1  z 2  32  13 and z 2 
 2 
2
 32  13
z1  z 2   2  3i    2  3i   6i
 z1  z 2  6i  6
14.
z1  z 2

F

Option (B).
z1  z 2

6
3

1
2 13
13
 
For a two-dimensional plane, the unit vectors,  eˆ r ,eˆ   of the polar coordinate system and ˆi, ˆj of the
cartesian system, are related by the following two equations
êr  cos ˆi  sin ˆj
ê   sin ˆi  cos ˆj

Which one of the following is the CORRECT value of
(A) 1
Key:
(D)
Sol:
Given
(B) ê
  eˆ r  eˆ  

?
(C) eˆ r  eˆ 
(D) eˆ r  eˆ 
êr  cos ˆi  sin ˆj and ê   sin ˆi  cos ˆj

 
 eˆ r  eˆ    cos   sin   ˆi   sin   cos   ˆj and eˆ r  eˆ    cos ˆi  sin ˆj   sin ˆi  cos ˆj

   cos   sin   ˆi    sin   cos   ˆj

 eˆ r  eˆ      sin   cos   ˆi   cos   sin   ˆj  eˆ r  eˆ 


15.
Option (D)
Which one of the following statements related to octane number is NOT correct?
(A) Linear alkanes with higher carbon number have higher octane number.
(B) Branching in linear alkanes increases their octane number.
(C) Catalytic reforming of hydrocarbons increases their octane number.
(D) Gasoline quality is measured in terms of octane number.
Key:
(A)
Sol:
Linear alkanes with higher carbon number have higher octane number.
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
7
|CH-GATE-2023|
16.
www.gateforumonline.com
Which one of the following options represents the major components of oleum?
(A) Sulfuric acid and nitric acid
(B) Concentrated sulfuric acid and petroleum jelly
(C) Sulfuric acid and hydrochloric acid
(D) Sulfuric acid and sulfur trioxide
Key:
(D)
Sol:
Oleum = Sulfuric acid + Sulphor trioxide
17.
For a reversible endothermic chemical reaction with constant heat of reaction over the operating
temperature range, K is the thermodynamic equilibrium constant. Which one of the following figures
shows the CORRECT dependence of K on temperature T?
(A)
(B)
n(K)
n(K)
1T
1T
(C)
(D)
n(K)
n(K)
1T
Key:
(A)
Sol:
Endothermic reversible reaction,
1T

R
A 
k1
k2
k1CA  k 2 C B  0
k1CA  k 2 C B
k
k
 E RT
kD
k1 k D1 e 1

 1 e E2  E1  RT
 E 2 RT
k 2 k D2 e
k D2
k D1
k D2
nk
e
nk  n
k D1
k D2

 E 2  E1 
1T
RT
For endothermic, if temp is increased forward reaction rate will increase and vice versa
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
8
|CH-GATE-2023|
18.
www.gateforumonline.com
Nitrile rubber is manufactured via polymerization process. Which one of the following options is the
CORRECT pair of monomers used in this process?
(A) Acrylonitrile and styrene
(B) Acrylonitrile and butadiene
(C) Butadiene and styrene
(D) Butadiene and isoprene
Key:
(B)
Sol:
Nitrile rubber = Acrylonitrile + butadiene
19.
John and Jane independently performed a thermodynamic experiment, in which X and Y represent the
initial and final thermodynamic states of the system, respectively. John performed the experiment under
reversible conditions, for which the change in entropy of the system was Srev . Jane performed the
experiment under irreversible conditions, for which the change in entropy of the system was Sirr .
Which one of the following relationships is CORRECT?
(A) Srev  Sirr
(B) Srev  Sirr
Key:
(A)
Sol:
Entropy is considered as point function.
(C) Srev  Sirr
(D) Srev  2Sirr
This it only depends on initial and final stage.

20.
s rev  s irrev
For a packed-bed comprising of uniform-sized spherical particles of diameter Dp , the pressure drop
across the bed is given by the Kozeny-Carman equation when the particle Reynolds number  Rep   1.
Under this condition, minimum fluidization velocity is proportional to D np . Which one of the following
is the CORRECT value of exponent n ?
(B) −1
(A) 2
Key:
(A)
Sol:
Kozney Carmann equation
P 150 D 1   

