AIEEE 2003 PHYSICS PAPER

AIEEE 2003 HINTS & SOLUTIONS
Section I - Physics & Chemistry
1. The physical quantities not having same dimensions are
(1) speed and (μ ο ε ο ) – 1 / 2
(2) torque and work
(3) momentum and Planck’s constant
(4) stress and Young’s modulus
sol (3) Momentum [p] = [MLT–1] and Plank’s constant [h] = [ML2T–1]
2. Three forces start acting simultaneously on a particle moving with velocity, v. These forces are represented in
magnitude and direction by the three sides of a triangle ABC (as shown). The particle will now move with
velocity
(1) v, remaining unchanged
(2) less than v
(3) greater than v
(4) | v | in the direction of the largest force BC
r
sol (1) Net force on the particle is zero so the v remains unchanged
The coordinates of a moving particle at any time ‘t’ are given by x = αt and y = βt 3 . The speed of the
3
3.
particle at time ‘t’ is given by
(α 2 + β 2 )
(1)
2) 3t (α 2 + β 2 )
(3) 3t 2 (α 2 + β 2 )
(4) t 2 (α 2 + β 2 )
2 2 2 2
x 2 + y 2 ⇒ r2 = ( α + β ) t6 ⇒ 2r (dr / dt) = ( α + β ) . 6t5 ⇒ (dr / dt) = 3t2 (α 2 + β 2 ) .
sol (3) r =
4. A car, moving with a speed of 50 km / hr, can be stopped by brakes after at least 6m. If the same car is
moving at a speed of 100 km / hr, the minimum stopping distance is
(1) 6 m
(2) 12 m
(3) 18 m
(4) 24 m
sol (4) 0 = (250 / 18)2 + 2a . 6; S = u2 / 2a = 24.
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5. A boy playing on the roof of a 10 m high building throws a ball with a speed of 10 m /s. at an angle of 30°
with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground ?
(1) 8.66 m
(2) 5.20 m
(3) 4.33 m
(4) 2.60 m
v 2 sin 2θ (10) 2 sin(2 × 30)
= = 5 3 = 8.66 m
sol (1) calculate the range of ball R =
g 10
The displacement of a particle varies according to the relation x = 4 (cos πt + sin πt ). The amplitude of the
6.
particle is
(1) 8
(2) – 4
(3) 4
(4) 4 2
sol (4) x = 4 2 (cos π t . sin 45o + sin π t . cos 45o) = 4 2 sin ( π t + 45o)
7. Consider the following two statements :
A. Linear momentum of a system of particles is zero.
B. Kinetic energy of a system of particles is zero. Then
(1) A implies B and B implies A
(2) A does not imply B and B does not imply A
(3) A implies B but B does not imply A
(4) A does not imply B but B implies A
sol (4) Momentum being a vector quantity where K.E. being a scalar and a +ve quantity
8. A horizontal force of 10 N necessary to just hold a block stationary against a wall. The coefficient of
friction between the block and the wall is 0.2. The weight of the block is
(1) 2 N
(2) 20 N
(3) 50 N
(4) 100 N
10N
sol (1) f = μ R = mg = 0.2 × 10 = 2 N
9. A marble block of mass 2 kg lying on ice when given a velocity of 6 m / s is stopped by friction in 10 s.
Then the coefficient of friction is
(1) 0.01
(2) 0.02
(3) 0.03
(4) 0.04
sol (2) The Correct answer is 0.06
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10. A light spring balance hangs from the hook of the other light spring balance and a block of mass M kg hangs
from the former one. Then the true statement about the scale reading is
(1) Both the scales read M / 2 kg each
(2) Both the scales read M kg each
(3) The scale of the lower one reads M kg and of the upper one zero
(4) The reading of the two scales can be anything but the sum of the reading will be M kg
sol (2) The mass is hanging from the lower spring. Since the springs are light, tension is constant throughout
the springs.
11. A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m. If a force P is applied
at the free end of the rope, the force exerted by the rope on the block is
(1) PM / (M + m)
(2) Pm / (M + m)10 N
(3) Pm / (M – m)
(4) P
sol (1) P’ = Ma; P = (m + M)a ⇒ P’ = PM / (m + M)
12. A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads 49
N, when the lift is stationary. If the lift moves downward with an acceleration of 5 m / s2, the reading of the
spring balance will be
(1) 49 N
(2) 24 N
(3) 74 N
(4) 15 N
sol (2) When the lift is stationary R= mg ⇒ 49 = m × 9.8 ⇒ m = 5kg
When the lift is moving downward with an acceleration R = m (9.8 - a) ⇒ R = 5 (9.8-5) = 24N
When a U238 nucleus originally at rest, decays by emitting an alpha particle having a speed ‘u’, the recoil
13.
speed of the residual nucleus is
(1) – 4u / 238
(2) 4u / 238
(3) – 4u / 234
(4) 4u / 234
sol (4) Initial momentum of the system = Mass × velocity of nucleus = 238 × 0 = 0
Final momentum of the system = Momentum of α particle + Momentum of residual nucleus = 4u + 34v
r 4u 4u
By equating 4u + 234v = 0 ⇒ v = . But speed = .
234 234
14. A body is moved along a straight line by a machine delivering a constant power. The distance moved by the
body in time ‘t’ is proportional to
(1) t1⁄2
(2) t3⁄4
(3) t3 / 2
(4) t1⁄4
sol (3) P = const. (d / dt)(1⁄2 mv2) = const.
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A rocket with a lift-off mass 3.5 × 104 kg is blasted upwards with an initial acceleration of 10 m / s2. Then the
15.
initial thrust of the blast is
(1) 1.75 × 105 N
(2) 3.5 × 105 N
(3) 7.0 × 105 N
(4) 14.0 × 105 N
sol (3) F– mg = ma F = m (g + a) F = 7 × 105
16. Two spherical bodies of mass M and 5M and radii R and 2R respectively are released in free space with initial
separation between their centres equal to 12R. If they attract each other due to gravitational force only, then the
distance covered by the smaller body just before collision is
(1) 1.5 R
(2) 2.5 R
(3) 4.5 R
(4) 7.5 R
sol (4) am = 5a5m. since the force on both the masses will be equal
17. A particle performing uniform circular motion has angular momentum L. If its angular frequency is doubled
and its kinetic energy halved, then the new angular momentum is
(1) L / 2
(2) L / 4
(3) 2L
(4) 4L
Ε
1
sol (2) E = Lω ∴ Lα so if its angular frequency is doubled and its kinetic energy is haved then new
ω
3
L
angular momentum will become times.
4
18. Let F be the force acting on a particle having position vector r, and T be the torque of this force about the
origin. Then
(1) r . T = 0 and F . T = 0
(2) r . T = 0 and F . T 1 0
(3) r . T 1 0 and F . T = 0
(4) r . T 1 0 and F . T 1 0
sol (1) since T = F × r, ∴ T is ⊥ r to F & r
19. A circular disc X of radius R is made from an iron plate of thickness t, and another disc Y of radius 4R is
made from an iron plate of thickness t / 4. Then the relation between the moment of inertia IX and IY is
(1) IY = 64 IX
(2) IY = 32 IX
(3) IY = 16 IX
(4) IY = IX
1
ρ π R 4T ⇒ I ∝ R 4 t (where ρ = density, R= Radius, t = thickness)
sol (1) Moment of inertia of circular disc I =
2
4
I Y  R2   t2 
  ⇒ (4) 4   ⇒ I Y = 64 I X
1
=  
I x  R1  t 
4
   1
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20. The time period of a satellite of earth is 5 hours. If the separation between the earth and the satellite is
increased to 4 times the previous value, the new time period will become
(1) 20 hours
(2) 10 hours
(3) 80 hours
(4) 40 hours
3/ 2
R 
= T1 (4) 3 / 2 = 8 × T1 = 8 × 5 = 40 hours .
sol (4) T ∝ R 3 / 2 ⇒ T2 = T1  2 
 R1 
21. The escape velocity for a body projected vertically upwards from the surface of earth is 11 km / s. If the body
is projected at an angle of 45° with the vertical, the escape velocity will be
(1) 11 / 2 km / s
(2) 11 2 km / s
(3) 22 km / s
(4) 11 km / s
2GM
sol (4) Escape velocity does not depends on angle of projection ve =
R
22. A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so
that the mass executes SHM of time period T. If the mass is increased by m, the time period becomes5T/
3.Then the ratio of m / M is
(1) 5 / 3
(2) 3 / 5
(3) 25 / 9
(4) 16 / 9
M / k ⇒ (5T / 3) = 2 π {(Μ + m)/k} = 16 / 9.
sol (4) T ∝
A spring of spring constant 5 × 103 N / m is stretched initially by 5 cm from the unstretched position. Then
23.
the work required to stretch it further by another 5 cm is
(1) 6.25 N - m
(2) 12.50 N - m
(3) 18.75 N - m
(4) 25.00 N - m
1 1
K ( x2 − x1 ) = 5 × 10 3 (10 2 − 5 2 ) × 10 − 4 = 18.75 N − m
2 2
sol (3) Work done =
2 2
24. A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to the lower end.
The weight stretches the wire by 1 mm. Then the elastic energy stored in the wire is
(1) 0.1 J
(2) 0.2 J
(3) 10 J
(4) 20 J
sol (1) E = 1⁄2 F . (dl) = 1⁄2 × 200 × 10–3 = 0.1 J
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25. A body executes simple harmonic motion. The potential energy (P.E.), the kinetic energy (K.E.) and total
energy (T.E.) are measured as a function of displacement x. Which of the following statements is true ?
(1) P.E. is maximum when x = 0
(2) K.E. is maximum when x = 0
(3) T.E. is zero when x = 0
(4) K.E. is maximum when x is maximum.
1
m ω2 (a 2 − x 2 )
sol (2) Kinetic Energy is maximum at mean position KE =
2
1
mω2 a 2 (when x = a i.e., mean position)
KE max =
2
26. The length of a simple pendulum executing simple harmonic motion is increased by 21%. The percentage
increase in the time period of the pendulum of increased length is
(1) 10%
(2) 11%
(3) 21%
(4) 42%
l ⇒ T’ / T = (1.21 l / l ) ⇒ T’ = 1.1 T ⇒ % increased in T = 10%.
sol (1) T ∝
27. Two particles A and B of equal masses are suspended from two massless springs of spring constants k1 and
k2,respectively. If the maximum velocities, during oscillation, are equal, the ratio of amplitudes of A and B
is
(1) k1 / k2
(2) ( k1 / k 2 )
(3) k2 / k1
(4) ( k 2 / k1 )
KA KB
sol (4) (Vmax ) A = (Vmax ) B ⇒ a A ω Α = a B ωΒ ⇒ a A m = a s m [m A = m B given]
A B
aA k2
a A k1 = a B k 2 ⇒ =
aB k1
28. A metal wire of linear mass density of 9.8 g / m is stretched with a tension of 10 kg-wt between two rigid
supports 1 metre apart. The wire passes at its middle point between the poles of a permanent magnet, and it
vibrates in resonance when carrying an alternating current of frequency n. The frequency n of the alternating
source is
(1) 25 Hz
(2) 50 Hz
(3) 100 Hz
(4) 200 Hz
sol (2) In condition of resonance frequency of A.C. will be equal to natural frequency of wire
10 × 9.8
1T 1 100
n= = = = 50 Hz
−3
2l μ 2 × 1 9.8 ×10 2
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The displacement y of a wave travelling in the x-direction is given by y = 10–4 sin (600 t – 2x + (p / 3))
29.
metres, where x is expressed in metres and t in seconds. The speed of the wave-motion, in ms– 1, is
(1) 200
(2) 300
(3) 600
(4) 1200
ω
sol (2) y = a sin (ωt − kx + φ) in above equation wave velocity v =
k
So by comparing this with given equation ω = 600, k = 2 so v = 300 m/s
30. A tuning fork of known frequency 256 Hz makes 5 beats per second with the vibrating string of a piano. The
beat frequency decreases to 2 beats per second when the tension in the piano string is slightly increased. The
frequency of the piano string before increasing the tension was
(1) 256 + 5 Hz
(2) 256 + 2 Hz
(3) 256 – 2 Hz
(4) 256 – 5 Hz
sol (4) 256 ~ fp = 5 Þ fp = (256 – 5) or (256 + 5); since f ∝ T ,∴ fp increases to f ‘p;since 256 ~ f ‘p = 2,
∴ fp = (256 – 5).
