AMITY INSTITUTE FOR COMPETITIVE EXAMINATIONS
Test Booklet Code
PAPER-1: CHEMISTRY, PHYSICS & MATHEMATICS
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P
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3.
The Test is of 3 hours duration.
4.
The Test Booklet consists of 90 questions. The maximum marks are 360.
5.
There are three parts in the question paper A, B, C consisting of Chemistry, Physics and Mathematics having
30 questions in each part of equal weightage. Each question allotted 4 (four) marks for each correct response.
6.
Candidates will be awarded marks as in Instruction No. 5 for correct response of each question.
1/4 (one fourth) marks will be deducted for indicated incorrect response of each question. No deduction
AMITY
from the total score will be made if no response is indicated for an item in the Answer Sheet.
7.
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will be treated as wrong response and marks for wrong response will be deducted accordingly as per instruction
6 above.
8.
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the Answer Sheet. Use of pencil is strictly prohibited.
Institute for Competitive Examinations
9.
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Amity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42
-1-
AMITY INSTITUTE FOR COMPETITIVE EXAMINATIONS
PART-A: CHEMISTRY
1.
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2.
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3.
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PART-B: PHYSICS
31.
(2)
32.
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33.
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51. (3)
52. (4)
53. (2)
54.
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55. (4)
Institute
for
Competitive
Examinations
56.
(3)
57.
(2)
58.
(4)
59.
(2)
60.
(4)
PART-C: MATHEMATICS
61.
(4)
62.
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63.
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64.
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90.
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Amity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42
-2-
Test Code-P
AIEEE - 2011
PART-A: CHEMISTRY
1.
The presence or absence of hydroxy group on which carbon atom of sugar differentiates RNA and
DNA?
(1) 1st
(2) 2nd
(3) 3rd
O
HOH 2 C
(4) 4th
O
HOH 2 C
OH
OH
1
H
H
1
H
H
H
Sol: Ans [2]
H
H
2
OH
2
H
OH
Deoxy Ribose Sugar
2.
H
OH
Ribose Sugar
Among the following the maximum covalent character is shown by the compound:
(1) FeCl2
(2) SnCl2
(3) AlCl3
(4) MgCl2
AMITY
Sol: Ans [3]
According to Fajan’s rule more is the polarising power of cation more will be covalent
character. Al3+ has more polarising power than. Fe2+, Sn2+, Mg2+.
3.
Which of the following statement is wrong?
(1) The stability of hydrides increases from NH3 to BiH3 in group 15 of the periodic table.
(2) Nitrogen cannot form d – p bond.
Institute
for Competitive Examinations
(3) Single N – N bond is weaker than the single P – P bond.
(4) N2O4 has two resonance structures.
Sol: Ans [1]
Stability of Hydrides of 15th group element decreases down the group due to decrease in
bond dissociation enthalpy.
4.
Phenol is heated with a solution of mixture of KBr and KBrO3. The major product obtained in the
above reaction is:
(1) 2-Bromophenol
(2) 3-Bromophenol
(3) 4-Bromophenol
(4) 2,4,6-Tribromophenol
Sol: Ans [4]
KBr and KBrO3 react and produce Br2 in water
OH
OH
Br
Br
 Br2 

H2O
Br
2,4,6-tribromophenol
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-3-
Chemistry, Physics & Mathematics (Test Code: P)
5.
AIEEE-2011
A 5.2 molal aqueous solution of methyl alcohol, CH3OH, is supplied. What is the mole fraction of
methyl alcohol in the solution?
(1) 0.100
Sol: Ans [3]
6.
(2) 0.190
xsolute =
(3) 0.086
(4) 0.050
5.2
= 0.086
55.55  5.2
The hybridisation of orbitals of N atom in NO3 , NO2 and NH4 are respectively:
(1) sp, sp2, sp3
(2) sp2, sp, sp3
(3) sp, sp3, sp2
H
O
O
Sol: Ans [2]
(4) sp2, sp3, sp
N
O
N
O
N
H
O
sp2
H
H
sp3
sp
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7.
Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be
added to 4 kg of water to prevent it from freezing at –6°C will be:
(Kf for water = 1.86 K kg mol–1, and molar mass of ethylene glycol = 62 g mol–1)
(1) 804.32g
Sol: Ans [1]
(2) 204.30 g
(3) 400.00g
(4) 304.60g
Wsolute  1000
Wsolute  Wsolvent
Tf  K f 
Institute for
Competitive
Examinations
W
 1000
6  1.86 
solute
62  4000
Wsolute = 804.32g
8.
The reduction potential of hydrogen half-cell will be negative if:
(1) p(H2) = 1 atm and [H+] = 2.0 M
(2) p(H2) = 1 atm and [H+] = 1.0 M
(3) p(H2) = 2 atm and [H+] = 1.0 M
(4) p(H2) = 2 atm and [H+] = 2.0 M
Sol: Ans [3]
2H (aq)  2e  
 H 2(g )
E H  /H  E 0H  /H 
2
2
0.591 [p(H 2 )]
log
2
[H  ]2
For negative reduction potential log
[pH 2 ]
must be positive
[H  ]2
0.591
[2]
log 2
2
[1]
0.0591
log 2
= 
2
EH / H  0 
2
= 
0.0591
 0.3010 = –0.0089V
2
Amity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42
-4-
Chemistry, Physics & Mathematics (Test Code: P)
AIEEE-2011
9.
Which of the following reagents may be used to distinguish between phenol and benzoic acid?
(1) Aqueous NaOH
Sol: Ans [4]
(2) Tollen’s reagent
(3) Molisch reagent
(4) Neutral FeCl3
Phenol on reaction with FeCl3 give violet colour of ferric phenoxide
3C6 H 5 OH  FeCl3 
 (C6 H 5O)3  3HCl
Ferric phenoxide
(Violet Colour )
10.
Trichloroacetaldehyde was subjected to Cannizzaro’s reaction by using NaOH. The mixture of the
products contains sodium trichloroacetate and another compound. The other compound is:
(1) 2, 2, 2-Trichloroethanol
(2) Trichloromethanol
(3) 2, 2, 2-Trichloropropanol
(4) Chloroform
Cl
Sol: Ans [1]
Cl
O
Cannizzar reaction

