aieee 2005 solved paper.pdf
76
www.aieeepage.com 1. A car, starting from rest, accelerates at the rate f through a distance S, then continues at constant speed for time t and then decelerates at the rate 2 f to come to rest. If the total distance traversed is 15 S, then 1) S = 6 1 ft 2 2) S = ft 3) S = 4 1 ft 2 4) S = 2 1 ft 2 Sol : f fl2 t 1 2t 1 t 15 S 15S 2t x 2 2 1 2 1 2 1 1 2 1 = + + f t f f ⇒ S S t 15 2 ft S 1 = + + ⇒ S t 12 ft 1 = (1) S ft = 2 1 2 1 (2) Dividing (1) by (2), we get 6 1 1 = t ⇒ 72 ft 6 2 1 2 2 = = t f S None of the given options is correct 2. A particle is moving eastwards with a velocity of 5 ms -1 . In 10 seconds the velocity charges to 5 ms -1 northwards. The average acceleration in this time is 1) 2 1 ms -2 towards north 2) 2 1 ms -2 towards north-east 3) 2 1 ms -2 towards north-west 4) zero Sol: (2) t v ∆ interval Time y in velocit Change on accelerati Average G = = cos90 v 2v v v ) v v ( v ∆ 2 1 2 2 2 1 1 2 + + = − = G G G 0 5 5 2 2 + + = [As | v 1 | = | v 2 | = 5m / s] s / m 2 5 = Average acceleration 2 m/s 2 1 10 2 5 t v ∆ = = = G towards North - West PHYSICS S W ° 90 1 v E N 2 v 1 v − ) v ( v v 1 2 − + = ∆ G
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Transcript of aieee 2005 solved paper.pdf
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1. A car, starting from rest, accelerates at the rate f through a distance S, then continues at constant speed for time
t and then decelerates at the rate 2f
to come to rest. If the total distance traversed is 15 S, then
1) S = 61
ft2 2) S = ft 3) S = 41
ft2 4) S = 21
ft2
Sol :f fl2t1 2t1t
15 S
15S 2t x 22
121 2
112
1 =++ ftff
⇒ SSt 152ft S 1 =++
⇒ St 12ft1 = (1)
Sft =212
1(2)
Dividing (1) by (2), we get
61 1 =t
⇒ 72ft
621 22
=
= tfS
None of the given options is correct
2. A particle is moving eastwards with a velocity of 5 ms-1. In 10 seconds the velocity charges to 5 ms-1 northwards.The average acceleration in this time is
1)21
ms-2 towards north 2) 21
ms-2 towards north-east
3) 21
ms-2 towards north-west 4) zero
Sol: (2) tv∆
interval Timeyin velocit Changeonaccelerati Average ==
cos90v2vvv)vv (v∆ 2122
2112 ++=−=
055 22 ++= [As | v1 | = | v2 | = 5m / s]
s/m25=
Average acceleration 2m/s2
110
25tv∆ === towards North - West
PHYSICS
S
W°90
1vE
N2v
1v−
)v(vv 12 −+=∆
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3. Out of the following pair, which one does NOT have identical dimensions is1) impulse and momentus 2) angular momentum and Planck’s constant3) work and torque 4) moment of inertia and moment of force
Sol: (4)4. The relation between time t and distance x is t = ax2 + bx where a and b are constants. The acceleration is
1) 2 b v3 2) -2 a b v2 3) 2 a v2 4) - 2 a v3
Sol: (4)t = ax2 + bx; diff. with respect to time (t)
dtdxbxt
dtd += )(
dtda 2 = ..2. vb
dtdxxa +
1 = 2axv + bv = v (2ax + b)
2ax + b = v1
.. diff.
2av = - fv21
f = -2av3
5. A smooth block is released at rest on a 450 incline and then slides a distance ‘d’. The time taken to slide is ‘n’ timesas much a slide on rough incline than on a smooth incline. The coefficient of friction is
1) 2n1-1 =kµ 2) 2n
1 - 1 =kµ 3) 28 n1 - 1 =µ 4) 28 n
1 - 1 =µ
Sol: 2)
θµθ c o ss in gg −
d
smooth rough
d = 21
(g sinθ )t2
d = 21
2 . sin tg θ
θµθθ cos - sin 2
sin 2 21 gg
dtg
dt ==
cosθµg-θsin2
θsin2
gd
gdn =
n = µ−11
µ−=
11 2n
21 1
n=−µ
21 - 1
n=µ
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6. A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m/s2. Hereaches the ground with a speed of 3 m/s. At what height, did he bail out?
a) 182 m 2) 91 m 3) 111 m 4) 293 mSol : (4)
50mv
3 m/s
a = -2 m/s
2gh =v
514 50 9.8 2 =××=v
m 244 4980 - 3
2 2u - v
222≈=
×=S
Initially he has fallen 50 m∴ Total hight from where he bailed out = 244 + 50 = 294 m
7. A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetratebefore coming to rest further it will penetrate before coming to rest assuming that it faces consistant resistance tomotion ?
1) 2.0 m 2) 3.0 m 3) 1.0 m 4) 1.5 mSol: (3)
uv = 0
x
2u
3cm
3 . a . 2 u - 2
22
=
u
2.a.3 3 2
=uu
8u
2=a
x. a . 2 2u - 0
2
=
x 8
u . 2 4
22×=u
x = 1 cm
8. An annular ring with inner and outer radii R1 and R2 is rolling without slipping with a uniform angular speed. The
ration of the forces experienced by the two particles situated on the inner and outer parts of the ring, 2
1
FF
is
1)2
2
1
RR
2)1
2
RR
3)2
1
RR
4) 1
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Sol: 3)
a2
a1
R2
R1
11 Rω=V22 Rω=V
12
21
2
1
21
1 1
v RRR
Ra ωω
===
22
2
22
2 v RR
a ω==
taking particle masses equal2
1
2
1
2
1
2
1
RR
aa
mama
FF ===
9. A projectile can have the same range ‘R’ for two angles of projection. If ‘t1’ and ‘t2’ be the times of flights in thetwo cases, then the product of the two time of flights is proportional to
1) 21
R2) 2R 3) R 4)
R1
Sol: 3)
T1 T2 = g
R2 (It is a formula)
T, T2 ∞ R
10. The upper half of an inclined plane with inclination φ is perfectly smooth while the lower half is rough. A bodystarting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half isgiven by1) 2 cos φ 2) 2 sin φ 3) tan φ 4) 2 tan φ
Sol: 4)Acclaration of block while sliding down upper half = g sinθ ;
retardation of block while sliding down lower half = - )cos - sin ( θµθ gg
for the block to come to rest at the bottom, acclaration in I half = retardation in II half.
g sinθ = - )cos - sin ( θµθ gg
φ=⇒ tan 2 µ
11. A mass ‘m’ moves with a velocity ‘v’ and collides inelastically with another identical mass. After collision the Ist
mass moves with velocity 3v
in a direction perpendicular to the initial direction of motion. Find the speed of the
2nd mass after collision.
3v
m mbeforecollision
aftercollision
1) v3 2) v 3) 3v
4) 32
v
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Sol: 4)
mm
u = V1
v3=
v
02 =u
In x direction : mv + O = m (O) + m(v2) x
In y direction : O + O = m )(v m 3
2+
v y is
3V)(V y2 = V)(V x2 =
34 v V
3V V
3V 2
22
2
2 =+=+
=∴ V
= 32V
12. A particle of mass 0.3 kg is subjected to a force F = - kx with k = 15 N / m. What will be its initial acceleration ifit is released from a point 20 cm away from the origin ?1) 15 m/s2 2) 3 m/s2 3) 10 m/s2 4) 5 m/s2
Sol: 3)m = 0.3 ⇒ F = m.a. = -15 x
50x- x 3
150 - x 0.315 ==−=a
2sm 10 0.2 x 50- a ==
13. The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant Kand compresses it by length L. The maximum momentum of the block after collision is
M
1)M
KL2
22) L MK 3)
KML2
4) Zero
Sol: 2)
M
L . mK V : kl
21 mv
21 22 ==
Momentam = m x v = L . mK x m
= L . km
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14. A block is kept on a frictionless inclined surface with angle of inclination ‘α ’. The incline is given an acceleration‘α ’ to keep the block stationary. Then α is equal to
1) g cosec α 2) g / tan α 3) g tan α 4) gSol: 3)
cosa
αsinga
a
α=α cos asin g
α= tan g a
15. A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m. It rolls down a smooth surface tothe ground, then climbs up another hill of height 30 m and finally rolls down to a horizontal base at a height of 20m above the ground. The velocity attained by the ball is
1) 20 m/s 2) 40 m/s 3) 3010 m/s 4) 10 m/s
Sol: 2)
100 3020
mghmghmv
21 2 +
20 x 10v21 100 x 10 2 +=
sm 40 1600 v800, v
21 2 ===
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16. A body A of mass M while falling vertically downwards under gravity breaks into two parts; a body B of mas 31
M and a body C of mass 32
M. The centre of mass of bodies B and C taken together shifts compared to that of
body A towards
1) does not shift 2) depends on height of breaking3) body B 4) body C
Sol: 1)does not shift as no external face acts.
17. The moment of inertia of a uniform semicircular disc of mass M and radius r about a line perpendicular to theplane of the disc through the centre is
1) 2
52 Mr 2) 2
41 Mr 3) 2
21 Mr 4) 2Mr
Sol: 3)
2Mr21
18. The change in the value of ‘g’ at a height ‘h’ above the surface of the earth is the same as at a depth ‘d’ below thesurface of earth. When both ‘d’ and ‘h’ are much smaler than the radius of earth, then which one of the followingis correct?
1) d = 2
3h2) d =
2h
3) d = h 4) d = 2 h
Sol. 4)
=
R2h - 1g g 1
h ;
=
Rd - 1g g 1
d
2h d =∴
19. A particle of mass 10 g is kept on the surface of a uniform sphere of mass 100 kg and radius 10 cm. Find the workto be done against the gravitational force between them to take the particle far away from the sphere
(you may take G = 6.67 x 10-11 Nm2 / Kg2)
1) 3.33 x 10-10 J 2) 13.34 x 10-10 J 3) 6.67 x 10-10 J 4) 6.67 x 10-9 JSol. 3)
RGMV −=
1.0100 x 10 x 6.67-
RGM-
-11
==V
= - 6.67 x 10-8
‘V’ at infinity (at large distance) = 0
8-10 x 6.67 x 1000
10 x vm ==ω
= 6.67 x 10-10 J
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20. A ‘T’ shaped object with dimensions shown in the figure, is lying on a smooth floor. A force ‘ F ’ is applied at thepoint P parallel to AB, such that the object has only the translational motion without rotation. Find the location ofP with respect to C.
l
2l
A
C
P
B
F
1) l 23
2) l 32
3) l 4) l 34
Sol. 4)
l
2l
A
C
P
B
F),0( l
To have linear motion. The force F has 0.2l to be applied at centre of mass.i.e. the paint ‘p’ has to centre of mass
= 2l l
2l x l l x 2l Y++= =
34l
l3l4 2
=
21. Which of the following is incorrect regarding the first law of thermodynamics?
1) It is a restatement of the principle of conservation of energy2) It is not applicable to any cyclic process3) It introduces the concept of the entropy4) It introduces the concept of the internal energy
Sol. 2)
22. Consider a car moving on a straight road with a speed of 100 m/s. The distance at which car can be stopped is[ ]0.5 =kµ
1) 1000 m 2) 800 m 3) 400 m 4) 100 mSol. 1)
2as u v 22 =−
s g)(- 2 u 0 2 µ=−
s x 10 x 21 - x 2 1002 =−
s = 1000 m
23. A body of mass m is accelerated uniformly from rest to a speed v is a time T. The instantaneous power deliveredto the body as a function of time is given by
1) 22
2.t
Tmv
2) .t2
2
Tmv
3) 22
2.t
21
Tmv
4) .t 21
2
2
Tmv
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Sol: 2)u = 0; v = u + aT; v = a.T.Instantaneous power = F x v = m.a.v = m.a.at = m.a2.t
∴ instantaneous power = tTv m 2
2
24. Average density of the earth
1) is a complex function of g 2) does not depend on g3) is inversely proportional to g 4) is directly proportional to g
Sol: 4)
22
x vG RGm
Rg ρ==
2
3
R
πR34PG
g××
=
G.R. 34 g ρπ=
→ρ average density
π=ρ
RG 43g
‘ρ ’ is directly profactional to (g)
25. If ‘S’ is stress and ‘Y’ is Young’s modulus of material of a wire, the enrgy stored in the wire per unit volume is
1)Y
S2
22) 2S2Y 3)
YS
24) 2
2SY
Sol. 1)Energy stored per unit volume
=
StrainStressY∵
=
YStressStrain∵
= strain x stress x 21
= Y
strain x stress x 21
= YS .
