AIEEE 2005 SOLVED PAPER - Parishram Publication

AIEEE 2005 SOLVED PAPER

Physics

l . A projectile can have the same range R for two angles of projection. If c j andt2 be the times of flights in the two cases, then the product of the two times o f flights is proportional to :

i 2

3.

(a) R (l>) 2 R

(d) R

2. An annular ring with inner and outer radii Rx and R 2 is rolling without slipping with a uniform angular speed. The ratio o f the forces experienced by the two particles situated on

the inner and outer parts o f the ring, •=*- is :

m 1(a)

(c)

R ,

1

(b)

(d)R-j

A smooth block is released at rest on a 45° incline and then slides a distance d. The time taken to slide is n times as much to slide on rough incline than on a smooth incline. The coefficient o f friction is :

(a) m* = 1 - "V ( b )n fc - v * _2

(C) (dl u r f ^

4. The upper half of an inclined plane with inclination 4> is perfectly smooth, while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom, if the coefficient of friction for the lower half is given b y :(a) 2 sin <t> (b) 2 cos <J>(c) 2 tan 4> (d) tan (j»

5. A bullet fired into a fixed target loses half of its velocity after penetrating 3cm. How much further it will penetrate before coming to rest, assuming that it faces constant resistance to motion ?

(a) 3.0 cm (b>_2.0_cm(c) 1.5 cm (d) 1.0 cm

6 . Out of the following pairs, which one does not have identical dimensions ?(a) Angular momentum and Planck’s constant(b) Impulse and momentum(c) Moment of inertia and moment of a force(d) Work and torque

7. The relation between time t and distance x is t = ax2 + bx, where a and b are constants. The

acceleration is :

8.

9.

(a) - 2 abv

(c) - 2 av3

(b) 2 bv3

(d) 2 av2

A car, starting from rest, accelerates at the rate / through a distance S, then continues at constant speed for time t and then decelerates at the rate f/2 to come to rest. If the total distance travelled is 15 S, then :

(a) S = /t (b) S = | / t 2

(d) S = i f t 2(c) S = i / r 2

A particle is moving eastwards with a velocity of 5 ms'1. In 10 s the velocity changes to 5 ms 1 northwards. The average acceleration in this time is :

(a)1

V2ms towards north-east

10.

1(b) ms 2 towards north

(c) zero

(d) ^ ms-2 towards north-west

A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2m/s2. He reaches the ground with a speed of 3 m/s. At what height, did he bail out ?(a) 91 m (b) 182 m(c) 293 m ( d ) l l l m

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11. A block is kept on a frictionless inclined surface with angle o f inclination a. The incline is given an acceleration a to keep the block stationary. Then a is equal to :

(a) g/ tan a (b) g cosec a(c) g (d) g tan a

12. A spherical ball o f mass 20 kg is stationary at the top of a hill of height 100 m. It rolls down a smooth surface to the ground, then climbs up another hill of height 30 m and finally rolls down to a horizontal base at a height of 20 m above the ground. The velocity attained by the ball is :(a) 40 m/s (b) 20 m/s(c) 10 m/s (d) 10V30 m/s

13. A body A of mass M while falling vertically downwards under gravity breaks into two

parts; a body B o f mass i M and, a body C of

2 , mass M. The centre o f mass of bodies B and C

taken together shifts compared to that of body A towards:(a) depends on height o f breaking(b) does not shift(c) bodyC(d) body B

14. The moment of inertia of uniform semicircular disc of mass M and radius r about a line perpendicular to the plane of the disc through the centre is :

(a) i Mr2

(c) Mr2

(b) | Mr2

(d) | Mr2

15. A particle of mass 0.3 kg is subjected to a force F = - k x with fc = 15N/m. What will be its initial acceleration, if it is released from a point20 cm away from the origin ?(a) 3m/s2 (b) ISm/s2

(c) 5 n v V (d) 10 m/s2

16. The block o f mass M moving on the frictionless horizontal surface collides with the spring of

spring constant k and compresses it by length L. The maximum momentum o f the block after collision is :

M

777777777

(a) yfMk L

(c) zero

(b)

ML2(d) n r

17. A mass m moves with a velocity v and collides inelastically with another identical mass. After

collision the Ist mass moves with velocity -7= inV3

a direction perpendicular to the initial direction of motion. Find the speed of the second mass after collision :(a) v (b) V3 v, , 2 W J 3 V

18. A 20 cm long capillary tube is dipped in water. The water rises upto 8 cm. If the entire arrangement is put in a freely falling elevator, the length of water column in the capillary tube will b e :(a) 8 cm (b) 10 cm(c) 4 cm (d) 20 cm

19. If S is stress and Y is Young’s modulus of material of a wire, the energy stored in the wire per unit volume is :

(a) 2 S2y

y >> 2Y(c) 7 7Sz

0 » f y

(d> A20. Average density o f the earth :

(a) does not depend on g(b) is a complex function of g(c) is directly proportional to g(d) is inversely proportional to g

21. A body o f mass m is accelerated uniformly from rest to a speed v in a time T. The instantaneous power delivered to the body as a function of time, is given b y :

r x mv »(a) Cb) ^ t 2

( d ) i ^ t 2

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22. Consider a car moving on a straight road with a speed of 100 m/s. The distance at which car can be stopped) is : [n k =0.5](a) 800 m (b) 1000 m(c) 100 m (d) 400 m

23. Which of the following is incorrect regarding the first law of thermodynamics ?(a) It is not applicable to any cyclic process(b) It is a restatement of the pnnciple of

conservation of energy(c) It introduces the concept of the internal

energy(d) It introduces the concept of the entropy

24. A T shaped object with dimensions shown hj the figure, is lying on a smooth floor. A force F is applied at the point P parallel to AS, such that the object has only the translational motion without rotation. Find the location o fP with respect to C :

25.

AC ] u

(a)

(c)

!•S'

< b )| i

(d) i

The change in the value of g at a height h above the surface of the earth is the same as at a depth d below the surface of earth. When both d and h arc much smaller than the radius of earth, then which one of the following is correct ?

(a) d = ~ (b), 3h

d ~ 2

26.

28.

(c) d = 2h (d) d = /tA particle of mass 10 g is kept on the surface of a uniform sphere o f mass 100 kg and radius 10 cm. Find the work to be done against the gravitational force between them, to take the particle far away from the sphere : (you may takeG = 6.67 x l0 ‘ n Nm2/kg2)(a) 13.34 x 10“10 J (b) 3.33 x 10“M J

(c) 6.67 x 10 9 J

29.

27. A gaseous mixture consists of 16 g of helium andC„

16 g of oxygen. The ratio of the mixture is : Cv

(a) 1.59 (b) 1.62(c) 1.4 (d) 1.54The intensity of gamma radiation from a given source is /. On passing through 36 mm of lead, it is reduced to 1/8. The thickness of lead, which will reduce the intensity to J/2 will be : (a) 6 mm (b) 9 mm(c) 18 mm (d) 12 mmThe electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480 nm, is incident on it. The band gap in (eV) for the semiconductor is :(a) 1.1 eV (b) 2.5 eV(c) 0.5 eV (d) 0.7 eVA Photocell is illuminated by a small brightsource placed 1 m away. When the same source

1of light is placed - m away, the number of

electrons emitted by photocathode would :(a) decrease by a factor of 4(b) increase by a factor of 4(c) decrease by a factor of 2(d) increase by a factor of 2 Starting with a sample of pure 6r>Cu, 7/8 of it decays into Zn in 15 min. The corresponding half-life is :

(b) 15 min

30.

31.

32.

33.

(a) 10 min

(c) 5 min (d)

If radius of the 2J A1 nucleus is estimated to be 3.6 fermi, then the radius of ^ T e nucleus be nearly:(a) 6 fermi (b) 8 fermi(c) 4 fermi (d) 5 fermi

The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is :

T .

2T„

K

(d) 6.67 x 10“10 J(a) 1/2 (c) 1/3

2S„

(b) 1/4 (d) 2/3

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34. The figure shows a system of two concentric 38.spheres of radii r, and r2 and kept at temperatures Tj andT2, respectively. The radial rate of flow of heat in a substance between the two concentric spheres, is proportional to :

(a)

(c)

</2 ~ 5 )(r,r2)

V i( r2 - rl )

(b) In (!)(d) (r2 - rj)

35. A system goes from A to B via two processes 1 and 11 as shown in figure. If Al/j and AU2 are the changes in internal energies in the processes I and II respectively, then :

(a) AU, = AU 2 '(b) relation between AU, and AU2 cannot be

determined(c) Al/ 2 > At/,(d) a a 2 < At/,

36. The function sin2 (cot)represents :

(a) a periodic, but not simple harmonic, motion with a period 2n/<o

(b) a periodic, but not simple harmonic, motion with a period n/<o

(c) a simple harmonic motion with a period271/u

(d) a simple harmonic motion with a periodn/w

37. A Young’s double slit experiment uses amonochromatic source. The shape of theinterference fringes formed on a screen is :(a) hyperbola (b) circle(c) straight line (d) parabola

Two simple harmonic motions are represented

by the equations y, = 0.1 sin ^100 nt + ^jand

y 2 = 0.1 cos irt. The phase difference of the velocity of particle 1, with respect to the velocity of particle 2 is :/• •» ~ n (a) 6

(c)- Tt~ T

39. A fish looking up through the water sees the outside world, contained in a circular horizon.

4If the refractive index of water is ~ and the fish

is 12 cm below the water surface, the radius of this circle in cm is :

. 36( b , 37

40.

41.

42.

(a) 36V7

(c) 36>/5 (d) 4v5

Two point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm. Approximately, what is the maximum distance at which these dots can be resolved by the eye ? [Take wavelength o f light = 500 nm] (a) 5 m (b) 1 m(c) 6 m (d) 3 m

A thin glass (refractive index 1.5) lens has optical power o f- 5 D in air. Its optical power in a liquid medium with refractive index 1.6 will be :(a) ID (b) - 1 D(c) 25 D (d) -2 5 D

The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy ?

— n = 4

1 11 111 IV

n -2

n - I

(a) III (b) IV(c) I (d) II

43. If the kinetic energy of a free electron doubles, its de-Broglie wavelength changes by the factor:

(b) 2(a) 1

10 72(d) y/2

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44. In a common base amplifier, the phase difference between the input signal voltage and output voltage is :

(b) 7T .71J

(c) zero w f

45. In a full wave rectifier circuit operating from 50 Hz mains frequency, the fundamental frequency in the ripple would be :(a) 50 Hz (b) 25 Hz(c) 100 Hz (d) 70.7 Hz

46. A nuclear transformation is denoted by A'(n, a )-> gLi. Which of the following is the

nucleus of element X ?(a) 562C

Vi(b) s°B

‘ Be(c) SB (d) 4

47. A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10 divisions per milliampere and voltage sensitivity is 2 divisions per millivolt. In order that each division reads 1 V, the resistance in Ohm’s needed to be connected in series with the coil will be :(a) 103 (b) 10s(c) 99995 (d) 9995

48. Two voltameters, one of copper and another of silver, are joined in parallel. When a total charge q flows through the voltameters, equal amount of metals are deposited. If the electrochemical equivalents of copper and silver are zx and z2 respectively, the charge which flows through the silver voltameter is :

.9 n * <J(a ) z

1 + — Z 2

(C) q —22

(b)1 + ^

(d) q f z \

49. In the circuit, the galvanometer G shows zero deflection. If the batteries A and B have negligible internal resistance, the value of the resistor R will be :

5 0 0 0-VWWW-

12V e

(a) 200 f i (b) 100 fi(c) 5000 (d) 1000Q

50. Two sources of equal emf are connected to an external resistance R. The internal resistances of the two sources are R, and R2 (R2 > * i ) If the potential difference across the source having internal resistance R2, is zero, then :

(a) R = -

rv.oj i; (R: + R2)

(R2 - R i )(b) R = R 2 - R,

CO R =R,R2

(R, + R2) * 1 * 2

cd) R (R2 - Rj )

51. A fully charged capacitor has a capacitance C. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity s and mass m. If the temperature of the block is raised by AT, the potential difference V across the capacitance is:

2mCA7' „ » mCAT(a)

(c)

I sms AT

lb)

(d) /2msATC v ' V C

52. One conducting t/-tube can slide inside another as shown in figure, maintaining electrical contacts between the tubes. The magnetic field B is perpendicular to the plane of the figure. If each tube moves towards the other at a constant speed v, then the emf induced in the circuit in terms of B, I and v, where ( is the width o f each tube, will be :

(a) Blv (b) - Blv(c) zero (d) 2Blv

53. A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now be :

(a) doubled (b) four times(c) one-fourth (d) halved

54. Two thin, long, parallel wires, separated by a distance d carry a current of i A in the same direction. They w ill:

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(a) attract each other with a force o f —-°t ,(2nd)

•2(b) repel each other with a force o f-- 1* -

(2nd)■ ikJ2(c) attract each other with a force of -^-2-—

(2tuH ,-2

(d) repel each other with a force of -i-5-—(2nd2)

55. When an unpolarized light o f intensity J0 is incident on a polarizing sheet, the intensity of the light which does not get transmitted is :

i/ o(a)

(c) zero(b)

(d) I056. A charged ball B hangs from

a silk thread S, which makes an angle G with a large charged conducting sheet P, as shown in the figure. The surface charge density a of the sheet is proportional to :(a) cos G (b) cot 0(c) sin 0 (d) tan 0

57. Two point charges + 8 q and - 2 q are located at x = 0 and x = L respectively. The location o f a point on the jc-axis at which the net electric field due to these two point charges is zero is :

(a) 2 L

(c) 8L (d) 4 L58. Two thin wire rings each having a radius R are

placed at a distance d apart with their axes coinciding. The charges on the two rings are + q and - q. The potential difference between the centres of the two rings is :

(a) ^

(b)

(d)

4jtt0cI2

q 12ne0 R

zero

<1 1

V« 2 + d2

47t£o V r 2 + d2

59. A parallel plate capacitor is made by stacking n equally spaced plates connected alternatively. If the capacitance between any two adjacent plates is C, then the resultant capacitance is :(a) (n - 1)C (b) ( n + 1 )C(c) C (d) nC

62.

