AIEEE-2010
CHEMISTRY, PHYSICS & MATHS
AIEEE-2010 QUESTION PAPER
1.
2.
3.
4.
PART-A : CHEMISTRY
The standard enthalpy of formation of NH3 is
–46.0kJmol–1. If the enthalpy of formation of
H2 from its atoms is – 436 kJmol–1 and that of
N2 is – 712 kJmol–1, the average bond enthalpy
of N–H bond in NH3 is
1) –964 kJ mol–1
2) +352 kJ mol–1
3) +1056 kJmol-1
4) –1102 kJ mol–1
The time for half life period of a certain reaction
A products is 1 hour. When the initial
concentration of the reactant ‘A’, is 2.0 molL–1,
how much time does it take for its concentration
to come from 0.50 to 0.25 mol L–1 if it is zero
order reaction?
1) 4h
2) 0.5h
3) 0.25h
4) 1h
7.
m
A solution containing 2.675 g of CoCl3. 6NH3
(molar mass = 267.5g mol–1) is passed through
a cation exchanger. The chloride ions obtained
in solution were treated with excess of AgNO3
to give 4.78g of AgCl (molar mass =
143.5 g mol–1). The formula of the complex is
(At. Mass of Ag = 108u)
1) [Co(NH3)6]Cl3
2) [CoCl 2(NH3)4]Cl
4) [CoCl(NH3)5]Cl2
3) [CoCl3(NH3)3]
Consider the reaction :
S(s) + 2H+(aq) + 2Cl–(aq)
Cl2(aq)+ H2S(aq)
The rate equation for this reaction is,
rate = k[Cl2][H2S]
Which of these mechanisms is /are consistent
with this rate equation ?
H+ + Cl– + Cl+ + HS– (slow)
(A) Cl2+ H2
+
–
H+ + Cl– + S (fast)
Cl + HS
H+ + HS– (fast equilibrium)
(B) H2S
–
Cl2 + HS
2Cl– + H+ + S (slow)
1) B only
2) Both A and B
3) Neither A nor B
4) A only
m
m
m
š
m
5.
6.
If 10 –4 dm 3 of water is introduced into a
1.0 dm3 flask to 300k, how many moles of water
are in the vapour phase when equilibrium is
established ? (Given : Vapour pressure of H2O
at 300 K is 3170 Pa; R = 8.314 JK–1 mol–1)
2) 1.53 × 10–2 mol
1) 5.56 × 10–3 mol
–2
4) 1.27 × 10–3 mol
3) 4.46 × 10 mol
AKASH MULTIMEDIA
8.
9.
One mole of a symmetrical alkene on ozonolysis
gives two moles of an aldehyde having a
molecular mass of 44u. The alkene is
1) propene
2) 1-butene
3) 2-butene
4) ethene
If sodium sulphate is considered to be completely
dissociated into cations and anions in aqueous
solution, the change in freezing point of water
%Tt , when 0.01 mol of sodium sulphate is
dissolved in 1 kg of water, is
(Kf = 1.86 K kg mol–1)
1) 0.0372 K
2) 0.0558 K
3) 0.0744 K
4) 0.0186 K
From amongst the following alcohols the one
that would react fastest with conc.HCl and
anhydrous ZnCl2, is
1) 2 – Butanol
2) 2- Methylpropan-2-ol
3) 2-Methylpropanaol 4) 1 -Butanol
In the chemical reactions,
NH2
}}}}m A }}}m B
NaNO2
HCl,278K
HBF4
the compound ‘A’ and ‘B’ respectively are
1) nitrobenzene and fluorobenzene
2) phenol and benzene
3) benzene
diazonium
chloride
and
fluorobenzene
4) nitrobenzene and chlorobenzene
10. 29.5 mg of an organic compound containing
nitrogen was digested according to Kjeldahl’s
method and the evolved ammonia was absorbed
in 20 mL of 0.1 M HCl solution. The excess of
the acid required 15mL of 0.1 M NaOH solution
for complete neutralization. The percentage of
nitrogen in the compound is
1) 59.0
2) 47.4
3) 23.7
4) 29.5
11. The energy required to break one mole of
Cl – Cl bonds in Cl2 is 242 kJ mol–1. The longest
wavelength of light capable of breaking a
single Cl – Cl bond is (c = 3 × 108 ms –1 and
NA = 6.02 × 1023 mol–1)
1) 594 nm 2) 640 nm 3) 700 nm 4) 494 nm
1
AIEEE-2010
CHEMISTRY, PHYSICS & MATHS
12. Ionisation energy of He+ is 19.6×10–18J atom–1.
The energy of the first stationary state (n = 1) of
Li2+ is
ii) H2PO4–
1) 4.41 × 10–16 J atom–1
3) – 2.2 × 10–15 J atom–1
4) 8.82 × 10–17 J atom–1
13. Consider the following bromides :
Me
Me
Me
Br
Br
(B) Br
(A)
(C)
The correct order of SN1 reactivity is
1) B > C > A
2) B > A > C
3) C > B > A
4) A > B > C
14. Which one of the following has an optical
isomer?
2) [Co(en)3]3+
1) [Zn(en) (NH3)2]2+
2) [Co(H 2O) 4(en)]3+
4) [Zn(en)2]2+
(en = ethylenediamine)
15. On mixing, heptane and octane form an ideal
solution. At 373 K, the vaopur pressure of the two
liquid components (heptane and octane) are
105 kPa and 45 kPa and 45 kPa respectively.
Vapour pressure of the solution obtained by mixing
25.0g of heptane and 35g of octane will be (molar
mass of heptane = 100g mol–1 and of octane =
114 g mol–1).
1) 72.0 kPa
2) 36.1 kPa
3) 96.2 kPa
4) 144.5 kPa
16. The main product of the following reaction is
C6 H 5 CH 2 CH(OH)CH CH 3 2
H 5 C6
1)
H
C6H5
2
CH3
C=C
2)
CH3
H
H5C6CH2CH2
CH(CH3)2
C=C
H
conc.H2 SO4
C6H5CH2
CH CH 3 2
3)
}}}}m ?
H
C=C
C = CH2
4)
H
m
+ H O m HPO + H O
+ OH m H PO + O
2
2
4
+
3
–
2–
iii) H2PO4–
3
4
In which of the above does H 2PO4– act as an
acid?
1) ii only 2) i and ii 3) iii only 4) i only
2) – 4.41 × 10–17 J atom–1
Me
17. Three reactions involving H 2PO 4– are given
below:
H3O+ + H2PO4–
i) H3PO4 + H2O
H3C
18. In aqueous solution the ionization constants for
carbonic acid are
K1 = 4.2 × 10–7 and K2 = 4.8 × 10–11
select the correct statement for a saturated 0.034
M solution of the carbonic acid.
1) The concentration of CO32 is 0.034 M.
2) The concentration of CO32 is greater than
that of HCO3 .
3) The concentration of H + and HCO 3 are
approximately equal.