 Cl
OH
Cl
H
Cl
O
 Cl
-
O
OH
Cl
2,2,2-trichloroethanol
Which one of the following orders presents the correct sequence of the increasing basic nature of the
Cl
Cl
AMITY
11.
given oxides?
(1) Al2O3  MgO  Na2O  K2O
(2) MgO  K2O  Al2O3  Na2O
(3) Na2O  K2O  MgO  Al2O3
(4) K2O  Na2O  Al2O3  MgO
Sol: Ans [1]
Basic strength of oxides increases with increase in metallic chracter.
Institute
forofCompetitive
Examinations
A gas absorbs a photon
355 nm and emits at two wavelengths.
If one of the emissions is at
12.
680 nm, the other is at:
(1) 1035 nm
(2) 325 nm
(3) 743 nm
(4) 518 nm
Sol: Ans [3] E total = E1 + E2
hC hC  hC
= 
2

1
1
1
1
=  

1
2
1
1
1
= 680  
355
2
13.
2 = 743
Which of the following statements regarding sulphur is incorrect?
(1) S2 molecule is paramagnetic
(2) The vapour at 200°C consists mostly of S8 rings
(3) At 600°C the gas mainly consists of S2 molecules
(4) The oxidation state of sulphur is never less than +4 in its compounds.
Amity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42
-5-
Chemistry, Physics & Mathematics (Test Code: P)
Sol: Ans [4]
14.
AIEEE-2011
Sulphur shows oxidation states from –2 to +6
The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a
volume of 10 dm3 to a volume of 100 dm3 at 27°C is:
(1) 38.3 J mol–1 K–1
Sol: Ans [1]
(2) 35.8 J mol–1 K–1
(3) 32.3 J mol–1 K–1
(4) 42.3 mol–1 K–1
v2
s = 2.303 nR log v
1
= 2.303 × 2 × 8.314 log
100
10
= 38.3 J mol–1 K–1
15.
Which of the following facts about the complex [Cr(NH3)6]Cl3 is wrong?
(1) The complex involves d2-sp3 hybridisation and is octahedral in shape.
(2) The complex is paramagnetic
AMITY
(3) The complex is an outer orbital complex
(4) The complex gives white precipitate with silver nitrate solution.
Sol: Ans [3]
Cr(NH)6]Cl3
Cr3+ – 3d34s04p0
d2sp3
Institute
for Competitive Examinations
[Cr(NH ) ]
3
3 6
xx xx
xx
3d
4s
NH3
NH3
NH3
xx
xx
xx
4p
NH3 NH3 NH3
Inner orbital complex
16.
The structure of IF7 is:
(1) square pyramid
(2) trigonal bipyramid
(3) octahedral
(4) pentagonal bipyramid
F
F
F
T
F Pentagonal bipyramidal
Sol: Ans [4]
F
F
F
Amity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42
-6-
Chemistry, Physics & Mathematics (Test Code: P)
AIEEE-2011
17.
The rate of a chemical reaction doubles for every 10°C rise of temperature. If the temperature is
raised by 50°C, the rate of the reaction increases by about:
(1) 10 times
Sol: Ans [3]
(2) 24 times
(3) 32 times
(4) 64 times
Temperature increase by 10°C, rate is double, its temperature increase 5 times, rate increase
ri (2)5 (raise to power 5 time) or 32 times.
18.
The strongest acid amongst the following compound is:
(1) CH3COOH
(2) HCOOH
(3) CH3CH2CH(Cl)CO2H
(4) ClCH2CH2CH2COOH
Sol: Ans [3]
19.
—Cl is electron withdrawing group, hence increase acidity of acid.
Identify the compound that exhibits tautomerism.
(1) 2-Butene
(2) Lactic acid
O
(3) 2-Pentanone
Tautomerisation

(4) Phenol
HO
AMITY
Sol: Ans [3]
H3C
CH3
H3C

Keto form
(excess)
CH3
enol
(Trace)
(Less stable)
A vessel at 1000 K contains CO with a pressure of 0.5 atm. Some of the CO is converted into CO
Institute
for Competitive Examinations
on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is:
20.
2
(1) 1.8 atm
Sol: Ans [1]
2
(2) 3 atm
CO 2(g )

(3) 0.3 atm
(4) 0.18 atm
 2CO(g )

C
0.5 atm
0
–x
+2x
_____________________________________
At equilibrium
0.5 – x
2x
Total pressure = 0.5 – x + 2x = 0.5 – x
0.5 + x = 0.8
x = 0.3
pCO2 = 0.5 – 0.3 = 0.2
pCO = 2 × 0.3 = 0.6
K
[0.6]2
= 1.8 atm
[0.2]
Amity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42
-7-
Chemistry, Physics & Mathematics (Test Code: P)
21.
AIEEE-2011
In context of the lanthanoids, which of the following statements is not correct?
(1) There is a gradual decrease in the radii of the members with increasing atomic number in the series.
(2) All the members exhibit +3 oxidation state.
(3) Because of similar properties the separation of lanthanoids is not easy.
(4) Availability of 4f electrons results in the formation of compounds in +4 state for all the members of
the series.
Sol: Ans [4] Factual
22.
‘a’ and ‘b’ are van der Waals’ constants for gases. Chlorine is more easily liquefied than ethane because
(1) a and b for Cl2  a and b for C2H6
(2) a and b for Cl2  a and b for C2H6
(3) a for Cl2  a for C2H6 but b for Cl2  b for C2H6
(4) a for Cl2  a for C2H6 but b for Cl2  b for C2H6
Sol: Ans [4]
Value of ‘a’ related to Interaction in between molecules. But for ‘b’ value depend size of
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molecule.
23.
The magnetic moment (spin only) of [NiCl4]2– is:
(1) 1.82 BM
Sol: Ans [3]
(2) 5.46 BM
(3) 2.82 BM
(4) 1.41 BM
[NiCl4]2–, Cl– are weak ligand.
 d – for Competitive Examinations
Institute
8
  n (n  2) B.M
  8 = 2.82 B.M.
24.
In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face
centre position. If one atom of B is missing from one of the face centred points, the formula of the
compound is:
(1) A2B
Sol: Ans [4]
(2) AB2
Total atom at corner (A) =
(3) A2B3
(4) A2B5
1
8 1
8
Total atom at face centre (B) =
1
6  3
2
If one atom remove from one face centred point
1 5
Total atom (B) = 5  
2 2
Formula A2B5
Amity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42
-8-
Chemistry, Physics & Mathematics (Test Code: P)
AIEEE-2011
25.
The outer electron configuration of Gd (Atomic No: 64) is:
(1) 4f3 5d5 6s2
Sol: Ans [4]
26.
(3) 4f4 5d4 6s2
(4) 4f7 5d1 6s2
Factual
Boron cannot form which one of the following anions?
(1)
BF63
Sol: Ans [1]
27.
(2) 4f8 5d0 6s2
BH 4
(2)
(3)
B(OH) 4
(4)
BO2
Boron cannot make hexavalent due to absence of vacant d-orbital.
Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the
presence of:
(1) two ethylenic double bonds
(2) a vinyl group
(3) an isopropyl group
(4) an acetylenic triple bond
Sol: Ans [2]
Factual
H
H
H
ozonolysis
O