63.

64.

65.

60. When two tuning forks (fork 1 and fork 2) are sounded simultaneously, 4 beats per second are heard. Now, some tape is attached on the prong of the fork 2. When the tuning forks are sounded again, 6 beats per second are heard. If the frequency of fork 1 is 200 Hz, then what was the original frequency of fork 2 ?(a) 200 Hz (b) 202 Hz(c) 196 Hz (d) 204 Hz

61. I f a simple harmonic motion is represented byd2* « • • • j .+ a x = 0, its time period is :

00 # (b) £(c) 2na (d) 2nJa

The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would :(a) first increase and then decrease to the

original value(b) first decrease and then increase to the

original value(c) remain unchanged(d) increase towards a saturation value

An observer moves towards a stationary source of sound, with a velocity one-fifth of the velocity o f sound. What is the percentage increase in the apparent frequency ?(a) Zero (b) 0.5%(c) 5% (d) 20%

If I0 is the intensity o f the principal maximum in the single slit diffraction pattern, then what will be its intensity when the slit width is doubled ?

(a) 22q (b) 410

(c) /„

Two concentric coils each of radius equal to 2ti cm are placed at right angles to each other.3 A and 4 A are the currents flowing in each coiL respectively. The magnetic induction in Wb/rh2 at the centre o f the coils will be :

(Ho = 47t x 10"7 Wb/Am)

(a) 1 2 x 10

(c) 5 x 10~5

i-S (b) 1 0 '5

(d) 7 xH T 5

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66.

67.

68.

69.

70.

71.

76.

77.

78.

A coil o f inductance 300 mH and resistance 20 is connected to a source of voltage 2V. The current reaches half of its steady state value in : (a) 0.05 s (b) 0.1s(c) 0.15 s (d) 0.3 sThe self-inductance o f the motor of an electric fan is 10 H. In order to impart maximum power at 50 Hz. it should be connected to a capacitance o f :(a) 4mF (b) 8 nF(c) 1 uF (d) 2nFAn energy source will supply a constant currentinto the load, if its internal resistance is :(a) equal to the resistance of the load(b) very large as compared to the load

resistance(c) zero(d) non-zero but less than the resistance of the

load

A circuit has a resistance of 12 ft and an impedance of 15 O. The power factor o f the circuit will be :(a) 0.8 (b) 0.4(c) 1.25 (d) 0.125The phase difference between the alternatingcurrent and emf is n/2. Which of the followingcannot be the constituent o f the circuit ?(a) C alone (b) R, L(c) I, C (d) L alone

A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected along the direction of the fields with a certain velocity, then :

Chemistry

(a) its velocity will decrease(b) its velocity will increase(c) it will turn towards right of direction of

motion(d) it will turn towards left of direction of

motion

72. A charged particle o f mass m and charge q travels on a circular path of radius r that is perpendicular to a magnetic field B. The time taken by the particle to complete one revolutionis :

2nmqB ' ' m(a) (b)

2tt q2B

(c)2nqB

m (d)2itmqB

73. In a potentiometer experiment the balancing with a cell is at length 240 cm. On shunting the cell with a resistance o f 2 0 , the balancing length becomes 120 cm. The internal resistance o f the cell i s :(a) lO (b) 0.5 0(c) 4 0 (d) 2 0

74. The resistance o f hot tungsten filament is about10 times the cold resistance. What will be the resistance o f 100 W and 200 V lamp, when not in use ? ,(a) 400 (b) 20 0(c) 4000 (d) 200 0

75. A magnetic needle is kept in a non-uniform magnetic field. It experiences :(a) a torque but not a force(b) neither a force nor a torque(c) a force and a torque(d) a force but not a torque

The oxidation state of Cr in [CrCNH;, )4C12]+ is :

(a) 0 (b) + 1(c) +2 (d) +3Which one of the following types of drugsreduces fever ?(a) Tranquiliser (b) Antibiotic(c) Antipyretic_ (d) AnalgesicWhich of the following oxides is amphoteric in character ?(a) Sn02 (b) Si02(c) C02 -- (d) CaO

79. Which one o f the following species is diamagnetic in nature ?(a) H2 (b) H2(c) H2 (d) HeJ

80. If a is the degree o f dissociation of Na2S04, the van’t Hoff factor (i) used for calculating the molecular mass is :(a) 1 - 2a (b) 1 + 2a(c) 1 - a (d ) 1 + a

81. Which of the following is a polyamide ?(a) Bakelite (b) Terylene(c) Nylon-66 (d) Teflon

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82. Due to the presence o f an unpaired electron, free radicals a re :(a) cations(b) anions(c) chemically inactive(d) chemically reactive

83. For a spontaneous reaction the AG, equilibrium constant (JC) and £c°dl will be respectively :

(a) -ve, > 1, -ve (b) -ve, < 1 , -ve(c) +ve, > 1, -ve (d) -ve, > 1, +ve

84. Hydrogen bomb is based on the principle o f :(a) artificial radioactivity(b) nuclear fusion(c) natural radioactivity(d) nuclear fission

85. An ionic compound has a unit cell consisting of A ions at the comers of a cube and fl ions on the centres of the faces of the cube. The empirical formula for this compound would be :(a) A3B (b) AB3(c) A2B (d) AB

8 6 . The highest electrical conductivity of the following aqueous solutions is of :(a) 0.1 M difluoroacelic acid(b) 0.1M fluoroacetic acid(c) 0.1 M chloroacetic acid(d) 0.1 M acetic acid

87. Lattice energy o f an ionic compound depends upon :(a) charge on the ion and size of the ion(b) packing of ions only(c) size o f the ion only(d) charge on the ion only

88. Consider an endothermic reaction X - » Y with the activation energies Eb and Ef for the backward and forward reactions respectively. In general:(a) there is no definite relation between Eb

and Ej(b) Eh =E f(c) Eb > Ef(d) E„ < E f

89. Aluminium oxide may be electrolysed at 1000 °C to furnish aluminium metal (Atomic mass=27 amu; 1 faraday= 96,500 Coulombs). The cathode reaction is

Al3+ + 3e~ —» Al°To prepare 5.12 kg o f aluminium metal by this method would require :

(a) 5.49 x 101 C of electricity

(b) 5.49 x 104 C of electricity(c) 1.83 x 107 C of electricity

(d) 5.49 x 107 C of electricity

90. The volume o f a colloidal particle, Vc as compared to the volume of a solute particle in a true solution V , could be :

( a ) £ » 103 ( b ) £ * 10-3vs vs

(c) ^ * 1023 ( C ) £ « lvs vs

91. Consider the reaction : N2 + 3H2 ---- > 2NH3

carried out at constant temperature and pressure. If AH and AU are the enthalpy and internal energy changes for the reaction, which of the following expressions is true ?(a) A H > A U (b) AH < AU(c) AH =AU (d) AH = 0

92. The solubility product o f a salt having general

formula MX2, in water is 4 x 10' 12 . The

concentration of M 2* ions in the aqueous solution o f the salt is :(a) 4.0 x lO ' 10 M (b) 1.6 x 10^ M(c) 1.0 x 10-4 M (d) 2.0 x 10‘ 6 M

93. Benzene and toluene form nearly ideal solutions. At 20° C, the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The partial vapour pressure of benzene at 20° C for a solution containing 78 g of benzene and 46 g of toluene in torr is :(a) 53.5 (b) 37.5(c) 25 (d) 50

94. Which one of the following statements is not true about the effect of an increase in temperature on the distribution o f molecular speeds in a gas ?(a) The area under the distribution curve

remains the same as under the lower temperature

(b) The distribution becomes broader(c) The fraction of the molecules with the

most probable speed increases(d) The most probable speed increases

95. For the reaction2N02 ( g ) ^ = h2 NO (g)+ 0 2 (g )

(Kc = 1.8 x 10~6 at 184°C)(R = 0.00831 k J/(mol. K))When K . and Kc are compared at 184° C it is found that:

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(a) whether K p is greater than, less than or equal to Kc depends upon the total gas pressure

(b) K p = Kc(c) Kt, is less than Kc(d) Kp is greater than Kc

96. The exothermic formation of ClFj is represented by the equation :Cl2(g )+ 3 F2( g )^ = * 2 ClF3(g); AHr=-329 kJ Which o f the following will increase the quantity of C1F3 in an equilibrium mixture of Cl2, F2 and C1F3 ?(a) Adding I-2(b) Increasing the volume o f the container(c) Removing Cl2(d) Increasing the temperature

97. Hydrogen ion concentration in mol/L in a solution of pH = 5.4 will be :(a) 3.98x10" (b) 3.68x10(c) 3.88 x 10 (d) 3.98 x10s

98. A reaction involving two different reactants can never be(a) bimolecular reaction(b) second order reaction(c) first order reaction(d) unimolecular reaction

99. Two solutions of a substance (non electrolyte) arc mixed in the following manner.480 ml. o f 1.5 M first solution +520 mL of 1.2 M second solution.What is the molarity of the final mixture ?(a) 2.70 M (b) 1.344 M(c) 1.50 M (d) 1.20 M

100. During the process o f electrolytic refining of copper, some metals present as impurity settle as 'anode mud’. These are :(a) Fe and Ni (b) Ag and Au(c) Pb and Zn (d) Se and Ag

Electrolyte KCI KNO-, HC1 NaOAc NaCl

Am(S cm2 m ol'1) 149.9 145.0 426.2 91.0 126.5

(a) be a function of the molecular mass of the substance

(b) remain unchanged(c) increase two fold(d) decrease twice

103. In a multi-electron atom, which of the following orbitals • described by the three quantum numbers will have the same energy in the absence of magnetic and electric fields ?(A) n = 1,1= 0, m = 0(B) n = 2, / = 0, m = 0(C) n = 2, { = 1, m = 1(D) n = 3 ,1 = 2, m = 1(E )n = 3 l= 2 ,m = 0(a) (D) and (E) ’ (b) (C) and (D)(b) (B) and (C) (d) (A) and (B)

104. Based on lattice energy and other considerationswhich one of the following alkali metal chlorides is expected to have the highest melting point ?(a) RbCl (b) KCi(c) NaCl (d) LiCl

105. A schematic plot of In K versus inverse of

temperature for a reaction is shown below :

6.0

InK.

2 01.5 x 10" 2 x 10"

Calculate A“ iioac using appropriate molar conductances of the electrolytes listed above at infinite dilution in H20 at 25° C :(a) 217.5 (b) 390.7(c) 552.7 (d) 517.2

102. If we consider that 1/6, in place of 1/12, mass o f carbon atom is taken to be the relative atomic mass unit, the mass o f one mole of a substance w ill :

1/T (K"')

The reaction must be :(a) highly spontaneous at ordinary temperature(b) one with negligible enthalpy change(c) endothermic(d) exothermic

106. Heating mixture ofCu20 and Cu2Swill give(a) CujSO, (b) CuO + CuS(c) Cu + S 0 3 (d) Cu + SO2

107. The molecular shapes o f SF4, CF4 and XeF4 are:(a) different with 1, 0 and 2 lone pairs of

electrons on the central atom, respectively(b) different with 0, 1 and 2 lone pairs of

electrons on the central atom, respectively(c) the same with 1, 1 and 1 lone pair of

electrons on the central atoms, respectively

(d) the same with 2, 0 and 1 lone pairs of electrons on the central atom, respectively

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108. The disperse phase in colloidal iron (III) hydroxide and colloidal gold is positively and negatively charged, respectively. Which of the following statements is not correct ?(a) Coagulation in both sols can be brought

about by electrophoresis(b) Mixing the sols has no effect(c) Sodium sulphate solution causes

coagulation in both sols(d) Magnesium chloride solution coagulates,

the gold sol more readily than the iron (III) hydroxide sol.