A Carnot engine takes 3 × 106 cal. of heat from a reservoir at 627° C, and gives it to a sink at 27° C. The
31.
work done by the engine is
(1) zero
(2) 4.2 × 106 J
(3) 8.4 × 106 J
(4) 16.8 × 106 J
T2 Q2 Q2
300
sol (3) T = Q ⇒ = ⇒ Q2 = 10 6 cal
900 3 × 10 6
1 1
Work done = Q1 − Q2 = 3 × 10 6 − 10 6 = 2 × 10 6 cal = 2 × 4.2 × 10 6 J = 8.4 × 10 6 J
32. “Heat cannot by itself flow from a body at lower temperature to a body at higher temperature” is a statement or
consequence of
(1) first law of thermodynamics
(2) second law of thermodynamics
(3) conservation of momentum
(4) conservation of mass
sol (2)
33. During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute
temperature. The ration Cp / Cv for the gas is
(1) 3 / 2
(2) 4 / 3
(3) 2
(4) 5 / 8
γ 3
sol (1) Adiabatic law P1− γT γ = constant P ∝ T γ/γ−1 given that Pα T ∴ =3⇒ γ =
3
γ −1 2
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34. Which of the following parameters does not characterize the thermodynamic state of matter?
(1) Volume
(2) Temperature
(3) Pressure
(4) Work.
sol 4 P, V & T are thermodynamic variables which characterize the thermodynamic state of matter.
According to Newton’s law of cooling, the rate of cooling of a body is proportional to (∆θ)n, where ∆θ is the
35.
difference of the temperature of the body and the surroundings, and n is equal to
(1) one
(2) two
(3) three
(4) four.
sol (1) Rate of cooling ∝ ∆θ, ∴ n = 1
36. The earth radiates in the infra-red region of the spectrum. The spectrum is correctly given by
(1) Wien’s law
(2) Rayleigh Jeans law
(3) Planck’s law of radiation
(4) Stefan’s law of radiation.
sol (3) Infra-red lies in longer wavelength side and Planck’s law is applicable both for short and long waves.
37. To get three images of a single object, one should have two plane mirrors at an angle of
(1) 30°
(2) 60°
(3) 90°
(4) 120°
 360  360
sol (3) By using n =  − 1 ⇒ 3 = − 1 ⇒ θ = 90°
θ θ

38. Consider telecommunication through optical fibers. Which of the following statements is not true?
(1) Optical fibers may have homogeneous core with a suitable cladding
(2) Optical fibers can be of graded refractive index
(3) Optical fibers are subject to electromagnetic interference from outside
(4) Optical fibers have extremely low transmission loss.
sol (2) If optical fibers are subjected to electromagnetic interference from outside then due to interference,
the transmission of a particular signal could never have been possible
39. The image formed by an objective of a compound microscope is
(1) virtual and enlarged
(2) virtual and diminished
(3) real and diminished
(4) real and enlarged
sol (4) since uo > fo, ∴ image (vo) is real and enlarged.
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40. To demonstrate the phenomenon of interference, we require two sources which emit radiation
(1) of the same frequency and having a definite phase relationship
(2) of nearly the same frequency
(3) of the same frequency
(4) of different wavelengths
sol 1 For interference, the sources should be coherent
Dimensions of (1 / μ ο ε ο ), where symbols have their usual meaning, are
41.
(1) [LT –1]
(2) [L– 1 T]
(3) [L– 2T2]
(4) [L– 2T –2]
sol (4) C = 1 / (μοεο ) , \ [1 / (μ ο ε ο ) ] = [C2] = [L2T–2]
Three charges –q1, + q2 and –q3 are placed as shown in the figure. The x-component of the force on –q1 is
42.
proportional to
(1) (q2 / b2) – (q3 / a2) sin θ
(2) (q2 / b2) – (q3 / a2) cos θ
(3) (q2 / b2) + (q3 / a2) sin θ
(4) (q2 / b2) + (q3 / a2) cos θ
sol (3) force on –q1 due to q2 ∝ (q2 / b2) along +ve X-direction and that due to q3 ∝ (q3 / a2) along the
direction making an angle of θ with –ve Y-axis & (90° – θ ) with +ve Xaxis
∴ Fnet ∝ {(q2 / b2) + (q3 / a2) sin θ }
43. A thin spherical conducting shell of radius R has a charge q. Another charge Q is placed at the centre of the
shell.The electrostatic potential at a point P a distance R / 2 from the centre of the shell is
ο
(1) ((q + Q) / (4 πε )) (2 / R)
−q 3
(2) 2Q / 4πε ο R
(3) (2Q / 4πε ο R) – (2q / 4πε ο R) aθ
b
(4) (2Q / 4πε ο R) + (q / 4πε ο R) x
+q 2
-q1
sol (4) Net potential at P
Q 1q 2Q q
1
V= + . ⇒V = +
4πε 0  R  4πε 0 R 4πε 0 R 4πε 0 R

2
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If the electric flux entering and leaving an enclosed surface respectively is φ1 and φ 2 , the electric charge
44.
inside the surface will be
(1) (φ1 + φ 2 )ε ο
(2) (φ 2 − φ1 )ε ο
(3) (φ1 + φ 2 ) / ε ο
(4) (φ 2 − φ1 ) / ε ο
sol (2) Electric flux entering the surface (φ1 ) taken negative while flux leaving the surface (φ 2 ) taken
positive and according to Gauss Law
1
φ Total = (Qenclosed ) ⇒ Qenclosed = Qenclosed = φ Τotal × ε 0 ⇒ Qenclosed = (φ 2 − φ1 ) × ε 0
ε0
The work done in placing a charge of 8 × 10–18 coulomb on a condenser of capacity 100 micro-farad is
45.
(1) 32 ×10– 32 joule
(2) 16 ×10– 32 joule
(3) 3.1 ×10– 26 joule
(4) 4 ×10– 10 joule.
(8 × 10 −18 ) 2
Q2
= 32 × 10 −32 J
sol (1) By using W = ⇒W = −6
2 × 100 × 0
2C
46. A sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. The
capacitance of the capacitor
(1) increases
(2) decreases
(3) Remains unchanged
(4) Becomes infinite.
ε0 A εA
. It t ≈ neglibible, then C ' = 0
sol (3) By using C ' =
d −t d
47. A 3volt battery with negligible internal resistance is connected in a circuit as shown in the figure. The
current, I , in the circuit will be
(1) 1 / 3A
(2) 1A
(3) 1.5A
(4) 2A
sol (3) c Req. = (3 × 6) / (3 + 6) = 2 Ω , ∴ i = (3 / 2) = 1.5 A
48. An ammeter reads upto 1 ampere. Its internal resistance is 0.81 ohm. To increase the range to 10 A the
value of the required shunt is
(1) 0.09 Ω
(2) 0.03 Ω
(3) 0.3 Ω
(4) 0.9 Ω
sol (1) ig G = (i – ig) . S ⇒ 1 × 0.81 = (10 – 1) S ⇒ S = 0.09 W
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49. The length of a wire of a potentiometer is 100 cm, and the e.m.f of its stand ard cell is E volt. It is employed
to measure the e.m.f. of a battery whose internal resistance is 0.5 Ω . If the balance point is obtained at
l = 30 cm.from the positive end, the e.m.f. of the battery is
(1) 30 E / 100
(2) 30 E / 100.5
(3) 30 E / (100 – 0.5)
(4) 30 (E – 0.5i) / 100, where i is the current in the potentiometer wire.
sol (1) e – ir = (E / 100) × 30; i = 0 at balance point.
50. The length of a given cylindrical wire is increased by 100 %. due to the consequent decrease in diameter the
change in the resistance in diameter the change in the resistance of the wire will be
(1) 300 %
(2) 200%
(3) 100 %
(4) 50%
sol (1) R = ρI / ( π D2 / 4) ⇒ R = KLD–2 & R’ = K . (2l) D’2, ∴ R’ / R = 2D2 / D’2 = 4
⇒ R’ = 4R, ∴ dR’ = 3R, ∴ % change in R = (3R / R) × 100 % = 300 %
51. A strip of copper and another of germanium are cooled from room temperature to 80 K. The resistance of
(1) Each of these increases
(2) Each of these decreases
(3) Copper strip increases and that of germanium decreases
(4) Copper strip decreases and that of germainium increases.
sol (4) Resistance of a conductor increases with temperature where as for a semiconductor the resistance de
creases with rise in temperature.