21 2
26. A 20 cm long capillary tube is dipped in water. The water rises up to 8 cm. If the entire arrangement is put in afreely falling elevator the length of water column in the capillary tube will be
1) 10 cm 2) 8 cm 3) 20 cm 4) 4 cm
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Sol. 3)Water fills the tube entirely in gravity less condition.∴ 20 cm.
27. The figure shows a system of two concentric spheres of radii r1 and r2 kept at temperatures T1 and T2, respectively.The radial rate of flow of heat in a substance between the two concentric spheres is proportional is
1r
2r
1T
2T
1) In
1
2
rr
2) (r2 - r1) / (r1r2) 3) (r2 - r1) 4) ( )12
21
r - r
rr
Sol: 4)
1r
2r
1T
2T
∆TT −dr
Consider a shall of thickness (dr) and of radii (r).It the temperature of inner and outer surfaces of this shell beT, (T -dT)
dtdθ
= rate of flow of heat though it
= ( )
drTdTkA −− )T(
= drdTkr 4-
drdt - kA 2π=−
2rdr
k4(dr/dT) - dt
π=
integrating between the proper limits
K41
dtdQ- dt
2
1
T
Tπ
=∫ ∫
2
1
r
r2r
dr
π=
212 r
1 - r1
k4 dt.dQ - )T - T(
1
12
2121
r - r)TT(rrK 4
dtdQ −π
=
12
21
r - rr r x
dtdQ∴
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28. A system goes from A to B via two processes I and II as shown in figure. If ∆ U1 and ∆U2 are the changes ininternal energies in the processes I and II respectively, then
p
V
A BI
II
1) relation between ∆U1 and ∆U2 can not be determined 2) ∆U1 = ∆U2
3) ∆U2 < ∆U1 3) ∆U2 > ∆U1
Sol: 2)As in a cyclic process, net internal energy is zero
21 U U ∆=∆
29. The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is
T
2T0
2S0
T0
S0 S<
<
<
<
<
1) 41 2) 2
1 3) 32 4) 3
1
Sol. 4)T
2T0
2S0
T0
S0 S
<
<
1Q
2Q
3Q
oooooo1 ST23ST
21STQ =+=
oo2 STQ =
0Q3 =
31
321
QQ1
QQQ
QW
1
2
1
21
1=−=−=
−==η
∴ (4) is correct
30. If radius of the 2713 Al nucleus is estimated to be 3.6 Fermi then the radius of 125
52 Te nucleus be nearly
1) 8 fermi 2) 6 fermi 3) 5 fermi 4) 4 fermiSol. 2)
53
12527
AA
RR 3/13/1
2
1
2
1 =
=
=
6 3.6 x 35 R 2 == fermi
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31. Starting with a sample of pure Cu,66 8
7 of it decays into Zn in 15 minutes. The corresponding half-life is
1) 15 minutes 2) 10 minutes 3) 7 21
minutes 4) 5 minutes
Sol: 4)
87
days ∴ remains undecayed = 87 - 1 =N
= 3
21
81
=
3 is the no. of half times.
n = Tt
; 3 = T15
T = half life period = 315
= 5 minutes
32. A photocell is illuminated by a small bright source placed 1 m away. When the same source of light is placed 21
m
away , the number of electrons emitted by photocathode would
1) increase by a factor of 4 2) decrease by a factor of 43) increase by a factor of 2 4) decrease by a factor of 2
Sol: 1)
2r1 I ∞ ; 4
1 rr
II
2
1
2
2
1 =
=
times 4I2 →
When internly is 4 times, No. of photoelectron limited would increase by 4 times.
33. The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorterthan 2480 nm is incident on it. The band gap in (eV) for the semiconductor is
1) 2.5 eV 2) 1.1 eV 3) 0.7 eV 4) 0.5 eVSol: 4)
ev10 x 1.61 x
10 x 298010 x 3 x 10 x 6.63 hc E 19-8-
8-34
=
λ=∆
= 0.5 ev
34. The intensity of gamma radiation from a given source is I. On passing through 36 mm of lead, it is reduced to 8I
.
The thickness of lead which will reduce the intensity to 2I
will be
1) 9 mm 2) 6 mm 3) 12 mm 4) 18 mmSol: 3)
µd0.eII −= , appling logarithm both sides
=−
0IIlogµd
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)2...(I2I
logdµ
)1...(I8I
log36µ
=×−
=×−
3
21log
21log3
21log
81log
36 =
=
=d
mm123
36d ==
35. A gaseous mixture consists of 16 g of helium and 16 g of oxygen. The ratio v
p
CC
of the mixture is
1) 1.62 2) 1.59 3) 1.54 4) 1.4Sol: 1)
21
2211
xxrxrx
CvCp
xm ++=
+
3216
416
57
3216
35
416
+
×+×=
= 1.63
36. In a common base amplifier, the phase difference between the input signal voltage and output voltage is
1) π 2)4π
3)2π
4) 0
Sol: 4)Zero
37. The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents theemission of a photon with the most energy?
n = 4
n = 3
n = 2
n = 1
I II III IV
1) IV 2) III 3) II 4) I
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Sol:.2)
−==
−= 2222
21
3)4(h 41
31hc
h1
h1hcE 0.05hc
1697hc =
×=
×=
×−=
−=
→= 16412hc
164416hc
41
21hcE 221)a(h hc=
163
0.5hc×=
0.75hc43hc
21
11hcE 21Lh
=
=
−=
→=
−=
→= 2231h 11
31hcE
= -energy is not enutted, but is obsanleed.∴ III transition gives most energy
38. If the kinetic energy of a free electron doubles, its deBroglie wavelength changes by the factor
1) 2 2)21
3) 2 4) 21
Sol: 4)
2.m.(KE)h
Phλ ==
KE1λ ∝=∴
If KE is doubled, λ becomes 2λ
39. A nuclear transformation is denoted by X(n, α ) Li73 . Which of the following is the nucleus of element X?
1) B105 2) 6
12C 3) Be114 4) B9
5
Sol: 1)4
27
31
0A
z HeLihX +→+on compansion, A = 10 , z = 5
It is boran 105 B
40. In a full wave rectifier circuit operating from 50 Hz mains frequency, the fundamental frequency in the ripplewould be
1) 25 Hz 2) 50 Hz 3) 70.7 Hz 4) 100 HzSol: 4)
Input frequency, Hz50=f 501T =⇒
For full wave rectifier, 1001
2TT1 ==
⇒ Hz1001 =f∴ (4) is correct
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41. Two simple harmonic motions are represented by the equations y1 = 0.1 sin
+
3 100 ππ t and y2 = cos t π . The
phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is
1) 3π
2) 6π−
3) 6π
4) 3π−
Sol: 2)
+×=
3ππt100cosπ1001.0v1
+=−=
2ππtcos0.1ππtsinπ1.02v
∴ Phase diff = 21 φ−φ
2π
3π −=
63π-2π=
6π−=
42. A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index
of water is 34
and the fish is 12 cm below the surface, the radius of this circle in cm is
1) 736 2) 736 3) 54 4) 536
Sol: 1)
43
µ1θsin c ==
1273θtan c
R==
CθCθ
R
cm12
cm7
36R =⇒
∴ (1) is correct
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43. Two point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm.Aproximately, what is the maximum distance at which these dots can be resolved by the eye? [Take wavelengthof light = 500 nm]1) 1 m 2) 5 m 3) 3 m 4) 6 m
Sol: 2)
dλ22.1≥
Dy
⇒ m51.6
30105)22.1(
10310λ)22.1( 7
33
≈=××××=≤ −
−−ydD
m5Dmax =∴∴ (2) is correct
44. A thin glass (refractive index 1.5) lens has optical power of -5 D in air. Its optical power in a liquid medium withrefractive index 1.6 will be1) -1 D 2) 1 D 3) -25 D 4) 25 D
Sol: 2)
−
−=
21
11115.11
RRfa...(1)
−
−=
21
1116.15.11
RRfm....(2)
Dividing (1) by (2)
−
−=1
6.15.1
15.1
a
m
ff
⇒ cm160=mf
D16.16.1µ
m
===f
Pm
∴ (2) is correct
45. The function sin2 t)(ω represents
1) a periodic, but not simple harmonic, motion with a period ωπ
2) a periodic, but not simple harmonic, motion with a period ωπ2
3) a simple harmonic motion with a period ωπ
4) a simple harmonic motion with a period ωπ2
Sol: 1)
cotsin2=f
−
−=t
2yωcos1
21
ωt)2cos(1
−= t.
1/2π2cos1
21
ωπT/2 =
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46. A Young’s double slit experiment uses a monochromatic source. The shape of the interference fringes formed ona screen is
1) circle 2) hyperbola 3) parabola 4) straight line
Sol: 2)
47. The bob of simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of theoscillating bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillationwould
1) first decrease and then increase to the original value2) first increase and then decrease to the original value3) increase towards a saturation value4) remain unchanged
Sol: 2)Centre of mass of comformation of liquid and hollow portion, first increases and when total water is drainedout, centre of mass regain its orginal position
∴ ‘T’ first increases and there decreases to orginal value
cc
48. If I0 is the intensity of the principal maximum in the single slit diffraction pattern, then what will be its intensitywhen the slit width is doubled?
1) 4 I0 2) 4 I0 3)20I 4) I0
Sol: 4)
2
osinII
φφ=
and
λπ=φ
Day
For principal maxima y = 0
0=φ∴
Hence intensity will remains same∴ (4) is correct
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49. When two tuning forms (fork 1 and fork 2) are sounded simultaneously, 4 beats per second are heard. Now, sometape is attached on the prong of the fork 2. When the tuning forks are sounded again, 6 beats per second areheard. If the frequency of fork 1 is 200 Hz, then what was the original frequency of fork 2?
1) 202 Hz 2) 200 Hz 3) 204 Hz 4) 196 HzSol: 4)
When 2n is taped,
6~ 21 =nn instead of 4 as previous
6n - 200 ; 6n 221 ==−∴ n
.19442 zn =
50. If a simple harmonic motion is represented by 0 2
2=+ x
dtxd α , its time period is
1) απ2
2) απ2
3) απ2 4) πα2
Sol: 1)
α ω ω- dx 22
2
=⇒== xdt
xd
∝=== 2π
ω2πT
51. An observer moves towards a stationary source of sound, with a velocity one-fifth of the velocity of sound. Whatis the percentage increase in the apparent frequency?1) 0.5% 2) zero 3) 20% 4) 5%
Sol: 3)
=
+=
+=
56n 5 2
v
vvn
vvovnn
%201005
56n ; 561
=×−=−=n
nnn
52. Two point charges +8q and -2q are located at x = 0 and x = L respectively. The location of a point on the x axis atwhich the net electric field due to these two point charges is zero is
1)4L
2) 2 L 3) 4 L 4) 8 L
Sol: 2)x
+ 8qx = 0 x = L
- 2q
q
222
)(8
xqq
xlqq ×=
+×
L x x L 2 12 =+==+
xxnL
from x = 0, total distance = L + L = 2L
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53. When an unpolarized light of intensity I0 is incident on a polarizing sheet, the intensity of the light which does notget transmitted is
1) 041 I 2) 02
1 I 3) 0I 4) zero
Sol: (2)
θcos2oII =
Intensity of polarized light = 2oI
⇒ Intensity of untransmitted light = 22oo
oIII =−
∴ (2) is correct
54. Two thin wire rings each having a radius R are placed in a distance d apart with their axes coinciding. The chargeson the two rings are +q and -q. The potential difference between the centres of the two rings is
1)
+∈ 220 R
1 - 1 2 dR
Qπ 2) 2
0 d 4 ∈πQR
3)
+∈ 220 R
1 - 1 4 dR
Qπ 4) zero
Sol: 1)
+ q - q
1 2
d
At (1) using, potential ( ) using 2 toself1 dueVVV +=
2200 R
q π41
dRq+
−=εε
At (2) using potential dueself2 VV)(V += to 1 using
2200 dRε
qRq-.
