VITEEE

SOLVED PAPER 2007

PART - I (PHYSICS) 4. Assuming f to be the frequency of first line in

Balmer series, the frequency of the immediate
1. The magnetic moment of the ground state of an next (i.e. second) line is
atom whose open sub shell is half filled with (a) 0.50 f (b) 1.35 f
five electrons is (c) 2.05 f (d) 2.70 f
(a) µB (b) 35 µB 5. The velocity of a particle at which the kinetic
35
energy is equal to its rest energy is
(c) 35 µB (d) µB 35 3c c
(a) (b) 3
2. Indicate which one of the following statements 2 2
is NOT CORRECT ?
1
(a) Intensities of reflections from different 3c 2 c 3
crystallographic planes are equal (c) (d)
2 2
(b) According to Bragg's law higher order of
reflections have high values for a given 6. One electron and one proton is accelerated by
equal potential. Ratio in their deBroglie
wavelength of radiation
wavelength is
(c) For a given wavelength of radiation there
is a smallest distance between the me
crystallographic planes which can be (a) 1 (b) mp
determined
(d) Bragg's law may predict a reflection from a mp me
crystallographic plane to be present but it (c) (d)
me mp
may be absent due to the crystal symmetry
3. Identify the graph which correctly represents 7. Two electrons one moving in opposite direction
the Moseley's law with speeds 0.8c and 0.4c where c is the speed
of light in vacuum. Then the relative speed is
about
f f (a) 0.4c (b) 0.8c
(a) (b) (c) 0.9c (d) 1.2 c
O z O z 8. A photo-sensitive material would emit electrons
if excited by photons beyond a threshold. To
overcome the threshold, one would increase
f f (a) the voltage applied to the light source
(c) (d) (b) the intensity of light
(c) the wavelength of light
O z O z
(d) the frequency of light
9. The radius of nucleus is 15. Identify the logic gate from the following TRUTH
(a) proportional to its mass number table
(b) inversely proportional to its mass number Inputs Output
(c) proportional to the cube root of its mass A B Y
number
0 0 1
(d) not related to its mass number
0 1 0
10. Radio carbon dating is done by estimating in
specimen 1 0 0
(a) the amount of ordinary carbon still present 1 1 0
(b) the amount of radio carbon still present (a) NOR gate (b) NOT gate
(c) the ratio of amount of 14C6 to 12C6 still (c) AND gate (d) NAND gate
present
(d) the ratio of amount of 12C6 to 14C6 still 16. In Boolean algebra, A B is equal to
present (a) A B (b) A B
11. Ionization power and penetration range of
(c) A.B (d) A + B
radioactive radiation increases in the order
(a) and respectively 17. Radar waves are sent towards a moving airplane
(b) and respectively and the reflected waves are received. When the
(c) and respectively airplane is moving towards the radar, the
(d) and respectively wavelength of the wave
12. The half life of a radioactive element is 3.8 days. (a) decrease
The fraction left after 19 days will be (b) increase
(a) 0.124 (b) 0.062 (c) remains the same
(c) 0.093 (d) 0.031 (d) sometimes increase or decrease
13. Two identical P-N junctions are connected in 18. The transmission of high frequencies in a coaxial
series in three different ways as shown below to cable is determined by
a battery. The potential drop across the P-N
junctions are equal in 1
(a) where L and C are inductance and
(LC)1/ 2
P N N P capacitance
(b) (LC )2
(c) the impedance L alone
1 (d) the dielectric and skin effiect
P N P P N P N 19. The output stage of a television transmitter is
most likely to be a
(a) plate-modulated class C amplifier
2 3 (b) grid-modulated class C amplifier
(c) screen-modulated class C amplifier
(a) in circuits 2 and 3
(d) grid-modulated class A amplifier
(b) in circuits 1 and 2
(c) in circuits 1 and 3 20. The antenna current of an AM transmitter is 8A
(d) in none of the circuit when only the carrier is sent, but it increases to
14. The temperature coefficient of a zener 8.93A when the carrier is modulated by a single
mechanism is sine wave. Find the percentage modulation.
(a) negative (b) positive (a) 60.1 % (b) 70.1%
(c) infinity (d) zero (c) 80.1% (d) 50.1%
21. Two point like charges Q1 and Q2 of whose 24. Theelectric field intensity E , due to an electric
strength are equal in absolute value are placed dipole of moment p , at a point on the equatorial
at a certain distance from each other. Assuming line is
the field strength to be posiive in the positive (a) parallel to the axis of the dipole and
direction of x-axis the signs of the charges Q1 opposite to the direction of the dipole
and Q2 for the graphs (field strength versus moment p
distance) shown in Figures 1,2,3 and 4 are (b) perpendicular to the axis of the dipole and
is directed away from it
(c) parallel to the dipole moment
E E
(d) perpendicular to the axis of the dipole and
x Q1 x is directed toward it
Q1
Q2 Q2 25. Twelve wires of each of resistance 6 ohms are
connected to form a cube as shown in the figure.
1 2 The current enters at a corner A and leaves at
the diagonally opposite corner G. The joint
resistance across the corners A and G are
E E
E F
Q1 x Q1 x A B
Q2 Q2
3 4

(a) Q1 positive, Q2 negative; both positive; H G

Q1 negative, Q2 positive; both negative
D C
(b) Q1 negative, Q2 positive;Q1 positive, Q2
(a) 12 ohms (b) 6 ohms
negative; both positive; both negative (c) 3 ohms (d) 5 ohms
(c) Q1 positive, Q2 negative; both negative; 26. A conductor and a semi-conductor are
connected in parallel as shown in the figure. At
Q1 negative, Q2 positive; both positive
a certain voltage both ammeters registers the
(d) Both positive; Q1 positive, Q2 negative; same current. If the voltage of the DC source is
Q1 negative, Q2 positive; both negative increased then
Conductor
22. ABCD is a rectangle. At corners B, C and D of
A1
the rectangle are placed charges + 10 × 10–12C,
–20×10–12C and 10 × 10–12C respectively. Semiconductor
Calculate the potential at the fourth corner. The A2
side AB = 4cm and BC = 3cm
(a) 1.65 V (b) 0.165 V
(a) the ammeter connected to the
(c) 16.5 V (d) 2.65 V semiconductor will register higher current
than the ammeter connected to the
23. A parallel plate capacitor of capacitance 100 pF
conductor
is to be constructed by using paper sheets of (b) the ammeter connected to the conductor
1 mm thickness as dielectric. If the dielectric will register higher current than the ammeter
constant of paper is 4, the number of circular connected to the semiconductor
(c) the ammeters connected to both
metal foils of diameter 2 cm each required for the
semiconductor and conductor will register
purpose is the same current
(a) 40 (b) 20 (d) the ammeter connected to both
semiconductor and conductor will register
(c) 30 (d) 10 no change in the current
27. A uniform copper wire of length 1m and cross- 32. A wire AB, in the shape of two semicircular
sectional area 5×10–7m2 carries a current of 1A . segments of radius R each and carrying a current
Assuming that there are 8×1028 free electrons/ I, is placed in a uniform magnetic field B directed
m3 in copper, how long will an electron take to into the page (see figure). The magnitude of the
drift from one end of the wire to the other force due to the field B on the wire AB is
(a) 0.8×103s (b) 1.6×103s
(c) 3.2×103s (d) 6.4×103s
28. The temperature coefficient of resistance of a
wire is 0.00125/K. At 300K its resistance is
I R R
1 .The resistance of the wire will be 2 at
A B
(a) 1154 K (b) 1100 K
(c) 1400 K (d) 1127 K (a) zero (b) 4 RIB
29. A rectangular coil ABCD which is rotated at a
(c) R2 I B (d) 2 RIB
constant angular velocity about an horizontal
as shown in the figure. The axis of rotation of 33. There are two solenoids of same length and
the coil as well as the magnetic field B are inductance L but their diameters differ to the
horizontal. Maximum current will flow in the extent that one can just fit into the other. They
circuit when the plane of the coil is are connected in three different ways in series.
1) They are connected in series but separated
B
by large distance 2) they connected in series
A with one inside the other and senses of the
turns coinciding 3) both are connected in series
C
with one inside the other with senses of the turns
D opposite as depicted in figures 1,2 and 3
respectively. The total inductance of the
(a) inclined at 30 degrees to the magnetic field
solenoids in each of the case 1, 2 and 3 are
(b) perpendicular to the magnetic field respectively
(c) inclined at 45 degrees to the magnetic field
(d) parallel to the magnetic field
30. If the total emf in a thermocouple is a parabolic
1 2 1
function expressed as E = at + bt , which of
2
the following relations does not hold good
(a) neutral temperature t n = –a/b
(b) temperature of inversion ti = –2a/b
2
(c) thermo-electric power p = a + bt
(d) tn = a/b
31. The proton of energy 1 MeV describes a circular
path in plane at right angles to a uniform
magnetic field of 6.28 ×10–4T. The mass of the 3
proton is 1.7 × 10–27 Kg.The cyclotron frequency
of the proton is very nearly equal to (a) 0, 4L0, 2L0 (b) 4L0, 2L0, 0
(a) 107 Hz (b) 105 Hz (c) 2L0, 0, 4L0 (d) 2 L0, 4L0, 0
6
(c) 10 Hz (d) 104 Hz
34. From figure shown below a series LCR circuit 38. Following diffraction pattern was obtained using
connected to a variable frequency 200V source. a diffraction grating using two different
L = 5H, C = 80 F and R = 40 . Then the source wavelengths and . With the help of the
figure identify which is the longer wavelength
frequency which drive the circuit at resonance
and their ratios.
is I

