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June 16th, 2014, 04:25 PM
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Past year question papers of AIPMT Main Exam in PDF format

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1. If the dimensions of a physical quantity are given by MaLbTc, then the physical quantity will be
(1) Force if a = 0, b = –1, c = –2 (2) Pressure if a = 1, b = –1, c = –2
(3) Velocity if a = 1, b = 0, c = –1 (4) Acceleration if a = 1, b = 1, c = –2
Sol. Answer (2)
Pressure [P] = [M1L–1T–2] F
A
ϒ ⁄ ′ ∞ = ≤ ƒ
2. A particle starts its motion from rest under the action of a constant force. If the distance covered in first
10 seconds is S1 and that covered in the first 20 seconds is S2, then
(1) S2 = S1 (2) S2 = 2S1
(3) S2 = 3S1 (4) S2 = 4S1
Sol. Answer (4)
2
1
1
10 50
2
S= a = a
2
2
1
(20) 200
2
S = a = a ® S2 = 4S1
3. A bus is moving with a speed of 10 ms–1 on a straight road. A scooterist wishes to overtake the bus in
100 s. If the bus is at a distance of 1 km from the scooterist, with what speed should the scooterist chase
the bus?
(1) 10 ms–1 (2) 20 ms–1
(3) 40 ms–1 (4) 25 ms–1
Sol. Answer (2)
rel
rel
1000m
10m/s
time 100s
S
V = = =
® VS – VB = 10
® VS = VB + 10 = 20 m/s
4. The mass of a lift is 2000 kg. When the tension in the supporting cable is 28000 N, then its acceleration
is
(1) 14 ms–2 upwards (2) 30 ms–2 downwards
(3) 4 ms–2 upwards (4) 4 ms–2 downwards
Sol. Answer (3)
T = m(g + a)
– 28000 – 20000 2
4 m/s
2000
a T mg
m
= = = upwards
5. An explosion blows a rock into three parts. Two parts go off at right angles to each other. These two
are,
1 kg first part moving with a velocity of 12 ms–1 and 2 kg second part moving with a velocity of 8 ms–1. If
the third part flies off with a velocity of 4 ms–1, its mass would be
(1) 3 kg (2) 5 kg
(3) 7 kg (4) 17 kg
Sol. Answer (2)
P1+P2+P3=0
_ _ _ _
, = ( + )
_ _ _
P3 P1 P2
®
+
=
2 2
1 2
3
3
P P
m
v
=
+ = 144 256 20
4 4
= 5 kg.
6. A block of mass M is attached to the lower end of a vertical spring. The spring is hung from a ceiling
and
has force constant value k. The mass is released from rest with the spring initially unstretched. The
maximum
extension produced in the length of the spring will be
(1) Mg/2k (2) Mg/k
(3) 2 Mg/k (4) 4 Mg/k
Sol. Answer (3)
Loss of gravitational potential energy = Gain of spring potential energy
Mgx = 1 2
2
kx
= x 2Mg
k
7. Two bodies of mass 1 kg and 3 kg have position vectors iˆ+2jˆ+kˆ and –3iˆ– 2jˆ+kˆ, respectively. The
centre
of mass of this system has a position vector
(1) –iˆ+jˆ+kˆ (2) –2iˆ+2kˆ
(3) –2iˆ–jˆ+kˆ (4) 2iˆ– jˆ–2kˆ
Sol. Answer (3)
1 1 2 2
1 2
cm
m r m r
r
m m
+
=
+
_ _
_
=
ˆ 2ˆ ˆ–9ˆ–6ˆ 3ˆ
4
i+ j+k i j+ k ˆ ˆ ˆ =–2i –j+k
(3)
8. Four identical thin rods each of mass M and length l, form a square frame. Moment of inertia of this
frame
about an axis through the centre of the square and perpendicular to its plane is
(1) 1 2
3
Ml (2) 4 2
3
Ml
(3) 2 2
3
Ml (4) 13 2
3
Ml
Sol. Answer (2)
A B
D C
z L/2
Iz = 4 × IAB
2 2 4 2
4
12 4 3
ML ML ML
. + =