109. The number of hydrogen atom(s) attached tophosphorus atom in hypophosphorous acid is : (a) three (b) one(c) two (d) zero

110. What is the conjugate base of OH' ?(a) O2- (b) O-(c) H20 (d) 0 2

111. Heating an aqueous solution of aluminium chloride to dryness will give :(a ) Al(OH)Cl, (b) A1,0,(c) A12C16 (d) A1C13

112. The correct order o f the thermal stability of hydrogen halides (H -.20 is :(a) HI > HC1 < HF > HBr(b) HC1 < HF > HBr < HI(c) HF > HC1 > HBr > HI(d) HI > HBr > HC1 > HF

113. Calomel (H g 2Cl2) on reaction with ammonium hydroxide gives:(a) HgO(b) Hg20(c) NH2—Hg—Hg—Cl(d) HgNH2Cl

114. The number and type of bonds between two carbon atoms in calcium carbide are :(a) two sigma, two pi(b) two sigma, one pi(c) one sigma, two pi(d) one sigma, one pi

115. The oxidation state of chromium in the final product formed by the reaction between KI and acidified potassium dichromate solution is :(a) +3 (b) +2(c) + 6 (d) +4

116. In silicon dioxide :(a) there are double bonds between silicon

and oxygen atoms(b) silicon atom is bonded to two oxygen

atoms

(c) each silicon atom is surrounded by two oxygen atoms and each oxygen atom is bounded to two silicon atoms

(d) each silicon atom is surrounded by four oxygen atoms and cach oxygen atom is bonded to two silicon atoms

117. The lanthanide contraction is responsible for the fact that:(a) Zr and Zn have the same oxidation state(b) Zr and H f have about the same radius(c) Zr and Nb have similar oxidation state(d) Zr and Y have about the same radius

118. The IUPAC name of the co-ordination compound K3[Fe(CN)6] is :(a) Tripotassium hexacyanoiron (II)(b) Potassium hexacyanoiron (II)(c) Potassium hexacyanoferrate (III)(d) Potassium hexacyanoferrate (II)

119. In which o f the following arrangements the order is not according to the property indicated against it ?(a) Li < Na < K < Rb : Increasing metallic

radius(b) I < Br < F < C l: Increasing electron gain

enthalpy (with negative sign)(c) B < C < N < O : Increasing . first

ionisation enthalpy(d) Al3+ < Mg2+ < Na+ < F' : Increasing

ionic size

120. Of the following sets which one does not contain isoelectronic species ?(a) BO j ', CO^, NO 3

(b) S O f, CO|', NOj

(c) CN', N2, C\-

(d) p o J", SO*-, c io ;

1 2 1 . 2-methylbutane on reacting with bromine in the presence o f sunlight gives mainly :(a) l-bromo-3-methylbutane(b) 2-bromo-3-methylbutanc(c) 2-bromo-2-methylbutane(d) l-bromo-2-methylbutane

122. Which of the following compounds shows optical isomerism ?(a) [Co(CN)5]3" (b) [Cr(C20 4)3]3~

(c) [ZnCl4]2- (d) [Cu(NH3)4]2+

123. Which one o f the following cyano complexes would exhibit the lowest value of paramagnetic behaviour ?

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(a) [Co(CN)6]3- (b) [Fe(CN)6] 3'

(c) [Mn(CN)6] 3~ (d) [Cr(CN)6]3“

(Atomic no. Cr = 24, Mn = 25, Fe = 26, Co = 27)

124. The best reagent to convert pent-3-en-2-ol into pent-3-en-2-one is :(a) pyridinium chloro-chromate(b) chromic anhydride in glacial acetic acid(c) acidic dichromate(d) acidic permanganate

125. A photon of hard gamma radiation knocks a proton out of j j Mg nucleus to form :

(a) the isobar of “ Na

(b) the nuclide i 3 Na

(c) the isobar of parent nucleus(d) the isotope of parent nucleus

126. Reaction of one molecule of HBr with one molecule of 1, 3-butadiene at 40° C gives predominantly :(a) l-bromo-2-butene under kinetically

controlled conditions(b) 3-bromobutene under thermodynamically

controlled conditions(c) l-bromo-2-butene under

thermodynamically controlled conditionsCd) 3-bromobutene under kinetically controlled

conditions

127. The decreasing order o f nucleophilicity among the nucleophiles :(A)CH3C — O-

IIO

(B)CH30-

(C) e rro

o(a) (C), (B), (A), (D)(b) (B), (C), (A ), (D)(c) (D), (C), CB), CA)Cd) CA), CB), CC), (D)

128. Tertiary alkyl halides are practically inert to substitution by SN 2 mechanism because of :(a) steric hindrance Cb) inductive effectCc) instability Cd) insolubility

129. In both DNA and RNA, heterocylic base and phosphate ester linkages are a t :

Ca) C5 and c { respectively of the sugarmolecule

Cb) c j and C5 respectively of the sugarmolecule

(c) C2 and C 5 respectively of the sugarmolecule

(d) C5 and C '2 respectively of the sugarmolecule

130. Among the following acids which has the lowest pKa value ?Ca) CH3CH2COOH Cb) (CH3)2CH— COOH(c) HCOOH Cd) CH3COOH

131. Of the five isomeric hexanes, the isomer which can give two monochlorinated compounds is : Ca) 2-methylpentaneCb) 2, 2-dimethylbutane(c) 2, 3-dimethylbutanc(d) n-hexane

132. Alkyl halides react with dialkyl copper reagents to g ive :Ca) alkenyl halides Cb) alkanesCc) alkyl copper halides(d) alkenes

133. Which one of the following methods is neither meant for the synthesis nor for separation of amines ?(a) Curtius reaction Cb) Wurtz reaction(c) Hofmann method(d) Hinsberg method

134. Which types of isomerism is shown by2, 3-dichlorobutane ?Ca) Structural Cb) GeometricCc) Optical (d) Diastereo

135. Amongst the following the most basic compound is :(a) p-nitroaniline Cb) acetanilideCc) aniline Cd) benzylamine

136. Acid catalyzed hydration of alkenes except ethene leads to the formation o f :Ca) mixture of secondary and tertiary alcohols Cb) mixture of primary and secondary alcohols(c) secondary or tertiary alcohol Cd) primary alcohol

137. Which of the following is fully fluorinated polymer ?Ca) PVC Cb) Thiokol(c) Teflon Cd) Neoprene

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138. Elimination of bromine from 2-bromobutane results in the formation o f :(a) predominantly 2-butyne Cb) predominantly 1-butene(c) predominantly 2-butene(d) equimolar mixture o f 1 and 2-butene

139. Equimolar solutions in the same solvent have:(a) different boiling and different freezing

points(b) same boiling and same freezing points(c) same freezing point but different boiling

point(d) same boiling point but different freezing

point

140. The reactionO O

11 . I'R — C — X + Nu —► R — C — Nu + X is fastest when X is :(a) OCOR (b) OC2H5(c) NH2 (d) Cl

141. The structure of diborane (B^-y contains :Ca) four 2C-2e~ bonds and four 3C-2e~ bonds Cb) two 2C-2e‘ bonds and two 3C-3e" bonds Cc) two 2C-2e~ bonds and four 3C-2e“ bonds(d) four 2C-2e" bonds and two 3C-2e~ bonds

142. Which o f the following statements in relation to the hydrogen atom is correct ?Ca) 3s, 3p and 3d orbitals all have the same

energyCb) 3s and 3p orbitals are o f lower energy than

3d orbital(c) 3p orbital is lower in energy than 3d orbital Cd) 3s orbital is lower in energy than 3p orbital

143. Which of the following factors may be regarded as the main cause of lanthanide contraction ?(a) Greater shielding o f 5d electron by 4f

electronsCb) Poorer shielding of 5d electron by 4/

electronsCc) Effective shielding o f one o f 4/electrons by

another in the sub-shell Cd) Poor shielding of one of 4f electron by

another in the sub-shell144. The value o f the ‘spin only* magnetic moment

for one o f the following configurations is 2.84 BM. The correct one is :Ca) d5 Cin strong ligand field)(b) d3 Cin weak as well as in strong fields)Cc) d4 Cin weak ligand field)Cd) d4 Cin strong ligand field)

145. Reaction of cyclohexanone withdimethylamine in the presence of catalytic amount of an acid forms a compound. Water during the reaction is continuously removed. The compound formed is generally known as : Ca) an amine (b) an imineCc) an enamine (d) a Schiff s base

146. p-cresol reacts with chloroform in alkaline medium to give the compound A which adds hydrogen cyanide to form, the compound B. The latter on acidic hydrolysis gives chiral carboxylic acid. The structure of the carboxylic acid is :

CH2COOH

CH,COOH OH ( 1

CH (OH)COOH

CHCOH)COOH OH OH

147. If the bond dissociation energies of XY, X 2 and Y2 Call diatomic molecules) are in the ratio o f 1 :1 : 0.5 and A/7f for the formation of XY is-200 kJ mol-1. The bond dissociation energy of X 2 will be :Ca) 400 kJ mol-1 Cb) 300 kJ mol' 1Cc) 200 kJ mol' Cd) none of these

148. An amount o f solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH3 and H2S gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm ? The equilibrium constant for NH4HS decomposition at this temperature is :

Ca) 0.11 (b) 0.17

Cc) 0.18 Cd) 0.30

149. An organic compound having molecular mass 60 is found to contain C = 20%, H = 6.67% and N = 46.67% while rest is oxygen. On heating it gives NH3 alongwith'a solid residue. The solid residue give violet colour with alkaline copper sulphate solution. The compound is :

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(a) CH3CH2CONH2 (b) (NH2)2CO

Cc) CH3CONH2 Cd) CH3NCO

150. r1/4 can be taken as the time taken for the3

concentration of a reactant to drop to — of its4

initial value. If the rate constant for a first order reaction is k, the t1/4 can be written as :

Ca) 0.75/k Cb) 0.69/kCc) 0.29/k Cd) 0.10/fc

Mathematics

2.

3.

1. If C is the mid point o f AB and P is any point outside AB, then :

(a) PA + PB + PC= O

Cb) PA + PB + 2 PC = 0

Cc) PA + PB = PC

Cd) PA + PB = 2 PCLet P be the point (1 ,0 ) and Q a point on the locus y 2 = &x. The locus of mid point of PQ is: Ca) x 2 - 4y + 2= 0 Cb) x 2 + 4y + 2= 0

(c) y 2 + 4 x + 2 = 0 (d) y 2 - 4 x + 2 = 0

If in a frequency distribution, the Mean and Median are 21 and 22 respectively, then its Mode is approximately :Ca) 24.0 Cb) 25.5Cc) 20.5 Cd) 22.0

Let R = {(3> 3), C6, 6 ), C9, 9), (12,12), C6, 12), (3, 9), (3> 12), (3, 6) } be a relation on the setA = 6, 9,12}. The relation is :(a) reflexive and symmetric only(b) an equivalence relation(c) reflexive only(d) reflexive and transitive only

If A 2 - A + / = 0, then the inverse af A is :(a ) I - A Cb) A - 1(c) A Cd) A + /If the cube roots o f unity are 1, ®, w2, then the roots o f the equation (x - 1 )3 + 8 = 0, are :Ca) - 1 , 1 + 2o), 1 + 2cu2

Cb) -1 ,1 - 2<o, 1 - 2<o2

(c) - 1, - 1, - 1 Cd) - 1, - 1 + 2(0, - 1 - 2w2

4.

5.

6 .

7. lim

n 2 ,+ -=■ sec 1 n

(a) ^ tan 1

(c) - cosec 1

(b) tan 1

(d) - sec 1

8 . Area of the greatest rectangle that can be

x 2 y2inscribed in the ellipse —„ + ^ = 1 is :

a2 b2

(a) - (b) Vab

(c) ab (d) 2ab9. The differential equation representing the

family of curves y 2 = 2 c(x + Vc), where c > 0, is

a parameter, is of order and degree as follows :(a) order 2, degree 2(b) order 1, degree 3(c) order 1, degree 1 .Cd) order 1, degree 2

10. ABC is a triangle. Forces ^ , 4 , R acting alongIA, IB and IC respectively are in equilibrium, where I is the incentre of A ABC. Then P : Q : is :Ca) cos A : cos B : cos C. . . A B C Cb) cos - - : cos - : cos -=

, . . A . B . C Cc) sin ^ : sin ^ : sin ^

Cd) sin A: sin B : sin C

11. If the coefficients o f r th, C r+ l)th and Cr+ 2)th terms in the binomial expansion of Cl + y)'" are in AP, then m and r satisfy the equation :Ca) m2 - mC4r - 1) + 4r2 + 2 = 0(b) m2 - mC4r + 1) + 4r2 - 2 = 0(c) m2 -m (4 r+ 1)+ 4r2 + 2 = 0(d) m2 - m (4r - l ) + 4^ - 2 = 0

r 1 2 1 2 2 4 ( Q \~2 ~2 + - j s e e 2 —j t a n m_ n n n n v 2 ;

12. In a triangle PQR, Z R If tan | —© and

are the roots of ax + fee + e = 0, a * 0 ,

then:

equals:(a) b= a + c (c) c = a + b

(b) b = c (d) a = b + c

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13. If the letters o f the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word SACHIN appears at serial number:(a) 602 (b) 603(c) 600 (d) 601

14. The value o f S0C4 + I S6 ' rC1 is :

(a) 56C4(c) ss,-

(b)(d)

55(55/-

IS. If A ■[' ?: and I = 1 0 0 1 , then which one

of the following holds for all n > 1, by the

principle o f mathematical induction ?

(a) A" = 2"~1A + ( n - 1)1

(b) An = nA + (n - 1)1

(c) A n = 2n~1A - (n - 1)/(d) An = nA - (n - 1)/

16. If the coefficient of x7 in

the coefficient of x "7 in

ax2 +

ax -

(*)] " |ual'

, then

(a) afc = 1

(c) a + b = 1 (d) a - b = l

17. Let / : ( - ! , ! ) —> B, be a function defined by

f ( x ) = tan _1 2x, then / is both one-one

l - * 2and onto when B is the interval:

<*>(-1-1) <■»[-§•! (c) r[°4 (d) N )

18. If Zj and z2 are two non-zero complex numbers such that|z, + z 2|=|z,| + |z2|, then

arg 2, - arg z2 is equal to :

(b) 0r \ W

~ 2

(c) - n

19. Ifw = -

( d ) |

and | w| = 1, then z lies on :

(a) a parabola (c) a circle

(b) a straight line (d) an ellipse

20 . If a2 + b2 + c2 = - 2 and

f ( x ) =1 + a2x (1 + b2)x (1 + c2)x

(1 + a )x l + b2x (1 + c2)x1 + C X

a and b satisfy the relation: .