L
Qs 2 DP 2 3
(C) −2
(D) 1
2
For minimum fluidization,
150VD 1   
P
  g  F  g 1    
L
Qs 2 Dp 2 3
2
 VD Dp 2
n2
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
9
|CH-GATE-2023|
21.
Match the quantities in Group 1 with their units in Group 2 listed in the table below.
Group 1
Key:
www.gateforumonline.com
Group 2
P.
Thermal conductivity
I.
W.m 2 K 1
Q.
Convective heat transfer coefficient
II.
W.m 1K 1
R.
Stefan-Boltzmann constant
III.
W.K 1
S.
Heat capacity rate
IV. W.m 2 K 4
(A) P-II, Q-I, R-IV, S-III
(B) P-I, Q-II, R-III, S-IV
(C) P-III, Q-IV, R-II, S-I
(D) P-IV, Q-I, R-III, S-II
(A)
Sol:
22.
P.
Thermal conductivity
II.
W / mK
Q.
Convective heat transfer coefficient
I.
W / m2 K
R.
Stefan-Boltzmann constant
IV. W / m 2 K 4
S.
Heat capacity rate
III.
W/K
A slab of thickness L, as shown in the figure below, has cross-sectional area A and constant thermal
conductivity k. T1 and T2 are the temperatures at x = 0 and x = L, respectively. Which one of the
following options is the CORRECT expression of the thermal resistance for steady-state onedimensional heat conduction?
T2
T1
0
L
x
(A)
L
kA
(B)
k
LA
(C)
kA  T1  T2 
L
(D)
A
Lk
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
10
|CH-GATE-2023|
Key:
(A)
Sol:
Q  kA
dT
dx
dT  T2  T1 
Q  kA

L
L kA
 Resistance 
23.
www.gateforumonline.com
L
kA
Spray dryers have many advantages. Which one of the following is NOT an advantage of a typical spray
dryer?
(A) Has short drying time
(B) Produces hollow spherical particles
(C) Has high heat efficiency
(D) Is suitable for heat sensitive materials
Key:
(C)
24.
Which one of the following quantities of a flowing fluid is measured using a rotameter?
(A) Static pressure
(B) Dynamic pressure
(C) Volumetric flow rate
(D) Viscosity
Key:
(C)
25.
A liquid surge tank has Fin and Fout as the inlet and outlet flow rates respectively, as shown in the figure
below. Fout is proportional to the square root of the liquid level ℎ. The cross-sectional area of the tank is
20 cm2. Density of the liquid is constant everywhere in the system. At steady state, Fin  Fout  10 cm 2s 1
and h = 16 cm. The variation of h with Fin is approximated as a first order transfer function. Which one
of the following is the CORRECT value of the time constant (in seconds) of this system?
Fin
h
Fout
(A) 20
(B) 20
(C) 64
(D) 128
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
11
|CH-GATE-2023|
Key:
(C)
Sol:
Fin  Fout  A
dh
ht
Adh
Fin  k n 
dt
 h
h
Fin  k  0 
 2
2 h0


dh
A

dt

Fin 
k h0
Fin 
k
dh k h 0
…(1)
hA 
dt
2
2 h0
2

kh
A
2 h0
h  hD   h  h 0  
h  h0 
h
2 hD
h  hD 
hD
h
2

2 hD
dh
dt
1
2 h0
hD

h
2 hD

hD

hD
h
2 hD
www.gateforumonline.com
2
2

h hD
2h D
Adh s
hD
h


0
2 2 hD
dt
At s . Fin  s  k  
…(2)
Substracting above two equation
Fin 
kh
dh
A
dt
2 hD
Fin  Fout  10 cm3 sec
h s  16 cm
k h s  10
k  2.5
Fin  Ash  s  
h s
Fin  s 

kh  s 
2 hD
1
k
As 
2 hD

1
2.5
20s 
8

p
p s  1
 p 
20  8
 64
2.5
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
12
|CH-GATE-2023|
26.
www.gateforumonline.com
A packed distillation column, with vapor having an average molecular weight of 45 kg.kmol1 , density
of 2 kg.m 3 and a molar flow rate of 0.1 kmol.s 1 , has a flooding velocity of 0.15 m.s 1. The column is
designed to operate at 60 % of the flooding velocity. Which one of the following is the CORRECT value
for the column diameter (in m)?
(A)
5

Key:
(D)
Sol:
 r  0.1 kmol sec
m
(B) 5 
(C) 4
(D)
10

M  45 kg kmol
Vf  0.15 m s
 v  2 kg m 3
VDp  0.6  0.15
 r  v  AN  Voperating
m
0.1