9. A thin circular ring of mass M and radius R is rotating in a horizontal plane about an axis vertical to its
plane
with a constant angular velocity ⎤. If two objects each of mass m be attached gently to the opposite ends
of
a diameter of the ring, the ring will then rotate with an angular velocity
(1)
M
M m

+
(2)
( –2 )
2
M m
M m

+
(3)
2
M
M m

+
(4)
(M 2m)
M
⎤ +
Sol. Answer (3)
Conservation of angular momentum
MR2⎤ = (M + 2m)R2⎤2
®
2
M
M m
⎤2 = ⎤
+
10. A body, under the action of a force =( + )
_ ˆ ˆ ˆ F 6i–8j 10kN, acquires an acceleration of 1 m/s2. The mass of
this body must be
(1) 10 2 kg (2) 2 10kg
(3) 10 kg (4) 20 kg
Sol. Answer (1)
= = + + =
_
_
| | 62 82 102
10 2 kg.
| | 1
m F
a
(4)
11. If F
_
is the force acting on a particle having position vector r
_ and ⎮
_
be the torque of this force about the
origin, then
(1) r ⊕ ⎮ = 0
_ _ and F ⊕ ⎮⎯0
_ _ (2) 0 r ⊕ ⎮ ⎯
_ _ and 0 F ⊕ ⎮=
_ _
(3) r ⊕ ⎮ > 0
_ _ and F ⊕ ⎮<0
_ _ (4) 0 r ⊕ ⎮ =
_ _ and 0 F ⊕ ⎮=
_ _
Sol. Answer (4)
⎮ =r.F
_ _ _
4 r ∞ ⎮
_ _ and F ∞ ⎮
_ _
12. The figure shows elliptical orbit of a planet m about the sun S. The shaded area SCD is twice the
shaded
area SAB. If t1 is the time for the planet to move from C to D and t2 is the time to move from A to B then
C
D
B
A
m v
S
(1) t1 = t2 (2) t1 > t2
(3) t1 = 4t2 (4) t1 = 2t2
Sol. Answer (4)
Areal velocity is constant
AreaABS = AreaCDS
® t1= 2t2
13. An engine pumps water continuously through a hose. Water leaves the hose with a velocity v and m
is the
mass per unit length of the water jet. What is the rate at which kinetic energy is imparted to water?
(1)
1 2 2
2
m v (2) 1 3
2
mv
(3) mv3 (4) 1 2
2
mv
Sol. Answer (2)
dK
dt = =
2
1 3
2 2
v dmdl mv
dt dl
14. A body of mass 1 kg is thrown upwards with a velocity 20 m/s. It momentarily comes to rest after
attaining
a height of 18 m. How much energy is lost due to air friction? (g = 10 m/s2)
(1) 10 J (2) 20 J
(3) 30 J (4) 40 J
Sol. Answer (2)
Loss of energy = Initial energy – final energy
1 2

2
= mv mgh
1 2
20 – 1 10 18
2
= . . .
= 200 – 180 = 20 J
(5)
15. The two ends of a rod of length L and a uniform cross-sectional area A are kept at two temperatures
T1 and
T2(T1 > T2). The rate of heat transfer,
dQ
dt , through the rod in a steady state is given by
(1) dQ kA(T1–T2)
dt L
= (2) dQ kL(T1–T2)
dt A
=
(3) dQ k(T1–T2)
dt LA
= (4) (1– 2) dQ kLA T T
dt
=
Sol. Answer (1)
Law of conduction = 1 2 dQ kA(T –T )
dt L
16. In thermodynamic processes which of the following statements is not true?
(1) In an adiabatic process PV© = constant
(2) In an adiabatic process the system is insulated from the surroundings
(3) In an isochoric process pressure remains constant
(4) In an isothermal process the temperature remains constant
Sol. Answer (3)
During isochoric process, volume is constant.
17. A black body at 227ºC radiates heat at the rate of 7cals/cm2s. At a temperature of 727ºC, the rate of
heat
radiated in the same units will be
(1) 80 (2) 60
(3) 50 (4) 112
Sol. Answer (4)
E = ⌠T4
®
4
2 2
1 1
E T
E T