(1 + a2)x (1 + b2)jc

then f ( x ) is a polynomial of degree :(a) 2 (b) 3(c) 0 (d) 1

21. The system of equations a x + y + z = a - lx + a y + z = a - l x + y + a z = a - l has no solution, if a is :(a) 1 (b) not - 2(c) either - 2 or 1 (d) - 2

22. The value of a for which the sum of the squares o f the roots o f the equation x 2 - (a - 2)x - a - 1 = 0 assume the least value is :(a) 2 (b) 3(c) 0 (d ) 1

23. I f the roots of the equation x 2 - bx + c = 0 be

two consecutive integers, then b2 - 4c equals:

(a) 1 (b) 2(c) 3 (d) - 2

24. Suppose f ( x ) is differentiable at x - 1 and1

25

t fO + h ) = 5, then / ’ (1) equals :—► 0 fl

(a) 6 (b) 5(c) 4 (d) 3

Let/be differentiable for all x. If / ( l ) = - 2 and / ' (x ) £ 2 for x e [ l , 6], then :Ca) / (6) = 5 (b) / (6) < 5(c) / (6) < 8 (d) / (6) S 8

26. If / is a real-valued differentiable function satisfying | f ( x ) - f (y ) ! * ( * - y )2, x ,y e R

and / (0 )= 0, then / ( l ) equals :(a) 1 (b) 2(c) 0 (d) - 1

27. If x is so small that x 3 and higher powers of xmay be neglected, then

(1 + * )>3/2 - H ')(1 - x ? /2

may be approximated

as :

(a)H * 2(c) 3x + | X 2

c b ) - | * ’

(d) 1 - f x 2

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28.

29.

30.

31.

32.

33.

If x = 2 a", y = 2 b", z =n-0 n = 0

I c"1 V 0

where a, b, c are in AP and j fll < 1 .1 b| < 1, | c| < 1, then x, y, z are in :(a) HP(b) Arithmetico-Geometric Progression(c) AP(d) GP

In a triangle ABC, let ZC = it/ 2 . if r is the inradius and R is the circumradius of the triangle ABC, then 2(r+ R ) equals:(a) c + a (b) a + b + c(c) a + b (d) b + c

< 1 yIf cos x - cos 2 ~ a ’ t*ien

4x2 - 4xy cos a + y 2 is equal to :(a) - 4 sin2 a (b) 4 sin2 a(c) 4 (d) 2 sin 2 a

If to a AABC, the altitudes from the verticesA. B, C on opposite sides are in HP, thensin A, sin B, sin C are in :(a) HP(b) Arithmetico-Geometric Progression(c) AP(d) GP

The normal to the curve x = a(cos G + 0 sin G), y = a (sin 0 - 0 cos G) at any point ‘ 0’ is such that:(a) it is at a constant distance from the origin(b) it passes through (o n/2 , - a)(c) it makes angle n/ 2 + 0 with the x-axis(d) it passes through the origin

A function is matched below against an interval where it is supposed to be increasing. Which of the following pairs is incorrectly matched ?

Interval Function(a) (-o o ,-4 ]

(b) ( - 00,1

(c) [ 2, co)(d) ( - co, co)

x + 6x + 6

3x2 - 2x + 1

2x3 - 3x2 - 12x + 6 x 3 - 3x2 + 3x + 3

34. Let a andp be the distinct roots of ax2 + bx + c = 0, then

1 - cos (cur2 + bx + c ) . ,hm ----------------- *------- - is equal to :

( * - « r

(a) ± (<x-p )2

(c) 0

(b) - (a - p)2

(d) ^ ( a - | ) ) 2

35. If x ^ = y (log y - log x + 1)

solution o f the equation is :

38.

then the

(a) log [ “ } = cy (b) log = c;

(c) * log = cy (d) y log = cx

36. The line parallel to the x-axis and passing through the intersection of the lines ax + 2by + 3b = 0 and bx - 2ay - 3a = 0, where (a, b) * (0, 0) is :(a) above the x-axis at a distance of (2/3)

from it.(b) above the x-axis at a distance of (3/2)

from it.(c) below the x-axis at a distance of (2/3)

from it.(d) below the x-axis at a distance of (3/2)

from it.

37. A spherical iron ball 10 cm in radius is coated with a layer of ice o f uniform thickness that melts at a rate of 50 cm3/min. When the thickness o f ice is 15 cm, then the rate at which the thickness o f ice decreases, is :

(a) y— cm/minD7T

(c) — cm/min

(log x - 1 )

+ (log x )2

(a)

(b) -=-r- cm/min 54n

1(d ) g g - cm/min

dx is equal to :

xe*

(c)

(log X )2 + 1

X

x 2 + l+ c

(b)

(d)

+ c1 + X

logx+ c

39. Let f : R - + R be a

(log x )2 + 1

differentiable function

having / (2) = 6, / ' (2) = j .

Then lim ^ dt equals:x -»2 J6 - x - 2 M

(a) 18 (b) 12(c) 36 (d) 24

40. Let / (x ) be a non-negative continuous function such that the area bounded by the curve y = / (x ), x-axis and the ordinates x = n/4and x = p > 7t / 4 is

P sin p + - cos p + V2p). Then / ( j j is:

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’•* O ) ( l - J + Va) (b) [ l - J - ^ 2]

. : (c) ( j - Z i + l j (d ) ^ . . , / 2 - l ]

41. If/, = j^ .d x , l2 = dx,i 0 0

2 2I3 - j 2r dx and l A - [ 2V dx, then :

1 1(a) / 3 > / 4 (b )/..,=/*(c) /, > I 2 (d ) I2 > Ix

42. The area enclosed between the curvey ~ log, ( * + e) and the co-ordinate axes is :(a) 4 (b) 3(c) 2 (d) 1

43. The parabolas y 2 - 4x and x 2 = 4y divide the square region bounded by the lines x = 4, y - 4 and the co-ordinate axes. If S,, S2, S3 are respectively the areas o f these parts numbered from top to bottom, then S, : S2 : S3 is :(a) 1 : 1 : 1 (b) 2 : 1 : 2(c) 1 : 2 : 3 (d) 1 : 2 : 1

44. If the plane 2ax - 3ay + 4az + 6 = 0 passesthrough the mid point of the line joining the centres of the spheres

48.

x + y + z + 6x - 8y - 2z = 13 andx + y + z - lOx + 4y - 2z = 8, then aequals:(a) 2(c) 1

45. The distance

(b) - 2 (d) - 1

between the line

r = 2 i - 2 j+ 3 k + X ( i - j + 4k ) and the

plane "r • ( i + 5j + k ) = 5 is :, , 10 <a) 1. , 10

3V3

(b) ro

46. For any vector a, the value of ( a x i )2 + ("a x j )2 + ( a x k )2 is equal to :

(a) 4 a >2

>2(b) 2 a

(d) 3a>2(c) a

47. If non-zero numbers a, b, c are in HP, then the x y 1

straight line — + — + - = 0 always passes

through a fixed point. That point is :

(b) (1, - 2)

(c) ( - 1, - 2) (d) ( - 1, 2)

If a vertex of a triangle is (1,1) and the mid points of two sides through this vertex ate ( - 1, 2) and (3, 2) , then the centroid of the triangle is:

« ( i 9 (b) f] 7- lC V

(d) U l )V. *>./

49.

7^3 ’ 3)

If the circles x 2 + y 2 + 2ax cy + a - 0 and x 2 + y 2 - 3ax + dy - 1 = 0 intersect in two distinct points P and Q, then the line 5x + by - a - 0 passes through P and Q for :(a) exactly two values of a(b) infinitely many values of a(c) no value of a(d) exactly one value of a

50. A circle touches the x-axis and also touches the circle with centrc at (0, 3) and radius 2. The locus of the centre o f the circle is :(a) a parabola (b) a hyperbola(c) a circle (d) an ellipse

If a circle passes through the point (a, b) and cuts the circle x 2 + y 2 = p2 orthogonally, then

the equation of the locus o f its centre is :(a) 2ax + 2by - (a2 + b2 + p2) = 0-

(b) x 2 + y 2 - 2ax - 3by + (a2 - b2 - p2) = 0

(c) 2ax + 2by - (a2 - b2 + p2) = 0

(d) x 2 -r y 2 - 3cx - 4by + (a2 + b2 - pA) = 0

52. An ellipse has OK as semi minor axis, F and F' its foci and the angle FBF'is a right angle. Then the eccentricity of the ellipse is :

(a, A oa I

< c , i

51.

53. The locus of a point P (« , |S) moving under the condition that the line y = u x + p lsa tangent

x 2 y2to the hyperbola —=■ - — ■ = 1 is :

a b(a) a hyperbola (b) a parabola(c) a circle (d) an ellipse

54. If the angle 0 between thex + 1 y - 1 z - 2_ 1 2” ” ~~2~ and the

line

plane

12x - y + -JXz i 4 = Ois such that sin 0 = The

value of X is :

(a)

C O , |

Cb) |

(d ) |

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55. The angle between the lines2x = 3y = - z and 6x = - y = - 4z is :(a) 30° (b) 45°(c) 90° (d) 0°

56. Let A and B be two events such that

57.

58.

P (A u B ) = |, P ( A n B ) = i and P (A ) = i

where A stands for complement o f event A. Then events A and B are :(a) mutually exclusive and independent(b) independent but not equally likely(c) equally likely but not independent(d) equally likely and mutually exclusive Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house, is :(a) 7/9 (b) 8/9Cc) 1/9 (d) .2/9

A random variable X has Poisson distribution with mean 2. Then P (X > 1.5)equals :

_2(a) (b) 1 - - J

(c) 0 « • ) 4e

59. Two points A and B move from rest along a straight line with constant acceleration / and / ' respectively. If A takes m sec more than B and describes 'n' unit more than B in acquiring the same speed, then :

(a) ( / ' ■ / )n = \ j f ™2

(b) f ' ) m = f f ' n 2

(c) ( / + / ’ )m2 =* jf f ' n

(d) i f - f W = f f ' n

60. A lizard, at an initial distance of 21 cm behind an insect, moves from rest with an acceleration of 2 cm/s2 and pursues the insect which is crawling uniformly along a straight line at a speed o f 20 cm/s. Then the lizard will catch the insect after:(a) 24 s (b) 21 s(c) 1 s (d) 20 s

61. The resultant R of two forces acting on a particle is at right angles to one o f them and its magnitude is one third o f the other force. The ratio of larger force to smaller one is :(a) 3: 2>/2 (b) 3 :2(c) 3 :V 2 (d) 2 : 1

62. Let a = i - k , b *= x i + j + ( l - x)fe and

l:= ,y i + .x j+ (l + jc r-y)k. Then [a T> 7?]

depends o n :

(a) neither x nor y (b) both x and y (c) on ly* (d) onlyy

63. Let a, b and cbe distinct non-negative numbers. If the vectors ai + aj + ck, i + k and

ci + cj + fck lie in a plane, then c is :

(a) the harmonic mean of a and b(b) equal to zero(c) the arithmetic mean of a and b(d) the geometric mean of a and b

—> —> —*64. If a , b , c are non-coplanar vectors and A. is

a real number, then

[X (a + T > ) X2 T> \~c] = [~ 2 1? + ^ 1?]

for:(a) exactly two values o f X(b) exactly three values o f X(c) no value of X(d) exactly one value o f X

65. A and B are two like parallel forces. A couple of moment H lies in the plane of A and B and is contained with them. The resultant of A and B after combining is displaced through a distance :

H (b) H(a)

(c)

A - B H

A + B (d)

2(A + B) 2 H

A - B

66 . The sum of the series 1 1 11 +

(a)

Cc)

4-2! 16-4!e+ 12/e

e+ 1■fe

64-6!

(b)

(d)

+ ...00 is :

e - 12Jee - 1

67. Let xlt x 2, ........x n be n observations such thatI x f - 400 and Z x t = 80. Then a possible value

o f n among the following is :(a) 12 (b) 9(c). 18 (d) 15

6 8 . A particle is projected from a point O with velocity u at an angle o f 60° with the horizontal. When it is moving in a direction at right angle to its direction atO, then its velocity is given b y :

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(a) V3

M J

<b> T

« s69. if both the roots o f the quadratic equation

x 2 - 2 kx + Jc2 + i fc -5=0 are less than 5, then

k lies in the interval:(a) (4. 5] (b) ( - oo, 4)(c) (6, oo) (d) (5, 6]

70. If aj.a2.a 3, ........a„,... are in GP, then thedeterminant

loga n - 1 tog0*, 2 log a,, * 4 lo g a „ .s log a,,* 7 log Op 4 g

A =lOgGft

log 0,1-3loga„ T6

is equal t o : (a) 2 (c) 0

(b) 4(d) 1

71. A real valued function / (x ) satisfies the functional equation

/ (* - y ) - / (* ) / (y ) - /(a - * ) / (° + y )where a is a given constant and / (0) = 1, / (2a - x ) is equal to :(a) / (- x ) (b) /(a) + /(a - x )(c) / (x ) (d) - / (x )

72. If the equationan x " + an_1xn~* + ...... + a,x = 0,

* 0, n > 2 , has a positive root x = a, then the equation

na^cn 1 + (n - l)a„ ,x" 2 + ... + a, = 0 has a

positive root, which is :(a ) equal to a(b) greater than or equal to a(c) smaller than «(d) greater than a

73. The value of

(a) 2n

, s 71M 5

JJcos2 Xdx, a > 0, is :

( b ) f

(d) ait

74.

75.