 d2
2  0.6  0.15 4
4  0.1
100
d2 

2  0.6  0.15  T1 T1
10
d

AN 
27.
An isothermal jacketed continous stirred tank reactor (CSTR) operating at 150°C is shown in the figure
below. The cold feed entering the system at 30°C is preheated to a temperature T  T  150C using a
heat exchanger HX1. This preheated feed is further heated to 150°C using the utility heater HX2 The
mass flow rate and heat capacity are same for all the process streams, and the overall heat transfer
coefficient is independent of temperature. Which one of the following statements is the CORRECT
action to take if it is desired to increase the value of T ?
Cold 30
feed
HX1
TC
HX2
150C
CSTR
150C
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
13
|CH-GATE-2023|
www.gateforumonline.com
(A) Increase both heat transfer area of HX1 and heat duty of HX2 .
(B) Decrease both heat transfer area of HX1 and heat duty of HX2
(C) Increase the heat transfer area of HX1 and decrease the heat duty of HX2
(D) Decrease the heat transfer area of HX1 and increase the heat duty of HX2
Key:
(C)
Sol:
If we increase HT area of Hx1 , cold feed outlet temperature (T) will increase and thus to maintained
increase (T) We need to reduce Heat duty of Hx 2 .
28.
Consider a system where a Carnot engine is operating between a source and a sink. Which of the
following statements about this system is/are NOT correct?
(A) This engine is reversible.
(B) The engine efficiency is independent of the source and sink temperatures.
(C) This engine has the highest efficiency among all engines that operate between the same source and
sink.
(D) The total entropy of this system increases at the completion of each cycle of the engine.
Key:
(B. D)
Sol:
The engines independent of source and sink temperatures.
Total entropy of system increase at the completion of each cycle of the engine.
29.
For a fully developed turbulent flow of an incompressible Newtonian fluid through a pipe of constant
diameter, which of the following statements is/are CORRECT?
(A) Reynolds stress, averaged over a sufficiently long time, is zero everywhere inside the pipe.
(B) Reynolds stress at the pipe wall is zero.
(C) Average velocity of the fluid is half of its center-line velocity.
(D) Average pressure gradient in the flow direction is constant.
Key:
(B, D)
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
14
|CH-GATE-2023|
30.
www.gateforumonline.com
Given that E  in W.m2  is the total hemispherical emissive power of a surface maintained at a certain
temperature, which of the following statements is/are CORRECT?
(A) E does not depend on the direction of the emission.
(B) E depends on the viewfactor.
(C) E depends on the wavelength of the emission.
(D) E does not depend on the frequency of the emission.
Key:
(A, D)
31.
The position x(t) of a particle, at constant , is described by the equation.
d2 x
 2 x.
dt 2
The initial conditions are x  t  0   1 and
dx
dt
t 0
 3 
 0. The position of the particle at t    is _______
 
(in integer).
Key:
(–1)
Sol:
D. E is
d2 x
d
 2 x   D2  2  x  0, where D 
2
dt
dt
A.E is m 2  2  0  m 2  2  n  i

  i

Solution is x  C.F  e0t  C1 cos t  C2 sin t 
 x  C1 cos t  C2 sin t
…(1)
dx
 C1 sin t  C2 cos t
dt
…(2)
Using the initial conditions x = 1 at t = 0 and
dx
 0 at t  0
dt
(1) and (2)  1  C1 and 0  C2  C2  0

x  cos t
At t 

3
 x  cos3  1

The position of the particle at t 
3
is –1.

© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
15
|CH-GATE-2023|
32.
www.gateforumonline.com
Burning of methane in a combustor yields carbon monoxide, carbon dioxide, and water vapor. Methane
is fed to the combustor at 100 mol.hr 1 , of which 50% reacts. The theoretical oxygen requirement (in
mol. hr 1 ) is _________ (rounded off to one decimal place).
Key:
(199.5 to 200.5)
Sol:
CH 4  2 2  CO 2  2H 2 O
3
CH 4  O 2  CO  2H 2 O
2
Theoretical oxygen requirement is amount of O2 needed for complete.
Combustion of full fed to combustion reaction.
For 100 mole of CH4  200 mole of O2 required.
33.
The viscosity of incompressible Newtonian fluid is measured using a capillary tube of diameter 0.5 mm
and length 1.5m. The fluid flow is laminar, steady and fully developed. For a flow rate of 1 cm3s 1 , the
pressure drop across the length of the tube is 1 MPa. If the viscosity of the fluid is k 103 Pa.s, the
value of k is ____________ (rounded off to two decimal places).
Key:
(1.02)
Sol:
d  0.5 mm
L  1.5 m
P  1 MPa
  1 cm3 s
  k  103 Pa.s
P 128

L
d 4
3
P d 4 1  106    0.5  10 




 1.022  103 Pa.s
L 128
1.5
128  1  106
4
34.
A liquid L containing a dissolved gas S is tripped in a countercurrent operation using a pure carrier gas
V. The liquid phase inlet and outlet mole fractions of S are 0.1 and 0.01, respectively. The equilibrium
distribution of S between V and L is governed by ye  x e , where xe are the mole fractions of S in V
and L, respectively. The molar feed rate of the carrier gas stream is twice as that of the liquid stream.
Under dilute solution conditions, the minimum number of ideal stages required is __________ (in
integer).
Key:
(3)
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
16
|CH-GATE-2023|
Sol:
www.gateforumonline.com
Stripping factor = 2
X0  0.1
 x D  y n 1 m  1  1 
1    
x y
s s
n
n 1 m 