=

®
4
2
727 273
7
227 273
E + = . +
cal/cm2s
= 112 cal/cm2s
18. The internal energy change in a system that has absorbed 2 kcals of heat and done 500 J of work is
(1) 7900 J (2) 8900 J
(3) 6400 J (4) 5400 J
Sol. Answer (1)
Q = ⊗U + W
⊗U = Q – W = 2 × 1000 × 4.2 – 500
= 8400 – 500 = 7900 J
19. The driver of a car travelling with speed 30 m/sec towards a hill sounds a horn of frequency 600 Hz. If
the
velocity of sound in air is 330 m/s, the frequency of reflected sound as heard by driver is
(1) 500 Hz (2) 550 Hz
(3) 555.5 Hz (4) 720 Hz
(6)
Sol. Answer (4)
n2 = 0 330 30
600
– s 330 – 30
v v
n
v v
+ + = .

= 720 Hz
20. A simple pendulum performs simple harmonic motion about x = 0 with an amplitude a and time period
T. The
speed of the pendulum at x = a/2 will be
(1)
a 3
T
(2)
3
2
a
T
(3)
a
T
(4)
3 2a
T
Sol. Answer (1)
v = ⎤a2 y2
=
2
2 2 3
4
a a a
T T
=
21. Which one of the following equations of motion represents simple harmonic motion?
(1) Acceleration = kx (2) Acceleration = k0x + k1x2
(3) Acceleration = –k(x + a) (4) Acceleration = k(x + a)
where k, k0, k1 and a are all positive
Sol. Answer (3)
In simple harmonic motion acceleration is directly proportional to displacement and is opposite to
displacement.
22. The electric field part of an electromagnetic wave in a medium is represented by Ex = 0;
ϒ ⁄ = ′ . . ∞
≤ ƒ
6rad 2rad
2.5 cos 2 10 10
y s m
E N t x
C , Ez = 0. The wave is
(1) Moving along –x direction with frequency 106 Hz and wavelength 200 m
(2) Moving along y direction with frequency 2 × 106 Hz and wavelength 200 m
(3) Moving along x direction with frequency 106 Hz and wavelength 100 m
(4) Moving along x direction with frequency 106 Hz and wavelength 200 m
Sol. Answer (4)
⎤ = 2 × 106 rad/s
k = × 10–2 rad/m
speed c = 2 108
k
⎤= . m/s along +ve x-axis

⎣ = = =
. 2
2 2
200 m
k 10
f = 106 Hz
2
⎤ =

(7)
23. A wave in a string has an amplitude of 2 cm. The wave travels in the +ve direction of x axis with a
speed of
128 m/s and it is noted that 5 complete waves fit in 4 m length of the string. The equation describing the
wave
is
(1) y = (0.02) m sin(7.85x – 1005t)
(2) y = (0.02) m sin(7.85x + 1005t)
(3) y = (0.02) m sin(15.7x – 2010t)
(4) y = (0.02) m sin(15.7x + 2010t)
Sol. Answer (1)
Amplitude A = 0.02 m
wavelength,
4
0.8 m
5
⎣ = =
2
k 7.85 rad/m = =

⎤ = vk = 7.85 × 128 rad/s
= 1004.8 = 1005 rad/s
4 equation is y = (0.02) m sin(7.85x – 1005t)
24. Each of the two strings of length 51.6 cm and 49.1 cm are tensioned separately by 20 N force. Mass
per
unit length of both the strings is same and equal to 1 g/m. When both the strings vibrate simultaneously
the
number of beats is
(1) 3 (2) 5
(3) 7 (4) 8
Sol. Answer (3)
Frequency
1
2
f T
L
=