 HCHO +
AMITY
H
28.
R
R
Sodium ethoxide has reacted with ethanoyl chloride. The compound that is produced in the above
reaction is:
(1) Diethyl ether
(2) 2-Butanone
(3) Ethyl chloride
(4) Ethyl ethanoate
O
O
Institute
for
Competitive
Examinations
Cl + Na O
H C
+ NaCl
+
3
Sol: Ans [4]
H3C
CH3
O
ethyl acetate
or ethyl
ethanoate
29.
CH3
The degree of dissociation () of a weak electrolyte, AxBy is related to van’t Hoff factor (i) by the
expression.
(1)

i 1
( x  y  1)

(2)
i 1
x  y 1
(3)

x  y 1
i 1
(4)

x  y 1
i 1
Sol: Ans [1]
 xA y  yB  x
A x B y 
At equilibrium
i
(1 – )
x  y  (1  )
1
x
 
y
i 1
x  y 1
Amity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42
-9-
Chemistry, Physics & Mathematics (Test Code: P)
30.
AIEEE-2011
Silver Mirror test is given by which one of the following compounds?
(1) Acetaldehyde
Sol: Ans [1]
(2) Acetone
(3) Formaldehyde
(4) Benzophenone
All aldehyde give silver mirror test by tollen’s reagnet. Except formaldehyde
O
H
-
OH
 H 3C
O 
Cannizzaro
-
OH  H
O
H
PART-B: PHYSICS
31.
100 g of water is heated from 30°C to 50°C. Ignoring the slight expansion of the water, the change in
its internal energy is (specific heat of water is 4184 J/kg/K)
(1) 4.2 kJ
Sol: Ans [2]
(2) 8.4 kJ
(3) 84 kJ
(4) 2.1 kJ
Change in internal energy = (ms T)
0.1 × 4184 × 20 = 8368 J  8.4 kJ
AMITY
32.
The half life of a radioacitve substance is 20 minutes. The approximate time interval (t2 – t1) between
the time t2 when
2
1
of it has decayed and time t1 when of it had decayed is
3
3
(1) 7 min
(2) 14 min
(3) 20 min
(4) 28 min
2N
 for
Ne
Institute
Competitive
Examinations
.....(i)
3
t1
0
Sol: Ans [3]
0
N0
 N0 e t
3
2
.....(ii)
Dividing (ii) by (i)
1
 e  ( t
2
2 t1 )
loge2 = (t2 – t1)
(t2 – t1) = 20 min.
33.
A mass M, attached to a horizontal spring, executes S.H.M with amplitude A1. When the mass M
passes through its mean position then a smaller mass m is placed over it and both to them move
 
together with amplitude A2. The ratio of  1  is
 A2 
(1)
M
Mm
(2)
1/ 2
Mm
M
(3)
 M 


Mm
1/ 2
(4)
Mm


 M 
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- 10 -
Chemistry, Physics & Mathematics (Test Code: P)
AIEEE-2011
Sol: Ans [4]

34.
Let v1 and v2 are velocities at mean position before and after putting mass, m
Mv1 = (M + m). v2
M . A1
K
k
 ( M  m) A2
M
M m

A2 
M
 A1
M m
or
A1

A2
M m
M
Energy required for the electron excitation in Li++ from the first to the third Bohr orbit is
(1) 12.1 eV
Sol: Ans [3]
(2) 36.3 eV
(3) 108.8 eV
(4) 122.4 eV
1 1 
8
E = 13.6 × (3)2    = 13.6 × 9 × = 108.8 eV
1
9
9


The transverse displacement y(x, t) of a wave on a string is given by y ( x, t )  e  ( ax
2 bt 2  2 ab xt
AMITY
35.
) . This
represents a
a
b
(1) wave moving in +x direction with speed
(3) standing wave of frequeny b
(2) wave moving in –x direction with speed
(4) standing wave of frequency
b
a
1
b
Institute for Competitive Examinations
Sol: Ans [2]
y(x, t) can be written as y(x, t) = e 

ax  bt

2
Direction of propogation of wave is negative x-axis
v  b/a
36.
A resistor ‘R’ and 2 F capacitor in series is connected through a switch to 200 V direct supply.
Across the capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb
light up 5s after the switch has been closed. (log102.5 = 0.4)
(1) 1.3 × 104 
Sol: Ans [3]
(2) 1.7 × 105 
(3) 2.7 × 106 
(4) 3.3 × 107 
Charge on capacitor when P.D. across it 120 V
q = 2 × 120 = 240 m
t



RC
Using q  q0  1  e 


5



R  210
240  400  1  e



6
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t
6
e R210  0.4
or
t
 log e 2.5  2.303  0.4
R  2  106
5