N P  log


log s




 0.1  0  1  1 
 0.01  0 1  2   2 



 log 
log 2





35.
log 5.5
log 2
yn 1  0
 2.46  3 stage
y0  X e , L 
XN  0.01
V
2
In binary gas-liquid system, N A,EMD is the molar flux of a gas A for equimolar counter diffusion with a
liquid B. N A,UMD is the molar flux of A for steady one-component diffusion through stagnant B. Using
the mole fraction of a in the bulk of the gas phase as 0.2 and that at the gas-liquid interface as 0.1 for
both the modes of diffusion, the ratio of N A,UMD to N A,EMD is equal to _______ (rounded off to two
decimal places).
Key:
Sol:
(1.18)
N A, UMD
N A EMD



D AB  y A1  y A2   P, RTZ
ZRT D AB P   y A1  y A2  y Bm
1
y Bm
1
1

 1.176  1.18
y B1  y B2 1  0.2   1  0.1
y
0.8
n B1
n
y B2
0.9
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
17
|CH-GATE-2023|
www.gateforumonline.com
Q. No. 36- 65 Carry Two Marks Each
36.
An exhibition was held in a hall on 15 August 2022 between 3 PM and 4 PM during which any person
was allowed to enter only once. Visitors who entered before 3:40 PM exited the hall exactly after 20
minutes from their time of entry. Visitors who entered at or after 3:40 PM, exited exactly at 4 PM. The
probability distribution of the arrival time of any visitor is uniform between 3 PM and 4 PM. Two
persons X and Y entered the exhibition hall independent of each other. Which one of the following
values is the probability that their visits to the exhibition overlapped with each other?
5
9
(A)
Key:
(B)
4
9
(C)
2
9
(D)
7
9
(A)
Sol:
15 Aug 2022
3 PM
4 PM
Visitors entry time
< 3.40 Pm / = 3.40 PM / > 3.40 PM
Leaves at 20 mins exactly
Total number of possible cases of X and Y entry
X 3.40
Y3.40
Y  3.40
Y  3.40
Similarly for X 3.40 and X 3.40 6 more cases will be there
Probability of entry overlapping 
8 1

9 3
Probability of their exits overlapping 
37.
4
9
Simpson‟s one-third rule is used to estimate the definite integral
1 
1
1
1  x  dx
2
with an interval length of 0.5. Which one of the following is the CORRECT estimate of I obtained using
this rule?
(A)
Key:
1 1

3
3
(B)
1 2

3
3
(C)
1 1

3
3
(D)
1 2

3
3
(B)
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
18
|CH-GATE-2023|
Sol:
www.gateforumonline.com
We have f  x   1  x 2 , a  1, b  1and h  0.5
 y 0  f  a   f  1  1   1  0
2
y1  f  a  h   f  0.5   1 
1
3
3


4
4
2
y 2  f  a  2h   f  0   1  0  1
y3  f  a  3h   f  0.5   1 
1
3

4
2
y 4  f  b   f 1  1  1  0

By Simpson‟s one-third rule,
1

1  x 2 dx 
1
h
 y 0  y 4   4  y1  y3   2y 2 
3

 3
1
3
  0  0   4 

  2 1 
6 
2 
 2

1
2 4 3
  4 3  2   
6
6
6
1 2 3 1 2
 
 
3
3
3
3

38.
Option (B).
Match the products in Group 1 with the manufacturing processes in Group 2 listed in the table below.
Group 1
Key:
Group 2
P.
Acetaldehyde
I.
Sulfate process
Q.
Sulfuric acid
II.
Electric furnace process
R.
Pulp
III. Wacker process
S.
Phosphorus
IV. Contact process
(A) P-III, Q-IV, R-I, S-II
(B) P-III, Q-I, R-IV, S-II
(C) P-IV, Q-I, R-II, S-III
(D) P-I, Q-IV, R-II, S-III
(A)
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
19
|CH-GATE-2023|
www.gateforumonline.com
Sol:
39.
P.
Acetaldehyde
III. Wacker process
Q.
Sulfuric acid
IV. Contact process
R.
Pulp
I.
Sulfate process
S.
Phosphorus
II.
Electric furnace process
Match the reactions in Group 1 with the catalysts in Group 2 listed in the table below.
Group 1
Group 2
P. C6 H6  C 2  Chlorobenzene  HC
I.
1
Q. H 2 C  CH 2  O 2  Ethylene oxide
2
II. V2 O5
1
R. CH3OH  O 2  Formaldehyde  H 2 O
2
III. FeC 3
9
S. Naphthalene  O2  Phthalic Anhydride  2H 2 O  2CO 2
2
IV. Ag2O
Mixed oxide of Mo and Fe
(A) P-III, Q-IV, R-II, S-I
(B) P-III, Q-IV, R-I, S-II
(C) P-IV, Q-II, R-I, S-III
(D) P-IV, Q-III, R-I, S-II
Key:
(B)
40.
Water in a container at 290 K is exposed to air containing 3 % CO2 by volume. Air behaves like an
ideal gas and is maintained at 100 kPa pressure. The liquid phase comprising of dissolved CO2 in water
behaves like an ideal solution. Use Henry‟s constant of CO2 dissolved in water at 290 K as 12 MPa.
Under equilibrium conditions, which one of the following is the CORRECT value of the mole fraction
of CO2 dissolved in water?
(A) 2.9  104
Key:
(B) 0.9  104
(C) 2.5  104
(D) 0.5  104
(C)
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
20
|CH-GATE-2023|
41.
www.gateforumonline.com
The enthalpy (H, in J.mol1 ) of a binary liquid system at constant temperature and pressure is given as
H  40x1  60x 2  x1x 2  4x1  2X2  ,
Where x1 and x 2 represent the mole fractions of species 1 and 2 in the liquid, respectively. Which one
of the following is the CORRECT value of the partial molar enthalpy of species 1 at infinite dilution,
H1 (in J.mol1 )?
(A) 100
(B) 42
Key:
(B)
Sol:
H  40x1  60x 2  x1x 2  4x1  2x 2 
H1  H 
(C) 64
(D) 40
x 2 dH
dx1
 d

