1. Five cells each of emfEand internal resistance rsend the same amount of current through an external resistance R whether the cells are connected in para llel or in series. Then the ratio R ris (a) 2 (b) 1 2 (c) 1 5 (d) 1 (e) 5 2. The power dissipated in the transmission cables carrying currentIand voltage Vis inversely proportional to (a) V(b) V2 (c) V(d) I(e) I3. A rigid container with thermally insulated walls co ntains a gas and a coil of resistance 50 , carrying a current of l A. The change in i nternal energy of the gas after 2 minutes will be (a) 6 kJ (b) 10 kJ (c) 3 kJ (d) 12 kJ (e) 1.5 kJ 4. The magnitude of the magnetic field inside a long solenoid is increased by (a) de cr ea si ng i ts radius (b ) decreas ing th e c urre nt thro ugh it (c) incr eas ing its area of cross-sec tio n (d ) intro duc ing a me dium o f higher perme abi lity (e) dec reasing the numb er of turns in it 5. A bar magnet of moment of inertia 9 × l0 –5 k g m 2 placed in a vibration magnetometer and oscillating in a uniform magnetic field l6 2 × l0 –5 T makes 20 oscillations in 15 s. The magnetic moment of the bar magnet i s (a) 3 Am 2 (b) 2 Am 2 (c) 5 Am 2 (d) 6 Am 2 (e) 4 Am 2 6. Identify the correctly matched pairMat er ial Ex a m pl e (a) Diamagnetic - Gadol iniu m (b) So ft ferro magnetic - Alnico (c) Hard ferromagne ti c - Copper (d) Paramagnetic - Sodi um (e) Permanent magnet - Al umi num 7. If the radius of the dees of cyclotron is r, then the kinetic energy of a proton of mass maccelerated by th e cyclotron at an oscillating frequency vis (a) 42 m 2 v 2 r2 (b) 42 mv 2 r2 (c) 22 mv 2 r2 (d) 2 mv 2 r2 (e) 2 m 2 v 2 r2 8. If a magnetic dipole of moment Msituated in the direction of a magnetic field Bis rotated by 180°, then th e amount ofwork done is ( a) MB ( b) 2MB (c) MB 2 ( d) 0 (e) MB 9. The polarit y of induced e mf is given by ( a) Ampe re 's circ ui ta l law (b ) Bio t-Sa v art law (c) Lenz’s la w (d ) Fleming's rig ht ha nd r ule (e ) Fl eming 's le ft han d rule 10. In an LCR series circuit, at resonance ( a) the c urre nt and vo ltag e ar e i n phas e (b ) the impedance is maxi mu m (c ) the curre nt i s mini mu m (d ) the qu ality fa ctor is inde pen dent of R(e) the c urre nt leads the voltage by /2 11. A conducting ring of radius l mkept in a u niform magnetic field Bof 0.01 T, rotates uniformly with an angula r velocity 100 rad s –1 with its axis of rotation perpendicular to B. The maximum induced emf in it is (a) 1.5V ( b) V (c) 2V ( d) 0.5 V (e) 4V 12. A step dow n tran sformer increase s the inp ut current 4 A to 24 A at th e secondary . If the number of turns in t he primar y coil is 330, the number of turns in th e secondary coil is (a) 6 0 ( b) 50 (c) 6 5 ( d) 45 (e) 5 5 13. In a plane electromagnetic wave, the electric field ofamplitude l V m –1 varies with time in free space. The average energy density of magnetic field is (in J m –2 ) (a) 8.86 × 10 –2 ( b) 4.43 × l0 –2 (c) 17.72 × 10 –2 ( d) 2.21 × 10 –2 (e) 1.11 × 10 –2 14. Which one of the following is the property of a monochromatic, pla ne electromagn etic wave in free space? (a) Electric and magnetic fields have a p hase diff erenc e of /2 (b ) The ene rgy c ontrib utio n of b oth electric and magnetic fields are equal (c ) The dire ct ion o f pro pagat ion is in the directio n ofelectric filed E (d ) The pressure exe rted by the wav e is the prod uct ofenergy density and the speed of the wave (e) The speed of the wave is B/E K ERAL A PET 2 0 1 4 - S OL VED PAPER Paper-I : Physics Chemistry
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
1. Five cells each of emf E and internal resistance r send thesame amount of current through an external resistance R
whether the cells are connected in parallel or in series. Then
the ratio R
r is
(a) 2 (b)1
2
(c)1
5(d) 1
(e) 5
2. The power dissipated in the transmission cables carryingcurrent I and voltage V is inversely proportional to(a) V (b) V 2
(c) V (d) I
(e) I
3. A rigid container with thermally insulated walls contains agas and a coil of resistance 50 , carrying a current of l A.The change in internal energy of the gas after 2 minutes will be(a) 6 kJ (b) 10 kJ(c) 3 kJ (d) 12 kJ(e) 1.5 kJ
4. The magnitude of the magnetic field inside a long solenoidis increased by(a) decreasing its radius(b) decreasing the current through it(c) increasing its area of cross-section(d) introducing a medium of higher permeability(e) decreasing the number of turns in it
5. A bar magnet of moment of inertia 9 × l0 –5 kg m2 placed ina vibration magnetometer and oscillating in a uniformmagnetic field l62 × l0 –5T makes 20 oscillations in 15 s.The magnetic moment of the bar magnet is(a) 3 Am2 (b) 2 Am2
(c) 5 Am2 (d) 6 Am2
(e) 4 Am2
6. Identify the correctly matched pair Material Example
7. If the radius of the dees of cyclotron is r , then the kineticenergy of a proton of mass m accelerated by the cyclotronat an oscillating frequency v is(a) 42m2v2r 2 (b) 42mv2r 2
(c) 22mv2r 2 (d) 2mv2r 2
(e) 2m2v2r 2
8. If a magnetic dipole of moment M situated in the directionof a magnetic field B is rotated by 180°, then the amount of work done is(a) MB (b) 2MB
(c)MB
2(d) 0
(e) MB9. The polarity of induced emf is given by
(a) Ampere's circuital law(b) Biot-Savart law(c) Lenz’s law(d) Fleming's right hand rule(e) Fleming's left hand rule
10. In an LCR series circuit, at resonance(a) the current and voltage are in phase(b) the impedance is maximum(c) the current is minimum(d) the quality factor is independent of R (e) the current leads the voltage by /2
11. A conducting ring of radius l m kept in a uniform magneticfield B of 0.01 T , rotates uniformly with an angular velocity100 rad s –1 with its axis of rotation perpendicular to B. The
maximum induced emf in it is(a) 1.5V (b) V(c) 2V (d) 0.5V(e) 4V
12. A step down transformer increases the input current 4 A to24 A at the secondary. If the number of turns in the primarycoil is 330, the number of turns in the secondary coil is(a) 60 (b) 50(c) 65 (d) 45(e) 55
13. In a plane electromagnetic wave, the electric field of amplitude l V m –1 varies with time in free space. The average
energy density of magnetic field is (in J m –2
)(a) 8.86 × 10 –2 (b) 4.43 × l0 –2
(c) 17.72 × 10 –2 (d) 2.21 × 10 –2
(e) 1.11 × 10 –2
14. Which one of the following is the property of amonochromatic, plane electromagnetic wave in free space?(a) Electric and magnetic fields have a phase difference
of /2(b) The energy contribution of both electric and magnetic
fields are equal(c) The direction of propagation is in the direction of
electric filed E
(d) The pressure exerted by the wave is the product of energy density and the speed of the wave
15. The apparent flattening of the sun at sunset and sunrise isdue to(a) refraction (b) diffraction(c) total internal reflection (d) interference(e) polarization
16. The polarising angle for a medium is found to be 60°. The
critical angle of the medium is
(a)1 1
sin2
(b)
1 3sin
2
(c)1 1
sin3
(d)
1 1sin
4
(e)1 2
sin3
17. Identify the mismatch in the following(a) Myopia - Concave lens(b) For rear view - Concave mirror
18. In Young's double slit experiment, to increase the fringewidth(a) the wavelength of the source is increased(b) the source is moved towards the slit(c) the source is moved away from the slit(d) the slit separation is increased(e) the screen is moved towards the slit
19. Light of wavelength 5000 A° is incident normally on a slit of width 2.5 × l0 –4 cm. The angular position of second minimum
from the central maximum is
(a) 1 1sin
5
(b) 1 2
sin5
(c)3
(d)6
(e)4
20. An electron of mass me and a proton of mass m p areaccelerated through the same potential. Then the ratio of their de Broglie wavelengths is
(a) 1 (b) e
p
m
m
(c) e
p
m
m (d) p
e
m
m
(e) p
e
m
m
21. The half-life of a radioactive substance is 20 minutes. Thetime taken between: 5% decay and 87.5% decay of thesubstance will be
22. The ratio of the surface area of the nuclei 52Te125 to that of
13Al27 is
(a)5
3(b)
125
17
(c)
1
4 (d)
25
9
(e)3
5
23. If the frequency of incident light falling on a photosensitivemetal is doubled, the kinetic energy of the emitted photoelectron is
(a) unchanged
(b) halved
(c) doubled
(d) more than twice its initial value
