General Aptitude Questions 4 1 2 6 3 π

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General Aptitude Questions
Q.No-1-5 Carry One Mark Each

1. 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
Answer: (B)
4
Exp: Volume of a sphere of diameter 1 unit is π(1 2)3 =π 6 and volume of a cube of side 1 unit is
3
13 = 1

2. The unruly crowd demanded that the accused be without trial.
(A) hanged (B) hanging (C) hankering (D) hung
Answer: (A)

3. 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
Answer: (D)

4 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)
Answer: (C)

5. 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) 1.43 (B) 2.06 (C) 2.68 (D) 2.88
Answer: (A)
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Exp: Area of the window = Area of square + Area of equilateral triangle
3 3
=12 + (12)=1+ =1.43
4 4
[Perimeter of the window is 6 m
⇒ 4x+2x = 6, where x is length of each side]
⇒ x = 1

Q.No-6-10 Carry Two Marks Each

6. 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.

7. 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) and (iii) only
(C) (iv) only (D) (i) only
Answer: (C)

8. Find the missing sequence in the letter series.
B, FH, LNP, _ _ _ _.
(A) SUWY (B) TUVW (C) TVXZ (D) TWXZ
Answer: (C)

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9. The binary operation □ is defined as a □ b = ab+(a+b), where a and b are any two real numbers.
he 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

Answer: (A)
Exp: a(cid:1)x=a⇒ax+(a+x)=a
⇒x(1+a)=0⇒x =0 is the identity element

( )
10. Which of the following curves represents the function y=1n esin(x) for x <2π?
Here, x represents the abscissa and ݕ represents the ordinate.

(A)

(B)

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(C)

(D)

Answer: (B)
( )
Exp: y=ln e1sinx is a periodic function having period π and maximum value of y is 1 and y > 0.

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Chemical Engineering
Q.No-1-25 Carry One Mark Each

1. Which one of the following is an iterative technique for solving a system of simultaneous linear
algebraic equations?
(A) Gauss elimination (B) Gauss-Jordan
(C) Gauss-Seidel (D) LU decomposition
Answer: (C)

2. The Laplace transform of eat sin eatsin(bt)is
b (s−a) (s−a) b
(A) (B) (C) (D)
(s−a)2 +b2 (s−a)2 +b2 (s−a)2 −b2 (s−a)2 −b2
Answer: (A)
b
Exp: L{eatsinbt}=
(s−a)2 +b2

3. What are the modulus (r) and argument (θ) of the complex number 3 + 4i?
4 3
(A) r= 7,θ=tan−1  (B) r= 7,θ=tan−1 
3 4
3 4
(C) r=5,θ=tan−1  (D) r=5,θ=tan−1 
4 3
Answer: (D)
Exp: Given, Z=3+4i
T= Z = (3)2 +(4)2 =5
y 4
θ=Tan−1   = Tan−1  
x 3

4. A liquid mixture of ethanol and water is flowing as inlet stream P into a stream splitter. It is split
into two streams, Q and R, as shown in the figure below.

Q

P

R

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The flow rate of P, containing 30 mass% of ethanol, is 100kg/h. What is the least number of
additional specification(s) required to determine the mass flow rates and compositions (mass %)
of the two exit streams?
(A) 0 (B) 1 (C) 2 (D) 3
Answer: (C)
Exp: Independent material balance equations
Given,M =0.3
EtoH,P
P =100 kg h
∴Q+R =100 ___(1)
m Q+m R=30 ___(2)
EtOH,Q EtoH,R
Now, to solve equations (1) & (2) so as to determine all flow rates and mass compositions, both
m Qandm R have to be specified.
EtoH,Q EtoH,R
∴ Least no. of additional specifications required = 2

5. The partial molar enthalpy (in kJ/mol) of species 1 in a binary mixture is given
by h =2−60x2 +100x x2, where x and x are the mole fractions of species 1 and 2,
1 2 1 2 1 2
respectively. The partial molar enthalpy (in kJ/mol, rounded off to the first decimal place) of
species 1 at infinite dilution is ________
Answer: (58)
Exp: h =2−60x2 +100x x2 _______(1)
1 2 1 2
∴ Partial molar enthalpy of species 1 at infinite dilution may be obtained from eqn. (1) by putting
x = 0 and x =1
1 2
∴h = 2−60= −58.0kJ mol
1infintedilution