4) The concentration of H+ is double that of CO32 19. The edge length of a face entered cubic cell of
an ionic substance is 508 pm. If the radius of
the cation is 110 pm, the radius of the anion is
1) 288 pm
2) 398 pm
3) 618 pm
4) 144 pm
20. The correct order of increasing basicity of the
given conjugate bases (R= CH3) is
1) RCOO HC C R NH 2
2) R HC y C RCOO NH 2
3) RCOO NH 2 HC y C R
4) RCOO HC y C NH 2 R
21. The correct order of increasing basicity of the
given conjugate bases (R = CH3) is
1) Al3+ > Mg2+ > Na+ > F– > O2–
2) Na+ > Mg2+ > Al3+ > O2– > F–
3) Na+ > F– > Mg2+ > O2– > Al3+
4) O2– > F– > Na+ > Mg2+ > Al3+
22. Solubility product of silver bromide is
5.0 × 10–13. The quantity of potassium bromide
(molar mass taken as 120g of mol–1) to be added
to 1 litre of 0.05 M solution of silver nitrate to
start the precipitation of AgBr is
2) 1.2 × 10–9g
1) 1.2 × 10–10g
4) 5.0 × 10–8 g
3) 6.2 × 10–5g
AKASH MULTIMEDIA
AIEEE-2010
CHEMISTRY, PHYSICS & MATHS
23. The Gibbs energy for the decomposition of
Al2O3 at 5000C is as follows :
2
4
Al 2 O3 m Al O2 , % r G 966kJmol 1
3
3
The potential difference needed for electrolytic
reduction of Al2O3 at 5000C is at least
1) 4.5 V
2) 3.0 V
3) 2.5 V
PART-B : PHYSICS
31. A rectangular loop has a sliding connector PQ
of length l and resistance R 8 and it is moving
with a speed v as shown . The set-up is placed
in a uniform magnetic field going into the plane
of the paper. The three currents I1,I2 and I are
2) 10
3) 11
4) 8
25. Percentage of free space in cubic close packed
structure and in body centered packed structure
are respectively
1) 30% and 26%
2) 26% and 32%
3) 32% and 48%
4) 48% and 26%
26. Out of the following, the alkene that exhibits
optical isomerism is
1) 3 – methyl– 2 – pentane
2) 4 – methyl – 1– pentane
3) 3 – methyl – 1 – pentane
4) 2 – methyl – 2 – pentane
27. Biuret test is not given by
1) carbohydrates
2) polypeptides
3) urea
4) proteins
0
28. The correct order of E M2 / M values with negative
sign for the four successive elements Cr, Mn,
Fe and Co is
1) Mn > Cr > Fe > Co 2) Cr > Fe > Mn > Co
3) Fe > Mn > Cr > Co 4) Cr > Mn > Fe > Co
29. The polymer containing strong intermolecular
forces e.g. hydrogen bonding, is
1) teflon
2) nylon 6, 6
3) polystyrene
4) natural rubber
30. For a particular reversible reaction at
temperature T, %H and %S were found to be
both +ve. If Te is the temperature at equilibrium,
the reaction would be spontaneous when
2) T > Te
1) Te > T
3) Te is 5 times T
4) T = Te
AKASH MULTIMEDIA
R8
R8
24. At 250C, the solubility product of Mg(OH)2 is
1.0 10–11. At which pH, will Mg2+ ions start
precipitating in the form of Mg(OH)2 from a
solution of 0.001 M Mg2+ ions ?
1) 9
l
P
4) 5.0 V
V
R8
I
I2
I1
1) I1 I 2 BlV
2BlV
,I R
R
BlV
2BlV
,I 3R
3R
BlV
3) I1 I 2 I R
BlV
BlV
,I 4) I1 I 2 6R
3R
2) I1 I 2 32. Let C be the capacitance of a capacitor
discharging through a resistor R. Suppose t1 is
the time taken for the energy stored in the
capacitor to reduce to half its initial value and t2
is the time taken for the charge to reduce to onefourth its initial value. Then the ratio t1/t2 will
be
1) 1
2)
1
2
3)
1
4
3) 2
Directions : Questions number 33 – 34 contain
Statement -1 and Statement-2. Of the four
choices given after the statements, choose the
one that best describes the two statements.
33. Statement -1 : Two particles moving in the same
direction do not lose all their energy in a
completely inelastic collision.
Statement-2 : Principle of conservation of
momentum holds true for all kinds of collisions.
1) Statement -1 is true, Statement -2 is true,
Statement-2 is the correct explanation of
Statement1
2) Statement -1 is true, Statement-2 is true;
Statement-2 is not the correct explanation of
Statement -1.
3) Statement -1 is false, Statement -2 is true.
4) Statement-1 is true, Statement -2 is false
3
AIEEE-2010
CHEMISTRY, PHYSICS & MATHS
34. Statement -1 : When ultraviolet light is incident
on a photocell, its stopping potential is V0 and
the maximum kinetic energy of the photoelectrons is Kmax. When the ultraviolet light is
replaced by X-rays , both V0 and Kmax increase
Statement-2 : Photoelectrons are emitted with
speeds ranging from zero to a maximum value
because of the range of frequencies present in
the incident light
1) Statement -1 is true, Statement -2 is true,
Statement-2 is the correct explanation of
Statement1
2) Statement -1 is true, Statement-2 is true;
Statement-2 is not the correct explanation of
Statement -1.
3) Statement -1 is false, Statement -2 is true.
4) Statement-1 is true, Statement -2 is false
35. A ball is made of a material of density S where
Soil S Swater with Soil and Swater representing
the densities of oil and water, respectively. The
oil and water are immiscible . If the above ball
is in equilibrium in a mixture of this oil and
water, which of the following pictures represents
its equilibrium position ?
oil
1)
2)
water
x’
x
d
x
2)
x’
d
3)
x’
x
d
oil
4) x
x’
d
water
36. A
JG
particle is moving with velocity
V K yiˆ xjˆ , where K is a constant . The
general equation for its path is
2) y x 2 constant
1) y 2 x 2 constant
3) y 2 x constant
4) xy = constant
37. Two long parallel wires are at a distance 2d apart.
They carry steady equal currents flowing out of
the plane of the paper as shown. The variation
of the magnetic field B along the line XX’ is given
by
4
d
38. In the circuit shown below, the key K is closed
at t = 0. The current through the battery is
v
k
L
R1
R2
1)
oil
4)
d
B
2)
3)
d
B
V R1 R 2 V
at t 0 and
at t d
R1 R 2
R2
VR1R 2
water
d
B
oil
water
1)
B
3)
R12 R 22
at t 0 and
V
at t d
R2
V R1 R 2 V
at t 0 and
at t d
R2
R1 R 2
VR1R 2
V
at t 0 and
at t d
4) R 2
R12 R 22
39. The figure shows the position -time (x -t) graph
of one-dimensional motion of a body of mass
0.4kg. The magnitude of each impulse is
2
x(m)
0
1) 0.4 Ns
2
4
6
8
t (s)
2) 0.8 Ns
10 12
14 16
3) 1.6 Ns
4) 0.2Ns
AKASH MULTIMEDIA
AIEEE-2010
CHEMISTRY, PHYSICS & MATHS
Directions : Questions number 40 -41 are based
on the following paragraph.
A nucleus of mass M + %m is at rest and decays
M
each,
into two daughter nuclei of equal mass
2
Speed of light is c.
45. A diatiomic ideal gas is used in a Car engine as
the working substance. If during the adiabatic
expansion part of the cycle, volume of the gas
increases from V to 32V the efficiency of the
engine is
1) 0.5
2) 0.75
3) 0.99
4) 0.25
40. The bindign energy per nucleon for the parent
nucleus is E1 and that for the daughter nuclei is E2.