1 1
2 2
1 1
C = 80 F L = 5H 1 2 2 1

R = 40
(a) is longer than and the ratio of the
longer to the shorter wavelength is 1.5
(b) is longer than and the ratio of the
V = 200 V longer to the shorter wavelength is 1.5
(c) and are equal and their ratio is 1.0
(d) is longer than and the ratio of the
25 longer to the shorter wavelength is 2.5
(a) 25 Hz (b) Hz
39. In Young's double slit experiment, the interference
pattern is found to have an intensity ratio
50 between bright and dark fringes is 9. This implies
(c) 50 Hz (d) Hz
the
(a) the intensities at the screen due to two slits
35. If the coefficient of mutual induction of the
are 5 units and 4 units respectively
primary and secondary coils of an induction coil (b) the intensities at the screen due to the two
is 5H and a current of 10A is cut off in 5×10–4 slits are 4 units and 1 units respectively
second, the emf induced(in volt) in the secondary (c) the amplitude ratio is 7
coil is (d) the amplitude ratio is 6
(a) 5×104 (b) 1×105 40. Rising and setting sun appears to be reddish
(c) 25×10 5 (d) 5×106 because
(a) Diffraction sends red rays to earth at these
36. A voltage of peak value 283 V and varying
times
frequency is applied to a series L, C, R
(b) Scattering due to dust particles and air
combination in which R = 3 ohm, L=25 mH and molecules are responsible
C= 400 F. The frequency (in Hz) of the source at (c) Refraction is responsible
which maximum power is dissipated in the above (d) Polarization is responsible
is
(a) 51.5 (b) 50.7 PART - II (CHEMISTRY)
(c) 51.1 (d) 50.3
37. Four independent waves are represented by 41. The catalyst used in Rosenmund reaction is
equations (a) Zn / Hg (b) Pd / BaSO4
(1) X1 = a1sin t (3) X2 = a2sin2 t (c) Raney Ni (d) Na in Ethanol
(2) X3 = a1sin t (4) X4 = a1sin( t+ ) H 2O
42. CH 2 CO 2
O RMgX ?
Interferen ce is possible between waves
represented by equations (a) ROOC(CH2)COOR
(b) RCOCH2CH2COOH
(a) 3 and 4 (b) 1 and 2 (c) RCOOR
(c) 2 and 3 (d) 1 and 4 (d) RCOOH
43. Identify, which of the below does not possess 51. In TeCl4 the central atom tellurium involves
any element of symmetry? (a) sp3 hybridization
(a) (+)– Tartaric acid (b) sp3d hybridization
(b) Carbon tetrachloride (c) sp3d2 hybridization
(c) Methane (d) dsp2 hybridization
(d) Mesotartatric acid 52. The purple colour of KMnO4 is due to the
transition
44. The weakest acid amongst the following is
(a) C.T. (L M) (b) C.T. (M L)
(a) ClCH2COOH (b) HCOOH
(c) d – d (d) p –d
(c) FCH2CH2COOH (d) CH2(I)COOH
53. A nuclear reaction of 235 with a neutron
92 U
45. HOOC CH 2 4 COOH 2C 2 H5OH
H2SO 4 produces 90
36 Kr and two neutrons. Other element
C2 H5OOC CH 2 COOC2 H 5
Toluene produced in this reaction is
The purpose of using toluene here is (a) 137 (b) 144
52Te 55 Cs
(a) to make both substances (acid & alcohol)
miscible (c) 137 (d) 144
56 Ba 56 Ba
(b) that the product is insoluble in toluene 54. AgCl dissolves in a solution of NH3 but not in
(c) that the reactants are insoluble in water water because
(d) because of the formation of low boiling (a) NH3 is a better solvent than H2O
azotrope (b) Ag+ forms a complex ion with NH3
46. Trans esterification is the process of (c) NH3 is a stronger base than H2O
(a) conversion of an aliphatic acid to ester (d) the dipole moment of water is higher than
(b) conversion of an aromatic acid to ester NH3
(c) conversion of one ester to another ester 55. Which of the following is hexadenate ligand?
(d) conversion of an ester into its components (a) Ethylene diamine
namely acid and alcohol (b) Ethylene diamine tetra acetic acid
(c) 1, 10- phenanthroline
47. The correct sequence of base strengths in
(d) Acetyl acetonato
aqueous solution is
56. A coordinate bond is a dative covalent bond.
(a) (CH3)2NH > CH3NH2>(CH3) 3N
Which of the below is true?
(b) (CH3)3N > (CH3)2 NH>CH3NH2 (a) Three atoms form bond by sharing their
(c) (CH3)3N > CH3 NH2= (CH3 )2 NH electrons
(d) (CH3)2N H > (CH3 )3 N> CH3 NH2 (b) Two atoms form bond by sharing their
48. When aqueous solution of benzene electrons
diazoniumchloride is boiled, the product formed (c) Two atoms form bond and one of them
is provides both electrons
(a) C6H5CH2OH (b) C6H6+N2 (d) Two atoms form bond by sharing electrons
(c) C6H5COOH (d) C6H5OH obtained from third atom
49. Carbylamine reaction is given by aliphatic 57. Which of the following complex has zero
(a) primary amine magnetic moment (spin only)?
(b) secondary amine (a) [Ni(NH3 )6 ]Cl2 (b) Na 3[FeF6 ]
(c) tertiary amine (c) [Cr(H 2O)6 ]SO4 (d) K 4 [Fe(CN)6 ]
(d) quaternary ammonium salt
58. The IUPAC name of [Ni(PPh3)2Cl2]2+ is
NH3 (a) bis dichloro (triphenylphosphine) nickel (II)
50. C6 H5 CHO ?
H 2 ,Ni (b) dichloro bis (triphenylphosphine) nickel (II)
(a) C6H5NH2 (b) C6H5NHCH3 (c) dichloro triphenylphosphine nickel (II)
(c) C6H5 CH2NH2 (d) C6H5NHC6H5 (d) triphenyl phosphine nickel (II) dichloride
59. Among the following the compound that is both 67. Given the equilibrium system:
paramagnetic and coloured is NH4Cl (s) NH4+(aq) + Cl–(aq)
(a) K2Cr2O7 (b) (NH4)2 [TiCl6] ( H = +3.5kcal/mol).
(c) VOSO4 (d) K3Cu (CN)4 What change will shift the equilibrium to the
60. On an X-ray diffraction photograph the intensity right?
of the spots depends on (a) Decreasing the temperature
(a) neutron density of the atoms/ions (b) Increasing the temperature
(b) electron density of the atoms/ions (c) Dissolving NaCl crystals in the equilibrium
(c) proton density of the atoms/ions mixture
(d) photon density of the atoms/ions (d) Dissolving NH 4NO 3 crystals in the
equilibrium mixture
61. An ion leaves its regular site occupy a position
68. According to Arrhenius equation, the rate
in the space between the lattice sites is called
constant (k) is related to temperature (T) as
(a) Frenkel defect (b) Schottky defect
(c) Impurity defect (d) Vacancy defect k2 Ea 1 1
(a) In –
62. The 8:8 type of packing is present in k1 R T1 T2
(a) MgF2 (b) CsCl
(c) KCl (d) NaCl k2 Ea 1 1
(b) In – –
63. When a solid melts reversibly k1 R T1 T2
(a) H decreases (b) G increases
(c) E decreases (d) S increases k2 Ea 1 1
(c) In
64. Enthalpy is equal to k1 R T1 T2