6. For a flow through a smooth pipe, the Fanning friction factor (f) is given by f =mRe−0.2 in the
turbulent flow regime, where Re is the Reynolds number and m is a constant. Water flowing
through a section of this pipe with a velocity 1m/s results in a frictional pressure drop of 10kPa. What
will be the pressure drop across this section (in kPa), when the velocity of water is 2 m/s?
(A) 11.5 (B) 20 (C) 34.8 (D) 40

Answer: (C)
4fL2V2ρ
Exp: For Turbulent flow we know that∆p=
2D
F is directly proportional to Re-0.2
Therefore ∆PαV1.8
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∆p V1.8 1
∴ 1 = 1 =
∆p V1.8 22.8
2 2
V1.8 21.8
∆∆pp == ∆∆pp ×× 2 ==1100××  ==3344..88kkPPaa
2 1 VV1.8 11
1

7. IIInnn aaa cccyyyccclllooonnneee ssseeepppaaarrraaatttooorrr uuussseeeddd fffooorrr ssseeepppaaarrraaatttiiiooonnn ooofff sssooollliiiddd pppaaarrrtttiiicccllleeesss fffrrrooommm aaa ddduuusssttt lllaaadddeeennn gggaaasss,,, ttthhheee ssseeepppaaarrraaatttiiiooonnn
ffaaccttoorr iiss ddeeffiinneedd aass tthhee rraattiioo ooff tthhee cceennttrriiffuuggaall ffoorrccee ttoo tthhee ggrraavviittaattiioonnaall ffoorrccee aaccttiinngg oonn tthhee
particle. S dddeeennnooottteeesss ttthhheee ssseeepppaaarrraaatttiiiooonnn fffaaaccctttooorrr aaattt aaa lllooocccaaatttiiiooonnn (((nnneeeaaarrr ttthhheee wwwaaallllll))) ttthhhaaattt iiisss aaattt aaa rrraaadddiiiaaalll dddiiissstttaaannnccceee rrr
r
fffrrrooommm ttthhheee ccceeennntttrrreee ooofff ttthhheee cccyyyccclllooonnneee... WWWhhhiiiccchhh ooonnneee ooofff ttthhheee fffooollllllooowwwiiinnnggg ssstttaaattteeemmmeeennntttsss iiisss IIINNNCCCOOORRRRRREEECCCTTT???
(A) S ddeeppeennddss oonn mmaassss ooff tthhee ppaarrttiiccllee
r
(B) S ddeeppeennddss oonn tthhee aacccceelleerraattiioonn dduuee ttoo ggrraavviittyy
r
(C) S ddeeppeennddss oonn ttaannggeennttiiaall vveelloocciittyy ooff tthhee ppaarrttiiccllee
r
(D) S ddeeppeennddss oonn tthhee rraaddiiaall llooccaattiioonn (( rr )) ooff tthhee ppaarrttiiccllee
r
Answer: (A)

8. AA vveerrttiiccaall ccyylliinnddrriiccaall vveesssseell hhaass aa llaayyeerr ooff kkeerroosseennee ((ooff ddeennssiittyy 880000 kkgg//mm33)) oovveerr aa llaayyeerr ooff wwaatteerr
((ooff ddeennssiittyy 11000000 kkgg//mm33)).. LL-sshhaappeedd ggllaassss ttuubbeess aarree ccoonnnneecctteedd ttoo tthhee ccoolluummnn 3300 ccmm aappaarrtt.. TThhee
iiinnnttteeerrrfffaaaccceee bbbeeetttwwweeeeeennn ttthhheee tttwwwooo lllaaayyyeeerrrsss llliiieeesss bbbeeetttwwweeeeeennn ttthhheee tttwwwooo pppoooiiinnntttsss aaattt wwwhhhiiiccchhh ttthhheee LLL-tubes are connected.
TTThhheee llleeevvveeelllsss (((iiinnn cccmmm))) tttooo wwwhhhiiiccchhh ttthhheee llliiiqqquuuiiidddsss rrriiissseee iiinnn ttthhheee rrreeessspppeeeccctttiiivvveee tttuuubbbeeesss aaarrreee ssshhhooowwwnnn iiinnn ttthhheee fffiiiggguuurrreee bbbeeelllooowww...