Then
2) E1 > E2
1) E2 = 2E1
3) E2 > E1
4) E1 = 2E2
46. If a source of power 4 kW produces
1020 photons/second, the radiation belong to a
part of the spectrum called
1) X-rays
2) ultraviolet rays
3) microwaves
4) H rays
47. The respective number of significant figures for
the numbers 23.023, 0.0003 and 2.1 s 10–3 are
1) 5, 1, 2
2) 5, 1, 5 3) 5, 5, 2 4) 4, 4, 2
41. The speed of daughter nuclei is
1) c
3) c
%m
2) c
M %m
2 %m
M
4) c
%m
M %m
%m
M
42. A radioactive nucleus (initial mass number A
and atomic number Z) emits 3B particles and
2 positrons. The ratio of number of neutrons to
that of protons in the final nucleus will be
1)
AZ4
Z2
2)
A Z 8
Z4
3)
AZ4
Z 8
4)
A Z 12
Z4
43. A thin semi-circular ring of radius r has a
positive chargeG q distributed uniformly over it.
The net field E at the centre O is
q
1) 2 Q2 F r 2 ĵ
0
q
2) 4 Q2 F r 2 ĵ
0
j
q
3) 2 2 ĵ
4Q F0 r
q
4) 2Q2 F r 2 ĵ
0
i
O
44. The combination of gates shown below yields
1) OR gate
A
X
2) NOT gate
3) XOR gate
48. In a series LCR circuit R = 200 8 and the
voltage and the frequency of the main supply
is 220V and 50Hz respectively. On taking out
the capacitance from the circuit the current lags
behind the voltage by 300. On taking out the
inductor from the circuit the current leads the
voltage by 300. The power dissipated in the LCR
circuit is
1) 305W 2) 210W
3) Zero W 4) 242W
49. Let there be a spherically symmetric charge
distribution with charge density varying as
¥5 r´
S r S0 ¦ µ upto r=R, and S r 0 for
§ 4 R¶
r > R, where r is the distance from the origin.
The electric field at a distance r(r < R) from the
origin is given by
1)
4 QS0 r ¥ 5 r ´
3F 0 ¦§ 3 R µ¶
2)
S0 r ¥ 5 r ´
4F 0 ¦§ 3 R µ¶
3)
4S0 r ¥ 5 r ´
3F 0 ¦§ 4 R µ¶
4)
S0 r ¥ 5 r ´
3F 0 ¦§ 4 R µ¶
50. The pontential energy function for the force
between two atoms in a diatomic molecule
is approximately given by
a
b
U (X) = U X 12 6 , where a and b are
X
X
constants and X is the distance between the
atoms. If the dissociation energy of the molecule
is D [U(X d) U at equilibrium ],D is
B
4) NAND gate
AKASH MULTIMEDIA
1)
b2
2a
2)
b2
12a
3)
b2
4a
4)
b2
6a
5
AIEEE-2010
CHEMISTRY, PHYSICS & MATHS
51. Two identical charged spheres are suspended
by strings of equal lengths. The strings make
an angle of 30 0 with each other. When
suspended in a liquid of density 0.8 g cm-3, the
angle remans the same. If density of the material
of the sphere is 1.6 g cm-3, the dielectric constant
of the liquid is
1) 4
2) 3
3) 2
4) 1
52. Two conductors have the same resistance at 00C
but their temperature coefficients of resistance
are
B1 and B 2 . The respective temperature
coefficients of their series and parallel
combinations are nearly
1)
B1 B 2
2
, B1 B 2
2) B1 B 2 ,
BB
1 2
3) B1 B 2 , B B
1
2
4)
B1 B 2
2
B1 B 2 B1 B 2
,
2
2
53. A point P moves in counter-clockwise direction
on a circular path as shown in the figure. The
movement of ‘P’ is such that it sweeps out a
length s = t3 + 5, where s is in metres and t is in
seconds. The radius of hte path is 20 m. The
acceleration of ‘P’ when t = 2 s is nearly.
1) 13 m/s2
y
2) 12 m/s2
B
P(x,y)
3) 7.2 m/s2
2
4) 14 m/s2
0m
O
x
A
54. Two fixed frictionless inclined planes making
an angle 300 and 600 with the vertical are shown
in the figure. Two blocks A and B are placed on
the two planes. what is the relative vertical
acceleration of A with respect to B?
A
B
55. For a particle in uniform circle motion, the
G
acceleration a at a point P R, R
on the circle of
radius R is (Here R is measured from the
X-axis)
v2
v2
v2
v2
1) cos R ˆi sin Rˆj 2) sin R ˆi cos Rˆj
R
R
R
R
2
2
2
2
v ˆ v ˆ
v
v
i
j
3) cos R ˆi sin Rˆj 4)
R
R
R
R
Directions : Questions number 56-58 are based
on the following paragraph
An initially parallel cylindrical beam travels in
a medium of refractive index N (I) N 0 N 2 I,
where N 0 and N 2 are positive constants and I is
the intensity of the light beam. The intensity of
the beam is decreasing with increasing radius.
56. As the beam enters the medium , it will
1) diverge
2) converge
3) diverge near the axis and converge near the
periphery
4) travel as a cylindrical beam
57. The initial shape of the wave front of the beam
is
1) convex
2) concave
3) convex near the axis and concave near the
periphery
4) planar
58. The speed of light in the medium is
1) minimum of the axis of the beam
2) the same everywhere in the beam
3) directly proportional to the intensity
4) maximum on the axis of hte beam
59. A small particle of mass m is projected at an
angle R with the X-axis with an intial velocity
O0 in the x-y plane as shown in the figure. At a
v0 sin R
time t ,the angular momentum of the
g
particle is
2) mg v t cos R kˆ
1) mg v t 2 cos R ˆj
0
0
y
1
2
V
3) mg v0 t cos R kˆ
2
1
2
4) mg v0 t cos R ˆi
R
2
x
ˆ
ˆ
ˆ
where i, j and k are unit vectors along x, y and
z– axis respectively.
0
600
1)
2)
3)
4)
6
30
0
4.9 ms-2 in vertical direction
9.8 ms-2 in vertical direction
Zero
4.9 ms-2 in horizontal direction
AKASH MULTIMEDIA
AIEEE-2010
CHEMISTRY, PHYSICS & MATHS
60. The equation of a wave on astring of linear mass
density 0.04 kg m–1 is given by
¨
¥ t
x ´·
§ 0.04(s) 0.50(m) µ¶ ¸¹
y 0.02(m)sin © 2 Q ¦
ª
The tension in the string is
1) 4.0N
2) 12.5N
3) 0.5 N
4) 6.25 N
PART-C : MATHEMATICS
61. Let cos(B C) where 0 b B, C b
56
1)
33
4
5
and let sin(B C) ,
5
13
Q,
4
20
3)
7
1) 4 2 2
25
4)
16
62. Let S be a non-empty subset of R. Consider the
following statement :
P : There is a rational number x  S such that
x > 0.
Which of the following statements is the
negation of the statement P ?
1) There is no rational number x  S such that
xb0
2) Every rational number x  S satisfies x r 0
3) x  S and x b 0  x is not rational
4) There is a rational number x  S such that
xb0
G
G
G
63. Let a ˆj kˆ and c iˆ ˆj kˆ. Then vector b
G
G
G
satisfying aG s b cG 0 and aG.b 3 is
1) 2iˆ ˆj 2 kˆ
2) iˆ ˆj 2 kˆ
3) iˆ ˆj 2 kˆ
4) iˆ ˆj 2 kˆ
2) 4 2 1
3) 4 2 1
4) 4 2 2
67. If two tangents drawn from a point P to the
parabola y2 = 4x are at right angles, then the
locus of P is
1) 2x + 1 = 0
2) x = –1
3) 2x – 1 = 0
4) x = 1
G
G ˆ ˆ
68. If the vectors a i j 2 kˆ, b 2iˆ 4 ˆj kˆ and
G
c Miˆ ˆj Nkˆ are mutually orthogonal, then
(M , N ) 1) (2, –3)
then tan 2B 19
2)
12
66. The area bounded by the curves y = cosx and
3Q
y = sinx between the ordinates x = 0 and x 2
is
2) (–2, 3)
3) (3, –2)
4) (–3, 2)
69. Consider the following relations :
R = {(x, y) |x, y are real numbers an x = wy for
some rational number w};
«®¥ m p ´
, µ m.n. p and q are integers
­®§ n q ¶
such that n, q x 0 and qm pn^.