The plane x + 2y - z = 4 cuts the sphere x 2 + y 2 + z2 - x + z - 2 = 0 in a circle of

radius:(a) J l (b) 2(c) 1 (d) 3If the pair o f lines ax2 + 2(a + b)xy + by2 = 0

lie along diameters of a circle and divide thecircle into four sectors such that the area of oneof the sectors is thrice the area o f anothersector, then:(a) 3a2 + 2ab + 3 b2 =-0

(b) 3a2 + lOab + 3b2 = 0

(c) 3a2 -2 a b + 3 b 2 =0

(d) 3a2 - lOab + 3b2 = 0

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A n s w e r s

PHYSICS AND CHEMISTRY1. (d) 2. (d) 3. (a) 4. N9. (d) 10 . (c) 1 1 . (d) 12. 0 )

17. (c) 18. (d) 19. (b) 20. (c)25. (c) 26. (d) 27. (b) 28. (d)33. (C) 34. (c) 35. (a) 36. (b)41. (a) 42. (a) 43. (c) 44. (c)49. (b) 50. (b) 51. (d) 52. (d)57. (a) 58. (b) 59. (a) 60. (c)65. (c) 66. (b) 67. (c) 68. (c)73. (d) 74. (a) 75. (c) 76. (d)81. (c) 82. (d) 83. (d) 84. (b)89. (d) 90. (a) 91. Cb) 92. (c)97. (a) 98. (d) 99. (b) 100. Cb)

105. (d) 106. (d) 107. (a) 108. CO113. (d) 114. (c) 115. (a) 116. (d)12 1 . (c) 122. (b) 123. (a) 124. (b)129. (a) 130. (c) 131. (c) 132. (b)137. (c) 138. (c) 139. (b) 140. (d)145. (c) 146. (c) 147. (d) 148. (a)

5. (d) 6. (c) 7. Cc) 8. (*>13. Cb) 14. Cd) 15. (d) 16. (a)2 1 . (a) 22. (b) 23. Ca,d) 24. Cc)29. (c) 30. Cb) 31. Cc) 32. (a)37. (d) 38. Ca) 39. (b) 40. (a)45. (c) 46. (b) 47. (d) 48. (b)53. (a) 54. Ca) 55. (a) 56. (d)61. (b) 62. (a) 63. Cd) 64. (c)69. (a) 70. (c) 71. Ca) 72. (d)77. Cc) 78. (a) 79. (C) 80. (b)85. (b) 86. (a) 87. (a) 88. (d)93. Cd) 94. <e> 95. Cd) 96. (a)

101. (b) 102. (b) 103. (a) 104. (c)109. (c) 110 . (a) 1 1 1 . Cb) 112. Cc)117. (b) 118. (c) 119. (c) 120. (b)125. (b) 126. (c> 127. Cb) 128. Ca)133. (b) 134. Cc) 135. (d) 136. Cc)141. (d) 142. Ca) 143. (d) 144. Cd)149. (b) 150. (c)

* No option is matching

*► MATHEMATICSi: Cd) 2 . Cd) 3.9. (b) 10. (b) 1 1 .

17. (a) 18. (b) 19.25. Cd) 26. Cc) 27.33. (b) 34. Cd) 35.41. (c) 42. (d) 43.49. (c) 50. Ca) 51.57. (c) 58. Cb) 59.65. (c) 66. (a) 67.73. (c) 74. Cc) 75.

(a) 4. (d) 5. (a)(b) 12. (c) 13. (d)(b) 20. (a) 21. (d)(b) 28. (a) 29. (c)(b) 36. (d) 37. (c)(a) 44. (b) 45. (c)(a) 52. (d) 53. (a)Ca) 60. (b) 61. (a)(c)(a)

68. (a) 69. Cb)

6. Cb) 7. (a) 8. Cd)14. (a) 15. (d) 16. (a)22. (d) 23. (a) 24. Cb)30. (b) 31. (c) 32. Ca.c)38. (a) 39. (a) 40. Ca)46. (b) 47. Cb) 48. (b)54. (d) 55. (c) 56. (b)62. (a) 63. <d) 64. Cc)70. (c) 71. (d) 72. (c)

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76. [Cr(NH3)4Cl2]+Let oxidation state of Cr - x

NH3 = 0 Cl = - 1

Net charge = + 1

[ Cr(NH3 )4Cl2]+ x + 4 x 0 + 2( - l ) = + l

x - 3

77. Antipyretic drugs reduce fever. Analgesic releives in pain, antibiotics act against bacterial infections while tranquilisers are used against mental disorders.

78. A species is amphoteric if it is soluble in acid (behaves as a base) as well as in base (behaves as an acid).

Sn02 +4HC1— -» SnCl4 + 2H20Basic Acid

Sn02 +2NaOH---- > Na2Sn03 + H20Acid Base

79. A species is said to be diamagnetic if it has all electrons paired.

Species ElectronsMO

Electronicconfiguration

Magneticbehaviour

3 a ls 2 a* Is1 Paramagnetic

h ; 1 crls1 Paramagnetic

H 2 2 als2 Diamagnetic

He 2' 3 o'ls2a*ls I Paramagnetic

80. Na2S04 2Na+ + S04~ van’t Hoff factor i = [1 + (y - l )a ]where y is the number of ions from one mole solute, (in this case =3 ), a the degree of dissociation. i = (1 + 2a )

81. Nylon-66 is a polyamide of hexamethylene diamine (CH2)6(NH2)2 and adipic acid (CH2)4(COOH)2.(Each reactant has six-carbon chain hence trade code (66) is used.)

82. Free radicals have unpaired electrons but are neutrals and are reactive.

CH3 + CH3 ---- > CHj — CH3

83. AG° = - 2.303RT log Keq AG° = - nFE°eU ..

If a cell reaction is spontaneous (proceeding in forward side), it means

Keq > 1 and E°eM =+ ve

Thus AG° = - ve84. 4 jH -> 2He+ 2$*e+ energy

It is the principal reaction of Hydrogen-bomb.

85. Unit cell consists of A ions at the comers.Thus number of ions of the type

Unit cell consists of B ions at the centre of the six faces.

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Thus number o f ions o f the type

8 = | . 3

[Each comers is shared by 8 cubes and each face is shared by 2 faces]Thus formula is AB3.

86 . Fluoro group causes negative inductive effect increasing ionisation, thus 0.1 M difluoroacetic acid has highest electrical conductivity.

H O

F—«- C —4-C — « - 0 — «-H

J87. Greater the charge, smaller the radius, greater

the polarising power and thus greater the covalent nature. This leads to increase in lattice energy.

88. X -> Y is an endothermic reaction AH = + ve

CT>

OCLU

u/ A

J 1AH

Ef — £(, + AH Ef > E b

Eb = energy o f activation o f backward reaction Ef = energy o f activation o f forward reaction

AH = heat of reaction Thus Thus

89. Al3+ + 3e‘ -► A1iv = zQ

where w = amount of metal= 5.12 kg= 5.12 x 103g

z = electrochemical equivalent _ Equivalent weight Atomic mass

96500 Electrons x 9650027

3 x 96500

-v3 275.12x 103 =3x96500

xQ

n _ 5 .1 2 x l0 3x 3x96500 „= 27 C= 5.49 x 107 C

90. Size o f colloidal particles = 1 to 100 nm (say 10* nm)

V'c = | jtr3 = | 7t(10)3

Size of true solution particles a 1 nm

Thus

V ^ T t d ) 3

77 = 103

91. AH = AU + AngRT

AH = enthalpy change (at constant pressure) AU = internal energy change (at constant

volume) (given reaction is exothermic) Ang = mole of (gaseous products - gaseous

reactants)= - ve •

Thus AH < AU.Note : Numerical value of AH < AU in exothermic reaction and when Ang < 0

■M2* + 2X2s

f2+ir v~"i2K ,p = lM z+] l X ' rIf solubility be s then

* Jp= (s )(2 s )2 =4s3

4s3 = 4 x 10"12

••• s = 1 x 10‘ 4 M

M 2+ = s = 1 x 10~4 M

93. Mixture contains 78g benzene = 1 mole benzene and 46 g toluene = 0.5 mole tolueneTotal mole of benzene and toluene = 1.5 mol

Mole fraction of benzene in mixture = — = -1.5 3

VP of benzene p i = 75 torr

.-. partial vapour pressure of benzene = pi Xb

= 75 x - = 50 torr

94. Distribution o f molecules (N) with velocity (u) at two temperatures 7\ and T2 (T2 >T ,) is shown below :

At both temperatures, distribution of molecules with increase in velocity first increases, reaches a maximum value and then decreases.

95. 2N02(g )^ = * 2 N 0 (s )+ o 2(* )

Kc = 1.8 x 10"6 at 184°C (= 457 K)

K = 0.00831 kJm or1 K_1

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Kp = K c ( R T p

where,. Ang = (gaseous products - gaseous reactants)

= 3 - 2 = 1 . . k p = 1.8 x l0 “6 x 0.00831 x 457

= 6.836 x 10-6 > 1.8 x 10~6

Thus K„ >K,.r L96. Reaction is exothermic. By Le-Chatelier

principle, a reaction is spontaneous in forward side (in the direction of formation of moreClF3) if F2 is added, temperature is lowered and C1F3 is removed.

97. pH =5.4

[H+]= 10 '54 = 10“6 100'6

Antilog of 0.6 is w 4[H+] « 4x 10~6 M

98. There are two different reactants (say A and B).

A + fl —- * product Thus it is a bimolecular reaction.

If £ = k [A ][B ]

it is second-order reaction

,f ( § ) - « *

or = k [B]it is first order reaction.Molecularity is independent of rate, but is the sum of the reacting substances thus it cannot be unimolecular reaction.^ , , ■ Af,V, + M 2V299. Total molarity = — — ..

vi + 21.5x480+ 1.2x520

~ 480+ 520= 1.344 M

100. During electrolysis, noble metals (inert metals) like Ag, Au and Pt are not affected and separate as anode mud from the impure anode.

101. AacOH = AcONa + AHC1 - A-NaCI= 91.0+ 426.2- 126.5 = 390.7

102. Remains unchanged.

103. (A) Is (B) 2s(C) 2p (D) 3d(E) 3dIn the absence of any field, 3 d in (D) and (E) will be of equal energy.

104. As we go down in the group, ionic character increases hence, melting point of halides should increase but NaCl has the highest melting point (800°C) due to its high lattice energy.

105. Variation of Keq with temperature T is given by van’t Hoff equation.

. v AH° A S°g er> 2.303 RT + R

A BSlope of the given line is + ve indicating that term A is positive thus AH° is - ve.Thus reaction is exothermic.

106. Following reaction takes place during bessemerisation :

2Cu20 + Cu2S---- > 6Cu + SO 2107.

Molecule Structure Hybridisation of central

atom

Lonepair

F

sf4J.

F — S— F1FF

sp3d one

CF, F— C— F1FF

sp3zero

XeF4 1F— :Xer— F

1F

sp3d2 two

108. Mixing the soles together can cause coagulation since the charges are neutralised.

109. Hypophosphorus acid (H3P 0 2) is a monobasic acid and has only one ionisable H, two H-atoms are directly attached to phosphorus. Thus the correct statement is (c).

H

H — P— O— HIIO

110. Conjugate base is formed by loss of H*.OH- ---- > 0 2' + H+

Conjugatebase

O2* is the conjugate base of OH“.

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111. Aqueous solution of A1C13 is acidic due to hydrolysis.

A1C13 + 3H20 Al(OH)3 + 3HC1

On strongly heating Al(OH)3 is converted into AljO j j

2A1(0H)3 A120 3 + 3H20

112. In a group, AG^° (fix ’) changes from - ve to + ve downwards.H F(g )AG = - 273 .20 kJ mol' 1 H l(g )A G = + 1.72 kJ mol"1

Thus HF is thermally stable and HI not.Thus HF > HCI > HBr > HI

113. Hg2Cl2 + 2NH3 - * HgNH2Cl + Hg + NH4C1white black

114. CaC2 (Calcium carbide) is ionic.CaC2 ---- > Ca2+ + C2“

"C = C“C?~ has one a and two n bonds.

120.

115. Cr2o|- + 14H+ + 61“ - » 2CrJ+ + 7H20 + 3I23+

Ci20 2' is reduced toCr Thus final state of Cr is + 3. Hence, (a)

3+-

116. IO

. Su

IO

0 A 0 A o '

118.

119.

Species Electron in central element

Electrons in other element

Chargegained

Total

BOf 5 3 x 8 = 24 + 3 32c o r . 6 3 x 8 = 24 + 2 32N03 7 3 x 8 = 24 + 1 32S05- 16 3 x 8 = 24 + 2 42CN‘ 6 7 1 . 14n 2 7 7 0 14p2r- 6 6 i + 2 14?o 34- IS 4 x 8 = 32 + 3 50so^- 16 . 4 x 8 = 32 + 2 SOCIO4 17 4 x 8 = 32 + 1 50

Thus (a) B03‘ , CO 3 , N 0 3 are isoelectronic. (b) SO^', CO 3", N 0 3 are not isoelectronic.

H H ,

3° (C—H) bond hasminimum bond energy hence easily cleaved giving 2 bromo 2-methyl butane

121. H3C— C— C— CH,I IH CH3

Br

CH, — CH, — C— CH,

CH,

117. Lanthanide contraction, cancels almost exacdy the normal size increase on descending a group o f transition elements, thus Nb and Ta and, Zr and Hf have same covalent and ionic radii.

K3 [Fe(CN)6]Cation AnionOxidation state o f Fe in anion = + 3 Thus it is potassium hexacyanoferrate (III).(a) Metallic radii increase in a group from top

to bottom.Thus Li < Na < K < Rb —True

(b) Electron gain of enthalpy of Cl > F and decreases along a group Thus I < Br < F < Cl is true.

(c) Ionisation enthalpy increases along a period left to right but due to presence of half-filled orbitals in N, ionisation enthalpy o f N > O.Thus B < C < N < O i s incorrect.

122.

C2O4

C2O43 -

123.

C204

Mirror image is not superimposable hence, optical isomerism is possible.

Hybridi- Unpaired Magneticsation electrons moment

(a) [Co(CN)6]3‘ d2sp3 0 0

(b) [Fe(CN)6]3~ d2sp3 I >/3BM

(c) [Mn(CN)6] 3' d2sp3 2 yfs BM

(d) [CrtCI'OJ3- d2sp3 3 VlS BM

Thus least paramagnetism is in (a).