G G/T
(a) –T 2 (b) –T 2 In
k2
–
Ea 1 1
T T (d) k1 R T1 T2
V P

G/T
69. Equivalent amounts of H2 and I2 are heated in a
G
(c) T2 (d) T 2 closed vessel till equilibrium is obtained. If 80%
T V T of the hydrogen can be converted to HI, the Kc
P
65. Condition for spontaneity in an isothermal at this temperature is
process is (a) 64 (b) 16
(c) 0.25 (d) 4
(a) A W 0 (b) G U 0
70. For the reaction H2(g)+I2(g) 2HI(g), the
(c) A U 0 (d) G–U 0 equilibrium constant Kp changes with
66. Given: 2C s 2O 2 g 2CO2 g ; (a) total pressure
(b) catalyst
H –787kJ (c) the amount H2 and I2
(d) temperature
1 71. How long (in hours) must a current of 5.0 amperes
H2 g O2 g H 2O( ); H –286kJ
2 be maintained to electroplate 60g of calcium from
molten CaCl2?
1
C2 H 2 g 2 O2 g 2CO 2 g H 2O( ) (a) 27 hours (b) 8.3 hours
2 (c) 11 hours (d) 16 hours
; H –1310kJ 72. For strong electrolytes the plot of molar
The heat of formation of acetylene is conductance vs C is
(a) –1802 kJ (b) +1802 kJ (a) parabolic (b) linear
(c) +237 kJ (d) –800 kJ (c) sinusoidal (d) circular
73. If the molar conductance values of Ca2+ and Cl– 80. In which of the below reaction do we find , -
at infinite dilution are respectively 118.88×10–4 unsaturated carbonyl compounds undergoing
m2 mho mol–1 and 77.33×10–4 m2 mho mol–1 then a ring closure reaction with conjugated dienes?
that of CaCl2 is (in m2 mho mol–1) (a) Perkin reaction
(a) 118.88 × 10–4 (b) 154.66 × 10–4 (b) Diels-Alder reaction
(c) 273.54 × 10–4 (d) 196.21× 10–4
(c) Claisen rearrangement
74. The standard reduction potentials at 298K for
(d) Hoffman reaction
the following half reactions are given against
each
Zn2+ (aq) + 2e– Zn(s) E0 = – 0.762 V PART - III (MATHEMATICS)
Cr3+ (aq) + 3e– Cr(s) E0 = – 0.740 V
2H+ (aq) + 2e– H2(g) E0 = 0.00 V
Fe3+ (aq) + 3e– Fe2+(aq) E0 = + 0.762 V 81. Let the pairs a , b and c , d each determine a
The strongest reducing agent is plane. Then the planes are parallel if
(a) Zn(s) (b) Cr(s)
(c) H2(g) (d) Fe2+(aq) (a) a c b d 0
75. The epoxide ring consists of which of the
following ? (b) a c b d 0
(a) Three membered ring with two carbon and
one oxygen (c) a b c d 0
(b) Four membered ring with three carbon and
one oxygen a b c d 0
(d)
(c) Five membered ring with four carbon and
one oxygen
82. The area of a parallelogram with 3iˆ ˆj – 2kˆ and
(d) Six membered ring with five carbon and
one oxygen ˆi – 3jˆ 4kˆ as diagonals is
76. In the Grignard reaction, which metal forms an
organometallic bond? (a) 72 (b) 73
(a) Sodium (b) Titanium
(c) Magnesium (d) Palladium (c) 74 (d) 75
77. Phenol is less acidic than 83. If cosx + cos2 x=1then the value of
(a) p-chlorophenol sin12 x+ 3sin10 x + 3sin8x + sin 6 x–1 is equal to
(b) p-nitrophenol (a) 2 (b) 1
(c) p-methoxyphenol (c) –1 (d) 0
(d) ethanol
84. The product of all values of (cos +i sin )3/5 is
78. Aldol condensation is given by
equal to
(a) trimethylacetaldehyde
(b) acetaldehyde (a) 1 (b) cos +i sin
(c) benzaldehyde (c) cos3 +i sin3 (d) cos5 +i sin5
(d) formaldehyde 2
79. Give the IUPAC name for 1 i
85. The imaginary part of is
O O i 2i –1
|| ||
H3C CH 2 C H 2 C CH 2 C OCH3 4
(a) (b) 0
(a) Ethyl-4-oxoheptonate 5
(b) Methyl-4-oxoheptonate
(c) Ethyl-4-oxohexonate 2 4
(c) (d) –
(d) Methyl-4-oxohexonate 5 5
 u u
86. If sin –1 x sin –1 y , then cos–1x + cos–1y is 93. If (x +y )sin u = x2y2, then x y
2 x y
equal to
(a) sin u (b) cosec u
(a) (b) (c) 2 tan u (d) tan u
2 4
94. The angle between the tangents at those points
3 on the curve x = t2 + 1 and y = t2 – t – 6 where it
(c) (d)
4 meets x-axis is
87. The equation of a directrix of the
4 5
x 2 y2 (a) tan –1 (b) tan –1
ellipse 1 is 29 49
16 25
(a) 3y = 5 (b) y = 5 10 8
(c) 3y = 25 (d) y = 3 (c) tan –1 (d) tan –1
49 29
88. If the normal at (ap2, 2ap) on the parabola
y2 = 4ax, meets the parabola again at (aq2, 2aq),
then 4
(a) p2 + pq + 2 = 0 (b) p2 – pq + 2 = 0 95. The value of x – 3 dx is equal to
(c) q2 + pq + 2 = 0 (d) p2 + pq + 1 = 0 1
89. The length of the straight line x – 3y = 1
intercepted by the hyperbola x2 – 4y2 = 1is 5
(a) 2 (b)
6 2
(a) 10 (b)
5
1 3
(c) (d)
1 6 2 2
(c) (d) 10
10 5 96. The area of the region bounded by the straight
90. The curve described parametrically by lines x = 0 and x = 2 and the curves y = 2x and
x = t2 + 2t–1, y = 3t + 5 represents y = 2x –x2 is equal to
(a) an ellipse (b) a hyperbola
(c) a parabola (d) a circle 2 4 3 4
91. If the normal to the curve y = f(x) at (3,4) makes (a) – (b) –
log 2 3 log 2 3
3
an angle with the positive x-axis, then f '(3)
4 1 4 4 3
(c) – (d) –
is equal to log 2 3 log 2 2
3
(a) – 1 (b)
4 dx
97. The value of 7
is equal to
3 0 a2 x2
(c) 1 (d) –
4
92. The function f(x) = x2 e–2x, x > 0. Then the
maximum value of f(x) is 231 1 235 1
(a) (b)
1 1 2047 a13 2048 a13
(a) (b)
e 2e
232 1 231 1
1 4 (c) (d)
(c) (d) 2047 a13 2048 a13
e2 e4
 101. The general solution of the differential equation
2
x 1– x d2 y dy
98. The value of the integral e 2
dx is 2 y 2e3x is given by
1 x dx 2 dx
e3x
ex
1– x
C (a) y c1 c 2 x e x
(a) 8
1 x2
e –3x
(b) y c1 c 2 x e – x
1 x 8
(b) ex C
1 x2 e3x
(c) y c1 c2 x e – x
8
ex e –3x
(c) C y c1 c2 x e x
1 x 2 (d)
8
(d) ex(1–x) + C 102. The solution of the differential equation
ydx+(x–y3)dy = 0 is
y y 1 3
99. If x sin dy y sin – x dx (a) xy y C (b) xy = y4 + C
x x 3
(c) y4 = 4xy + C (d) 4y = y3 + C
103. The number of positive integral solutions of the
y equation x1 x2x3x4x5=1050 is
and y 1 , then the value of cos is
2 x (a) 1870 (b) 1875
(c) 1865 (d) 1880
equal to 104. Let A= {1,2,3,....., n}and B ={a,b,c}, then the
number of functions from A to B that are onto is
1 (a) 3n – 2n (b) 3n – 2n –1
(a) x (b) n
x (c) 3 (2 – 1) (d) 3n – 3(2n – 1)
105. Everybody in a room shakes hands with
(c) log x (d) ex everybody else. The total number of hand
100. The differential equation of the system of all shakes is 66. The total number of persons in the
room is
circles of radius r in the XY plane is (a) 9 (b) 12
(c) 10 (d) 14
2 2
dy
3
d2 y 106. If(G,*) is a group and the order of an element
(a) 1 r2 a G is 10, then the order of the inverse of a* a
dx dx 2 is
1
(a) 10 (b)
3 2 3 10
dy 2 d2 y
(b) 1 r 1
dx dx 2 (c) 5 (d)
5
107. A box contains 9 tickets numbered 1 to 9
inclusive. If 3 tickets are drawn from the box one
2 3 2
dy 2 d2 y at a time, the probability that they are
(c) 1 r alternatively either {odd, even, odd} or {even,
dx dx 2 odd, even} is
5 4
(a) (b)
2 3 3 17 17
dy d2 y
(d) 1 r2 5 5
dx dx 2 (c) (d)
16 18
 1 5 1 –1 2 5
108. If P(A) = , P(B) = and P(B/A)= then
12 12 15 114. If the rank of the matrix 2 –4 a – 4 is 1,
p A B is equal to 1 –2 a 1