(e) reduced to 14
th
24. The significant result deduced from the Rutherford'sscattering experiment is that
(a) whole of the positive charge is concentrated at thecentre of atom
(b) there are neutrons inside the nucleus
(c) -particles are helium nuclei
(d) electrons are embedded in the atom
(e) electrons are revolving around the nucleus
25. On an average, the number of neutrons and the energy of a
neutron released per fission of a uranium atom arerespectively
(a) 2.5 and 2 keV (b) 3 and l keV
(c) 2.5 and 2 MeV (d) 2 and 2 keV
(e) l and 2 MeV
26. The inputs A, B and C to be given in order to get an outputY = 1 from the following circuit are
(a) 0, 1, 0
(b) 1, 0, 0
(c) 1, 0, l
A
B
C
Y
(d) 1, 1, 0
(e) 0 , 0, 1
27. The collector resistance and the input resistance of a CEamplifier are respectively 10 k and 2 k . If of the transistor is 49, the voltage gain of the amplifier is
(a) 125 (b) 150
(c) 175 (d) 200
(e) 24528. The light emitting diode (LED) is
(a) a heavily doped p-n junction with no external bias
(b) a heavily doped p-n junction with reverse bias
(c) a heavily doped p-n junction with forward bias
(d) a lightly doped p-n junction with no external bias
(e) a lightly doped p-n junction with reverse bias
29. A point-to-point communication mode is seen in
(a) Satellite cable communication(b) Television transmission
(c) FM radio transmission
(d) AM radio transmission
(e) Fax transmission30. If the heights of transmitting and the receiving antennasare each equal to h, the maximum line-of-sight distance between them is (R is the radius of earth)
(a) 2R h (b) 4R h
(c) 6R h (d) 8R h
(e) R h
31. The ionospheric layer acts as a reflector for the frequencyrange
(a) l kHz to 10 kHz (b) 3 MHz to 30 MHz
(c) 3 kHz to 30 kHz (d) 100 kHz to 1 MHz(e) 3 GHz to 30 GHz
32. ln a simple pendulum experiment, the maximum percentageerror in the measurement of length is 2% and that in theobservation of the time-period is 3%. Then the maximum percentage error in determination of the acceleration due togravity g is(a) 5% (b) 6%
(c) 1% (d) 8%(e) 10%
33. The pitch and the number of circular scale divisions in ascrew gauge with least count 0.02 mm are respectively
(a) l mm and 100 (b) 0.5 mm and 50(c) 1 mm and 50 (d) 0.5 mm and 100
(e) l mm and 20034. A ball is dropped from the top of a tower of height 100 m
and at the same time another ball is projected verticallyupwards from ground with a velocity 25 ms –1. Then thedistance from the top of the tower, at which the two ballsmeet is(a) 68.4 m (b) 48.4 m
(c) 18.4 m (d) 28.4 m
(e) 78.4 m
35. The ratio of distances traversed in successive intervals of time when a body falls freely under gravity from certainheight is
(a) l : 2 : 3 (b) l : 5 : 9
(c) 1 : 3 : 5 (d) 1 : 2 : 3
(e) l : 4 : 936. A particle starting with certain initial velocity and uniform
acceleration covers a distance of 12 m in first 3 seconds anda distance of 30 m in next 3 seconds. The initial velocity of the particle is
(a) 3 ms –1 (b) 2.5 ms –1
(c) 2 ms –1 (d) 1.5 ms –1
(e) 1 ms –1
37. A ball of mass 10 g moving perpendicular to the plane of thewall strikes it and rebounds in the same line with the samevelocity. If the impulse experienced by the wall is 0.54 Ns,the velocity of the ball is
(a) 27 ms –1 (b) 3.7 ms –1
(c) 54 ms –1 (d) 37 ms –1
(e) 5.4 ms –1
38. A particle has the position vector ˆˆ ˆ2r i j k
and the
linear momentum ˆˆ ˆ2 p i j k
. Its angular momentum
about the origin is
(a) ˆˆ ˆ 3i j k (b) ˆˆ ˆ 3i j k
(c) ˆˆ ˆ 3i j k (d) ˆˆ ˆ 5i j k
(e) ˆˆ ˆ 5i j k
39. The vertical component of velocity of a projectile at itsmaximum height (u - velocity of projection, -angle of projection) is
(a) u sin (b) u cost
(c)sin
u
(d) 0
(e)cos
u
40. The coordinates of a particle moving in x–y plane at anyinstant of time t are x = 4t2; y = 3t2. The speed of the particleat that instant is
(a) 10 t (b) 5 t(c) 3 t (d) 2 t
(e) 13t
41. A cyclist bends while taking turn in order to
(a) reduce friction(b) provide required centripetal force
(c) reduce apparent weight(d) reduce speed
(e) sit comfortably
42. Two blocks of masses 2 kg and 4 kg are attached by an
inextensible light string as shown in the figure. If a force of 120 N pulls the blocks vertically upward, the tension in thestring is (take g = 10 ms –2)
(a) 20 N
(b) 15 N
(c) 35 N
F = 120 N
4 kg
2 kg(d) 40 N(e) 30 N
43. The total energy of a solid sphere of mass 300 g which rollswithout slipping with a constant velocity of 5 ms –1 along astraight line is
44. A bullet when fired into a target loses half of its velocityafter penetrating 20 cm. Further distance of penetration before it comes to rest is
(a) 6.66 cm (b) 3.33 cm
(c) 12.5 cm (d) 10 cm(e) 5 cm
45. In elastic collision(a) both momentum and kinetic energy are conserved
(b) neither momentum nor kinetic energy is conserved(c) only momentum is conserved
(d) only kinetic energy is conserved(e) forces involved in the interaction are non-conservative
46. Two discs rotating about their respective axis of rotationwith angular speeds 2 rads –1 and 5 rads –1 are brought intocontact such that their axes of rotation coincide. Now, theangular speed of the system becomes 4 rads –1. If the momentof inertia of the second disc is 1 × 10 –3 kg m2, then the
moment of inertia of the first disc (in kg m2) is(a) 0.25 × 10 –3 (b) 1.5 × 10 –3
(c) 1.25 × 10 –3 (d) 0.75 × 10 –3
(e) 0.5 × 10 –3
47. A wheel is rotating at 1800 rpm about its own axis. Whenthe power is switched off, it comes to rest in 2 minutes.Then the angular retardation in rad s –1 is
(a) 2 (b)
(c)2
(d)
4
(e)6
48. If the angular momentum of a particle of mass m rotatingalong a circular path of radius r with uniform speed is L, thecentripetal force acting on the particle is
(a)2
2
L
mr (b)
2L
mr
(c)L
mr (d)
2L m
r
(e) 2Lmr
49. Pick out the wrong statement from the following(a) The Sl unit of universal gravitational constant is
Nm2kg –2
(b) The gravitational force is a conservative force
(c) The force of attraction due to a hollow spherical shellof uniform density on a point mass inside it is zero
(d) The centripetal acceleration of the satellite is equal toacceleration due to gravitational potential
(e) Gravitational potential energy
=gravitational potential
mass of the body
50. If a body of mass m has to be taken from the surface of earthto a height h = R, then the amount of energy required is(R : radius of earth)
(a) mgR (b)R
3
mg
(c)R
2
mg(d)
R 12
mg
(e)R
9
mg
51. The total energy of an artificial satellite of massm revolvingin a circular orbit around the earth with a speed v is
(a) 21
2mv (b) 21
4mv
(c) 21
4
mv (d) – mv2
(e)21
2 mv
52. Two soap bubbles each with radius r 1 and r 2 coalesce invacuum under isothermal conditions to form a bigger bubbleof radius R. Then R is equal to
(a) 2 21 2r r (b) 2 2
1 2r r
(c) r 1 – r 2 (d)2 2
1 2
2
r r
(e) 2 21 22 r r
53. The ratio of hydraulic stress to the corresponding strain isknown as(a) compressibility (b) bulk modulus
54. A boy can reduce the pressure in his lungs to 750 mm of mercury. Using a straw he can drink water from a glass uptothe maximum depth of (atmospheric pressure = 760 mm of mercury; density of mercury = 13.6 gcm –3)
(a) 13.6 cm (b) 9.8 cm(c) 10 cm (d) 76 cm
(e) 1.36 cm55. A spring stores l J of energy for a compression of l mm. The
additional work to be done to compress it further by l mm is(a) l J (b) 2 J(c) 3 J (d) 4 J
(e) 0.5 J56. If m represents the mass of each molecule of a gas and T, its
absolute temperature then the root mean square velocity of the gaseous molecule is proportional to
57. A Carnot engine operating between temperatures T1 andT2 has efficiency 0.2. When T2 is reduced by 50 K, itsefficiency increases to 0.4. Then T1 and T2 are respectively(a) 200 K, 150 K (b) 250 K, 200 K (c) 300 K, 250 K (d) 300 K, 200 K (e) 300 K, 150 K
58. A molecule of a gas has six degrees of freedom. Then themolar specific heat of the gas at constant volume is
(a)R
2(b) R
(c)3R
2(d) 2 R
(e) 3 R 59. Total number of degrees of freedom of a rigid diatomic
molecule is(a) 3 (b) 6(c) 5 (d) 2(e) 7
60. If the differential equation for a simple harmonic motion is2 2
2
d y
dt + 2 y = 0, the time-period of the motion is
(a) 2s (b)2s
s
(c)2
s
(d) 2s
(e)2
s
61. Identify the wrong statement from the following