52. A 220 volt, 1000 watt bulb is connected across a 110 volt. mains supply. The power consumed will be
(1) 1000 watt
(2) 750 watt
(3) 500watt
(4) 250 watt
2
V 
2
 110 
sol (4) By using Pconsumed =  A  × PR =   × 1000 = 250 watt
V   220 
 R
The thermo e.m.f of a thermo-couple is 25 μ V/ °C at room temperature. A galvanometer of 40 ohm resis
53.
tance, capable of detecting current as low as 10–5 A, is connected with the thermocouple. The smallest
temperature difference that can be detected by this system is
(1) 20° C
(2) 16°C
(3) 12° C
(4) 8° C
sol (2) According to ohm’s law V=iR so here if temperature difference is ∆θ then
25 × 10 −6 × ∆θ = 10 −5 × 40 ⇒ ∆θ = 16°C
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54. The negative Zn pole of a Daniell cell, sending a constant current through a circuit, decreases in mass by 0.13g
in 30 minutes. If the electrochemical equivalent of Zn and Cu are 32.5 and 31.5 respectively, the increase in the
mass of the positive Cu pole in this time is
1) 0.242 g
(2) 0.180 g
(3) 0.141g
(4) 0.126 g.
sol (4) m1 / m2 = Z1 / Z2 = E1 / E2 ⇒ m2 = (0.13 × 31.5) / 32.5 = 0.126 g
55. A thin rectangular magnet suspended freely has a period of oscillation equal to T. Now it is broken into two
equal halves (each having half of the original length) and one piece is made to oscillate freely in the same
field. If the period of oscillation is T ‘, the ratio T ‘ / T is
(1) 1/4
(2) 1 / 2 2
(3) 1/2
(4) 2
T' 1 T' 1
T
sol (3) By using T ' = ⇒ =⇒ =
2
n T n T
≈ (Ml2 / 12) / (Ml2 / (8 × 12)) = 8 & M / M’ = (ml / (m (l / 2)) = 2
{IM ' / I ' M } = 2 ⇒ T’ / T = 1⁄2
∴ T / T’ =
56. A particle of mass M and charge Q moving with velocity v describes a circular path of radius R when subjected
to a uniform transverse magnetic field of induction B. The work done by the field when the particle completes
one full circle is
(1) B Q v 2 π R
(2) (Mv2 / R) 2 π R
(3) zero
(4) BQ2 πR
sol (3) c W = F . d = Fd cos 90° = 0
A particle of charge–16 × 10–18 coulomb moving with velocity 10 ms–1 along the x-axis enters a region
57.
where a magnetic field of induction B is along the y-axis, and an electric field of magnitude 104 V/ m is
along the negative z-axis. If the charged particle continues moving along the x-axis, the magnitude of B is
(1) 10–3 Wb/m2
(2) 103 Wb/m2
(3) 105 Wb/m2
(4) 1016 Wb/m2
sol (2) qE = qvB sin 90° ⇒ B = E / v = 103 wb / m2
58. Curie temperature is the temperature above which
(1) a paramagnetic material becomes ferromagnetic
(2) a ferromagnetic material becomes paramagnetic
(3) a paramagnetic material becomes diamagnetic
(4) a ferromagnetic material becomes diamagnetic.
sol (2). χ ∝ (1 / T); χ
ferro
>> χ para , both of +ve valve. As is increased and so above curie temperature,
ferro-magnetic material behaves like paramagnetic material.
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59. A manetic needle lying parallel to a magnetic field requires W units of work to turn it through 60°. The
torque needed to maintain the needle in this position will be
(1) 2W
(2) 3W
(3) W
(4) ( 3 / 2)W
sol ( 2) Work done in deflecting magnetic needle through an angle θ W = MB (1- cos θ ) and torque in this
position ; τ = ÌÂsinθ
MB
so W-MB (1 − cos60°) ⇒ W = ...(1)
2
3MB
and τ = ÌÂ × sin 60° ⇒ τ = ...(2)
2
from equation (1) ⇒ τ = 3 W
60. The magnetic lines of force inside a bar magnet
(1) are from south-pole to north-pole of the magnet
(2) are from north-pole to south-pole of the magnet
(3) do not exist
(4) depend upon the area of cross-section of the bar magnet
sol (2) Lines of force inside the magnet are form south pole to north pole of the magnet
61. Two coils are placed close to each other. The mutual inductance of the pair of coils depends upon
(1) the currents in the two coils
(2) the rates at which currents are changing in the two coils
(3) relative position and orientation of the two coils
(4) the materials of the wires of the coils.
sol (3) Mutual inductance depends on the linkage of flux between the two coils hence on relative position &
orientation of the two coils
62. The core of any transformer is laminated so as to
(1) increase the secondary voltage
(2) reduce the energy loss due to eddy currents
(3) make it light weight
(4) make it robust and strong
sol (2) loss due to eddy current is reduced
63. When the current changes from +2A to – 2A in 0.05 second, an e.m.f. of 8V is induced in a coil. The
coefficient of self-induction of the coil is
(1) 0.1 H
(2) 0.2H
(3) 0.4 H
(4) 0.8 H
sol (1) e = – L (di / dt) ⇒ 8 = 4L / 0.5 ⇒ L = 0.1 H
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64. In an oscillating LC circuit the maximum charge on the capacitor is Q. The charge on the capacitor when the
energy is stored equally between the electric and magnetic field is
(1) Q
(2) Q / 2
(3) Q / 3
(4) Q / 2
sol (4) Max ES energy = Q2 / 2c; if EES = EM then EES = Q2 / 4C ⇒ Q’2 / 2C = Q2 / 4C ⇒ Q’ = Q / 2
65. Which of the following radiations has the least wavelength ?
(1) X-rays
(2) γ -rays
(3) β -rays
(4) α -rays
sol (2) λ r is least among the wavelengths of the given rays.
66. Two identical photocathodes receive light of frequencies f1 and f2. If the velocities of the photo electrons
(of mass m) coming out are respectivley v1 and v2, then
(1) v1 – v2 = [(2h / m) (f1 – f2)]1⁄2
(2) v1 − v2 = (2h / m) (f1 – f2)
2 2
(3) v1 + v2 = [(2h / m) (f1 + f2)]1⁄2
(4) v1 + v2 = (2h / m) (f1 + f2)
2 2
t
sol (2) hf1 = φ 0
2
mv1 ...(1)
2
12
and hf 2 = φ 0 + mv2 ....(2)
2
2h
Solving equation (1) and (2) we get v1 − v2 = ( f1 − f 2 )
2 2
m
67. Which of the following cannot be emitted by radioactive substances during their decay?
(1) electrons
(2) Protons
(3) Neutrinoes
(4) Helium nuclei
sol (2) Neutrinoes are emitted in β -decay along with e– & e+
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68. A radioactive sample at any instant has its disintegration rate 5000 disintegrations per minute. After 5
minutes, the rate is 1250 disintegrations per minute. Then, the decay constant (Per minute) is
(1) 0.8 ln 2
(2) 0.4 ln 2
(3) 0.2 ln 2
(4) 0.1 ln 2
sol (2) R ∝ Nt; 1250 = 500 (1⁄2)n ⇒ n = (t / T) = 2 ⇒ T = t / N = 5 / 2 = 2.5 min ; l = 0.693 / 2.5 = 0.4 ln2
∴
A nucleus with Z = 92 emits the following in a sequence : α, β − ,β− , α, α, α, α, α, β− ,β− , α, β + ,b+,
69.
α. The Z of the resulting nucleus is
(1) 74
(2) 76
(3) 78
(4) 82.
sol (3) DZ = – (2 × 8) + (4 × 1) – (2 × 1) = – 14 ∴ Z = 92 – 14 = 78
70. Which of the following atoms has the lowest ionization potential ?
(1) 8O16
(2) 7N14
(3) 55Cs133
(4) 18Ar40
sol (3) 55Cs133 has least stable configuration
2
The wavelengths involved in the spectrum of deuterium ( 1 D) are slightly different from that of hydrogen
71.
spectrum, because
(1) the attraction between the electron and the nucleus is different in the two cases
(2) the size of the two nuclei are different
(3) the nuclear forces are different in the two cases
(4) the masses of the two nuclei are different
sol (3) in H-spectra, wavelengths are determined on the basis of ES coulomb attraction between e– and +ve
nucleus (P+). In deutron nucleus, there is a neutron in addition to proton and nuclear force is
different which gives rise to different energy level.
In the nuclear fusion reaction 1 H + 1 H →
3
2 4
72. He + n, given that the repulsive potential energy between the two
2
nuclei is Na + , Ca 2+ , Mg 2+ the temperature at which the gases must be heated to initiate the reaction is nearly
[Boltzmann’s constant k = 1.38 × 10–23 J/ K]
(1) 109 K
(2) 107 K
(3) 105 K
(4) 103 K
sol (1)
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73. If the binding energy of the electron in a hydrogen atom is13.6 eV, the energy required to remove the
electron from the first excited state of Li++ is
(1) 122.4 eV
(2) 30.6 eV
(3) 13.6 eV
(4) 3.4 eV
13.6 × 32
13.6 z 2
sol (2) Ei = here n = 2; z = 3 so Ei = = 30.6 eV
2
4
n
74. The difference in the variation of resistance within temperature in a metal and a semiconductor arises
essentially due to the difference in the
(1) variation of scattering mechanism with temperature
(2) crystal structure
(3) variation of the number of charge carries with temperature
(4) type of bonding.
sol (3) With increase in temp., more electrons are released due to breaking of bonds in a semiconductor
hence conductivity increases. In metals, due to raise in temp., nucleus vibrate and obstruct the
passage of e– & hence conductivity decreases
75. In the middle of the depletion layer of a reverse-biased p-n junction, the
(1) potential is zero
(2) electric field is zero
(3) potential is maximum
(4) electric field is maximum.
sol (2) During reversed bias current through P–N junction is zero. since i ∝ E, ∴ E = 0
76. In Bohr series of lines hydrogen spectrum, the third line from the red end corresponds to which one of the
following inter–orbit jumps of the electron for Bohr orbits in an atom of hydrogen ?
(1) 3 → 2
(2) 5 → 2
(3) 4 → 1
(4) 2 → 5
sol (2) IIIrd line of visible region of H2 spectrum jumps from 5 → 2
77. The de Broglie wavelength of a tennis ball of mass 60 g moving with a velocity of 10 metres per second is
approximately [Planck’s constant = 6.63 × 10–34 Js]
(1) 10–33 metre
(2) 10–31 metre
(3) 10–16 metre
(4) 10–25 metre
sol (1) λ = h / mv = (6.63 × 10–34) / (60 × 10–3 × 10) = 1.105 × 10–33 M
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78. The orbital angular momentum for an electron revolving in an orbit is given by Öl(l + 1).(h / 2p). This
momentum for an s–electron will be given by
(1) +1⁄2.(h / 2 π )
(2) zero
(3) (h / 2 π )
(4) 2.(h / 2 π )
h
sol (2) mvΩ = l (l + 1) , thus for a ‘S’ orbital Angular momentum = 0
2π
79. How many unit cells are present in a cube–shaped ideal crystal of NaCl of mass 1 g ?
(1) 2.57 × 1021 unit cells
(2) 5.14 × 1021 unit cells
(3) 1.28 × 1021 unit cells
(4) 1.71 × 1021 unit cells
sol (1) No. of unit cells = 1⁄4 × (1 / 58.5) × 6.02 × 1023 = 2.57 × 1021
80. Glass is a
(1) micro–crystalline solid
(2) super–cooled liquid
(3) gel
(4) polymeric mixture
sol (2) super–cooled liquid of silicates
81. Which one of the following statements is correct ?
(1) Manganese salts give a violet borax bead test in the reducing flame
(2) From a mixed ppt. of AgCl and AgI, ammonia solution dissolves only AgCl
(3) Ferric ions give a deep green ppt. on adding potassium ferrocyanide solution
−
(4) On boiling a solution having K+, Ca2+ & HCO 3 ions we get a ppt. of K2Ca(CO3)2
sol (2) AgCl dissolves in NH3 giving colourless complex of [Ag(NH3)2]Cl.
82. According to the Periodic law of elements, the variation in properties of elements is related to their
(1) atomic masses
(2) nuclear masses
(3) atomic numbers
(4) nuclear neutron–proton number ratios
sol (3) Modern periodic law
83. Graphite is a soft solid lubricant extremely difficult to melt. The reason for this anomalous behaviour is that
graphite
(1) is a non–crystalline substance
(2) is an allotropic form of diamond
(3) has molecules of variable molecular masses like polymers
(4) has carbon atoms arranged in large plates of rings of strongly bound carbon atoms with weak interplate
bonds
sol (4) Graphite has carbon atoms arranged in large plates of rings of strongly bond carbon atoms with
weak interplate bonds.