πε41
+
220
22o00
21dRε
qdRε
qRq
ε4π1
Rq.
ε4π1vv∆v
+−
+−+=−=
2200 drε
2qRq.
ε2π1
+−=
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55. A parallel plate capacitor is made by stacking n equally spaced plates connected alternatively. If the capacitancebetween any two adjacent plates is ‘C’ then the resultant capacitance is
1) (n + 1)C 2) (n - 1)C 3) nC 4) CSol: 2)
(n-1) spacing between n capcities∴ resultant capacity (n-1) c
56. A charged ball B hangs from a silk thread S, which makes an angle θ with a large charged conducting sheet P, asshown in the figure. The surface charge density σ of the sheet is proportional to
++++++++++P
S
B
θ
1) cotθ 2) cosθ 3) tanθ 4) sinθSol: 3)
θ θ
θ
mg
T sin
T cosT
qEqF .0εσ
==
mgT
qT
=
=
θεσθ
cos
.sin0
.mgεσqtanθ
0=
θ tan is σ ∝∴
57. One conducting U tube can slide inside another as shown in figure, maintaining electrical contacts between thetubes. The magnetic field B is perpendicular to the plane of the figure. If each tube moves towards the other at aconstant speed V, then the emf induced in the circuit in terms of B, l and V where l is the width of each tube, willbe
XX X
Xv
vX X
XA B
XX
XX X
X
X
X
CD
1) -Blv 2) Blv 3) 2 Blv 4) zero
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Sol: 3)Relative velocity = V + V = 2V∴ emf. = B.L (2V)
58. A fuly charged capacitor has a capacitance ‘C’. It is discharged through a small coil of resistance wire embeddedin a thermally insulated block of specific heat capacity ‘s’ and mass ‘m’. If the temperature of the block is raisedby ‘∆T’, the potential difference ‘V’ across the capacitance is
1) sTmC∆
2) sTmC∆2
3) CTms∆2
4) CTms∆
Sol: 3) csm t 2.m.s. V ;t .CV
21 2 ∆=∆=
59. Two thin, long, parallel wires, separated by a distance ‘d’ carry current of ‘i’ a in the same direction. They will
1) repel each other with a force of )2(2
0d
iπ
µ 2) attract each other with a force of )2(2
0d
iπ
µ
3) repel each other with a force of )2( 2
20
di
πµ
4) attract each other with a force of )2( 2
20
di
πµ
Sol: 2)
F
)attractive(πd2
ii,µlight
F 20=
)attractive(πd2iµ 2
0=
60. A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will nowbe
1) four times 2) doubled 3) halved 4) one fourthSol: 2)
J
2
RtVH = (as the same heater is used this formula has to be adopted, because it works at same voltage)
Resestance of part 2R →
∴ ‘H’ will be doubled.
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61. In the circuit, the galvanometer G shows zero deflection. If the batteries A and B have negligible internal resistance,the value of the resistor R will be -
G
A
2V
12V B R
Ω500
1) 100 Ω 2) 200 Ω 3) 1000Ω 4) 500Ω
Sol: 1)
ai
R2v
12 v
5ω
12 - 2 = ( )i ω 5
501 i =
RI
+===
ω512
51
ω512
0
5 ω + R = 6 ω = R = 100 Ω62. A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10 divisions per milliampere and
voltage sensitivity is 2 divisions per millivolt. In order that each division reads 1 volt, the resistance in ohms neededto be connected in series with the coil will be -
1) 105 2) 103 3) 9995 4) 99995Sol: 3)
Ω==⇒= 52
10 G Ger galvanomet of Resistance
ysensitivitCurrent ysensitivit Voltage
Here gi = Full scale deflection current = mA1510150 =
V = voltage to be measured = Ω=−× − 99955
1015150
3
63. Two sources of equal emf are connected to an external resistance R. The internal resistances of the two sourcesare R1 and R2 (R2 > R1). If the potential difference across the source having internal resistance R2 is zero, then
1) R = R2 - R1 2) R = R2 x (R1 + R2) / (R2 - R1)
3) R = R1 R2 / (R2 - R1) 4) R = R1 R2 / (R1 + R2)
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Sol: 1)
I
2v
R1 R2
R
ε ε
21R R2 I
R++= ε
Potential difference across L cell
0 2 =−== RiV ε
0.22
21=
++− R
RRRεε
02 221 =−++ RRRR
21 RRR −+
12 R RR −=∴
64. Two voltameters, one of copper and another of silver, are joined in parallel. When a total charge q flows throughthe voltameters, equal amount of metals are deposited. If the electrochemical equivalents of copper and silver arez1 and z2 respectively the charge which flows through the silver voltameter is
1)1
21zz
q
+ 2)2
11zz
q
+ 3)1
2
zzq 4)
2
1
zzq
Sol: 1)
).......(z q1 z
1
2
2
1 iqq
zzqm =⇒∝⇒=
)...(1
q 1qq q q
2
12
2
1
221 ii
qqqqalso
+=⇒+=⇒+=
From equation (1) and (ii) 1
22
1zz
qq+
=
65. Two concentric coils each of radius equal to 2π cm are placed at right angles to each other. 3 ampere and 4ampere are the currents flowing in each coil respectively. The magnetic induction in Weber/m2 at the centre of thecoils will be
( -70 10 x 4 πµ = Wb / A..m.)
1) 10-5 2) 12 x 10-5 3) 7 x 10-5 4) 5 x 10-5
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Sol: 4)
(2)
(1)
ππ 43µ
)2(2i µ 010
1×==B
5.4µ 02
22
12 π=+= BBB
27 10510 ××= −
25 /nωb 105 −×=
66. In a potentiometer experiment the balancing with a cell is at length 240 cm. On shunting the cell with a resistanceof Ω2 , the balancing length becomes 120 cm. The internal resistance of the cell is
1) Ω 5.0 2) Ω 1 3) Ω 2 4) Ω 4Sol: 3)
2120
120240
2
21 ×−=×
−= Rl
llr
Ω= 267. A charged particle of mass m and charge q travels on a circular path of radius r that is perpendicular to a magnetic
field B. The time taken by the particle to complete one revolution is
1)m
B 2 2qπ 2)B
q 2 mπ3) qB
2 mπ4)
m 2 Bqπ
Sol: 3)
qvB 2=
rmv
qBm 2
v 2 ππ == rT
68. A magnetic needle is kept in a non-uniform magnetic field. It experiences
1) neither a force for a torque 2) a torque but not a force3) a force but not a torque 4) a force and a torque
Sol: 4)A magnetic needle kept in non-uniform magnetic field experience face and torque due to unequal forcesacting on poles
69. The resistance of hot tungsten filament is about 10 times the cold resistance. What will be the resistance of 100W and 200 V lamp when not in use ?
1) Ω 20 2) Ω 40 3) Ω 002 4) Ω 004
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Sol: 2)
2
RVViP ==
Ω=×== 400100
2002002
PVR
Ω== 40 10
400 cold R
70. The self inductance of the motor of an electric fan is 10 H. In order to impart maximum power at 50 Hz, it shouldbe connected to a capacitance of
1) F 8 µ 2) F 4 µ 3) F 2 µ 4) F 1µ
Sol: 4)
2c 21
π== nP
c 50 2150
×=
π
C = 1 uf.
71. An energy source will supply a constant current into the load if its internal resistance is
1) very large as compared to the load resistance 2) equal to the resistance of the load3) non-zero but less than the resistance of the load 4) zero
Sol: 4) ,rR
EI+
= Internal resistance (r) is
zero, ttanconsREI ==
72. The phase difference between the alternating current and emf is 2π
. Which of the following cannot be the
constituent of the circuit?
1) R, L 2) C alone 3) L alone 4) L, CSol: 4)73. A circuit has a resistance of 12 ohm and an impedance of 15 ohm. The power factor of the circuit will be -
1) 0.4 2) 0.8 3) 0.125 4) 1.25Sol:.2)
Power factor = 8.054
1512cos ====φ
ZR
74. A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If anelectron is projected along the direction of the fields with a certain velocity then -
1) its velocity will increase 2) its velocity will decrease3) it will turn towards left of direction of motion 4) it will turn towards right of direction of motion
Sol:.2)due to electricfield, it experiences force and accelerates i.e, its velocity decreases.
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75. A coil of inductance 300 mH and resistance Ω2 is connected to a source of voltage 2 V. The current reaches halfof its steady state value in
1) 0.1 s 2) 0.05 s 3) 0.3 s 4) 0.15 sSol: 1)
−=
−−LIt
eii 10
)1(2 00 eii −=
t0.693RL =×
210300693.0
3−××=
= 0.1 see
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CHEMISTRY76. The oxidation state of Cr in [Cr(NH3)4 Cl2]
+ is
1) 0 2) +1 3) +2 4) +3
Ans. (4)
77. Which one of the following types of drugs reduces fever?
1) Tranquiliser 2) Antibiotic 3) Antipyretic 4) Analgesic
Ans. (3)
78. Which of the following oxides is amphoteric in character?
1) SnO2 2) SiO2 3) CO2 4) Cao
Ans. (1)
2SnO is an amphoteric oxide since it can reacts both with acid and bases.
79. Which one of the following species is diamagnetic in nature?
1) −2H 2) +
2H 3) 2H 4) +2He
Ans. (3) There is no unpaired electron in H2
80. If α is the degree of dissociation of Na2SO4, the vant Hoff’s factor (i) used for calculating the molecular mass is
1) 1 - 2α 2) 1 + 2α 3) 1 - α 4) 1 + α
Ans. (2) −− +→ 24
)(42 SONa2SONa
1 - α 2α αi = l - α + 2α + αi = l + 2α
81. Which of the following is a polyamide ?
1) Bakelite 2) Terylene 3) Nylon-66 4) Teflon
Ans. (3)
– C – – C – NH – – NH –
O O
42CH 62CH
82. Due to the presence of an unpaired electron, free radicals are:
1) Cations 2) Anions3) Chemically inactive 4) Chemically reactive
Ans. (4)
83. For a spontaneous reaction the ∆G, equilibrium constant (K) and 0CellE will be respectively
1) -ve, >1, -ve 2) -ve, <1, - ve 3) +ve, >1, -ve 4) -ve, >1, +ve
Ans. (4)
For Spontaneous process,
1K,veG >−=∆ and veEocell +=
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84. Hydrogen bomb is based on the principle of
1) artificial radioactivity 2) nuclear fusion 3) natural radioactivity 4) nuclear fission
Ans. (2)
85. An ionic compound has a unit cell consisting of A ions at the corners of a cube and B ions on the centres of thefaces of the cube. The empirical formula for this compound would be
1) A3B 2) AB3 3) A2B 4) ABAns. (2)
Contribution of atom of ‘A’ at 8 corners of cube = 81
x 8 = 1
Contribution of atom of ‘B’ at six faces of cube = c x 21
= 3
∴ Formula of compound = 34B
86. The highest electrical conductivity of the following equeous solutions is of
1) 0.1 M difluoroacetic acid 2) 0.1 M fluoroacetic acid2) 0.1 M chloroacetic acid 4) 0.1 M acetic acid
Ans. (1)Difluoroacetic acid will be strongest acid due to electron withdrawing effect of two fluorine atoms so as it willshow maximum electrical conductivity.