(a) If the length of a spring is halved, the time period of
each part becomes1
2 times the original
(b) The effective spring constant K of springs in parallel
is given by1 2
1 1
K K K
(c) The time period of a stiffer spring is less than that of asoft spring
(d) The spring constant is inversely proportional to thespring length
(e) The unit of spring constant is Nm –1
62. The total energy of the particle executing simple harmonicmotion of amplitude A is 100 J. At a distance of 0.707 A fromthe mean position, its kinetic energy is(a) 25 J (b) 50 J(c) 100 J (d) 12.5 J(e) 70 J
63. Two travelling waves, yl = A sin [k ( x + ct ) ] and y2 = A sin[k ( x – ct )] are superposed on a string. The distance betweenadjacent antinodes is
(a) c
(b)
2
ct
(c)2k
(d)
k
(e)
k
64. If a stretched wire is vibrating in the second overtone, thenthe number of nodes and antinodes between the ends of the string are respectively(a) 2 and 2 (b) l and 2(c) 4 and 3 (d) 2 and 3(e) 3 and 4
65. Pick out the correct statement in the following with referenceto stationary wave pattern(a) In a tube closed at one end, all the harmonics are
present(b) In a tube open at one end, only even harmonics are
present(c) The distance between successive nodes is equal to
the wavelength(d) In a stretched string, the first overtone is the same as
the second harmonic(e) Reflection of a wave from a rigid wall changes the
phase by 45°
66. A plane square sheet of charge of side 0.5 m has uniformsurface charge density. An electron at 1 cm from the centreof the sheet experiences a force of l.6 × 10 –2 N directedaway from the sheet. The total charge on the plane squaresheet is(0 = 8.854 × 10 –12 C2m –2 N –1)(a) 16.25 µC (b) –22.15 µC(c) – 44.27 µC (d) 144.27 µC(e) 8.854 µC
67. The energy stored in a capacitor of capacitance C having acharge Q under a potential V is
(a)
21
Q V2 (b)
21
C V2
(c)21 Q
2 V(d)
1QV
2
(e)1
CV2
68. The electrostatic force between two point charges is directly proportional to the(a) sum of the charges(b) distance between the charges(c) permittivity of the medium(d) square of the distance between the charges
(e) product of the charges69. The time period of revolution of a charge q1 and of mass m
moving in a circular path of radius r due to Coulomb forceof attraction with another charge q2 at its centre is
70. A point charge of 2 C experiences a constant force of 1000 Nwhen moved between two points separated by a distanceof 2 cm in a uniform electric field. The potential difference between the two points is(a) 12 V (b) 8 V(c) 10 V (d) 16 V
(e) 5 V71. In the network shown below, if potential across XY is 4 V,
then the input potential across AB is
A
B
2
4
4
8
2
X
Y
4 4 V
8
(a) 16 V (b) 20 V(c) 8 V (d) 12 V(e) 24 V
72. If the ammeter A shows a zero reading in the circuit shown below, the value of resistance R is
10 2 V
500 R A
(a) 500 (b) 125 (c) 100 (d) 41.5(e) 4
73. In a reaction 2A + B 3C, the concentration of A decreasesfrom 0.5 mol L –1 to 0.3 mol L –1 in 10 minutes. The rate of production of 'C' during this period is(a) 0.01 mol L –1 min –1 (b) 0.04 mol L –1 min –1
(c) 0.05 mol L –1 min –1 (d) 0.03 mol L –1 min –1
(e) 0.02 mol L –1 min –1
74. Ammonium ion (NH4+) reacts with nitrite ion (NO2
– ) in
aqueous solution according to the equation NH4
+ (aq) + NO2 – (aq) N2(g) + 2H2O(l)
The following initial rates of reaction have been measuredfor the given reactant concentrations.Expt. No. [NH4
+], (M) [NO2 – ], (M) Rate (M/hr)
1 0.010 0.020 0.0202 0.015 0.020 0.0303 0.010 0.010 0.005Which of the following is the rate law for this reaction?(a) rate = k [NH4
+ ] [NO2 – ]4
(b) rate = k [NH4+] [NO2
– ]
(c) rate =k [NH4
+
] [NO2
–
]
2
(d) rate = k [NH4+]2 [NO2
– ]
(e) rate = k [NH4+]1/2 [NO2
– ]1/4
75. Gold sol can be prepared by(a) hydrolysis of gold (III) chloride(b) oxidation of gold by aquaregia(c) peptization(d) treating gold (III) chloride with metallic zinc
(e) reduction of gold (III) chloride with formalin solution76. The IUPAC name of the complex [Co(NH3)2(H2O)4]Cl3 is
(e) Tetraaquadiaminecobalt (II) chloride77. The products obtained by the ozonolysis of 2-ethylbut-
l-ene are
(a) propanone and ethanol(b) ethanal and 3-pentanone(c) butanal and ethanol(d) methanal and 2-pentanone(e) methanal and 3-pentanone
78. When but-2-yne is treated with Na in liquid ammonia(a) cis-2-butene is obtained(b) trans-2-butene is formed(c) n-butane is the major product(d) it rearranges to but-l-yne(e) there is no reaction
79. The correct decreasing order of reactivity for a given alkyl(R) group in both S Nl and S N2 reaction mechanisms is(a) R-I > R-Br > R-C1 > R-F
80. The compound of molecular formula C5H10O(A) reacts withTol1en's reagent to give silver mirror but does not undergoaldol condensation. The compound A is(a) 3-pentanone (b) 2,2-dimethylpropanal
81. When n-hexane is heated with anhydrous AlCl3 and HCl
gas the major product obtained is(a) l-chlorohexane (b) 2-chlorohexane(c) 3-chlorohexane (d) hex-3-ene(e) mixture of 2-methylpentane and 3-methylpentane
82. How many monochloro structural isomers are expected infree radical monochlorination of 2-methylbutane?(a) 2 (b) 3(c) 4 (d) 5(e) 6
83. Chloroform reacts with oxygen in the presence of light togive
84. Which one of the following is not expected to undergoiodoform reaction?(a) Propan-2-ol (b) 1-Phenylethanol(c) 2-Butanol (d) Ethanol(e) Diphenyl methanol
85. Identify the combination of compounds that undergo
Aldol condensation followed by dehydration to produce but-2-enal(a) methanal and ethanol(b) two moles of ethanal(c) methanal and propanone(d) ethanal and propanone(e) two moles of ethanol
86. The correct increasing order of the acid strength of benzoicacid (I), 4-nitrobenzoic acid (II), 3,4-dinitrobenzoic acid (III)and 4-mcthoxybenzoic acid(IV) is(a) I < II < III < IV (b) II < I < IV < III(c) IV < I < II < III (d) IV < II < I < III
(e) I < IV < II < III87. An organic compound with the molecular formula C8H8O
forms 2,4-DNP derivative, reduces Tollen's reagent andundergoes Cannizzaro reaction. On vigorous oxidation, itgives l, 2-benzenedicarboxylic acid. The organic compoundis(a) 2-ethylbenzaldehyde (b) 2-methylbenzaldehyde(c) acetophenone (d) 3-methylbenzaldehyde(e) phenylacetaldehyde
88. Phenyl isocyanide is prepared from aniline by(a) Rosenmund's reaction(b) Kolbe's reaction
89. Choose the correct order of decreasing basic strength of the following compounds in aqueous solution(i) C6H5 NH2 (ii) C2H5 NH2(iii) NH3 (iv) (CH3)2 NH(a) (i) > (ii) > (iii) > (iv) (b) (iv) > (ii) > (iii) > (i)(c) (ii) > (i) > (iii) > (iv) (d) (iv) > (iii) > (ii) > (i)(e) (ii) > (iv) > (iii) > (i)
90. Gabriel's phthalimide synthesis can be used to prepare(a) ethanamine
91. The sugar moiety present in RNA molecule is(a) -D-2-deoxyribose (b) -D-galactose(c) -D-fructofuranose (d) -D-ribose(e) -D-glucopyranose
92. Novolac, the linear polymer used in paints is(a) copolymer of 1,3-butadiene and styrene(b) obtained by the polymerization of methyl methacrylate(c) initial product obtained in the condensation of phenol
and formaldehyde in the presence of acid catalyst(d) obtained by the polymerization of caprolactum(e) copolymer of melamine and formaldehyde
93. The carbohydrate used as storage molecules in animals is(a) sucrose (b) glycogen
(c) maltose (d) glucose(e) fructose
94. Green chemistry deals with
(a) study of plant physiology
(b) study of extraction of natural products from plants(c) detailed study of reactions involved in the synthesis
of chlorophyll
(d) utilization of existing knowledge base for reducing thechemical hazards along with developmental activities
(e) synthesis of chemical compounds using green light
95. A 250 W electric bulb of 80% efficiency emits a light of 6626Å wavelength. The number of photons emitted per second by the lamp is (h = 6.626 × 10 –34 J s)
(a) 1.42 × 1017 (b) 2.l8 × 1016
(c) 6.66 × 1020 (d) 2.83 × 1016
(e) 4.25 × 1016
96. The shortest wavelength of the line in hydrogen atomicspectrum of Lyman series when R H = 109678 cm –1 is
(a) 1002.7 Å (b) 1215.67 Å
(c) 1127.30 Å (d) 911.7 Å(e) 1234.7 Å
97. The work function of a metal is 5 eV. What is the kineticenergy of the photoelectron ejected from the metal surfaceif the energy of the incident radiation is 6.2 eV?