R

6
 2  10  2.303  0.4 
R = 2.7 × 106 
37.
A current I flows in an infinitely long wire with cross section in the form of a semicircular ring of
radius R. The magnitude of the magnetic induction along its axis is
(1)
o I
2 R
Sol: Ans [1]
o I
2 2 R
(2)
(3)
o I
2 R
(4)
o I
4 R
Let curent is going inward
Due to symmetry By = 0
x
dBx  dB.sin 
AMITY

o.dl
.sin 
2 R
( 
I
)
R
dB
y

 o .R.sin .d 
2 R

Bx 
o 
2o
 sin  d  
2 0
2
or
By 
o I
2 R
Institute for Competitive Examinations
1
38.
A Carnot engine operating between temperatures T1 and T2 has efficiency
62 K, its efficiency increases to
(1) 372 K and 310 K
Sol: Ans [1]
. When T2 is lowered by
1
. Then T1 and T2 are, respectively
3
(2) 372 K and 330 K
T
T 5
1
1 2  2 
6
T1
T1 6
(3) 330 K and 268 K
(4) 310 K and 248 K
... (i)
T  62
T2  62 2
1
1 2

or
3
T1
T1
3
Dividing,
6
...(ii)
T2
5/6 5


T2  62 2 / 3 4
4T2 = 5T2 – 310
T2 = 310 k
From (i) T1 
6
 310  372 k
5
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39.
dv
 2.5 v
dt
where v is the instantanous speed. The time taken by the object, to come to rest, would be
An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by:
(1) 1 s
(2) 2 s
(3) 4 s
(4) 8 s
0
Sol: Ans [2]
t
dv


2.5

 dt
v
u
0
0
 v1/ 2 

  2.5 t
 1/ 2 u
–2 v1/2 = –2.5 t
 t
40.
2  6.25
 2 sec.
2.5
The electrostatic potential inside a charged spherical ball is given by  = a r2 + b where r is the
distance from the centre, a, b are constants. Then the charge density inside the ball is
(1)
(2)
24  a o r
(3)
6 a  o r
24 a  o
(4)
6 a  o
AMITY
Sol: Ans [4]
E 
d
 2ar
dr

qi
2
flux = ( 2ar.4r )   ( r  R )
0

qi = –8  0 ar3
80 ar 3
 60 a
Charge density = 4
2
r
3
A car is fitted with a convex side view mirror of focal length 20 cm. A second car 2.8 m behind the
Institute for Competitive Examinations
41.
first car is overtaking the first car at a relative speed of 15 m/s. The speed of the image of the second
car as seen in the mirror of the first one is
(1)
1
m/s
10
Sol: Ans [2]
(2)
1
m/s
15
(3) 10 m/s
(4) 15 m/s
1 1 1
 
v u f

dv
du
 m 2
dt
dt
Also, m 
...(i)
f
20
1


u  f 280  20
15
2

dv
1
 1
      15   m / sec.
dt
15
 15 
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42.
If a wire is stretched to make it 0.1% longer, its resistance will
(1) increase by 0.05%
(2) increase by 0.2%
(3) decrease by 0.2%
(4) decrease by 0.05%
Sol: Ans [2]
43.
AIEEE-2011
R
l
l2
= .
(V = volume of wire)
A
V
100 
R
l
 2  100  0.2% (increases)
R
l
Three perfect gases at absolute temperatures T1, T2 and T3 are mixed. The masses of molecules are
m1, m2 and m3 and the number of molecules are n1, n2 and n3 respectively. Assuming no loss of
energy, the final temperature of the mixture is
(1)
(T1  T2  T3 )
3
(2)
n1T1  n2T2  n3T3
n1  n2  n3
AMITY
(3)
n1T12  n2T22  n3T32
n1T1  n2T2  n3T3
Sol: Ans [2]
(4)
n12T12  n22T22  n32T32
n1T1  n2T2  n3T3
Let final temperature of mixture is T, n1Cv (T  T1 )  n1Cv (T  T2 )  n3 (T  T3 )  0

T
n1T1  n2T2  n3T3
(n1  n2  n3 )
Institute
for Competitive Examinations
Two identical charged spheres suspended from a common point by two massless strings of length l are
44.
initially a distance d(d << l) apart because of their mutual repulsion. The charge begins to leak from
both the spheres at a constant rate. As a result the charges approach each other with a velocity v.
Then as a function of distance x between them,
(1)
v  x 1/ 2
Sol: Ans [1]
(2)
v  x 1
(3)
v  x1/ 2
(4)
vx
T sin   Fe
T cos   mg
Fe
 tan   mg
for small ,
x3 
T
Fe
x
kq 2

2l mg.x 2
x/2
mg
2lkq 2
 cq 2
mg
x/2
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differentiating w.r.t. time
3x2
dx
dq
 2cq.
dt
dt
1/ 2
 x3 
3 x  v  2c  
 c 
2
.
dq
dt
 v  x 1/ 2
45.
Work done in increasing the size of a soap bubble from a radius of 3 cm to 5 cm is nearly (surface
tension of soap solution = 0.03 Nm–1)
(2) 0.2  mJ
(1) 4 mJ
Sol: Ans [4]
(3) 2  mJ
(4) 0.4  mJ
W  S A
 0.03  2  4 [r22  r12 ]
 0.24[25  9]  104
AMITY
 0.384  103 J
= 0.4  mJ
46.
A fully charged capacitor C with initial charge q0 is connected to a coil of self inductane L at t = 0.
The time at which the energy is stored equally between the electric and the magnetic field is

LC
(1)  LC
(3) 2 LC
(4)
LC
Institute
for(2)Competitive
Examinations
4
Sol: Ans [2]
1 q2 1 2
 Li or q  LC i
2C 2
 q0 cos(wt) =
LC lo sin( wt )
also q0  LC i0

tan( wt )  1
wt = /4

47.
t

LC
4
Two bodies of masses m and 4m are placed at a distance r. The gravitational potential at a point on
the line joining them where the gravitational field is zero is
(1) zero
Sol: Ans [4]
(2)