87. Lattice energy of an ionic compound depends upon
1) Charge on the ion and size of the ion 2) Packing of ions only3) Size of the ion only 4) Charge on the ion only
Ans. (1)
88. Consider an endothermic reaction YX → with the activation energies Eb and Ef for the backward and forwardreactions, respectively. In general
1) there is no definite relation between Eb and Ef 2) Eb + Ef
3) Eb>Ef 4) Eb< Ef
Ans. (4)
Ef
Eb
E
Reactions
Products
Reaction Progress
89. Aluminium oxide may be electrolysed at 10000C to furnish aluminium metal (At. Mass = 27 amu; 1 Faraday =96,500 Coulombs). The cathode reaction is 03 3 AleAl →+ −+ .
To prepare 5.12 kg of aluminium metal by this method would require
1) 5.49 x 101 C of electricity 2) 5.49 x 104 C of electricity3) 1.83 x 107 C of electricity 4) 5.49 x 107 C of electricity
Ans. (4)9 gm of Al requires 96500 C5120 gms of Al requires ......?(5.12 kg)
= 996500 x 5120
= 5.49 x 107 compounds
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90. The volume of a colloidal particle, VC as compared to the volume of a solute particle in a true solution VS, could be
1)S
C
VV ~ 103 2)
S
C
VV
~ 10-3 3)S
C
VV
~ 1023 4)S
C
VV
~ 1
Ans. (1)For true solution the diameter range is 1 to < 10Å and for colloidal solution diameter range is 10 to 1000Å. Takingthe lower limits.
3
S
C
3S
3C
S
C
rr
πr34
πr34
VV
==
Putting ratio of diameters in place of S
C
rr
∵ Ratio of diameters = 33
101
10 =
∴3
S
C 10~VV
−
91. Consider the reaction : 322 23 NHHN →+ carried out at constant temperature and pressure. If ∆H and ∆U arethe enthalpy and internal energy changes for the reaction, which of the following expressions is true?
1) ∆H > ∆U 2) ∆H < ∆U 3) ∆H = ∆U 4) ∆H = 0Ans. (2)
For the reaction
2(g))(2 3H+gN )(32 gNH
∆ n = -2∆H = ∆ v + ∆ n RT∆H < ∆ v ∴ ∆ n = -2
92. The solubility product of a salt having general formula MX2, in water : 4 x 10-12. The concentration of M2+ ions inthe aqueous solution of the salt is
1) 4.0 x 10-10 M 2) 1.6 x 10-4 M 3) 1.0 x 10-4 M 4) 2.0 x 10-6 MAns. (3)
Let ‘S’ be the solubility of salt in mol. Lt–1
(-)22 2x M +→ +Mx
s s 2s2)(2 ][x ][ −+= MK
PS
S . (2S)2
-123 10 x 4 4 == SKPS
M10 S ][ -42 ==+M
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93. Benzene and toluene form nearly ideal solutions. At 200C, the vapour pressure of benzene is 75 torr and that oftoluene is 22 torr. The partial vapour pressure of benzene at 200C for a solution containing 78 g of benzene and 46g of toluene in torr is
1) 53.5 2) 37.5 3) 25 4) 50Ans. (4)
Moles of benzene = 7878
= 1
Moles of Toluene = 9246
= 0.5
Mole fraction of Benzene XB = 5.011
+
XB = 32
Partial Vapour pressure of benzene, B0BB x P P ×=
P = 75 x 32
= 50 torr.
94. Which one of the following statements is NOT true about the effect of an increase in temperature on the distributionof molecular speeds in a gas?
1) The area under the distribution curve remains the same as under the lower temperature2) The distribution becomes broader3) The fraction of the molecules with the most probable speed increases4) The most probable speed increases
Ans. (3)
95. For the reaction
)(22 gNO ),(2)(2 gg ONO +
(Kc = 1.8 x 10-6 at 1840 C)(R = 0.0831 kJ / (mol.K))When Kp and Kc are compared at 1840 C it is found that1) Whether Kp is greater than, less than or equal to Kc depends upon the total gas pressure2) Kp = Kc
3) Kp is less than Kc4) Kp is greater than Kc
Ans. (4)Kp & Kc are related
as, nc (RT) K ∆=Κ p
for the reaction,
)(22 gNO )(2)(2 gg ONO +
n∆ = 11
cp (RT) K K =∴
cp K K >∴
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96. The exothermic formation of ClF3 is represented by the equation :
)(2)(2 3 gg FCl + )(32 gClF ; ∆ rH = –329 kJWhich of the following will increase the quentity of ClF3 in an equilibrium mixture of Cl2, F2 and ClF3?
1) Adding F2 2) Increasing the volume of the container3) Removing Cl2 4) Increasing the temperature
Ans. (1)Increase in the concentration of reactants at equilibrium into shift the equilibrium products side.
97. Hydrogen ion concentration in mol / L in a solution of pH = 5.4 will be :
1) 3.98 x 10-6 2) 3.68 x 10-6 3) 3.88 x 106 4) 3.98 x 108
Ans. (1)-pH10 ][ =+H
= -5.410
= -0.4-5 10 x 10
= 0.6-6 10 x 10= (antilog 0.6) x 10-6
= 3.98 x 10-6 M
98. A reaction involving two different reactants can never be
1) bimolecular reaction 2) second order reaction3) first order reaction 4) unimolecular reaction
Ans. (4)The molecularity of such reaction can’t be less than 2
99. Two solutions of a substance (non electrolyte) are mixed in the following manner. 480 ml of 1.5 M first solution +520 mL of 1.2 M second solution. What is the molarity of the final mixture?
1) 2.70 M 2) 1.344 M 3) 1.50 M 4) 1.20 MAns. (2)
The value lies between M1 & M2
100. During the process of electrolytic refining of copper, some metals present as impurity settle as ‘anode mud’.These are
1) Fe and Ni 2) Ag and Au 3) Pb and Zn 4) Sn and AgAns. (2)
101.Electrolyte : KCl
149.9KNO3
145.0HCl
426.2NaOAc
91.0NaCl126.5:)cm( 12 −∞Λ molS Calculate cHOA
∞Λ using appropriate molar conductances of the
electrolytes listed above at infinite dilution in H2O at 250C
1) 217.5 2) 390.7 3) 552.7 4) 517.2Ans. (2)
NaClHCl∞∞∞∞ ΛΛ+Λ=Λ - NaOAcHOAc
= 91 + 426.2 - 126.5= 390.7
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102. If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the massof one mole of a substance will
1) be a function of the molecular mass of the substance 2) remain unchanged3) increase two fold 4) decrease twice
Ans. (4)1 atomic mass unit on the scale of 1/6 of C-12 = 2 amu on the scale of 1/12 of C-12Numerically the mass of a substance will become half of the normal scale.
103. In a multi-electron atom, which of the following orbitals described by the three quantum members will have thesame energy in the absence of magnetic and electric fields ?
a) n = 1, l = 0, m = 0 b) n = 2, l = 0, m = 0 c) n = 2, l = 1, m = 1d) n = 3, l = 2, m = 1 e) n = 3, l = 2, m = 01) (d) and (e) 2) (c) and (d) 3) b) and c) 4) a) and b)
Ans. (1)∵ ‘n’ & ‘l’ values are same
104. Based on lattice energy and other considerations which one of the following alkali metal chlorides is expected tohave the highest melting point?
1) RbCl 2) KCl 3) NaCl 4) LiClAns. (3)
LiCl have lowest melting point due to covalent nature and thereafter melting point decreases going down thegroup because lattice enthalpies decreases and size of alkali metal increases.
105. A schematic plot of lnKeq l versus inverse of temperature for a reaction is shown below 6.0
2.0 x 10-31.5 x 10-32.0
6.0
In K
ey
)(1 1−KT
The reaction must be
1) highly spontaneous at ordinary temperature 2) one with negligible enthalpy change3) endothermic 4) exothermic
Ans. (4)
In
−∆=
211
2
T1
T1
RH
kk
In ]102105.1[RH
26 33 −− ×−×∆=
H∆ of reaction comes out to be -ve. Hence reaction is exothermic )4(∴
106. Heating mixture of Cu2O and Cu2S will give
1) Cu2SO3 2) CuO + CuS 3) Cu + SO3 4) Cu + SO2
Ans. (4)
222 6Cu 2 SOCuSOCu +→+
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107. The molecular shapes of SF4, CF4 and XeF4 are
1) different with 1, 0 and 2 lone pairs of electrons on the central atom, respectively2) different with 0, 1 and 2 lone pairs of electrons on the central atom, respectively3) the same with 1, 1 and 1 lone pair of electrons on the central atoms, respectively4) the same with 2, 0 and 1 lone pairs of electrons on the central atom, respectively
Ans. (1)
In ,4SF
n = 6 + 4 = 10
52n =
∴ Hybridization of sulphur is dsp3
Thus, shape is trigonal bipyramidal with one lone pair (See-saw shaped)In ,CF4
n = 4 + 4 = 8
42n =
∴ Hybridization of carbon is 3sp
Thus shape is tetrahedral with zero lone pairIn ,XeF4
n = 8 + 4 = 12
62n =
∴ Hybridization of Xe is 23dsp
Thus, shape is octahedral with 2 lone pairs (Square planar).
108. The disperse phase in colloidal iron (III) hydroxide and colloidal gold is positively and negatively charged, respectively.Which of the following statements is NOT correct?
1) Coagulation in both sols can be brought about by electrophoresis2) Mixing the sols has no effect3) Sodium sulphate solution causes coagulation in both sols4) Magnesium chloride solution coagulates, the gold sol more readily than the iron (III) hydroxide sol
Ans. (2)By adding oppositely charged solution mutual coagulation of two sols can be done.
109. The number of hydrogen atom(s) attached to phosphorus atom in hypophosphorous acid is
1) three 2) one 3) two 4) zeroAns. (3)
Molecular formula of the Acid is 23POH
Structural formula is
PH
OHH
O
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110. What is the conjugate base of OH– ?
1) O2– 2) O– 3) H2O 4) O2
Ans. (1)-2)()( O H - =+−OH
111. Heating an aqueous solution of aluminium chloride to dryness will give
1) Al (OH)Cl2 2) AL2O3 3) Al2Cl6 4) AlCl3
Ans. (3)Aluminium chloride in aqueous solution exists as ions pair
OH6AlCl2 23 + )aq(])OH(AlCl[)aq(])OH(AlCl[ 224422−+ +
The crystallization of 3AlCl from aqueous solution therefore yields an ionic solid having composition−+ ])OH(AlCl[])OH(AlCl[ 624622 which on heating gives 62ClAl .
OH12ClAl])OH(AlCl[])OH(AlCl[ 262K460
624622 + →−+
112. The correct order of the thermal stability of hydrogen halides (H - X) is
1) HI > HCl < HF > HBr 2) HCl < HF > DBr < HI3) HF > HCl > HBr > HI 3) HI > HBr > HCl > HF
Ans. 3)Thermal stability is direictly proportional to bond dissociation energy and bond dissociation is inversely proportionalto bond length. The decreasing order of bond dissociation energy is HF > HCl > HBr > HI∴ Order of thermal stability is HF > HCl > HBr > HI
113. Calomel (Hg2Cl2) on reaction with ammonium hydroxide gives
1) HgO 2) Hg2O 3) NH2 - Hg - Hg - Cl 4) HgNH2ClAns. (4)
ClNHCl)NH(HgHgNH2ClHg 42322 +↓↓→+chlorideamido)II(Mercury
114. The number and type of bonds between two carbon atoms in calcium carbide are1) Two sigma, two pi 2) Two sigma, one pi3) One sigma, two pi 4) One sigma, one pi
Ans. (3)
CaC
C
115. The odixation state of chromium in the final product formed by the reaction between Kl and acidified potassiumdichromate solution is1) +3 2) +2 3) +6 4) +4
Ans. (1)
OHICrH I OCr 223)((-)2
72 ++→++ ++−
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116. In silicon dioxide1) there are double bonds between silicon and oxygen atoms2) silicon atom is bonded to two oxygen atoms3) each silicon atom is surrounded by two oxygen atoms and each oxygen atom is bonded to two silicon
atoms.4) each silicon atom is surrounded by four oxygen atoms and each oxygen atom is bonded to two silicon
atomsAns. (4)
Silicon dioxide exhibits polymorphism. It is a network solid in which each Si atom is surrounded tetrahedral byfour oxygen atoms.