H1   40x1  60x 2  x1x 2  4x1  2x 2    x 2 
40x1  60 1  x 2   x1 1  x1  4x1  2  2x1   

 dx1

dH
 40  60  1  2x1  2  2x1    x1  x12   2 
dx1
dH
dx1
 40  60  1  2  0  20  2  18
x1  0
H1  40x1  60x 2  x1x 2  4x1  2x 2   x 2  18
Put x1  0, x 2  1
H1  60  18  42
42.
Which one of the following represents the CORRECT effects of concentration polarization in a reverse
osmosis process?
(A) Reduced water flux and reduced solute rejection
(B) Increased water flux and increased solute rejection
(C) Reduced water flux and increased solute rejection
(D) Increased water flux and reduced solute rejection
Key:
(A)
Sol:
Concentration polarization reduces flux of water because the increase in osmotic pressure reduces the
driving force for water transport and solute rejection because of lower water flux.
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
21
|CH-GATE-2023|
43.
www.gateforumonline.com
CO and H 2 participate in a catalytic reaction. The partial pressures (in atm) of the reacting species CO
and H 2 in the feed stream are pCO and pH2 , respectively. While CO undergoes molecular adsorption, H2
adsorbs via dissociative adsorption, that is, as hydrogen atoms. The equilibrium constants (in atm 1 )
corresponding to adsorption of CO and H 2 to the catalyst sites are KCO and KH2 , respectively. Total
molar concentration of active sites per unit mass of the catalyst is Ct (in mol. (g cat) 1 ) . Both the
adsorption steps are at equilibrium. Which one of the following expressions is the CORRECT ratio of
the concentration of catalyst sites occupied by CO to that by hydrogen atoms?
(A)
K CO pCO
K CO
(B)
K H2 p H2
(C)
K H2
pCO
(D)
p H2
K CO p CO
K H2 p H2
Key:
(A)
44.
A cascade control strategy is shown in the figure below. The transfer function between the output (y)
and the secondary disturbance (d2 ) is defined as
G d2  s  
y s
d2 s 
.
Which one of the following is the CORRECT expression for the transfer function Gd2  s  ?
yset (s)



Key:


1

10
d 2 (s)
d1 (s)
1
0.1s  1
1
2s  1


1
s 1
(A)
1
11s  21 0.1s  1
(B)
1
 s  1 0.1s  1
(C)
 s  1
 s  2  0.1s  1
(D)
 s  1
 s  1 0.1s  1


y(s)
(A)
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
22
|CH-GATE-2023|
45.
www.gateforumonline.com
Level (ℎ) in a steam boiler is controlled by manipulating the flow rate (F) of the make-up (fresh) water
using a proportional (P) controller. The transfer function between the output and the manipulated input is
h s
Fs

0.25 1  s 
s  2s  1
.
The measurement and valve transfer functions are both equal to 1. A process engineer wants to tune the
controller so that the closed-loop response gives decaying oscillations under servo mode. Which one of
the following is the CORRECT value of the controller gain to be used by the engineer?
(A) 0.25
(B) 2
(C) 4
Key:
(B)
46.
Which of the following statements is/are CORRECT?
(D) 6
(A) Bond number includes surface tension.
(B) Jakob number includes latent heat.
(C) Prandtl number includes liquid-vapor density difference.
(D) Biot number includes gravity.
Key:
47.
(A, B)
10 6 
3
If a matrix M is defined as M  
 , the sum of all the eigen values of M is equal to ________
6
10


(in integer).
Key:
(4160)
Sol:
The characteristic equation of M is
10  
6
0
6
10  
 10     36  0
2
 10     36  10    6
2
 10    6;10    6
   4,16 are the eigen values of M
 Eigen values of M3 are 43 ,163  64, 4096
(If  is an eigen value of matrix A than 3 is an eigen value of matrix A3 )