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84. The IUPAC name of CH3COCH(CH3)2 is
(1) Isopropylmethyl ketone
(2) 2–methyl–3–butanone
(3) 4–methylisopropyl ketone
(4) 3–methyl–2–butanone
CH 3
O
|
||
−
CH 3 − C CH − CH
sol (4) 1 4 3
3
2
85. When CH2=CH–COOH is reduced with LiAlH4, the compound obtained will be
(1) CH3–CH2–COOH
(2) CH2=CH–CH2OH
(3) CH3–CH2–CH2OH
(4) CH3–CH2–CHO
LiAlH 4
sol (2) CH 2 = CH = COOH CH 2 = CH 2 − CH 2 − OH + H 2 O
2H
86. According to the kinetic theory of gases, in an ideal gas, between two successive collisions a gas molecule
travels
(1) in a circular path
(2) in a wavy path
(3) in a straight line path
(4) with an accelerated velocity
sol (3) kinetic assumption
87. A reduction in atomic size with increase in atomic number is a characteristic of elements of
(1) high atomic masses
(2) d–block
(3) f–block
(4) radioactive series
sol (3) Lanthanide contraction.
88. The general formula CnH2nO2 could be for open chain
(1) diketones
(2) carboxylic acids
(3) diols
(4) dialdehydes
sol (2) General formula for carboxylic acids
89. An ether is more volatile than an alcohol having same molecular formula. This is due to
(1) dipolar character of ethers
(2) alcohols having resonance structures
(3) inter–molecular hydrogen bonding in ethers
(4) inter–molecular hydrogen bonding in alcohols
sol (4) Due to inter molecular H 2 bond in alcohols. B.P. of Alcohols in much then ether.
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90. Among the following four structures I to IV, it is true that,
H
|
CH3 CH3
O H −C ⊕
CH || | |
3 |
CH 3 − C − CH − C2 H 5 C2 H 5 − CH − C2 H 5
||
C 2 H 5 − CH − C3 H 7 H
(I) (II) (III) (IV)
(1) all four are chiral compounds
(2) only I and II are chiral compounds
(3) only III is a chiral compound
(4) only II and IV are chiral compounds
sol (2) Because the carbon contains different substitutent groups.
91. Which one of the following processes will produce hard water ?
(1) saturation of water with CaCO3
(2) saturation of water with MgCO3
(3) saturation of water with CaSO4
(4) addition of Na2SO4 to water
sol (3) Presence of sulphates of Ca & Mg causes hardness in water.
92. Which one of the following compounds has the smallest bond angle in its molecule ?
(1) SO2
(2) OH2
(3) SH2
(4) NH3
sol (3) Almost pure p orbitals of Sulphur are involved
93. Which one of the following pairs of molecules will have permanent dipole moments for both members ?
(1) SiF4 & NO2
(2) NO2 & CO2
(3) NO2 & O3
(4) SiF4 and CO2
sol (3) Both are bent molecules
94. Which one of the following groupings represents a collection of isoelectronic species ?
(atomic numbers : Cs : 55, Br : 35)
(1) Na+, Ca2+, Mg2+
(2) N3–, F–, Na+
(3) Be, Al3+, Cl–
(4) Ca2+, Cs+, Br
sol (2) No. of electron is 10 in N 3− , F − , Na + , N 3− = 7 + 3, F − = 9 + 1, Na + = 11 − 1
In the anion HCOO– the two carbon–oxygen bonds are found to be of equal length. What is the reason for it ?
95.
(1) Electronic orbitals of carbon atom are hybridised
(2) The C=O bond is weaker than the C–O bond
(3) The anion HCOO– has two resonating structures
(4) The anion is obtained by removal of a proton from the acid molecule
sol (3) FaCl
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96. The pair of species having identical shapes for molecules of both species is
(1) CF4, SF4
(2) XeF2, CO2
(3) BF3, PCl3
(4) PF5, IF5
sol (2) XeF2 & CO2 is linear structure
97. The atomic numbers of vanadium (V), chromium (Cr), manganese (Mn) and iron (Fe)are respectively 23,
24,25 and 26. Which one of these may be expected to have the highest second ionization enthalpy?
(1) V
(2) Cr
(3) Mn
(4) Fe
sol (2) Elimination of first electron gives a stable cation, so it has a high second I.E.
Consider the reaction equilibrium, 2SO2(g) + O2(g) ↔ 2SO3(g); DHo = –198 KJ. On the basis of Le
98.
Chatelier’s principle, the condition favourable for forward reaction is :
(1) lowering of temperature as well as pressure
(2) increasing temperature as well as pressure
(3) lowering the temperature and increasing the pressure
(4) any value of temperature and pressure
sol (3) According to Lechataliear principle, lowering the temperature & increasing the pressure is
favourable condition for the reaction.
99. What volume of hydrogen gas, at 273 K and 1 atm pressure will be consumed in obtaining 21.6 g of
elemental boron (atomic mass = 10.8) from the reduction of boron trichloride by hydrogen ?
(1) 89.6 L
(2) 67.2 L
(3) 44.8 L
(4) 22.4 L
sol (3) 2BCl3 + 3H2 → 2B + 6HCl; use (moles of H2) / 3 = (moles of B) / 2
For the reaction equilibrium, N2O4 (g) ↔ 2NO2 (g); the concentrations of N2O4 and NO2 at equilibrium are
100.
4.8 × 10–2 and 1.2 × 10–2 mol L–1 respectively. The value of Kc for the reaction is :
(1) 3.3 × 102 mol L–1
(2) 3 × 10–1 mol L–1
(3) 3 × 10–3 mol L–1
(4) 3 × 103 mol L–1
sol (3) Kc = [NO2]2 / [N2O4] = (1.2 × 10–2 )2 / (4.8 × 10–2) = 3 × 10–3 mol L–1
The solubility in water of a sparingly soluble salt AB2 is 1.0 × 10–5 mol L–1. Its solubility product number
101.
will be
(1) 4 × 10–15
(2) 4 × 10–10
(3) 1 × 10–15
(4) 1 × 10–10
sol 1 KSP (AB2) = S (2S)2 = 4S3 = 4 × (1.0 × 10–5)3 = 4 × 10–15
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102. When during electrolysis of a solution of AgNO3 9650 coulombs of charge pass through the electroplating
bath, the mass of silver deposited on the cathode will be
(1) 1.08 g
(2) 10.8 g
(3) 21.6 g
(4) 108 g
sol (2) 9650 C = 1 / 10 eq. of electrons
For the redox reaction : Zn (s) + Cu2+ (0.1 M) → Zn2+ (1 M) + Cu (s) taking place in a cell, Eocell is 1.10
103.
Volt. Ecell for the cell will be [2.303 (RT / F) = 0.0591]
(1) 2.14 volt
(2) 1.80 volt
(3) 1.07 volt
(4) 0.82 volt
( Zn + + )
0.059
ο
sol (3) ε = ε − log
(Cu + + )
n
0.059 1
log
1.10 -
2 0.1
1.10 - 0.0295 log 10 = 1.07 volt
104. In a 0.2 molal aqueous solution of a weak acid HX the degree of ionization is 0.3. Taking Kf for water as
1.85,the freezing point of the solution will be nearest to
(1) –0.480oC
(2) –0.360oC
(3) –0.260oC
(4) +0.480oC
sol (1)
The rate law for a reaction between the substances A and B is given by Rate = k [A]n [B]m. On doubling the
105.
concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the
reaction will be as
1
(1) 2 ( m+ n )
(2) (m + n)
(3) (n – m)
(4) 2(n–m)
sol (4) Rate1 = K xn ym; Rate2 = K (2x)n (y / 2)m = 2(n–m)K xn ym; Ratio = 2(n–m)
106. 25 ml of a solution of barium hydroxide on titration with 0.1 molar solution of hydrochloric acid gave a
litre value of 35 ml. The molarity of barium hydroxide solution was
(1) 0.07
(2) 0.14
(3) 0.28
(4) 0.35
.1× 35
sol (2) m1v1 = m2 v2, m1 = = .14
25
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107. The correct relationship between free energy change in a reaction and the corresponding equilibrium constant
Kc is
(1) ∆G =RT ln Kc
(2) − ∆G =RT ln Kc
(3) ∆G ° =RT ln Kc
(4) − ∆G ° =RT ln Kc
sol (4) ∆G ° =RT ln Kc or ∆G ° = - RT In Kc
If at 298 K the bond energies of C–H, C–C, C=C and H–H are respectively 414, 347, 615, and 435 KJ mol–1,
108.
the value of enthalpy change for the reaction H2C=CH2 (g) + H2 (g) → H3C–CH3 (g) at 298 K will be ,
(1) +250 KJ
(2) –250 KJ
(3) +125 KJ
(4) –125 KJ
∆H
sol (4) = Σ B.E. (reactants) – Σ B.E. (products)
109. The enthalpy change for a reaction does not depend upon
(1) the physical states of reactants and products
(2) use of different reactants for the same product
(3) the nature of intermediate reaction steps
(4) the differences in initial or final temperatures of involved substances
sol (3) According to Hess low, enthalpy change for a reaction does not depend on the nature of inter
mediate reaction steps.
110. A pressure cooker reduces cooking time for food because
(1) heat is more evenly distributed in the cooking space
(2) boiling point of water involved in cooking is increased
(3) the higher pressure inside the cooker crushes the food material
(4) cooking involves chemical changes helped by a rise in temperature
sol (2) B.pt of H2O involved in cooking is increased.
111. If liquids A and B form an ideal solution,
(1) the enthalpy of mixing is zero
(2) the entropy of mixing is zero
(3) the free energy of mixing is zero
(4) the free energy as well as the entropy of mixing are each zero
sol (1)For ideal solution ∆H (mixing) = 0
For the reaction system : 2NO (g) + O2 (g) → 2NO2 (g) volume is suddenly reduced to half its value by
112.
increasing the pressure on it. If the reaction is of first order with respect to O2 and second order with respect to
NO, the rate of reaction will
(1) diminish to one–fourth of its initial value
(2) diminish to one–eighth of its initial value
(3) increase to eight times of its initial value
(4) increase to four times of its initial value
sol (3) Rate = K [O2][NO]2
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113. For a cell reaction involving a two–electron change, the standard e.m.f. of the cell is found to be 0.295 V at
25oC.The equilibrium constant of the reaction at 25oC will be
(1) 1 × 10–10
(2) 29.5 × 10–2
(3) 10
(4) 1 × 1010
sol (4) E° = (0.059 / 2) log10 KC
114. In an irreversible process taking place at constant T and P and in which only pressure – volume work is
being done, the change in Gibbs freee energy (dG) and change in entropy (dS), satisfy the criteria
(1) (dS)V, E < 0, (dG)T, P < 0
(2) (dS)V, E > 0, (dG)T, P < 0
(3) (dS)V, E = 0, (dG)T, P = 0
(4) (dS)V, E = 0, (dG)T, P > 0
sol (2) The value of (dS) is always greater than zero for an irreversible process.