® 2 1
2 1
1 1 1
2
f f T
L L
ϒ ⁄
= . ′ ∞ ⎧ ≤ ƒ
= 3
20 1 1 1
10 2 49.1 51.6
ϒ ⁄ .′ ∞ ≤ ƒ
=
1 51.6 49.1
20000
2 51.6 49.1
. ϒ ⁄ ′≤ . ∞ƒ
= 6.97 = 7
25. Three capacitors each of capacitance C and of breakdown voltage V are joined in series. The
capacitance and
breakdown voltage of the combination will be
(1) 3C, 3V (2) ,
3 3
C V
(3) 3 ,
3
C V (4) , 3
3
C V
Sol. Answer (4)
,
eq 3
C = C V2 = V + V + V = 3V
(8)
26. A wire of resistance 12 ohms per meter is bent to form a complete circle of radius 10 cm. The
resistance
between its two diametrically opposite points, A and B as shown in the figure, is
A B
(1) 6 ∧ (2) 0.6 ∧
(3) 3 ∧ (4) 6 ∧
Sol. Answer (2)
A B
C
D
RACB = 1.2 ∧
RADB = 1.2 ∧
RAB = 0.6 ∧
27. A bar magnet having a magnetic moment of 2 × 104 JT–1 is free to rotate in a horizontal plane. A
horizontal
magnetic field B = 6 × 10–4 T exists in the space. The work done in taking the magnet slowly from a
direction
parallel to the field to a direction 60° from the field is
(1) 2 J (2) 0.6 J
(3) 12 J (4) 6 J
Sol. Answer (4)
W = MB [cos⎝1 – cos⎝2]
= 2 × 104 × 6 × 10–4 [cos0 – cos60]
=
1
12 6 J
2
. =
28. The magnetic force acting on a charged particle of charge –2 ⎧C in a magnetic field of 2T acting in y
direction,
when the particle velocity is (2iˆ+3jˆ).106ms 1, is
(1) 8 N in z direction (2) 8 N in –z direction
(3) 4 N in z direction (4) 8 N in y direction
Sol. Answer (2)
F
__
= q(v .B)
__ __
= 2.10 6[(2iˆ+3jˆ)106ms 1].2jˆT
= 8kˆ ®8N in –ve z direction
29. A conducting circular loop is placed in a uniform magnetic field 0.04 T with its plane perpendicular to
the
magnetic field. The radius of the loop starts shrinking at 2 mm/s. The induced emf in the loop when the
radius
is 2 cm is
(1) 1.6 ⎧V (2) 3.2 ⎧V
(3) 4.8 ⎧V (4) 0.8 ⎧V
(9)
Sol. Answer (2)
⎞ = B r2cos0 = B r 2
| | d
dt
∑ = ⎞ = 2 B rdr
dt

= 0.04 × × 2 × 2 × 10–2 × 2 × 10–3 V
= 3.2 ⎧V
30. The electric potential at a point (x, y, z) is given by V = –x2y – xz3 + 4.
The electric field
__
E at that point os
(1) E=iˆ(2xy z3)+jˆxy2+kˆ3z2x
__
(2) =ˆ(2 +3)+ˆ2+ˆ3 2
__
E i xy z jx k xz
(3) =ˆ2 +ˆ(2+ 2)+ˆ(3 2)
__
E i xy j x y k xz y
(4) =ˆ3+ˆ +ˆ2
__
E iz jxyz kz
Sol. Answer (2)
E
__
= ˆ ˆ ˆ dVi dVj dVk
dx dy dt

= iˆ(2xy+z3)+jˆx2+kˆ 3xz2
31. See the electrical circuit shown in this figure. Which of the following equations is a correct equation for
it?
R
r1
r2
i2
i1 ∑1
∑2
(1) ∑1 – (i1 + i2) R + i1r1 = 0 (2) ∑1 – (i1 + i2) R – i1r1 = 0
(3) ∑2 – i2 r2 – ∑1 – i1r1 = 0 (4) –∑2 – (i1 + i2) R + i2r2 = 0
Sol. Answer (2)
R
r1
r2
i2
i1
∑1
∑2
i1 + i2
A
E
D
B
F
C
KVL to the ABCD
∑1 – i1r1 – R (i1 + i2) = 0
® ∑1 – (i1 + i2) R1 – i1 r1 = 0
(10)
32. A galvanometer having a coil resistance of 60 ∧ shows full scale deflection when a current of 1.0 amp
passes
through it. It can be converted into an ammeter to read currents upto 5.0 amp by
(1) Putting in parallel a resistance of 15 ∧
(2) Putting in parallel a resistance of 240 ∧
(3) Putting in series a resistance of 15 ∧
(4) Putting in series a resistance of 240 ∧
Sol. Answer (1)
Ammeter
60
15
1 5 1
S G
n
= = = ∧