4Gm
r
(3)

6Gm
r
(4)

9Gm
r
Let gravitational field will be zero at distance x from mass m

Gm G  4m

x 2 (r  x) 2
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x

1
2

x rx
C
m
4m
2x = r – x or x = r/3
V at C  

48.
Gm G  4m

r / 3 2r / 3
9Gm
r
A thin horizontal circular disc is rotating about a vertical axis passing through its centre. An insect is at
rest at a point near the rim of the disc. The insect now moves along a diameter of the disc to reach its
other end. During the journey of the insect, the angular speed of the disc
(1) remains unchanged
(2) continuously decreases
(3) continuously increases
(4) first increases and then decreases
Sol: Ans [4]
I   constant
I first decreases then increases
AMITY

49.
 first increses then decreas0es.
Let the x – z (correct is x – y plane) plane be the boundary between two transparent media. Medium 1
in z  0 has a refractive index of 2 and medium 2 with z < 0 has a refractive index of 3. A ray of

light in medium 1 given by the vector A  6 3 iˆ  8 3 ˆj  10 kˆ is incident on the plane of separation.
Institute
forin medium
Competitive
Examinations
The angle of refraction
2 is
(1) 30°
Sol: Ans [2]
50.
(2) 45°
(3) 60°
(4) 75°
Let angle of incident = i
10
1


cos i =
36  3  64  3  100 2
3
sin i 

2
3
2
 3 sin r
2
1
sin r 

or r =45°
2
Two particles are executing simple harmonic motion of the same amplitude A and frequency  along
the x-axis. Their mean position is separated by distance X0 (X0 > A). If the maximum separation
between them is (X0 + A), the phase difference between their motion is
(1)

2
(2)

3
(3)

4
(4)

6
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AIEEE-2011
M1
Sol: Ans [1]
M2
Possibility:
x0
Phse difference = /2
51.
Direction: The question has a paragraph followed by two statements, Statement-1 and Statement-2.
Of the given four alternatives after the statements, choose the one that describes the statements.
A thin air film is formed by putting the convex surface of a plane-convex lens over a plane glass plate.
With monochromatic light, this film gives an interference pattern due to light reflected from the top
(convex) surface and the bottom (glass plate) surface of the film.
Statement-1:
When light reflects from the air-glass plate interface, the reflected wave suffers a phase
change of .
Statement-2:
The centre of the interference pattern is dark.
(1) Statement-1 is true, Statement-2 is false
AMITY
(2) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation of
Statement-1
(3) Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation of
Statement-1
(4) Statement-1 is false, Statement-2 is true.
Institute
forTrueCompetitive Examinations
Statement-1-
Sol: Ans [3]
Reflection at denser surface surffers phase change of .
Statement-2 - True
52.
A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats . It
is moving with speed v and is suddenly brought to rest. Assuming no heat is lost to the surroundings,
its temperature increases by
(1)
(   1)
Mv 2 K
2(   1) R
Sol: Ans [4]
(2)
Kinetic energy of
(   1) 2
Mv K
2 R
(3)
 Mv 2
K
2R
(4)
(   1) 2
Mv K
2R
1
M 0VC2 change into heat where VC is velocity of centre of mass of gas.
2

M  R 
1
M 0VC2  o 
 .T
2
M   1

1   1
2
T  
 Mv
2 R 
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53.
AIEEE-2011
A screw gauge gives the following reading when used to measure the diameter of a wire.
Main scale reading :
0 mm
Circular scale reading:
52 divisions
Given that 1 mm on main scale corresponds to 100 divisions of the circular scale.
The diameter of wire from the above data is
(1) 0.52 cm
Sol: Ans [2]
(2) 0.052 cm
(4) 0.005 cm
Diameter = Main scale reacding + L.C × (circular scale reading) – zero error.
0
54.
(3) 0.026 cm
1
 52  0.52 mm = 0.052 cm
100
A boat is moving due east in a region where the earth’s magnetic field is 5.0 × 10–5 NA–1 m–1 due
north and horizontal. The boat carries a vertical aerial 2 m long. If the speed of the boat is 1.50 ms–1, the
magnitude of the induced emf in the wire of aerial is
(1) 1 mV
Sol: Ans [4]
(2) 0.75 mV
emf  lvB
N
(3) 0.50 mV
(4) 0.15 mV
AMITY
B
 2  1.5  5  10
5
vB=1.5 m/sec
W
E
 15  105
S
= 0.15 mV
55.
Direction: The question has a Statement-1 and Statement-2. Of the four choices given after the
statements, choose the one that best describes the statements.
Institute
for Competitive Examinations
Statement-1: Sky wave signals are used for long distance radio communication. These signals are in
general, less stable than ground wave signals.
Statement-2:
The state of ionosphere varies from hour to hour, day to day and season to season.
(1) Statement-1 is true, Statement-2 is false
(2) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation of
Statement-1
(3) Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation of
Statement-1
(4) Statement-1 is false, Statement-2 is true.
Sol: Ans [4]
56.
A mass m hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley
has mass m and radius R. Assuming pulley to be a perfect uniform circular disc, the acceleration of the
mass m, if the string does not slip on the pulley is
(1)
3
g
2
(2) g
(3)
2
g
3
(4)
g
3
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Chemistry, Physics & Mathematics (Test Code: P)
AIEEE-2011
Sol: Ans [3]
mg – T = ma
... (i)
mR 2