115. Which one of the following characterstics is not correct for physical adsorption
(1) Adsorption on solids is reversible
(2) Adsorption increases with increase in temperature
(3) Adsorption is spontaneous
(4) Both enthalpy and entropy of adsorption are negative.
sol (2) Physical, adsorption decreases with temperature
In respect of the equation K = Ae –Ea / RT in chemical kinetics, which one of the following statements is correct?
116.
(1) k is equilibrium constant
(2) A is adsorption factor
(3) Ea is energy of activation
(4) R is Rydberg’s constant
sol (3) Ea is activation energy
117. Standard reduction electrode potentials of three metals A, B and C are respectively + 0.5 V, – 3.0 V and –
1.2 V. The reducing powers of these metals are
(1) B > C > A
(2) A > B > C
(3) C > B > A
(4) A > C > B
sol (1) High reduction potential implies strong O.A.
118. Which one of the following substances has the highest proton affinity?
(1) H2O
(2) H2S
(3) NH3
(4) PH3
sol (3) NH3 is a strong base compared to others and hence will be protonated faster.
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119. Which one of the following is an amphoteric oxide?
(1) ZnO
(2) Na2O
(3) SO2
(4) B2O3
sol (1) ZnO is amphoteric & dissolved in both acid & base
120. A red solid is insoluble in water. However it become soluable if some KI is added to water. Heating the red
solid in a test tube results in liberation of some violet coloured fumes and droplets of a metal appear on the
cooler parts of test tube. The red solid is
(1) (NH4)2Cr2O7
(2) HgI2
(3) HgO
(4) Pb3O4
sol (2) HgI2(scarlet red) + KI → K2[HgI4] (soluble); HgI2 → Hg + I2 (violet fumes);
121. Concentrated hydrochloric acid when kept in open air sometimes produces a cloud of white fumes. The
explanation for it is that.
(1) concentrated hydrochloric acid emits strongly smelling HCl gas all the time
(2) oxygen in the air reacts with the emitted HCl gas to form a cloud of chlorine gas
(3) strong affinity of HCl gas for moisture in air results in forming of droplets of liquid solution which
appears like cloudy smoke
(4) due to strong affinity for water, concentrated hydrocholoric acid pulls moisture of air towards itself. This
moisture forms droplets of water and hence the cloud.
sol (3)
122. What may be expected to happen when phospine gas is mixed with chlorine gas?
(1) The mixture only cools down
(2) PCl3 and HCl are formed and the mixture warms up
(3) PCl5 and HCl are formed and the mixture cools down
(4) PH3 . Cl2 is formed with warming up.
sol (2) Phosphine reacts violently with chlorine forming PCl3 and HCl.
The number of d-electrons retained in Fe2+ (At. no. of Fe = 26) ion is
123.
(1) 3
(2) 4
(3) 5
(4) 6
sol (4) Configuration for Fe2+ will be 4d6 4s0
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124. What would happen when a solution of potassium chormate is treated with an excess of dilute nitric acid?
(1) Cr3+ and Cr2O 7 − are formed
2
(2) Cr2O 7 − and H2O are formed
2
(3) CrO 2− is reduced to +3 state of Cr
4
(4) CrO 2− is oxidized to +7 state of Cr
4
2− + −2
sol (2) 2CrO4 + 2 H → Cr2 O7 + 2 H 2O
Acidic
Alkaline
125. In the coordination compound, K4[Ni(CN)4], the oxidation state of nickel is
(1) – 1
(2) 0
(3) + 1
(4) + 2
sol (3) 4 + x – 4 = 0 ⇒ x = 0
Ammonia forms the complex ion [Cu(NH3)4]2+ with copper ions in alkaline solutions but not in acidic
126.
solutions. What is the reason for it?
(1) In acidic solutions hydration protects copper ions
+
(2) In acidic solutions protons coordinate with ammonia molecules forming NH 4 ions and NH3 molecules are
not available
(3) In alkaline solutions insoluable Cu(OH)2 is precipitated which is sloluable in excess of any alkali
(4) Copper hydroxide is an amphoteric substance.
sol (2) NH 3 + H + ( Acid ) → NH 4 ; NH 3 not available as a ligand
+
127. One mole of the complex compound Co(NH3)5Cl3 gives 3 moles of ions on dissolutions in water. Ohe mole
of the same complex reacts with two moles of AgNO3 solution to yeild two moles of AgCl (s). The structure
of the complex is.
(1) [Co(NH3)5Cl] Cl2
(2) [Co(NH3)3Cl3].2 NH3
(3) [Co(NH3)4Cl2] Cl. NH3
(4) [Co(NH3)4Cl]Cl2. NH3
sol (1) As it gives three ions on dissolution and two moles of AgCl are precipitated.
The radius of La3+ (Atomic number of La = 57) is 1.06 Å. Which one of the following given values will be
128.
clolsest to the radius of Lu3+ (Atomic number of Lu = 71)
(1) 1.60 Å
(2) 1.40 Å
(3) 1.06 Å
(4) 0.85 Å
sol (4) Lu +4 = 0.85 Å
+3 +3
La57 → Lu71 size decreases due to lanthanide contraction
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The radionucleide 90Th234 undergoes two successive β -decays followed by one α - decay. The atomic
129.
number and the mass number respectively of the resulting radionucleide are
(1) 92 and 234
(2) 94 and 230
(3) 90 and 230
(4) 92 and 230
X 234 − α
Th234 Y 230
sol (3) −2β
→
 →

90 92 90
130. The half life of a radioactive isotope is three hours. If the initial mass of the isotope were 256 g, the mass of
it remaining undecayed after 18 hours would be.
(1) 4.0 g
(2) 8.0 g
(3) 12.0 g
(4) 16.0 g
sol (1) No. of half-lives = 18 / 3 = 6 hours; Mass undecayed = (1 / 26) × 256 = 4 g
131. Several blocks of magnesium are fixed to the bottom of a ship to
(1) keep away the sharks
(2) Make the ship lighter
(3) Prevent action of water and salt
(4) Prevent puncturing by under sea rocks
.
sol (3) Prevent action of water and salt
132. In curing cement plasters water is sprinkled from time to time. This helps in
(1) keeping it cool
(2) developing interlocking needle-like crystals of hydrated silicates
(3) hydrating sand and gravel mixed with cement
(4) converting sand into silicic acid
sol (2)
133. Which one of the following statements is not true?
2−
(1) The conjugate base of H2PO − is HPO 4
4
(2) pH + pOH = 14 for all aqueous solutions
(3) The pH of 1 × 10–8 M HCl is 8
(4) 96,500 couloumbs of electricity when passed through CuSO4 solution deposits 1 gram equivalent of
copper at the chathode.
sol (3) Being acidic solution pH should be less than 7
134. The correct order of increasing basic nature for bases NH3, CH3NH2 and (CH3)2NH is
(1) CH3NH2 < NH3 < (CH3)2NH
(2) (CH3)2NH < NH3 < CH3NH2
(3) NH3 < CH3NH2 < (CH3)2NH
(4) CH3NH2 < (CH3)2NH < NH3
sol (3) The presence of lone pair of electrons increases according to number of CH3 groups present due to +I
effect and also there is no steric factor in operation.
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135. Butane - 1 may be converted to butane by reaction with.
(1) Zn – HCl
(2) Sn – HCl
(3) Zn – Hg
(4) Pd / H2
sol (4) CH 3 − CH 2 − CH = CH 2 + H 2 → CH 3 − CH 2 − CH 2 − CH 3
pd
136. The solubilities of carbonates decrease down the magnesium group due to a decrease in
(1) lattice energies of solids
(2) hydration energies of cations
(3) inter-ionic attraction
(4) entropy of solution formation
sol (2) Due to decrease in hydration energy of cation & lattice energy remains almost unchanged.
During dehydration of alcohols to alkenes by heating with conc H2SO4 the initiation step is
137.
(1) protonation of alcohol molecule
(2) formation of carbocation
(3) elimination of water
(4) formation of an ester.
⊕
..
Protonetion
H 2 SO4 → H + + HSO4 ; C2 H 5OH + H +
−
of alcohol C 2 H 5 − ↓ - H Protoneted Alcohol
O
sol (1)
H
138. Which one of the following nitrates will leave behind a metal on strong heating ?
(1) Ferric nitrate
(2) Copper nitrate
(3) Manganese nitrate
(4) Silver nitrate
sol (4) 2AgNO3 → 2Ag + 2NO2 + O2
139. When rain is accompanied by a thunderstorm, the collected rain water will have a pH value
(1) slightly lower than that of rain water without thunderstorm
(2) slightly higher than that when the thunderstorm is not there
(3) uninfluenced by occurrence of thunderstorm
(4) which depends on the amount of dust in air.
sol (1)
140. Complete hydrolysis of cellulose gives
(1) D–fructose
(2) D–ribose
(3) D–glucose
(4) L–glucose
sol (3) Cellulose is a polymer of β − D - glucose
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141. For making good quality mirrors, plates of float glass are used. These are obtained by floating molten glass
over a liquid metal which does not solidify before glass. The metal used can be
(1) mercury
(2) tin
(3) sodium
(4) magnesium
sol (1)
142. The substance not likely to contain CaCO is
(1) a marble statue
(2) calcined gypsum
(3) sea shells
(4) dolomite
sol (2)
143. The reason for double helical structure of DNA is operation of
(1) van der Waal’s forces
(2) dipole–dipole interaction
(3) hydrogen bonding
(4) electrostatic attractions
sol (3)
144. Bottles containing C6H5I and C6H5CH2I lost their original labels. They were labelled A and B for testing. A
and B were separately taken in test tubes and boiled with NaOH solution. The end solution in each tube was
made acidic with dilute HNO3 and then some AgNO3 solution was added. Substance B gave a yellow
precipitate. Which one of the following statements is true for this experiment ?
(1) A was C6H5I
(2) A was C6H5CH2I
(3) B was C6H5I
(4) Addition of HNO3 was unnecessary
sol (1)
145. Ethyl isocyanide on hydrolysis in acidic medium generates
(1) ethylamine salt and methanoic acid
(2) propanoic acid and ammonium salt
(3) ethanoic acid and ammonium salt
(4) methylamine salt and ethanoic acid
sol (1) C2H5NC → C2H5NH2 + HCOOH
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146. The internal energy change when a system goes from state A to B is 40 KJ / mole. If the system goes from A to
B by a reversible path and returns to state A by an irreversible path what would be the net change in internal
energy ?