(1 eV = 1.6 × 10 –19 J)
(a) 6.626 × 10 –19 J (b) 8.01 × 10 –19J
(c) 1.92 × 10 –18 J (d) 8.010 × 10 –18J
(e) 1.92 × 10 –19J98. The lattice energy of NaCl is 788 kJ mol –1. This means that
788 kJ of energy is required
(a) to separate one mole of solid NaCl into one mole of Na(g) and one mole of Cl(g) to infinite distance
(b) to separate one mole of solid NaCl into one mole of Na+(g) and one mole of Cl – (g) to infinite distance
(c) to convert one mole of solid NaCl into one mole of gaseous NaCl
(d) to convert one mole of gaseous NaCl into one mole of solid NaCl
(e) to separate one mole of gaseous NaCl into one moleof Na+(g) and one mole of Cl – (g) to infinite distance
99. Arrange the following species in the correct order of their stability C2, Li2, O2
+, He2+
(a) Li2 < He2+ < O2
+ < C2 (b) C2 < O2+ < Li2 < He2
+
(c) He2+ < Li2 < C2 < O2
+ (d) O2+ < C2 < Li2 < He2
+
(e) C2 < Li2 < He2+ < O2
+
100. Molecular formulae and shapes of some molecules are given below. Choose the incorrect match
103. At 273 K, the density of a certain gaseous oxide at 2atmosphere is same as that of dioxygen at 5 atmosphere.The molecular mass of the oxide (in g mol –1) is(a) 80 (b) 64(c) 32 (d) 160(e) 70
104. The reaction of H2 is given belowH2 + CO + R – CH = CH2 R – CH2 – CH2 – CHO
is specifically called as(a) hydrogenation (b) reduction(c) hydroformylation (d) dehydration(e) formylation
105. Which of the following are isoelectronic species?(i) NH3 (ii) CH3
+
(iii) NH2 – (iv) NH4
+
Choose the correct answer from the codes given below(a) (i), (ii), (iii) (b) (ii), (iii), (iv)(c) (i), (ii), (iv) (d) (i), (iii), (iv)(e) (ii), (iii)
106. The salt of an alkali metal gives violet colour in the flametest. Its aqueous solution gives a white precipitate with barium chloride in hydrochloric acid medium. The salt is(a) K 2SO4 (b) KCl(c) Na2SO4 (d) K 2CO3(e) Li2SO4
107. In which one of the following the central atom is sp3
hybridized?(a) NH4
+ (b) BF3(c) SF6 (d) PCl5(e) XeF4
108. Which one of the following statements is not true in respect
of properties of interhalogen compounds?(a) They are all covalent compounds(b) They are volatile solids or liquids except ClF(c) IF5 has square pyramidal structure(d) They are all paramagnetic in nature(e) BrF3 is used in the preparation of UF6 in the
enrichment of 235U109. Which one of the following is an incorrect statement?
(a) O3 oxidises PbS to PbSO4(b) O3 dioxide nitric oxide to nitrogen dioxide(c) O3 oxidises aqueous KI at pH = 9.2(d) The two oxygen-oxygen bond lengths in O3 are
different(e) O3 is used as an oxidizing agent in the manufacture of KMnO4
110. The correct descending order of oxidizing power of thefollowing is(a) Cr 2O7
2– > MnO4 – > VO2
+
(b) MnO4 – > Cr 2O7
2– > VO2+
(c) VO2+ > MnO4
– > Cr 2O72–
(d) MnO4 > VO2+ > Cr 2O7
2–
(e) Cr 2O72– > VO2
+ > MnO4 –
111. The number of electrons that are involved in the reductionof permanganate to manganese(II) salt, manganate andmanganese dioxide respectively are(a) 5, 1, 3 (b) 5, 3, 1(c) 2, 7, 1 (d) 5, 2, 3(e) 2, 3; 1
112. The calculated magnetic moment of a divalent ion of anatom with atomic number 24 in aqueous solution is(a) 4.90 BM (b) 5.92 BM(c) 3.87 BM (d) 2.84 BM
(e) 1.73 BM113. The entropy of vaporization of a liquid is 58 JK –1 mol –1.
If 100 g of its vapour condenses at its boiling point of l23°C,the value of entropy change for the process is(Molar mass of the liquid = 58 g mol –1)(a) –100 JK –1 (b) 100 JK –1
(c) –123 JK –1 (d) 123 JK –1
(e) 1230 JK –1
114. The values of limiting ionic conductance of H+ and HCOO –
ions are respectively 347 and 53 Scm2mol –1 at 298 K, If themolar conductance of 0.025M methanoic acid at 298 K is 40Scm2mol –1, the dissociation constant of methanoic acid at
298 K is(a) l × l0 –5 (b) 2 × 10 –5
(c) l.5 × l0 –4 (d) 2.5 × l0 –5
(e) 2.5 × l0 –4
115. In a closed cylinder of capacity 24.6 L the following reactionoccurs at 27°C
A2(s) B2(s) + 2C(g). At equilibrium lg of B2(s)(molar mass = 50 g mol –1) is present. The equilibriumconstant K p for the equlibrium in atm2 unit is(R = 0.082 L atm K –1 mol –1)(a) 1.6 × 10 –2 (b) 1.6 × 10 –5
(c) 1.6 × l0 –3
(d) 1.6 × 10 –4
(e) l.6 × 10 –1
116. The pH of a saturated solution of a metal hydroxide of formula X(OH)2 is 12.0 at 298 K. What is the solubility product of the metal hydroxide at 298 K (in mol3 L –3)?(a) 2 × 10 –6 (b) 1 × l0 –7
(c) 5 × 10 –5 (d) 2 × 10 –5
(e) 5 × l0 –7
117. An aqueous solution containing 3g of a solute of molar mass 111.6 g mo1 –1 in a certain mass of water freezes at – 0.125°C. The mass of water in grams present in the solutionis (K f = 1.86 K kg mol –1)
118. A sample of sea water contains 5 × 10 –3 g of dissolvedoxygen in 1 kilogram of the sample. The concentration of O2 in that sea water sample in ppm is
(a) 5 × 10 –4 (b) 5 × l0 –3
(c) 5 × 10 –2 (d) 5 × 10 –1
(e) 5119. The change in potential of the half-cell Cu2+|Cu, when
aqueous Cu2+ solution is diluted 100 times at 298 K?