2
... (ii)
T.R 
a
...(iii)
a  R
2
a  .g
3

57.
T
mg
A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the
fountain is v. the total area around the fountain that gets wet is
(1)

v2
g
Sol: Ans [2]
(2)

v4
g2
(3)
 v4
2 g2
(4)

v2
g2
Maximum area =  (max. range)2
 v2 
=  
g
2
AMITY

58.
v 4
g2
Direction: This question has a Statement-1 and Statement-2. Of the four choices given after the
statements, choose the one that best describes the two statements.
Statement-1: A metallic surface is irradiated by a monochromatic light of frequency v > v (the
Institute
for Competitive Examinations
threshold frequency). The maximum kinetic energy and the stopping potential are K
0
max
and V0 respectively. If the frequency incident on the surface is doubled, both the Kmax
and V0 are also doubled.
Statement-2:
The maximum kinetic energy and the stopping potential of photoelectrons emitted from
a surface are linearly dependent on the frequency of incident light.
(1) Statement-1 is true, Statement-2 is false
(2) Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation of
Statement-1
(3) Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation of
Statement-1
(4) Statement-1 is false, Statement-2 is true.
Sol: Ans [4]
Statement-1- False
K max  hv  
Vo 
hv 

e e
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AIEEE-2011
'
K max
 2hv    2[hv  ]  
= 2 K max  
Vo' 
2hv 
 hv   
  2   
e
e
 e e e
 2Vo 

e
Statement-2 - True
59.
A pulley of radius 2m is rotated about its axis by a force F = (20 t – 5t2) newton (where t is measured
in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is
10 kg m2, the number of rotations made by the pulley before its direction of motion if reversed. is
(1) less than 3
(2) more than 3 but less than 6
(3) more than 6 but less than 9
(4) more than 9
AMITY
Sol: Ans [2]
Using   I 
(20t  5t 2 ). R  I .

t
0
0
d
dt
I .  d   R  (20t  5t 2 ) dt
Institute
for5Competitive
Examinations
I .  R (10t  t )
...(i)
2
3
3
for  = 0

5
10t 2  t 3  0
3
5
t 2 (10  t )  0
3

Using (i)
t = 6 sec

6
5
I . d   R  (10t 2  t 3 ) dt
3
0
0
6
5 
 10
I   R  t3  t 4 
12 0
 3


2 10 3 5 4 
 6  6   36 rad .
10  3
12 
no. of rotations 
36
2
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Chemistry, Physics & Mathematics (Test Code: P)
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60.
Water is flowing continuously from a tap having an internal diameter 8 × 10–3 m. The water velocity
as it leaves the tap is 0.4 ms–1. The diameter of the water stream at a distance 2 × 10–1 m below the
tap is close to
(1) 5.0 × 10–3 m
Sol: Ans [4]
(2) 7.5 × 10–3 m
(3) 9.6 × 10–3 m
(4) 3.6 × 10–3 m
1 2
Using P  v  gh  const.
2
1
1
(0.4)2   g  2  101  v 2  0
2
2

v  4.16
Using A1v1  A2 v2
or
D12v1  D22 v2

D2 

v1
0.4
 D1 
 8  103
v2
2.04
8  103
 3.6  103 m
2.45
PART-C: MATHEMATICS
AMITY
61.
Let ,  be real and z be a complex number. If z2 + z + b = 0 has two distinct roots on the line
Re z = 1, then it is necessary that
(1)   (0, 1)
Sol: Ans [4]
(2)   (–1, 0)
(3) || = 1
(4)   (1, )
Let roots be (1 + ki) and (1 + li)
Institute
for
Examinations
The sum
= 2 + (kCompetitive
+ l) i = (real)
k=–l
Product =  = 1 + k2  (1, ).
1
62.
8log(1  x )
dx is
1  x2
0
The value of 
(1)  log 2
Sol: Ans [1]
(2)

log 2
8
(3)

log 2
2
(4) log 2
Put x = tan  then
 /4
I = 8  log(1  tan )d 
....(i)
0
 /4



I = 8  log  1  tan      d 
4


0
.....(ii)
 /4
Adding we get
2I = 8  log 2 d   I = log2
0
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- 21 -
Chemistry, Physics & Mathematics (Test Code: P)
63.
d 2x
equals
dy 2
(1)
 d2y 
 2
 dx 
Sol: Ans [4]
1
1
 d 2 y   dy 
 2   
 dx   dx 
(2)
3
(3)
 d 2 y   dy 
 2  
 dx   dx 
2
(4)
 d 2 y   dy 
  2  
 dx   dx 
3
d 2 x d  dx  d  1 


dy 2 dy  dy  dy  dy / dx 
= 
64.
AIEEE-2011
1
d  dy 
  =–
 dy  dy  dx 
 dx 
 
.
2
3
 dx  d 2 y
 dy  . dx 2
 
Let I be the purchase value of an equipment and V(t) be the value after it has been used for t years.
The value V(t) depreciates at a rate given by differential equation
dV (t )
 k (T  t ), where k > 0 is a
dt
AMITY
constant and T is the total life in years of the equipment. Then the scrap value V(T) of the equipment
is
2
(1) T 
I
k
I
(2)
kT 2
2
(3)
I
k (T  t ) 2
2
(4) e–kT
dv
= –for
k (T – t)Competitive Examinations
Institute
dt
Sol: Ans [2]

t2 
Tt


 V = – k (
2 +c

I=0+cc=I
 2 T2 
kT 2
 V (T) = – k  T  2  + I = I –


2
65.
The coefficient of x7 in the expansion of (1 – x – x2 + x3)6 is
(1) 144
Sol: Ans [3]
(2) –132
(3) –144
(4) 132
(1 – x)6 (1 – x2)6
Tr, s = 6Cr (– x)r 6Cs (–x2)s
 r + 2s = 7
 coefficient
= 6 C1 6 C3 – 6 C3 6 C2 + 6 C5 6 C1
= 6 × 20 – 20 × 15 + 36 = – 144.
Amity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42
- 22 -
Chemistry, Physics & Mathematics (Test Code: P)
AIEEE-2011
66.
x
 5 
For x   0,  , define f ( x )   t sin t dt . Then f has
 2 
0
(2) local minimum at  and 2
(1) local maximum at p and 2
(3) local minimum at  and local maximum at 2 (4) local maximum at  and local minimum at 2
Sol: Ans [4]
f ´(x) =
x sin x = 0  x = , 2
at x = , local maximum and at x = 2 local minimum.
67.
The area of the region enclosed by the curves y = x, x = e, y = 1/x and the positive x-axis is
(1) 1/2 square units
(2) 1 square units
(3) 3/2 square units
(4) 5/2 square units
y
e
Sol: Ans [3]
Area = 1 +
x
y=
1
 x dx
1
= 1
1 3
 .
2 2
1
x
x=e
AMITY
68.
The lines L1 : y – x = 0 and L2 : 2x + y = 0 intersect the line L3 : y + 2 = 0 at P and Q respectively.
The bisector of the acute angle between L1 and L2 intersects L3 at R.
Statement-1:
The ratio PR : RQ equals 2 2 : 5
Statement-2:
In any triangle, bisector of an angle divides the triangle into two similar triangles.
(1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for
Statement-1.for Competitive Examinations
Institute
(2) Statement-1 is true, Statement-2 is true; Statement-2 is not correct explanation of Statement-1.
(3) Statement-1 is true, Statement-2 is false.
(4) Statement-1 is false, Statement-2 is true
Sol: Ans [3]
Solving lines P is (–2, –2) & Q is (1, –2)  OP = 2 2 and OQ =
5
so statement 1 is true and 2 is false.
69.
 sin( p  1) x  sin x
,x0