117. The lanthanide contraction is responsible for the fact that1) Zr and Zn have the same oxidation state2) Zr and Hf have about the same radius3) Zr and Nb have similar oxidation state4) Zr and Y have about the same radius
Ans. (2)As a result of lanthanide contraction there is a close similarity in size. Zr and Hf have nearly same atomic radiusi.e. 160 and 158 pm respectively.
118. The IUPAC name of the coordination compound K3[Fe(CN)6] is1) Tripotassium hexacyanoiron (II)2) Potassium hexacyanoiron (II)3) Potassium hexacyanoferrate (III)4) Potassium hexacyanoferrate (II)
Ans. (3)
119. In which of the following arrangements the order is NOT according to the property indicated against it ?1) Li < Na < K < Rb :
Increasing metallic radius2) I < Br < F < Cl :
Increasing electron gain enthalpy(with negative sign)
3) B < C < N < O :Increasing first ionization enthalpy
4) Al3+ < Mg2+ < Na+ < F– :Increasing ionic size
Ans. (3)Ionization Enthalpy of Nitrogen is greater than that of Oxygen due to stable half filled configuration of Nitrozen.
120. Of the following sets which one does NOT contain isoelectronic species ?
1) -3
-23
33 NO ,CO ,−BO 2) -
3-2
323 NO ,CO ,−SO
3) -222 C ,N ,−CN 4) -
4-2
434 ClO ,SO ,−PO
Ans. (2)
121. 2-Methylbutane on reacting with bromine in the presence of sunlight gives mainly1) 1-bromo-3-methylbutane 2) 2-bromo-3-methylbutane3) 2-bromo-2-methylbutane 4) 1-bromo-2-methylbutane
Ans. (3)The reaction proceeds through free radical mechanism.
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122. Which of the following compounds shows optical isomerism ?
1) −36 ])([ CNCo 2) −3
342 ])([ OCCr 3) −24 ][ZnCl 4) +2
43 ])([ NHCuAns. (2)
Due to lack of symmetry −3342 ])OC(Cr[ shows optical isomerism.
123. Which one of the following cyano complexes would exhibit the lowest value of paramagnetic behaviour?
1) −36 ])([ CNCo 2) −3
6 ])([ CNFe 3) −36 ])([ CNMn 4) −3
6 ])([ CNCr(At. Nos : Cr = 24, Mn = 25, Fe = 26, Co = 27)
Ans. (1)
In −36 ])([ CNCo , Co is
in +3 state and has d6 configuration, the sixelectrons get paired-up in presence of a strong filed ligand like CN(-)
124. The best reagent to convert pent-3-en-2-ol into pent-3-in-2-one is1) Pyridinium chloro-chromate2) Chromic anhydride in glacial acetic acid3) Acidic dichromate4) Acidic permanganate
ans. (1)Pyridinium chloro-chromate oxidizes an alcoholic group selectively in the presence of carbon-carbon double bond.
125. A photon of hard gamma radiation knocks a proton out of Mg 2412 nucleus to form
1) the isobar of Na 2311 2) the nuclide Na 23
11
3) the isobar of parent nucleus 4) the isotope of parent nucleusAns. (2)
pNaMg 11
2311
2412 +→γ+
126. Reaction of one molecule of HBr with one molecule of 1,3-butadiene at 400C gives predominantly1) 1-bromo-2-butene under kinetically controlled conditions2) 3-bromobutene under thermodynamically controlled conditions3) 1-bromo-2-butene under thermodynamically controlled conditions4) 3-bromobutene under kinetically controlled conditions
Ans. (3) additionHBrCHCHCHCH
4,1
22 →+=−=
BrCHCHCHCH −−=− 23
1- bromo-2-Butene
127. The decreasing order of nucleophilicity among the nucleophiles
a)CH C – O3
–
Ob) CH3O
–
c) CN–
d) H C3 S – O– isO
O1) (c), (b), (a), (d) 2) (b), (c), (a), (d)3) (d), (c), (b), (a) 4) (a), (b), (c), (d)
Ans. (2)∵ strong bases are generally good nucleophiles
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128. Tertiary alkyl halides are practically inert to substitution by 2NS mechanism because of
1) steric hindrance 2) inductive effect3) instability 4) insolubility
Ans. (1)In 2SN reaction mechanism reactivity of allkyl halide depends on stability of transition state which decreases withsteric hindrance in allkyl hallide.
129. In both DNA and RNA, heterocylic base and phosphate ester linkages are at -
(1) '5C and '
1C respectively of the sugar molecule
2) '1C and '
5C respectively of the sugar molecule
3) '2C and '
5C respectively of the sugar molecule
4) '5C and '
2C respectively of the sugar molecule
Ans. (2)
130. Among the following acids which has the lowest apK value ?
1) CH3CH2COOH 2) (CH3)2) CH – COOH3) HCOOH 4) CH3COOH
Ans. (3)Strong acids have low value of apK
131. Of the five isomeric hexanes, the isomer which can give two monochlorinated compounds is
1) 2-methylpentane 2) 2, 2-dimethylbutane 3) 2, 3-dimethylbutane 4) n-hexaneAns. (3)
3
||
3
33
CHCHCHCH
CHCH
−−−
The molecule contains two types of hydrogens only.Therefore can give two mono chlorinated compounds.
132. Alkyl halides react with dialkyl copper reagents to give
1) alkenyl halides 2) alkanes 3) alkyl copper halides 4) alkenesAns. (2)
21
212 CuXRRCuRxR +−→+−
(Corey - House synthesis)
133. Which one of the following methods is neither meant for the synthesis nor for separation of amines ?
1) Curtius reaction 2) Wurtz reaction3) Hofmann method 4) Hinsberg method
Ans. (2) Wartz’s rxn gives alkanes.
134. Which types of isomerism is shown by 2, 3- dichlorobutane ?
1) Structural 2) Geometric 3) Optical 4) DiastereoAns. (3)
The molecule has two chirality centres.
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135. Amongst the following the most basic compound is
1) p-nitroaniline 2) acetanilide 3) aniline 4) benzylamineAns. (4)
Due to resonance of electron pair in aniline basic strength decreases. In benzylamine electron pair is not involvein resonance.
2NH2NH2CH 2NH
2NO
3NHCOCH
(I) (II) (III) (IV)Decreasing order of basic strength is (II) > (I) > (IV) > (III)
136. Acid catalyzed hydration of alkenes except ethene leads to the formation of
1) mixture of secondary and tertiary alcohols2) mixture of primary and secondary alcohols.3) secondary or tertiary alcohol4) primary alcohol
Ans. (3)Any alkene expect ethylene on acid catalyzed hydration gives secondary or tertiary alcohol because reactionproceed according to electrophilic addition reaction mechanism.
137. Which of the following is fully fluorinated polymer?
1) PVC 2) Thiokol 3) Teflon 4) NeopreneAns. (3)
Thiokol
−−−−−−−− 222222 CHCHSSCHCHSSCHCH
S S S Sn
Neoprene :
−=−=− CHCHCCH
Cln
PVC:
−−− CC
H
Cl
H
H n
Teflon :
−−− CC
n
F F
FF
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138. Elimination of bromine from 2-bromobutane results in the formation of -
1) predominantly 2-butyne 2) predominantly 1-butene3) predominantly 2-butene 4) equimolar mixture of 1 and 2-butene
Ans. (3)In elimination reaction of allkyl halide major product obtain according to Saytzef’s rule (highly substituted alkeneis the major product).
33HBr
inationlime323 CHCHCHCHCHCH)Br(CHCH = →
−
139. Equimolar solutions in the same solvent have
1) Different boiling and different freezing points2) Same boiling and same freezing points3) Same freezing point but different boiling point4) Same boiling point but different freezing point
Ans. (2)van’t Hoff factor (i) is not known so it is impossible to predict relation between colligative properties of equimolarsolutions in the same solvent.If, i = 1Then Equimolar solutions in the same solvent will have same boiling and same freezing points.
140. The reaction
R – C + Nu + X X
OR – C
Nu
O
is fastest when X is
1) OCOR 2) OC2H5 3) NH2 4) ClAns. (4)
−Cl is the best leaving group among the given option.
141. The structure of diborane (B2H6) contains
1) four 2c-2e bonds and four 3c-2e bonds2) two 2c-2e bonds and two 3c-3e bonds3) two 2c-2e bonds and four 3c-2e bonds4) four 2c-2e bonds and two 3c-2e bonds
Ans. (4)Diborane, 62HB has hydrogen bridge structure B atoms and 4 terminal H atoms lie in the same plane; the terminalB-H bonds may be regarded as conventional 2 centre-2 electron bonds i.e. 2c-2e bonds. The two bridging Hatoms lie symmetrically above and below this plane. Here three centers are united by two paired electrons(3c-2e) bonding called banana bond.
142. Which of the following statements in relation to the hydrogen atom is correct?
1) 3s, 3p and 3d orbitals all have the same energy2) 3s and 3p orbitals are of lower energy than 3d orbital3) 3p orbital is lower in energy than 3d orbital4) 3s orbital is lower in energy than 3p orbital
Ans. (1)For hydrogen the energy order of orbital is1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f
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143. Which of the following factors may be regarded as the main cause of lanthanide contraction?
1) Greater shielding of 5d electron by 4f electrons2) Poorer shielding of 5d electrons by 4f electrons3) Effective shielding of one of 4f electrons by another in the subshell4) Poor shielding of one of 4f electron by another in the subshell
Ans. (4)
144. The value of the ‘spin only’ magnetic moment for one of the following configurations is 2.84 BM. The correct oneis
1) d5 (in strong ligand field)2) d3 (in weak as well as in strong fields)3) d4 (in weak ligand field)4) d4 (in strong ligand field)
Ans. (4)d4 in presence of strong field ligand, contains two unpaired electrons
145. Reaction of cyclohexanone with dimethylamine in the presence of catalytic amount of an acid froms a compoundif water during the reaction is continuously removed. The compound formed in generally known as
1) an amine 2) an imine3) an enamine 4) a Schiff’s base
Ans. (3)
O + HN (CH )3 2
O N(CH )3 2
146. P-cresol reacts with chloroform in alkaline medium to give the compound A which adds hydrogen cyanide to form,the compound B. The latter on acidic hydrolysis gives chiral carboxylic acid. The structure of the carboxylic acidis
1)
CH3
CH COOH2
OH
2)
CH3
CH COOH2
OH
3)
CH3
CH(OH)COOH
OH
4)
CH3
CH(OH)COOH
OH
Ans. (3)
+ CHCl3
CH3 CH3
CHOOH OH
OH(-)
CH3
CH(OH) COOHOH
i) HON
ii) H O3(+)
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147. If the bond dissociation energies of XY, X2 and Y2, (all diatomic molecules) are in the ratio of 1:1:0.5 and Hf∆
for the formation of XY is -200 kJ mole–1. The bond dissociation energy of X2 will be
1) 400 kJ mol–1 2) 300 kJ mol–1 3) 200 kJ mol–1 4) 100 kJ mol–1
Ans. No correct option
148. An amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50atm pressure. Ammonium hydrogen sulphide decomposes to yield NH3 and H2S gases in the flask. When thedecomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm. ? The equilibriumconstant for NH4HS decomposition at this temperature is
1) 0.11 2) 0.17 3) 0.18 4) 0.30Ans. (1)
)g(SH)g(NH 23 +)s(HSNH4
Initial 0 5.0
.equiAt p5.0 + p0
0.press
0.5 + 2p = 0.84∴ p = 0.17 atm
SHNHp 23ppK ×=
)17.0()67.0( ×=
2atm11.0=
149. An organic compound having molecular mass 60 is found to contain C = 20%, H = 6.67% and N = 46.67% whilerest is oxygen. On heating it gives NH3 alongwith a solid residue. The solid residue give violet colour with alkalinecopper sulphate solution. The compound is
1) 223 CONHCHCH 2) CO22 )(NH
3) 23CONHCH 4) CO3NCH
Ans. (2)
Element Percentage atomsof.nolativeRe ratioSimplest
C
N
O
20
67.6
67.46
66.26 67.1
67.6
33.3
67.1
1
4
2
1
H
The molecular formula is ONCH 24
So, the compound is 22NCONHH
↑+→∆32222 NHCONHCONHNHCONHNH2
BiuretBiuret gives violet colour with alkaline copper sulphate solution
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150.41t can be taken as the time taken for the concentration of a reactant to drop to 4
3 of its initial value. If the rate
constant for a first order reaction is K, the 41t can be written as
1) 0.75 / K 2) 0.69 / K 3) 0.29 / K 4) 0.10 / KAns. (3)
34log
K303.2t 4/1 =
K29.0=
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MATHEMATICS1. If A2 - A+ I = 0, then the inverse of A is
1) A + I 2) A 3) A -1 4) I - A
Ans.(4) A2 - A+ I = 0
AIA 2 −=⇒
AIAA −=⇒
AI A -1 −=∴
2. If the cube roots of unity are I, W, W2 then the roots of the equation (x-1)3+8 = 0, are
1) -1, -1+2w, -1 -2w2 2) -1, -1, -13) -1, 1-2w, 1-2w2 4) -1, 1+2w, 1+2w2
Ans.(3) (x-1)3 + 8 = 0Put x -1 = y
08y 3 =+→
2 2- , 2- 2,- y ωω=⇒
2 2-1 , 2-1 1,- x ωω=⇒
3. Let R = (3, 3) (6, 6) (9, 9) (12, 12), (6, 12) (3, 9) (3, 12), (3, 6) be a relation on the set A = 3, 6, 9, 12. Therelation is
1) reflexive and transitive is 2) reflexive only3) an equivalence relation 4) reflexive and symmetric only
Ans.(1) R is reflexive and transitive but not symmetric because (9, 3) ∉ R
4. Area of the greatest rectangle that can be inscribed in the ellipse is 1by
ax
2
2
2
2=+
1) 2ba 2) ab 3) ab 4) ba
Ans.(1) Area of greatest rectangle
= 4ab21 ×
= 2ab
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5. The differential equation representing the family of curves ( ), 22 cxcy += where c > 0, is a parameter, is oforder and degree as follows :
1) order 1, degree 2 2) order 1, degree13) order 1, degree 3 4) order 2, degree 2
Ans.(3) y2 = 2c(x + c ) c < 0since there is only one arbitary constantorder = 1
2cdxdy2y =
dxdyyc =⇒
2/32
dxdyy 2 x 2
+
=⇒
dxdyyy
Elimainating radical powers,We get, degree = 3
6.