33. Under the influence of a uniform magnetic field, a charged particle moves with constant speed V in a
circle
of a radius R. The time period of rotation of the particle
(1) Depends on both V and R
(2) Depends on V and not on R
(3) Depends on R and not on V
(4) Is independent of both V and R
Sol. Answer (4)
T 2r 2 mv 2m
v v qB qB
= = =

Independent of both V and R only depends on specific charge.
34. Power dissipated in an LCR series circuit connected to an a.c. source of emf ∑ is
(1)

+ ⎤ ⎤
2
2
2 1
R
R L
C
(2)

ϒ ⁄ ′ + ⎤ ∞ ′≤ ⎤ ∞ƒ
2
2
2 1
R
R L
C
(3)
∑ + ⎤ ⎤
2
2 2 1 R L
C
R (4)
ϒ ⁄ ∑′ + ⎤ ∞ ′≤ ⎤ ∞ƒ
2
2 2 1 R L
C
R
Sol. Answer (2)
=∑ = ∑
+ ⎤ ⎤
2 2
2 2
2 1
av
P R R
Z
R L
C
35. Three concentric spherical shells have radii a, b and c (a < b < c) and have surface charge densities
⌠, –⌠
and ⌠ respectively. If VA, VB and VC denote the potentials of the three shells, then, for c = a + b, we have
(1) VC = VB = VA (2) VC = VA ⎯ VB
(3) VC = VB ⎯ VA (4) VC ⎯ VB ⎯ VA
(11)
Sol. Answer (2)
VA =
0 0 0 0 0
2
a b c (a b c) a ⌠ ⌠ +⌠ =⌠ + = ⌠
∑ ∑ ∑ ∑ ∑
VB =
2
0
a b c
b
⌠ ϒ ⁄ ′ + ∞
∑ ′≤ ∞ƒ
=
2
0 0
a b a b a(a b)
b b
⌠ϒ ⁄ ⌠ + ′ + + ∞=
∑′≤ ∞ƒ ∑
VC =
2 2
0
a b c
c c
⌠ ϒ ⁄ ′ + ∞
∑ ′≤ ∞ƒ
=
2 2 2
0
a b c
c
⌠ϒ + ⁄
′ ∞
∑ ′≤ ∞ƒ
=
2
0
(a b)(a b) (a b)
a b
⌠ϒ + + + ⁄
′ ∞
∑′≤ + ∞ƒ
=
0 0
2
[( ) ] a b a b a ⌠ + + =⌠
∑ ∑
36. A student measures the terminal potential difference (V) of a cell (of emf ∑ and internal resistance r)
as a
function of the current (I ) flowing through it. The slope and intercept of the graph between V and I, then,
respectively, equal
(1) –∑ and r (2) ∑ and –r
(3) –r and ∑ (4) r and – ∑
Sol. Answer (3)
V = E – Ir
V
E
I
y = mx + C (x = I, y = V)
Slope m = –r, intercept C = ∑
37. A rectangular, a square, a circular and an elliptical loop, all in the (x – y) plane, are moving out of a
uniform
magnetic field with a constant velocity = ˆ.
__
V vi The magnetic field is directed along the negative z axis direction.
The induced emf, during the passage of these loops, out of the field region, will not remain constant for
(1) Any of the four loops (2) The rectangular, circular and elliptical loops
(3) The circular and the elliptical loops (4) Only the elliptical loop
Sol. Answer (3)
In square loop effective length inside magnetic field remains constant.
(12)
38. If a diamagnetic substance is brought near the north or the south pole of a bar magnet, it is
(1) Attracted by both the poles
(2) Repelled by both the poles
(3) Repelled by the north pole and attracted by the south pole
(4) Attracted by the north pole and repelled by the south pole
Sol. Answer (2)
Diamagnetic substance always repeled by magnet.
39. The number of photo electrons emitted for light of a frequency ⎨ (higher than the threshold frequency
⎨0) is
proportional to
(1) Frequency of light (⎨) (2) ⎨ – ⎨0
(3) Threshold frequency (⎨0) (4) Intensity of light
Sol. Answer (4)
Photoelectric current intensity.
40. Monochromatic light of wavelength 667 nm is produced by a helium neon laser. The power emitted is
9 mW. The number of photons arriving per sec. on the average at a target irradiated by this beam is
(1) 3 × 1019 (2) 9 × 1017
(3) 3 × 1016 (4) 3 × 1015
Sol. Answer (3)
–3 –9
16
–34 8
9 10 667 10
3 10
6.63 10 3 10
n p
hc
= = . . . = .
. . . ⎣
41. The figure shows a plot of photo current versus anode potential for a photosensitive surface for three
different
radiations. Which one of the following is a correct statement?
Retarding potential Anode potential
a
b
c
Photo current
(1) Curves (b) and (c) represent incident radiations of same frequency having same intensity
(2) Curves (a) and (b) represent incident radiations of different frequencies and different intensities
(3) Curves (a) and (b) represent incident radiations of same frequency but of different intensities
(4) Curves (b) and (c) represent incident radiations of different frequencies and different intensities
Sol. Answer (3)
Current intensity and Einstein photoelectric equation eV0 = h⎨ – ⎞.
42. The number of beta particles emitted by a radioactive substance is twice the number of alpha
particles emitted
by it. The resulting daughter is an
(1) Isotope of parent (2) Isobar of parent
(3) Isomer of parent (4) Isotone of parent
Sol. Answer (1)
Atomic number remains unchanged.
(13)
43. The ionization energy of the electron in the hydrogen atom in its ground state is 13.6 eV. The atoms
are excited
to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength of emitted radiation
corresponds to the transition between
(1) n = 4 to n = 3 states (2) n = 3 to n = 2 states
(3) n = 3 to n = 1 states (4) n = 2 to n = 1 states
Sol. Answer (1)
( –1)
6 4
2
n n = ® n =
Maximum wavelength corresponds to lowest energy.
44. In a Rutherford scattering experiment when a projectile of charge z1 and mass M1 approaches a target
nucleus
of charge z2 and mass M2, the distance of closest approach is r0. The energy of the projectile is
(1) Directly proportional of mass M1 (2) Directly proportional of M1 × M2
(3) Directly proportional of z1z2 (4) Inversely proportional to z1
Sol. Answer (3)
2 1 2
0 0
1
2 4
q q
mV
r
=