x

, x 0
The values of p and q for which the function f ( x )  q

2
 xx  x
, x0

x3 / 2
is continuous for all x in R, are
(1)
1
3
p  ,q  
2
2
(2)
5
1
p  ,q 
2
2
(3)
3
1
p   ,q 
2
2
(4)
1
3
p  ,q 
2
2
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- 23 -
Chemistry, Physics & Mathematics (Test Code: P)
Sol: Ans [3]
lim
x 0
sin( p  1) x  sin
1
=p+2=q=
x
2
 p= 
70.
AIEEE-2011
3
1
,q= .
2
2
If the angle between the line x 
 5 
1
y 1 z  3

and the plane x + 2y + 3z = 4 is cos  14  , then 
2



equals :
(1) 2/3
(2) 3/2
1  4  3
Sol: Ans [1]
sin  =
 =
5
2
(3) 2/5
(4) 5/3
3
14

14
2
.
3
AMITY
71.
(1) (–, )
Sol: Ans [3]
72.
1
is :
| x | x
The domain of the function f ( x) 
(2) (0, )
(3) (–, 0)
(4) (–, ) – {0}
For domain |x| > x  x < 0
The shortest distance between line y – x = 1 and curve x = y2 is :
Institute
for Competitive
Examinations
8
4
3
3 2
(1)
(2)
4
Sol: Ans [2]
(3)
8
The equation of parallel tangent is y = x +
3 2
(4)
3
1
4
1
4
3 2
11 = 8
1
 shortest distance =
73.
A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent
months his saving increases by Rs. 40 more than the saving of immediately previous month. His total
saving from the start of service will be Rs. 11040 after :
(1) 18 months
Sol: Ans [4]
(2) 19 months
(3) 20 months
(4) 21 months
n
[200 × 2 + (n –1)40] = 11040– 400 = 10,640
2
 n = 19
so total time taken = 19 + 2 = 21 months.
Amity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42
- 24 -
Chemistry, Physics & Mathematics (Test Code: P)
AIEEE-2011
74.
Consider the following statements
P : Suman is brilliant
Q : Suman is rich
R : Suman is honest
The negation of the statement “Suman is brilliant and dishonest if and only if Suman is rich” can be
expressed as :
(1) ~ P ^ (Q  ~ R)
Sol: Ans [2]
(2) ~ (Q  (P ^ ~ R)) (3) ~ Q  ~ P ^ R
(4) ~ (P ^ ~ R)  Q
The statement is Q  (P ~ R)
so its negation is ~ (Q  (P ~ R))
75.
If (1) is a cube root of unity, and (1 + )7 = A + B. Then (A, B) equals :
(1) (0, 1)
Sol: Ans [2]
(2) (1, 1)
(3) (1, 0)
(4) (–1, 1)
7
(1 + ) = A + B
(–2)7 = A + B
–2 = A + B = 1 + 
 A = 1 and B = 1
AMITY
76.

 1
 
 


1
3iˆ  kˆ and b  2iˆ  3 ˆj  6kˆ , then the value of 2a  b  a  b  a  2b  is :
If a 


10
7


(1) –5

Sol: Ans [1]


(2) –3
 
(3) 5
 

(4) 3
(2a – b) . [(a × b) × (a + 2b)]
= [2a – b a × b a + 2b]
Institute
for Competitive Examinations
= [a × b
a + 2b 2a – b]
= – 5 | a  b |2 = – 5
77.
If
dy
 y  3  0 and y(0) = 2, then y(ln 2) is equal to
dx
(1) 7
Sol: Ans [1]
(2) 5
(4) –2
dy
= dx  ln (y + 3) = x + c
y3

78.
(3) 13
y3
 e x  y (ln 2) = 7.
5
Equation of the ellipse whose axes are the axes of coordinates and which passes through the point
(–3, 1) and has eccentricity
(1) 3x2 + 5y2 – 32 = 0
2 / 5 is :
(2) 5x2 + 3y2 – 48 = 0
(3) 3x2 + 5y2 – 15 = 0
(4) 5x2 + 3y2 – 32 = 0
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- 25 -
Chemistry, Physics & Mathematics (Test Code: P)
Sol: Ans [1]
9
1
2
b2