2.303 RT0.06
F
(a) increases by 120 mV (b) decreases by 120 mV(c) increases by 60 mV (d) decreases by 60 mV(e) no change
120. Consider the following electrolytic cells(i) M(s) | M2+(aq), 0.1M || X2+(aq), 0.01M | X(s)(ii) M(s) | M2+(aq), 0.1M || X2+(aq), 0.1M | X(s) and
(iii) M(s) | M2+(aq), 0.011M || X2+(aq), 0.1M | X(s)The cell EMF of the above cells are E1, E2 and E3respectively. Which one of the following is true?(a) E1> E2 > E3 (b) E2 >E3 > E1(c) E3> E1> E2 (d) E1 >E3 > E2(e) E3 >E2 > E1
Paper-II : Mathematics
1. If the operation is defined by a b = a2 +b2 for all real
numbers a and b, then (2 3) 4 =
(a) 120 (b) 185(c) 175 (d) 129
(e) 3122. The number of students who take both the subjects
mathematics and chemistry is 30. This represents 10% of the enrolment in mathematics and 12% of the enrolment inchemistry. How many students take at least one of thesetwo subjects?(a) 520 (b) 490
(c) 560 (d) 480(e) 540
3. Let f(x) = |x – 2| , where x is a real number. Which one of thefollowing is true?(a) f is Periodic. (b) f (x + y) = f (x) + f (y)
(c) f is an odd function (d) f is not a 1–l function
(e) f is an even function4. lf A = {l,3,5,7} and B = {l,2,3,4,5,6, 7,8}, then the number of
one–to–one functions from A into B is
(a) 1340 (b) 1860(c) 1430 (d) 1880 (e) 1680
5. The range of the function f (x) = x2 + 2x+ 2 is
(a) (1, ) (b) (2, )
(c) (0, ) (d) [1, ) (e) ( – , )
6. If f (x) = x and g(x) = 2x – 3, then domain of (f o g) (x) is
(a) ( , 3) (b)3
,2
(c)3
,02
(d)3
0,2
(e)3
,2
7. If z =3
2
( 3 ) (3 4)
(8 6 )
i i
i, then |z| is equal to
(a) 8 (b) 2(c) 5 (d) 4(e) 10
8. Let 1 be a complex number. If || = l and z =1
1
,
then Re(z) is equal to
(a) l (b)1
| 1|
(c) Re() (d) 0
(e)
9. If z =2 i3e
, then l + z + 3z2 + 2z3 + 2z4 +3z5 is equal to(a) –3ei/3 (b) 3e i/3
(c) 3e 2i/3 (d) –3e2i/3
(e) 0
10. If z1= 2 2 (l + i) and z2 =l + i 3 , then zl2 z2
3 is equal to(a) 128 i (b) 64i(c) – 64 i (d) –128 i(e) 256
11. If the complex numbers zl, z2 and z3 denote the vertices of an isosceles triangle, right angled at zl, then
47. The value of sec2 (tan –13) + cosec2 (cot –1 2) is equal to(a) 5 (b) 13(c) 15 (d) 23(e) 25
48. If sin + cosec = 2, then the value of sin 6 + cosec6 isequal to
(a) 0 (b) 1(c) 2 (d) 23
(e) 26
49. If 0 < x < , then
sin8 7 sin 6 18sin 4 12sin 2
sin 7 6sin 5 12sin 3
x x x x
x x x =
(a) 2 sin x (b) sin x(c) sin 2x (d) 2 cos x
(e) cos x50. The points (2,5) and (5,l) are the two opposite vertices of arectangle. If the other two vertices are points on the straightline y = 2x + k, then the value of k is(a) 4 (b) 3(c) – 4 (d) –3(e) 1
51. The circumcentre of the triangle with vertices (8, 6), (8,–2)and (2, –2) is at the point(a) (2, –1) (b) (1, –2)(c) (5, 2) (d) (2, 5)(e) (4, 5)
52. The ratio by which the line 2x + 5y – 7 = 0 divides thestraight line joining the points (–4, 7) and (6,–5) is(a) 1 : 4 (b) 1 : 2(c) 1 : 1 (d) 2 : 3(e) 1 : 3
53. The number of points (a,b), Where a and b are positiveintegers lying on the hyperbola x2 – y2 = 512 is(a) 3 (b) 4(c) 5 (d) 6(e) 7
54. If p is the length of the perpendicular from the origin to the
line whose intercepts with the coordinate axes are1
3
and
1
4 then the value of p is
(a)3
4(b)
1
12(c) 5 (d) 12
(e)1
555. The slope of the straight line joining the centre of the
circle x2 + y2 –8x+ 2 y = 0 and the vertex of the parabolay = x2 – 4x +10 is
(a) 52
(b) 72
(c)3
2
(d)
5
2
(e)7
256. A straight line perpendicular to the line 2x + y = 3 is passing
57. If p and q are respectively the perpendiculars from the originupon the straight lines whose equations arex sec + ycosec = a and xcos –ysin = a cos 2, then4p2 + q 2 is equal to(a) 5a2 (b) 4a2
(c) 3a2 (d) 2a2
(e) a2
58. The shortest distance between the circles(x–1)2 +(y+2)2 =1 and (x + 2)2 + (y – 2)2 = 4 is(a) 1 (b) 2(c) 3 (d) 4(e) 5
59. The centre of the circle whose radius is 5 and which touchesthe circle x2 + y2 – 2x – 4y – 20 = 0 at (5, 5) is(a) (l0, 5) (b) (5 , 3)(c) (5 , 10) (d) (8 , 9)(e) (9, 8)
60. A circle passes through the points (0,0) and (0,l) and alsotouches the circle x2 + y2 = 16. The radius of the circle is(a) 1 (b) 2(c) 3 (d) 4(e) 5
61. A circle of radius 8 is passing through origin and the
point (4, 0). If the centre lies on the line y = x, then theequation of the circle is(a) (x–2)2 + (y–2)2 = 8 (b) (x+2)2 + (y+2)2 = 8(c) (x–3)2 + (y–3)2 =8 (d) (x+3)2 + (y+3)2 = 8(e) (x–4)2 + (y–4)2 = 8
62. The parametric form of the ellipse 4(x+ l)2 + (y –l)2 = 4 is
(a) x = cos – 1, y = 2 sin – 1(b) x = 2cos – 1, y = 2 sin +1(c) x = cos – 1, y = 2 sin + 1(d) x = cos + 1, y = 2 sin + 1(e) x = cos + 1, y = 2 sin – 1
63. A point P on an ellipse is at a distance of 6 units from a
focus. lf the eccentricity of the ellipse is3
5, then the distance
of P from the corresponding directrix is
(a)8
5
(b)5
8(c) 10 (d) 12(e) 15
64. If the length of the latus rectum and the length of transverse
axis of a hyperbola are 4 3 and 2 3 respectively, then
the equation of the hyperbola is
(a)2 2
13 4
x y
(b)2 2
13 9
x y
(c)2 2
16 9
x y
(d)2 2
16 3
x y
(e)2 2
13 6
x y
65. lf the eccentricity of the hyperbola2 2
2 21
x y
a bis 5
4 and
2x + 3y – 6 = 0 is a focal chord of the hyperbola, then thelength of transverse axis is equal to
(a)
12
5 (b) 6
(c)24
7(d)
24
5
(e)12
7
66. The length of the transverse axis of a hyperbola is 2 cos . The foci of the hyperbola are the same as that of the ellipse9x2 + 16y2 = 144. The equation of the hyperbola is
(a)2 2
2 21
cos 7 cos
x y
(b)2 2
2 21
cos 7 cos
x y
(c)2 2
2 21
1 cos 7 cos
x y
(d)2 2
2 21
1 cos 7 cos
x y
(e)2 2
2 21
cos 5 cos
x y
67. lf ˆˆ ˆ2 2 ,| |
a i j k b = 5 and the angle between a and
b
is6
, then the area of the triangle formed by these two
vectors as two sides is
(a)15
4(b)
15
2
(c) 15 (d)15 3
2
(e) 15 3
68. If .
a b = 0 and
a b makes an angle of 60° with
a , then(a) | | 2 | |
a b (b) 2 | | | |
a b
(c) | | 3 | |
a b (d) | | | |
a b
(e) 3 | | | |
a b
69. If ˆ ˆˆ ˆ ˆ ˆ, , i j j k i k are the position vectors of the vertices
of a triangle ABC taken in order, then A is equal to
x98. A straight line parallel to the line 2x – y + 5 = 0 is also a
tangent to the curve y2 = 4x + 5. Then the point of contactis(a) (2, l) (b) ( –1, 1)(c) (1, 3) (d) (3, 4)(e) (– 1, 2)
99. The function f (x) = 2x3 – l5x2 + 36x + 6 is strictly decreasingin the interval(a) (2, 3) (b) (– , 2)(c) (3, 4) (d) (– , 3) (4, )
(e) (– , 2) (3, )100. The slope of the tangent to the curve y2exy = 9e –3x2 at( –1, 3) is
(a)15
2
(b)
9
2
(c) 15 (d)15
2
(e)9
2
101. The radius of a cylinder is increasing at the rate of 5 cm/min
so that its volume is constant. When its radius is 5 cm andheight is 3 cm, the rate of decreasing of its height is(a) 6 cm/min (b) 3 cm/min(c) 4 cm/min (d) 5 cm/min(e) 2 cm/min
102. The function f (x) =22 1 1 4
151 30 4 5
x if x
x if x is not suitable
to apply Rolle’s theorem since(a) f (x) is not continuous on [1, 5]
(b) f (1) f (5)
(c) f (x) is continuous only at x = 4(d) f (x) is not differentiable in (4, 5)(e) f (x) is not differentiable at x = 4
103. The slope of the normal to the curve y = x2 –2
1. (d) Given : Number of cells, n = 5, emf of each cell = E Internal resistance of each cell = r
In series, current through resistance R
I =5
5
nE E
nr R r R
In parallel, current through resistance R
I =5
5
E nE E
r r nR r R R
n
According to question, I = I'
5 5
5 5 5
E E
r R r R
5r + R = r + 5 R
or R = r 1 R
r
2. (b) If V is the voltage across R and I the current through it,then power P delivered via transmission cables
P = VI or, I =P
V The power dissipated in the transmission cables
Pc = I 2 R =2
2
P R
V i.e., Pc
2
1
V
3. (a) Change in internal energy of the gas = Heat produced
due to current flowingGiven, I = 1A, R = 5t = 2 min = 120 sU = I 2 Rt = (1A)2 (50 ) (120 s) = 6 × 103 J = 6 kJ
4. (d) Inside a long solenoid, magnitude of magnetic field isincreased by introducing higher permeability medium.