+++
∞→1sec
n1...
n4sec
n2
n1sec
n1 Lim 2
22
222
2n is
1) 1 sec 21
2) 1 cosec 21
3) tan 1 4) 1 tan 21
Ans.(4) 1tan21
+++
∞→ 2
22
222
222
2n sec.....4sec.21sec
n1 Lt
nn
nn
nnn
22
n
1r n n
rsec nr
n1 Lt
⇒ ∑
=∞→
)dn(xsecx 221
0∫⇒
dtt sec 21 21
0∫⇒ tlet x2 =
dtdx x2 =
1 tan 21⇒
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7. ABC is a triangle. Forces R , , QP acting along IA, IB and IC respectively are in equilibrium, where I is theincentre of ABC ∆ . Then P : Q : R is
1) sin A : sin B : sin c 2)2
sin:2Bsin :
2Asin C
3)2
cos:2B cos :
2A cos C
4) cos A : cos B : cos C
Ans.(3)
C/2B/2
B
P
R
+− 2
cBQπ
Applying Lami’s theorem
+
=
+
=
+
2sin
2sin
2sin BA
RCA
QCB
P
2C cos :
2B cos :
2A cos R : Q: P =⇒
8. If in a frequency distribution, the mean and median are 21 and 22 respectively, then its mode is approximately
1) 22.0 2) 20.5 3) 25.5 4) 24.0Ans.(4) Mode = 3 median - 2 mean
= 66 - 42= 24
9. Let P be the point (1, 0) and Q a point on the locus xy 82 = . The locus of mid point PQ is
1) 0242 =+− xy 2) 0242 =++ xy 3) 0242 =++ yx 4) 0242 =+− yx
Ans.(1) Let the mid point be = (h, k)
Let Q = )y ( 11x
khx =+
=+
2y0
21 11
1h 2 x1 −=⇒ ; ky 21 =
x8yon lies y x 211 =
)12(84 2 −=⇒ hk
)12( 22 −=⇒ hk
∴ The eqn. of locus is
0242 =+− xy
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10. If C is the mid point of AB and P is any point outside AB, then
1) PC 2PB =+PA 2) PC PB =+PA
3) 0PC 2PB =++PA 4) 0PC PB =++PA
Ans.(1) A C B P
CAPCPA += ..(1)
BCPCPA += ...(2)
Adding BC 2 2 ++= CAPCPA
PC 2 PB PA =+⇒
11. If the coeffieients of rth, (r + 1)th and (r+2)th terms in the binomial expansion of (1+y)m are in A.P., then m and rsatisfy the equation
1) 024)14( 22 =−+−− rrmm 2) 024)14( 22 =+++− rrmm
3) 024)14( 22 =−++− rrmm 4) 024)14( 22 =++−− rrmm
Ans.(3) Coefficients of r, (r+1)th or (r+2)th terms are in A.P.
1m
1-rm
r C C m 2 ++=⇒ rC
r 1)(r1
r)-(m 1)r-(m1
r)-r(m2
++
+=⇒
024)14(m 22 =−++−⇒ rrm
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12. In a triangle PQR, 2π R =∠ . If tan
2P
and tan
2Q
are the roots of 0 a ,02 ≠=++ cbxax then
1) a = b + c 2) c = a + b 3) b = c 4) b = a + c
Ans.(2)
2π
P
QR
/2 Q P π=+⇒
4/2
QP π=+⇒
122
P tan =
+⇒ Q
12
Q tan 2tan12tan2
p tan=
−
+p
Q
2Q tan & 2
P tan ∴ are the root of the eqn. 02 =++ cbxax
ac−=⇒=⇒ 1
ab- 1
ac-1a
-b
c - a b- =⇒
b a c +=⇒
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13. The systems of equations
α x + y + z = α - 1
x + α y + z = α -1
x + y + α z = α - 1
has no solution, if α is
1) -2 2) either -2 or 13) not -2 4) 1
Ans.(1) ∆ = 0
0 11
1111
=α
αα
From options, it can be easily verified that α = -2
14. The value of α for which the sum of the squares of the roots of the equation 01)2(x2 =−−−− axa assume theleast value is
1) 1 2) 0 3) 3 4) 2
Ans.(1) Let , βα be the roots
2- a 2)-(a =+=+ βα
1)(a +−=βα
βα−β+α=β+α 2)( 222
)1( 2)2( 2 ++−= aa
22442 +++−= aaa
622 +−= aa
5)1( 2 +−= a
22 βα +∴ is least value a = 1
15. If the roots of the equation 0x2 =+− cbx be two consecutive integers, then b2 - 4c equals
1) -2 2) 3 3) 2 4) 1Ans.(4) Let α , α +1 be the roots
α + α + 1 = b
b 12 =+⇒ α
c 1)( =+αα
cb 42 −
)( 4 )12( 22 α+α−+α=
( )α+α−+α+α= 44144 22 = 1
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16. If the letters of the words SACHIN are arrange in all possible ways and these words are written out as indictionary, then the words are SACHIN appears at serial number.
1) 601 2) 600 3) 603 4) 602Ans.(1) S A C H I N
A - order : - A C H I N SRank = (5!) + 0 (4!) + 0(3!)+ 0 (2!) + 0(1!) + 1= 11205 +×= 601
17. The value of 3r-56
6
14
50 C ∑=
+r
C is
1) 455C 2) 3
55C 3) 356C 4) 4
56C
Ans.(4) 3
6
14
50 rc - 56 ∑=
+r
C
350
351
352
353
354
355
450 C C C C C C C ++++++
= 456C
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18. If A = , 1001
I and 1101
=
then which of the following holds for all ,1≥n by principle of mathematical induction
1) 1)I(nnAA n −−= 2) 1)I(nA2A 12 −−= −n
3) 1)I(nAn −+= nA 4) 1)I(n2A 1n −+= − An
Ans.(1) A =
1101
I =
1001
=
1101
1101
A2
++++
=10110001
=
1201
=
=
1301
1101
1201
A3
IA2A2 −=
=
−
=
1201
1001
2202
I2A3A3 −=
=
−
=
1301
2002
3303
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19. If the coefficient of 7x in 11
2 1
+
bxax equals the coefficient of 7−x in
11
21
−
bxax then a and b satisfy the
relation
1) a - b = 1 2) a + b = 1 3) 1=ba
4) ab = 1
Ans.(4)11
2
bx1x
+
53
7112r =−×=
coefficient of 56
5117 1a
bCx =
11
21
−
bxax
63
711 =+=r
Coefficeint of 65
6117
b1 a C x =−
65
611
56
611
b1 a C
b1 a C =
1 ab =⇒
20. Let f : (-1, 1) → B, be a function defined by f(x) = 21
12tan
xx
−− then f is both one-one and onto when B is the
interval
1)
2,0 π
2)
2,0 π
3)
−
2,
2ππ
4)
−
2,
2ππ
Ans.(4) Since B,1)1(f →−→ &
−= −
21
x12xtanf(x)
if is both one - one & onto i , e ( bijective )
⇒ if ( x) inverse tangent of n whose Range is always
−
2,
2ππ
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21. If 1z and 2z are two non-zero comple numbers such that 2121 z zzz +=+ , then arg 1z - arg 2z is equal to
1)2π
2) π− 3) 0 4)2π−
Ans.(3) [ ]2212
21 z +=+ zzz
)θ(θ cos z z 2z z 21212
22
1 −++⇒
= 212
22
1 z z 2z z ++
cos 1 )θ(θ 21 =−
0 21 =−⇒ θθ
0 )(z Arg - )(z Arg 21 =⇒
22. If iz31
z w−
= and ,1=w then z lies on
1) an ellipse 2) a circle 3) a straight line 4) a parabola
Ans.(3) i31z
zw−
=
i 31
z
−=
zw
zi =⇒31-z
r on the lies z ⊥⇒ bisector of the line segment joining ( )0 0, and 31,0
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23. If 2222 −=++ cba and
xcxbxaxcxbxaxcxbxa
xf222
222
222
1)1()1()1(1)1()1()1(1
)(+++++++++
=
then )(xf is a polynomial of degree
1) 1 2) 0 3) 3 4) 2
Ans.(4) 2cba 222 −=++
In 3211 ccc c ),x(f ++⇒
xc1 x)b(12x x)cba(1 x )c(1 xb12x x)cba(1 x )c(1 x)b(12xx)cba(1
f(x) 22222
22222
22222
++++++++++++++++++
=⇒
xc1 x)b(11 x )c(1 xb11 x )c(1 x)b(11
22
22
22
++++++
⇒
211 R RR −→
xc1x)b1(1x)c1(xb11
01-x0
22
22
++++⇒
[ ]xcxxc1 1)-(x - 22 −−+⇒
2 degree x)-(1 2 =⇒⇒
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24. The normal to the curve θ) cos θθ(sin ay θ),sin θθ (cos x −=+=α at any point ‘ θ ’ is such that
1) It passes through the origin
2) It makes angle θ2π + with the x - axis
3) It passes through
−a,
2πa
4) It is at a constant distance from the origin
Ans.(2 , 4) θsinθ)a(cosθx +=
θ) cos θθa(sin y −=
θθ
θθθθθθθθ
cossin
sincossin( cossin(cos =++−−+=
aa
dxdy
∴ slope of the normal
θθ
sincos−=−⇒
dydx
⇒ Slope of the normal is θ2π +
∴ Eqn. of the normal
)θsinθaθcosa(xθsin
cosθ- cosθ θ asinθ a y −−=+−⇒
θsin θ cos θ aθcos a θ cos x - θsin θ cos θ a θsin aθsin y 22 ++=+−⇒
aθsin y θ cosx =+⇒
⇒ It is at a constant distance from the origin.
25. A function is matched below against an interval where it is supposed to be increasing. Which of the following pairsis incorrectly matched?
