Chemical Engineering GATE 2016 Solution · Chemical Engineering GATE 2016 Solution Visit us at | New Batches are starting for GATE 2017 from Feb 14th & March 5th 1 General Aptitude

Chemical Engineering GATE 2016 Solution

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Visit us at wwwthegatecoachcom| New Batches are starting for GATE 2017 from Feb 14th amp March 5th 1

General Aptitude

Q 1 ndash Q 5 carry one mark each

Q1 The volume of a sphere of diameter 1 unit is ________ than the volume of a cube of side 1 unit

(A) least (B) less (C) lesser (D) low

Ans (B) less

Q2 The unruly crowd demanded that the accused be _____________ without trial

(A) hanged (B) hanging (C) hankering (D) hung

Ans (A) hanged

Q3 Choose the statement(s) where the underlined word is used correctly

(i) A prone is a dried plum

(ii) He was lying prone on the floor

(iii) People who eat a lot of fat are prone to heart disease

(A) (i) and (iii) only (B) (iii) only (C) (i) and (ii) only (D) (ii) and (iii) only

Ans (D) (ii) and (iii) only

Q4 Fact If it rains then the field is wet

Read the following statements

(i) It rains

(ii) The field is not wet

(iii) The field is wet

(iv) It did not rain

Which one of the options given below is NOT logically possible based on the given fact

(A) If (iii) then (iv) (B) If (i) then (iii)

(C) If (i) then (ii) (D) If (ii) then (iv)

Ans (C) if (i) then (ii)



Q5 A window is made up of a square portion and an equilateral triangle portion above it The base of

the triangular portion coincides with the upper side of the square If the perimeter of the window

is 6 m the area of the window in m2 is ___________

(A) 143 (B) 206 (C) 268 (D) 288

Ans (B) 206

Solution

2 2

6

5

3 3 6 6 6 6 3 36 362066

4 4 5 5 5 5 4 25 25

side

total area a a

Q 6 ndash Q 10 carry two marks each

Q6 Students taking an exam are divided into two groups P and Q such that each group has the same

number of students The performance of each of the students in a test was evaluated out of 200

marks It was observed that the mean of group P was 105 while that of group Q was 85 The

standard deviation of group P was 25 while that of group Q was 5 Assuming that the marks were

distributed on a normal distribution which of the following statements will have the highest

probability of being TRUE

(A) No student in group Q scored less marks than any student in group P

(B) No student in group P scored less marks than any student in group Q

(C) Most students of group Q scored marks in a narrower range than students in group P

(D) The median of the marks of group P is 100

Ans (C) Most students of group Q scored marks in a narrower range than students in group P



Q7 A smart city integrates all modes of transport uses clean energy and promotes sustainable use of

resources It also uses technology to ensure safety and security of the city something which critics

argue will lead to a surveillance state

Which of the following can be logically inferred from the above paragraph

(i) All smart cities encourage the formation of surveillance states

(ii) Surveillance is an integral part of a smart city

(iii) Sustainability and surveillance go hand in hand in a smart city

(iv) There is a perception that smart cities promote surveillance

(A) (i) and (iv) only (B) (ii) and (iii) only

(C) (iv) only (D) (i) only

Ans (B) (ii) and (iii) only

Q8 Find the missing sequence in the letter series

B FH LNP _ _ _ _

(A) SUWY (B) TUVW (C) TVXZ (D) TWXZ

Ans (C) TVXZ

Q9 The binary operation is defined as a b = ab+(a+b) where a and b are any two real numbers

The value of the identity element of this operation defined as the number x such that a x = a

for any a is

(A) 0 (B) 1 (C) 2 (D) 10

Ans (A) 0



Q10 Which of the following curves represents the function

Here x represents the abscissa and represents the ordinate

(A)

(B)

(C)

(D) Ans (C)



Chemical Engineering


Ans (C) Gauss - Seidel

Ans (A)

Solution

sinatL e bt

= 2 2( )

b

s a b

[By first shifting property]

Ans (D)

Solution

2 2 5r z a b and 1 1 4tan tan

3

b

a



Ans (B)

Solution

Stream P is splitting into Q amp R hence their composition will be same

Overall mass balance

P = Q + R

Q + R = 100 (1)

Doing component mass balance

Ethanol in P = Ethanol in Q + Ethanol in R

P X 03 = Q X 03 + R X 03

P = Q + R (2)

So it is obvious we cannot use component balance over splitter so least

parameter required is flow rate of any of 2 exit streams

So ie = 1

Ans ndash 58 kJ mol



Solution

For infinitely dilute solution of component (1)

1 20 amp 1x x

So 2 21 2 60 (1) 100 0 (1)h x x x

1 58h

Ans (C) 348

Solution We know that

211 4

2

f VP

g gd

21 1

1

02

2222

2 2 2022

1 1 1 211

2 022 2

2 021 1

18

22

1

4

2

348

f VP

d

V dm V

P f V

P f V V dm V

P V

P V

VP P Kpa

V



Q7 In a cyclone separator used for separation of solid particles from a dust laden gas the separation

factor is defined as the ratio of the centrifugal force to the gravitational force acting on the particle Sr

denotes the separation factor at a location (near the wall) that is at a radial distance r from the centre of

the cyclone Which one of the following statements is INCORRECT

(A) Sr depends on mass of the particle

(B) Sr depends on the acceleration due to gravity

(C) Sr depends on tangential velocity of the particle

(D) Sr depends on the radial location ( r ) of the particle

Ans (A) Sr depends on mass of the particle

Q8 A vertical cylindrical vessel has a layer of kerosene (of density 800 kgm3) over a layer of water (of

density 1000 kgm3) L-shaped glass tubes are connected to the column 30 cm apart The interface

between the two layers lies between the two points at which the L-tubes are connected The levels (in

cm) to which the liquids rise in the respective tubes are shown in the figure below

The distance (x in cm rounded off to the first decimal place) of the interface from the point at

which the lower L-tube is connected is _______

Ans 10 cm

Solution

We know that

1 2P P (1)

Kerosene

Water

x

20

30 42



Where

1

1

[20 (30 )]

800 981 [20 (30 )] 1000 981

800 981[50 ] 9810

392400 7848 9810

392400 1962

k waterP g x g x

x x

x x

x x

P x

And

P2 = water x g x 42

= 1000 x 981 x 42

= 412 020 Pa

Putting P1 and P2 in equation (1)

412020 = 392400+1962 x

1962 x = 19620

X = 10 cm

Q9 A composite wall is made of four different materials of construction in the fashion shown below

The resistance (in KW) of each of the sections of the wall is indicated in the diagram

3

025

1

07 Direction of heat flow Direction of heat flow



The overall resistance (in KW rounded off to the first decimal place) of the composite wall in the

direction of heat flow is _______

Ans (A) 39

Solution Overall resistance = 3 + 02 + 07 = 39 KW

Q10 Steam at 100oC is condensing on a vertical steel plate The condensate flow is laminar The average

Nusselt numbers are Nu1 and Nu2 when the plate temperatures are 10oC and 55oC respectively Assume

the physical properties of the fluid and steel to remain constant within the temperature range of interest

Using Nusselt equations for film-type condensation what is the value of the ratio Nu2 Nu1

(A) 05 (B) 084 (C) 119 (D) 141

Ans (C) 119


1 13 34 4

1

4

( ) ( )0943 0943

( ) ( )

1

( )

l l v fg l l v fg

g w g w

g w

gh k g h LL LNu h

h L T T k k T T

NuT T

1

4

1

1

4

1

1

(100 10)

1

90

Nu

Nu

and

1

4

2

1

4

2

1

(100 55)

1

45

Nu

Nu

2

1

1

490119

45

u

u

N

N

Q11 A binary liquid mixture of benzene and toluene contains 20 mol of benzene At 350 K the vapour

pressures of pure benzene and pure toluene are 92 kPa

and 35 kPa respectively The mixture follows Raoultrsquos

law The equilibrium vapour phase mole fraction



(rounded off to the second decimal place) of benzene in contact with this liquid mixture at 350 K is

_______

Ans 0396

Solution

From Raoultrsquos law

vi ii

p x p

amp wersquo i i

i

p pyi

p p

Mol fraction of benzene in vapor phase is given by

02 920396

02 92 08 35

VB B B

B V TB B T T

i

x ppy

x P x pp

Q12 Match the dimensionless numbers in Group-1 with the ratios in Group-2

Ans (D) P ndash II Q ndash III R ndash II

Q13 For what value of Lewis number the wet-bulb temperature and adiabatic saturation temperature

are nearly equal

(A) 033 (B) 05 (C) 1 (D) 2

Ans (C) 1



Ans 2

Solution Upper limit of reaction of reaction rate will be at

0

0 0

00

0

10 (10)2

1 5 15

A A

A AA

AA

A

C C

C Cr

CC

C

Q15 The variations of the concentrations (CA CR and CS) for three species (A R and S) with time in an

isothermal homogeneous batch reactor are shown in the figure below

Select the reaction scheme that correctly represents the above plot The numbers in the reaction

schemes shown below represent the first order rate constants in unit of s‒1



Ans (C)

Q16 Hydrogen iodide decomposes through the reaction 2HI H2 + I2 The value of the universal gas

constant R is 8314 J mol‒1K‒1 The activation energy for the forward reaction is 184000 J mol‒1 The ratio

(rounded off to the first decimal place) of the forward reaction rate at 600 K to that at 550 K is _______

Ans 285

Solution

From Arrhenius law

2

1 1 2

1 1Ekln

k R T T

1 2550 amp 600where T K T K

So 2

1

184000 1 1

8314 550 600

kln

k

2 1285Ans k k



Ans (D) P ndash III Q ndash I R ndash II

Q18 What is the order of response exhibited by a U-tube manometer

(A) Zero order (B) First order

(C) Second order (D) Third order

Ans (C) Second Order

Q19 A system exhibits inverse response for a unit step change in the input Which one of the following

statement must necessarily be satisfied

(A) The transfer function of the system has at least one negative pole

(B) The transfer function of the system has at least one positive pole

(C) The transfer function of the system has at least one negative zero

(D) The transfer function of the system has at least one positive zero

Ans (D) The transfer function of the system has at least one positive zero

Q20 Two design options for a distillation system are being compared based on the total annual cost

Information available is as follows

Option P Option Q

Installed cost of the system (Rs in lakhs) 150 120

Cost of cooling water for condenser (Rs in lakhsyear) 6 8



Cost of steam for reboiler (Rs in lakhsyear) 16 20

The annual fixed charge amounts to 12 of the installed cost Based on the above information what is

the total annual cost (Rs in lakhs year) of the better option

(A) 40 (B) 424 (C) 92 (D) 128

Ans (A) 40

Q21 Standard pipes of different schedule numbers and standard tubes of different BWG numbers are

available in the market For a pipe tube of a given nominal diameter which one of the following

statements is TRUE

(A) Wall thickness increases with increase in both the schedule number and the BWG number

(B) Wall thickness increases with increase in the schedule number and decreases with increase

in the BWG number

(C) Wall thickness decreases with increase in both the schedule number and the BWG number

(D) Neither the schedule number nor the BWG number has any relation to wall thickness

Ans (B)

Q22 Terms used in engineering economics have standard definitions and interpretations Which one of

the following statements is INCORRECT

(A) The profitability measure lsquoreturn on investmentrsquo does not consider the time value of money

(B) A cost index is an index value for a given time showing the cost at that time relative to a

certain base time

(C) The lsquosix-tenths factor rulersquo is used to estimate the cost of an equipment from the cost of a

similar equipment with a different capacity

(D) Payback period is calculated based on the payback time for the sum of the fixed and the

working capital investment

Ans (D)

Q23 India has no elemental sulphur deposits that can be economically exploited In India which one of

the following industries produces elemental sulphur as a by-product

(A) Coal carbonisation plants (B) Petroleum refineries

(C) Paper and pulp industries (D) Iron and steel making plants



Ans (B)

Q24 Two paper pulp plants P and Q use the same quality of bamboo as a raw material The chemicals

used in their digester are as follows

Plant P Plant Q

NaOH Yes No

Na2S Yes No

Na2CO3 Yes Yes

NaHCO3 No Yes

Na2SO3 No Yes

Which one of the following statements is CORRECT

(A) Plant P and Plant Q both use the Sulfite process

(B) Plant P and Plant Q both use the Kraft process

(C) Plant P uses Sulfite process

(D) Plant P uses Kraft process

Ans (D)

Q25 Match the industrial processes in Group-1 with the catalyst materials in Group-2

Group-1 Group-2

P Ethylene polymerisation I Nickel

Q Petroleum feedstock cracking II Vanadium pentoxide

R Oxidation of SO2 to SO3 III Zeolite

S Hydrogenation of oil

IV Aluminium triethyl with titanium chloride

promoter

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

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

Ans (A) P ndash IV Q ndash III R ndash II S ndash I



26 ndash Q 55 carry two marks each

Ans x3 = 15

Solution

4 x4 + 13 x5 = 46

4 x4 + 5 x5 = 30

2 x3 + 5 x4 + 3x5 = 61

By solving we get x3 = 15

Ans (D)

Solution

D2 + 1 = 0

D = i

Y = C1 cos x + C2 sinx

By using the boundary condition

5cos 10siny x x



Ans 255

Ans 265

Solution

f (x) = x3 + 5 Interval [1 4]

By applying mean value theorem

1 ( ) ( )( )

f b f af c

b a

X=265

Ans (A) 9467

Solution



Put h = 1 a = 0 b = 4

y0 = 3 yn = 55 y1 = 10 y2 = 21 y3 = 36

Ans (A) 1802

Solution At steady state energy balance equation can be written as

Energy IN ndash Energy OUT = heat solution + energy removed

Energy removed = heat of solution + energy OUT ndash energy IN

We have

Heat of solution = 2 4

2 4

650 42600

kJ kg H SO kJx

kg H SO h h

Energy out = 28

10 (40 25) 420kJ kJ

kg solution Kkg solution K h

Energy In = 2

2

286 (25 10) 378

kJ kJkg H O K

kg H O K h



So Energy removed

= 2600 + 378 + 420

= 1802 kJ

h

Energy removed = 1802 kJ

h





Ans 373 K


15irr revW W

(Change in IE)irr = 15 (change in IE)rev

1

2

1

2771277

2 1 2 1

2 1 2 1

2 1 1 1

2

2

( ) 15 ( )

15 ( )

15 ( )

6300 15 300 1 73

3

373

V V

V

P

P

C T T C T T

T T C T T

T T T T

T

T K

Ans 400


RTP

v b



0 1

1 0

1

1 0 1 0

( )

P

T P

P v b RT

RTv b

PRT

b b TP

Rb T b

P

v Rb

T P

h vv T

P T

R Rh T b T b b

P P

0

0

5 6

( )

4 10 (15 5) 10 400

T T

f i f i TT

dh b dP

h h b P P

J mol

Ans - 165



Ans - (C) 24

Solution

3 3

1

2

sec

120

240

36 001m mhr

P kPa

P kPa

Q

Required head to develop 2 12 1

P Py y

g

= 3(240 120) 10

129810

x

= 12232+12

= 24232 m

Power required = 4Q

= 9810 X 001 X 24232

= 2377 kW



Ans (C) 252 rpm

Solution

We have

Power No = 3 5

i

P

P N D

For constant power no

3 5

3 5 ( )

i

i

Pconst

PN D

P N D I

Volume of tank = 2

4VD H

( )VD H given

3 ( )VV D II

From equation (I )and (II)

3 5

3i

v

N DP

V D

For same 1 2

2

3 5 3 51 2

3 31

i i

v V

N D N DP

V D D

If all linear dimensions have to be doubled



2 2

1 1

2i V

i V

D D

D D

1

2

1

2

2

3 32 1

23

2 1 1

23

2

( 4 )

14 252

2

V

V

V

V

DN N

D

DN N N rpm

D

N rpm

Ans 215 mm

Solution -

We know that for Re( 01)N

terminal setting velocity under stokes law regime is given by

2

2

2

( )