Required sum is 64 + 4096 = 4160
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
23
|CH-GATE-2023|
48.
www.gateforumonline.com
The first derivative of the function
 1 12  1 6 
U  r   4      
 r  
 r 
Evalued at r = 1 is ___________ (in integer).
Key:
(–24)
Sol:
 1 12  1 6 
u  r   4        4  r 12  r 6 
 r  
 r 
Differentiate both sides w.r.t „r‟ we get
 u '  r   4  12  r 13  6r 7 
 d n
n 1 
  x   nx 
 d

At r =1, u '  r   4  12  6  24

49.
The first derivative of the given function at r = 1 is –24.
Wet air containing 10 mole percent water vapor is dried by continuously passing it through a column of
CaC 2 pellets. The pellets remove 50 percent of water from wet air entering the column. The mole
percent of water vapor in the product stream exiting the column is (rounded off to two decimal places).
Key:
(5.3)
Sol:
Basis, 100 moles of wet air is taken initially
H 2 O  10 moles
Air  90 moles
After removes of 50% of moisture total moles at exit
= 5 moles H2O  90 mole air
= 95
Mole fraction of water 
5
 0.053  5.3%
95
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
24
|CH-GATE-2023|
50.
www.gateforumonline.com
Orsat analysis showing the composition (in mol %, on a dry basis) of a stack gas is given in the table
below. The humidity measurement reveals that the mole fraction of H2O in the stack gas is 0.07. The
mole fraction of N2 calculated on a wet basis is (rounded off to two decimal places).
Species
N2
CO2
CO
O2
mol %
65
15
10
10
Key:
(0.61)
Sol:
Basis, 100 moles
N 2  65
CO2  15
CO  10
O 2  10 moles
x  moles of H 2 O
Mole fraction of water 
x
 0.07
100  x
x 1  0.07   7
x
7
 7.527 moles
0.93
Mole fraction of N 2 
51.
65
 0.6045  0.61
100  7.527
A pump draws water (density  1000 kg.m 3 ) at a steady rate of 10 kg.s 1. The pressures at the suction
and discharge sides of the pump are −20 kPa (gauge) and 350 kPa (gauge), respectively. The pipe
diameters at the suction and discharge side are 70 mm and 50 mm, respectively. The suction and
discharge lines are at the same elevation, and the pump operates at an efficiency of 80 %. Neglecting
frictional losses in the system, the power (in kW) required to drive the pump is __________ (rounded off
to two decimal places).
Key:
(4.74)
Sol:
  10 kg s
m
 w  1000 kg m 3
PS  20 kPa
PD  350 kPa
z1  z 2
d1  70 mm
d 2  50 mm
  0.8
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
25
|CH-GATE-2023|
www.gateforumonline.com
  A1V1  2 A 2 V2
m
1  2
A1V1  A 2 V2
2
10

  70  103   V1
1000 4
V1  2.6 m s
2
10

  50  103   V2
1000 4
V2  5.1 m s
P1 V12
P V2

 z1  Hp  2  2  z 2
g 2g
g 2g
20  103  350  103  2.6    5.1

 H p
1000  3.81
2  9.81
379.625
Hp 
9.81
gHpQ 1000  379.625  10
Power 

 4.74 kW

1000  0.8
2
52.
2
A cylindrical tank with a diameter of 500 mm contains water (density  1g.cm3 ) upto a height ℎ. A 5
mm diameter round nozzle, whose center is 1 cm above the base of the tank, has its exit open to the
atmosphere as shown in the schematic below. The pressure above the water level in the tank is
maintained at 2 bar (absolute). Neglect all frictional and entry/exit losses. Use acceleration due to
dh
(in mm.s 1 ) when
gravity as 10 ms 2 and atmospheric pressure as 1 bar. The absolute value of
dt
h = 51 cm is equal to __________ (rounded off to two decimal places).
2 bar (abs)
500 mm
h cm
5 mm
1 cm
Not to scale
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
26
|CH-GATE-2023|
Key:
(1.38)
Sol:
w  1 gm cm3
h  51 cm
g  10
V  2gh
Patm  1 bar
V  P1.Patm  2gh
www.gateforumonline.com
V  2  1  2  10  0.51
V  6.162 m s
dh
A
dt
0.121
dh


2
dt
 0.51
4
AV 

53.
dh
 1.381 mm sec
dt
A large tank is filled with water (density  1 g.cm3 ) upto a height of 5 m. A 100 m diameter solid
spherical particle (density  0.8 g.cm3 ) is released at the bottom of the tank. The particle attains its
terminal velocity  vt  after traveling to a certain height in the tank. Use acceleration due to gravity as
103 m.s 2 and water viscosity as 103 Pa.s. Neglect wall effects on the particle. If Stokes law is
applicable, the absolute value of (in mm.s 1 ) ____________ is (rounded off to two decimal places).
Key:
(1.1)
Sol:
Using formula of terminal velocity in stokes region
Vt 
Vt 
d 2p  p  f  g
18 f
100 10  100  800   9.81