Interval Function
1) ( )∞∞− , 333 23 ++− xxx
2) ) ,2[ ∞ 61232 23 ++− xxx
3)
∞−
31, 123 23 +− xx
4) ( ]4,−∞− 66 23 +− xx
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Ans.(3)
1x2x3)x(f 23 +−=
3/1x02x6)x('f ≥⇒≥−=
Option (3) is incorrect. Checking other function similarly we find that they are correctly matched.
26. Let α and β be the distinct roots of ,0ax2 =++ cbx then
( )( )2
2
αx axcbxaxcos-1 lim
−++
→ is euql to
1) ( )22
2βα −a
2) 0
3) ( )22
2βα −−a
4) ( )221 βα −
Ans.(1) We have 2
2
αx α)x()cbxaxcos(1lim
−++−
→
2
2
αx2
22
αx )x(2
β)x()x(asinlim2
α)x(
2cbxax(sin
lim2α−
−α−
=−
++
=→→
β−α−=++∴=++β)x)(x(acbxax
0cbxaxofroots areα,2
2∵
22
2
ax)x(
4a.
2)x)(x(a
2)x)(x(asin
lim2 β−
β−α−
β−α−
=→
22
22
2 )(2
a)(4
a)1(2 β−α=β−α=
27. Suppose f(x) is differentiable at x = 1 and 5)1(h1 lim
0=+
→hf
h then f(1) equals
1) 3 2) 4 3) 5 4) 6Ans.(4) Applying L’ Hospital rule
51
)1(' 0
=+→
hfLimh
5 (1)f' =⇒
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28. Let f be differentiable for all x. If f(1) = -2 and 2)(' ≥xf for x ∈ [1, 6], then
1) f(6) ≥ 8 2) f(6) < 83) f(6) < 5 4) f(6) = 5
Ans.(2) Here from LMVT
2 5
)1()6( ≥− ff
.8)6( ≥f
29. If f is a real-valued differentiable function satisfying ∈−≤− yxyxyfxf ,,)()()( 2 and f(0) = 0, then f(1) equals
1) -1 2) 0 3) 2 4) 1
Ans.(2) y-x lim y-x
- limyxyx →→
≤f(y)f(x)
or 0 (x) ' ≤f
0 (x) ' f =⇒
(x) f⇒ is constant, As 0(0) =⇒ f
0 =∴ f(1)
30. If x is so small that x3 and higher powers of x may be neglected, then ( )2
1
323
1
211)1(
x
xx
−
+−+
may be approximated
as
1) 2
831 x− 2) 2
833 xx +
3) 2
83 x− 4) 2
83
2xx −
Ans.(3)( )
( ) 2/1
32/3
12111
x
xx
−
+−+
( ) 2/1
22
1
4x
22.3
231
221.
23
231
x
xxx
−
++−++
=
( )( ) 2/12
2/1
2
183
183
−−−=−
−= xx
x
x
22
83...
21
83 xxx −=
++−=
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31. If ∑∑∑∞
=
∞
=
∞
=
===0n
n
0n
n
0n
n cz,b y,ax where a, b, c are in A.P and 1c 1, b 1, a <<< then x, y, z are in
1) GP 2) AP3) Arithmetic - Geometric Progression 4) HP
Ans.(4) c-11 z
b-11 y
11 ==−
=a
x
a, b, c, are in A.P
cb −−∴
11,
11,
a-11 are in H -P
z y, x,⇒ are in H-P
32. In a triangle ABC, let .2πC =∠ If r is the inradius and R is the circumradius of the triangle ABC, then 2(r+R)
equials
1) b + c 2) a + b 3) a + b + c 4) c + a
Ans.(2)
c a B
A
b
A
c
Let a = b = 1
2=⇒ cr = (s-c) tan c/2
/4 tan 22
22 Π
−+=
22222 −+⇒
211
222 −=−⇒r
11214
2 1 1 4
c b aR×××
××=∆
=
21=
22
12
1-1 2 R) (r 2 =
+=+∴
= a + b
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33. If ,2
cos- xcos 1-1 α=− y then 22 y cos 44 +− αxyx is equal to
1) 2 sin 2 α 2) 4
3) αsin 4 2 4) αsin 4- 2
Ans.(3) α 2y coscos 11 =− −− x
α=
−+⇒ −
4y-1 1
2yx cos
221 x
α cos2
y-4 x-1
2xy
22
=+⇒
( )( ) ( )222 cos2y-4 x-1 xy−=⇒ α
α−+α=+−− cosxy4yxcos4yxyx44 2222222
4cos 44x 22 =+−⇒ yxy α - α4sinαcos4 22 =
34. If in a ABC ∆ , the altitudes from the vertices A, B, C on opposite sides are in H.P., then sin A, sin B, sin C are in
1) G.P. 2) A.P.3) Arithmetic - Geometric - Progression 4) H . P.
Ans.(2) In a ABC ∆ ,
Let the altitudes be 321 p,p,p from the vertices A, B, C respectively then
cba∆=∆=∆= 2P ,2P ,2P 321
cba∆∆∆⇒ 2 ,2 ,2 are in HP
c1 ,
b1 ,
a1 ⇒ are in HP
⇒ a, b, c are in AP⇒ 2 R sin A, 2 R sin B, 2 R sin C are in AP⇒ sin A, Sin B, sin C in AP.
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35. If dx, 2 I dx, 232
1
02
1
01
xxI ∫∫ ==
dx, 2 I anddx 232
2
14
2
13
xxI ∫∫ == then
1) 12 I >I 2) 21 I >I 3) 43 I =I 4) 43 I >I
Ans.(2) ∫ ∫==1
0
1
0
x2
x1 dx 2 I dx, 2
32I
∫ ∫==1
0
2
0
x4
x3 dx 2 I dx, 2
32I
For 1) (0, ∈x
32 x >x32 x2 2 >⇒ x
dx 2 dx 2 32 1
0
21
0
x∫∫ >⇒
21 I I >=
For 2) (1, ∈x
23 x >x23
22 xx >
∫ ∫>2
1
x2
1
x dx 2 dx 2 23
34 I >I
36. The area enclosed between the curve y = loge (x+e) and the coordinate axes is
1) 1 2) 2 3) 3 4) 4Ans.(1) Required area
dx )log(0
1
exe
+= ∫−
Put x + e = t
dt log1
te
∫=
units sq. 1 x]- x log [ e1 == x
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37. The parabolas xy 42 = and yx 42 = divide the square region bounded by the lines x = 4, y = 4 and the coordinate
axes. If 321 S ,S ,S are respectively the area of these parts numbered from top to bottom; then 321 S :S :S is
1) 1 : 2 : 1 2) 1 : 2 : 1 3) 2: 1 : 2 4) 1: 1 : 1Ans.(4)
y = 4 (4 , 4)
x = 4
S3
(0, 0)
S1
S2
dx 4
x2 24
02
−= ∫ xS
= 3/16
16 4 4 S 322 =×=++∴ SS
332
316 16 S 32 =−=+∴ S
316S ,3
16 S 31 ==⇒
1 : 1 :1 S : S : S 321 =∴
38. If ydxdyx = (log y - log x + 1), then the solution of the equation is
1) cxy =
yx log 2) cyx =
xy log
3) cx=
xy log 4) cy=
yx log
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Ans.(3) If 1) x log -y (logy +=dxdyx
+=⇒ 1
xy log
xy
dxdy
Let y = vx
1) v(log V dxdvxv +=+⇒
v) v(log V dxdvxv +=+⇒
∫ ∫=⇒x
dx vlog v
dv
c log x log v)(log log +=⇒
log (log v) = log (cx)
xc xy log =⇒
39. The line parallel to the x-axis and passing through the intersection of the lines ax + 2by + 3b = 0 and bx - 2ay - 3a=0 where (a, b) ≠ (0, 0) is
1) below the x-axis at a distance of 23
from it
2) below the x-axis at a distance of 32
from it
3) above the x-axis at a distance of 23
from it
4) above the x-axis at a distance of 32
form it
Ans.(1) ax + 2by = -3 bbx - 2ay = 3a
Solving these two equations we get y = 23−
40. A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thickness that melts at a rate of 50cm3/min. When the thickness of ice is 5 cm, then the rate at which the thickness of ice decreases, is
1) 36π1
cm/min. 2) 18π1
cm/min. 3) 54π1
cm/min. 4) 6π5
cm/min.
Ans.(2) 310)(x π34V += Where x is thickness of ice.
dtdxx 2)10(4
dtdV +=∴ π
5at x dtdx =
∴ ; cm/min.
π181=
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41. ∫
+− dx
)(log1)1(log
2
2xx
is equal to
1) C1 x)(log
xlog2 ++ 2) C
1x x2 ++
3) Cx1
xe2
x+
+4) C
1 x)(log x
2 ++
Ans.(4) ∫
+− dxx
xxfLet2
2)(log11 log)(
Differentiating from the optins Q(x)
+
+c
1 x)(logx dx
d2
[ ] [ ][ ]1)(log
x1 x)log (2 1 x)(log
2
2
+
−+⇒
x
x
( )[ ]22
2
1)(log
1- x log +
⇒x
42. Let f : R →R be a differentiable function having f(2) = 6,
=
481)2('f . Then ∫ −→
f(x)
6
3
2dt
24t x
Ltx equals
1) 24 2) 36 3) 12 4) 18
Ans.(4) form) (0/0 2-x
dt 4
lim
3)(
62 x
txf
∫→
18)2('))2( ( 41
)(') 4 lim 33
2=×=×=
→ffxf (x) (f
x
43. Let f(x) be a non-negative continuous function such that the area bounded by the curve y = f(x), x-axis and the
ordinates 4π=x and
4πβ >=x is
++ β 2β cos
4πβsin β . Then
2πf is
1)
−+ 12
4π
2)
+− 12
4π
3)
−− 2
41 π
4)
+− 2
41 π
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Ans.(4) ββπββ 2cos4
sin)(β
π/4
++=∫ dxxf
2β cos4π - β cos β βsin )(β' ++=∴ f
24π1
2π +−=
f
44. The locus of a point P ),( βα moving under the condition that the line βαxy += is a tangent to the hyperbola
12
2
2
2=−
by
ax
is
1) an ellipse 2) a circle 3) a parabola 4) a hyperbola
Ans.(4) 2222 ba −=⇒ αβ
∴ the locus is 2222 byxa =−
22
2
2
11x by
a
=−⇒
∴ the locus is a- hyperbola
45. If the angle θ between the line 2
22
11
1x −=−=+ zy and the plane 04zλy2x =++− is such that 3
1 sin =θ the
value of λ is
1) 35
2) 5-3
3)43
4) 3-4
Ans.(1) 2) 2, (1, )c ,b ,(a 2
1z2
1y1
1x111 =−=−=+
0 4 z yx2 =+λ+−
),1,2()c ,b ,a( 222 λ−=
λλ
++=53
22-2 θsin
λ+λ=⇒
5 32
31
λλ 4 5 =+⇒
/35 =⇒ λ
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46. The angle between the lines 2x = 3y = -z and 6x = - y = - 4z is
1) 00 2) 090 3) 045 4) 030
Ans.(2) ,32 zyx −==
13/12\1x
−==⇒ zy
−= 1,
31 ,
21)c,b ,a( 111
6x = -y = -4z
4/1161x
−=
−=⇒ zy
−−=
41,1,
61)c b ( 222a
212121 ccbbaa ++
41
31
121 +−=
= 0
47. If the plane 2ax - 3ay +4az + 6 = 0 passes through the midpoint of the line joining the centres of the spheres
132z8x6xzyx 222 =−−+++ and 82z4x01zyx 222 =−+−++ y then α equals
1) -1 2) 1 3) -2 4) 2
Ans.(3) Centre of )1 ,4 ,3(1 −=S
Centre )1 ,2 ,5(2 −=S
Mid pt. of 21SS (1, 1, 1)2ax - 3ay + 4az + 6 = 0passes through (1, 1, 1)2a- 3a + 4a = -6
6- 3a =⇒
2- a =⇒
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48. The distance between the line )k4j-i( k3j2-i2 r +++= λ and the plane 5 )kj5i( .r =++ is
1) 910
2) 3310
3) 103
4) 310
Ans.(2) 4k) j - (i 3k 2j - i 2 r +++= λ
5 k) 5j i ( . r =++
the line passes through (2, -2, 3)
∴ the reqd distance = 125153102
++−+−
2710=
3310=
49. For any vector a , the value 222 )ˆ()ˆ()i( kajaa ×+×+× is equal to
1) 23a 2) 2a
3) 22a 4) 24a
Ans.(3) Let zkyjxia ++=
izkyjxiia ×++=× )(ˆ
jzk y +−=
( ) 222ˆ zyia +=×
|| ly ( ) 222ˆ xzia +=×
( ) 222 yxka +=×∴ The required sum
)(2 222 zyx ++=
22a=
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50. If non-zero numbers a, b, c are in H.P., then the straight line 01 =++cb
yax
always passes through a fixed point.