18

10 (1180 1000)

18 01

1000

P p f

t

f

p

t

t p

d gv

d x xv

x

v d

In Stokersquos law max value of particle Reynolds number is 01 and it is given by

Re 01p f

p

f

d vN

Putting values



2 3(1000 ) 10

0101

p pd x d x

3

3 3

8

1 1010

10 1

215

p

p

mmd m x mm

m

d mm

Ans 49 m3 h

Ans 0125 m s


2 2

fU

3

3 2

1 10 2005

005 10 1000 1

0125 sec

u

y

u

u m



Ans (B) 842


13

08 13

08

0023 Re Pr

0023 Pr

Nu

G d

mG

A

Since mass flow rate in the tubes is doubled Nu and there by hi gets affected

081

3

08

2023 Pr

2 i

i

GdNu

hNu

Nu h

We know that

0

0

1 1 1 1 i

i i

h hU h h h



ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025

Solution - Given that 1 201 02r m r m

11 120 1F F

We know that A1 F12 = A2 F21 hence 121

2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1

Solution Minimum reflux ratio is given by

R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution

Selectivity is given by

βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that

According to Langmuir adsorption isotherm

1A A

A

A A

K PQ

K P

Amount of Arsquo adsorbed

Kg of Adsorbentmax

Maximum amount of a can be adsorbed

Adsorbent

A

m

qQ

qimum amount of A can be

kg of

1A A

m A

K Pq

q K

for case I at PA = 10 mm Hg q = 01

1001(1)

1 10A

m A

K

q K

For case II PA 100 mm Hg Q = 04

1004(2)

1 100A

m A

KO

q K

On solving we get ampA mK q

So for PA = 50 mn Hg therefore 03 dsorbed

dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P

Ans 02667 secndash 1

Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )

OA A A B A AC C X and C C X So 1 2(1 )

Am

A A

X

k X k X

For case ndash I 1sec 8molm AC

81 02

10AX



For case ndash II 5sec 5molm AC

5

1 0510

AX

So for case ndash I 1 2

1 2

021 4 1 ( )

(1 02) (02)k k I

k k

For case ndash ii 1 2

1 2

055 02 ( )

(1 05) (05)k k II

k k

On solving equation (I) and (II) 11 02667 seck

Ans 8 m3



Solution From the equation

T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K

So from curve at XA = 08 amp T =370 K 3

10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)

Solution Thiele modulus is given by

LengthticCharactersratediffusionpore

ratereactionIntrinsic

So for Large value of Thiele modulus reaction rate will be high

Therefore we will prefer Scenario 2 over Scenario 1 as we donrsquot required that total pellet is covered by

catalyst



Ans 025

Solution

It is obvious there is delay so it can be assumed to be PFR amp CSTR in series so

the delay is equal to P (space time of PFR)

So here delay is 5 min

P = 5 min

And we know that for CSTR

( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -

For equal percentage value

f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3

by using boundary conditions f0 amp B can be solved now at x = 05

f = 1013 m3hr

Ans 25

Solution Characterized equation is given by

1 + G open loop P = 0



2

1 101 1 0

15 25 100cK

s

On solving by using routh Array and putting bL = 0 we get 25cK

Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate

Now the value of T = 100 seconds

2

2

1dt

td and is known can be calculated

159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m

x x of the vapor space

x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution

I Given process can be written as

C6H12 rarr C6H6 + 6 H2

Cyclo-hexane Benzene Hydrogen



II proposed process can be written as

C6H12 + 12 H2O rarr 6CO2 + 18H2

Cyclo-hexane Steam Carbon di-oxide Hydrogen

Maximum ratio will be obtained for 100 conversion of cyclohexane

2

2

18

3

H formed in NewprocessSo

H formed in old process = 6

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General Aptitude


Q1 The volume of a sphere of diameter 1 unit is ________ than the volume of a cube of side 1 unit

(A) least (B) less (C) lesser (D) low

Ans (B) less

Q2 The unruly crowd demanded that the accused be _____________ without trial

(A) hanged (B) hanging (C) hankering (D) hung

Ans (A) hanged

Q3 Choose the statement(s) where the underlined word is used correctly

(i) A prone is a dried plum

(ii) He was lying prone on the floor

(iii) People who eat a lot of fat are prone to heart disease

(A) (i) and (iii) only (B) (iii) only (C) (i) and (ii) only (D) (ii) and (iii) only

Ans (D) (ii) and (iii) only

Q4 Fact If it rains then the field is wet

Read the following statements

(i) It rains

(ii) The field is not wet

(iii) The field is wet

(iv) It did not rain

Which one of the options given below is NOT logically possible based on the given fact

(A) If (iii) then (iv) (B) If (i) then (iii)

(C) If (i) then (ii) (D) If (ii) then (iv)

Ans (C) if (i) then (ii)






(A) 143 (B) 206 (C) 268 (D) 288

Ans (B) 206

Solution

2 2

6

5

3 3 6 6 6 6 3 36 362066

4 4 5 5 5 5 4 25 25

side

total area a a



























B FH LNP _ _ _ _


Ans (C) TVXZ



for any a is

(A) 0 (B) 1 (C) 2 (D) 10

Ans (A) 0





(A)

(B)

(C)

(D) Ans (C)






Ans (A)

Solution

sinatL e bt

= 2 2( )

b

s a b


Ans (D)

Solution


3

b

a



Ans (B)

Solution



P = Q + R

Q + R = 100 (1)



P X 03 = Q X 03 + R X 03

P = Q + R (2)



So ie = 1

Ans ndash 58 kJ mol



Solution


1 20 amp 1x x

So 2 21 2 60 (1) 100 0 (1)h x x x

1 58h

Ans (C) 348


211 4

2

f VP

g gd

21 1

1

02

2222

2 2 2022

1 1 1 211

2 022 2

2 021 1

18

22

1

4

2

348

f VP

d

V dm V

P f V

P f V V dm V

P V

P V

VP P Kpa

V


















Ans 10 cm

Solution

We know that

1 2P P (1)

Kerosene

Water

x

20

30 42



Where

1

1

[20 (30 )]

800 981 [20 (30 )] 1000 981

800 981[50 ] 9810

392400 7848 9810

392400 1962

k waterP g x g x

x x

x x

x x

P x

And

P2 = water x g x 42

= 1000 x 981 x 42

= 412 020 Pa


412020 = 392400+1962 x

1962 x = 19620

X = 10 cm



3

025

1






Ans (A) 39






(A) 05 (B) 084 (C) 119 (D) 141

Ans (C) 119


1 13 34 4

1

4

( ) ( )0943 0943

( ) ( )

1

( )

l l v fg l l v fg

g w g w

g w

gh k g h LL LNu h

h L T T k k T T

NuT T

1

4

1

1

4

1

1

(100 10)

1

90

Nu

Nu

and

1

4

2

1

4

2

1

(100 55)

1

45

Nu

Nu

2

1

1

490119

45

u

u

N

N








_______

Ans 0396

Solution


vi ii

p x p

amp wersquo i i

i

p pyi

p p


02 920396

02 92 08 35

VB B B

B V TB B T T

i

x ppy

x P x pp




are nearly equal

(A) 033 (B) 05 (C) 1 (D) 2

Ans (C) 1



Ans 2


0

0 0

00

0

10 (10)2

1 5 15

A A

A AA

AA

A

C C

C Cr

CC

C







Ans (C)




Ans 285

Solution

From Arrhenius law

2

1 1 2

1 1Ekln

k R T T


So 2

1

184000 1 1

8314 550 600

kln

k

2 1285Ans k k

















Option P Option Q








(A) 40 (B) 424 (C) 92 (D) 128

Ans (A) 40



statements is TRUE



in the BWG number



Ans (B)





certain base time





Ans (D)







Ans (B)



Plant P Plant Q

NaOH Yes No

Na2S Yes No

Na2CO3 Yes Yes

NaHCO3 No Yes

Na2SO3 No Yes






Ans (D)


Group-1 Group-2






promoter







Ans x3 = 15

Solution

4 x4 + 13 x5 = 46

4 x4 + 5 x5 = 30

2 x3 + 5 x4 + 3x5 = 61


Ans (D)

Solution

D2 + 1 = 0

D = i



5cos 10siny x x



Ans 255

Ans 265

Solution

f (x) = x3 + 5 Interval [1 4]


1 ( ) ( )( )

f b f af c

b a

X=265

Ans (A) 9467

Solution



Put h = 1 a = 0 b = 4

y0 = 3 yn = 55 y1 = 10 y2 = 21 y3 = 36

Ans (A) 1802




We have


2 4

650 42600

kJ kg H SO kJx

kg H SO h h

Energy out = 28

10 (40 25) 420kJ kJ


Energy In = 2

2

286 (25 10) 378

kJ kJkg H O K

kg H O K h



So Energy removed

= 2600 + 378 + 420

= 1802 kJ

h


h





Ans 373 K


15irr revW W


1

2

1

2771277

2 1 2 1

2 1 2 1

2 1 1 1

2

2

( ) 15 ( )

15 ( )

15 ( )

6300 15 300 1 73

3

373

V V

V

P

P

C T T C T T

T T C T T

T T T T

T

T K

Ans 400


RTP

v b



0 1

1 0

1

1 0 1 0

( )

P

T P

P v b RT

RTv b

PRT

b b TP

Rb T b

P

v Rb

T P

h vv T

P T

R Rh T b T b b

P P

0

0

5 6

( )

4 10 (15 5) 10 400

T T

f i f i TT

dh b dP

h h b P P

J mol

Ans - 165



Ans - (C) 24

Solution

3 3

1

2

sec

120

240

36 001m mhr

P kPa

P kPa

Q


P Py y

g

= 3(240 120) 10

129810

x

= 12232+12

= 24232 m

Power required = 4Q

= 9810 X 001 X 24232

= 2377 kW



Ans (C) 252 rpm

Solution

We have

Power No = 3 5

i

P

P N D


3 5

3 5 ( )

i

i

Pconst

PN D

P N D I

Volume of tank = 2

4VD H

( )VD H given

3 ( )VV D II


3 5

3i

v

N DP

V D

For same 1 2

2

3 5 3 51 2

3 31

i i

v V

N D N DP

V D D




2 2

1 1

2i V

i V

D D

D D

1

2

1

2

2

3 32 1

23

2 1 1

23

2

( 4 )

14 252

2

V

V

V

V

DN N

D

DN N N rpm

D

N rpm

Ans 215 mm

Solution -



2

2

2

( )

18

10 (1180 1000)

18 01

1000

P p f

t

f

p

t

t p

d gv

d x xv

x

v d


Re 01p f

p

f

d vN

Putting values



2 3(1000 ) 10

0101

p pd x d x

3

3 3

8

1 1010

10 1

215

p

p

mmd m x mm

m

d mm

Ans 49 m3 h

Ans 0125 m s


2 2

fU

3

3 2

1 10 2005

005 10 1000 1

0125 sec

u

y

u

u m



Ans (B) 842


13

08 13

08

0023 Re Pr

0023 Pr

Nu

G d

mG

A


081

3

08

2023 Pr

2 i

i

GdNu

hNu

Nu h

We know that

0

0

1 1 1 1 i

i i

h hU h h h



ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587










(A) 143 (B) 206 (C) 268 (D) 288

Ans (B) 206

Solution

2 2

6

5

3 3 6 6 6 6 3 36 362066

4 4 5 5 5 5 4 25 25

side

total area a a



























B FH LNP _ _ _ _


Ans (C) TVXZ



for any a is

(A) 0 (B) 1 (C) 2 (D) 10

Ans (A) 0





(A)

(B)

(C)

(D) Ans (C)






Ans (A)

Solution

sinatL e bt

= 2 2( )

b

s a b


Ans (D)

Solution


3

b

a



Ans (B)

Solution



P = Q + R

Q + R = 100 (1)



P X 03 = Q X 03 + R X 03

P = Q + R (2)



So ie = 1

Ans ndash 58 kJ mol



Solution


1 20 amp 1x x

So 2 21 2 60 (1) 100 0 (1)h x x x

1 58h

Ans (C) 348


211 4

2

f VP

g gd

21 1

1

02

2222

2 2 2022

1 1 1 211

2 022 2

2 021 1

18

22

1

4

2

348

f VP

d

V dm V

P f V

P f V V dm V

P V

P V

VP P Kpa

V


















Ans 10 cm

Solution

We know that

1 2P P (1)

Kerosene

Water

x

20

30 42



Where

1

1

[20 (30 )]

800 981 [20 (30 )] 1000 981

800 981[50 ] 9810

392400 7848 9810

392400 1962

k waterP g x g x

x x

x x

x x

P x

And

P2 = water x g x 42

= 1000 x 981 x 42

= 412 020 Pa


412020 = 392400+1962 x

1962 x = 19620

X = 10 cm



3

025

1






Ans (A) 39






(A) 05 (B) 084 (C) 119 (D) 141

Ans (C) 119


1 13 34 4

1

4

( ) ( )0943 0943

( ) ( )

1

( )

l l v fg l l v fg

g w g w

g w

gh k g h LL LNu h

h L T T k k T T

NuT T

1

4

1

1

4

1

1

(100 10)

1

90

Nu

Nu

and

1

4

2

1

4

2

1

(100 55)

1

45

Nu

Nu

2

1

1

490119

45

u

u

N

N








_______

Ans 0396

Solution


vi ii

p x p

amp wersquo i i

i

p pyi

p p


02 920396

02 92 08 35

VB B B

B V TB B T T

i

x ppy

x P x pp




are nearly equal

(A) 033 (B) 05 (C) 1 (D) 2

Ans (C) 1



Ans 2


0

0 0

00

0

10 (10)2

1 5 15

A A

A AA

AA

A

C C

C Cr

CC

C







Ans (C)




Ans 285

Solution

From Arrhenius law

2

1 1 2

1 1Ekln

k R T T


So 2

1

184000 1 1

8314 550 600

kln

k

2 1285Ans k k

















Option P Option Q








(A) 40 (B) 424 (C) 92 (D) 128

Ans (A) 40



statements is TRUE



in the BWG number



Ans (B)





certain base time





Ans (D)







Ans (B)



Plant P Plant Q

NaOH Yes No

Na2S Yes No

Na2CO3 Yes Yes

NaHCO3 No Yes

Na2SO3 No Yes






Ans (D)


Group-1 Group-2






promoter







Ans x3 = 15

Solution

4 x4 + 13 x5 = 46

4 x4 + 5 x5 = 30

2 x3 + 5 x4 + 3x5 = 61


Ans (D)

Solution

D2 + 1 = 0

D = i



5cos 10siny x x



Ans 255

Ans 265

Solution

f (x) = x3 + 5 Interval [1 4]


1 ( ) ( )( )

f b f af c

b a

X=265

Ans (A) 9467

Solution



Put h = 1 a = 0 b = 4

y0 = 3 yn = 55 y1 = 10 y2 = 21 y3 = 36

Ans (A) 1802




We have


2 4

650 42600

kJ kg H SO kJx

kg H SO h h

Energy out = 28

10 (40 25) 420kJ kJ


Energy In = 2

2

286 (25 10) 378

kJ kJkg H O K

kg H O K h



So Energy removed

= 2600 + 378 + 420

= 1802 kJ

h


h





Ans 373 K


15irr revW W


1

2

1

2771277

2 1 2 1

2 1 2 1

2 1 1 1

2

2

( ) 15 ( )

15 ( )

15 ( )

6300 15 300 1 73

3

373

V V

V

P

P

C T T C T T

T T C T T

T T T T

T

T K

Ans 400


RTP

v b



0 1

1 0

1

1 0 1 0

( )

P

T P

P v b RT

RTv b

PRT

b b TP

Rb T b

P

v Rb

T P

h vv T

P T

R Rh T b T b b

P P

0

0

5 6

( )