89 90 then the value of a is

(a) (b) (a) –1 (b) 2
180 180
(c) –6 (d) 4
91 92 115. If b2 4ac for the equation ax4 + bx2 + c = 0, then
(c) (d) all the roots of the equation will be real if
180 180
109. If the probability density function of a random (a) b>0, a<0, c>0 (b) b<0, a>0, c>0

x (c) b>0, a>0, c>0 (d) b>0, a>0, c<0

variable X is f(x) = in 0 x 2, then
2
116. If x > 0 and log3x + log3( x ) + log3 ( 4 x ) +
P(X > 1.5 X > 1) is equal to
8 16
7 3 log3 x + log3 x +....= 4, then x equals
(a) (b)
16 4 (a) 9 (b) 81
7 21 (c) 1 (d) 27
(c) (d) 117. The number of real roots of the equation
12 64
3
110. If X is a poisson variate such that 1 1
2P(X = 0)+ P(X = 2 ) = 2P(X = 1) then E(X) is x x 0 is
x x
equal to
(a) 0 (b) 2
(a) 1 (b) 2
(c) 4 (d) 6
(c) 1.5 (d) 1.75
1 tan 118. If H is the harmonic mean between P and Q, then
111. If A and AB = I, then
– tan 1 H H
the value of is
(cos2 )B is equal to P Q

(a) A( ) (b) A PQ
2 (a) 2 (b)
P Q
– 1 P Q
(c) A(– ) (d) A (c) (d)
2 2 PQ

2x 1 4 8 119. If a and b are two unit vectors, then the vector

112. If x = –5 is a root of 2 2x 2 = 0, then ( a + b ) × ( a × b ) is parallel to the vector
7 6 2x
(a) a +b (b) 2 a + b
the other roots are
(c) a – b (d) 2 a – b
(a) 3, 3.5 (b) 1, 3.5
120. If is the angle between the lines AB and AC
(c) 1, 7 (d) 2, 7
where A, B and C are the three points with
113. The simultaneous equations Kx + 2y–z =1,
coordinates (1,2,–1), (2,0,3), (3,–1,2) respectively,
(K –1)y–2z = 2 and (K + 2)z = 3 have only one
solution when then 462 cos is equal to
(a) K = –2 (b) K = –1 (a) 20 (b) 10
(c) K = 0 (d) K = 1 (c) 30 (d) 40
 2007 SOLUTIONS

PART - I (PHYSICS) 3 3 36
2 Rc f 1.35f .
16 16 5
1. (d) The magnetic moment of the ground state 5. (d) Here, Kinetic energy = Rest energy
of an atom whose open sub-shell is half we know that
filled with n electrons is given by Kinetic energy (Relativistic)
= (m – m0)c2,
n(n 2). B and Rest energy = m0c2,
where B is the gyromagnetic moment of where m0 = rest mass, c = velocity of light
the atom. Also, mass (m) of a particle moving with
Here, n = 5. velocity v is given by
5(5 2). B 35 B m0
m .
2. (a) Bragg's law gives 2dsin = n , n= order of v2
reflection, d= distance between planes. For 1
c2
same and d, n for a given , smallest
d for least n, can be found. If crystal is (m – m0)c2 = m0c2 provides
symmetric reflections from different planes
may cancel out.
3. (b) According to Moseley's law, square root
of frequency of X-ray is plotted against m0
m0 c 2 m 0c2
atomic number it gives straight line, the v 2
relation is 1
c2
f c Z – 1 where c = constant
1
(for f 0, Z c , for Z = 0, f –c ) or, 1 1
Option (b) is correct. as Z can not be v2
1
negative. c2
4. (b) Balmer series is given for n 1 = 2 and
n2 = 3, 4, ... v2 1
1 1
or, 2 or, 2
1 1 c 4
Rc – v2
1
22 n 22 c2
For Ist line in spectrum n2 = 3
v2 1 3
or, 2
1
1 1 1 1 c 4 4
1 Rc – Rc – f
2 2 4 9
2 n2 3
v c.
5 36 2
f Rc Rc f
36 5 v2 1
1– 2
=
For second line n 2 = 4 c 4
1 1 1 1 v2 1 3 3 2
Rc – Rc –
2
2 2
4 2 4 16 2 =1 = v2 = c
c 4 4 4
4 –1 3
Rc v= c
16 2
6. (d) de-Broglie wavelength ( ) of a particle of 10. (c) Radio carbon dating is done by measuring
mass m and moving with a velocity v is ratio of 14C present in the sample, since
proportion of 14C and 12 C in a body is same;
h
given by, , where h is Planck's but after death 14C decays. Hence knowing
mv the present ratio of 14C to 12 C, sample can
constant. be dated.
When a particle having charge q is 11. (b) particles are positively charged He
accelerated through a potential V then nucleus, it can accept 2e– , rays are
1 negatively charged which are similar to e–,
qV mv 2 can donate 1e–, are radiations. Hence
2
ionisation power of is maximum. are
m 2 v2 most energetic and is least energetic
or, qV mv 2mqV Penetration power of is maximum
2m
12. (d) Given half life T= 3.8 days; t = 19 days
h N t 19
2mqV ? now 5
N0 T 3.8
Hence, de-Broglie wavelength of electron,
t
5
h N 1 T 1 1
e
2me eV N0 2 2 32

and de-Broglie wavelength of proton, N0

N 0.03 N 0
h 32
p
, where e represents the
2m p eV 13. (a) If the p-n junctions are identical, then their
resistances would be same. In circuit 1, the
charge of electron (or proton) first p-n function is forward biased and
second is reverse biased. Hence the
mp
e voltages across them would be different.
p me In circuit 2 both are reverse biased. So
potential drop would be same. In circuit 3
7. (c) Relative speed is given by
both are forward biased; again voltages
u' v would be same.
u 14. (a) Zener diode is a semiconductor device and
u 'v
1 for semiconductors, temperature coefficient
c2
is negative
Here, u' = 0.8 c and v = 0.4 c 15. (a) The truth table corresponds to the logic
0.8c 0.4c 1.2 c A B hence it is NOR gate.
u 0.9c.
(0.8c)(0.4c) 1.32
1 16. (d) A.B in Boolean algebra, can be written as
c2
8. (d) By Einstein's equation, photoemission A.B A B A B
occurs when h > 0 or h > h 0 i.e., 17. (a) By Dopplers effect, if source moves
frequency of incident photon is greater than towards the observer, frequency received
threshold frequency. will increase.