TTThhheee dddiiissstttaaannnccceee (((xxx iiinnn cccmmm,,, rrrooouuunnndddeeeddd oooffffff tttooo ttthhheee fffiiirrrsssttt dddeeeccciiimmmaaalll ppplllaaaccceee))) ooofff ttthhheee iiinnnttteeerrrfffaaaccceee fffrrrooommm ttthhheee pppoooiiinnnttt aaattt
which the lower L-ttuubbee iiss ccoonnnneecctteedd iiss _________
Answer: (10)

Exp: Given condition is pp ==pp
11 22
From given diagram
pp ==ρρgg(2200++(3300−−xx)) ++ρρ ggxx
11 wwaatteerr
==880000××99..88112200++(3300−−xx) ++11000000××99..8811xx
=392400+1962x
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PP ==ρρ ××gg××4422 ==11000000××99..8811××4422==441122002200
2 water
⇒ 441122002200 ==339922440000++11996622xx
x=10cm

9. AA ccoommppoossiittee wwaallll iiss mmaaddee ooff ffoouurr ddiiffffeerreenntt mmaatteerriiaallss ooff ccoonnssttrruuccttiioonn iinn tthhee ffaasshhiioonn sshhoowwnn bbeellooww..
TTThhheee rrreeesssiiissstttaaannnccceee (((iiinnn KKK///WWW))) ooofff eeeaaaccchhh ooofff ttthhheee ssseeeccctttiiiooonnnsss ooofff ttthhheee wwwaaallllll iiisss iiinnndddiiicccaaattteeeddd iiinnn ttthhheee dddiiiaaagggrrraaammm...

TTThhheee ooovvveeerrraaallllll rrreeesssiiissstttaaannnccceee (((iiinnn KKK///WWW,,, rrrooouuunnndddeeeddd oooffffff tttooo ttthhheee fffiiirrrsssttt dddeeeccciiimmmaaalll ppplllaaaccceee))) ooofff ttthhheee cccooommmpppooosssiiittteee wall, in
tthhee ddiirreeccttiioonn ooff hheeaatt ffllooww,, iiss________________
Answer: (3.9)
Exp: Total Resistance = RR ++RR ++RR
1122 2233 3344
0.25
11 11 11 11 11
== ++ == ++ 3 0.7
RR RR RR 00..2255 11
23 2 3
⇒ R =0.3 1
23
RR =33+00..22+00..77 =33..99kkWW
total 1 2 3 4

10. SSSttteeeaaammm aaattt 111000000oooCCC iiisss cccooonnndddeeennnsssiiinnnggg ooonnn aaa vvveeerrrtttiiicccaaalll sssttteeeeeelll ppplllaaattteee... TTThhheee cccooonnndddeeennnsssaaattteee ffflllooowww iiisss lllaaammmiiinnnaaarrr... TTThhheee
aavveerraaggee NNuusssseelltt nnuummbbeerrss are Nu and Nu ,, wwhheenn tthhee ppllaattee tteemmppeerraattuurreess aarree 1100oC and 55oC,
1 2
rrreeessspppeeeccctttiiivvveeelllyyy... AAAssssssuuummmeee ttthhheee ppphhhyyysssiiicccaaalll ppprrrooopppeeerrrtttiiieeesss ooofff ttthhheee fffllluuuiiiddd aaannnddd sssttteeeeeelll tttooo rrreeemmmaaaiiinnn cccooonnnssstttaaannnttt wwwiiittthhhiiinnn ttthhheee
tteemmppeerraattuurree rraannggee ooff iinntteerreesstt.. UUssiinngg NNuusssseelltt eeqquuaattiioonnss ffoorr ffiillmm-ttyyppee ccoonnddeennssaattiioonn,, wwhhaatt iiss tthhee
Nu
value of the ratio 2 ??
Nu
1
(((AAA))) 000...555 (((BBB))) 000...888444 (((CCC))) 111...111999 (((DDD))) 111...444111
Answer: (C)
Exp: WWWeee hhhaaavvveee aaavvveeerrraaagggeee hhheeeaaattt tttrrraaannnsssfffeeerrr cccoooeeeffffffiiiccciiieeennnttt fffooorrr LLLaaammmiiinnnaaarrr ffflllooowww ttthhhrrrooouuuggghhh vvveeerrrtttiiicccaaalll plate condensation as
1
ρ (ρ −ρ )h g K3 4
h = 0.943  LL LL VV ffgg 
 µµLL(TT −−TT ) 
ssaatt ww
IItt iiss ggiivveenn tthhaatt,, aallll tthhee pphhyyssiiccaall pprrooppeerrttiieess rreemmaaiinn ccoonnssttaanntt,, ssoo tthhaatt
1
h α
(T −T )14
sat w
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Where T =saturation temperature = 100OC
sat
T = Wall temperature
W
T =10;T =50
w1 w2
hL
Nu  k  h (T −T )14
To find 2 = 2 = 2 = sat w1
Nu1 hL h1 (T −T )14
sat w
 k  2
1
h (100−10)14
∴ 2 = =1.19
h1 (100−55)14