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124. CH, —CH — CH = CH — CH,IOH

CH3 — C— CH = C H — CH3

Only suitable reagent is chromic anhydride in glacial acetic acid. Other will also effect (C =C )b on d .24.125. £ Mg + y -----► { H + j2,3 Na

Thus nuclide o f ?3Na is formed.

126. CH2 = CHCH = CH2 + HBr---- »

CH3 CHCH = CH2 + CH3CH = = CHCH2Br

Br1,2-addition product 1 ,4-addition product

Addition is through the formation of allylic carbocation

CH2=CHCHCH3 <­(2* allylic)

(more stable)

c h 3c h = c h c h 2(1* allylic)

(less stable)

Under mild conditions (temperature « - 80° C) kinetic product is the 1, 2-addition product and under vigorous conditions (temp.» 40° C) thermodynamic product is the 1, 4-addition product.Thus l-bromo-2-butenc is the major product under given condition.

127. If acid is weak, its conjugate base (nucleophile) is strong and vice versa.(A) O

I!CH3 — C— O- is a conjugate base of

OII ■

CH3COH(I)

(B) CH30 ' is a conjugate base of CH3OH (II)(C) CN~ is a conjugate base o f HCN (III)

(D) H,C SO 3 is a conjugate base

S03H(IV)

Acidic nature of IV > I > III > II and nucleophiliciiy o f B > C > A > D

128. In Sjvj 2 reaction, nucleophile and alkyl halide react in one step.

R — C— Br+ Nu"I

\ / *N u ....... C ........Br

K RThus tertiary carbon is under steric hindrance thus reaction does not take place until (C—Br) bond breaks

R R

R — C— Brslow

» R— C + Br'

R Rwhich is then SN1 reaction.

129. Synthesis o f RNA/DNA from phosphoric acid, ribose and cytosine is given below Thus ester linkages are at Cs’ and C / of sugar molecule.

DehydrationOH / \ 5 \I ' *-

0=P-OH

i)H

OH ..I 5

0 =»p-0-CH, IOH

OH OH

130. Out of the given acids, strongest is HCOOH. (highest Ka value)Since pKa = - log KaThus lowest pKa is of HCOOH.

cl 2131. (A ) 2-methyl pentane------- > five types of

monochlorinated compounds

(B) 2, 2-dimethylbutanecl,

-> three types....

(C) 2, 3-dimethyIbutane -» two types .....

(D) n-hexane ■cl,

-> three types....

132. It is Corey House synthesis o f alkanes.

133. Wurtz reaction is used to prepare alkanes from alkyl halides :

2 R — X + 2 N a -d? e— >R — R + 2NaX

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134. CH3— CH— CH— CH3

Cl Cl

There are two chiral C-atoms (* ). Thus optical isomerism is possible.

135. (a) I QNC>2 group (electron with

-drawing) decreases basic nature o f aniline.

(b) CH3CNH2oII

CH3C group is also electron withdrawing.

— NH2 phenyl group is also electron

withdrawing.

(d) Benzyl is electron-repelling group increases basic nature.

Thus most basic compound is benzylamine. h2o/h*

136. CH2 = CH2 ----------- > CH3CH2OH

n2o/irCH3— CH = CH2 ----------->CH3CHCH3

IOH

$(2° alcohol) through 2° carbocation CH-.CHCH,

HjO/II*CH3 — c = CH2 ---------- »(CH3)3COH

CH3

(3° alcohol) through 3° carbocation (CH3)3C CH, •

CH3 — CH— CH = CH2

CH3

— > Cf UCH— CH— CH,3 © 3 2° carbocation

I, 2 H~

CH

shift

I

^2° > 2° alcohol

CH3— C— c h 2— c h 393

3°carbocation Thus best alternate is (c)

H ,0► 3° alcohol

137. Teflon is -fCF2 — CF2-}-n

138. CH3CH2CHCH3 ---- > CH3CH2CHCH3 •

Br

CH3CH2CH — CH2 + CH3CH = CHCH3Less substituted More substituted

II IStability o f I > II hence I is predominant.

139. Boiling point and freezing point depend on Kb (molal elevation constant) and K; (molal depression constant) of the solvent. Thus equimolar solution (o f the non-electrolyte) will have same boiling point and also same freezing point.

A7y - K f x molality ATb = K b x molality

Note : In question (139) set C, nature of solute has not been mentioned. Hence, we have assumed that solute is non-electrolyte.

O OII II

140. rt— C— X + N u '----- ->R — C— Nu + X~

Best leaving group (poorest nucleophile) is Cl0, thus fastest reaction is with Cl.

141. B2H6 has structure

Bridging bonds (3C-2e~)

Terminalbond2C-2o~

• Boron electron ° Hydrogen electron

Bridging bonds (3C-2e-)

142. Hydrogen atom is in Is' and these3s, 3p and 3d orbitals will have same energy w.r.t. Is orbital. .

143. Lanthanide contraction is due to poor shielding of one o f 4/ electron by another in the sub-shell.

144. (a) ds in strong field

n - unpaired electron = 1

n u b 2 g

Magnetic moment = Jn(n + 2)BM

= a/3 BM = 1.73 BM

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(b) d3 in strong/weak field

n = 3Magnetic moment = Vl5 = 3.87 BM

(c) d4 in weak field

MU e«O l i l 2 9

n = 4

Magnetic moment = V24 = 4.90 BM(d) d4 in strong field

U 'riM i n

n = 2Magnetic moment = V8 =2.83 BM

145.

3 . CH3enamme

ch 3

CH,

146. O + c h c 1 3 + O H_ Reimer-Ticmann reaction

p-cresol

(—OH is more activating than — CH3 ino, p-directing, thus — CHO goes to ortho w.r.t — OH)

147. Formation of XY is shown as

X 2 + Y2 -----> 2 XYAH = (BE)X_ X + (BE)y_ y - 2(BE)x_ y

If (BE) of X — Y = a then (BE) of {X — X ) = a

and (BE) of (Y — Y) = |

.-. AHf ( X - Y ) = - 200 kJ

.-. - 400 (for 2 mol XY) = a + - - 2a

148.

-4 0 0 = - |

a = + 800 kJ The bond dissociation energy of X 2 = 800 kJ mol-1

NH.,HS(s)*=^ NH3fe ) + H2S(g) initially 1 o.5 0

at equilibrium a - x) (05 + x) x

Total pressure at equilibrium

= Pn h 3 + P h 2s

= 0.5+ x + x = 0.84 x = 0.17 atm

Pnh3 = 0.50+ 0.17= 0.67 atm P h 2s =0.17 atm

■ - K p ~ Pn h 3 -Ph2s

149. Element Percentage Percentage SimpleRatioat. wt.

C 20.0 20 0 - i -53— 1.66 1.66 1.66 "

H 6.67 = 667 667 „ 1.66 =

N 46.67 * f = 3 . 3 3 3331.66 = z

O 26.66 2 6 6 6 1AA 16 - 1-66

1.661.66

Empirical formula = CH4N20 Empirical formula weight

= 12+ ( 4 x l ) + (2x14 )+ 16 = 60

. _ Mol, formula weight _ 60'' n Emp. formula weight 60 - *

Molecular formula = CH4N20 Given compound gives biuret test. Thus given compound is urea (NH2)2C0.

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Mathematics

1. Key Id ea : I f C is mid point ofAB and P be the- » —> PA + PB

origin, then PC=

Let P be the origin outside of AB and C is mid point of AB, then

pfc PA + PB

=> 2PC = P A + P B

2. The co-ordinates o f f are (1, 0). A general point Q on y 2 = 8x is (212, 4t> Mid point of PQ is (h, k), so

2fc = 2 ta + l ...(i)

and 2fc = 4 t = > t = f c / 2 ...(ii)On putting the value o f t from Eq. (ii) in Eq. (i), we get

2k2 2h = ^ - + l

4=> 4 h = k2 + 2

So the locus o f (ft, fc) is y 2 - 4x + 2= 0.

3. Key Idea : The relation among Mean, Mode and Median of any frequency distribution is

Mode = 3 Median - 2 Mean .

Given that Mean = 21 and Median = 22 v Mode = 3 Median - 2 Mean

Mode = 3 (2 2 )-2 (21).= 66 - 42 = 24

4. Since, (3, 3), (6, 6), (9, 9), (12, 12) eR=> R is reflexive relation.Now (6,12) e R but (12, 6) (R , so it is not a

symmetric relation.

Also (3, 6), (6,12) eR => (3, 12) 6 R => R is transitive relation.Note : Any relation is said to be an equivalence relation, if it is reflexive, symmetric and transitive relation simultaneously.

5. Key Id ea : I f A is any square matrix, then

A A 1 = / and A 1! = A 1.

Since, A 2 - A + J = 0

=> A -1 A2 - A~lA + A -1 7 = 0

(Pre multiply given relation by A-1)(A _1A ) A - (A -1 A ) + A~‘ = 0

A - / + A -1 = 0

i-i

A 1 = I - A

6. Since (x - 1)J + 8 = 0

=> ( x - 1 ) 3 = - 8 = (-2 )3\ 3

= 1m-2n = ( l ) I/3

roots o f | | are 1, © and ©2.

=> roots o f (x - 1 ) are -2, - 2© and - 2©2

=> roots roots of x are -1 ,1 -2 © and 1 - 2© .

Note : If 1, o) and o>2 are cube roots of unity, then1 + to + to2 = 0 and <a3 = 1

7. Let A sec1

sec

+ ...+ i sec2 l l

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f 1 2 ( i f 2 2 ( 2\- sec -- + - sec

[n n J n

.....+ 5r f(5)'= lira - Z U ) sec2 ( t f

n j) ti r - i \ ny

A = J x sec2 (x 2) dx

Let x 2 = f

2x dx = dc xdx = dt

A = ^ f1 sec2 t dc2 Jo

= | [tan c]Jj = - tan 1

(-acosO, bsinO) B A (a cosfl, b sinO)

(- a cosO. - b sinO) (a cos8. - b sinO)

^ = 2 x 2 ab cos 20 aO

On puiting = 0 for maxima or minima.

— = 0 dO

cos 20 = 0

Now

Now

2 0=2 => 0 = £ 2 4

= - 8ab sin2 0dO2

d VdO2

<0

8. Key Id ea : The parametric co-ordinates of ax 2 y 2point that lies on an ellipse —5- + ^ = 1 area b

(a cos 0, b sin 0) .

Let the co-ordinates o f the vertices o f rectangle ABCD are A (a cos 0, b sin 0),B ( - a cos 0, b sin 0), C ( - a cos 0, - b sin 0) and D (a cos 0 , -b sin 0), then length of rectangle, All - 2a cos 0 and breadth of rectangle, AD -2 b sin 0

Area of rectangle = AB x AD- 2a cos 0 x 2b sin 0

Area of rectangle, A = 2ab sin 20 ...(i)

Area is maximum at 0 = —.4

=> Maximum Area o f rectangle = 2 absq unit.(From (i))

Alternate SolutionFrom Eq. (i)

Area of rectangle, A = 2ab sin 20 v A <x sin 20 and - 1 5 sin 20 5 1

A is maximum when sin 20 = 1 => Maximum area o f rectangle = 2absq unit.

9. Key Idea : The differential equation of a family of curves ofn parameters is a differential equation of n maximum order.Equation o f family of curves is

y 2 = 2 c (x + Vc) ...(i)

On differentiating Eq. (i) with respect tox, then

2yy-i = 2c

=> c ~yy\On putting the value of c in Eq. (i), we get

y 2 = 2 >yi (x + yfyy^)

=> Cy2 - 2yyj x )2 - 4 (yyxf

:. The degree and order o f above equation are 3 and 1 respectively.

10. Key Idea : I f three forces acting on a particle keep it in equilibrium, each is proportional to the sine of the angle between the other two.• P = Q «

sin a sin p ~ sin y

This theorem is known as Lami's theorem.

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14. Key Idea : nCr + nCr . 1 = m" C r t l .

Now, »°C4 + SSC3 + S4,c , + 53c , + 52c ,+ S1t 3c , + soc 3

= S0c 3 + S0c 4 + 5,c 3 + 52c 3 + S3C+ 54 3c , + s5c .

= 51c„ + Slc 3 + S2c 3 + 53c 3 + S4c 3 + ssc ,(v ' t r + l'Cr-i - n+1C ,)

= 52C< + S2C, + WC , + 54C3 + “ C;SS/-

= 53c 4 + S3c 3 + S4c 3 + S5C: = S4c 4 + 54c 3 + 5Sc 3= ssc . + 5SC, = S6C,

15. A = [\ O'

a 2 =

a 2 =

a 3 =

1 0 1 11 0 2 11 0 2 1

?]

[!?1

f. ?]~ n ? ] can vcr’ iec by induction.

Now, go option by option

® [1 ?M n ^ [ " o 1 n - , ]

(d) n A - (n - 1 )/ = [ ” „ ° 1

. r i ° i* Ln 1J

16. Key Idea : The (r + l ) th term in the expansion

o/(x + a)n is "Cr x n r ar.

Let x7 is contained in (r+ l)th term in the

( 2 1 V*■expansion of I ax + .

22 - 3 r

Similarly, coefficient of x 7 in the expansion of

( 1 V1 - Hrr ' b * 2J " V

According to question,„b „s

ii/i ^ _i i f QC s ¥ ~ 6 ¥

a6 _ as

** b5 b6=$ ab = 1

17. Since x e (-1 ,1 )

tan 1 x e

=> 2 tan 1

and f ( x ) = tan-1

( — C 4 ’ 4j

H ’i)2x

1 - x J= 2 tan x, (x 2 < 1) .

So, / £ * ) « ( - 5 . 5 )

Function is one-one onto.