5. (e) Given, I = 9 × 10 –5 kg m2, B = 162 × 10 –5 T
T =15 3
s20 4
In a vibration magnetometer
Time period, T = 2 I
MB or M =
2
2
4 I
BT
2 5
22
2 5
4 9 104 A m
316 10
4
6. (d)7. (c) Kinetic energy of the proton accelerated by cyclotron,
K =2 2 2
2
q B r
m
As =
2
qB
m
or qB = 2m
K =(2 )
2
m r
m
2mr
8. (b) Work done in rotating the magnetic dipole from position 1= 0° to 2= 180°
W = MB (cos1 – cos2) W = MB (cos – cos 180°) = 2 MB
9. (c) Lenz's law gives the polarity of induced emf.10. (a) As we know,
At resonance, inductive reactance X L = capacitivereactance X C
In an LCR series circuit, the phase difference betweenthe current and voltage
tan C L X X
R
[ Here X C – X L = 0]
i.e., the current and voltage are in phase.11. (b) Given, B = 0.01 T , A = R2 = × (1 m)2 = m2
= 100 rads –1
The maximum induced emf max= BA= 0.01××100 V = V
12. (e) Given I p = 4 A, I s, = 24 A, N p = 330, N s = ?
As p ps
s p p s s
N I I N N
I N I
N s =4
330 5524
13. (d) The average energy density due to magnetic field
u B =22
rms 0
0 0
1 1
2 2 2
B B
0rms
2
B B
=2 20 0 0
00 0
( / )1 1
4 4
B E c E B
c
=2 20 0
0 0 0 0 00
1 1 1
4 4 (1/ )
E E c
c
= 2 12 20 0
1 18.854 10 (1)
4 4 E
= 2.21 × 10 –12 J m –3
14. (b) The energy is equally divided between electric andmagnetic fields.The electric and magnetic field are in the same phase.The direction of propagation is perpendicular to bothelectric and magnetic fields. The pressure exerted bythe wave is equal to the energy density of the wave.The speed of the wave Vwave = E/B.
15. (a) The apparent flattening (oval shape) of the sun atsunset and sunrise is due to refraction.
16. (c) Given,polarising angle i p = 60°We have to find critical angle C
= 2 half life= 2 × 20 = 40 minutes22. (d) As we know, radius of nucleus, R = R0 A
1/3
where R0 is a constant and A is the mass number
1/3 1/3Te Te
Al Al
125 5
27 3
R A
R A
2 2
Te Te
Al Al
5 25
3 9
S R
S R
23. (d) According to Einstein's photoelectric equation, the
kinetic energy (k) of the emitted photoelectronK = h – 00 is a work function of the metal.
If the frequency of incident radiation () is doubled,thenKinetic energy K = 2hv – 0 = 2(hv – 0) + 0= 2K + 0i.e., K > 2K
24. (a) The significant result deduced from the Rutherford's
scattering is that whole of the positive charge isconcentrated at the centre of atom i.e. nucleus.
25. (c) On an average 2.5 neutrons are released per fission of the uranium atom.And the energy of the neutron released per fission of the uranium atom is 2 MeV.
26. (e) A
B
C
Y
1
1
1
0
1
0
OR GateAND Gate
27. (e) According to question, output resistance, R0 = 10 k
Input resistance, Ri = 2 k and = 49
Voltage gain, AV = × 0 1049 245
2i
R
R
28. (c) The light emitting diode is a heavily doped p-n junctionwhich emits spontaneous radiation under forward bias.
29. (e) A point-to-point communication mode is seen in faxtransmission. This mode of communication takes placeover a link between a single transmitter and a receiver.
30. (d) The maximum line-of-sight distance between thetransmitting and receiving antennas is
d M = 2 2T R Rh Rhwhere hT and h R are the heights of transmitting andreceiving antennas respectively.
d M = 2 2 2 2 8 Rh Rh Rh Rh
( hT = h R = h)
31. (b) The ionospheric layer acts as a reflector for a certainrange of frequencies (3 MHz to 30 MHz). Beyond 30MHz em. waves penetrate the ionosphere and escape.
32. (d) As we know, time period of a simple pendulum
T = 22
2
4 L Lg
g T
The maximum percentage error in g
100 100 2 100g L T
g L T
= 2% + 2(3%) = 8%33. (c) Least count of a screw gauge
=Pitch
Number of circular scale divisions
=1 mm
50 = 0.02 mm
Therefore the pitch and no. of circular scale divisionsare 1mm and 50 respectively.
34. (e) Let the two balls Pand Q meet at height x m from theground after time t s from the start.We have to find distance, BC = (100 – x)
100 m
A
B
x m
(100 – ) x m
C
25 m s –1
P
Q
For ball PS = x m, u = 25 m s –1, a = – g
From S = ut + 21
2at
x = 25t – 21
2 gt .......... (i)
For ball QS = (100 – x) m, u = 0, a = g
100 – x = 0 + 21
2 gt .......... (ii)
Adding eqns. (i) and (ii), we get100 = 25t or t = 4 sFrom eqn. (i),
x = 25 × 4 – 219.8 (4) 21.6 m
2
Hence distance from the top of the tower = (100 – x) m = (100 – 21.6 m) = 78.4 m
35. (c) As we know, distance traversed in nth second
S n = u +1
(2 1)2 a n
Here, u = 0, a =g
S n =1
(2 1)2
g n
Distance traversed in 1st second i.e., n = 1
S 1 =1 1
(2 1 1)2 2
g g
Distance traversed in 2nd second i.e., n = 2
S 2 =1 3
(2 2 1)2 2
g g
Distance traversed in 3rd second i.e., n = 3
S 3 =1 5
(2 3 1)2 2
g g
S 1 : S 2 : S 3 =1 3 5
: : 1:3 : 52 2 2
g g g
36. (e) Let u be the initial velocity that have to find and a bethe uniform acceleration of the particle.For t = 3s, distance travelled S = 12 m andfor t = 3 + 3 = 6 s distance travelled S = 12 + 30 = 42 mFrom, S =ut + 1/2 at 2
12 = u × 3 +21
32 a or 24 = 6u + 9a ...(i)
Similarly, 42 = u × 6 + 216
2 a
or 42 = 6u + 18a ...(ii)On solving, we get u = 1 m s –1
37. (a) As the ball, m = 10 g = 0.01 kg rebounds after striking
the wall Change in momentum = mv – (– mv) = 2 mv
Inpulse = Change in momentum = 2mv
= 1Impulse 0.54 N s2 7 m s
2 2 × 0.01 kgm
38. (b) Given, position vector ˆˆ ˆ2r i j k
and momentum ˆˆ ˆ2 p i j k
Angular momentum L r p
= ˆ ˆˆ ˆ ˆ ˆ(2 ) ( 2 )i j k i j k
= ˆˆ ˆ( 2 1) (2 1) ( 1 4)i j k
= ˆˆ ˆ 3i j k 39. (d) At maximum height ( H ) i.e. at point P the vertical
component of the projectile u sin = 0 whereas itshorizontal component i.e. u cos remains the same.
H
u
xu cos
u sin
u cosP
y
40. (a) According to the question, at any instant t , x = 4t 2, y = 3t 2
v x = 2(4 ) 8dx d
t t dt dt
and v y = 2(3 ) 6dy d
t t dt dt
The speed of the particle at instant t.
v =2 2 2 2(8 ) (6 ) 10 x yv v t t t
41. (b) The cyclist bends while taking turn in order to providenecessary centripetal force.
47. (c) Given, no. of rotation n = 1800 rpm = 1800 rpsTime, t = 2 minutes = 120s
Initial angular speed 0= –12 1800
rad s60
= 60 rad s –1
Final angular speed (as wheel comes to rest) = 0
Angular retardation = 0 –
t
= –260 0
rad s
120 2
48. (a) From the relation, angular momentum, L = mvr
v = L
mr
Centripetal force acting on the particle
F =
2
2 2
3
Lm
mv Lmr
r r mr
49. (e) Gravitational potential energy= gravitation potential × mass of body.Hence statement (e) is wrong.
50. (c) Gravitational potential energy of the body on thesurface of earth.
U s =GMm
R
where M = mass of the earthm = mass of the bodyGravitational potential energy of the body at a heighth (= R) form the surface of earth
U h =2
GMm GMm GMm
R h R R R
Required amount of energy = U h – U s
= 112 2
GMm GMm GMm R R R
=2 2
GMm mgR
R 2
GM g
R
51. (e) Total energy of the satellite = – Kinetic energy of thesatelite
= 21mv
2
52. (a) Volumes of two soap bubbles
3 31 1 2 2
4 4 and
3 3V r V r
where r 1 and r 2 are the radii of soap bubbles.Let s be the surface tension of the soap solution. Theexcess pressure inside the two soap bubbles, then
1 21 2
4 4 and
S S P P
r r
When these two bubbles coalesce under isothermalconditions a bigger bubble of radius R is formed. If V
and P be the volume and excess pressure inside this bigger bubble, then
34 4and
3
S V R P
R
Here Boyle's law holds as the bigger bubble is formedunder isothermal conditionsi.e., P1V 1 + P2V 2 = PV
53. (b) Bulk modulus is the ratio of hydraulic stress to the
corresponding strain.54. (a) Pressure difference between lungs and atmosphere
= (760 –750) mm of Hg= 10 mm of Hg = 1 cm of HgLet the boy can suck water from depth h. ThenPressure difference = hwater g = 1 cm of Hg
or, h × 1g cm –3 × 980 cm s –2
= 1 cm × 13.6g cm –3 × 980 cm s –2
h = 13.6 cmThe boy can suck water from the depth of 13.6 cm
55. (c) As we know, energy stored in a spring
21
2U kx x = extension (or compression) in the spring.k = spring constant of the springAs per question, for x = 1mm = 1 × 10 –3m
–3 21(1 10 m) 1 J
2U k .......(i)
If spring is further compressed by 1 mm then
3 21(2 10 m)
2U k
.......(ii)
Dividing eqn. (ii) by (i), we get
4 or 4
U
U U U
Work doneW = U – U = 4U – U
= 3U = 3 × 1 J = 3J56. (d) Root mean square (rms) velocity of gas molecules
rms3 Bk T
vm
where m = mass of each molecule of gas, T = absolutetemperature and k B = Boltzmann constant.