S ¬¦
Then
1) neither R nor S is an equivalence relation
2) S is an equivalence relation but R is not an
equivalence relation
3) R and S both are equivalence relations
4) R is an equivalence relation but S is not an
equivalence relation
f :R
70. Let
mR
be
defined
by
« k 2 x, if
­ 2 x 3, if
x b 1
. If f has a local
x 1
minimum at x = –1, then a possible value of k is
1
1) 0
2) 3) –1
4) 1
2
f ( x) ¬
64. The equation of the tangent to the curve
4
y x 2 , that is parallel to the x-axis, is
x
1) y = 1
2) y = 2
3) y = 3
4) y = 0
71. The number of 3 × 3 non-singular matrices, with
four entries as 1 and all other entries as 0, is
1) 5
2) 6
3) at least 7
4) less than 4
65. Solution of the differential equation
72. Four numbers are chosen at random (without
replacement) from the set {1,2,3....,20}.
Statement-1 : The probability that the chosen
numbers when arranged in some order will form
1
.
an AP is
85
cos x dy = y(sinx – y)dx, 0 x 1) y sec x = tan x + c
3) tanx = (secx + c)y
AKASH MULTIMEDIA
Q
is
2
2) y tan x = sec x + c
4) sec x = (tanx + c)y
7
AIEEE-2010
CHEMISTRY, PHYSICS & MATHS
Statement-2 : If the four chosen numbers from
an AP, then the set of all possible values of
common difference is {p1, p 2, p 3, p 4, p 5}.
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 for
Statement-1
3) Statement-1 is true, Statement-2 is false
4) Statement-1 is false, Statement-2 is true
73. Statement-1 : The point A(3, 1, 6) is the mirror
image of the point B(1,3,4) in the plane
x–y+z = 5.
Statement-2 : The plane x – y + z = 5 bisects the
line segment joining A(3, 1, 6) and B(1, 3, 4).
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 for
Statement-1
3) Statement-1 is true, Statement-2 is false
4) Statement-1 is false, Statement-2 is true
74. Let
S3 S1 ¤j
¤ j( j 1)
10
j 1
10
C j , S2 ¤j
10
j 1
10
Cj
and
10
j 1
2 10
Cj.
Statement-1 : S3 = 55×29
Statement-2 : S1 = 90 × 28 and S2 = 10×28.
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 for
Statement-1
3) Statement-1 is true, Statement-2 is false
4) Statement-1 is false, Statement-2 is true
75. Let A be a 2 × 2 matrix with non-zero entries
and let A2 = I, where I is 2 × 2 identity matrix.
Define Tr(A) = sum of diagonal elements of
A and |A| = determinant of matrix A.
8
Statement-1 : Tr(A) = 0
Statement-2 : |A| = 1
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 for
Statement-1
3) Statement-1 is true, Statement-2 is false
4) Statement-1 is false, Statement-2 is true
76. Let R
mR
be a continuous funciton defined
1
.
by f ( x ) x
e 2e x
1
Statement-1 : f (c ) , for some c  R.
3
1
, for all x  R
Statement-2 : 0 f ( x ) b
2 2
1) Statement-1 is true, Statement-2 is true;
Statement-2 is not the correct explanation for
Statement-1
2) Statement-1 is true, Statement-2 is false
3) Statement-1 is false, Statement-2 is true
4) Statement-1 is true, Statement-2 is true;
Statement-2 is the correct explanation for
Statement-1
77. For a regular polygon, let r and R be the radii of
the inscribed and the circumscribed circles. A
false statement among the following is
1) There is a regular polygon with
r
1
R
2
2) There is a regular polygon with
r 2
R 3
r
3
R
2
r 1
4) There is a regular polygon with
R 2
3) There is a regular polygon with
78. If B and C are the roots of the equation
x2–x+1=0, then B 2009 C2009 1) –1
2) 1
3) 2
4) –2
79. The number of complex number z such that
|z–1| = |z + 1| = |z–i| equals
4) 0
1) 1
2) 2
3) d
AKASH MULTIMEDIA
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CHEMISTRY, PHYSICS & MATHS
80. A line AB in three-dimensional space makes
angles 450 and 1200 with the positive x-axis and
the positive y-axis respectively. If AB makes an
acute angle R with the positive z-axis, then R
equals
2) 600
3) 750
4) 300
1) 450
x y
81. The line L given by 1 passes through
5 b
the point (13, 32). The line K is parallel to L
x y
1. Then the
and has the equation
c 3
distance between L and K is
17
23
23
1) 17
2)
3)
4)
15
17
15
82. A person is to count 4500 currency notes. Let
an denote the number of notes he counts in the
nth minute. If a1 = a2 = .....=a10 = 150 and a10,
a11,....are in A.P. with common difference –2,
then the time taken by him to count all notes is
1) 34 minutes
2) 125 minutes
3) 135 minutes
4) 24 minutes
83. Let f : R
m R be a positive increasing function
f (3 x )
f (2 x )
1. Then lim
with xlim
md f ( x )
x md f ( x )
2
3
1)
2)
3) 3
4) 1
3
2
84. Let p(x) be a function defined on R such that
p'(x) = p'(1–x), for all x [0,1], p(0) 1 and
1
p(1) = 41. Then ° p( x )dx equals
0
1) 21
2) 41
3) 42
4)
41
85. Let f : (–1, 1)
R be a differentiable function
with f(0) = –1 and f'(0) = 1.
m
Let g(x) = [f(2f(x) + 2)]2. Then g'(0) =
1) –4
2) 0
3) –2
4) 4
86. There are two urns. Urn A has 3 distinct red
balls and urn B has 9 distinct blue balls. From
each urn two balls are taken out at random and
then transferred to the other. The number of
ways in which this can be done is
1) 36
2) 66
3) 108
4) 3
87. Consider the system of linear equations :
x1 + 2x2 + x3 = 3
2x1 + 3x2 +x3 = 3
3x1 + 5x2 + 2x3 = 1
The system has
1) exactly 3 solutions
2) a unique solution
3) no solution
4) infinite number of solutions
88. An urn contains nine balls of which three are
red, four are blue and two are green. Three balls
are drawn at random without replacement from
the urn. The probability that the three balls have
different colour is
1)
2
7
2)
1
21
3)
2
23
4)
1
3
89. For two data sets, each of size 5, the variances
are given to be 4 and 5 and the corresponding
means are given to be 2 and 4, respectively.