18. Let Zj = x, + iyi and z2 = x 2 + iy2 Now given that

|z, + z2| =| zx| + 1 z2|

V(x, + x2)2 + (y , + y 2)2

= V*? + y? + V^I + y f

=> x f + x£ + 2 x , x 2 + yi2 + y l + 2y ty 2

= x 2 + y f + x| + y j + 2 V (x f + y 2) ( x f + y|)

=> (X,x2 + y xy 2) = + y l ) ( x j + y l )

x M + y,2y 2 + 2xjx2y ]y 2

= *i2* I + + *?y2 + y f r l

* iy 2 - y i* 2 = ° >2 _ y\

C' V '=> 22 - 3r = 7=* 3r = 15=> r = 5

. r - nc — x7•• Jfi- L 5 ^ s - X

Coefficient o f x7 in the expansion of

f 2 1 V * _ 1 1 ^ fl6

+ to j “ C5F

<X2J V*l.=> arg z2 = arg zt=> arg z2 - arg z, = 0

Alternate Solutionv |z, + z 2|=|zj| + |z2|On squaring, we get|Z i + Z 2|2 =(| z,| +|z2|) 2 _

=> | Zj |2 + 1 z2|2 + 2Re (ZjZ2)

= |z, |2 + | z 2|2 + 2 |zt||z2|

=> R e(z1z2)=|z, ||z2|=> | Zi 11 z2 I cos (0j - 02) = | Zil | z21

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0, - 02 - 0 a rg (z , )-a r g (z 2)= 0

19. Key Idea : Ifz - z,

= k, then

2 - 3

= 1 => |z| =Z 3

=> z lies on X bisector of (0, 0) and

So, z lies on a straight line. Alternate Solution

K)-

W = -I

Z ~ 3

and|w | = 1

= 1

=> 31 z | = | 3z — i |Let z = x + iy/. 3| x + iy | =| 3 (x + ty) - 1|

=* 3V (x2 + y 2) = V (3*)2 + (3y - 1)2

=* 9x2 + 9y2 = 9x2 + 9y2 + 1 - 6y1

=* y = 6 /. Which shows that z lies on a straight line.

1 + G2 X (1 + b2)x (1 + c2)x20. /(x ) = f l + a2)x 1 + b2 x ( l + c2)x

(1 + a2)x (1 + b2)x 1 + c2 x

Applying C, - » + C2 + C3

/ (* ) =

I T U2 X + X +' b2 X + X + c2 A

x + a2 x + l + b2 x + x + c2 x

x + a2 x + x + b2x + 1 + c2 x

(1 + b2) x

1 + b2 x

(1 + b2) x

a + O x

(1 + c2) X

1 + c2 X

1 + (a + b + c + 2) x (1 + b2) x (1 + c2)x1 + (a2 + b2 + c2 + 2) x 1 + b2 x (1 + <?)x1 + (a2 + b2 + c2 + 2) x (2 + b2)x 1 + c2x

0 + b2) x

1 + b2 x

(1 + b2) x

(1 + c2) x

a + c2)x

1 + c2 x

z lies on a circle o f k * 3 and z lies on a perpendicular bisector of segment with end points Zj and z2, i f k = 1.

Given that w = ——r and I w I = 1i

Ki —> R, - K3, R2~* R2 - R30 0 x — 1

0 1 - x x - 1

1 (1 + b2) x l + c2 x0 x - 1

1 - x x -1

= ( x - l ) 2=> / (x ) is of degree 2.

21. The system of given equation has no solution, if

= 0

a 1 1

1 a 1

1 1 aa + 2 1a + 2 aa + 2 1

= 0

(C; - » + C2 + C3)

=> (a + 2)1

a - 1 0

1 0

a -1= 0

(R2 —> R2 — Rj, R j —> R3 - Rj)=J. (a + 2) (a - 1)2 = 0

a = 1, - 2But a = 1 makes above three equations same. So, the system of equation have infinite solution. So answer is a = - 2 for which the system of equations has no solution.

22. Let a and p be the roots o f equation

x 2 - (a - 2) x - a - 1 = 0, then

a + P = a - 2 and ap = - a - 1 Now, a 2 + p2 = (a + P)2 - 2ap

=> a 2 + p2 = (a - 2)2 + 2(a + 1)

=> a 2 + p2 = a 2 + 4 - 4 a + 2 a + 3=» a 2 + p2 = a2 - 2a+ 6

=> a 2 + p2 = (a - l)2 + 5The value o f a 2 + p2 will be least, if a - 1 = 0.

=> a = 1Alternate Solution

v a + p = (a -2 )an d ap = - a - l

/(a) = a 2 + p2 = (a + P)2 - 2aP = (a - 2)2 + 2(a + 1)

= a2 - 2a + 6

=> / '(a ) = 2 a - 2On putting / ’ (a) = 0 for maxima or minima.

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2a - 2 = 0 => a = 1Now f " (a ) = 2 => / " (1)= 2> 0 /. /(a) is minimum at a = 1.

23. Key Idea : If the roots of the equationx 2 + bx + c = 0 are consecutive integers, then

the value ofb2 - 4 cis always I.

Letn and (n + l)b c the roots o f x 2 - bx + c = 0 then n + (n + 1) = b and n(n + 1) = c .-. b2 - Ac = (2n + l ) 2 - 4n(n + 1)

= 4n2 + 4n + 1 - 4nz - 4n

24. / ' ( ! ) = t a k ± « z J ® )h-»o n

- t o Z fi + t t - Bm ^#1 /i->o n

Now, lim /i —♦ o

/> ->0

/a + « f d )= 5, so lim must be &->o h/ a )finite as / ’ (1) exist and lim

/t-*0 **

only, if / ( l ) = 0 and lim = 0.

So / ’(I)

can be finite

lim ^ = 5. /<->0 ft

25. Given that, / { ! ) = - 2 and / '(x )> 2<Jy „-3— 2 2dx

=> dy > 2dx

=> \fW dy Z 2 [ \ lx J/(D Ji

=2> / (6 )-/ (l )> 1 0/(6) £ 10 + / (l)

=> /(6) > 8Alternate Solution

/ (6 ) - / 0 K „6 - 1

=> / ( 6 ) - / a ) 2 i o=> / (6 )+ 2 ^ 1 0

=* /(6) 2: 8

26. lim l / W .- / ( > ) U lim | x -y |x - » y | J f - y | x - » j r

=> I /'Cy)I ^ o=* / ' ( y ) = o=> / (y ) = constantas /(0 ) = 0

=> / (y ) = o=> / (D = o

27.a + x ) 3/2- ( i + i x j

( i - x )

f

d - * r 2

= - ^ - ( i - x y 1/2

3x28

+ higher powers of x.

(v higher powers of x greater than x 2 can be neglected).

28. v x = 2 a" =1 + a + a2 + ...RsO

1= > X =

1 - a

Similarly,

y - l h a n d 2 = I ^

Now, a,b, c arc in AP=> - a, - b, - c are also in AP.=> 1 - a, 1 - b, 1 - c are also in AP.

1 1 11 - a ’ l - b ' l - c

=> x , y , z are in HP. Alternate Solution

are in HP.

1 1x = ~— -,y = Z - '

\ - a ' J 1 -fa ’ * 1 - c, y - 1 z - 1

, b = - — •, c = ------y ’ z

. x - 1 a = ------

V a, b, c are in AP2 b = a+ c

2 f > l z r L £ z I + £ z II y ) x z

2— — * 1 - — + 1 - — y x z

1 = 1 Iy ~ x z

=* x, y, z are in HP.

29. We know that C-^- = 2R ' sin C

i =$ c = 2R ...0) (v C = 90°)

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andC r

tan -7: -------2 s - cn r

tan -t =4 s - c

r = s - c a+ b+ c

=* r = -----2 ~ ~ ~ C=> 2r = a + b - cOn adding Eqs. (i) and (ii)=> 2 ( r + R ) = a + f c

...-i .. j. ....• i

...(ii)

30. Key Idea : cos' 1 x + cos 1 y

= cos' 1 (xy + - x 2 yfl - y 2) .• * V

We have, cos x - cos" ^ = a

^ - + yf l - X 2 f ^

- + i / l - x 2 = cos (X

=> 2V i~ -x^ •yl - = 2cos a - xy

4(1 - x 2) ( 4 - y 2) . 2 2 2=> —------- ^ -----—— = 4 cos a + x y

- 4xy cos a

=A - 4x2 - y 2 + x 2y 2 = 4 cos'1 a + x 2y 2

- 2xy cos a

=> 4x2 - 4xy cos a + y 2 = 4 sin2 a.

31. Since in A ABD

/ Ic/

90*-UV\ E

/^90’-C rD a

AD - c cos (90° - B)AD = c sin B Similarly,BE = a sin C and CF =b sin Av AD, B£, CF arc in HP.

c sin fl, a sin C, b sin A are in I IP.1 1 1

sin C sin B ’ sin A sin C ’ sin B sin A

are in AP.=> sin A, sin B, sin C are in AP.

Alternate Solution

ar(& ABC ) = ^ x B C x A D

1 2A=> A = - xa x AD => AD = —

Similarly, BE = ^ and CF =

v AD, BE, CF are in HP.I l l • uo.•. r . - m HP.a b c

=> a, b, c are in AP .=> sin A, sin B, sin C are in AP.

32. Key Idea : Equation of normal at any point(x ,, y , ) on any curve is _

1 ( X - X j )

Given that x = a (cos 0 + 0 sin 0)

—jrT = a (— sin 0 + sin 0 + 0 cos 0) d0

dxdO

= a 0 cos 0

and y = a (sin 0 - 0 cos 0) .dy

=> = a (cos 0 - cos 0 + 0 sin 0)al)

dy d0

dy dy/dO

= a 0 sin 0

= tan 0” dx dx/dO

Slope of normal

= - ^ = - cot 0 = tan dy

So equation of normal is y - a sin 0 + a 0 cos 0 =

(H

- -j! (x - a cos 0 - a 0 sin 0)sin 0

=> sin O y- a sin2 0+ a 0 cos Osin 0

= - x cos 0 + a cos2 0 + a 0 sin 0 cos 0

=> x cos 0 + y sin 0 = aIt is always at a constant distance V fromorigin.

33. (a ) / (x ) = x 3 + 6x 2 + 6

/ '( x )= 3x2 + 12x = 3x (x + 4)

+

, - 4 I 0 t

x e (- co, - 4) u (0, oo)

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(b ) / (x ) = 3x2 - 2x + 1

A . f- 1 +

- / '(x ) , = 6x - 2

/ ’ ( x ) > 0 V x e (| , « o )

34.

This is wrong... 1 -f cos (ax2 + bx + c)lim

= lim

= lim

(x - a )2

2 sin2 |f ax2 + bx + c ]I J

rs<'aiX

2 sin 2 |\ ( X - « ) ( X - P ) j

(x - a )2 (x - P)2

= lim 4 ( x - p ) 2= ^ ( a - p )2x —>a ^ +*

> sin2 f ~ ( x - a ) ( x ~ P ) l lim -

a

►“ ( 'a , [,2 ) ( x - p )

2 =T ( a - P r

Note : If a and p are the roots of the equation ax2 + bx + c = 0, then this equation can be rewritten,as a (x - a ) (x - p) = 0.

dy35. x “ = y (log y - log x + 1)

Now put — - t x

dy dt=> y = t x = > s ? " e + x s••• t + x ^ = t l o g t + t

=> t log t dx = x dtdt _ dx

t log t ~ x

=* ' ° * { x ) - CX

36. Key Idea : I f Lj and L2 are two lines, then the

equation of line passing through the point of intersection o/L, and h2 will be L, + XL2 = 0.

Equation of a line passing through the intersection of lines ax + 2by + 3b = 0 and bx - 2 ay - 3a = 0 is

(ax + 2by + 3b) + \(bx - 2ay - 3a) = 0 ...(i) Now, this line is parallel to x-axis, so coefficient

o f x = 0=> a + Xb=0=>X = - ~ .b

On putting this value in Eq. (i), we get b(ax + 2by + 3b) - a(bx - 2ay - 3a) = 0

=> 2b2y + 3b2 + 2a2y + 3a2 = 0

=> 2(b2 + a2)y + 3(b2 + a2) = 0

3** y = ~ t

Therefore the required line is below x-axis, at

a distance ^ from it.

Alternate SolutionEquations of given lines are

ax + 2by + 3b = 0 ...(i)and bx - 2ay - 3a = 0 ...(ii)On solving Eqs. (i) and (ii), we get

x = ° - y = - 2.•. Point o f intersection o f lines is

( * - 1)Also required line is parallel to x-axis,

m = 0Equation of required line is

fy + | ] = 0 ( x - ° )

3=> y = - 2Thus the required line is below x-axis at a

distance | from x-axis.Jl/

37. Given that - j- = 50 cm3/min at

aHH- 50

dr3 150 dt 4n

„ 2 d r _150 dt 4n dr 50

4nr

38. Let / •

4n x 225

2

( S „ - -

- j f e ? ) ’ *r (lo gx )2 + l - 21o g x dy

J (1 + (log x ) )

cm/min.

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45. Line is parallel to plane as

{ i - j + 4 < 0 ( i + 5 j + f c ) = 0.General point on the line is (a . + 2, •• X - 2, 4X + 3) . For X = 0 a point on this line is (2, - 2, 3) and distance from

"r • ( i + 5j + £ ) = 5 or x + 5y + z = S

d =

d =

2+ 5 (- 2)+ 3 -5

^ 1 + 2 5 + 1

-10

3V3

10y/3

46. Since (la x i ) - ( a x i ) =a a a l

i a 1

- = |a|2 - a f

Similarly, ( l i x j )2 =| |2 - af

and (1» x k )2 = 11» |2 - af

.-. (a x i)2 + ("a x j)2 + ("a x k )2

= 3| a|2 - ( a f + a f+ a|)

= 3|t|2 -| t| 2

= 2|"a|2 = 2af2

Alternate Solution

Let ~a = a,i + a2j + a3k

|a|2 = a f + a f + af

and ^ x i = (a,i + a2j + a3fc) x ( i )

= - a2fc + a3 j

( ^ x i )2 = af + af

Similarly, (1? x j )2 = af + af

and (a xk )2 = af + a2

(a x i)2 + ("a x j)2 + ("a x ft)2= (af + af + a2 + af + af + a f)

\ - 2 (af + a2 + a2)

= 2 ( a ) 2

47. v a,b, c are in HP2 _ 1 1 b a + c

1 2 1 n=> ------ r + -Ua b c

So straight line — + x + - = 0 always passes a b c

through a fixed point (1 ,-2 ).