–1/2 1/2rms rmsor
T v v m T
m
57. (b) When efficiency of carnot engine, = 0.2Efficiency of a Carnot engine,
2 2
1 11– or, 0.2 1–
T T
T T
2
1or 0.8
T
T ...(i)
When T 2 is reduced by 50 K, its efficiency becomes 0.4
2
1
– 500.4 1–
T
T
2
1
– 50or 0.6
T
T ...(ii)
Dividing eqn. (i) by (ii)
2
2
0.8 4
– 50 0.6 3
T
T
3T 2 = 4T 2 – 200 or T 2 = 200 K
From eqn. (ii), 21 – 50 200 – 50
250K 0.6 0.6
T T
58. (e) Molar specific heat of the gas at constant volume,
63 [ 6 given]
2 2V f
C R R R f
59. (c) Total 5 degrees of freedom : 3 translational and 2rotational for a rigid diatomic molecule.
60. (a) The differential equation of simple harmonic motion is
2 2
2 22 0 or –2
d y d y y y
dt dt ...(i)
Standard equation of simple harmonic motion is
22
2 –
d y y
dt ...(ii)
Comparing eq. (i) and (ii),
2 = 2 or = 2
As we know,2
T
Time period,2 2
22
T s
61. (b) In parallel combination, the effective spring constant(K) of springsK = K 1 + K 2 + ...Hence, option (b) is wrong.
62. (b) In SHM, Total energy, 2 2total
1E
2m A
and, Kinetic energy, 2 2 2K
1E ( – )
2 m A x
where x is the distance from the mean position.At x = 0.707 A
2 2 2 2 2K
1 1E ( – (0.707 ) (0.5 )
2 2
m A A m A
As per question, Etotal = 100 J
2 2K
1E 0.5 m A 0.5 100J 50J
2
63. (e) Given : y1 = A sin [k ( x + ct )] = A sin (kx – kct )and, y2 = A sin [k ( x – ct )] = A sin (kx – kct )According to principle of superposition, the resultantwave on the string y = y1 + y2= A sin (kx + kct ) + A sin (kx – kct )
= A [sin (kx + kct ) + sin (kx – kct )]= 2 A sin kx cos kct
The positions of antinodes (where amplitude ismaximum) are given by|sin kx| = 1
1
2kx n
or 1
where 0,1,2,...2
x n nk
3 5
, , ,....2 2 2
xk k k
Therefore the distance between adjacent antinodes is/k.
64. (d) In the third harmonics or second overtone the vibrationof a stretched wire is as shown in the figure.
A N
A N
ATotal number of nodes N = 2Total number of antinodes A = 3
65. (d) In a tube closed at one end, only odd harmonics are present but in a tube open at both ends all theharmonics are present.The distance betwen successive nodes is equal to half the wavelength i.e. /2.Reflection of a wave from a rigid wall changes the phase by 180º or .
66. (c) Electric field due to the sheet02
E
Here = uniform surface density of charge.Force experienced by the electron
0
0
2or
2
F eF eE
e
–12 –12
–19
2 1.6 10 8.854 10
1.6 10
= 17.708 × 10 –5C m –2 = 177.08 × 10 –6 C m –2
= 177.08 C m –2
Area of the sheet, A = (0.5m)2 = 0.25m2 (given)Total charge on the sheetQ = A = (177.08 C m –2) (0.25 m2) = 44.27C
As the electron experiences the force directed awayfrom the sheet therefore the sheet must be negativelycharged.
–44.27 CQ
67. (d) Energy stored in a capacitor
221 1 1
2 2 2
QU QV CV
C ( Q = CV )
68. (e) According to Coulomb's law the electrostatic force between two point charges
1 21 22
q q 1F i.e. F q q
4 r
where is the permittivity of the medium, q1 and q2 aretwo point charges and r distance between two charges.
69. (d) The force of attraction between two charges q1 and q2 provides the required centripetal force for circular motion.
21 2
20
1. .,
4
q qmvi e
r r
O
q m1,
r
q2
F
or 1 2
0
1
4
q qv
mr
Time period of revolution
1 2
0
2 2
1
4
r r T
v q q
mr
3 30 0
1 2 1 2
4 162
mr mr r
q q q q
70. (c) Given, F = 1000 N, q = 2C and d = 2 cm = 2 × 10 –2 m
–11000500NC
2
F N E
q C
Also or V
E V Ed d
V = (500 NC –1) (2 × 10 –2m) = 10V
71. (a) The equivalent circuit of given network is as shown inthe figure.
86. (c) Presence of electron withdrawing group (e.g. –NO2group) in benzoic acid increases the acid strength while presence of electron donating group (e.g. –OCH3)decreases the acid strength.
COOH COOH COOH
NO2 NO2
NO2
COOH
OCH3
( )IV (I) (II) (III)
< < <
87. (b) Given organic compound will be 2-methylbenzaldehyde as it gives the following reactions.
CHOCH3
+ 2[Ag(NH3) ] + 3OH2+ –
COO –
CH3
+ 2Ag + 4NH3+ 2H O2
C H–
O
CH3+ H N2 NO2 –H2O
NO2 C H–
N NO2
NO2
CHO COOH
CH3 [O] COOH
2-methyl benzaldehyde
1, 2-benzenedicarboxylic acid
88. (e) 6 5 2 3C H NH CHCl 3KOH
6 5 2(carbylamine
reaction)
C H NC 3KCl 3H O
89. (b) All aliphatic amines are stronger than NH3. Aromaticamines are less basic then aliphatic amines. Amongaliphatic amines 2° amines are more basic than 1° and
3° amines hence the correct order of basic strengthwill be
(CH3)2 NH > C2H5 NH2 > NH3 > C6H5 NH2
82. (c) H C3 CH2 CH CH3
CH32-Methylbutane
H C3 CH2 CH CH2Cl
CH32-Methyl-1-chlorobutane
H C3 CH2 C CH3
CH32-Methyl-2-chlorobutane
Cl
H C3 CH CH CH3
CH3
3-Methyl-2-chlorobutane
Cl
H C3 CH CH2 CH2Cl
CH33-Methyl-1-chlorobutane
h
Cl2
83. (b) On exposure to air and sunlight, chloroform, acolourless heavy liquid, oxidises to carbonyl chloride
(phosgene), a highly poisonous gas used in warfare.
light3 2 3 2
unstable
1CHCl O CCl (OH) COCl HCl
2
84. (e) Diphenyl methanol (C6H5CHOHC6H5) does not haveCH(OH)CH3 group hence it will not form yellow
precipitate with an alkaline solution of iodine (iodoform reaction)
85. (b)|| || | ||
OH3 3 3 2
Aldol
O O OH O
CH CH H C CH CH CHCH C H
23H O
But-2-enal
CH CH CH CHO
It is the - H that adds on the carbonyl oxygen of thesecond carbonyl compound.
90. (a) Gabriel's phthalimide synthesis is used for synthesis of 1° amines,
93. (b) Glycogen is known as animal starch. This remain storedin liver and muscles and acts as a reserve food materialin animals.
94. (e) Green chemistry may be defined as the programme of developing new chemical products and chemical processes or making improvements in the already
existing compounds and processes so as to make lessharmful to human health and environment. This meansthe same as to reduce the use and production of hazardous chemicals.
95. (c) E =34 6
1910
hc 6.626 10 3 103 10 J
6626 10
Energy emitted by bulb =250 80
200J100
n × 3 × 10 –19 = 200 n = 6.66 × 1020 (where n = no. of photons)
32. (d) Given system of equations are3x + y – z = 2x + 0 y – z = 12x + 2y + az = 5Since system has unique solution thereforedeterminant is non-zero.
3 1 1
1 0 1 0
2 2 a
3(0 + 2) – 1(a + 2) – 1(2 – 0) 0 6 – a – 2 – 2 0 2 – a 0 a 2.