(1) 40 KJ
(2) > 40 KJ
(3) < 40 KJ
(4) Zero
sol (4) A → B
40
A −40 → B

∆H = zero
147. The reaction of chloroform with alcoholic KOH and p–toluidine forms
(1)H3C CN
(2)H3C N2Cl
(3)H3C NHCHCl2
(4) H3C NC
sol (4) Carbylamine reaction
148. Nylon threads are made of
(1) polyvinyl polymer
(2) polyester polymer
(3) polyamide polymer
(4) polyethylene polymer
sol (3)
149. On mixing a certain alkane with chlorine and irradiating it with ultraviolet light, it forms only one
monochloroalkane.This alkane could be
(1) propane
(2) pentane
(3) isopentane
(4) neopentane
sol (4) Neo-pentane gives only mono chloro pentane because of only 10 hydrogen atoms
CH 3
|
−C − CH 3
CH 3
|
CH 3
150. Which of the following could act as a propellant for rockets ?
(1) Liquid hydogen + liquid nitrogen
(2) Liquid oxygen + liquid argon
(3) Liquid hydrogen + liquid oxygen
(4) Liquid nitrogen + liquid oxygen
sol (3)
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Sectoin II - Maths
The foci of the ellipse (x2 / 16) + (y2 / b2) = 1 and the hyperbola (x2 / 144) – (y2 / 81) = (1 / 25) coincide. Then
1.
the value of b2 is
(1) 5
(2) 7
(3) 9
(4) 1
sol (2) (25x2 / 144) – (25y2 / 81) = 1 ⇒ {x2 / (12 / 5)2 – (y2 / (9 / 5)2} = 1 ; (9 / 5)2 = (12 / 5)2 (1 – e2)
⇒ 1 – e2 = 81 / 144 Foci ( ± ae , 0) ≡ ( ± a / e , 0) ; Hence b2 = 7
The normal atg the point (bt1 2 , 2bt1) on a parabola meets the parabola again in the point (bt2 2 , 2bt2), then
2.
(1) t2 = – t1 + (2 / t1)
(2) t2 = t1 – (2 / t1)
(3) t2 = t1 + (2 / t1)
(4) t2 = t1 – (2 / t1)
sol (2) Parabola : y2 = 4bx { from parameter)
MN = –1 / (dy / dx) = – y / 2b = – t1
⇒ N : Y – 2 bt1 = – t1 ( X – bt12) .....(i)
(bt22 , 2bt2) on this ⇒ t2 = (– t1 – 2 / t1)
If the two circles (x – 1)2 + (y – 3)2 = r2 and x2 + y2 – 8x + 2y + 8 = 0 intersect in two distinct points, then
3.
(1) r < 2
(2) r = 2
(3) r > 2
(4) 2 < r < 8
sol (4) For S1 : C1 (1 , 3) , R1 = r. For S2 : C2 (4 , –1) , R2 = 3.
Condition R – r < CC1 < R + r = | R2 – r | < 5 < | R2 + r | ⇒ 2 < r < 8
4. The degree and order of the differential equation of the family of all parabolas whose axis is x–axis , are
respectively.
(1) 1 , 2
(2) 3 , 2
(3) 2 , 3
(4) 2 , 1
sol (1) y 2 = 4a( x − h)
2 yy1 = 4a ⇒ yy1 = 2a ⇒ y1 + yy2 = 0
2
Degree =1, order = 2
⇒ y2 = 2yxdy / dx ⇒ y = 2x (dy / dx)
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5. Consider points A , B , C and D with position vectors 7 i – 4 j + 7 k , i – 6j + 10 k, –i , 3j + 4k and 5i – j +
5k respectively. Then ABCD is a
(1) rhombus
(2) rectangle
(3) parallelogram but not a rhombus
(4) square
sol . None of the option is correct
A = (7, -4, 7), B = (1, -6, 10), C(-1, -3, 4) and D (5, -1, 5)
(7 − 1) 2 + (−4 + 6) 2 + (7 − 10) 2 = 36 + 4 + 9 = 7
AB =
(1 + 1) 2 + (−6 + 3) 2 + (10 − 4) 2 = 4 + 9 + 36 = 7
BC =
(−1 − 5) 2 + (−3 + 1) 2 + (4 − 5) 2 = 36 + 4 + 1 = 41
CD =
(5 − 7) 2 + (−1 + 4) 2 + (5 − 7) 2 = 4 + 9 + 4 = 17
DA =
rr r
6. If u , v and w are three non–coplanar vectors , then (u + v – w) , (u – v) (v – w) equals
rrr
(1) u. v w
r rr
(2) u , w v
rrr
(3) 3 u. v w
(4) 0
rrr rrr rrr rr r rrr rrr rrr rrr
sol (1) Expn. = [u u v ] – [u v v ] – [u u w] + [u v w] – [ v u v ] – [ v u w] – [ v v v ] + [ v v w] –
rrr rrr rrr rrr rr r
[ w u v ] + [ w u w] + [ w v v ] – [ w v w] = [ u v w]
7. Two systems of rectangular axes have the same origin. If a plane cuts them at distances a , b , c and
a’ , b’ , c ‘ from the origin, then
(1) 1 / a2 + 1 / b2 – 1 / c2 + 1 / a’2 + 1 / b’2 – 1 / c’2 = 0
(2) 1 / a2 – 1 / b2 – 1 / c2 + 1 / a’2 – 1 / b’2 – 1 / c’2 = 0
(3) 1 / a2 + 1 / b2 + 1 / c2 – 1 / a’2 – 1 / b’2 – 1 / c’2 = 0
(4) 1 / a2 + 1 / b2 + 1 / c2 + 1 / a’2 + 1 / b’2 + 1 / c’2 = 0
sol (3) Sum of reciprocal of squares of intercepts = const.
The trigonometric equation sin–1 x = 2 sin–1 a , has a solution for
8.
(1) all real values of a
(2) | a | < 1⁄2
(3) | a | 3 1 / 2
(4) 1⁄2 < | a | < 1 / 2
sol (2) | sin (2 sin–1 a) | < 1 ⇒ 2a (1 − a 2 ) < 1 ⇒ a2 (1 – a2) < 1 / 4. ⇒ 2 | a | < 1⁄2
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9. Events A , B , C are mutually exclusive events such that P(A) = {(3x + 1) / 3}, P(B) = {(1 – x) / 4} and P(C)
= {(1 – 2x) / 2}. The set of possible values of x are in the interval
(1) [1 / 3 , 2 / 3 ]
(2) [1 / 3 , 13 / 3 ]
(3) [0 , 1 ]
(4) [1 / 3 , 1 / 2]
sol (4) P (A) , P (B) , P (C) all lie between 0 & 1
10. Five horses are in a race. Mr A selects two of the horses at random and bets on them. The probability that
Mr.A selected the winning horse is
(1) 3 / 5
(2) 1 / 5
(3) 2 / 5
(4) 4 / 5
sol (3) n( S ) =5 C2
n( E ) = 2 C1 + 2 C1
n( E ) 2 C1 + 2 C1 2
p( E ) = = =
5
n( S ) 5
C2
The value of ‘a’ for which one root of the quadratic equation (a2 – 5a + 3) x2 + (3a – 1) x + 2 = 0 is twice as
11.
large as the other , is
(1) – 2 / 3
(2) 1 / 3
(3) – 1 / 3
(4) 2 / 3
sol (4) (3 α )2 = {(1 – 3a) / (a2 – 5a + 3)}2 ; 2α 2 = {2 / (a2 – 5a + 3)}
⇒ (a / 2) = {(1 – 3a)2 / 2 (a2 – 5a + 3)} ⇒ a = 2 / 3
12. A body travels a distance s in t seconds. It starts from rest and ends at rest. In the first part of the journey, it
moves with constant acceleration f and in the second part with constant retardation r. The value of t is given
by
(1) {2s / (1 / f) + (1 / r)}
{2 s ( f + r )}
(2)
{2s (1 / f + 1 / r )}
(3)
(4) 2s (1 / f + 1 / r)
sol (3) S = 1⁄2 f t12 + 1⁄2 r t2 2 ...... (i)
Also f t1 = r t2 = Vmax ...........(ii)
⇒ From (ii) in (i) , t2= {2 s / r (1 + r / f )} ⇒ t = t2 {(1 + (r / f ) = {2 S (1 / f + 1 / r )
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13. Two stones are projected from the top of a cliff h metres high, with the same speed u so as to hit the ground
at the same spot. If one of the stones is projected horizontally and the other is projected at an angle q to the
horizontal then tan q equals
(1) 2g (u / h )
(2) 2h (u / g )
(3) u (2 / gh)
(4) ( 2u / gh)
sol (3) u cos θ ́ t1 = u ́ t2
(2 h/g ) . t1 = (2u sin θ / g) + {2/ g(u 2 sin 2θ/2g) + h}
t2 =
Note : Only (3) is dimensionless.
ω2n
ωn
1
ω2 n
If 1, w , w2 are the cube roots of unity, then, ∆ = ωn 1
14. is equal to
ω2 n ω n
1
(1) 1
(2) w
(3) w2
(4) 0
sol (4) Applying R1 → R1 + R2 + R3
As, 1 + ωn + ω2n = 0 (if you don’t remember this, put n = 1 and you will get the answer)
∴∆ = 0
15. Let u = i + j , v – j and w = i + 2j + 3k. If n is a unit vector such that u , n = 0 and v . n = 0, then | w . n | is
equal to
(1) 1
(2) 2
(3) 3
(4) 0
sol (3) Should hold for all n . Observe n = k is a possible solution. ⇒ | w . n | = 3
16. A particle acted on by constant forces 4i + j – 3k and 3i + j – k is displaced from the point i + 2j + 3k to the point
5i + 4j + k. The total work done by the forces is
(1) 30 units
(2) 40 units
(3) 50 units
(4) 20 units
sol (2) W = F . S = (7i + 2j – 4k) . (4i + 2j – 2k) = 40 Units
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17. The vectors AB = 3i + 4k , and AC = 5i – 2j + 4k are the sides of a triangle ABC. The length of the median
through A is
(1) 72
(2) 33
(3) 288
(4) 18
sol (2) Choose A as origin ⇒ B ≡ 3i + 4k ; C = 5i – 2j + 4k
⇒ P.V. of mid pt. of BC is M (4i – j + 4 k)
⇒ Vector AM = 4i – j + 4k ⇒ length = |AM | = 33.
18. The area of the region bounded by the curves y = | x – 1 | and y = 3 – | x | is
(1) 3 sq. units
y
(2) 4 sq. units
(3) 6 sq. units
(4) 2 sq. units
sol (2) x
1
O
Curve 1 : y = x – 1, x > 1 & 1 – x, x < 1
Curve 2 : y = 3 + x, x > 0 = 3 – x, x < 0 ⇒ Area = 4
The solution of the differential equation (1 + y2) + (x – ea y) (dy / dx) = 0. is (where tan–1 = a)
19.
(1) 2x ea y = e2a y + k
(2) x ea y = tan–1 y + k
(3) x e2a y = e tan–1 y + k
(4) (x – 2) = ke–a y
sol (1) Linear diff. eqn Bemoulli’s form
20. Let f (x) be a function satisfying f ‘ (x) = f (x) with f (0) = 1 and g (x) be a function that y x 1 O satisfies
f (x) + g (x) = x2 . Then the value of the integral 0 ∫1 f (x) g (x) dx , is
(1) e + (e2 / 2) – (3 / 2)
(2) e – (e2 / 2) – (3 / 2)
(3) e + (e2 / 2) + (5 / 2)
(4) e – (e2 / 2) – (5 / 2)
sol (2) f (x) = ex ⇒ g (x) = x2 – ex ⇒ I = 0 ∫1 ex (x2 – ex) dx = e – (e2 / 2) – 3 / 2
21. The lines 2x – 3y = 5 and 3x – 4y = 7 are diameters of a circle having area as 154 sq. units. Then the
equation of the circle is
(1) x2 + y2 + 2x – 2y = 47
(2) x2 + y2 – 2x + 2y = 47
(3) x2 + y2 – 2x + 2y = 62
(4) x2 + y2 + 2x – 2y = 62
sol (2) Intensection at centre (1, – 1) . pr2 = 154 ⇒ r = 7 ⇒ circle : (x – 1)2 + (y + 1)2 = 49
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If f (y) = e y , g (y) = y ; y > 0 and F(t) = 0 ∫ t f (t – y) g(y) dy , then
22.