4 10 (15 5) 10 400

T T

f i f i TT

dh b dP

h h b P P

J mol

Ans - 165



Ans - (C) 24

Solution

3 3

1

2

sec

120

240

36 001m mhr

P kPa

P kPa

Q


P Py y

g

= 3(240 120) 10

129810

x

= 12232+12

= 24232 m

Power required = 4Q

= 9810 X 001 X 24232

= 2377 kW



Ans (C) 252 rpm

Solution

We have

Power No = 3 5

i

P

P N D


3 5

3 5 ( )

i

i

Pconst

PN D

P N D I

Volume of tank = 2

4VD H

( )VD H given

3 ( )VV D II


3 5

3i

v

N DP

V D

For same 1 2

2

3 5 3 51 2

3 31

i i

v V

N D N DP

V D D




2 2

1 1

2i V

i V

D D

D D

1

2

1

2

2

3 32 1

23

2 1 1

23

2

( 4 )

14 252

2

V

V

V

V

DN N

D

DN N N rpm

D

N rpm

Ans 215 mm

Solution -



2

2

2

( )

18

10 (1180 1000)

18 01

1000

P p f

t

f

p

t

t p

d gv

d x xv

x

v d


Re 01p f

p

f

d vN

Putting values



2 3(1000 ) 10

0101

p pd x d x

3

3 3

8

1 1010

10 1

215

p

p

mmd m x mm

m

d mm

Ans 49 m3 h

Ans 0125 m s


2 2

fU

3

3 2

1 10 2005

005 10 1000 1

0125 sec

u

y

u

u m



Ans (B) 842


13

08 13

08

0023 Re Pr

0023 Pr

Nu

G d

mG

A


081

3

08

2023 Pr

2 i

i

GdNu

hNu

Nu h

We know that

0

0

1 1 1 1 i

i i

h hU h h h



ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587



















B FH LNP _ _ _ _


Ans (C) TVXZ



for any a is

(A) 0 (B) 1 (C) 2 (D) 10

Ans (A) 0





(A)

(B)

(C)

(D) Ans (C)






Ans (A)

Solution

sinatL e bt

= 2 2( )

b

s a b


Ans (D)

Solution


3

b

a



Ans (B)

Solution



P = Q + R

Q + R = 100 (1)



P X 03 = Q X 03 + R X 03

P = Q + R (2)



So ie = 1

Ans ndash 58 kJ mol



Solution


1 20 amp 1x x

So 2 21 2 60 (1) 100 0 (1)h x x x

1 58h

Ans (C) 348


211 4

2

f VP

g gd

21 1

1

02

2222

2 2 2022

1 1 1 211

2 022 2

2 021 1

18

22

1

4

2

348

f VP

d

V dm V

P f V

P f V V dm V

P V

P V

VP P Kpa

V


















Ans 10 cm

Solution

We know that

1 2P P (1)

Kerosene

Water

x

20

30 42



Where

1

1

[20 (30 )]

800 981 [20 (30 )] 1000 981

800 981[50 ] 9810

392400 7848 9810

392400 1962

k waterP g x g x

x x

x x

x x

P x

And

P2 = water x g x 42

= 1000 x 981 x 42

= 412 020 Pa


412020 = 392400+1962 x

1962 x = 19620

X = 10 cm



3

025

1






Ans (A) 39






(A) 05 (B) 084 (C) 119 (D) 141

Ans (C) 119


1 13 34 4

1

4

( ) ( )0943 0943

( ) ( )

1

( )

l l v fg l l v fg

g w g w

g w

gh k g h LL LNu h

h L T T k k T T

NuT T

1

4

1

1

4

1

1

(100 10)

1

90

Nu

Nu

and

1

4

2

1

4

2

1

(100 55)

1

45

Nu

Nu

2

1

1

490119

45

u

u

N

N








_______

Ans 0396

Solution


vi ii

p x p

amp wersquo i i

i

p pyi

p p


02 920396

02 92 08 35

VB B B

B V TB B T T

i

x ppy

x P x pp




are nearly equal

(A) 033 (B) 05 (C) 1 (D) 2

Ans (C) 1



Ans 2


0

0 0

00

0

10 (10)2

1 5 15

A A

A AA

AA

A

C C

C Cr

CC

C







Ans (C)




Ans 285

Solution

From Arrhenius law

2

1 1 2

1 1Ekln

k R T T


So 2

1

184000 1 1

8314 550 600

kln

k

2 1285Ans k k

















Option P Option Q








(A) 40 (B) 424 (C) 92 (D) 128

Ans (A) 40



statements is TRUE



in the BWG number



Ans (B)





certain base time





Ans (D)







Ans (B)



Plant P Plant Q

NaOH Yes No

Na2S Yes No

Na2CO3 Yes Yes

NaHCO3 No Yes

Na2SO3 No Yes






Ans (D)


Group-1 Group-2






promoter







Ans x3 = 15

Solution

4 x4 + 13 x5 = 46

4 x4 + 5 x5 = 30

2 x3 + 5 x4 + 3x5 = 61


Ans (D)

Solution

D2 + 1 = 0

D = i



5cos 10siny x x



Ans 255

Ans 265

Solution

f (x) = x3 + 5 Interval [1 4]


1 ( ) ( )( )

f b f af c

b a

X=265

Ans (A) 9467

Solution



Put h = 1 a = 0 b = 4

y0 = 3 yn = 55 y1 = 10 y2 = 21 y3 = 36

Ans (A) 1802




We have


2 4

650 42600

kJ kg H SO kJx

kg H SO h h

Energy out = 28

10 (40 25) 420kJ kJ


Energy In = 2

2

286 (25 10) 378

kJ kJkg H O K

kg H O K h



So Energy removed

= 2600 + 378 + 420

= 1802 kJ

h


h





Ans 373 K


15irr revW W


1

2

1

2771277

2 1 2 1

2 1 2 1

2 1 1 1

2

2

( ) 15 ( )

15 ( )

15 ( )

6300 15 300 1 73

3

373

V V

V

P

P

C T T C T T

T T C T T

T T T T

T

T K

Ans 400


RTP

v b



0 1

1 0

1

1 0 1 0

( )

P

T P

P v b RT

RTv b

PRT

b b TP

Rb T b

P

v Rb

T P

h vv T

P T

R Rh T b T b b

P P

0

0

5 6

( )

4 10 (15 5) 10 400

T T

f i f i TT

dh b dP

h h b P P

J mol

Ans - 165



Ans - (C) 24

Solution

3 3

1

2

sec

120

240

36 001m mhr

P kPa

P kPa

Q


P Py y

g

= 3(240 120) 10

129810

x

= 12232+12

= 24232 m

Power required = 4Q

= 9810 X 001 X 24232

= 2377 kW



Ans (C) 252 rpm

Solution

We have

Power No = 3 5

i

P

P N D


3 5

3 5 ( )

i

i

Pconst

PN D

P N D I

Volume of tank = 2

4VD H

( )VD H given

3 ( )VV D II


3 5

3i

v

N DP

V D

For same 1 2

2

3 5 3 51 2

3 31

i i

v V

N D N DP

V D D




2 2

1 1

2i V

i V

D D

D D

1

2

1

2

2

3 32 1

23

2 1 1

23

2

( 4 )

14 252

2

V

V

V

V

DN N

D

DN N N rpm

D

N rpm

Ans 215 mm

Solution -



2

2

2

( )

18

10 (1180 1000)

18 01

1000

P p f

t

f

p

t

t p

d gv

d x xv

x

v d


Re 01p f

p

f

d vN

Putting values



2 3(1000 ) 10

0101

p pd x d x

3

3 3

8

1 1010

10 1

215

p

p

mmd m x mm

m

d mm

Ans 49 m3 h

Ans 0125 m s


2 2

fU

3

3 2

1 10 2005

005 10 1000 1

0125 sec

u

y

u

u m



Ans (B) 842


13

08 13

08

0023 Re Pr

0023 Pr

Nu

G d

mG

A


081

3

08

2023 Pr

2 i

i

GdNu

hNu

Nu h

We know that

0

0

1 1 1 1 i

i i

h hU h h h



ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587









(A)

(B)

(C)

(D) Ans (C)






Ans (A)

Solution

sinatL e bt

= 2 2( )

b

s a b


Ans (D)

Solution


3

b

a



Ans (B)

Solution



P = Q + R

Q + R = 100 (1)



P X 03 = Q X 03 + R X 03

P = Q + R (2)



So ie = 1

Ans ndash 58 kJ mol



Solution


1 20 amp 1x x

So 2 21 2 60 (1) 100 0 (1)h x x x

1 58h

Ans (C) 348


211 4

2

f VP

g gd

21 1

1

02

2222

2 2 2022

1 1 1 211

2 022 2

2 021 1

18

22

1

4

2

348

f VP

d

V dm V

P f V

P f V V dm V

P V

P V

VP P Kpa

V


















Ans 10 cm

Solution

We know that

1 2P P (1)

Kerosene

Water

x

20

30 42



Where

1

1

[20 (30 )]

800 981 [20 (30 )] 1000 981

800 981[50 ] 9810

392400 7848 9810

392400 1962

k waterP g x g x

x x

x x

x x

P x

And

P2 = water x g x 42

= 1000 x 981 x 42

= 412 020 Pa


412020 = 392400+1962 x

1962 x = 19620

X = 10 cm



3

025

1






Ans (A) 39






(A) 05 (B) 084 (C) 119 (D) 141

Ans (C) 119


1 13 34 4

1

4

( ) ( )0943 0943

( ) ( )

1

( )

l l v fg l l v fg

g w g w

g w

gh k g h LL LNu h

h L T T k k T T

NuT T

1

4

1

1

4

1

1

(100 10)

1

90

Nu

Nu

and

1

4

2

1

4

2

1

(100 55)

1

45

Nu

Nu

2

1

1

490119

45

u

u

N

N








_______

Ans 0396

Solution


vi ii

p x p

amp wersquo i i

i

p pyi

p p


02 920396

02 92 08 35

VB B B

B V TB B T T

i

x ppy

x P x pp




are nearly equal

(A) 033 (B) 05 (C) 1 (D) 2

Ans (C) 1



Ans 2


0

0 0

00

0

10 (10)2

1 5 15

A A

A AA

AA

A

C C

C Cr

CC

C







Ans (C)




Ans 285

Solution

From Arrhenius law

2

1 1 2

1 1Ekln

k R T T


So 2

1

184000 1 1

8314 550 600

kln

k

2 1285Ans k k

















Option P Option Q








(A) 40 (B) 424 (C) 92 (D) 128

Ans (A) 40



statements is TRUE



in the BWG number



Ans (B)





certain base time





Ans (D)







Ans (B)



Plant P Plant Q

NaOH Yes No

Na2S Yes No

Na2CO3 Yes Yes

NaHCO3 No Yes

Na2SO3 No Yes






Ans (D)


Group-1 Group-2






promoter







Ans x3 = 15

Solution

4 x4 + 13 x5 = 46

4 x4 + 5 x5 = 30

2 x3 + 5 x4 + 3x5 = 61


Ans (D)

Solution

D2 + 1 = 0

D = i



5cos 10siny x x



Ans 255

Ans 265

Solution

f (x) = x3 + 5 Interval [1 4]


1 ( ) ( )( )

f b f af c

b a

X=265

Ans (A) 9467

Solution



Put h = 1 a = 0 b = 4

y0 = 3 yn = 55 y1 = 10 y2 = 21 y3 = 36

Ans (A) 1802




We have


2 4

650 42600

kJ kg H SO kJx

kg H SO h h

Energy out = 28

10 (40 25) 420kJ kJ


Energy In = 2

2

286 (25 10) 378

kJ kJkg H O K

kg H O K h



So Energy removed

= 2600 + 378 + 420

= 1802 kJ

h


h





Ans 373 K


15irr revW W


1

2

1

2771277

2 1 2 1

2 1 2 1

2 1 1 1

2

2

( ) 15 ( )

15 ( )

15 ( )

6300 15 300 1 73

3

373

V V

V

P

P

C T T C T T

T T C T T

T T T T

T

T K

Ans 400


RTP

v b



0 1

1 0

1

1 0 1 0

( )

P

T P

P v b RT

RTv b

PRT

b b TP

Rb T b

P

v Rb

T P

h vv T

P T

R Rh T b T b b

P P

0

0

5 6

( )

4 10 (15 5) 10 400

T T

f i f i TT

dh b dP

h h b P P

J mol

Ans - 165



Ans - (C) 24

Solution

3 3

1

2

sec

120

240

36 001m mhr

P kPa

P kPa

Q


P Py y

g

= 3(240 120) 10

129810

x

= 12232+12

= 24232 m

Power required = 4Q

= 9810 X 001 X 24232

= 2377 kW



Ans (C) 252 rpm

Solution

We have

Power No = 3 5

i

P

P N D


3 5

3 5 ( )

i

i

Pconst

PN D

P N D I

Volume of tank = 2

4VD H

( )VD H given

3 ( )VV D II


3 5

3i

v

N DP

V D

For same 1 2

2

3 5 3 51 2

3 31

i i

v V

N D N DP

V D D




2 2

1 1

2i V

i V

D D

D D

1

2

1

2

2

3 32 1

23

2 1 1

23

2

( 4 )

14 252

2

V

V

V

V

DN N

D

DN N N rpm

D

N rpm

Ans 215 mm

Solution -



2

2

2

( )

18

10 (1180 1000)

18 01

1000

P p f

t

f

p

t

t p

d gv

d x xv

x

v d


Re 01p f

p

f

d vN

Putting values



2 3(1000 ) 10

0101

p pd x d x

3

3 3

8

1 1010

10 1

215

p

p

mmd m x mm

m

d mm

Ans 49 m3 h

Ans 0125 m s


2 2

fU

3

3 2

1 10 2005

005 10 1000 1

0125 sec

u

y

u

u m



Ans (B) 842


13

08 13

08

0023 Re Pr

0023 Pr

Nu

G d

mG

A


081

3

08

2023 Pr

2 i

i

GdNu

hNu

Nu h

We know that

0

0

1 1 1 1 i

i i

h hU h h h



ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587










Ans (A)

Solution

sinatL e bt

= 2 2( )

b

s a b


Ans (D)

Solution


3

b

a



Ans (B)

Solution



P = Q + R

Q + R = 100 (1)



P X 03 = Q X 03 + R X 03

P = Q + R (2)



So ie = 1

Ans ndash 58 kJ mol



Solution


1 20 amp 1x x

So 2 21 2 60 (1) 100 0 (1)h x x x

1 58h

Ans (C) 348


211 4

2

f VP

g gd

21 1

1

02

2222

2 2 2022

1 1 1 211

2 022 2

2 021 1

18

22

1

4

2

348

f VP

d

V dm V

P f V

P f V V dm V

P V

P V

VP P Kpa

V


















Ans 10 cm

Solution

We know that

1 2P P (1)

Kerosene

Water

x

20

30 42



Where

1

1

[20 (30 )]

800 981 [20 (30 )] 1000 981

800 981[50 ] 9810

392400 7848 9810

392400 1962

k waterP g x g x

x x

x x

x x

P x

And

P2 = water x g x 42

= 1000 x 981 x 42

= 412 020 Pa


412020 = 392400+1962 x

1962 x = 19620

X = 10 cm



3

025

1






Ans (A) 39






(A) 05 (B) 084 (C) 119 (D) 141

Ans (C) 119


1 13 34 4

1

4

( ) ( )0943 0943

( ) ( )

1

( )

l l v fg l l v fg

g w g w

g w

gh k g h LL LNu h

h L T T k k T T

NuT T

1

4

1

1

4

1

1

(100 10)

1

90

Nu

Nu

and

1

4

2

1

4

2

1

(100 55)

1

45

Nu

Nu

2

1

1

490119

45

u

u

N

N








_______

Ans 0396

Solution


vi ii

p x p

amp wersquo i i

i

p pyi

p p


02 920396

02 92 08 35

VB B B

B V TB B T T

i

x ppy

x P x pp




are nearly equal

(A) 033 (B) 05 (C) 1 (D) 2

Ans (C) 1



Ans 2


0

0 0

00

0

10 (10)2

1 5 15

A A

A AA

AA

A

C C

C Cr

CC

C







Ans (C)