The variance of the combined data set is
1)
11
2
2) 6
3)
13
2
4)
5
2
90. The circle x2 + y2= 4x + 8y + 5 intersects the line
3x – 4y = m at two distinct points if
1) –35 < m < 15
2) 15 < m < 65
3) 35 < m < 85
4) –85 < m < –35
123456789012345678901234
123456789012345678901234
123456789012345678901234
AIEEE 2010 ANSWERS
PART-A : CHEMISTRY
1) 2
2) 3
3) 1
4) 4
5) 4
6) 3
7) 2
8) 2
9) 3
10) 3
11) 4
12) 2
13) 1
14) 2
15) 1
AKASH MULTIMEDIA
16) 1
17) 1
18) 3
19) 4
20) 4
21) 4
22) 2
23) 3
24) 2
25) 2
26) 3
27) 1
28) 1
29) 2
30) 2
9
AIEEE-2010
CHEMISTRY, PHYSICS & MATHS
PART-B : PHYSICS
PART-C : MATHEMATICS
31) 2
32) 3
33) 1
34) 4
35) 2
61) 1
62) 4
63) 4
64) 3
65) 4
36) 1
37) 1
38) 3
39) 2
40) 3
66) 4
67) 2
68) 4
69) 2
70) 3
41) 3
42) 2
43) 3
44) 3
45) 1
71) 3
72) 3
73) 2
74) 3
75) 3
46) 2
47) 1
48) 1
49) 4
50) 2
76) 4
77) 2
78) 2
79) 1
80) 2
51) 3
52) 3
53) 4
54) 4
55) 4
81) 3
82) 1
83) 4
84) 1
85) 1
56) 3
57) 2
58) 4
59) 1
60) 3
86) 3
87) 3
88) 1
89) 1
90) 1
.AIEEE 2010 HINTS AND SOLUTIONS.
.
PART-A : CHEMISTRY
1.
6.
(2) Enthalpy of formation of NH3= –46 kJ/mole
(3) 2-butene is symmetrical alkene
O
CH 3 CH H CH 3 }}}}
m 2CH 3 CHO
Zn / H O
= N 2 3H 2 m 2NH3 %H f 2 s 46kJmol
Molar mass of CH3CHO is 44u.
Bond breaking is endothermic and Bond
formation is exothermic
Assuming ‘x’ is the bond energy of N–H bond
(kJ mol–1)
= 712 3 s 436
6x 46 s 2
= x = 352 kJ/mol
2.
(3) For a zero order reaction k 7.
(2) Vant Hoff’s factor (i) for Na2SO4 = 3
3 s 1.80 s
8.
2t 1
0.01
0.0558K
1
(2) 30 alcohols react fastest with ZnCl2/conc. HCl
due to formation of 30 carbocation and
= 2– methyl propan – 2 – ol is the only
30 alcohol
NH2
9.
(3)
F
N2 Cl
}}}m
}}}}m
................. (2)
10.
Since [A0] = 2M, t1/2 = 1 hr; k = 1
= from equation (1)
(1) CoCl3.6NH3
m
t
xCl–
0.25
0.25hr
1
}}}m
AgNO3
x AgCl n
+ N2 + BF3 + HCl
HBF4
NaNO 2
HCl ,278 K
(B)
(A)
< A >0
2
3.
2
x
...... (1)
t
Where x = amount decomposed
k = zero order rate constant for a zero order
reaction
k
3
(3) Moles of HCl reacting with ammonia
= (moles of HCl absorbed) – (moles of NaOH
solution required)
= (20 × 0.1 × 10–3) – (15 × 0.1 × 10–3)
= moles of NH3 evolved
= moles of nitrogen in organic compound
n(AgCl) = xn (CoCl3.6NH3)
= wt. of nitrogen in org.comp = 0.5 × 10
–3
4.78
2.675
s
143.5
267.5
=x 3
= 7 × 10 g
= The complex is [Co(NH ) ]Cl
3 6
4.
5.
10
3
(4) Rate equation is to be derived with respect
to slow step, from mechanism (A)
Rate = k[Cl2][H2S]
PV
128 s 10 5 moles
(4) n RT
× 14
–3
3170 s 10 5 atm s 1L
= 1.27 × 10–3 mol
0.0821L atm k 1mol 1 s 300K
% wt =
7 s 10 3
23.7%
29.5 s 10 3
11. (4) Energy required for 1 Cl2 molecule =
242 s 10 3
joules.
NA
This energy is contained in photon of
wavelength ‘ M ’.
AKASH MULTIMEDIA
AIEEE-2010
hc
M
E
CHEMISTRY, PHYSICS & MATHS
6.626 s 10 34 s 3 s 108
M
242 s 10 3
6.022 s 10 23
XHeptane = 0.45
= Mole fraction of octane = 0.55 = X
0
M 4947 A 494nm
Total pressure =
= 72.0 kPa
Hence 13.6 x Z2He = 19.6 × 10–18 J atom–1.
13.6Z Li2 s
2
9
= - 4.41 × 10–17 J / atom
4
= – 19.6 × 10–18 ×
13.
¨ Z2Li2 ·
1
2
s
13.6Z
© 2 ¸
He
12
ª© Z He ¹¸
OH CH3
conc. H2SO4
16. (1)
CH2-C-CH-CH3
CH3
loss of proton
CH3
CH=CH-HC
CH3
>
Me
(2) Only option (2) is having non-super
imposable mirror image & hence one optical
isomer
en
Zn+2
C=C
H
PO 4 H 2 O m HPO24 H3O ii) H 2acid
conjugate base
PO4 OH m H3 PO4 O2
iii) H 2acid
conjugate base
No optical isomer. It is
Tetrahedral with a plane of symmetry
2)
en
Only in reaction (ii) H 2 PO 4 acts as ‘acid’
en
Co+3
Co+3
en
Optical isomer
H2O
[H+]A z ¨ªHCO3 ·¹ z ¨ªH ·¹ total
en
H2O
H2O
Horizontal plane is plane of symmetry
Zn+2
4)
No optical isomer. It is
Tetrahedral with a plane of symmetry
(1) Mole fraction of Heptane =
25 /100
0.25
0.45
25 35 0.557
100 114
AKASH MULTIMEDIA
2
¨CO3 · is
ª
¹
negligible compared to ¨ªHCO3 ·¹ or ¨ªH ·¹total
19. (4) For an ionic substance in FCC arrangement,
2 r r edge length
2(110 + r–) = 508
r– = 144 pm
en
15.
(3) A m H 2 CO 3 U H HCO 3 K1 = 4.2 × 10–7
As K2 << K1
All major [H+]total = [H+]A and from I equilibrium,
H2O
3)
18.
B m HCO 3 U H CO32
en
en
Co+3
(trans isomer)
PO4 H 2 m H3 O H 2 PO4
17. (1) i) H3acid
conjugate base
NH3
en
CH(CH3)2
H
NH3
1)
(conjugated
system)
Trans isomers is more stable & main product here
Me
>
Me
14.
CH2-CH-CH-CH3
(1) SN1 proceeds via carbocation intermediate,
the most stable one forming the product faster.
Hence reactivity order for A, B, C depends on
stability of carbocation created.
Me
0
i i
= (105 × 0.45) + (45 × 0.55) kPa
¨1 1 ·
2
12. (2) IE He 13.6Z He © 2 2 ¸ where Z He 2
ª1 d ¹
E1 Li 2
¤X P
octane
20. (4) Correct order of increasing basic strength is
R – COO(–) < CH y C(–) < NH 2 < R(–)
21. (4) For isoelectronic species higher the Z/e ratio,
smaller the ionic radius. Z/e values for,
11
AIEEE-2010
CHEMISTRY, PHYSICS & MATHS
28. (1) Mn2+ is more stable.
8
9
0.8; F 0.9
10
10
O2 Na 11
12
1.1; Mg2+ =
1.2
10
10
Al3+ =
13
1.3
10
29. (2) nylon 6, 6 is a polymer of adipic acid and
hexamethylene diamine
O
O
||
||
C CH 2 4 C NH CH 2 6 NH n
30. (2) %G %H T%S
ZZX AgBr
22. (2) Ag Br YZZ
at equilibrium,
Precipitation starts when ionic product just
exceeds solubility product
Ksp = [Ag+][Br–]
K sp
–
[Br ] = ¨ Ag ·
ª
¹
5 s 10 13
10 11
0.05
Precipitation just starts when 10 moles of KBr
is added to 1L of AgNO3 solution.