(1) F (t) = et – (1 + t)
(2) F (t) = t e t
(3) F (t) = t e –1
(4) F(t) = 1 – e–1(1 + t)
sol (1) F (t) = 0 ∫ t et – y . ydy = et 0 ∫ t ye– ydy = et – (1 + t)
x 2 + 1)} , is
23. The function f (x) = log {x +
(1) an odd function
(2) a periodic function
(3) neither an even nor an odd function
(4) an even function
sol (1) f (– x)= log ( x 2 + 1) – x)1 f (x). Also f (x) + f (– x) = log {x2 + 1 – x2}= 0 ⇒ Odd function
If the sum of the roots of the quadratic equation ax2 + bx + c = 0 is equal to the sum of the squares of their
24.
reciprocals, then (a / c), (b / a) and (c / b) are in
(1) Geometric Progression
(2) Harmonic Progression
(3) Arithmetic Geometric Progression
(4) Arithmetic Progression
sol (2) a + b = – b / a, For reciprocal roots : eqn : cy2 + by + a = 0 ⇒ (1 / a) + (1 / b) = – (b / c)
⇒ (1 / a2) + (1 / b2) = (2ac – b2) / c2 ⇒ from cond n : (a / c), (b / a), (c / b) in HP
25. If the system of linear equations
x + 2ay + az = 0 x + 3by + bz = 0 x + 4cy + cz = 0
has a non–zero solution, then a , b , c
(1) are in G.P.
(2) aqre in H.P.
(3) satisfy a + 2b + 3c = 0
(4) are in A.P.
sol (2) Non-trivial sol. ⇒ D = |[1, 2a, a][1, 3b, b][1, 4c, c]| = 0 ⇒ a, b, c in H.P
26. If {(1 + i) / (1 – i)}x = 1, then
(1) x = 2n , where n is any positive integer
(2) x = 4n + 1, where n is any positive integer
(3) x = 2n + 1 , where n is any positive integer
(4) x = 4n , where n is any positive integer
sol (4) (eip /4 / e–ip / 4)x = 1 ⇒ ei x p / 2 = 1 ⇒ x = 4n
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27. A function f from the set of natural numbers to integers defined by f (n) = {{(n – 1) / 2}, when n is odd
{ – (n / 2) , when n is even is
(1) onto but not one–one
(2) one–one and onto both
(3) neither one–one nor onto
(4) one–one but not onto
sol (4) Not all integer covered
28. Let f (x) be a polynomial function of second degree. If f (1) = f (–1) f and a , b , c are in A.P., then
f ‘ (a) , f ‘(b) and f ‘(c) are in
(1) G.P.
(2) H.P.
(3) Arithmetic–geometric Progression
(4) A.P
sol (4) f (x) = Px2 + R ⇒ f ‘ (x) = 2Px
Y is equal to
29. The sum of the series (1 / 1.2) – (1 / 2.3) + (1 / 3.4) – .....upto
(1) log e 2 – 1
(2) log e 2
(3) log e (4 / e)
(4) 2 loge 2
sol (1) S = lt n → ∞ S n ((– 1)r +1) / r (r + 1)) = loge 2 – 1
r=1
30. A square of side a lies above the x–axis and has one vertex at the origin. The side passing through the origin
makes an angle a{0 < a < (p / 4)} withs the positive direction of x–axis. The equation of its diagonal not
passing through the origin is
(1) y (cos α + sin α ) + x (sin α – cos α ) = a B
(2) y (cos α + sin α ) + x (sin α + cos α ) = a A
(3) y (cos α + sin α ) + x (cos α – sin α ) = a M
(4) y (cos α – sin α ) – x (sin α – cos α ) = a O
1
C
2a cos a, 2a sin a)
sol (3) D1 : y1 = x tan a; Ao (–
⇒ M o (– a / 2 cos a, (a / 2 ) sin a) MBC = – cot a & D2 OBACMthrough M
If the pair of straight lines x2 – 2pxy – y2 = 0 and x2 – 2qxy – y2 = 0 be such that each pair bisects the angle
31.
between the other pair, then
(1) p = – q
(2) pq = 1
(3) pq = –1
(4) p = q
sol (1) P = – q
32. Locus of centroid of the triangle whose vertices are (a cos t , a sin t) , (b sin t , – b cos t) and (1 , 0), where t
is a parameter, is
(1) (3x – 1)2 + (3y)2 = a2 + b2
(2) (3x + 1)2 + (3y)2 = a2 + b2
(3) (3x + 1)2 + (3y)2 = a2 – b2
(4) (3x – 1)2 + (3y)2 = a2 – b2
sol (1) 3x = a cos t + b sin t + 1; 3y = a sin t – b cos t Eliminate t
AIEEE-2003 www.aieeepage.com P 36 of 45
If limx → 0 {log (3 + x) – log (3 – x) / x } = k , the value of k is
33.
(1) – (1 / 3)
(2) (2 / 3)
(3) – (2 / 3)
(4) 0
sol (2) L.H. rule ⇒ limit = 2 / 3
34. The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit
together is given by
(1) 30
(2) ∠5 × ∠4
(3) ∠7 × ∠5
(4) ∠6 × ∠5
sol (4) First arrange the men in 5! ways & them select 5 out of 6 gaps to arrange the women. for women
⇒ 6! ways. ⇒ 6! ways ⇒ total = 6! × 5!
35. A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the
first five questions. The number of choices available to him is
(1) 196
(2) 280
(3) 346
(4) 140
sol (1) Either 4 from 1st fine or 5 from first five ⇒ 5C4 × 8C6 + 5C5 × 8C5 = 196
If A = {[a , b] [b , a]} and A2 = [a , b ] [ b , a], then
36.
(1) α = a2 + b2 , β = 2ab
(2) α = a2 + b2 , β = a2 – b2
(3) α = 2ab , β = a2 + b2
(4) α = a2 + b2 , β = ab
sol (1) On multiplying & equating corresponding elements
In an experiment with 15 observations on x, the following results were available : Σx2 = 2830 , Σx = 170
37.
One observation that was 20 was found to be wrong and was replaced by the correct value 30. Then the
corrected variance is
(1) 188.66
(2) 177.33
(3) 8.33
(4) 78.00
sol (4) x1 +.... + xn – 1 = 170 – 20 = 150; x1 2 + .... + xn – 1 2 = 2830 – 400 = 2430
New S1 xi = 180. New Sxi2 = 3330 ⇒ New variance = 78; New mean = 12
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The shortest distance from the plane 12 x + 4 y + 3 z = 327 to the sphere x2 + y2 + z2 + 4 x – 2 y – 6z = 155 is
38.
(1) 11(4 / 13)
(2) 13
(3) 39
(4) 26
sol (2) Shortest dist. ⇒ perp.dist. ⇒ dist. from centre to plane = 13 units
39. The two lines x = ay + b, z – cy + d and x = a’ y + b’, z = c’y + d’ will be perpendicular, if and only if
(1) aa’ + bb’ + cc’ = 0
(2) (a + a’) (b + b’) + (c + c’) = 0
(3) aa’ + cc’ + 1 = 0
(4) aa’ + bb’ + cc’ + 1 = 0
sol (3) Apply. perp. condn for planes
40. The lines (x – 2 ) / 1 = (y – 3) / 1 = (z – 4) / – k and (x – 1) / k = (y – 4) / 2 = (z – 5) / 1 are coplanar if
(1) k = 1 or – 1
(2) k = 0 or – 3
(3) k = 3 or – 3
(4) k = 0 or – 1
sol (2) D = 0 for co-plannar ⇒ k = 0 or – 3
If f (a + b – x) = f (x), then a ∫ b x f (x) dx is equal to
41.
(1) (a + b) / 2 a ∫ b f (x) dx
(2) (b – a) / 2 a ∫ b f(x) dx
(3) (a + b) / 2 a ∫ b f (a + b – x) dx
(4) (a + b) / 2 a ∫ b f (b – x) dx
sol (1) I = a ∫ b xf (x) dx = a ∫ b (a + b – x) f (a + b – x) dx ⇒ 2I = (a + b) a ∫ b f (x) dx.
42. A couple is of moment G and the force forming the couple is P. If P is turned through a right angle, the
moment of the couple thus formed is H. If instead, the forces P are turned through and angle a,
then the moment of couple be comes
(1) H cos α + G sin α
(2) G cos α + H sin α
(3) H sin α – G cos α
(4) H sin α – G cos α
sol (2) For α = 0, moment = G ; α = 90°, moment = H
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43. The resultant of forces P and Q is R. If Q doubled then R is doubled. If the direction of Q is reversed, then
R is again doubled. Then P2 : Q2 : R2 is
(1) 2 : 3 : 2
(2) 1 : 2 : 3
(3) 2 : 3 : 1
(4) 3 : 1 : 1
sol (2) P + Q = R; |P + 2Q| = 2|R| ; |P – Q| = 2|R| Hence P2 : Q2 : R2
44. The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively, then
P (X = 1) is
(1) 1 / 16
(2) 1 / 8
(3) 1 / 4
(4) 1 / 32
sol (4) np = 4 ; npq = 2 ⇒ 1 – p = 1 / 2 ⇒ p = 1 / 2, q = 1 / 2, n = 8
p(x = 1) = 8 × p (1 – p)7 = 23 / 28 = 1 / 32
If f (x) = xn, then the value of f (1) – f ‘(1) / 1! + f ‘’ (1) / 2 ! – f ‘’’ (1) / 3! + ...... + (– 1)n f n (1) / n! is
45.