48. Key Id ea : If A ^ y i ) , B (x2,y 2) and

C (x3, y 3) are the vertices of a trcang/e, then the

co-ordinates of the centroid will be( x, + x2 + x 3 y, + y 2 + y 3~\{ 3 * 3 J'

Let D and E are mid points of AB and AC. So co-ordinates o f B and C are ( - 3, 3) and (5, 3) respectively.

A (1,1)

(-1.2)0 E (3.2)

B (-3,3)

Centroid o f triangle

C (5.3)

( Xj + x 2 + x 3 y, + y 2 + y3}{ 3 ’ 3 J

f l - 3 + 5 1 + 3+3^= l 3 > 3 J

-(>• 349. Key Idea : J/S, = 0 and S2 = 0 are two circles,

then the equation of line passes through thepoints of intersection of S, = 0 and S2 =0, is

S, - S2 = a

Equation of circles areSj = x2 + y 2 + 2ax + cy + a = 0

and S2 s x2 + y 2 - 3ax + dy -1 = 0

respectively.Chord through intersection points P and Q of the circles S, = 0 and S2 = 0 is S, - S2 = 0 .

5ax + (c - d)y + a + 1 = 0

On comparing it with 5x + by - a = 0, we get 5a c - d a + 1 T = b ~ - a

=> a(- a) = a + 1 => a2 + a + 1 = 0

Which gives no real value of a.The line will passes through P and Q for no

value of a

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50. Since circle touches the x-axis and also touches circle with the centre at (0, 3) and radius 2, then. y

52. ZFBF ' = 90° =* ZOBF ' = 45° V

/(o.3)( c,

__ /Ch.k)A )

N JO

- C,C2 = »i + r2h + (k - 3)2 = (| k\ + 2)2

ft2 + <c2 + 9 - 6/c = fc2 + 4 + 4|Locus of centre of circle is

x 2 = - 5+ 6y + 4\y\

x 2 = lOy - S (v y > 0)

This equation represents a parabola. Thus locus of the centre o f ihe circle is a parabola.

51. Key Idea : Two circles

x 2 + y 2 + 2g}x + 2/,y + c, = 0

and x2 + y 2 + 2g2x + 2/2y + c2 = 0 cuts orthogonally, then

2g >82 + 2/,/2 = c, + c2.Let the circle is x 2 + y 2 + 2gx + 2fy + c = 0,

It cuts circle x 2 + y 2 = p2 orthogonally.

So, c = p2 and it passes through (a, b) .

a2 + b2 + 2 ga + 2fb + p2 = 0

So, locus of ( - g, - f ) is

a2 + b2 - 2ax - 2by + p2 = 0

=> 2ax + 2by - (a2 + b2 + p2) = 0

Alternate SolutionLet the centre o f required circle is (-g, - /). This circle cut the circle x 2 + y 2 = p2

orthogonally. The centre and radius o f circle x 2 + y 2 - p2 are (0, 0) and p respectively.

••• g 2 + f 2 = p 2 + (a + g )2 + ( b+ f ) 2

=> g 2 + f 2 = p2 + a2 + g2 + 2as + b2

+ f 2 + 26/

=> p2 + a2 + b2 + 2ag + 2bf = 0

Thus the locus o f ccntre is2ax + 2by - (a2 + b2 + p2) = 0

EUOb)

( /4 5 ”

V l[-ae. 0)o F )a (a. 0)

(ae, 01/ 1 '

*0IIa

and^ - > - 5

aN<K>If—CirU

=> 2e2 =1

e = 72Alternate Solution

.-. F and F' are foci o f an ellipse, whose co-ordinatcs arc (ae, 0)and(-ae, 0) respectively and co-ordinatcs of B are (0, b).

Slope o f BF = —-ae ■

and slope of BF' = — ae

ZFBF' = 90°

ae \ae)

=> b2 = a2e2

e2 = 1 - — = 1 - 2^ .2 ?a2 a2

=> 2e2 = l = » e = -i=v2

53. Key Idea : fine y = mx + c is tangent to

hyperbola ~ ~ ~ = 1, if c2 = a2m2 - b2. a b

Line y = ax + p is tangent, if p2 = a2a 2 - b2.So locus of (a, p) is y 2 = a2* 2 - b2

=> a2x 2 - y 2 - b2 = 0

Since this equation represents a hyperbola, solocus of a point P (a, P) is a hyperbola.

54. Since the angle between the line and plane is same as angle between the line and normal to the plane

cin n = f l + 2 J + 2 f c ) - (21-J+ Vlfe) 1 3^4 + 1 + X " 3

=> 2 - 2+ 2yfX=yj5+ X

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4X = 5 + X 3X= 5

X - 1

Note: Thex - _ y - fl

ai " h =

angle

Cl

betweenand

a line plane

a2x + i^y + c2z + d2 = 0 is given by

sin9 =+ bfiz C;C2

J a f + l i f T t f ^af + fc jT c f

55. The given equations can be rewritten as x y z i x y z 3 ~ 2 = - 6 2 = -12/= -3

Angle between the lines is given by6 -2 4 + 1 8

cos 9 = , -------- , -■■-■■■—-------V9 + 4 + 36 V4+144+9

= ____ 9____= oV49V157

=» 0= 90°Note : If a,, fy, q and are d.r’s. of twolines then the angled between them is given by

aia2 + bp2 + cp2cos 0 =

56. Given that

V9? + + q2 -Ja2 + + c

P(A u B) =

l - P ( A u B ) = J

=> 1 - P ( A ) - P ( B ) + P ( A n B ) = -g

=» P ( A ) - P ( B ) + i = |

=

=> P(B ) = | and P (A ) = |

Clearly P(A n B ) = P (A )P (B )so the events A and B are independent eventsbut not equally likely.Note : Two events A and B are said to equally likely, if P(A)~ P(B) and known as independent events, if P(A n B) = P(A) P(B).

57. Person 1 has three options to apply.Similarly, person 2 has three options to apply and person 3 has three options to apply.Total cases = 33Now, favourable cases = 3 (An either all hasapplied for house 1 or 2 or 3)

' 3 1So probability = — = —.33 9

58. Key Idea : /n a poisson distribution,

e~xXrP(X - r ) ~ - r l(X = mean) .

P (X = r> 1.5)= P (2 )+ P (3 )+ ... a>= 1 - ( (P (0 )+ P (1 ) )

• f - 2x e‘ 2 x 2l -1 3[ 1 J e2

59. Key Idea : If u is initial velocity of any particle and a is a acceleration, then

v = u + atwhere vis a speed of particle after t seconds.We have vA = /(t + m)

vB= / ' tNow, vA = vB f t = /(t + m)

S + n = i / ( t + m)2

S = | / ’ t 2

n=-|/ (t + m)2 - i / ' t 2

...(0...(ii)...(iii)

...(iv)

...(v)

- (v i )

_ 1 / (/ 'O 2 2 /2

- i / ' t 2 2

2n/ = ( ( / ' ) - / / ' ) t

t 2 =2nf

From Eq. (iii)fin

1 ( / ' - / ) So from Eqs. (vii) and (viii)

f 2m2 2n/

...(vii)

...(viii)

( Z ' - / ) 2 / ' ( / ' - / )

=> f ' f m 2 = 2n ( f ' - f )60. Let lizard catch the insect C

and distance covered by insect = S

time taken by insect, t =

Distance covered by lizard = 21 + S

21 + S = ^ (2)-t2

...(0

21 + 20t = t2 t2 - 2 a - 2 1 = 0

...00

(using Eq. (i))

=> t - -21t + r -21 = 0

=> t(t -2 1 )+ l ( t -2 1 ) = 0=> ( t + l ) ( t - 2 1 ) = 0=> t = 21 s.

61. Key Idea :IfR is a resultant of two forces P and Q acting on a particle, then

_ R sin B . _ K sin a _ sin(a + P) an ~ sin(a + P ) '

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Given that R = -

P = R sin 6 sin (90° + 0) and Q R sin 90°

sin (90° + 0)

r Q = 73 cos 0

cos 0 =

3R sin 0 3R - 2-J21 3

=> P = 2V2 R• P _ 2j 2 R r-

Q 3R ~ '

= > j = 3 : 2s/2

62. Key Idea : If "a =aji + a2j + a3k,

b = bji + b2j + bjii and ~c = q i + c2j + Cjfc

- + , * - » ai °2[a b c ]= b, b2 b3

C1 c2 c311 0 - 1

[a b c ]= x 1 1 - xy x l + x - y

Applying C3 -> C3 + C,U 0 0 j

- I * 1 1i y x 1 + x I

= 1 1| X 1 + X= 1(1 + x ) - x = 1

Thus [I? 1j lc ] does not depend upon neither* nor^'.

63. Key Idea : I f three vectors a*, 1) and c are

coplanar, then [ 1* l> ] = OlThe given points lies in a plane, if

=>

a a c1 0 1 = 0c c b0 a c1 0 1 = 00 c b

- l l “ c U b = 0

(C, -> C, - C 2)

• c A - ah = 0, Therefore c is GM of a and b.

64.U ^ + M X(a2 + b2) X(a3 + b3)

X2b, X %XCj Xc2

X2b3Xc3

02bj+Cj 2 + 2 + C3

0] a2 «3 a2 “ 3!

b2 - - h b2 *>3Cl C2 c3 <1 c2 C3)

=> X4 = - 1

So, no real value of X exists.Alternate Solution

[ X ( a + b ) X2 T> X "<?]=[li b> + 7 T>]

=> X(a + Tj ) (X2T> x X ' c ) = 'a .((T ) + " c ) x ! > )

=* X4 ( a + t ) ( t x - ? ) 4 . ( c x t )

=> X4 { t . t f * t ) } = - { t - t f x ^ ) >

=> X4 = -1 ( v [ a ^ ] . 0)=> No real value of X exists.

65. Let the resultant of A and B is displaced by x. Then change in net momentum of A and B = applied momentum

••• Ax + B x = HHx =

A + B

66. We know that

Put

* V 2 r 3 v-4e ' = l - r + i ___ , x2! 3! 47

■ 1+ 2 l + T ! + 6T + - - "

1X ~ 2

el/2 + e-l/2= 1 +

~ * £ = 1 + ® r . + (i) h +—67. I f a> 0, i =* 1, 2, 3,..., 11 which are not identical,

then

AM of mth power > mth power of AM, if m > 1x i + ... + xJL > + X 2 + ...+.X.V

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* 2 1 1

( n Ii= 1

n n

k

400n > ( ? ) s„ 2 n 802n > 400

n > 16

68. Given u 1 vTi is making an angle 60° with horizontal (upwards).

Therefore ~v should be at 30° with horizontal(downward) as shown in fig.

>90"

v cos 30° = it cos 60° >/3v 1

2 ~ U 2

V' V 369. Key Idea : I f both roots o f ax 2 + bx + c = 0 are

less than <x, then b2 - 4ac > 0, / ( « )> 0 and - b 2a

Let / (x ) = x 2 - 2kx + k2 + k - 5

D = 4k2 - 4 (k2 + k - 5)

= - 4k + 20 St 0 k<,5

Also - tt- < 5 ...(ii)2a

and / ( 5 ) > 0 ...(iii)From (ii), k < 5 ...(iv)From (iii), 25 - 10k + k2 + k - 5 > 0

=> k2 - 9k + 20 > 0

=> k2 - 5k - 4k + 20 > 0

=> (fc - 5) (fc - 4) > 0=* k < 4 and fc > 5 ...(v)From (i), (iii) and (iv)=> k < 4

70. Since, a, ,a2,. . . , a,,,... are in GP.thenloga„, loga71+i , logan, 2, . . . , logan)8 are in AP.

a a + d a+ 2d\So, A = a + 3d a + 4d a + 5d |

a+ 6d a + 7d a+ 8d|

. 1 . Co —► C’j -C ],Applying c 2 ^ c2 _ c;

l a d 2d A = a+3d d 2d =0

| a + 6d d 2d

(Since two columns are similar)71. Given / (x - y ) = / (x ) f { y ) - /(a - x ) /(a + y )

Let x = 0 = y=> /(0 ) = (/ (0 ))2 - (/ (a))2

=>. 1 = 1 - (/ (a ))2

=> /(a) = 0.-. / ( 2 a - x ) = / { a - ( x - a ))

= /(a) / (x - a) - / (« - a) / (* )= 0 - / ( x ) 1 = - /CO

72. Let / (x ) = a„ x'* + a„_, x n~1 + . . . + a, x = 0,

a, * 0It have roots x = 0 and a > 0

/ ' ( x ) = na,^"-1 + ( n - l ) a n. 1 x " ' 2

+ ... + Gj = 0So definitely its derivative is zero between 0 and a.So, / ' (x ) has a positive root smaller than a .

73. Let / = f * — — - dx, a > 0 ■'-*1 + o'

/ = r dx1 + a~

Eqs. (i) a

2/= V71

...0)

...(0

• ••(ii)

On adding Eqs. (i) and (ii)(1 + o ') cos2 x

(1 + a*)dx

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