® r0 q1q2
® r0 z1z2
45. In the nuclear decay given below
A A A–4 A–4
ZX Z+1Y Z–1B Z–1B
The particles emitted in the sequence are
(1) 〈, ®, © (2) ®, 〈, ©
(3) ©, ®, 〈 (4) ®, ©, 〈
Sol. Answer (2)
– A A A–4 A–4
ZX Z1Y Z–1B Z–1B ® 〈 ©
↓↓↓ + ↓↓ ↓↓
46. The mean free path of electrons in a metal is 4 × 10–8 m. The electric field which can give on an
average
2 eV energy to an electron in the metal will be in units of V/m
(1) 5 × 107 (2) 8 × 107
(3) 5 × 10–11 (4) 8 × 10–11
Sol. Answer (1)
W = F.S
® W = qE.S
® 2 eV = eE × 4 × 10–8 m
® 7 –1
–8
2
5 10 Vm
4 10
E V
m
= = .
.
(14)
47. Sodium has body centred packing. Distance between two nearest atoms is 3.7 Å. The lattice
parameter is
(1) 8.6 Å (2) 6.8 Å
(3) 4.3 Å (4) 3.0 Å
Sol. Answer (3)
In bcc
4r= 3a
® 2(d) = 3a
® 2 2 3.7Å
4.3 Å
3 1.73
a d = = . =
48. A p-n phot2odiode is fabricated from a semiconductor with a band gap of 2.5 eV. It can detect a signal
of
wavelength
(1) 4000 Å (2) 6000 Å
(3) 4000 nm (4) 6000 nm
Sol. Answer (1)
Maximum wavelength
124000eV Å
4960 Å
2.5 eV
hc
E
⎣ = = =
49. The symbolic representation of four logic gates are given below
(i) (ii)
(iii) (iv)
The logic symbols for OR, NOT and NAND gates are respectively
(1) (i), (iii), (iv) (2) (iii), (iv), (ii)
(3) (iv), (i), (iii) (4) (iv), (ii), (i)
Sol. Answer (4)
50. A transistor is operated in common-emitter configuration of VC = 2 V such that a change in the base
current
from 100 ⎧A to 200 ⎧A produces a change in the collector current from 5 mA to 10 mA. The current gain is
(1) 50 (2) 75
(3) 100 (4) 150
Sol. Answer (1)