Number of moles of KBr to be added = 10
–11
Weight of KBr = 10–11 × 120
= 1.2 × 10–9g
%g nFE
E
WG
nF
for a reaction to be spontaneous
negative.
Therefore T > Te
31. (2) A moving conductor is equivalent to a
battery of emf = v B l (motion emf)
Equivalent circuit l = l1 + l2
applying kirchoff’s law
l1R + lR – vBl = 0
l2R + lR – vBl = 0
960000
2.5V
4 s 96500
2
K sp ¨ª Mg ·¹ ¨ªOH ·¹
.................. (1)
.................. (2)
R
R
2
24. (2) Mg 2OH U Mg OH 2
2
%G should be
PART-B : PHYSICS
–11
23. (3)
%G = 0
I2
I1
adding (1) & (2)
2|R + |R = 2vBl
K
sp
¨ªOH ·¹ 10 4
¨ª Mg2 ·¹
=p
OH
= 4 and pH = 10.
25. (2) packing fraction of cubic close packing and
body centered packing are 0.74 and 0.68
respectively.
H
26.(3)
C2H5
H2C=HC
CH3
Only 3 – methyl – 1 – pentene has a chiral
carbon
27. (1) It is a test characteristic of amide linkage.
Urea also has amide linkage like proteins.
12
|
2vBl
vBl
; |1 |2 3R
3R
32. (3) U q2
1 q2
1
(q 0 e t / T )2 0 e 2t / T
2 C 2C
2C
(where U =CR)
U = Ui e–2t/ U
1
U i U i e 2t1/ U
2
1
T
e 2t l / U  t1 | n2
2
2
q0
Now q =
= q0 e–t/2T
4
t1 1
t2 = T|n4 = 2T|n2 = t 4
2
AKASH MULTIMEDIA
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CHEMISTRY, PHYSICS & MATHS
m2
m1
for x > d, Bin between = (ˆj)
33. (1)
v2
v1
If it is a completely inelastic collision then
m1v1 + m2v2 = m1v + m2v
m1v1 m 2 v2
p
p
K.E m1 m 2
2m1 2m 2
G
G
as p1 and p 2 both simultaneously cannot be zero
therefore total KE cannot be lost.
2
1
v
K.Ephotoelectron = hO K
Stopping potential is to stop the fastest
photoelection
hO K
V0 e e
so, K.Emax and V0 both increases.
But K.E ranges from zero to K.Emax because fo
loss of energy due to subsequent collisions
before getting ejected and not due to range of
frrequencies in the incident light.
(2) Soil S Swater
Oil the least dense of them so it should settle at
the top with water at the base. Now the ball is
denser than oil but less denser than water. So, it
will sink through oil but will not sink in water.
So it will stay at the oil-water interface.
G
36. (1) v Kyiˆ Kxjˆ
dy dy dt Kx
dx
dy
s Ky,
Kx;
dx dt dx Ky
dt
dt
y dy = x dx
y2 = x2 + c
37. (1) The magnetic field in between because of
each will be in opposite direction
Bin between
N0i ˆ
N0i
j
( ˆj)
=
2 Qx
2 Q(2d x)
N0 i ¨ 1
1 · ˆ
( j)
©
2Q ª x 2d x ¸¹
towards x’ net magnetic field will add up and
direction will be (ˆj)
2
2
34. (4) Since the frequency of ultraviolet light is
less than the frequency of X–rays, the energy
of each incident photon will be more for X–
rays
35.
towards x net magnetic field will add up and
direction will be (ˆj)
38. (3) At t = 0, inductor behaves like an infinite
resistance
V
So at t = 0, i = R and at t = d , inductor behaves
2
like a conducting wire
V(R1 R 2 )
V
i
R eq
R1R 2
39. (2) From the graph, it is a straight line so, uniform
motion. Because of impulse direction of velocity
changes as can be seen from the slope of the
graph.
2
Initial velocity = = 1 m/s
2
2
Final velocity = 1m / s
2
G
Pi 0.4 N s
JJG
Pj1 0.4 N s
G G G
J Pf Pi 0.4 0.4
G
0.8N s(J impulse)
G
J 0.8 N s
40.
41.
(3) After decay, the daughter nuclei will be
more stabel hence binding energy per nucleon
will be more than that of their parent nucleus.
(3) Conserving the momentum
M
M
V1 V2
2
2
V1 = V2
............ (1)
0
1 M 2 1 M 2
V1 . .V2 ................ (2)
2 2
2 2
%mc2 .
%mc 2 M 2 2 %mc 2
2 %m
V1 ;
V12 ; V1 c
2
M
M
42. (3) In positive beta decay a proton is
transformed into a neutron and a positron is
emitted.
at x = d, Bin between = 0
p }}
m n0 e for x < d, Bin between = (ˆj)
no. of neutrons initially was A – Z
AKASH MULTIMEDIA
13
AIEEE-2010
CHEMISTRY, PHYSICS & MATHS
no. of neutrons after decay (A – Z) – 3 × 2
(due to alpha particles) + 2 x 1 (due to positive
beta decay)
The no. of proton will reduce by 8. [as 3 × 2
(due to alpha particles) +2 (due to positive beta
decay)]
Hence atomic number reduces by 8.
43.
¥ q´
§ Qr µ¶
(3) Linear charge density M ¦
y
dR
R
R
45. (2) The efficiency of cycle is
T
I 1 2
T1
for adiabatic process
= constant
7
5
For diatomic gas H T1V1H 1 T2 V2
¥V ´
T1 T2 ¦ 2 µ
§ V1 ¶
T1 T2 (32)
H 1
H 1
7
1
5
4 s 103
10 s 6.023 s 10 34
20
f = 6.03 s 1016 Hz
The obtained frequency lies in the band of
X-rays.
47.
(1)
48. (4) The given circuit is under resonance as
XL = XC
Hence power dissipated in the circuit is
49.
V2
242W
R
(2) Apply shell theorem the total charge upto
distance r cab be calculated as followed
dq = 4 Qr 2 .dr.S
¨5 r ·
4Qr 2 .dr.S0 © ¸
ª4 R¹
¨5
r3 ·
4 QS0 © r 2 dr dr ¸
R ¹
ª4
¥5 2
r3 ´
QS
dq
q
4
r
dr
dr
0°¦
°
4
R µ¶
0§
r
¨ 5 r3 1 r 4 ·
4QS0 ©
¸
ª4 3 R 4 ¹
E
3
4
kq 1 1 .4 QS ¨ 5 ¥ r ´ r ·
©
¸
0
¦ µ
4 QF 0 r 2
r2
ª 4 § 3 ¶ 4R ¹
E
S0 r ¨ 5 r ·
4 F 0 ©ª 3 R ¸¹
a
b
6
12
x
x
U(x d) 0
dU
¨12a 6b ·
© 13 7 ¸
as, F dx
x ¹
ªx
50. (3) U(x) 6
at equilibrum, F = 0 = x T2 (2 )
5 2/5
T1 4T2
¥ 1´ 3
I ¦ 1 µ 0.75
§ 4¶ 4
14
f
P
44. (1)
Truth table for given combination is
A
B
X
0
0
0
0
1
1
1
0
1
1
1
1
This comes out to be truth of OR gate
TV
(1) 4 s 10 3 10 20 s hf
x
K.dq
E ° dE sin R( ˆj) ° 2 sin R( ˆj)
r
K qr
E 2 ° dR sin R( ˆj)
r Qr
Q
q
Kq
ˆ
( ˆj)
= 2 ° sin R( j)
2
2Q F 0 r 2
r Q0
H 1
46.