33. (e) Since matrix is singular therefore2 2
01 3
k
k
(2 – k) (3 – k) – 2 = 0 6 – 5k + k 2 – 2 = 0
k 2 – 5k + 4 = 0 – k 2 + 5k – 4 = 0 5k – k 2 = 4.
34. (a) Consider
log 1 log log
1 1log loglog 1
log log 11log
a a a
a ab
a ca
b c
b c
c
c
=
log1 log log
log log1 log
log log log1
b c
b c
c c
=
0 log log
log 0 log
log log 0
b c
b c
c c
= – log b [– log c × log c] + log c [– log b × log c]= 2 log b log c – 2 log b log c = 0Alternatively:
On solving determinant, we get
0 log log
log 0 log
log log 0
b c
b c
c c
which is the determinant of a skew symmetric matrixof odd order.Hence, its value is zero.
35. (a) Given system of equations can be re-written as2x + y – 4 = 0 ......... . (1)3x + 2y – 2 = 0 ........ (2)x + y + 2 = 0 ............ (3)
Consider
2 1 4
3 2 2
1 1 2
= 2 (4 + 2) – 1 (6 + 2) – 4 (3 – 2)
= 12 – 8 – 4 = 0Since, determinant is zero therefore system of equations has unique solution.
36. (d) Given : |x – 2| + | x + 2| < 4
|x – 2| =( 2) if 2
( 2) if 2
x x
x x
| x + 2| =( 2) if 2
( 2) if 2
x x
x xSolution for x > 2 :
( x – 2) + ( x + 2) < 42 x < 4 x < 2 (Contradiction)
Solution for x < –2 –( x – 2) – ( x + 2) < 4 – x + 2 – x – 2 < 4 –2 x < 4 x > – 2 (Contradiction)
Hence, there is no solution of the given inequation.37. (e) From the figure it is clear that there are 3 lines.
Line which passes from (0, 14) and (19, 14) is y = 14 In the shaded region 0 y 14Line which passes from (5, 0) and (0, 14) is14 x + 5y = 70 In the shaded region 14 x + 5 y 70Line which passes from (5, 0) and (19, 14) isx – y – 5 = 0 In the shaded region x – y 5Thus, inequations are 14 x + 5y 70, x – y 5, y 14.
38. (c) Statement given in option (c) is correct. [p (~ q) ] = (~ p) ~ (~ q)= (~ p) q
50. (c) Let ABCD be a rectangle with vertices A (2, 5) andC (5, 1).
D
AM
B
C
(2, 5)
(5, 1)
Now, M is the mid point of (Diagonal) line AC.
M is2 5 5 1 7
, ,32 2 2
Since M is on the given straight line y = 2 x + k
therefore M satisfies the equation of straight line.
Put x = 72
and y = 3 in y = 2 x +k
we get 3 =7
2 3 7 42
k k
51. (c) Let PQR be a triangle with vertices P(2, – 2),Q(8, – 2) and R(8, 6).
C
P Q
R
(8, – 2)
(8, 6)
(2, – 2)
Since circumcentre (C) is the mid point of Hypotenusetherefore
C =2 8 2 6
, (5,2)2 2
52. (c) Given equation is 2 x + 5 y – 7 = 0 and given points are(– 4, 7) and (6, – 5).Thus a = 2, b = 5, c = – 7and x1 = –4, y1 = 7
x2 = 6, y2 = – 5The required ratio is
= – 1 1
2 2
2( 4) 5 (7) 7
2(6) 5( 5) 7
ax by c
ax by c
= 1: 1
53. (b) a2 – b2 = 512 (a + b) (a – b) = 29
(a + b, a – b) = (28, 2), (27, 22), (26, 23), (25, 24)Since, a > b, a + b > a – b therefore the other combinations like (24, 25) etc cannot be accepted.(29, 1) also cannot be accepted since a and b are positive integers.
54. (e) Equation of line in intercept form is 1
1 13 4
x y
3 x + 4 y – 1 = 0
p = 2 2
3(0) 4(0) 1 1.
53 4
55. (b) Given eq n of circle is x2 + y2 – 8 x + 2 y = 0
So, centre: (4, – 1)Given eq n of parabola can be rewritten as y = ( x – 2)2 + 6
Two circles do not intersect dmin = C1 C2 – (r 1 + r 2)
= 5 – (1 + 2) = 2.59. (e) Let the required centre be (h, k )
5
(1, 2)
5
( , )h k
1 2, (5, 5)
2 2
h k
(By using mid-point formula) h = 9 and k = 8.
60. (b) Let centre of bigger circle be (0, 0) and centre of smaller circle be (h, k )
(0, 1)
(0, 0)
(h, k)
x y+ = 1622
The two circles touch internally
C1 C2 = |r 1 – r 2|
2 2 2 24h k h k
2 2 2 22 4 2h k h k
r = 2.61. (a) Let the required circle be
x2 + y2 + 2gx + 2 fy + c = 0 .... (1)
Since circle is passing through origin therefore c = 0Radius = 2 2 8g f c g2 + f 2 = 8 .... (2)
Since circle is passing through (4, 0) therefore we have16 + 8g = 0 g = – 2 f 2 = 4 f = – 2 Required circle is x2 + y2 – 4 x – 4 y = 0or ( x – 2)2 + ( y – 2)2 = 8
62. (c) Given equation of ellipse can be re-written as2 2
2
( 1) ( 1)1
1 2
x y
x + 1 = 1 cos and y – 1 = 2 sin
x = cos – 1 and y = 2 sin + 163. (c) Let P be any ponit on ellipse and F be the focus. Then
66. (b) Length of the transverse axis of a hyperbola = 2 cos Þ 2a = 2 cos a = cos Given equation of ellipse can be rewritten as
2 2
116 9
x y
Foci = 16 9,0 7,0
Since focus of hyperbola = 7,0
e cos =7
7cos
e
Now, a2 + b2 = 7 7 = b2 + cos2 Now, required equation of hyperbola is
2 2
2 21
cos 7 cos
x y
67. (a) Let a
= ˆˆ ˆ2 2i j k
1 4 4a = 3
Given b
= 5
Area =1 1 1 1
| | | | . | | sin 3 52 2 2 2
a b a b
=15
4
6
68. (e) .
a b a = cos60
a b a
cos 60° =1
2 =
.
a b a
a b a =
2
a
a b a
1
2=
a
a b...(1)
as . 0, a b a b
. a b b = cos30
a b b
cos 30° =
3 .
2
a b b
a b b
a b
60°
30°
b
a
3
2=
2
b
a b b=
b
a b...(2)
Dividing (2) by (1)
32
12
=
b
a
3 =
b
ab = 3
a
69. (e)
A B
C
j + k i + j
i + k
Now, ˆ ˆAC k j
and ˆ ˆAB k i
Let be the angle between AC
and AB.
cos =1 0 0 0 1
22 2
= 60° =3
70. (a) Consider ˆˆ| | 7a b
Squaring on both the sides, we get2 2 2
2 .a b a a b b
7 = 14 – 22 2
b b
21 4 9 and .
a a b b
2
7b
7b 71. (a) Consider
2 2 2 2
2 . . .a b c a b c a b a c b c
Since |a| b + c i. e. a. (b + c) = 0 .......... (1)Similarly b. (c + a) = 0 .......... (2)and c. (a + b) = 0 .......... (3)(1) + (2) + (3) gives2 (a. b + a. c + b. c) = 0
2 2 2 2
a b c a b c
72. (b) We know2 2 2 2
u v w u v w
+ 2 [u. v +v. w + u. w] = 0
9 + 16 + 25 = – 2 [u. v +v. w +u. w]
[u. v +v. w + u. w] = 50 252
73. (a) Since ˆˆ ˆ3 2 6i j k is a unit vector therefore
98. (b) Given curve is y2 = 4 x + 5on differentiating, we get
22 4
dy dy y
dx dx y
Given line is 2 x – y + 5 = 0 y = 2 x + 5
slope of line is 2. Therefore,
22 1 y
y
put y = 1 in the equation of curve, we get1 = 4 x + 5 x = – 1Hence, point of contact is (– 1, 1)
99. (a) Let f ( x) = 2 x3 – 15 x2 + 36 x + 6 f ( x) = 6 x2 – 30 x + 36= 6( x2 – 5 x + 6)
= 6( x – 2) (x – 3) < 0 x (2, 3)Hence, f ( x) is strictly decreasing in the interval (2, 3)
100. (c) Given curveis y2 e xy = 9e –3 x2
on differentiating both side, we get
y2 e xy 3.2 9 . 2 xydy dy x y e y e x
dx dx
9.e –3
3 3
3 .6 18
dy dye e
dx dx
3 3 3 39 27 6 18
dy dye e e e
dx dx
45.e –3 = 3 39 6 dy
e edx
15 =dy
dx
101. (a) Volume of cylinder = V = r 2h
2 .2 0
dV dh dr r h r
dt dt dt
Given 5,dr
dt height (h) = 3cm
25. 30.(5) 0dh
dt
150
625
dh
dt
102. (e) Since left hand and right hand derivative are not equalat x = 4 f ( x) is not differentiable at x = 4.Hence, Rolle’s theorem is not suitable on f ( x).