9. (c) Radius of nucleus is given by R A1/ 3 1

As
R R 0 A1/ 3 , where A = mass number
wavelength will decrease.
18. (d) A coaxial cable consists of a conducting
wire surrounded by a dielectric space, over AC 42 32 25 5cm.
which there is a sleeve of coppermesh Potential at A = VB + VC + VD.
covered with a shield of PVC insulation.
The power transmission is regulated by 1 q1 q2 q3
VA
dielectric. At high frequencies energy loss 4 0 AB AC AD
due to mesh is significant (called skin
effect). 10 10 –12 –20 10 –12 10 10 –12
VA 9 109
–2 –2
19. (b) Due to their inherent distortion, plate- 4 10 5 10 3 10 –2
modulated class C amplifiers are not used
as audio amplifiers. Also, class A amplifiers 10 –20 10
are not used owing to low efficiency. A grid- 9 109 10 –10
4 5 3
modulated amplifier has very high
frequency. Therefore, in output stage of a 150 – 240 200
TV transmitter, grid-modulated class C 0.9
60
amplifier is used.
20. (b) The rms value of carrier current is Ic = 8A. 0.9 110
The rms value of modulated current 1.65V
60
= Ic + Im = 8.93 A = It
Percentage modulation = ma × 100 23. (d) The capacitance (C) of a parallel plate
The current relation in AM wave is capacitor with dielectric is given by
A
It 2 ma 2 It2 C 0
1 ma –1 2 1
Ic2 2 Ic2 d – t 1–
K

8.93
2 C = 100 pF = 100 × 10–12 F, = 8.85 × 10–12,
Modulation index, m a –1 2 A = r2 = 3.14×(10–2)2 (r = lcm)
82
t = 10–3m, d = t, K = 4
79.7
–1 2 1.24 –1 2
64 8.85 10 –12 3.14 10 –4
C1
= 0.701 1
10 –3 –10 –3 (1 – )
Percentage modulation = ma × 100 =70.1% 4
1
21. (d) For a point charge, E . For positive 27.79 10 –16
x2
10 –3 – 0.75 10 –3
charges, electric field will decrease in
positive direction as distance increases,
27.79 10 –13
(case 1), for negative charge, as distance
increases field will increase (case 4). As we 0.25
move from positive charge to negative
charge, field will keep on decreasing (case 27.79 10 –13
111.16 10 –13 F
2), As we move from negative to positive 0.25
charge, field will keep on increasing (case 3) (for 1 set)
A 4 cm B(q 1) Required C = 100 10 –12
F
22. (a) 3 cm 3 cm
C 100 10 –12
n 10
D(q 3 ) 4 cm C(q 2) C1 111.16 10 –13
24. (a) The resultant intensity at a point on 26. (c) Since conductor and semiconductor are
connected in parallel hence voltage across
equatorial line is E . E is parallel and them is same. If ammeters show same
opposite to direction of p . V
reading hence their resistances R are
I
E2 same. If voltage is increased by small value
E 3 cm
then following the same relation V I for
E1 constant R, both conductor and
semiconductor show same current.
27. (d) Time taken by free electrons to cross the
–q +p
O conductor
I
25. (d) Let a total current of 6I enter at A. It divides t where drift velocity v d
vd neA
into three equal parts, each of 2I, along AE,
AB and AD. At E, B and D each the current
1
2I divides into two equal parts, each of I, vd 28
along EF, EH, BF, BC, DH and DC. 8 10 1.6 10 –19 5 10 –7
At F, H and C, the two currents each of I, 10 –2
combine together to give a current of 2I at m/s .
64
each corner. Thus, at G we get the same
1 64
current 6I as shown in the figure. t 64 102 s 6.4 103 sec
vd 10–2
E I F
2I
R 2 – R1
I I 28. (b) Temperature coefficient,
A B 2I R1 t 2 – t1
2I I 2I G 6I
2I H
R 2 – R1
6I
I 2I t 2 – t1
D I C R1

2 –1
E t 2 – t1
1 0.00125
Let r be the value of resistance of each arm 1
of the cube and R be the joint resistance t2 t1 800 300 1100K
0.00125
across the conners A and G.
29. (d) Torque on the coil is = nIB A cos . If coil
Applying Kirchhoff's law along the loop is set with its plane parallel to direction of
AEFGA, we get magnetic field B, then =0°, cos = 1
2Ir + Ir + 2Ir = E = nIBA.1 = nIBA = maximum.
or, 5Ir = E ........... (1) Hence, I = maximum (as n,B, A are constant)
Also, by Ohm's law, 1 2
6I × R = E 30. (d) E at
bt is the parabolic equation for
2
6IR = 5Ir, using (1) thermo emf. The thermoelectric power is
5 dE
or, R r. S
6 dt
Here, r = 6 dE
S = a + bt. The graph between S
5 dt
R 6 5
6 and straight line.
 When t = 0, S = a (intercept).
1 1
dE –6
At neutral temperature 0 and 2 5 80 10 2 400 10 –6
dt
1
–a
t = tn 0 = a + btn tn and at cold 2 20 10 –3
b

–2a 103 1000 25

junction t i 2t n
b 40 40
31. (d) Energy of proton = K.E. = 1MeV = 106 eV. 35. (b) M = 5 H, I = 10 A, t = 5×10–4 s.
Now, emf induced in secondary is
Bq
Frequency
2 m dI
e –M , (–ve sign shows direction)
dt
6.28 10 –4 1.6 10 –19
2 3.14 1.7 10 –27 10
e 5 –4
1 10 5 V
5 10
0.9 10 4 Hz 10 4 Hz
36. (d) Average power of an LCR circuit is
32. (d) Force on a conductor of length due to a P = EvIvcos . Maximum power is dissipated
magnetic field of strength B is given by at resonance when XL =XC. The resonance
frequency is given by
F I( B)
1
F I B sin
2 LC
Here, = 90°
F I B I R B RIB 1
Therefore, a force of magnitude ( RIB) will 2 3.14 25 10 –3 400 10 –6
act on each wire. The direction of the forces
on each wire will be same. Thus, 1
total force on the wire AB = RIB + RIB 2 3.14 5 20 10 –5 10
= 2 RIB
33. (d) In first case, the two inductances are in 105
series hence total inductance = L0 + L0 3.14 200 10
= 2L0
In second case, the current is same, but no. 103 159.2
50.31 Hz
of turns has doubled since the sense of 6.28 10 3.16
turning is same.
37. (d) Coherent sources sh ould have same
2 2 frequency and wavelength and constant
Thus L n L 2n
or zero phase difference. Frequency is same
hence inductance L = 4L0
only for wave X1 and X4 and
In third case, the sense of turns is opposite.
phase difference =
So net inductance cancel each other
38. (c) Since position of central maximum is same,
L= 0
hence asin = is same for both
34. (b) At resonance, impedance of circuit is
wavelengths. i.e, a = constant, = same,
minimum.When impedance of capacitor and
is same for both waves
inductor is same, XL = XC Resonance
1 1
frequency and 1
2 LC 2
39. (b) For bright fringes, I max = (a + b)2 43. (a) (+) – Tartaric acid does not have element of
For dark fringes, Imin = (a – b)2 symmetry.

COOH COOH COOH

2 | | |
Im ax a b a b
Now 9 3 H C OH HO C H H C OH
Imin 2 a–b
a–b | | |
HO C H H C OH H C OH
| | |
a + b = 3(a – b) a + b = 3a–3b 2a = 4b COOH COOH COOH

a a2 (+) – Tartaric acid (A) Meso tartaric acid (D)

2 2
4.
b b (Plane of symmetry)