= T2 s 4
= U at equilibrium a
¥ 2a ´
¦§ b µ¶
2
2a
b
b
b2
¥ 2a ´ 4a
¦§ b µ¶
= D ¨ª U(x d) U at equilibrium ·¹ b2
4a
AKASH MULTIMEDIA
AIEEE-2010
CHEMISTRY, PHYSICS & MATHS
51. (3) From F.B.D fo sphere, using Lami’s theorem
F
tan R
mg
............... (i)
T
mg
F
¥
S´
mg K ¦ 1 µ
§ d¶
or K 1
1
S
2
d
52. (4) Let R 0 be the initial resistance of both
conductors
= At temperature R
their resistance will be,
R1 R 0 (1 B1R) and R 2 R 0 (1 B 2 R)
for , series combination, Rs = R1 + R2
R s0 (1 B s R) R 0 (1 B1R) R 0 (1 B 2 R)
where R so R 0 R 0 2R 0
= 2R (1 B R) 2R
0
s
or B s 0
R 0 R (B 1 B 2 )
B1 B 2
2
for parallel combination, R p R1R 2
R1 R 2
=
at = 6 s 2 = 12 m/s2
at t = 2s, v=3(2)2 = 12 m/s
centripetal acceleration,
=
ac =
R0
R 2 (1 B1R B 2 R B1B 2 R)
(1 B p R) 0
2
R 0 (2 B1R B 2 R)
as
B1
=B B
1
2
B2
are small quantities
is negligible
AKASH MULTIMEDIA
v 2 144
m / s2
R
20
net acceleration a 2t a 2i
z 14m / s2
54. (4) mg sin R = ma
= a gsin R
where a is along the inclined plane
vertical component of acceleration is g sin2 R
relative vertical acceleration of A with respect
to B is
g
2
g[sin2 60 –sin2 30] 4.9m / s
2
in vertical direction.
=
=
55. (3) For a particle in uniform circular motion,
G v2
a
towards centre of circle
R
G
=a v2
( cos Rˆi sin Rˆj)
R
y
P(R,R)
ac
R0R0
R
0
R0 R0
2
=
and
2
ds
3t 2
dt
dv
6t
and rate of change of speed =
dt
tangential acceleration at t = 2s,
R (1 B1R)R 0 (1 B 2 R)
R p0 (1 B pR) 0
R 0 (1 B1R) R 0 (1 B 2R)
where, R p0 B1 B 2
53. (4) S = t3 + 5 = speed,v when suspended in liquid, as R remains same,
F’
=
tan R
¥ S´
............. (ii)
mg ¦ 1 µ
§ d¶
using (i) and (ii)
F
F’
F
mg
¥ S´
mg ¦ 1 µ where, F ’ K
§ d¶
F
mg
B1 B2
B B
1 2 [1 (B1 B2 )R]
2 (B1 B2 )R
2
as (B1 B 2 )2 is negligible
= Bp R
F
=
or B p ac
x
56. (2) As intensity is maximum at axis,
=
µ will be maximum an speed will be
minimum on the axis of the beam
= beam will converge.
15
AIEEE-2010
CHEMISTRY, PHYSICS & MATHS
57.
(4) For a parallel cylinderical beam, wavefront
will be planar.
65.
dy
y tan x y2 sec x
dx
1 dy 1
tan x sec x
y 2 dx y
58. (1)
G
G G
59. (3) L m(r s v)
G
1
¨
·
L m © vo cos Rt ˆi (v0 sin Rt gt 2 )ˆj ¸ s
2
ª
¹
Let
¨ v 0 cos Rˆi (v 0 sin R gt)ˆj ·
ª
¹
60. (4) T Nv N
X2
k2
1 dy dt
y 2 dx dx
dy
dt
t tan x sec x  (tan x)t sec x
dx
dx
I .F . e
(2 Q / 0.004)2
6.25N
=0.04
(2 Q / 0.50)2
(1) cos(B C) sin(B C) 4
5
5
13
 tan(B C)  tan(B C) 3
4
5
12
°
tan xdx
Q
66.
4
5Q
4
0
Q
(cos x sin x )dx ° (sin x cos x )dx (4) °
3Q
2
3 5
4
12 56
tan 2B tan(B C B C) 3 5 33
1
4 12
°
5Q
4
62. (4) P : there is a rational number x  S such
that x > 0
P : Every rational number x  S satisfies
xb0
G
G
63. (4) cG b s aG  b .cG 0
 b1iˆ b2 ˆj b3 kˆ . iˆ ˆj kˆ 0
b1 – b2 – b3 = 0
G
and aG.b 3
b
2
(3) Parallel to x-axis 
 x2  y3
8
dy
0 1 3 0
dx
x
Equation of tangent is y – 3 = 0(x–2)  y 3 0
16
4
(cos x sin x ) 4 2 2
cos x
0
Q
sin x
Q
5Q
4
3Q
2
2Q
4
67. (2) The locus of perpendicular tangents is
directrix
i.e., x = –a; x = –1
G
G
GG
68. (4) aG.b 0, b.cG 0,
c .a 0
 2M 4 N 0
– b3 = 3
b1 = b2 + b3 = 3 + 2b3
take b3 = –2
G
b (3 2b3 )iˆ (3 b3 ) ˆj b3 kˆ.
64.
sec x
Solution is t ( I .F ) ° ( I .F )sec xdx
1
sec x tan x c
y
PART-C : MATHEMATICS
61.
1
t
y
¨ 1 ·
mv 0 cos Rt © gt ¸ kˆ 1 mgv0 t 2 cos Rkˆ
2
ª 2 ¹
2
(4) cos x dy = y(sinx – y)dx
M 1 2N 0
solving we get M 3; N 2
69. (2) xRy need not implies yRx
S:
m p
s š qm pn
n q
m m
s
reflexive
n n
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AIEEE-2010
m p
s
n q
CHEMISTRY, PHYSICS & MATHS
 qp s mn
74. (3) S1 symmetric
m p p r
s , s  qm pn, ps rq  ms rn transitive.
n q q s
if x b 1
if x > –1
(3) f(x) = k–2x
= 2x +3
f(x) = k-2x
= 2x+3
k-2x
S2 1
-1
Thi is true
where k = -1
S3 x m1 71. (3) First row with exactly one zero; total number
of cases = 6
First row 2 zeros we get more cases
Total we get more than 7.
Directions : Questions Number 72 to 76 are
Assertion-Reason type questions. Each of these
questions contains two statements.
Statement-1:(Assertion) and Statement-2
:(Reason)
Each of these questions also has four alternative
choices, only one of which is the correct answer.
You have to select the correct choice.
72. (3) N(S) = 20C4
Statement-1:
common difference is 1; total number of cases = 17
common difference is 2; total number of cases = 14
common difference is 3; total number of cases = 11
common difference is 4; total number of cases = 8
common difference is 5; total number of cases = 5
common difference is 6; total number of cases = 2
Prob. =
17 14 11 8 5 2
20
C4
1
.
85
73. (2) A=(3, 1, 6); B = (1, 3, 4)
Mid-point of AB = (2, 2, 5) lies on the plane.
and d.r's of AB = (2, –2, 2)
d.r's of normal to plane = (1, –1, 1)
AB is perpendicular bisector
A is image of B
Statement-2 is correct but it is not correct
explanation.