Ans 285

Solution

From Arrhenius law

2

1 1 2

1 1Ekln

k R T T


So 2

1

184000 1 1

8314 550 600

kln

k

2 1285Ans k k

















Option P Option Q








(A) 40 (B) 424 (C) 92 (D) 128

Ans (A) 40



statements is TRUE



in the BWG number



Ans (B)





certain base time





Ans (D)







Ans (B)



Plant P Plant Q

NaOH Yes No

Na2S Yes No

Na2CO3 Yes Yes

NaHCO3 No Yes

Na2SO3 No Yes






Ans (D)


Group-1 Group-2






promoter







Ans x3 = 15

Solution

4 x4 + 13 x5 = 46

4 x4 + 5 x5 = 30

2 x3 + 5 x4 + 3x5 = 61


Ans (D)

Solution

D2 + 1 = 0

D = i



5cos 10siny x x



Ans 255

Ans 265

Solution

f (x) = x3 + 5 Interval [1 4]


1 ( ) ( )( )

f b f af c

b a

X=265

Ans (A) 9467

Solution



Put h = 1 a = 0 b = 4

y0 = 3 yn = 55 y1 = 10 y2 = 21 y3 = 36

Ans (A) 1802




We have


2 4

650 42600

kJ kg H SO kJx

kg H SO h h

Energy out = 28

10 (40 25) 420kJ kJ


Energy In = 2

2

286 (25 10) 378

kJ kJkg H O K

kg H O K h



So Energy removed

= 2600 + 378 + 420

= 1802 kJ

h


h





Ans 373 K


15irr revW W


1

2

1

2771277

2 1 2 1

2 1 2 1

2 1 1 1

2

2

( ) 15 ( )

15 ( )

15 ( )

6300 15 300 1 73

3

373

V V

V

P

P

C T T C T T

T T C T T

T T T T

T

T K

Ans 400


RTP

v b



0 1

1 0

1

1 0 1 0

( )

P

T P

P v b RT

RTv b

PRT

b b TP

Rb T b

P

v Rb

T P

h vv T

P T

R Rh T b T b b

P P

0

0

5 6

( )

4 10 (15 5) 10 400

T T

f i f i TT

dh b dP

h h b P P

J mol

Ans - 165



Ans - (C) 24

Solution

3 3

1

2

sec

120

240

36 001m mhr

P kPa

P kPa

Q


P Py y

g

= 3(240 120) 10

129810

x

= 12232+12

= 24232 m

Power required = 4Q

= 9810 X 001 X 24232

= 2377 kW



Ans (C) 252 rpm

Solution

We have

Power No = 3 5

i

P

P N D


3 5

3 5 ( )

i

i

Pconst

PN D

P N D I

Volume of tank = 2

4VD H

( )VD H given

3 ( )VV D II


3 5

3i

v

N DP

V D

For same 1 2

2

3 5 3 51 2

3 31

i i

v V

N D N DP

V D D




2 2

1 1

2i V

i V

D D

D D

1

2

1

2

2

3 32 1

23

2 1 1

23

2

( 4 )

14 252

2

V

V

V

V

DN N

D

DN N N rpm

D

N rpm

Ans 215 mm

Solution -



2

2

2

( )

18

10 (1180 1000)

18 01

1000

P p f

t

f

p

t

t p

d gv

d x xv

x

v d


Re 01p f

p

f

d vN

Putting values



2 3(1000 ) 10

0101

p pd x d x

3

3 3

8

1 1010

10 1

215

p

p

mmd m x mm

m

d mm

Ans 49 m3 h

Ans 0125 m s


2 2

fU

3

3 2

1 10 2005

005 10 1000 1

0125 sec

u

y

u

u m



Ans (B) 842


13

08 13

08

0023 Re Pr

0023 Pr

Nu

G d

mG

A


081

3

08

2023 Pr

2 i

i

GdNu

hNu

Nu h

We know that

0

0

1 1 1 1 i

i i

h hU h h h



ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587







Ans (B)

Solution



P = Q + R

Q + R = 100 (1)



P X 03 = Q X 03 + R X 03

P = Q + R (2)



So ie = 1

Ans ndash 58 kJ mol



Solution


1 20 amp 1x x

So 2 21 2 60 (1) 100 0 (1)h x x x

1 58h

Ans (C) 348


211 4

2

f VP

g gd

21 1

1

02

2222

2 2 2022

1 1 1 211

2 022 2

2 021 1

18

22

1

4

2

348

f VP

d

V dm V

P f V

P f V V dm V

P V

P V

VP P Kpa

V


















Ans 10 cm

Solution

We know that

1 2P P (1)

Kerosene

Water

x

20

30 42



Where

1

1

[20 (30 )]

800 981 [20 (30 )] 1000 981

800 981[50 ] 9810

392400 7848 9810

392400 1962

k waterP g x g x

x x

x x

x x

P x

And

P2 = water x g x 42

= 1000 x 981 x 42

= 412 020 Pa


412020 = 392400+1962 x

1962 x = 19620

X = 10 cm



3

025

1






Ans (A) 39






(A) 05 (B) 084 (C) 119 (D) 141

Ans (C) 119


1 13 34 4

1

4

( ) ( )0943 0943

( ) ( )

1

( )

l l v fg l l v fg

g w g w

g w

gh k g h LL LNu h

h L T T k k T T

NuT T

1

4

1

1

4

1

1

(100 10)

1

90

Nu

Nu

and

1

4

2

1

4

2

1

(100 55)

1

45

Nu

Nu

2

1

1

490119

45

u

u

N

N








_______

Ans 0396

Solution


vi ii

p x p

amp wersquo i i

i

p pyi

p p


02 920396

02 92 08 35

VB B B

B V TB B T T

i

x ppy

x P x pp




are nearly equal

(A) 033 (B) 05 (C) 1 (D) 2

Ans (C) 1



Ans 2


0

0 0

00

0

10 (10)2

1 5 15

A A

A AA

AA

A

C C

C Cr

CC

C







Ans (C)




Ans 285

Solution

From Arrhenius law

2

1 1 2

1 1Ekln

k R T T


So 2

1

184000 1 1

8314 550 600

kln

k

2 1285Ans k k

















Option P Option Q








(A) 40 (B) 424 (C) 92 (D) 128

Ans (A) 40



statements is TRUE



in the BWG number



Ans (B)





certain base time





Ans (D)







Ans (B)



Plant P Plant Q

NaOH Yes No

Na2S Yes No

Na2CO3 Yes Yes

NaHCO3 No Yes

Na2SO3 No Yes






Ans (D)


Group-1 Group-2






promoter







Ans x3 = 15

Solution

4 x4 + 13 x5 = 46

4 x4 + 5 x5 = 30

2 x3 + 5 x4 + 3x5 = 61


Ans (D)

Solution

D2 + 1 = 0

D = i



5cos 10siny x x



Ans 255

Ans 265

Solution

f (x) = x3 + 5 Interval [1 4]


1 ( ) ( )( )

f b f af c

b a

X=265

Ans (A) 9467

Solution



Put h = 1 a = 0 b = 4

y0 = 3 yn = 55 y1 = 10 y2 = 21 y3 = 36

Ans (A) 1802




We have


2 4

650 42600

kJ kg H SO kJx

kg H SO h h

Energy out = 28

10 (40 25) 420kJ kJ


Energy In = 2

2

286 (25 10) 378

kJ kJkg H O K

kg H O K h



So Energy removed

= 2600 + 378 + 420

= 1802 kJ

h


h





Ans 373 K


15irr revW W


1

2

1

2771277

2 1 2 1

2 1 2 1

2 1 1 1

2

2

( ) 15 ( )

15 ( )

15 ( )

6300 15 300 1 73

3

373

V V

V

P

P

C T T C T T

T T C T T

T T T T

T

T K

Ans 400


RTP

v b



0 1

1 0

1

1 0 1 0

( )

P

T P

P v b RT

RTv b

PRT

b b TP

Rb T b

P

v Rb

T P

h vv T

P T

R Rh T b T b b

P P

0

0

5 6

( )

4 10 (15 5) 10 400

T T

f i f i TT

dh b dP

h h b P P

J mol

Ans - 165



Ans - (C) 24

Solution

3 3

1

2

sec

120

240

36 001m mhr

P kPa

P kPa

Q


P Py y

g

= 3(240 120) 10

129810

x

= 12232+12

= 24232 m

Power required = 4Q

= 9810 X 001 X 24232

= 2377 kW



Ans (C) 252 rpm

Solution

We have

Power No = 3 5

i

P

P N D


3 5

3 5 ( )

i

i

Pconst

PN D

P N D I

Volume of tank = 2

4VD H

( )VD H given

3 ( )VV D II


3 5

3i

v

N DP

V D

For same 1 2

2

3 5 3 51 2

3 31

i i

v V

N D N DP

V D D




2 2

1 1

2i V

i V

D D

D D

1

2

1

2

2

3 32 1

23

2 1 1

23

2

( 4 )

14 252

2

V

V

V

V

DN N

D

DN N N rpm

D

N rpm

Ans 215 mm

Solution -



2

2

2

( )

18

10 (1180 1000)

18 01

1000

P p f

t

f

p

t

t p

d gv

d x xv

x

v d


Re 01p f

p

f

d vN

Putting values



2 3(1000 ) 10

0101

p pd x d x

3

3 3

8

1 1010

10 1

215

p

p

mmd m x mm

m

d mm

Ans 49 m3 h

Ans 0125 m s


2 2

fU

3

3 2

1 10 2005

005 10 1000 1

0125 sec

u

y

u

u m



Ans (B) 842


13

08 13

08

0023 Re Pr

0023 Pr

Nu

G d

mG

A


081

3

08

2023 Pr

2 i

i

GdNu

hNu

Nu h

We know that

0

0

1 1 1 1 i

i i

h hU h h h



ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587







Solution


1 20 amp 1x x

So 2 21 2 60 (1) 100 0 (1)h x x x

1 58h

Ans (C) 348


211 4

2

f VP

g gd

21 1

1

02

2222

2 2 2022

1 1 1 211

2 022 2

2 021 1

18

22

1

4

2

348

f VP

d

V dm V

P f V

P f V V dm V

P V

P V

VP P Kpa

V


















Ans 10 cm

Solution

We know that

1 2P P (1)

Kerosene

Water

x

20

30 42



Where

1

1

[20 (30 )]

800 981 [20 (30 )] 1000 981

800 981[50 ] 9810

392400 7848 9810

392400 1962

k waterP g x g x

x x

x x

x x

P x

And

P2 = water x g x 42

= 1000 x 981 x 42

= 412 020 Pa


412020 = 392400+1962 x

1962 x = 19620

X = 10 cm



3

025

1






Ans (A) 39






(A) 05 (B) 084 (C) 119 (D) 141

Ans (C) 119


1 13 34 4

1

4

( ) ( )0943 0943

( ) ( )

1

( )

l l v fg l l v fg

g w g w

g w

gh k g h LL LNu h

h L T T k k T T

NuT T

1

4

1

1

4

1

1

(100 10)

1

90

Nu

Nu

and

1

4

2

1

4

2

1

(100 55)

1

45

Nu

Nu

2

1

1

490119

45

u

u

N

N








_______

Ans 0396

Solution


vi ii

p x p

amp wersquo i i

i

p pyi

p p


02 920396

02 92 08 35

VB B B

B V TB B T T

i

x ppy

x P x pp




are nearly equal

(A) 033 (B) 05 (C) 1 (D) 2

Ans (C) 1



Ans 2


0

0 0

00

0

10 (10)2

1 5 15

A A

A AA

AA

A

C C

C Cr

CC

C







Ans (C)




Ans 285

Solution

From Arrhenius law

2

1 1 2

1 1Ekln

k R T T


So 2

1

184000 1 1

8314 550 600

kln

k

2 1285Ans k k

















Option P Option Q








(A) 40 (B) 424 (C) 92 (D) 128

Ans (A) 40



statements is TRUE



in the BWG number



Ans (B)





certain base time





Ans (D)







Ans (B)



Plant P Plant Q

NaOH Yes No

Na2S Yes No

Na2CO3 Yes Yes

NaHCO3 No Yes

Na2SO3 No Yes






Ans (D)


Group-1 Group-2






promoter







Ans x3 = 15

Solution

4 x4 + 13 x5 = 46

4 x4 + 5 x5 = 30

2 x3 + 5 x4 + 3x5 = 61


Ans (D)

Solution

D2 + 1 = 0

D = i



5cos 10siny x x



Ans 255

Ans 265

Solution

f (x) = x3 + 5 Interval [1 4]


1 ( ) ( )( )

f b f af c

b a

X=265

Ans (A) 9467

Solution



Put h = 1 a = 0 b = 4

y0 = 3 yn = 55 y1 = 10 y2 = 21 y3 = 36

Ans (A) 1802




We have


2 4

650 42600

kJ kg H SO kJx

kg H SO h h

Energy out = 28

10 (40 25) 420kJ kJ


Energy In = 2

2

286 (25 10) 378

kJ kJkg H O K

kg H O K h



So Energy removed

= 2600 + 378 + 420

= 1802 kJ

h


h





Ans 373 K


15irr revW W


1

2

1

2771277

2 1 2 1

2 1 2 1

2 1 1 1

2

2

( ) 15 ( )

15 ( )

15 ( )

6300 15 300 1 73

3

373

V V

V

P

P

C T T C T T

T T C T T

T T T T

T

T K

Ans 400


RTP

v b



0 1

1 0

1

1 0 1 0

( )

P

T P

P v b RT

RTv b

PRT

b b TP

Rb T b

P

v Rb

T P

h vv T

P T

R Rh T b T b b

P P

0

0

5 6

( )

4 10 (15 5) 10 400

T T

f i f i TT

dh b dP

h h b P P

J mol

Ans - 165



Ans - (C) 24

Solution

3 3

1

2

sec

120

240

36 001m mhr

P kPa

P kPa

Q


P Py y

g

= 3(240 120) 10

129810

x

= 12232+12

= 24232 m

Power required = 4Q

= 9810 X 001 X 24232

= 2377 kW



Ans (C) 252 rpm

Solution

We have

Power No = 3 5

i

P

P N D


3 5

3 5 ( )

i

i

Pconst

PN D

P N D I

Volume of tank = 2

4VD H

( )VD H given

3 ( )VV D II


3 5

3i

v

N DP

V D

For same 1 2

2

3 5 3 51 2

3 31

i i

v V

N D N DP

V D D




2 2

1 1

2i V

i V

D D

D D

1

2

1

2

2

3 32 1

23

2 1 1

23

2

( 4 )

14 252

2

V

V

V

V

DN N

D

DN N N rpm

D

N rpm

Ans 215 mm

Solution -



2

2

2

( )

18

10 (1180 1000)

18 01

1000

P p f

t

f

p

t

t p

d gv

d x xv

x

v d


Re 01p f

p

f

d vN

Putting values



2 3(1000 ) 10

0101

p pd x d x

3

3 3

8

1 1010

10 1

215

p

p

mmd m x mm

m

d mm

Ans 49 m3 h

Ans 0125 m s


2 2

fU

3

3 2

1 10 2005

005 10 1000 1

0125 sec

u

y

u

u m



Ans (B) 842


13

08 13

08

0023 Re Pr

0023 Pr

Nu

G d

mG

A


081

3

08

2023 Pr

2 i

i

GdNu

hNu

Nu h

We know that

0

0

1 1 1 1 i

i i

h hU h h h



ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587






















Ans 10 cm

Solution

We know that

1 2P P (1)

Kerosene

Water

x

20

30 42



Where

1

1

[20 (30 )]

800 981 [20 (30 )] 1000 981

800 981[50 ] 9810

392400 7848 9810

392400 1962

k waterP g x g x

x x

x x

x x

P x

And

P2 = water x g x 42

= 1000 x 981 x 42

= 412 020 Pa


412020 = 392400+1962 x

1962 x = 19620

X = 10 cm



3

025

1






Ans (A) 39






(A) 05 (B) 084 (C) 119 (D) 141

Ans (C) 119


1 13 34 4

1

4

( ) ( )0943 0943

( ) ( )