Ratio of intensities of the two slits, Cl H

| |
I1 a2 4 Cl C Cl H C H
2 | |
I2 b 1 Cl H
Carbon tetracholride (C)
40. (b) Rising and setting sun appears red because (B) Methane
light from the sun travels slightly more
distance from the horizon, than when it is 44. (b) Proton doner s are acids. Electr on
overhead. Hence blue light is scattered by withdrawing groups (like halogens)
dust in the atmosphere because scattering increase the acidity of carboxylic acids.
Therefore, HCOOH is weakest acid among
1 the given choices.
and b < r. Red colour is less
4 45. (a) Toluene is used in the reaction to make both
substances (acid and alcohol) miscible.
scattered and reaches us. 46. (c) Trans esterification is the process of
conversion of one ester to another ester.
PART - II (CHEMISTRY) RCOOR' + R''OH RCOOR'' + R'OH
47. (a) It is expected that the basic nature of
41. (b) Roseumund's reaction – amines should be in order tertiary >
secondary > primary but the observed
Pd / BaSO4
RCOCl H 2 RCHO HCl order in the case of lower members is found
to be as secondary > Primary > tetriary. This
CH2CO anomalous behaviour of tetriary amines is
42. (b) O + RMg X
CH2CO due to steric factors i.e crowding of alkyl
groups cover nitrogen atom from all sides
thus makes the approach and bonding by a
O Mg X R proton relatively difficult which results the
maximum steric strain in tetiary amines. The
CH2 C H2O electrons are there but the path is blocked,
O
CH2 C – Mg(OH) X resulting the reduction in basicity.
Thus the correct order is
O R2 NH > R NH2 > R3N.
Boiling
R C CH2 CH2 COOH 48. (d) C6H5N2+Cl– + H2O
C6H5OH + N2 + HCl
O
Phenol
49. (a) Carbylamine test is given by aliphatic 235 144 92 1
1
primary amine. 53. (d) 92U n 56 BaKr 2 0n
0 36
Aliphatic (or aromatic) 1° amine + CHCl3 + Sum of atomic number of reactants = sum
KOH Bad smelling isocyanide. of atomic masses of products.
54. (b) AgCl dissolves in a solution of NH3 but
50. (c) Benzaldehyde reacts with ammonia to form
not in water because Ag+ forms soluble
hydrobenzamide.
complex ion with NH3.
C6H5 CH = O H2 NH 55. (b) EDTA is hexadentate ligand. 4 oxygen and
+ + O = CH C6H5 2 nitrogen atoms act as donars.
C6H5 CH = O H2 NH – OOCH C
2
CH2COO–
C 6H 5 CH = N
CH C 6H5 – OOCH C CH2COO–
2
C 6H 5 CH = N
56. (c) A co-ordinate bond is a dative covalent
Hydrobenzamide.
bond in which two atoms form a bond and
Under one of them provides both electrons.
H2, Ni
Pressure 57. (d) [Ni(NH3)6]Cl2 sp3d2 hybridisation
2 C6H5 CH2 NH2 2 unpaired electrons
Na3[FeF6] sp3d2 hybridisation
1 3 unpaired electrons
51. (b) Hybridisation [Number of valence
2 [Cr(H2O)6]SO4 d2sp3 hybridisation
electrons of central atom + no. of 3 unpaired electrons
monovalent atoms attached to it + negative K4[Fe(CN)6] d2sp3 hybridisation
charge if any – positive charge if any] No unpaired electrons
Zero magnetic momoment means all the
1
[6 + 4 + 0 – 0] = 5 = sp3 d electrons paired.
2
58. (c) The IUPAC name is dichloro triphenyl
phosphine nickel (II).
Te 52 Kr 4d10 5s 2 5p4
59. (c) (d) In K4[Fe(CN)6] Cu is in + 1 oxidation
52. (a) The permanganate ion has an intense state hence has no unpaired electron
purple colour. Mn (+ VII) has a d 0 hence colourless and diamagnetic.
configuration. So the colour arises from (b) In (NH4)2 [TiCl6] Ti is in + 4 oxidation
charge transfer and not from d—d spectra. state. hence has no unpaired electron
In MnO4 – an electron is momentarily hence colourless and diamagnetic.
changing O– – to O– and reducing the (c) In VOSO4, V is in +4 oxidation state
oxidation state of the metal from Mn(VII) to hence has one unpaired electron, thus
Mn (VI). Charge transfer requires that the it is coloured and paramagnetic.
energy levels on the two different atoms (a) In K2Cr2O7, Cr is in +6 oxidation. hence
are fairly close. has no unpaired electron and thus it
is diamagnetic. Though K2Cr2O7 has
O = (8) = 2, 6 no unpaired electron but it is coloured.
K L
This is due to charge transfer
Mn (25) = 2 , 8, 15 spectrum.
K L M
60. (b) On an X-ray diffraction photograph the
hence the charge transfer occurs from
intensity of the spots depends on electron
L M. density of atoms/ions.
61. (a) In Frankels defect an ion leaves its regular 65. (a) Since G A P. V
site and occupy a position in the space For a spontaneous process G should be
between the lattice sites.
negative which is possible only if
+ – + –
A P. V or A W 0 .
A B A B 66. (c) We have to find
+
A
– – +
B B A 2C(s) H 2(g) C2 H2(g) H ?
This is the equation for formation of
+ – + –
A B A B acelytene Given
– + – + 2C (s) 2O 2(g) 2CO 2(g) ; H 787 kJ
B A B A
....(1)
62. (b) 8 : 8 type of packing is present in CsCl. 1
H 2(g) O2(g) H 2 O( ) ; H 286 kJ
6 : 6 type of packing is present in NaCl and 2
KCl. ...(2)
63. (d) When solid melts S increases. because 1
when solid changes into liquid randomness C2 H 2(g) 2 O2(g)
2
increases.
64. (d) Gibbs Helmholtz Equation– 2CO 2(g) H 2 O( ) ;
G H – T S .......(1)
H 1310 kJ ....(3)
differentiate this equation w.r.t. temperature
at constant pressure Add eq.(1) and (2)
1
G Gy Gx 2C(s) H 2(g) 2 O 2(g) 2CO 2(g)
– ....(2) 2
T P T T P
P H 2 O( ) ; H 1073kJ ......(4)
= – Sy – ( – Sx)
Subtract eq. (3) form eq. (4) we get
= – ( Sy – Sx) = S ......(3)
where S change in entropy 2C(s) H 2(g) C 2 H 2(g) ;
on combining equation (1) & (3) we get H 237 kJ
G 67. (b) Endothermic reactions are favoured at high
G H T ....(4) temperature. Therefore, increasing the
T temperature will shift equilibrium to the
P

Equation (4) is on alternative form of Gibbs right.

Helmholtz equation k2 Ea 1 1
Dividing equ. (4) by T2, we get 68. (a) ln
k1 2.303R T1 T2
G H 1 G
2 2 69. (a) H 2 I2 2HI
T T T T
P Initial moles 1 1 0
At equilibrium (1 x) (1 x) 2x
on rearrangement, we get
G H [HI]2 (2x) 2
– Kc =
T T2 (1 x)2
[H 2 ][I2 ]
P
Given x = 80% = 0.80
G
H –T 2 (2 0.80) 2
T Kc = 64
P (1 0.80) 2
70. (d) The equilibrium constant Kp will change 74. (a) Lower the value of reduction potential,
with temperature for the reaction greater will be the reducing power of
H2(g) + I2(g) 2HI(g). element.
Since Zn has lowest reduction potention
Catalyst does not alter the state of
hence Zn is the strongest reducing agent.
equilibrium.
75. (a) Th e epoxide ring consists of thr ee
Equilibrium constant depends only upon
membered ring with two carbon atoms and
temperature. The relation of k with temp
one oxygen.
can be shown as
O
k H 1 1
log 2 – CH2
k1 2.303R T1 T2
76. (c) Grignard reagent is a sigma bonded
Mol.mass 40 organometallic compound in which Mg is
71. (d) Eq. mass of Ca++ = 20
2 2 bonded with one alkyl and one halogen
group. This can be prepared as
Eq.mass of water 20
Z= R – X Mg
ether
R – Mg – X
96500 96500
Given w = 60 g, i = 5 amp. 77. (a,c) Presence of electron attracting group like –
w = zit NO2 , – Cl increases the acidity of phenol
20
as it enables the ring to draw more electrons
or 60 = 5 t from the phenoxy oxygen and thus
96500 releasing easily the proton. Presence of
96500 60 electron releasing group e.g. – OCH3 on
57900 sec = 16 h.
t= benzene sing decreases the acidity of
20 5
72. (b) According to Debye – Huckel – Onsagar phenol as it strengthens the negative charge
on phenoxy oxygen and thus proton release
o o
equation m m – A B m C becomes difficult.
Further phenols are much more acidic than
where A and B are the Debye – Huckel
alcohols. The acidic nature of phenol is due
constants. If we plot a graph between molar
to the formation of stable phenoxide ion in
conductance m against the square solution.

roots of the concentration C a straight C6 H5 OH H 2 O C6 H5O – H3O

No resonance is possible in alkoxide ions
line is obtained
(RO–) derived from alcohols. The negative
charge is localized on oxygen atom. Thus
alcohols are not acidic.
Molar Conductance