6
18  103
108  200  9.81
 1.09 mm sec  1.1 mm sec
18  103
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
27
|CH-GATE-2023|
54.
www.gateforumonline.com
A fluid is flowing steadily under laminar conditions over a thin rectangular plate at temperature Ts as
shown in the figure below. The velocity and temperature of the free stream are u  and T , respectively.
When the fluid flow is only in the x-direction, h x is the local heat transfer coefficient. Similarly, when
the fluid flow is only in the y-direction, h y is the corresponding local heat transfer coefficient, where,
Nu, Re, and Pr, respectively are the appropriate Nusselt, Reynolds and Prandtl numbers. The average
1 
1 w
heat transfer coefficients are defined as h    h x dx and h w   h y dy. If w = 1 m and   4m, the
 0
w 0
value of the ratio of h w to h  is ___________ (in integer).
y
w
TS

x
Key:
(2)
Sol:
NU x  0.332 Re x1/ 2 Pr1/3
1/ 2
1/3
 V 
h xx
 0.332   x  Pr
k
  
1
hx 
x
A
hx 
x
hx 
1 x A
dx
x 0 x
1/2
1
x
2A
hx  A 

x
1/ 2
x
  1m
  4m
h   2A
h 
2A
A
2
h
2
h
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
28
|CH-GATE-2023|
55.
www.gateforumonline.com
A perfectly insulated, concentric tube countercurrent heat exchanger is used to cool lubricating oil using
water as a coolant (see figure below). Oil enters the outer annulus at a mass flow rate of 2 kg.s 1 with a
temperature of 100°C and leaves at 40°C. Water enters the inner tube at a mass flow rate of 1 kg.s 1 with
a temperature of 20°CC and leaves at 80°C. Use specific heats of oil and water as 2089 J.kg 1 K 2 and
4178 J.kg 1K 1 , respectively. There is no phase change in both the streams. Under steady-state
conditions, the number of transfer units (NTU) is ________ (in integer).
40C
100C
20C
80C
Oil
Water
Key:
(3)
Sol:
   1 kg sec
m
Cp  4178 kg k
UA
Cmin
NTU 
From energy balance
  100  40    mC
   80  20 
 mC
  100  40   uAT
 mC
p n
p c
p n
m
 p
mC
uA  20
60
uA 60

3
 p 20
mC
NTU  3
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
29
|CH-GATE-2023|
56.
www.gateforumonline.com
Partially saturated air at 1 bar and 50°C is contacted with water in an adiabatic saturator. The air is
cooled and humidified to saturation, and exits at 25°C with an absolute humidity of 0.02 kg water per kg
dry air. Use latent heat of vaporization of water as 2450 kJ.kg 1 , and average specific heat capacity for
dry air and water, respectively as 1.01 kJ.kg 1K 1 and 4.18 kJ.kg 1K 1. If the absolute humidity of air
entering the adiabatic saturation is H  103 kg water per kg dry air, the value of H is _________
(rounded off to two decimal places).
Key:
(9.29)
Sol:
TDB  50C
TW  25
A  2450 kJ kg
C pw  4.18
C p air  1.01
Yw  0.02 kgH 2 O kg dry air
For the humid heat we have.
Cs  1.01  4.18 y 2
TD  Tw 
50  25 
  yw  y ' 
CS
2450  0.02  y '
1.01  4.18y '
Solving for y '  9.29  10 3
H  9.29
Value vary depend upon assumption Cpw .
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
30
|CH-GATE-2023|
57.
www.gateforumonline.com
Distillation of a non-reactive binary mixture with components A and B is carried out in a batch still as
shown in the figure below. The initial charge of the mixture in the still is 1 kmol. The initial and final
amounts of A in the still are 0.1 kmol and 0.01 kmol, respectively. Use a constant relative volatility of
4.5. The mole fraction of B remaining in the vessel is _______ (rounded off to three decimal places).
Condenser
Product receiver
Key:
(0.97)
Sol:
F  1 kmol
Batch still
X F  0.1 k.mol
X W  0.01 kmol
 AB  4.5
n
F 1  X P 
FX E
  n
WX W
W 1  X W 
n
1  0.1
1  0.9
 4.5 n
W  0.01
W 1  0.01
10 0.65122

W
W 4.5
W 3.5  0.065122
W  0.4582
WX W  0.01
0.01
 0.02182
0.4582
X B  1  X W  0.97817
XW 
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
31
|CH-GATE-2023|
58.
www.gateforumonline.com
Fresh catalyst is loaded into a reactor before the start of the following catalytic reaction.
A  products
The catalyst gets deactivated over time. The instantaneous activity (𝑡), at time 𝑡, is defined as the ratio of
the rate of reaction rA (t) (mol.(g cat) 1 hr 1 ) to the rate of reaction with fresh catalyst. Controlled
experimental measurements led to an empirical correlation rA (t)  0.5t  10 where t is in hours. The
activity of the catalyst at t = 10 hr is _______ (rounded off to one decimal place).
Key:
(0.5)
Sol:
A  Products
 rA '  0.5 t  10
Activity 
 A 0.5t  10