1

( )

l l v fg l l v fg

g w g w

g w

gh k g h LL LNu h

h L T T k k T T

NuT T

1

4

1

1

4

1

1

(100 10)

1

90

Nu

Nu

and

1

4

2

1

4

2

1

(100 55)

1

45

Nu

Nu

2

1

1

490119

45

u

u

N

N








_______

Ans 0396

Solution


vi ii

p x p

amp wersquo i i

i

p pyi

p p


02 920396

02 92 08 35

VB B B

B V TB B T T

i

x ppy

x P x pp




are nearly equal

(A) 033 (B) 05 (C) 1 (D) 2

Ans (C) 1



Ans 2


0

0 0

00

0

10 (10)2

1 5 15

A A

A AA

AA

A

C C

C Cr

CC

C







Ans (C)




Ans 285

Solution

From Arrhenius law

2

1 1 2

1 1Ekln

k R T T


So 2

1

184000 1 1

8314 550 600

kln

k

2 1285Ans k k

















Option P Option Q








(A) 40 (B) 424 (C) 92 (D) 128

Ans (A) 40



statements is TRUE



in the BWG number



Ans (B)





certain base time





Ans (D)







Ans (B)



Plant P Plant Q

NaOH Yes No

Na2S Yes No

Na2CO3 Yes Yes

NaHCO3 No Yes

Na2SO3 No Yes






Ans (D)


Group-1 Group-2






promoter







Ans x3 = 15

Solution

4 x4 + 13 x5 = 46

4 x4 + 5 x5 = 30

2 x3 + 5 x4 + 3x5 = 61


Ans (D)

Solution

D2 + 1 = 0

D = i



5cos 10siny x x



Ans 255

Ans 265

Solution

f (x) = x3 + 5 Interval [1 4]


1 ( ) ( )( )

f b f af c

b a

X=265

Ans (A) 9467

Solution



Put h = 1 a = 0 b = 4

y0 = 3 yn = 55 y1 = 10 y2 = 21 y3 = 36

Ans (A) 1802




We have


2 4

650 42600

kJ kg H SO kJx

kg H SO h h

Energy out = 28

10 (40 25) 420kJ kJ


Energy In = 2

2

286 (25 10) 378

kJ kJkg H O K

kg H O K h



So Energy removed

= 2600 + 378 + 420

= 1802 kJ

h


h





Ans 373 K


15irr revW W


1

2

1

2771277

2 1 2 1

2 1 2 1

2 1 1 1

2

2

( ) 15 ( )

15 ( )

15 ( )

6300 15 300 1 73

3

373

V V

V

P

P

C T T C T T

T T C T T

T T T T

T

T K

Ans 400


RTP

v b



0 1

1 0

1

1 0 1 0

( )

P

T P

P v b RT

RTv b

PRT

b b TP

Rb T b

P

v Rb

T P

h vv T

P T

R Rh T b T b b

P P

0

0

5 6

( )

4 10 (15 5) 10 400

T T

f i f i TT

dh b dP

h h b P P

J mol

Ans - 165



Ans - (C) 24

Solution

3 3

1

2

sec

120

240

36 001m mhr

P kPa

P kPa

Q


P Py y

g

= 3(240 120) 10

129810

x

= 12232+12

= 24232 m

Power required = 4Q

= 9810 X 001 X 24232

= 2377 kW



Ans (C) 252 rpm

Solution

We have

Power No = 3 5

i

P

P N D


3 5

3 5 ( )

i

i

Pconst

PN D

P N D I

Volume of tank = 2

4VD H

( )VD H given

3 ( )VV D II


3 5

3i

v

N DP

V D

For same 1 2

2

3 5 3 51 2

3 31

i i

v V

N D N DP

V D D




2 2

1 1

2i V

i V

D D

D D

1

2

1

2

2

3 32 1

23

2 1 1

23

2

( 4 )

14 252

2

V

V

V

V

DN N

D

DN N N rpm

D

N rpm

Ans 215 mm

Solution -



2

2

2

( )

18

10 (1180 1000)

18 01

1000

P p f

t

f

p

t

t p

d gv

d x xv

x

v d


Re 01p f

p

f

d vN

Putting values



2 3(1000 ) 10

0101

p pd x d x

3

3 3

8

1 1010

10 1

215

p

p

mmd m x mm

m

d mm

Ans 49 m3 h

Ans 0125 m s


2 2

fU

3

3 2

1 10 2005

005 10 1000 1

0125 sec

u

y

u

u m



Ans (B) 842


13

08 13

08

0023 Re Pr

0023 Pr

Nu

G d

mG

A


081

3

08

2023 Pr

2 i

i

GdNu

hNu

Nu h

We know that

0

0

1 1 1 1 i

i i

h hU h h h



ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587







Where

1

1

[20 (30 )]

800 981 [20 (30 )] 1000 981

800 981[50 ] 9810

392400 7848 9810

392400 1962

k waterP g x g x

x x

x x

x x

P x

And

P2 = water x g x 42

= 1000 x 981 x 42

= 412 020 Pa


412020 = 392400+1962 x

1962 x = 19620

X = 10 cm



3

025

1






Ans (A) 39






(A) 05 (B) 084 (C) 119 (D) 141

Ans (C) 119


1 13 34 4

1

4

( ) ( )0943 0943

( ) ( )

1

( )

l l v fg l l v fg

g w g w

g w

gh k g h LL LNu h

h L T T k k T T

NuT T

1

4

1

1

4

1

1

(100 10)

1

90

Nu

Nu

and

1

4

2

1

4

2

1

(100 55)

1

45

Nu

Nu

2

1

1

490119

45

u

u

N

N








_______

Ans 0396

Solution


vi ii

p x p

amp wersquo i i

i

p pyi

p p


02 920396

02 92 08 35

VB B B

B V TB B T T

i

x ppy

x P x pp




are nearly equal

(A) 033 (B) 05 (C) 1 (D) 2

Ans (C) 1



Ans 2


0

0 0

00

0

10 (10)2

1 5 15

A A

A AA

AA

A

C C

C Cr

CC

C







Ans (C)




Ans 285

Solution

From Arrhenius law

2

1 1 2

1 1Ekln

k R T T


So 2

1

184000 1 1

8314 550 600

kln

k

2 1285Ans k k

















Option P Option Q








(A) 40 (B) 424 (C) 92 (D) 128

Ans (A) 40



statements is TRUE



in the BWG number



Ans (B)





certain base time





Ans (D)







Ans (B)



Plant P Plant Q

NaOH Yes No

Na2S Yes No

Na2CO3 Yes Yes

NaHCO3 No Yes

Na2SO3 No Yes






Ans (D)


Group-1 Group-2






promoter







Ans x3 = 15

Solution

4 x4 + 13 x5 = 46

4 x4 + 5 x5 = 30

2 x3 + 5 x4 + 3x5 = 61


Ans (D)

Solution

D2 + 1 = 0

D = i



5cos 10siny x x



Ans 255

Ans 265

Solution

f (x) = x3 + 5 Interval [1 4]


1 ( ) ( )( )

f b f af c

b a

X=265

Ans (A) 9467

Solution



Put h = 1 a = 0 b = 4

y0 = 3 yn = 55 y1 = 10 y2 = 21 y3 = 36

Ans (A) 1802




We have


2 4

650 42600

kJ kg H SO kJx

kg H SO h h

Energy out = 28

10 (40 25) 420kJ kJ


Energy In = 2

2

286 (25 10) 378

kJ kJkg H O K

kg H O K h



So Energy removed

= 2600 + 378 + 420

= 1802 kJ

h


h





Ans 373 K


15irr revW W


1

2

1

2771277

2 1 2 1

2 1 2 1

2 1 1 1

2

2

( ) 15 ( )

15 ( )

15 ( )

6300 15 300 1 73

3

373

V V

V

P

P

C T T C T T

T T C T T

T T T T

T

T K

Ans 400


RTP

v b



0 1

1 0

1

1 0 1 0

( )

P

T P

P v b RT

RTv b

PRT

b b TP

Rb T b

P

v Rb

T P

h vv T

P T

R Rh T b T b b

P P

0

0

5 6

( )

4 10 (15 5) 10 400

T T

f i f i TT

dh b dP

h h b P P

J mol

Ans - 165



Ans - (C) 24

Solution

3 3

1

2

sec

120

240

36 001m mhr

P kPa

P kPa

Q


P Py y

g

= 3(240 120) 10

129810

x

= 12232+12

= 24232 m

Power required = 4Q

= 9810 X 001 X 24232

= 2377 kW



Ans (C) 252 rpm

Solution

We have

Power No = 3 5

i

P

P N D


3 5

3 5 ( )

i

i

Pconst

PN D

P N D I

Volume of tank = 2

4VD H

( )VD H given

3 ( )VV D II


3 5

3i

v

N DP

V D

For same 1 2

2

3 5 3 51 2

3 31

i i

v V

N D N DP

V D D




2 2

1 1

2i V

i V

D D

D D

1

2

1

2

2

3 32 1

23

2 1 1

23

2

( 4 )

14 252

2

V

V

V

V

DN N

D

DN N N rpm

D

N rpm

Ans 215 mm

Solution -



2

2

2

( )

18

10 (1180 1000)

18 01

1000

P p f

t

f

p

t

t p

d gv

d x xv

x

v d


Re 01p f

p

f

d vN

Putting values



2 3(1000 ) 10

0101

p pd x d x

3

3 3

8

1 1010

10 1

215

p

p

mmd m x mm

m

d mm

Ans 49 m3 h

Ans 0125 m s


2 2

fU

3

3 2

1 10 2005

005 10 1000 1

0125 sec

u

y

u

u m



Ans (B) 842


13

08 13

08

0023 Re Pr

0023 Pr

Nu

G d

mG

A


081

3

08

2023 Pr

2 i

i

GdNu

hNu

Nu h

We know that

0

0

1 1 1 1 i

i i

h hU h h h



ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587









Ans (A) 39






(A) 05 (B) 084 (C) 119 (D) 141

Ans (C) 119


1 13 34 4

1

4

( ) ( )0943 0943

( ) ( )

1

( )

l l v fg l l v fg

g w g w

g w

gh k g h LL LNu h

h L T T k k T T

NuT T

1

4

1

1

4

1

1

(100 10)

1

90

Nu

Nu

and

1

4

2

1

4

2

1

(100 55)

1

45

Nu

Nu

2

1

1

490119

45

u

u

N

N








_______

Ans 0396

Solution


vi ii

p x p

amp wersquo i i

i

p pyi

p p


02 920396

02 92 08 35

VB B B

B V TB B T T

i

x ppy

x P x pp




are nearly equal

(A) 033 (B) 05 (C) 1 (D) 2

Ans (C) 1



Ans 2


0

0 0

00

0

10 (10)2

1 5 15

A A

A AA

AA

A

C C

C Cr

CC

C







Ans (C)




Ans 285

Solution

From Arrhenius law

2

1 1 2

1 1Ekln

k R T T


So 2

1

184000 1 1

8314 550 600

kln

k

2 1285Ans k k

















Option P Option Q








(A) 40 (B) 424 (C) 92 (D) 128

Ans (A) 40



statements is TRUE



in the BWG number



Ans (B)





certain base time





Ans (D)







Ans (B)



Plant P Plant Q

NaOH Yes No

Na2S Yes No

Na2CO3 Yes Yes

NaHCO3 No Yes

Na2SO3 No Yes






Ans (D)


Group-1 Group-2






promoter







Ans x3 = 15

Solution

4 x4 + 13 x5 = 46

4 x4 + 5 x5 = 30

2 x3 + 5 x4 + 3x5 = 61


Ans (D)

Solution

D2 + 1 = 0

D = i



5cos 10siny x x



Ans 255

Ans 265

Solution

f (x) = x3 + 5 Interval [1 4]


1 ( ) ( )( )

f b f af c

b a

X=265

Ans (A) 9467

Solution



Put h = 1 a = 0 b = 4

y0 = 3 yn = 55 y1 = 10 y2 = 21 y3 = 36

Ans (A) 1802




We have


2 4

650 42600

kJ kg H SO kJx

kg H SO h h

Energy out = 28

10 (40 25) 420kJ kJ


Energy In = 2

2

286 (25 10) 378

kJ kJkg H O K

kg H O K h



So Energy removed

= 2600 + 378 + 420

= 1802 kJ

h


h





Ans 373 K


15irr revW W


1

2

1

2771277

2 1 2 1

2 1 2 1

2 1 1 1

2

2

( ) 15 ( )

15 ( )

15 ( )

6300 15 300 1 73

3

373

V V

V

P

P

C T T C T T

T T C T T

T T T T

T

T K

Ans 400


RTP

v b



0 1

1 0

1

1 0 1 0

( )

P

T P

P v b RT

RTv b

PRT

b b TP

Rb T b

P

v Rb

T P

h vv T

P T

R Rh T b T b b

P P

0

0

5 6

( )

4 10 (15 5) 10 400

T T

f i f i TT

dh b dP

h h b P P

J mol

Ans - 165



Ans - (C) 24

Solution

3 3

1

2

sec

120

240

36 001m mhr

P kPa

P kPa

Q


P Py y

g

= 3(240 120) 10

129810

x

= 12232+12

= 24232 m

Power required = 4Q

= 9810 X 001 X 24232

= 2377 kW



Ans (C) 252 rpm

Solution

We have

Power No = 3 5

i

P

P N D


3 5

3 5 ( )

i

i

Pconst

PN D

P N D I

Volume of tank = 2

4VD H

( )VD H given

3 ( )VV D II


3 5

3i

v

N DP

V D

For same 1 2

2

3 5 3 51 2

3 31

i i

v V

N D N DP

V D D




2 2

1 1

2i V

i V

D D

D D

1

2

1

2

2

3 32 1

23

2 1 1

23

2

( 4 )

14 252

2

V

V

V

V

DN N

D

DN N N rpm

D

N rpm

Ans 215 mm

Solution -



2

2

2

( )

18

10 (1180 1000)

18 01

1000

P p f

t

f

p

t

t p

d gv

d x xv

x

v d


Re 01p f

p

f

d vN

Putting values



2 3(1000 ) 10

0101

p pd x d x

3

3 3

8

1 1010

10 1

215

p

p

mmd m x mm

m

d mm

Ans 49 m3 h

Ans 0125 m s


2 2

fU

3

3 2

1 10 2005

005 10 1000 1

0125 sec

u

y

u

u m



Ans (B) 842


13

08 13

08

0023 Re Pr

0023 Pr

Nu

G d

mG

A


081

3

08

2023 Pr

2 i

i

GdNu

hNu

Nu h

We know that

0

0

1 1 1 1 i

i i

h hU h h h



ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587








_______

Ans 0396

Solution


vi ii

p x p

amp wersquo i i

i

p pyi

p p


02 920396

02 92 08 35

VB B B

B V TB B T T

i

x ppy

x P x pp




are nearly equal

(A) 033 (B) 05 (C) 1 (D) 2

Ans (C) 1



Ans 2


0

0 0

00

0

10 (10)2

1 5 15

A A

A AA

AA

A

C C

C Cr

CC

C







Ans (C)




Ans 285

Solution

From Arrhenius law

2

1 1 2

1 1Ekln

k R T T


So 2

1

184000 1 1

8314 550 600

kln

k

2 1285Ans k k

















Option P Option Q








(A) 40 (B) 424 (C) 92 (D) 128

Ans (A) 40



statements is TRUE



in the BWG number



Ans (B)





certain base time





Ans (D)







Ans (B)



Plant P Plant Q

NaOH Yes No

Na2S Yes No

Na2CO3 Yes Yes

NaHCO3 No Yes

Na2SO3 No Yes






Ans (D)


Group-1 Group-2






promoter







Ans x3 = 15

Solution

4 x4 + 13 x5 = 46

4 x4 + 5 x5 = 30

2 x3 + 5 x4 + 3x5 = 61


Ans (D)