® = = =
⊗ ⎧
(10 – 5)
50
(200 – 100)
C
B
I mA
I A

Last edited by Neelurk; March 11th, 2020 at 09:41 AM.
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Re: Past year question papers of AIPMT Main Exam in PDF format

51. 10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water
produced in
this reaction will be
(1) 1 mol (2) 2 mol
(3) 3 mol (4) 4 mol
(15)
Sol. Answer (4)
2 2 2
5 2 4mol
2H +O 2H O
Here, limiting reagent is O2
52. Oxidation numbers of P in PO4
3–, of S in SO4
2– and that of Cr in Cr2O7
2– are respectively
(1) –3, +6 and +6 (2) +5, +6 and +6
(3) +3, +6 and +5 (4) +5, +3 and +6
Sol. Answer (2)
+5, +6 and +6
53. Maximum number of electrons in a subshell of an atom is determined by the following
(1) 2n2 (2) 4l + 2
(3) 2l + 1 (4) 4l – 2
Sol. Answer (2)
4l + 2
54. Which of the following is not permissible arrangement of electrons in an atom?
(1) n = 3, l = 2, m = –2, s = –1/2 (2) n = 4, l = 0, m = 0, s = –1/2
(3) n = 5, l = 3, m = 0, s = +1/2 (4) n = 3, l = 2, m = –3, s = –1/2
Sol. Answer (4)
The value of ‘m’ varies from –1, 0, +1
55. From the following bond energies
H–H bond energy : 431.37 kJ mol–1
C=C bond energy : 606.10 kJ mol–1
C–C bond energy : 336.49 kJ mol–1
C–H bond energy : 410.50 kJ mol–1
Enthalpy for the reaction,
C = C + H – H H – C – C – H
H
H
H
H
H
H
H
H
will be
(1) 553.0 kJ mol–1 (2) 1523.6 kJ mol–1
(3) –243.6 kJ mol–1 (4) –120.0 kJ mol–1
Sol. Answer (4)
⊗H = (HC=C + HH–H) – (2 × HC–H + 1 × HC–C)
= (606.1 + 431.37) – (2 × 410.5 + 1 × 336.49)
= 1037.47 – 1157.49
= –120.02 kJ mol–1
Η –120.0 kJ mol–1
(16)
56. The ionization constant of ammonium hydroxide is 1.77 × 10–5 at 298 K. Hydrolysis constant of
ammonium
chloride is
(1) 5.65 × 10–12 (2) 5.65 × 10–10
(3) 6.50 × 10–12 (4) 5.65 × 10–13
Sol. Answer (2)
–14
w –10
H –5
b
K 10
K 5.65 10
K 1.77 10
= = = .
.
57. Given
(i) Cu2+ + 2e– Cu, Eo = 0.337 V
(ii) Cu2+ + e– Cu+, Eo = 0.153 V
Electrode potential, Eo for the reaction, Cu+ + e– Cu, will be
(1) 0.38 V (2) 0.52 V
(3) 0.90 V (4) 0.30 V
Sol. Answer (2)
2 – o o
1 1
2 – o o
2 2
– o o
3 3
Cu 2e Cu, E 0.337 V, G – 2 F 0.337 0.674 F
Cu Cu e , E –0.153 V, G –1 F – 0.153 0.153 F
Cu e Cu, E ?, G – 0.521F
+
+ +
+
+ = ⊗ = . . =
+ = ⊗ = . . =+
+ = ⊗ =
o
o 3
3
G –0.521F
E 0.521 V
–nF –1 F