 0
10
At t = 10
Activity 
a
59.
0.5  10  10
10
10  5
 0.5
10
A unimolecular, irreversible liquid-phase reaction
AP
Was carried out in an ideal batch reactor at temperature T. The rate of the reaction (rA ) measured at
different conversions XA is given in the table below. This reaction is also carried in an ideal continuous
stirred tank reactor (CSTR) at the same temperature T with a feed concentration of 1 mol.m3 , under
steady-state conditions. For a conversion of 0.8, the space time (in s) of the CSTR is ___________ (in
integer).
XA
0
0.1
0.2
0.4
0.6
0.8
-rA (mol.m-3s-1 )
0.45
0.35
0.31
0.18
0.11
0.05
Key:
(16)
Sol:
FADX A   A X V

V CAD X A

VD
 A
1  0.8
0.05
  16 sec

© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
32
|CH-GATE-2023|
60.
www.gateforumonline.com
An irreversible liquid-phase second-order reaction
k
A 
B
With rate constant k  0.2 liter.mol1 min 1 , is carried out in an isothermal non-ideal reactor. A tracer
experiment conducted on this reactor resulted in a residence time distribution (E-curve) as shown in the
figure below. The areas of the rectangles (i), (ii) and (iii) are equal. Pure A at a concentration of
1.5 mol.liter 1 is fed to the reactor. The segregated model mimics the nonideality of this reactor. The
percentage conversion of A at the exit of the reactor is _____ (rounded off to the nearest integer).
E(t) (min 1 )
0
5
10
15
t(min)
Key:
(71 to 73)
61.
The outlet concentration C A of a plug flow reactor (PFR) is controlled by manipulating the inlet
concentration CA0 . The following transfer function describes the dynamics of this PFR.
CA  s 
 V

 exp      k  s 
CA0  s 
 F

In the above equation, V  1 m3 , F  0.1 m3 min 1 and k  0.5 min 1 . The measurement and valve transfer
functions are both equal to 1. The ultimate gain, defined as the proportional controller gain that produces
sustained oscillations, for this system is ____________ (rounded off to one decimal places).
Key:
(148.0 to 148.8)
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
33
|CH-GATE-2023|
62.
www.gateforumonline.com
The transfer function of a measuring instrument is
Gm s 
1.05
exp  s  .
2s  1
At time t = 0, a step change of +1 unit is introduced in the input of this instrument. The time taken by the
instrument to show an increase of 1 unit in its output is ________ (rounded off to two decimal palces).
Key:
(6.99 to 7.19)
Sol:
Gm  
1.05
exp  s 
2s  1
Y  s  1.05

exp  s 
X  s  2s  1
1  1.05   s
Y s   
e
s  2s  1 
Taking inverse Laplace
y  t   1.05  1 1  e  t  t  h 
1
 1  e  t  t  2
1.05
1
e  t  t  2  1 
1.05
t  1
 1.05  1 
 ln 

2
 1.05 
 0.05  t
0.5n 

 1.05  2
t  7.073
63.
Key:
A design engineer needs to purchase a membrance module (M) for a plant. Details about the two
available options, M1 and M2, are given in the table below. The overall plant has an expected life to 7
years. If the interest rate is 8% per annum, compounded annually, the difference in the net present value
(NPV) of these two options, in lakhs of rupees, is ________ (rounded off to one decimal place).
M1
M2
Purchase cost (in lakhs of rupees)
10
5
Expected life (years)
5
3
(4.7)
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
34
|CH-GATE-2023|
www.gateforumonline.com
64.
The purchase cost of a new distillation column is Rs. 10 lakhs with an installation factor of 5.8. The cost
of the capital is to be annualized over a period of 6 years at a fixed rate of interest of 5 % per annum,
compounded annually. The annual cost (in lakhs of rupees) of the installed capital is _______ (rounded
off to one decimal place).
Key:
(11.72)
Sol:
Annual cost  10 
65.
Pumps A and B are being considered for purchase in a chemical plant. Cost details for these two pumps
are given in the table below. The interestl rate is 10% per annum, compounded annually. For both the
pumps to have the same capitalized cost, the salvage value (in Rs.) of pump B should be ________
(rounded off to the nearest integer).
10
 11.72 lakhs
58
Item
Installed cost (Rs)
Uniform end of
maintenance (Rs.)
Salvage value (Rs)
Service life (year(s))
year
Pump A
Pump B
16000
32000
2400
1600
1000
?
1
2
Key:
(2180)
Sol:
Calculating capital cost for both A and B


V V V 
V V
V 
 V0  D 1S  m    V0  0  S  m 
i 

1  i 
1  i   1 i  B

n
16000 
32000  Vs 1600
16000  1000 2400

 32000 

2
1.1  1
0.1
1.1  1 0.1
16000  150000  24000  32000  16000 
32000  VS
0.21
32000  VS
 142000
0.21
VS  2180
© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission
35