That point is
1) (-1, 2) 2) (-1, -2)
3) (1, -2) 4)
−
21,1
Ans.(3) Let a, b and c be 1, 31 ,
21
01by =++
cax
0 3 2y x =++⇒
(1, -2) satisfies the equation
51. If a vertex of a triangle is (1, 1) and the mid points of two sides through this vertex are (-1, 2) and (3, 2), then thecentroid of the traingle is
1)
−
37,1
2)
−
37,
31
3)
37,1
4)
37,
31
Ans.(3)
B E(1, 3)
(3, 2)(-1, 2)
(1, 1)
C
FD
A
E = -1 + 3 -1, 2+ 2 -1= (1, 3)Centried of =∆ ABC centroid of DEF ∆
++++=
3322,
3311-
=
37,1
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52. If the circles 0222 =++++ acyaxyx and 01322 =−+−+ dyaxyx intersect in two distinct points P and Q thenthe line 5x + by - a = 0 passes through P and Q for
1) exactly one value of α2) no value of α3) infinitely many values of α4) exactly two values of α
Ans.(2) Radical axis is5a x + (c-d) y + (a+1) = 0 ....(1)5 x + by - a = 0 .......(2)
a-1a
55a +=
1a- 2 +=⇒ a
01a 2 =++⇒ aNo real value of a satisfies
53. A circle touches the x-axis and also touches the circle with centre at (0, 3) and radius 2. The locus of the centreof the circle is
1) an ellipse 2) a circle3) a hyperbola 4) a parabola
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Ans.(4)
(0,3)2
(h, k)
k
222 )()( kkyhx =−+−
4)3()0( 22 =−+− yx
[ ] kkhd += 2),(),3,0(
2)(k )3k(h 22 +=−+⇒
222 )2k()3k(h +=−+⇒
4k4kk69kh 222 ++=−++⇒
05k10h 2 =+−⇒
∴ locus of its centre is
05k10h2 =++
i.e., 05102 =++ yx
5y10x2 −−=
)12( 52 +−= yx
∴ The locus is a parabola
54. If a circle passes through the point (a, b) and cuts the circle 222 pyx =+ orthogonally, then the equation of thelocus of its centre is
1) ( ) 043 22222 =−++−−+ pbabyaxyx
2) ( ) 0- 22 222 =+−+ pbabyax
3) ( ) 03by-2ax- 22222 =−−+ pbayx
4) ( ) 0 22 222 =++−+ pbabyax
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Ans.(4) It the equation be
02222 =++++ cfygxyx
passes through (a, b)
02222 =++++ cbfagba ....(1)
It cuts 222 pyx =+ orthogonally2(g.0 + f.0) = c-p2
02 =−⇒ pc2pc =
Substituting the value of c in 1we get
022 222 =++++ pbfagba
0a2bf 2ag 222 =++++⇒ pblocus of its centre (-g, -f) is
022 222 =+++−− pbabfag
0)(a - 2bf 2ag 222 =+++⇒ pb
0)(22 222 =++−+ Pbabyax
55. An ellipse has OB as semi minor axis, F and F' its focil and the angle FBF' is a right angle. Then the eccentricityof the ellipse is
1) 21
2)21
3)41
4) 31
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Ans.(1)
I (a e,o)(-ae, o)
(0,1) B
F1F ae2
Slope of F B = aeb
Slope of aebBF
−=1
1aeb- −=×
aeb
222 eab =⇒
eaab 2222 −=222 aea2 =
21e2 =⇒
21e ==
56. Let a, b, and c be distinct non-negative numbers. If the vectors ki ,kcja ia +++ and kbjc ic ++ lie in a plane, then c is
1) the Geometric Mean of a and b2) the Arithmetic Mean of a and b3) equal to zero4) the Harmonic Mean of a and b
Ans.(1) 0 101 =bcc
caa
ab=⇒ 2c ⇒ c is the G.M. of b & a
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57. If c ,b ,a are non-coplanar vectors and λ is a real number then [ ] [ ]b c b a c b )b a( 2 +=+ λλλ for
1) exactly one value of λ2) no value of λ3) exactly three values of λ4) exactly two values of λ
Ans.(2)
]b)cb[(.a]cb[).baλ( 2 ×+=λ×λ+
]bcbb[.a)cb(.b)cb(.a[λ4 ×+×=×+×⇒
)bc(.a)]cb(.a[λ4 ×=×⇒
1λ4 −=⇒
no real value of λ is possible
58. Let kxjixbkia )1( , =++=−= and .)1( kyxjxiyc −+++= then ],,[ cba depends on
1) only y 2) only x3) both x and y 4) neither x nor y
Ans.4) ]c b [a
yxxyxx−+
−−
=1
11101
[ ] )y(x - )x1(x)yx1( 2 −−−−+=
yxxxyx +−+−−+= 221
= 1
59. Three houses are available in a locality. Three persons apply for the houses. Each applies for one house withoutconsulting others. The probability that all the three apply for the same house is
1) 92
2) 91
3) 98
4) 97
Ans.(2) Three person A, B, C& 3 houses a, b, c
91
31
31
33 =××
⇒
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60. A random variable X has Poisson distribution with mean 2. Then P(x > 1.5) equals
1) 22e
2) 0
3) 231
e− 4) 2
3e
Ans.(3)! r e )rx(P
rλ==λ−
2=λ
1)] P(x 0)p(x [ - 1)5.1x(P =+==>
p(x= 0) +p (x=1)
+=
−−
! 12e
! 02 e 1202
22
e3)21(e =+= −
2e3-1)5.1x(P =>
61. In the circuit, the galvanometer G shows zero deflection. If the batteries A and B have negligible internal resistance,the value of the resistor R will be -
G
A
2V
12V B R
Ω500
1) 100 Ω 2) 200 Ω 3) 1000Ω 4) 500Ω
Sol: 1)
ai
R2v
12 v
5ω
12 - 2 = ( )i ω 5
501 i =
RI
+===
ω512
51
ω512
0
5 ω + R = 6 ω = R = 100 Ω
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62. A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10 divisions per milliampere andvoltage sensitivity is 2 divisions per millivolt. In order that each division reads 1 volt, the resistance in ohms neededto be connected in series with the coil will be -
1) 105 2) 103 3) 9995 4) 99995Sol: 3)
Ω==⇒= 52
10 G Ger galvanomet of Resistance
ysensitivitCurrent ysensitivit Voltage
Here gi = Full scale deflection current = mA1510150 =
V = voltage to be measured = Ω=−× − 99955
1015150
3
63. Two sources of equal emf are connected to an external resistance R. The internal resistances of the two sourcesare R1 and R2 (R2 > R1). If the potential difference across the source having internal resistance R2 is zero, then
1) R = R2 - R1 2) R = R2 x (R1 + R2) / (R2 - R1)
3) R = R1 R2 / (R2 - R1) 4) R = R1 R2 / (R1 + R2)
Sol: 1)
I
2v
R1 R2
R
ε ε
21R R2 I
R++= ε
Potential difference across L cell
0 2 =−== RiV ε
0.22
21=
++− R
RRRεε
02 221 =−++ RRRR
21 RRR −+
12 R RR −=∴
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64. Two voltameters, one of copper and another of silver, are joined in parallel. When a total charge q flows throughthe voltameters, equal amount of metals are deposited. If the electrochemical equivalents of copper and silver arez1 and z2 respectively the charge which flows through the silver voltameter is
1)1
21zz
q
+ 2)2
11zz
q
+ 3)1
2
zzq 4)
2
1
zzq
Sol: 1)
).......(z q1 z
1
2
2
1 iqq
zzqm =⇒∝⇒=
)...(1
q 1qq q q
2
12
2
1
221 ii
qqqqalso
+=⇒+=⇒+=
From equation (1) and (ii) 1
22
1zz
qq+
=
65. Two concentric coils each of radius equal to 2π cm are placed at right angles to each other. 3 ampere and 4ampere are the currents flowing in each coil respectively. The magnetic induction in Weber/m2 at the centre of thecoils will be
( -70 10 x 4 πµ = Wb / A..m.)
1) 10-5 2) 12 x 10-5 3) 7 x 10-5 4) 5 x 10-5
Sol: 4)
(2)
(1)
ππ 43µ
)2(2i µ 010
1×==B
5.4µ 02
22
12 π=+= BBB
27 10510 ××= −
25 /nωb 105 −×=
66. In a potentiometer experiment the balancing with a cell is at length 240 cm. On shunting the cell with a resistanceof Ω2 , the balancing length becomes 120 cm. The internal resistance of the cell is
1) Ω 5.0 2) Ω 1 3) Ω 2 4) Ω 4Sol: 3)
2120
120240
2
21 ×−=×
−= Rl
llr
Ω= 2
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67. A charged particle of mass m and charge q travels on a circular path of radius r that is perpendicular to a magneticfield B. The time taken by the particle to complete one revolution is
1)m
B 2 2qπ 2)B
q 2 mπ3) qB
2 mπ4)
m 2 Bqπ
Sol: 3)
qvB 2=
rmv
qBm 2
v 2 ππ == rT
68. A magnetic needle is kept in a non-uniform magnetic field. It experiences
1) neither a force for a torque 2) a torque but not a force3) a force but not a torque 4) a force and a torque
Sol: 4)A magnetic needle kept in non-uniform magnetic field experience face and torque due to unequal forcesacting on poles
69. The resistance of hot tungsten filament is about 10 times the cold resistance. What will be the resistance of 100W and 200 V lamp when not in use ?
1) Ω 20 2) Ω 40 3) Ω 002 4) Ω 004
Sol: 2)
2
RVViP ==
Ω=×== 400100
2002002
PVR
Ω== 40 10
400 cold R
70. The self inductance of the motor of an electric fan is 10 H. In order to impart maximum power at 50 Hz, it shouldbe connected to a capacitance of
1) F 8 µ 2) F 4 µ 3) F 2 µ 4) F 1µ
Sol: 4)
2c 21
π== nP
c 50 2150
×=
π
C = 1 uf.
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71. An energy source will supply a constant current into the load if its internal resistance is
1) very large as compared to the load resistance 2) equal to the resistance of the load3) non-zero but less than the resistance of the load 4) zero
Sol: 4) ,rR
EI+
= Internal resistance (r) is
zero, ttanconsREI ==
72. The phase difference between the alternating current and emf is 2π
. Which of the following cannot be the
constituent of the circuit?
1) R, L 2) C alone 3) L alone 4) L, CSol: 4)73. A circuit has a resistance of 12 ohm and an impedance of 15 ohm. The power factor of the circuit will be -
1) 0.4 2) 0.8 3) 0.125 4) 1.25Sol:.2)
Power factor = 8.054
1512cos ====φ
ZR
74. A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If anelectron is projected along the direction of the fields with a certain velocity then -
1) its velocity will increase 2) its velocity will decrease3) it will turn towards left of direction of motion 4) it will turn towards right of direction of motion
Sol:.2)due to electricfield, it experiences force and accelerates i.e, its velocity decreases.
75. A coil of inductance 300 mH and resistance Ω2 is connected to a source of voltage 2 V. The current reaches halfof its steady state value in
1) 0.1 s 2) 0.05 s 3) 0.3 s 4) 0.15 sSol: 1)
−=
−−LIt
eii 10
)1(2 00 eii −=
t0.693RL =×
210300693.0
3−××=
= 0.1 see