(1) 2n – 1
(2) 0
(3) 1
(4) 2n
sol (2) Taylor expansion with x = 1. = zero
46. The sum of the radii of inscribed and circumscribed circles for an n sides regular polygon of side a, is
(1) (a / 2) cot ( π / 2)
(2) a cot ( π / 2)
(3) (a / 4) cot ( π / 2n)
(4) a cot ( π / n)
sol (1) Draw fig. a2 = 2Rc
2 {1 – cos (2 π / 2n)} & ri = (a / 2) cot ( π / 2n)
⇒ Rc + ri = (a / 2) cot ( π / 2n)
47. If x1, x2, x3 and y1, y2, y3 are both in G.P with the same common ratio, then the point (x1, y1), (x2, y2) and (y3 , y3)
(1) lie on an ellipse
(2) lie on a circle
(3) are vertices of a triangle
(4) lie on a straight line
sol (4) All lie on the line y = rx r : common ratio
If z and ω are two non-zero complex numbers such that zω = 1 Arg(z) – Arg ( ω ) = ( π / 2), then zω is equal
48.
to
(1) – 1
(2) i
(3) – i
(4) 1
sol (3) Consider z = i, ω = 1 ⇒ zω = – i
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49. Let z1 and z2 be two roots of the equation z2 + az + b = 0, z being complex. Further assume that the origin, z1
and z2 form and equilateral triangle. Then
(1) a2 = 2b
(2) a 2 = 3b
(3) a2 = 4b
(4) a2 = b
sol (4) z1 2 + z22 = z1 z2 (eq. D) ⇒ a2 – 2b = b ⇒ a2 = b
The upper (3 / 4)th portion of a vertical pole subtends an angle tan–1 (3 / 5) at a point in the horizontal plane
50.
through its foot and at a distance 40 m from the foot. A possible height of the vertical
(1) 40 m
(2) 60 m
(3) 80 m
(4) 20 m
sol (1) (l / 4) cot θ = 40 & l cot (tan– 1 3 / 5 + θ ) = 40, Solving, we get l = 40
In a triangle ABC, medium AD and BE are drawn. If AD = 4, 51.
are of D ABC is A
(1) 16 / 3
(2) 32 / 3 E
F
30°
(3) 64 / 3 8/3
(4) 8 / 3 60°
4/3
B C
D
sol (4) Apply sine rule to find OB, & then AO / OD = BO / OE = CO / OF
Area = D AOB + D AOE + D BOD + D CEOD = 32 / 3 3
If in a triangle ABC a cos2 (C / 2) + c cos2 (A / 2) = 3b / 2, then the sides a, b and c
52.
(1) are in G.P
(2) are in H.P
(3) satisfy a + b = c
(4) are in A.P
sol (4) (a / 2) (1 + cos C) + (c / 2) (1 + cos A) = 3b / 2 But a cos C + c cos A = b
(Napier) ⇒ (a + c) + (a cos C + c cos A) = 3b
53. a, b, c are 3 vectors, such that a + b + c = 0.|a| . |b| + |b| . |c| + |c| . |a| = 3, then a . b + b . c + c . a is equal to
(1) – 7
(2) 7
(3) 1
(4) 0
sol (1) Choose a = i, b = 2i, c = – 3i ⇒ – 7
If x is positive, the first negative term in the expansion of (1+ x) (27 /5) is
54.
(1) 5th terms
(2) 8th
(3) 6th term
(4) 7 th terms
sol (2) Tr + 1 = —— ⇒ 8th term negative
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The number of integral terms in the expansion of ( 3 + 8 5) 256 is
55.
(1) 33
(2) 34
(3) 35
(4) 32
sol (1) Integral term ⇒ ( 3 )256 – r × (5)r / 8 to be integral ⇒ r = 0, 8, . . ., 256 = {33} terms
If nCr denotes the number of combinations of n things taken r at a time, then the expression
56.
n
Cr +1 + nCr – 1 + 2 × nCr equals
(1) n +2Cr +1
(2) n +1Cr
(3) n +1Cr +1
(4) n +2Cr
sol (1) nCr – 1 + nCr = n + 1Cr ; nCr + 1 + nCr = n + 1Cr + 1 n + 1Cr = n + 1Cr + 1 = n + 1Cr + 1
57. Two particles start simultaneously from the same point and move along two straight lines, one with uniform
velocity u and the other from rest with uniform acceleration f. Let a be the angle between their directions
of motion. The relative velocity of the second particle w.r.t. the first is least after a time
(1) f cos α / u
(2) u sin α
(3) u cos α / f ft
(4) u cos α / f
|Vrel|2 = (u – ft cos α )2 + (ft sin α )2 ⇒ (d / dt) {|Vrel|2} = 0
sol (3)
If | [a a2 1 + a3] [b b2 1 + b3] [c c2 1 + c3] | = 0 and vectors (1, a, a2), (1, b, b2) and (1, c, c2) are
58.
non-coplanar , then the product abc equals
(1) – 1
(2) 1
(3) 0
(4) 2
sol (1) D = |[1, a, a2][1, b, b2][1, c, c2]| + abc |[1, a, a2][1, b, b2][1, c, c2]| = 0
⇒ abc = – 1 u ft
If the function f(x) = 2x3 – 9ax2 + 12 a2 x + 1, where a > 0, attains its maximum and minimum at p and q
59.
respectively such that p2 = q, then a
(1) 1
(2) 2
(3) 1⁄2
(4) 3
sol (2) f’(x) = 0 ⇒ x2 – 3ax + 2a2 = 0 ⇒ (x – 2a) (x – a) = 0
f’’(x) = 12x – 18a > 0 for x = 2a ⇒ min at 2a ∴ p2 = q ⇒ a2 = 2a ⇒ a = 2
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If f (x) = {xe – ((1 / |x| + ( 1 / x)| , x ≠ 0 ; {0 , x = 0 then f (x) is
60.
(1) continuous for all x but not differentiable at x = 0
(2) neither differentiable nor continuous at x = 0
(3) discontinuous everywhere
(4) continuous as well as differentiable for all x
sol (2) f(x) = { x e– 0, x → 0–
{ x e– 2 / x, x → 0+ (undefined 1 / x) ⇒ NC, ND at O
The number of real solutions of the equation x2 – 3 | x | + 2 = 0 is
61.
(1) 4
(2) 1
(3) 3
(4) 2
sol (1) (|x| – 1) (|x| – 2) = 0 ⇒ 4 real soln.
The value of the integral I = ∫1 x (1 – x)n dx is
62.
0
(1) 1 / (n + 2)
(2) 1 / (n + 1) – 1 / (n + 2)
(3) 1 / (n + 1) – 1 / (n + 2)
(4) 1 / (n + 1)
sol (2) Put 1 – x = t ⇒ I = 0 ∫ 1 tn × (1 – t)dt = (1 / (n + 1)) – (1 / (n + 2))
2
The value of lim x → 0 { 0 ∫ x sec2 t d t / x sin x} is
63.
(1) 2
(2) 1
(3) 0
(4) 3
sol (2) LH rule L = (sec2 (x2) × 2x) / (x cos x + sin x);
again LH rule ⇒ (2 sec2 (x2) + 0) / (2 cos x – x sin x) ⇒ L = 1.
The radius of the circle in which the sphere x2 + y2 + z2 + 2x – 2y – 4z – 19 = 0 is cut by the plane
64.
x + 2y + 2z + 7 = 0 is
(1) 2
(2) 3
(3) 4
(4) 1
1
sol (2) Centre (–1, 1, 2) ⇒ distance from plane = 4.
4
5
Radius of sphere = 5 ⇒ radius of circle = (5 − 4 ) = 3
2 2
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65. A tetrahedron has vertices at O (0, 0, 0), A (1, 2, 1), B (2, 1, 3) and C ( – 1, 1, 2). Then the angle between
the faces OAB and ABC will be
(1) cos–1 (17 / 31)
(2) 30°
(3) 90°
(4) cos–1 (19 / 35)
sol (4) Use vectors. Find normals to OAB and ABC
Angle between faces = angle between normals
Domain of definition of the function f(x) = 3 / (4 – x2) + log10 (x3 – x), is
66.
(1) (– 1, 0) ∪ (1, 2)
(2) (1, 2) ∪ (2, ∞ )
(3) ( –1 , 0) ∪ (1, 2) ∪ (2 , ∞ )
(4) (1, 2)
sol (3) x3 – x > 0 & 4 – x2 ≠ 0
r =1 Σ
If f : R → R satisfies f (x + y) = f(x) + f (y), for all x, y ∈ R and f (1) = 7, n
f (r) is
67.
(1) 7(n + 1) / 2
(2) 7n (n + 1)
(3) 7n(n + 1)
(4) 7n / 2
sol (3) f (x) = k x ⇒ r =1 Σ n n f (r) = k r =1 Σ
n r = kr (n +1) / 2
f (1) = 7 ⇒ k = 7
68. The real number x when added to its inverse gives the minimum value of the sum at x equal to
(1) 1
(2) – 1
(3) – 2
uθ
(4) 2
sol (4) S = x + (1 / x) Smin = 2 (AM 3 GM)
x )2 ≥ 0 x + (1 / x) 3 2
or ( x – (1 /
69. Let R1 and T2 respectively be the maximum ranges up and down an inclined plane and R be the maximum range
on the horizontal plane. Then R1, R, R3 are in
(1) A.P
(2) G.P
(3) H.P
(4) Arithmetic-Geometric Progression (A.G.P) 45
sol (2) R2 = Rmax × Rmin
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Let (d / dx) F(x) = (e sin x / x), x > 0 If 1 ∫ 4 (3 / x) esin x3 dx = F (k) – F(1) then one of the possible values of k,is
70.
(1) 16
(2) 63
(3) 64
(4) 15
sol (2) 1 ∫ x (3 / x) e sin x3 dx = F (f (x)) – F (l)
Diff. both sides f ‘ (x) × e sin f (x) / f (x) = (3 / x) e sin x3× 1
⇒ f (x) = x3 ⇒ k = f (4) = 64
lim n→∞ 1+ 24 + 34 + .... + n4 / n5– lim n→∞ 1 + 23 + 33 + ..... + n3 / n5 is
71.
(1) zero
(2) 1 / 4
(3) 1 / 5
(4) 1 / 30
n →∞ r =1 Σ
n
r =1 Σ
n
(1 / n) (r / n)4 – (1 / n) lt (1 / n) (r / n)3
sol (3) L = lt
= 1 ∫ 0 x4 dx = 1 / 5.
72. The median of a set of 9 distinct observation is 20.5. If each of the largest 4 observations of the set is
increased by 2, then the median of the new set
(1) is decreased by 2
(2) is two times the original median
(3) remains the same as that of the original set
(4) is increased by 2.
sol (3) Median = x5 ; (Median)New = (Median)old
lim n→( π/2) [1 – tan (x / 2)] [1 – sin x] / [1 + tan (x / 2)] [ π – 2x]3 is
73.
(1) 0
(2) 1 / 32
(3) ∞
(4) 1 / 8
sol (2) ( π / 4) – (x / 2) = h ⇒ L = It h → 0 (tan h / 64 h) × (1 – cos 2h) / h2
= (1 / 64) × 1 × (22 / 2) = 1 / 32.
74. Let f (a) = g (a) = k and their nth derivatives f n (a), gn (a) exist and are not equal for some n. Further if lim
x → a f (a) g(x) – f(a) – g (a) f (x) + g(a) / g(x) – f(x) = 4 Then the value of k is
(1) 2
(2) 1
(3) 0
(4) 4
sol (4) Applying LH rule successively n times α θ u
⇒ f (a) g(n) (a) – g(a) f (n) (a) / g(n) (a) – f (n) (a) = 4
f (a) = g (a) = k ⇒ k = 4
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75. If the equation of the locus of a point equidistant from the point (a1, b1) and (a2, b2) is (a1 – a2) x + (b1 – b2)
y + c = 0. then the value of ‘c’ is
(1) a12 – a22 + b12 – b2 2
(2) 1⁄2 (a12 + a2 2 + b12 + b22)
a12 + b12 − a 2 2 − b2 2
(3)
(4) 1⁄2 (a22 + b22 – a12 – b12)
sol (4) (a1 + a2 / 2, b1 + b2 / 2) lies on the locus
⇒ (a12 – a22) / 2 + (b12 – b22) / 2 + c = 0
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