11. 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 Raoult’s 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 ________.
Answer: (0.40)
x =0.2
Exp: Bezene
x =0.8
Tolerance
P =92KPa
vap,Benzene

P =35KPa
vap
,Tolerance
According to Raoult’s law:
y P =x P
Benzere total Benzere vap,Benzene
x .P
⇒y = B vap,B
B P
total
x P
= B vap,B
x P +x P
B vap,B t vap,T
=0.3965≈0.40

12. Match the dimensionless numbers in Group-1 with the ratios in Group-2.
Group-I Group-II
P Biot number I Buoyancyforce

Viscousforce
Q Schmidt number II Internalthermalresistanceof asolid

Boundarylayerthermalresistance
R Grashof number III Momentumdiffusivity

Massdiffusivity
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(A) P-II, Q-I, R-III (B) P-I, Q-III, R-II
(C) P-III, Q-I, R-II (D) P-II, Q-III, R-I
Answer: (D)

13. FFoorr wwhhaatt vvaalluuee ooff LLeewwiiss nnuummbbeerr,, tthhee wweett-bbuullbb tteemmppeerraattuurree aanndd aaddiiaabbaattiicc ssaattuurraattiioonn tteemmppeerraattuurree
are nearly equal?
(A) 000...333333 (((BBB))) 000...555 (((CCC))) 111 (((DDD))) 222
Answer: (C)
Exp: WWhheenn lleewwiiss nnuummbbeerr iiss uunniittyy,, tthhee wweett-bbuullbb tteemmppeerraattuurree aanndd aaddiiaabbaattiicc ssaattuurraattiioonn tteemmppeerraattuurree aarree
equal.

14. For a non-cataallyyttiicc hhoommooggeenneeoouuss rreeaaccttiioonn AA ⟶ BB,, tthhee rraattee eexxpprreessssiioonn aatt 330000 KK
1100CC
is −rr (mmoollmm−3ss−1)= A ,, where C is the concentration of AA ((iinn mmooll//mm3). Theoretically,
A 11++55CC A
A
tthhee uuppppeerr lliimmiitt ffoorr tthhee mmaaggnniittuuddee ooff tthhee rreeaaccttoorr rate (−r in mol mm−33ss−11,, rounded off to the first
A
ddeecciimmaall ppllaaccee)) aatt 330000 KK iiss ____________________
Answer: (2.0)
10C 10
Exp: −r = A =
A 1+5C  1 
A  ++5
C 
A
 1 
NNooww,, tthheeoorreettiiccaallllyy tthhee uuppppeerr lliimmiitt ffoorr (−r ) is reached when  →→00,,ii..ee..,,CC →→∞∞
A C  A
A
1100
In that case, (−−rr )== ==22..00
A 5

15. TThhee vvaarriiaattiioonnss ooff tthhee ccoonncceennttrraattiioonnss ((CC , C and C )) ffoorr tthhrreeee ssppeecciieess ((AA,, RR aanndd SS)) wwiitthh ttiimmee,, iinn
A R S
aann iissootthheerrmmaall hhoommooggeenneeoouuss bbaattcchh rreeaaccttoorr aarree sshhoowwnn iinn tthhee ffiigguurree bbeellooww..

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