1

1

and
a 2 b2
5
a2
 a2 =
79.
AIEEE-2011
32
32
and b2 =
3
5
If the mean deviation about the median of the numbers a, 2a, ....., 50a is 50, then |a| equals :
(1) 2
Sol: Ans [3]
(2) 3
a
(3) 4
(4) 5
51a
51a
51a
 2a 
 ....  50a 
= 50 × 50
2
2
2

|a|
[49 + 47 + 45 + ... + 5 + 3 + 1 + 3 + ... + 49] = 2500
2

|a|
[2 × 25 × 25] = 2500
2
AMITY
|a| = 4.
80.
 1  cos{2( x  2)} 
lim 

x 2 
x2


(1) does not exist
(2) equals
(3) equals – 2
2
(4) equals
1
2
Institute for Competitive Examinations
Sol: Ans [1]
81.
lim
x 2
Statement-1:
 | sin( x  2) | 
2sin 2 ( x  2)
| 2 || sin( x  2) |
 lim
= 2 lim

 which does not exist.
x

2
x2 
x 2

x2
x2
The number of ways of distributing 10 identical balls in 4 distinct boxes such that no
box is empty is 9C3.
Statement-2:
The number of ways of choosing any 3 places from 9 different places is 9C3.
(1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for
Statement-1.
(2) Statement-1 is true, Statement-2 is true; Statement-2 is not correct explanation of Statement-1.
(3) Statement-1 is true, Statement-2 is false.
(4) Statement-1 is false, Statement-2 is true
Sol: Ans [2]
The number of ways of distributing n identical balls is r distinct boxes such that no box
remains empty is
n –1
Cr –1.
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- 26 -
Chemistry, Physics & Mathematics (Test Code: P)
AIEEE-2011
82.
Let R be the set of real numbers
Statement-1:
A = {(x, y)  R × R: y – x is an integer) is an equivalence reaction on R.
Statement-2:
B = {(x, y)  R × R : x = y for some rational number } is an equivalence relation on
R.
(1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for
Statement-1.
(2) Statement-1 is true, Statement-2 is true; Statement-2 is not correct explanation of Statement-1.
(3) Statement-1 is true, Statement-2 is false.
(4) Statement-1 is false, Statement-2 is true
Sol: Ans [3]
83.
Statement-1 is true, Statement-2 is false.
Consider 5 independent bernoulli’s trials each with probability of success p. If the probability of at
least one failure is greater than or equal to
 1 3
 , 
 2 4
31
, then p lies in the interval:
32
 3 11 
 , 
 4 12 
 1
 0, 2 
 11 
 ,1
 12 
AMITY
(1)
Sol: Ans [3]
84.
(2)
1 – (p)5 
31
1

 p5 
32
32
(3)
(4)
 1
 0, 2 
The two circles x2 + y2 = ax and x2 + y2 = c2 (c> 0) touch each other if
(1) 2 |a| = c
|a| = c
(3) a = 2c
(4) |a| = 2c
Institute
for(2)Competitive
Examinations
Sol: Ans [2]
85.
|r1 – r2| = C1C2
Let A and B be two symmetric materices of order 3.
Statement-1:
A(BA) and (AB)A are symmetric materices.
Statement-2:
AB is symmetric matrix if matrix multiplication of A with B is commutative.
(1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for
Statement-1.
(2) Statement-1 is true, Statement-2 is true; Statement-2 is not correct explanation of Statement-1.
(3) Statement-1 is true, Statement-2 is false.
(4) Statement-1 is false, Statement-2 is true
Sol: Ans [2]
[A(BA)]T = (BA)TAT = (ATBT)A
= (AB)A
= A(BA) since A and B are symmetric
 Statement 1 is true
(AB)T = BTAT = BA = AB  statement 2 is true
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- 27 -
Chemistry, Physics & Mathematics (Test Code: P)
86.
AIEEE-2011
If C and D are two events such that C  D and P (D)  0, then the correct statement among the
following is:
(2) P(C|D)  P(C)
(1) P(C|D) = P(C)
Sol: Ans [2]
(3) P(C|D) < P(C)
(4) P(C|D)=
P(D)
P(C)
(C D) P(C)
C

P  = P
C  D.
P(D) P(D)
 D
P(D)  0  0 < P(D)  1
1
<
P(D)
1
 P(C) 
P(C)
P(D)
C
 P(C)  P  
D
AMITY
87.



 


The vectors a and b are not prependicular and c and d are two vectors satisfying: b  c  b  d


and a.d  0. Then the vector d is equal to:
(1)

  b .c  
b     c
 a .b 
(2)

  a.c  
c     b
 a.b 
(3)

  b .c  
b     c
 a.b 
(4)

  a .c  
c     b
 a.b 
Institute
for Competitive
Examinations
 
  
a  ( b  c )  a  (b  d )
Sol: Ans [4]
   
      
 (a .c )b  ( a.b )c  ( a.d )b  ( a.b ) d [a.d  0]

88.
Statement-1:
  (a.c ) 
d c   b
(a .b )
The point A(1, 0, 7) is the mirror image of the point B (1, 6, 3) in the line:
x y 1 z  2


1
2
3
Statement-2:
The line :
x y 1 z  2


bisects the line segment joining A(1, 0, 7) and B(1, 6, 3).
1
2
3
(1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for
Statement-1.
(2) Statement-1 is true, Statement-2 is true; Statement-2 is not correct explanation of Statement-1.
(3) Statement-1 is true, Statement-2 is false.
(4) Statement-1 is false, Statement-2 is true
Amity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42
- 28 -
Chemistry, Physics & Mathematics (Test Code: P)
AIEEE-2011
Sol: Ans [2]
Line
x y 1 z  2


is  to the line segment AB. Also the line passes through the mid
1
2
3
point of AB.
89.
If A = sin2x + cos4x, then for all real x:
(1)
3
A 1
4
Sol: Ans [1]
(2)
13
A 1
16
(3) 1  A  2
(4)
3
13
A
4
16
A = sin2x + cos4x = cos4x – cos2x + 1
= t2 – t + 1 for 0  t  1
3
 t2 – t +1  1
4
90.
The number of values of k for which the linear equations
4x + ky + 2z = 0
kx + 4y + z = 0
AMITY
2x + 2y + z = 0
possess a non-zero solution is:
(1) 3
(2) 2
(3) 1
4 k
k
(4) zero
2
4 1 0
For non-zero solution  = 0 
Institute
for Competitive
Examinations
2 2 1
Sol: Ans [2]

4(4 – 2) – k(k –2) + 2 (2k – 8) = 0

+8 – k2 + 2 k + 4k – 16 = 0

– k2 + 6k – 8 = 0

k2 – 6 k + 8 = 0

k = 2, or 4.

Amity Institute for Competitive Examinations : Phones: 26963838, 26850005/6, 25573111/2/3/4, 95120-2431839/42
- 29 -