Solution

D2 + 1 = 0

D = i



5cos 10siny x x



Ans 255

Ans 265

Solution

f (x) = x3 + 5 Interval [1 4]


1 ( ) ( )( )

f b f af c

b a

X=265

Ans (A) 9467

Solution



Put h = 1 a = 0 b = 4

y0 = 3 yn = 55 y1 = 10 y2 = 21 y3 = 36

Ans (A) 1802




We have


2 4

650 42600

kJ kg H SO kJx

kg H SO h h

Energy out = 28

10 (40 25) 420kJ kJ


Energy In = 2

2

286 (25 10) 378

kJ kJkg H O K

kg H O K h



So Energy removed

= 2600 + 378 + 420

= 1802 kJ

h


h





Ans 373 K


15irr revW W


1

2

1

2771277

2 1 2 1

2 1 2 1

2 1 1 1

2

2

( ) 15 ( )

15 ( )

15 ( )

6300 15 300 1 73

3

373

V V

V

P

P

C T T C T T

T T C T T

T T T T

T

T K

Ans 400


RTP

v b



0 1

1 0

1

1 0 1 0

( )

P

T P

P v b RT

RTv b

PRT

b b TP

Rb T b

P

v Rb

T P

h vv T

P T

R Rh T b T b b

P P

0

0

5 6

( )

4 10 (15 5) 10 400

T T

f i f i TT

dh b dP

h h b P P

J mol

Ans - 165



Ans - (C) 24

Solution

3 3

1

2

sec

120

240

36 001m mhr

P kPa

P kPa

Q


P Py y

g

= 3(240 120) 10

129810

x

= 12232+12

= 24232 m

Power required = 4Q

= 9810 X 001 X 24232

= 2377 kW



Ans (C) 252 rpm

Solution

We have

Power No = 3 5

i

P

P N D


3 5

3 5 ( )

i

i

Pconst

PN D

P N D I

Volume of tank = 2

4VD H

( )VD H given

3 ( )VV D II


3 5

3i

v

N DP

V D

For same 1 2

2

3 5 3 51 2

3 31

i i

v V

N D N DP

V D D




2 2

1 1

2i V

i V

D D

D D

1

2

1

2

2

3 32 1

23

2 1 1

23

2

( 4 )

14 252

2

V

V

V

V

DN N

D

DN N N rpm

D

N rpm

Ans 215 mm

Solution -



2

2

2

( )

18

10 (1180 1000)

18 01

1000

P p f

t

f

p

t

t p

d gv

d x xv

x

v d


Re 01p f

p

f

d vN

Putting values



2 3(1000 ) 10

0101

p pd x d x

3

3 3

8

1 1010

10 1

215

p

p

mmd m x mm

m

d mm

Ans 49 m3 h

Ans 0125 m s


2 2

fU

3

3 2

1 10 2005

005 10 1000 1

0125 sec

u

y

u

u m



Ans (B) 842


13

08 13

08

0023 Re Pr

0023 Pr

Nu

G d

mG

A


081

3

08

2023 Pr

2 i

i

GdNu

hNu

Nu h

We know that

0

0

1 1 1 1 i

i i

h hU h h h



ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587







Ans 2


0

0 0

00

0

10 (10)2

1 5 15

A A

A AA

AA

A

C C

C Cr

CC

C







Ans (C)




Ans 285

Solution

From Arrhenius law

2

1 1 2

1 1Ekln

k R T T


So 2

1

184000 1 1

8314 550 600

kln

k

2 1285Ans k k

















Option P Option Q








(A) 40 (B) 424 (C) 92 (D) 128

Ans (A) 40



statements is TRUE



in the BWG number



Ans (B)





certain base time





Ans (D)







Ans (B)



Plant P Plant Q

NaOH Yes No

Na2S Yes No

Na2CO3 Yes Yes

NaHCO3 No Yes

Na2SO3 No Yes






Ans (D)


Group-1 Group-2






promoter







Ans x3 = 15

Solution

4 x4 + 13 x5 = 46

4 x4 + 5 x5 = 30

2 x3 + 5 x4 + 3x5 = 61


Ans (D)

Solution

D2 + 1 = 0

D = i



5cos 10siny x x



Ans 255

Ans 265

Solution

f (x) = x3 + 5 Interval [1 4]


1 ( ) ( )( )

f b f af c

b a

X=265

Ans (A) 9467

Solution



Put h = 1 a = 0 b = 4

y0 = 3 yn = 55 y1 = 10 y2 = 21 y3 = 36

Ans (A) 1802




We have


2 4

650 42600

kJ kg H SO kJx

kg H SO h h

Energy out = 28

10 (40 25) 420kJ kJ


Energy In = 2

2

286 (25 10) 378

kJ kJkg H O K

kg H O K h



So Energy removed

= 2600 + 378 + 420

= 1802 kJ

h


h





Ans 373 K


15irr revW W


1

2

1

2771277

2 1 2 1

2 1 2 1

2 1 1 1

2

2

( ) 15 ( )

15 ( )

15 ( )

6300 15 300 1 73

3

373

V V

V

P

P

C T T C T T

T T C T T

T T T T

T

T K

Ans 400


RTP

v b



0 1

1 0

1

1 0 1 0

( )

P

T P

P v b RT

RTv b

PRT

b b TP

Rb T b

P

v Rb

T P

h vv T

P T

R Rh T b T b b

P P

0

0

5 6

( )

4 10 (15 5) 10 400

T T

f i f i TT

dh b dP

h h b P P

J mol

Ans - 165



Ans - (C) 24

Solution

3 3

1

2

sec

120

240

36 001m mhr

P kPa

P kPa

Q


P Py y

g

= 3(240 120) 10

129810

x

= 12232+12

= 24232 m

Power required = 4Q

= 9810 X 001 X 24232

= 2377 kW



Ans (C) 252 rpm

Solution

We have

Power No = 3 5

i

P

P N D


3 5

3 5 ( )

i

i

Pconst

PN D

P N D I

Volume of tank = 2

4VD H

( )VD H given

3 ( )VV D II


3 5

3i

v

N DP

V D

For same 1 2

2

3 5 3 51 2

3 31

i i

v V

N D N DP

V D D




2 2

1 1

2i V

i V

D D

D D

1

2

1

2

2

3 32 1

23

2 1 1

23

2

( 4 )

14 252

2

V

V

V

V

DN N

D

DN N N rpm

D

N rpm

Ans 215 mm

Solution -



2

2

2

( )

18

10 (1180 1000)

18 01

1000

P p f

t

f

p

t

t p

d gv

d x xv

x

v d


Re 01p f

p

f

d vN

Putting values



2 3(1000 ) 10

0101

p pd x d x

3

3 3

8

1 1010

10 1

215

p

p

mmd m x mm

m

d mm

Ans 49 m3 h

Ans 0125 m s


2 2

fU

3

3 2

1 10 2005

005 10 1000 1

0125 sec

u

y

u

u m



Ans (B) 842


13

08 13

08

0023 Re Pr

0023 Pr

Nu

G d

mG

A


081

3

08

2023 Pr

2 i

i

GdNu

hNu

Nu h

We know that

0

0

1 1 1 1 i

i i

h hU h h h



ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587







Ans (C)




Ans 285

Solution

From Arrhenius law

2

1 1 2

1 1Ekln

k R T T


So 2

1

184000 1 1

8314 550 600

kln

k

2 1285Ans k k

















Option P Option Q








(A) 40 (B) 424 (C) 92 (D) 128

Ans (A) 40



statements is TRUE



in the BWG number



Ans (B)





certain base time





Ans (D)







Ans (B)



Plant P Plant Q

NaOH Yes No

Na2S Yes No

Na2CO3 Yes Yes

NaHCO3 No Yes

Na2SO3 No Yes






Ans (D)


Group-1 Group-2






promoter







Ans x3 = 15

Solution

4 x4 + 13 x5 = 46

4 x4 + 5 x5 = 30

2 x3 + 5 x4 + 3x5 = 61


Ans (D)

Solution

D2 + 1 = 0

D = i



5cos 10siny x x



Ans 255

Ans 265

Solution

f (x) = x3 + 5 Interval [1 4]


1 ( ) ( )( )

f b f af c

b a

X=265

Ans (A) 9467

Solution



Put h = 1 a = 0 b = 4

y0 = 3 yn = 55 y1 = 10 y2 = 21 y3 = 36

Ans (A) 1802




We have


2 4

650 42600

kJ kg H SO kJx

kg H SO h h

Energy out = 28

10 (40 25) 420kJ kJ


Energy In = 2

2

286 (25 10) 378

kJ kJkg H O K

kg H O K h



So Energy removed

= 2600 + 378 + 420

= 1802 kJ

h


h





Ans 373 K


15irr revW W


1

2

1

2771277

2 1 2 1

2 1 2 1

2 1 1 1

2

2

( ) 15 ( )

15 ( )

15 ( )

6300 15 300 1 73

3

373

V V

V

P

P

C T T C T T

T T C T T

T T T T

T

T K

Ans 400


RTP

v b



0 1

1 0

1

1 0 1 0

( )

P

T P

P v b RT

RTv b

PRT

b b TP

Rb T b

P

v Rb

T P

h vv T

P T

R Rh T b T b b

P P

0

0

5 6

( )

4 10 (15 5) 10 400

T T

f i f i TT

dh b dP

h h b P P

J mol

Ans - 165



Ans - (C) 24

Solution

3 3

1

2

sec

120

240

36 001m mhr

P kPa

P kPa

Q


P Py y

g

= 3(240 120) 10

129810

x

= 12232+12

= 24232 m

Power required = 4Q

= 9810 X 001 X 24232

= 2377 kW



Ans (C) 252 rpm

Solution

We have

Power No = 3 5

i

P

P N D


3 5

3 5 ( )

i

i

Pconst

PN D

P N D I

Volume of tank = 2

4VD H

( )VD H given

3 ( )VV D II


3 5

3i

v

N DP

V D

For same 1 2

2

3 5 3 51 2

3 31

i i

v V

N D N DP

V D D




2 2

1 1

2i V

i V

D D

D D

1

2

1

2

2

3 32 1

23

2 1 1

23

2

( 4 )

14 252

2

V

V

V

V

DN N

D

DN N N rpm

D

N rpm

Ans 215 mm

Solution -



2

2

2

( )

18

10 (1180 1000)

18 01

1000

P p f

t

f

p

t

t p

d gv

d x xv

x

v d


Re 01p f

p

f

d vN

Putting values



2 3(1000 ) 10

0101

p pd x d x

3

3 3

8

1 1010

10 1

215

p

p

mmd m x mm

m

d mm

Ans 49 m3 h

Ans 0125 m s


2 2

fU

3

3 2

1 10 2005

005 10 1000 1

0125 sec

u

y

u

u m



Ans (B) 842


13

08 13

08

0023 Re Pr

0023 Pr

Nu

G d

mG

A


081

3

08

2023 Pr

2 i

i

GdNu

hNu

Nu h

We know that

0

0

1 1 1 1 i

i i

h hU h h h



ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587





















Option P Option Q








(A) 40 (B) 424 (C) 92 (D) 128

Ans (A) 40



statements is TRUE



in the BWG number



Ans (B)





certain base time





Ans (D)







Ans (B)



Plant P Plant Q

NaOH Yes No

Na2S Yes No

Na2CO3 Yes Yes

NaHCO3 No Yes

Na2SO3 No Yes






Ans (D)


Group-1 Group-2






promoter







Ans x3 = 15

Solution

4 x4 + 13 x5 = 46

4 x4 + 5 x5 = 30

2 x3 + 5 x4 + 3x5 = 61


Ans (D)

Solution

D2 + 1 = 0

D = i



5cos 10siny x x



Ans 255

Ans 265

Solution

f (x) = x3 + 5 Interval [1 4]


1 ( ) ( )( )

f b f af c

b a

X=265

Ans (A) 9467

Solution



Put h = 1 a = 0 b = 4

y0 = 3 yn = 55 y1 = 10 y2 = 21 y3 = 36

Ans (A) 1802




We have


2 4

650 42600

kJ kg H SO kJx

kg H SO h h

Energy out = 28

10 (40 25) 420kJ kJ


Energy In = 2

2

286 (25 10) 378

kJ kJkg H O K

kg H O K h



So Energy removed

= 2600 + 378 + 420

= 1802 kJ

h


h





Ans 373 K


15irr revW W


1

2

1

2771277

2 1 2 1

2 1 2 1

2 1 1 1

2

2

( ) 15 ( )

15 ( )

15 ( )

6300 15 300 1 73

3

373

V V

V

P

P

C T T C T T

T T C T T

T T T T

T

T K

Ans 400


RTP

v b



0 1

1 0

1

1 0 1 0

( )

P

T P

P v b RT

RTv b

PRT

b b TP

Rb T b

P

v Rb

T P

h vv T

P T

R Rh T b T b b

P P

0

0

5 6

( )

4 10 (15 5) 10 400

T T

f i f i TT

dh b dP

h h b P P

J mol

Ans - 165



Ans - (C) 24

Solution

3 3

1

2

sec

120

240

36 001m mhr

P kPa

P kPa

Q


P Py y

g

= 3(240 120) 10

129810

x

= 12232+12

= 24232 m

Power required = 4Q

= 9810 X 001 X 24232

= 2377 kW



Ans (C) 252 rpm

Solution

We have

Power No = 3 5

i

P

P N D


3 5

3 5 ( )

i

i

Pconst

PN D

P N D I

Volume of tank = 2

4VD H

( )VD H given

3 ( )VV D II


3 5

3i

v

N DP

V D

For same 1 2

2

3 5 3 51 2

3 31

i i

v V

N D N DP

V D D




2 2

1 1

2i V

i V

D D

D D

1

2

1

2

2

3 32 1

23

2 1 1

23

2

( 4 )

14 252

2

V

V

V

V

DN N

D

DN N N rpm

D

N rpm

Ans 215 mm

Solution -



2

2

2

( )

18

10 (1180 1000)

18 01

1000

P p f

t

f

p

t

t p

d gv

d x xv

x

v d


Re 01p f

p

f

d vN

Putting values



2 3(1000 ) 10

0101

p pd x d x

3

3 3

8

1 1010

10 1

215

p

p

mmd m x mm

m

d mm

Ans 49 m3 h

Ans 0125 m s


2 2

fU

3

3 2

1 10 2005

005 10 1000 1

0125 sec

u

y

u

u m



Ans (B) 842


13

08 13

08

0023 Re Pr

0023 Pr

Nu

G d

mG

A


081

3

08

2023 Pr

2 i

i

GdNu

hNu

Nu h

We know that

0

0

1 1 1 1 i

i i

h hU h h h



ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587










(A) 40 (B) 424 (C) 92 (D) 128

Ans (A) 40



statements is TRUE



in the BWG number



Ans (B)





certain base time





Ans (D)







Ans (B)



Plant P Plant Q

NaOH Yes No

Na2S Yes No

Na2CO3 Yes Yes

NaHCO3 No Yes

Na2SO3 No Yes






Ans (D)


Group-1 Group-2






promoter







Ans x3 = 15

Solution

4 x4 + 13 x5 = 46

4 x4 + 5 x5 = 30

2 x3 + 5 x4 + 3x5 = 61


Ans (D)

Solution

D2 + 1 = 0

D = i



5cos 10siny x x



Ans 255

Ans 265

Solution

f (x) = x3 + 5 Interval [1 4]


1 ( ) ( )( )

f b f af c

b a

X=265

Ans (A) 9467

Solution



Put h = 1 a = 0 b = 4

y0 = 3 yn = 55 y1 = 10 y2 = 21 y3 = 36

Ans (A) 1802




We have


2 4

650 42600

kJ kg H SO kJx

kg H SO h h

Energy out = 28

10 (40 25) 420kJ kJ


Energy In = 2

2

286 (25 10) 378

kJ kJkg H O K

kg H O K h



So Energy removed

= 2600 + 378 + 420

= 1802 kJ

h


h





Ans 373 K


15irr revW W


1

2

1

2771277

2 1 2 1

2 1 2 1

2 1 1 1

2

2

( ) 15 ( )