78. (b) Aldol condensation is given by aldehydes

Strong electrolyte which have -H atoms. So acetaldehyde
gives this reaction.
O O
|| ||
(CH 3 )3 C C H CH3 C H
C Trimethyl acetaldehyde
acetaldehyde
73. (c) Molar conductance of CaCl2
= Molar conductance of Ca2++ + 2 × (molar O O
conductance of Cl–) || ||
= 118.88 × 10 + 2 (77.33 × 10–4)
–4 C6 H 5 C H H C H
= 273.54 × 10–4 m2 mho mol–1 benzaldehyde formaldehyde
 O O 83. (d) Given : cos x + cos2x = 1
|| || cos x =1– cos2x cos x = sin 2 x
79. (d) H3C CH 2 C H 2 C CH 2 C OCH3 2
cos x = sin x4 1– sin2 x = sin 4x
6 5 4 3 2 1 4 2
sin x + sin x =1
Methyl 4 oxohexanoate
cubic both sides, we have
80. (b) It is cycloaddition reaction between a sin12x + sin6x + 3 sin6x (sin4 x + sin2x) =1
conjugated diene and subsituted alkene or sin12 x + sin6x + 3 sin10x + 3sin8x =1
, unsaturated carboxyl compound. sin12x + 3sin10x + 3 sin8x + sin6 x–1= 0
CH2 84. (c) (cos + i sin )3/5 = (cos 3 + i sin 3 )1/5
= [cos (2k + 3 ) + i sin (2k + 3 )]1/5
+ CHO
(acrolein – CHO 2k 3 2k 3
dienophile) = cos i sin ,
diene adduct 5 5
where k = 0, 1,2,3,4
PART - III (MATHEMATICS) Product of all values.
3 3 2 3
81. (c) Since a and b are coplanar, therefore = cos i sin . cos
5 5 5
a × b is a vector perpendicular to the plane 2 3 4 3 4 3
i sin . cos i sin
containing a and b . Similarly c × d is a 5 5 5
vector perpendicular to the plane
4 3 4 3 6 3 6 3
cos
containing c and d . Thus, the two planes i sin . cos i sin .
5 5 5 5
will be parallel if their normals i.e. a × b
8 3 8 3
and c × d are parallel. cos i sin
5 5
(a ×b )× (c ×d )=0
3 2 3 4 3 6 3 8 3
82. (d) Given : diagonals d1 = 3iˆ ˆj 2kˆ and = cos
5 5 5 5 5
d2 ˆi 3jˆ 4kˆ
3 2 3 4 3 6 3 8 3
i sin
5 15 5 5 5
Area of a parallelogram = d1 d 2
2
ˆi ˆj kˆ
Now, d1 × d 2 3 1 2 55 33 22
cos
= cos 2.2. 55 11. .
1 3 4 22 55 55

= ˆi(4 6) ˆj (12 2) kˆ ( 9 1) 55 33 22
i isin
sin 2.2. 55 11 ..
22 55 55
d1 × d 2 = 2iˆ 14ˆj 10kˆ
5 6 8 5 6 8
d1 d 2 = ( 2) 2 ( 14) 2 ( 10) 2 = cos . i sin
2 5 5 2 5 5
= 4 196 100 300 2 75
5 6 2 5 6 8
1 cos . i sin
Area of parallelogram = d1 d 2 2 5 5 2 5 5
2
= cos (3 + 4 ) + i sin(3 + 4 )
1 = cos (4 + 3 ) + i sin (4 + 3 )
= × 2 75 = 75 square units
2 = cos 3 + i sin 3
 2
1 i 1 i2 2i
85. (d) 2 Y
i 2i 1 2i i y= a/e

directrix
1 1 2i 2i 2 i 4i 2i 2 4i 2 2 4i (0,5)
2 i 2 i 2 i 4 i2 4 1 5

[ i2 = –1]
F (0,c)
(–4,0) (4,0)
2i 2i 4i 2i 2 4i 2 2 4i 2 4i X
2 1 2 i 4 i 2 4 1 5 5 5 X'
F' (0, – c)
4
The imaginary part = –
5 (0,–5) y= – a/e

86. (a) Given sin –1x + sin –1y = (i) Y'

2

we know that sin–1x + cos–1x =

2 88. (a) Since the normal at (ap2, 2ap) on y2 = 4ax
sin–1 x = /2 – cos–1x meets the curve again at (aq2, 2aq), therefore
Equation (1) becomes. px + y = 2ap + ap3 passes through (aq2,2aq)
paq2 + 2aq = 2ap + ap3
– cos–1x + – cos–1y = p(q2–p2) = 2(p – q)
2 2 2 p (q + p) = –2
p2 + pq + 2 = 0
cos–1x + cos–1y = 89. (d) Given : equation of line, x–3y=1 (1)
2 and hyperbola x2 – 4y2 = 1 (2)
putting x = 1 + 3y in equation (2), we get
x2 y2 (1 + 3y)2 – 4y2 =1 1+ 9y2 + 6y – 4y2 = 1
87. (c) Equation of ellipse 1 , where
b2 a2 2
5y + 6y = 0 y(5y + 6) = 0
a > b. 6
y = 0 or y =
x2 y2 5
Given, 1 b = 4, a = 5
16 75 18 13
x = 1 for y = 0 & x =1 =
5 5
b2 16
But e = 1 2
1
a 25 for y = –6/5
the line (1) cuts the hyperbola (2) in at
3 most two point.
e= co-ordinates of points are P(1,0) &
5
13 6
a Q ,
equation of directrix y = 5 5
e

5 2 2
13 6
y= 3y = ± 25 PQ = 1 0
3/ 5 5 5
 2 U U sin U
18 36 x y tan U
324 36 360 x y cos U
5 25
25 25 U U
x y tan U
6 10 x y
length of straight line PQ = units.
5 94. (c) Equation of the given curve in parametric
90. (c) Given x = t2 + 2t–1 & y = 3t + 5 form,
x = t2 + 1 and y = t2 – t – 6
x = t2 + 2t + 1–2 & y = 3t + 3+2
Y-coordinate of the point, where the given
x = (t+1)2 –2 & y = 3(t + 1) + 2 (2)
curve meets X-axis is 0.
(t+1) = x 2 .......... (1) When y = 0, then t2 – t – 6 = 0
Equation (2) becomes [using equation (1)] t 2 3t 2t 6 0
y= 3 x 2 2 t(t 3) 2(t 3) 0
y–2 = 3 x 2 (t 3)(t 2) 0
squaring both sides, we get t 3 or 2
(y–2)2 = 9 (x + 2)
when t = 3, then x = 10
Y2 =9X where Y = y – 2 & X = x + 2
when t = –2, then x = 5
This equation represents a parabola Hence, the points where the curve meets
91. (c) Slope of the normal at (3,4) is the value of the X-axis are (10, 0) and (5, 0).
1 1 3 dy
at x = 3 or = tan =–1
f '(x) f '(3) 4 dy dt 2t 1
f`'(3) = 1 Now, dx dx 2t
92. (c) Given : f (x) = x2 e–2x, x > 0 dt
f '(x) = x2.e–2x(–2) + e–2x.2x Slope of the tangent at point (10, 0)
put f '(x) = 0 2e–2x. x (–x + 1) = 0 dy dy 5
x = 1 or x = 0 m1
dx x 10 dx t 3 6
f"(x) = (–4x2 – 6x + 1)e–2x
f"(1) = –9e–2x < 0 Slope of the tangent at point (5, 0),
f"(0) = e–2x > 0 dy dy 5 5
value of f(x) is maximum at x = 1 m2
dx x 5 dx t 2 4 4
1 If be the angle beween two tangents, then
f(x) = x2.e–2x f(1) = e–2 = 2
e 55
93. (d) Given : (x + y) sin U = x2y2 m 2 m1 64
tan
1 m1m 2 5 5
x 2 y2 1
sinU = = v (let) 6 4
x y
Here n = 2 – 1 = 1 15 10 5
12 12 10
v v
Euler's theorem x. y. nv 24 25 49 49
x y
24 24
sin U sin U 10
x y sin U tan
x y 49
U U 1 10 1 10
x.cos U y.cos U. sin U tan tan
x y 49 49
 4 3 4
dx
95. (b) | x 3 | dx (x 3)dx (x 3)dx 97. (d) Let I 2
1 1 3 0 (a x 2 )7

x2
3
x2
4
Put x a tan dx a sec2 d
3x 3x
2 2
1 3
limit at x 0 0&x
2
1 2 2 1 2 2
[3 1 ] 3[3 1] .[4 3 ] 3[4 3]
2 2