=
AKASH MULTIMEDIA
j 2
8
10
j 1
10
2x+3
¤ ( j 2)! 88! ( j 2) ! 90.2
10
¤ j j( j 1)! 10!
9 ( j 1) !
if x < -1
if x > -1
Lt f x b 1
j 1
90
S is an equivalence relation.
70.
¤ j( j 1) j( j 1)( j 10!
2)!(10 j )!
10
¤ ( j 1)! 99! ( j 1) ! 10.2 .
10
9
j 1
¤ j( j 1) j
10
<
>
j 1
¤
10
j 1
10!
j !10 j !
j( j 1) 10C j ¤j
10
10
j 1
C j 90.28 10.2 9
= 90.28 + 20.28 = 110.28 = 55.29.
75.
¥ a b´
, abcd x 0
(3) Let A ¦
§ c d µ¶
¥ a b´ ¥ a b´
A2 ¦
.
§ c d µ¶ ¦§ c d µ¶
¥ a 2 bc ab bd ´
 A2 ¦
µ
§ ac cd bc d 2 ¶
 a 2 bc 1, bc d 2 1
ab + bd = ac + cd = 0
c x 0 and b x 0  a d 0
Trace A = a + d = 0
|A| = ad – bc = –a2 – bc = –1.
76. (4) Let f : R
mR
defined by f ( x ) f ’( x ) e
2x
be a continuous function
1
ex
e x 2e x
e2 x 2
2 e x 2e2 x .e x
e
2 x2
2
f'(x) = 0
 e 2 x 2 2e 2 x
e2x = 2
 ex 2
maximum f ( x ) 2
1
4
2 2
17
AIEEE-2010
CHEMISTRY, PHYSICS & MATHS
0 f ( x) b
x  R
1
2 2
1
1
Since 0 3 2 2
f (c ) 77.
Now A 2 m 2 n 2 1
1 1
 cos2 R 1
2 4
1
cos2 R 4
1
Q
 cos R ( R Being acute)  R .
2
3
x y
81. The line L given by 1 passes through
5 b
the point (13, 32). The line K is parallel to L
x y
1. Then the
and has the equation
c 3
distance between L and K is
 for some c  R
1
3
Q
a
cot
2
n
'a' is side of polygon
(2) r R
Q
a
cosec
2
n
cot
Q
n cos Q
Q
n
cosec
n
Q 2
cos for some n  N .
n 3
r
R
78. (2) x2 – x + 1 = 0  x 1
2
B i
3i
,
2
1p 1 4
1 p 3i
; x
2
2
1
2
i 3
2
3
3
¥ Q´
B 2009 C2009 2cos2009 ¦ µ
§ 3¶
2Q ·
2Q ´
¨
¥
2 cos ©668Q Q ¸ 2 cos ¦ Q µ
§
3 ¹
3¶
ª
2Q
¥ 1´
2cos 2 ¦ µ 1
§ 2¶
3
79. (1) Let z = x + iy
Re z = 0
x=0
|z–1| = |z + 1|
x=y
|z – 1| = |z –i|
y = –x
|z + 1| = |z – i|
Only (0,0) will satisfy all conditions.
Number of complex number z = 1





0
80. (2) A cos 45 1
2
1
m cos1200 ; n cos R
2
where R is the angle which line makes with
positive z-axis.
18
17
2)
3)
15
23
4)
17
23
15
b
5
3
c
Line L is parallel to line k.
Slope of line K b 3
 bc 15
5 c
(13, 32) is a point on L.
 Q
Q
C cos i sin
3
17
Sol. (3) Slope of line L C Q
Q
B cos i sin ,
3
1)
13 32
32
8
1  5 b
b
5
3
 b 20  c 4
Equation of K : y – 4x = 3

Distance between L and K 52 32 3
17
23
17
82. A person is to count 4500 currency notes. Let
an denote the number of notes he counts in the
nth minute. If a1 = a2 = .....=a10 = 150 and a10,
a11,....are in A.P. with common difference –2,
then the time taken by him to count all notes is
1) 34 minutes
2) 125 minutes
3) 135 minutes
4) 24 minutes
Sol. (1) Till 10th minute number of
counted notes = 1500
n
3000 [2 s 148 (n 1)(2)] n[148 n 1]
2
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AIEEE-2010
CHEMISTRY, PHYSICS & MATHS
n2 – 149n + 3000 = 0
n = 125, 24
n = 125, 24
n = 125 is not possible.
Total time = 24 + 10 = 34 minutes
83.
88. An urn contains nine balls of which three are
red, four are blue and two are green. Three balls
are drawn at random without replacement from
the urn. The probability that the three balls have
different colour is
2)
1
21
3)
2
23
4)
1
3
Sol. (1) n(S) = 9C3
f (2 x ) f (3 x )
 0 1
f ( x)
f ( x)
n(E) = 3C1 × 4C1 × 2C1
f (2 x )
f (3 x )
b lim
x md
x md f ( x )
x md f ( x )
By sandwich theorem
f (2 x )
 lim
1
x md f ( x )
 lim 1 b lim
84.
2
7
1)
(4) f(x) is a positive increasing function
 0 f ( x ) f (2 x ) f (3 x )
Probability =
3s4s2
9
C3
24 s 3!
24 s 6 2
s 6! .
9!
9s8s 7 7
89. For two data sets, each of size 5, the variances
are given to be 4 and 5 and the corresponding
means are given to be 2 and 4, respectively.
The variance of the combined data set is
(1) p'(x) = p'(1–x)
 p( x ) p(1 x ) c
at x = 0
p(0) = –p(1) + c
 42 c
now p(x) = –p(1–x) + 42
p(x) + p(1–x) = 42

1
¤ xi
1
5
0
0
¥d
´
85. (1) g ’( x) 2( f (2 f ( x) 2)) ¦§ f (2 f ( x ) 2)
µ¶ 4)
5
2
2, ¤ xi 10, ¤ yi 20,
¤ xi2 40; ¤ yi2 105
dx
2f(2f(x)+2)f'(2f(x) +2).(2f'(x))
T 2z
 g ’(0) 2 f (2 f (0) 2). f ’(2 f (0) 2).
2( f ’(0) 4 f (0) f ’(0)
= 4(–1) (1)
= –4
90.
86. (3) Total number of ways = 3C2 × 9C2
9s8
3 s 36 108
2
1 2 1
87. (3) D 2 3 1 0 ; D1 3 5 2
Given system, does not
No solution.
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13
2
1
¥1
´
T 2x ¦ ¤ xi2 µ ( x )2 (¤ yi2 ) 16
§2
¶
5
2 I ° (42)dx  I 21.


3)
x 2; y 4
1
3s
2) 6
Sol. (1) T 2x 4; T 2y 5
I ° p( x )dx ° p(1 x )dx
0
11
2
1)
1
¥ x y´
¤ xi2 ¤ yi2 ¦
§ 2 µ¶
10
2
1
145 90 55 11
(40 105) 9 10
10
10 2
(1) Circle x2 + y2– 4x – 8y – 5 = 0
4 16 5 5
If circle is intersecting line 3x – 4y = m
at two distinct points.
length of perpendicular from centre < radius
Centre = (2, 4), Radius =
3 2 1
3 3 1 x0
P
1 5 2
have any solution.


6 16 m
5  10 m 25
5
 35 m 15.
These Solutions Prepared by Srichaitanya educational Institutions, Vijayawada
19