15 ( )

15 ( )

6300 15 300 1 73

3

373

V V

V

P

P

C T T C T T

T T C T T

T T T T

T

T K

Ans 400


RTP

v b



0 1

1 0

1

1 0 1 0

( )

P

T P

P v b RT

RTv b

PRT

b b TP

Rb T b

P

v Rb

T P

h vv T

P T

R Rh T b T b b

P P

0

0

5 6

( )

4 10 (15 5) 10 400

T T

f i f i TT

dh b dP

h h b P P

J mol

Ans - 165



Ans - (C) 24

Solution

3 3

1

2

sec

120

240

36 001m mhr

P kPa

P kPa

Q


P Py y

g

= 3(240 120) 10

129810

x

= 12232+12

= 24232 m

Power required = 4Q

= 9810 X 001 X 24232

= 2377 kW



Ans (C) 252 rpm

Solution

We have

Power No = 3 5

i

P

P N D


3 5

3 5 ( )

i

i

Pconst

PN D

P N D I

Volume of tank = 2

4VD H

( )VD H given

3 ( )VV D II


3 5

3i

v

N DP

V D

For same 1 2

2

3 5 3 51 2

3 31

i i

v V

N D N DP

V D D




2 2

1 1

2i V

i V

D D

D D

1

2

1

2

2

3 32 1

23

2 1 1

23

2

( 4 )

14 252

2

V

V

V

V

DN N

D

DN N N rpm

D

N rpm

Ans 215 mm

Solution -



2

2

2

( )

18

10 (1180 1000)

18 01

1000

P p f

t

f

p

t

t p

d gv

d x xv

x

v d


Re 01p f

p

f

d vN

Putting values



2 3(1000 ) 10

0101

p pd x d x

3

3 3

8

1 1010

10 1

215

p

p

mmd m x mm

m

d mm

Ans 49 m3 h

Ans 0125 m s


2 2

fU

3

3 2

1 10 2005

005 10 1000 1

0125 sec

u

y

u

u m



Ans (B) 842


13

08 13

08

0023 Re Pr

0023 Pr

Nu

G d

mG

A


081

3

08

2023 Pr

2 i

i

GdNu

hNu

Nu h

We know that

0

0

1 1 1 1 i

i i

h hU h h h



ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587







Ans (B)



Plant P Plant Q

NaOH Yes No

Na2S Yes No

Na2CO3 Yes Yes

NaHCO3 No Yes

Na2SO3 No Yes






Ans (D)


Group-1 Group-2






promoter







Ans x3 = 15

Solution

4 x4 + 13 x5 = 46

4 x4 + 5 x5 = 30

2 x3 + 5 x4 + 3x5 = 61


Ans (D)

Solution

D2 + 1 = 0

D = i



5cos 10siny x x



Ans 255

Ans 265

Solution

f (x) = x3 + 5 Interval [1 4]


1 ( ) ( )( )

f b f af c

b a

X=265

Ans (A) 9467

Solution



Put h = 1 a = 0 b = 4

y0 = 3 yn = 55 y1 = 10 y2 = 21 y3 = 36

Ans (A) 1802




We have


2 4

650 42600

kJ kg H SO kJx

kg H SO h h

Energy out = 28

10 (40 25) 420kJ kJ


Energy In = 2

2

286 (25 10) 378

kJ kJkg H O K

kg H O K h



So Energy removed

= 2600 + 378 + 420

= 1802 kJ

h


h





Ans 373 K


15irr revW W


1

2

1

2771277

2 1 2 1

2 1 2 1

2 1 1 1

2

2

( ) 15 ( )

15 ( )

15 ( )

6300 15 300 1 73

3

373

V V

V

P

P

C T T C T T

T T C T T

T T T T

T

T K

Ans 400


RTP

v b



0 1

1 0

1

1 0 1 0

( )

P

T P

P v b RT

RTv b

PRT

b b TP

Rb T b

P

v Rb

T P

h vv T

P T

R Rh T b T b b

P P

0

0

5 6

( )

4 10 (15 5) 10 400

T T

f i f i TT

dh b dP

h h b P P

J mol

Ans - 165



Ans - (C) 24

Solution

3 3

1

2

sec

120

240

36 001m mhr

P kPa

P kPa

Q


P Py y

g

= 3(240 120) 10

129810

x

= 12232+12

= 24232 m

Power required = 4Q

= 9810 X 001 X 24232

= 2377 kW



Ans (C) 252 rpm

Solution

We have

Power No = 3 5

i

P

P N D


3 5

3 5 ( )

i

i

Pconst

PN D

P N D I

Volume of tank = 2

4VD H

( )VD H given

3 ( )VV D II


3 5

3i

v

N DP

V D

For same 1 2

2

3 5 3 51 2

3 31

i i

v V

N D N DP

V D D




2 2

1 1

2i V

i V

D D

D D

1

2

1

2

2

3 32 1

23

2 1 1

23

2

( 4 )

14 252

2

V

V

V

V

DN N

D

DN N N rpm

D

N rpm

Ans 215 mm

Solution -



2

2

2

( )

18

10 (1180 1000)

18 01

1000

P p f

t

f

p

t

t p

d gv

d x xv

x

v d


Re 01p f

p

f

d vN

Putting values



2 3(1000 ) 10

0101

p pd x d x

3

3 3

8

1 1010

10 1

215

p

p

mmd m x mm

m

d mm

Ans 49 m3 h

Ans 0125 m s


2 2

fU

3

3 2

1 10 2005

005 10 1000 1

0125 sec

u

y

u

u m



Ans (B) 842


13

08 13

08

0023 Re Pr

0023 Pr

Nu

G d

mG

A


081

3

08

2023 Pr

2 i

i

GdNu

hNu

Nu h

We know that

0

0

1 1 1 1 i

i i

h hU h h h



ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587








Ans x3 = 15

Solution

4 x4 + 13 x5 = 46

4 x4 + 5 x5 = 30

2 x3 + 5 x4 + 3x5 = 61


Ans (D)

Solution

D2 + 1 = 0

D = i



5cos 10siny x x



Ans 255

Ans 265

Solution

f (x) = x3 + 5 Interval [1 4]


1 ( ) ( )( )

f b f af c

b a

X=265

Ans (A) 9467

Solution



Put h = 1 a = 0 b = 4

y0 = 3 yn = 55 y1 = 10 y2 = 21 y3 = 36

Ans (A) 1802




We have


2 4

650 42600

kJ kg H SO kJx

kg H SO h h

Energy out = 28

10 (40 25) 420kJ kJ


Energy In = 2

2

286 (25 10) 378

kJ kJkg H O K

kg H O K h



So Energy removed

= 2600 + 378 + 420

= 1802 kJ

h


h





Ans 373 K


15irr revW W


1

2

1

2771277

2 1 2 1

2 1 2 1

2 1 1 1

2

2

( ) 15 ( )

15 ( )

15 ( )

6300 15 300 1 73

3

373

V V

V

P

P

C T T C T T

T T C T T

T T T T

T

T K

Ans 400


RTP

v b



0 1

1 0

1

1 0 1 0

( )

P

T P

P v b RT

RTv b

PRT

b b TP

Rb T b

P

v Rb

T P

h vv T

P T

R Rh T b T b b

P P

0

0

5 6

( )

4 10 (15 5) 10 400

T T

f i f i TT

dh b dP

h h b P P

J mol

Ans - 165



Ans - (C) 24

Solution

3 3

1

2

sec

120

240

36 001m mhr

P kPa

P kPa

Q


P Py y

g

= 3(240 120) 10

129810

x

= 12232+12

= 24232 m

Power required = 4Q

= 9810 X 001 X 24232

= 2377 kW



Ans (C) 252 rpm

Solution

We have

Power No = 3 5

i

P

P N D


3 5

3 5 ( )

i

i

Pconst

PN D

P N D I

Volume of tank = 2

4VD H

( )VD H given

3 ( )VV D II


3 5

3i

v

N DP

V D

For same 1 2

2

3 5 3 51 2

3 31

i i

v V

N D N DP

V D D




2 2

1 1

2i V

i V

D D

D D

1

2

1

2

2

3 32 1

23

2 1 1

23

2

( 4 )

14 252

2

V

V

V

V

DN N

D

DN N N rpm

D

N rpm

Ans 215 mm

Solution -



2

2

2

( )

18

10 (1180 1000)

18 01

1000

P p f

t

f

p

t

t p

d gv

d x xv

x

v d


Re 01p f

p

f

d vN

Putting values



2 3(1000 ) 10

0101

p pd x d x

3

3 3

8

1 1010

10 1

215

p

p

mmd m x mm

m

d mm

Ans 49 m3 h

Ans 0125 m s


2 2

fU

3

3 2

1 10 2005

005 10 1000 1

0125 sec

u

y

u

u m



Ans (B) 842


13

08 13

08

0023 Re Pr

0023 Pr

Nu

G d

mG

A


081

3

08

2023 Pr

2 i

i

GdNu

hNu

Nu h

We know that

0

0

1 1 1 1 i

i i

h hU h h h



ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587







Ans 255

Ans 265

Solution

f (x) = x3 + 5 Interval [1 4]


1 ( ) ( )( )

f b f af c

b a

X=265

Ans (A) 9467

Solution



Put h = 1 a = 0 b = 4

y0 = 3 yn = 55 y1 = 10 y2 = 21 y3 = 36

Ans (A) 1802




We have


2 4

650 42600

kJ kg H SO kJx

kg H SO h h

Energy out = 28

10 (40 25) 420kJ kJ


Energy In = 2

2

286 (25 10) 378

kJ kJkg H O K

kg H O K h



So Energy removed

= 2600 + 378 + 420

= 1802 kJ

h


h





Ans 373 K


15irr revW W


1

2

1

2771277

2 1 2 1

2 1 2 1

2 1 1 1

2

2

( ) 15 ( )

15 ( )

15 ( )

6300 15 300 1 73

3

373

V V

V

P

P

C T T C T T

T T C T T

T T T T

T

T K

Ans 400


RTP

v b



0 1

1 0

1

1 0 1 0

( )

P

T P

P v b RT

RTv b

PRT

b b TP

Rb T b

P

v Rb

T P

h vv T

P T

R Rh T b T b b

P P

0

0

5 6

( )

4 10 (15 5) 10 400

T T

f i f i TT

dh b dP

h h b P P

J mol

Ans - 165



Ans - (C) 24

Solution

3 3

1

2

sec

120

240

36 001m mhr

P kPa

P kPa

Q


P Py y

g

= 3(240 120) 10

129810

x

= 12232+12

= 24232 m

Power required = 4Q

= 9810 X 001 X 24232

= 2377 kW



Ans (C) 252 rpm

Solution

We have

Power No = 3 5

i

P

P N D


3 5

3 5 ( )

i

i

Pconst

PN D

P N D I

Volume of tank = 2

4VD H

( )VD H given

3 ( )VV D II


3 5

3i

v

N DP

V D

For same 1 2

2

3 5 3 51 2

3 31

i i

v V

N D N DP

V D D




2 2

1 1

2i V

i V

D D

D D

1

2

1

2

2

3 32 1

23

2 1 1

23

2

( 4 )

14 252

2

V

V

V

V

DN N

D

DN N N rpm

D

N rpm

Ans 215 mm

Solution -



2

2

2

( )

18

10 (1180 1000)

18 01

1000

P p f

t

f

p

t

t p

d gv

d x xv

x

v d


Re 01p f

p

f

d vN

Putting values



2 3(1000 ) 10

0101

p pd x d x

3

3 3

8

1 1010

10 1

215

p

p

mmd m x mm

m

d mm

Ans 49 m3 h

Ans 0125 m s


2 2

fU

3

3 2

1 10 2005

005 10 1000 1

0125 sec

u

y

u

u m



Ans (B) 842


13

08 13

08

0023 Re Pr

0023 Pr

Nu

G d

mG

A


081

3

08

2023 Pr

2 i

i

GdNu

hNu

Nu h

We know that

0

0

1 1 1 1 i

i i

h hU h h h



ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587







Put h = 1 a = 0 b = 4

y0 = 3 yn = 55 y1 = 10 y2 = 21 y3 = 36

Ans (A) 1802




We have


2 4

650 42600

kJ kg H SO kJx

kg H SO h h

Energy out = 28

10 (40 25) 420kJ kJ


Energy In = 2

2

286 (25 10) 378

kJ kJkg H O K

kg H O K h



So Energy removed

= 2600 + 378 + 420

= 1802 kJ

h


h





Ans 373 K


15irr revW W


1

2

1

2771277

2 1 2 1

2 1 2 1

2 1 1 1

2

2

( ) 15 ( )

15 ( )

15 ( )

6300 15 300 1 73

3

373

V V

V

P

P

C T T C T T

T T C T T

T T T T

T

T K

Ans 400


RTP

v b



0 1

1 0

1

1 0 1 0

( )

P

T P

P v b RT

RTv b

PRT

b b TP

Rb T b

P

v Rb

T P

h vv T

P T

R Rh T b T b b

P P

0

0

5 6

( )

4 10 (15 5) 10 400

T T

f i f i TT

dh b dP

h h b P P

J mol

Ans - 165



Ans - (C) 24

Solution

3 3

1

2

sec

120

240

36 001m mhr

P kPa

P kPa

Q


P Py y

g

= 3(240 120) 10

129810

x

= 12232+12

= 24232 m

Power required = 4Q

= 9810 X 001 X 24232

= 2377 kW



Ans (C) 252 rpm

Solution

We have

Power No = 3 5

i

P

P N D


3 5

3 5 ( )

i

i

Pconst

PN D

P N D I

Volume of tank = 2

4VD H

( )VD H given

3 ( )VV D II


3 5

3i

v

N DP

V D

For same 1 2

2

3 5 3 51 2

3 31

i i

v V

N D N DP

V D D




2 2

1 1

2i V

i V

D D

D D

1

2

1

2

2

3 32 1

23

2 1 1

23

2

( 4 )

14 252

2

V

V

V

V

DN N

D

DN N N rpm

D

N rpm

Ans 215 mm

Solution -



2

2

2

( )

18

10 (1180 1000)

18 01

1000

P p f

t

f

p

t

t p

d gv

d x xv

x

v d


Re 01p f

p

f

d vN

Putting values



2 3(1000 ) 10

0101

p pd x d x

3

3 3

8

1 1010

10 1

215

p

p

mmd m x mm

m

d mm

Ans 49 m3 h

Ans 0125 m s


2 2

fU

3

3 2

1 10 2005

005 10 1000 1

0125 sec

u

y

u

u m



Ans (B) 842


13

08 13

08

0023 Re Pr

0023 Pr

Nu

G d

mG

A


081

3

08

2023 Pr

2 i

i

GdNu

hNu

Nu h

We know that

0

0

1 1 1 1 i

i i

h hU h h h



ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587







So Energy removed

= 2600 + 378 + 420

= 1802 kJ

h


h





Ans 373 K


15irr revW W


1

2

1

2771277

2 1 2 1

2 1 2 1

2 1 1 1

2

2

( ) 15 ( )

15 ( )

15 ( )

6300 15 300 1 73

3

373

V V

V

P

P

C T T C T T

T T C T T

T T T T

T

T K

Ans 400


RTP

v b



0 1

1 0

1

1 0 1 0

( )

P

T P

P v b RT

RTv b

PRT

b b TP

Rb T b

P

v Rb

T P

h vv T

P T

R Rh T b T b b

P P

0

0

5 6

( )

4 10 (15 5) 10 400

T T

f i f i TT

dh b dP

h h b P P

J mol

Ans - 165



Ans - (C) 24

Solution

3 3

1

2

sec

120

240

36 001m mhr

P kPa

P kPa

Q


P Py y

g

= 3(240 120) 10

129810

x

= 12232+12

= 24232 m

Power required = 4Q

= 9810 X 001 X 24232

= 2377 kW



Ans (C) 252 rpm

Solution

We have

Power No = 3 5

i

P

P N D


3 5

3 5 ( )

i

i

Pconst

PN D

P N D I

Volume of tank = 2

4VD H

( )VD H given

3 ( )VV D II


3 5

3i

v

N DP

V D

For same 1 2

2

3 5 3 51 2

3 31

i i

v V

N D N DP

V D D




2 2

1 1

2i V

i V

D D

D D

1

2

1

2

2

3 32 1

23

2 1 1

23

2

( 4 )