1 1
2
a sec 2
(8) 3(2) .(7) 3(1) I d
2 2 0 a14 (1 tan 2 )7
7 5
4 6 3
2 2
2 2
1 1 1
2 d . cos12 .d
96. (b) Required area (y 2 y1 )dx a13 0 sec12 a13 0
0
But

y 2 1
x
y=2 (2,4) sin 2m 1 .cos 2n 1 . d B(m, n)
2
(x2y2) 0

(0,1) Q
(1,1) m n 1
& B(m, n) &
m n 2
P

2
1
x' I sin 0 . cos12 d
o
x a13 0
(2,0)
2
–x
2x

1 13
m , n
y=

2 2

1 13
.
1 2 2
2
I
x 2 2a13 1 13
(2 2x x )dx
2 2
0

2 11 9 7 5 3 1
2x x3 . . . . . . .
1
x2 . 2 2 2 2 2 2
log 2 3 2a13 6.5.4.3.2.1
0

4 8 1 3 4 231 1
4 .
log 2 3 log 2 log 2 3 2048 a13
 (1 x)2
98. (c) Let I ex dx cos log(1) c c 0
(1 x 2 )2 2
y
x 1 x 2 2x cos log x
I e dx x
(1 x 2 )2 100. (c) The equation of the family of circles of
radius r is
1 x2 2x (x – a)2 + (y – b)2 = r2 ...(1)
ex dx
2 2
(1 x ) (1 x 2 )2 Where a & b are arbitrary constants.
Since equation (1) contains two arbitrary
1 x constants, we differentiate it two times w.r.t
I ex dx 2 e x . dx
(1 x 2 ) (1 x 2 )2 x & the differential equation will be of
second order.
1 Differentiating (1) w.r.t. x, we get
I .e x ex . (1 x 2 ) 2 2xdx
(1 x 2 )
dy
2(x a) 2(y b) 0
ex x dx
2 dx
(1 x 2 )2 dy
(x a) (y b) 0 ...(2)
ex x dx
ex dx
I 2 Differentiating (2) w.r.t. x, we get
1 x2 (1 x 2 )2
2
d2 y dy
ex x 1 (y b) 0 ...(3)
2
2 2
dx dx 2 dx
(1 x )
2
x dy
e 1
I +C dx
1 x2 (y b)
d2 y ...(4)
y y dx 2
99. (c) Given : x sin dy y sin x dx
x x On putting the value of (y – b) in equation
(2), we get
y
y sin x
dy x dy
2
dy
dx y 1
x sin dx dx
x x a
d2 y ...(5)
y dy dz dx 2
Put z z.1 x.
x dx dx Substituting the values of (x – a) & (x – b)
dz zx sin z x in (1), we get
x. z z cosec z
dx x sin z
2 2 2 2 2
dy dy dy
dz dx 1 1
x cosec z sin z dz dx dx dx
dx x r2
2 2
cos z log x c d2 y d2 y
2
y dx dx 2
cos log x c
x
2 3 2
dy 2 d2 y
1 r
But y(1) x 1, y dx dx 2
2 2
 104. (d) Number of onto functions: If A & B are two
d2 y dy 3x sets having m & n elements respectively
101. (c) Given : 2 y 2e
dx 2 dx such that 1 n m then number of onto
The auxiliary equation is functions from A to B is
D2 + 2D + 1 = 0 or m2 + 2m + 1 = 0 n
n r n
(m 1)(m 1) 0 m 1, 1 –1 . Cr r n
r 1
i.e., repeated roots
Given A = {1, 2, 3, ---- n} & B = {a, b, c}
Complementary function = (c1 + c2x)e–x Number of onto fun ction s
Now Particular Integral (P.I.)
3
3–r 3 n
1 –1 . Cr r
. 2e3x [D = 3] r 1
D2 2D 1
3 13 n 3 2 3 n
1 C1 1 –1 C2 2
1 3x 2e3x e3x
P.I. . 2.e
32 2.3 1 16 8 3
C3 3
n
–1
3 3

Solution y = C. F. + P. I.
3 3
C1 C2 2n 3
C3 3n
3x
e
y c1 c2 x e – x 3! 3! n 3! n
8 2 3
2!1! 2!1! 3! 0!
102. (c) Given ydx + (x – y3) dy = 0
dx dx 1 3 3. 2n 3n
y x y3 0 x y2
dy dy y
3n 3 2n 1
compare this equation to general equation
105. (b) Let there are n persons in the room. The
dx 1 2 total number of hand shakes is same as the
i.e. Px Q P ,Q y
dy y number of ways of selecting 2 out of n.

1 n n n 1
Pdy
dy C2 66 66
I.f . e e y
e log y
y 2!

Q I.f .dy C1 n2 n 132 0

x × I.f.
n 12 n 11 0 n 12
y4 106. (a) We know that, let (G, 0) be a group & e be
x.y y 2 .y dy C1 xy C1
4 the identity then
(a * a)–1 = a–1 o a–1
4xy y4 4C1 = (a–1)–1 = a.
107. (d) No. of tickets = 9
y4 4xy 4C1 y4 4xy C , No. of odd numbered tickets = 5
where C = –4C1 No. of even numbered tickets = 4
103. (b) Given, x1x 2 x3 x 4 x 5 Required probability = P {odd, even, odd}
1050
+ P (even, odd, even)
x1 x 2 x 3 x 4 x 5 2 3 52 7 5
C1 4
C1 4
C1 4
C1 5
C1 3
C1
Each of 2, 3 or 7 can take 5 places and 52 9
C1 8
C1 7
C1 9
C1 8
C1 7
C1
can be disposed in 15 ways.
Hence, number of positive integral solution 5 4 4 4 5 3 5
= 53 × 15 = 1875 9 8 7 9 8 7 18
108. (a) Given
1 1 1 tan
1 5 1 A 2
.
P A ,P B ,P B/ A sec tan 1
12 12 15
P A B 1 10 1 tan
We know that P B / A B 2
.
P A sec 01 tan 1

1 P A B 1 1 tan
B 2
.
15 1/12 sec tan 1
1 1
P A B 1
15 12 180 B .A
sec2
But,
P A B P A P B P A B sec 2 .B A

1 5 1 89 1
P A B cos 2 .B A
12 12 180 180
2 x
109. (a) f x dx , where f x 112. (b) Given : 2x 1 4 8
1.5 2
2 2x 2 0
2
2 x 1 2 1 x 2 7 6 2x
dx . xdx
1.5 2 2 1.5 2 2
1.5
2x 1 4x 2 12 4 4x 14
1 1 175 7
4 2.25 1.75
4 4 400 16 8 12 14x 0
110. (b) If 2 P(X = 0 ) + P (X = 2) = 2 P (X = 1)
Let probability distribution of X be given 8x 3 24x 4x 2 12 16x
by 56 96 112 x 0
r m
m .e 8x 3 4x 2 152x 140 0
P(X = r) where r = 0, 1, 2, .....
r! (x + 5) is a factor of above equation
m0 e m m2 .e m
m.e m
8x 3 40x 2 36x 2 180x
2 2
0! 2! 1!
28x 140 0
2
m2 8x x 5 36x x 5 28 x 5 0
2 2m m2 4m 4 0
2
x 5 8x 2 36x 28 0
2
m 2 0 m 2
x 5 4 2x 2 9x 7 0
1 tan
111. (c) A & AB I
tan 1
4 x 5 2x 2 7x 2x 7 0
1
B I A , A11 1, A12 tan ,
4 x 5 x 2x 7 1 2x 7 0
A 21 tan , A 22 1
4 x 5 2x 7 x 1 0
1 tan 2 2
|A| 1 tan sec x 5, 3.5,1
tan 1
113. (b) For only one solution | A | 0 118. (a) Given : H is the harmonic mean between
P&Q
k 2 1
2PQ 1 P Q
0 k 1 2 0 H
P Q H 2PQ
0 0 k 2
2 1 1 H H
k k 1 k 2 0 2
H Q P P Q
k 0, k 1, k 2. k 1
119. (c) We have a b a b
114. (c) Let
1 2 5 1 2 5 a a b b a b
A 2 4 a 4 0 0 a 6
1 2 a 1 0 0 a 6 a .b a a .a b b.b a b.a b

R2 R2 2R1 , R 3 R 3 R1 a .b a b
clearly rank of A is 1 if a = –6
2 2
115. (c) Given : ax 4 bx 2 c 0 a b b.b b 1, a .a a 1

Equation will be real if D 0

a .b 1 a b
b2 4ac 0 b2 4ac
4 x a b where x a.b 1 is a scalar
116. (a) Given : log 3 x log 3 x log 3 x
The given vector is parallel to a b .
log 3 8 x log 3 16
x 4
120. (a) Given : A, B & C are three points with co-
1
1 1 1 1 ordinates (1, 2, –1), (2, 0, 3) & (3, –1, 2)
log3 x 2 4 8 16 4 respectively.
Now, direction ratio's of
1
1
AB = 2 – 1, 0 – 2, 3 + 1 = 1, – 2, 4 &
1 a direction ratio's of
log3 x 2 4 S
1 r AC = 3 –1, –1 –2, 2 + 1 = 2, – 3, 3
log3 x 2 4 x2 34 x 9 we know that

3
a1a 2 b1b 2 c1c 2
1 1 cos
117. (a) x x 0 a12 b12 c1 . a 2 2 b 2 2
2
c22
x x

2
1 2 2 3 4 3
1 1 cos
x x 1 0 1 4 16 . 4 9 9
x x
2 6 12 20
1
x 0 x2 1 0 x i 21 . 22 462
x
Thus, the given equation has no real roots. 462 cos 20