14 252

2

V

V

V

V

DN N

D

DN N N rpm

D

N rpm

Ans 215 mm

Solution -



2

2

2

( )

18

10 (1180 1000)

18 01

1000

P p f

t

f

p

t

t p

d gv

d x xv

x

v d


Re 01p f

p

f

d vN

Putting values



2 3(1000 ) 10

0101

p pd x d x

3

3 3

8

1 1010

10 1

215

p

p

mmd m x mm

m

d mm

Ans 49 m3 h

Ans 0125 m s


2 2

fU

3

3 2

1 10 2005

005 10 1000 1

0125 sec

u

y

u

u m



Ans (B) 842


13

08 13

08

0023 Re Pr

0023 Pr

Nu

G d

mG

A


081

3

08

2023 Pr

2 i

i

GdNu

hNu

Nu h

We know that

0

0

1 1 1 1 i

i i

h hU h h h



ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587









Ans 373 K


15irr revW W


1

2

1

2771277

2 1 2 1

2 1 2 1

2 1 1 1

2

2

( ) 15 ( )

15 ( )

15 ( )

6300 15 300 1 73

3

373

V V

V

P

P

C T T C T T

T T C T T

T T T T

T

T K

Ans 400


RTP

v b



0 1

1 0

1

1 0 1 0

( )

P

T P

P v b RT

RTv b

PRT

b b TP

Rb T b

P

v Rb

T P

h vv T

P T

R Rh T b T b b

P P

0

0

5 6

( )

4 10 (15 5) 10 400

T T

f i f i TT

dh b dP

h h b P P

J mol

Ans - 165



Ans - (C) 24

Solution

3 3

1

2

sec

120

240

36 001m mhr

P kPa

P kPa

Q


P Py y

g

= 3(240 120) 10

129810

x

= 12232+12

= 24232 m

Power required = 4Q

= 9810 X 001 X 24232

= 2377 kW



Ans (C) 252 rpm

Solution

We have

Power No = 3 5

i

P

P N D


3 5

3 5 ( )

i

i

Pconst

PN D

P N D I

Volume of tank = 2

4VD H

( )VD H given

3 ( )VV D II


3 5

3i

v

N DP

V D

For same 1 2

2

3 5 3 51 2

3 31

i i

v V

N D N DP

V D D




2 2

1 1

2i V

i V

D D

D D

1

2

1

2

2

3 32 1

23

2 1 1

23

2

( 4 )

14 252

2

V

V

V

V

DN N

D

DN N N rpm

D

N rpm

Ans 215 mm

Solution -



2

2

2

( )

18

10 (1180 1000)

18 01

1000

P p f

t

f

p

t

t p

d gv

d x xv

x

v d


Re 01p f

p

f

d vN

Putting values



2 3(1000 ) 10

0101

p pd x d x

3

3 3

8

1 1010

10 1

215

p

p

mmd m x mm

m

d mm

Ans 49 m3 h

Ans 0125 m s


2 2

fU

3

3 2

1 10 2005

005 10 1000 1

0125 sec

u

y

u

u m



Ans (B) 842


13

08 13

08

0023 Re Pr

0023 Pr

Nu

G d

mG

A


081

3

08

2023 Pr

2 i

i

GdNu

hNu

Nu h

We know that

0

0

1 1 1 1 i

i i

h hU h h h



ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587







Ans 373 K


15irr revW W


1

2

1

2771277

2 1 2 1

2 1 2 1

2 1 1 1

2

2

( ) 15 ( )

15 ( )

15 ( )

6300 15 300 1 73

3

373

V V

V

P

P

C T T C T T

T T C T T

T T T T

T

T K

Ans 400


RTP

v b



0 1

1 0

1

1 0 1 0

( )

P

T P

P v b RT

RTv b

PRT

b b TP

Rb T b

P

v Rb

T P

h vv T

P T

R Rh T b T b b

P P

0

0

5 6

( )

4 10 (15 5) 10 400

T T

f i f i TT

dh b dP

h h b P P

J mol

Ans - 165



Ans - (C) 24

Solution

3 3

1

2

sec

120

240

36 001m mhr

P kPa

P kPa

Q


P Py y

g

= 3(240 120) 10

129810

x

= 12232+12

= 24232 m

Power required = 4Q

= 9810 X 001 X 24232

= 2377 kW



Ans (C) 252 rpm

Solution

We have

Power No = 3 5

i

P

P N D


3 5

3 5 ( )

i

i

Pconst

PN D

P N D I

Volume of tank = 2

4VD H

( )VD H given

3 ( )VV D II


3 5

3i

v

N DP

V D

For same 1 2

2

3 5 3 51 2

3 31

i i

v V

N D N DP

V D D




2 2

1 1

2i V

i V

D D

D D

1

2

1

2

2

3 32 1

23

2 1 1

23

2

( 4 )

14 252

2

V

V

V

V

DN N

D

DN N N rpm

D

N rpm

Ans 215 mm

Solution -



2

2

2

( )

18

10 (1180 1000)

18 01

1000

P p f

t

f

p

t

t p

d gv

d x xv

x

v d


Re 01p f

p

f

d vN

Putting values



2 3(1000 ) 10

0101

p pd x d x

3

3 3

8

1 1010

10 1

215

p

p

mmd m x mm

m

d mm

Ans 49 m3 h

Ans 0125 m s


2 2

fU

3

3 2

1 10 2005

005 10 1000 1

0125 sec

u

y

u

u m



Ans (B) 842


13

08 13

08

0023 Re Pr

0023 Pr

Nu

G d

mG

A


081

3

08

2023 Pr

2 i

i

GdNu

hNu

Nu h

We know that

0

0

1 1 1 1 i

i i

h hU h h h



ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587







0 1

1 0

1

1 0 1 0

( )

P

T P

P v b RT

RTv b

PRT

b b TP

Rb T b

P

v Rb

T P

h vv T

P T

R Rh T b T b b

P P

0

0

5 6

( )

4 10 (15 5) 10 400

T T

f i f i TT

dh b dP

h h b P P

J mol

Ans - 165



Ans - (C) 24

Solution

3 3

1

2

sec

120

240

36 001m mhr

P kPa

P kPa

Q


P Py y

g

= 3(240 120) 10

129810

x

= 12232+12

= 24232 m

Power required = 4Q

= 9810 X 001 X 24232

= 2377 kW



Ans (C) 252 rpm

Solution

We have

Power No = 3 5

i

P

P N D


3 5

3 5 ( )

i

i

Pconst

PN D

P N D I

Volume of tank = 2

4VD H

( )VD H given

3 ( )VV D II


3 5

3i

v

N DP

V D

For same 1 2

2

3 5 3 51 2

3 31

i i

v V

N D N DP

V D D




2 2

1 1

2i V

i V

D D

D D

1

2

1

2

2

3 32 1

23

2 1 1

23

2

( 4 )

14 252

2

V

V

V

V

DN N

D

DN N N rpm

D

N rpm

Ans 215 mm

Solution -



2

2

2

( )

18

10 (1180 1000)

18 01

1000

P p f

t

f

p

t

t p

d gv

d x xv

x

v d


Re 01p f

p

f

d vN

Putting values



2 3(1000 ) 10

0101

p pd x d x

3

3 3

8

1 1010

10 1

215

p

p

mmd m x mm

m

d mm

Ans 49 m3 h

Ans 0125 m s


2 2

fU

3

3 2

1 10 2005

005 10 1000 1

0125 sec

u

y

u

u m



Ans (B) 842


13

08 13

08

0023 Re Pr

0023 Pr

Nu

G d

mG

A


081

3

08

2023 Pr

2 i

i

GdNu

hNu

Nu h

We know that

0

0

1 1 1 1 i

i i

h hU h h h



ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587







Ans - (C) 24

Solution

3 3

1

2

sec

120

240

36 001m mhr

P kPa

P kPa

Q


P Py y

g

= 3(240 120) 10

129810

x

= 12232+12

= 24232 m

Power required = 4Q

= 9810 X 001 X 24232

= 2377 kW



Ans (C) 252 rpm

Solution

We have

Power No = 3 5

i

P

P N D


3 5

3 5 ( )

i

i

Pconst

PN D

P N D I

Volume of tank = 2

4VD H

( )VD H given

3 ( )VV D II


3 5

3i

v

N DP

V D

For same 1 2

2

3 5 3 51 2

3 31

i i

v V

N D N DP

V D D




2 2

1 1

2i V

i V

D D

D D

1

2

1

2

2

3 32 1

23

2 1 1

23

2

( 4 )

14 252

2

V

V

V

V

DN N

D

DN N N rpm

D

N rpm

Ans 215 mm

Solution -



2

2

2

( )

18

10 (1180 1000)

18 01

1000

P p f

t

f

p

t

t p

d gv

d x xv

x

v d


Re 01p f

p

f

d vN

Putting values



2 3(1000 ) 10

0101

p pd x d x

3

3 3

8

1 1010

10 1

215

p

p

mmd m x mm

m

d mm

Ans 49 m3 h

Ans 0125 m s


2 2

fU

3

3 2

1 10 2005

005 10 1000 1

0125 sec

u

y

u

u m



Ans (B) 842


13

08 13

08

0023 Re Pr

0023 Pr

Nu

G d

mG

A


081

3

08

2023 Pr

2 i

i

GdNu

hNu

Nu h

We know that

0

0

1 1 1 1 i

i i

h hU h h h



ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587







Ans (C) 252 rpm

Solution

We have

Power No = 3 5

i

P

P N D


3 5

3 5 ( )

i

i

Pconst

PN D

P N D I

Volume of tank = 2

4VD H

( )VD H given

3 ( )VV D II


3 5

3i

v

N DP

V D

For same 1 2

2

3 5 3 51 2

3 31

i i

v V

N D N DP

V D D




2 2

1 1

2i V

i V

D D

D D

1

2

1

2

2

3 32 1

23

2 1 1

23

2

( 4 )

14 252

2

V

V

V

V

DN N

D

DN N N rpm

D

N rpm

Ans 215 mm

Solution -



2

2

2

( )

18

10 (1180 1000)

18 01

1000

P p f

t

f

p

t

t p

d gv

d x xv

x

v d


Re 01p f

p

f

d vN

Putting values



2 3(1000 ) 10

0101

p pd x d x

3

3 3

8

1 1010

10 1

215

p

p

mmd m x mm

m

d mm

Ans 49 m3 h

Ans 0125 m s


2 2

fU

3

3 2

1 10 2005

005 10 1000 1

0125 sec

u

y

u

u m



Ans (B) 842


13

08 13

08

0023 Re Pr

0023 Pr

Nu

G d

mG

A


081

3

08

2023 Pr

2 i

i

GdNu

hNu

Nu h

We know that

0

0

1 1 1 1 i

i i

h hU h h h



ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587







2 2

1 1

2i V

i V

D D

D D

1

2

1

2

2

3 32 1

23

2 1 1

23

2

( 4 )

14 252

2

V

V

V

V

DN N

D

DN N N rpm

D

N rpm

Ans 215 mm

Solution -



2

2

2

( )

18

10 (1180 1000)

18 01

1000

P p f

t

f

p

t

t p

d gv

d x xv

x

v d


Re 01p f

p

f

d vN

Putting values



2 3(1000 ) 10

0101

p pd x d x

3

3 3

8

1 1010

10 1

215

p

p

mmd m x mm

m

d mm

Ans 49 m3 h

Ans 0125 m s


2 2

fU

3

3 2

1 10 2005

005 10 1000 1

0125 sec

u

y

u

u m



Ans (B) 842


13

08 13

08

0023 Re Pr

0023 Pr

Nu

G d

mG

A


081

3

08

2023 Pr

2 i

i

GdNu

hNu

Nu h

We know that

0

0

1 1 1 1 i

i i

h hU h h h



ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587







2 3(1000 ) 10

0101

p pd x d x

3

3 3

8

1 1010

10 1

215

p

p

mmd m x mm

m

d mm

Ans 49 m3 h

Ans 0125 m s


2 2

fU

3

3 2

1 10 2005

005 10 1000 1

0125 sec

u

y

u

u m



Ans (B) 842


13

08 13

08

0023 Re Pr

0023 Pr

Nu

G d

mG

A


081

3

08

2023 Pr

2 i

i

GdNu

hNu

Nu h

We know that

0

0

1 1 1 1 i

i i

h hU h h h



ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587







Ans (B) 842


13

08 13

08

0023 Re Pr

0023 Pr

Nu

G d

mG

A


081

3

08

2023 Pr

2 i

i

GdNu

hNu

Nu h

We know that

0

0

1 1 1 1 i

i i

h hU h h h



ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587







ie iU h

and 082 iU h

0

S i

S P

T T UAnT T mC

0 2S i

S p

T T U AnT T m C

Dividing 08

0

135 20

2135 90

2135 20

135

n

nT

0

0938211487

115

135n

T

0

00

11508167

135

842

nT

T C



Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587







Ans (A) 26

Ans 025


11 120 1F F


2

AF

A

22 21 1

21 22 22 2

014 1025

4 402

r rF

r r



Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587







Ans 1


R min = 1

1 1

9 71

7 5DX Y

Y x

Ans (C) 105

Solution


βCA = 119884119888119884119886

119883119888119883119886 =

0301

0207= 105



Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587







Ans (C) 03

Solution -

We know that


1A A

A

A A

K PQ

K P


Kg of Adsorbentmax


Adsorbent

A

m

qQ


kg of

1A A

m A

K Pq

q K


1001(1)

1 10A

m A

K

q K


1004(2)

1 100A

m A

KO

q K



dsorbed

kg Aq

kg of A



Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587







Ans (B) 108

Solution -

1

2

1108

BLM BLM

NA P

NA Y P


Solution -

For reaction

1 2A A Br k C k C

For MFR 1 2( )

O OA A A A

m

A A B

C X C X

r k C k C

0(1 )


Am

A A

X

k X k X


81 02

10AX




5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587








5

1 0510

AX


1 2

021 4 1 ( )

(1 02) (02)k k I

k k


1 2

055 02 ( )

(1 05) (05)k k II

k k


Ans 8 m3




T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587








T = 350 + 25 XA

At XA = 08

T = 350 + 25 x 08 = 370 K


10

A

molr

m s

For MFR 0

30 100 088

( ) 100

m A

Am

A

V X

FA rA

FA XV m

r

Ans (B)






catalyst



Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587







Ans 025

Solution




P = 5 min


( )mt

m

eE t

At t = 0 E (0) = 1

m



So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587







So at t = 5 1

(0 5) 005 20 minm

m

E

So 5

02520

p

m

Ans 1013 m3 hr

Solution -


f = f0 times eBx

at x = 01 f = 2

at x = 02 f = 3


f = 1013 m3hr

Ans 25





2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587







2

1 101 1 0

15 25 100cK

s


Ans (A) 159




Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587







Overshoot = 2

80exp

100 1

B

A

can be calculate


2

2

1dt


159 sec

Ans (D) 78

Solution Given that

3

3

150

10150 [10 ]

100

1666

volume of liquid m


x m v



2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587







2

2000 4000 (6000)4

106000

100

T

dC dh

dh

V = 2

2

1666

4

4

d hh

d

2

2

2

2

3

3

16666000 (6600)

4

4

( ) 6000 2 4 (1666) (6600)

( ) 4

4 (1666) (6600)3000

4 (1666) (6600)

3000

4663

775

T

T

dC d

d

d C d

d d d

dd

d

d

d m

Ans (D) 6

Solution







C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587








C6H12 + 12 H2O rarr 6CO2 + 18H2



2

2

18

3



For more details


Call us at (+91) ndash 9873452122 9818652587







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