= = =
.
58. What is the [OH–] in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of
0.10 M
Ba(OH)2?
(1) 0.12 M (2) 0.10 M
(3) 0.40 M (4) 0.0050 M
Sol. Answer (2)
20 ml . 0.5 N HCl
30 ml . 2 N Ba(OH)2
=
=
1 meq
6 meq
5 meq
5 = 50 × N
N = 0.1 M
59. The energy absorbed by each molecule (A2) of a substance is 4.4 × 10–19 J and bond energy per
molecule
is 4.0 × 10–19 J. The kinetic energy of the molecule per atom will be
(1) 4.0 × 10–20 J (2) 2.0 × 10–20 J
(3) 2.2 × 10–19 J (4) 2.0 × 10–19 J
Sol. Answer (2)
Ek per atom = Eabsorbed–EB.E.
2
=
4.4 10–19 –4 10–19
2
. .
= 2 × 10–20
(17)
60. For the reaction, N2 + 3H2 2NH3, if 3 4 d[NH ]
2 10
dt
= . mol L–1 s–1, the value of d[H2 ]
dt

would be:
(1) 1 × 10–4 mol L–1 s–1 (2) 3 × 10–4 mol L–1 s–1
(3) 4 × 10–4 mol L–1 s–1 (4) 6 × 10–4 mol L–1 s–1
Sol. Answer (2)
2 NH3 1dH 1d
3 dt 2 dt
=+
H2 4 4 d 3
2 10 3 10
dt 2
= . . = .
61. For the reaction A + B products, it is observed that:
(a) On doubling the initial concentration of A only, the rate of reaction is also doubled and
(b) On doubling the initial concentrations of both A and B, there is a change by a factor of 8 in the rate of
the reaction
The rate of this reaction is given by:
(1) rate = k [A] [B] (2) rate = k [A]2 [B]
(3) rate = k [A] [B]2 (4) rate = k [A]2 [B]2
Sol. Answer (3)
Rate [A]x
(2)1 (2)x
Rate [A]x[B]y
8 (2)1 (2)y
y = 2
Hence Rate = K[A] [B]2
62. The equivalent conductance of
M
32
solution of a weak monobasic acid is 8.0 mhos cm2 and at infinite dilution is
400 mhos cm2. The dissociation constant of this acid is
(1) 1.25 × 10–4 (2) 1.25 × 10–5
(3) 1.25 × 10–6 (4) 6.25 × 10–4
Sol. Answer (2)
8 1
0.02
400 50
〈 = = =
21 1 4 5
k C. 0.02 0.02 10 1.25 10
32 8
= 〈 = . . = . = .
63. A 0.0020 m aqueous solution of an ionic compound Co(NH3)5 (NO2) Cl freezes at –0.00732°C.
Number of moles
of ions which 1 mol of ionic compound produces on being dissolved in water will be (kf = – 1.86°C/m)
(1) 1 (2) 2
(3) 3 (4) 4
(18)
Sol. Answer (2)
0.00732 = i × 1.86 × 0.002
0.00732
i 2
1.860.002
= _
64. In the reaction
BrO3(aq) 5Br (aq) 6H 3Br2(l) 3H2O(l) + + + +
The rate of appearance of bromine (Br2) is related to rate of disappearance of bromide ions as following :
(1) 2 d(Br ) 3 d(Br )
dt 5 dt

= (2) d(Br2 ) 3 d(Br )
dt 5 dt

=
(3) 2 d(Br ) 5 d(Br )
dt 3 dt

= (4) d(Br2 ) 5 d(Br )
dt 3 dt

=
Sol. Answer (2)
(Br2 ) [Br ] 1d 1d
3 dt 5 dt

=
(Br2 ) [Br ] d 3 d
dt 5 dt

=
65. Lithium metal crystallises in a body centred cubic crystal. If the length of the side of the unit cell of
lithium
is 351 pm, the atomic radius of the lithium will be:
(1) 300.5 pm (2) 240.8 pm
(3) 151.8 pm (4) 75.5 pm
Sol. Answer (3)
4r = a 3
a 3 351 1.732
r
4 4
= = . = 151.983 pm
66. The dissociation constants for acetic acid and HCN at 25°C are 1.5 × 10–5 and 4.5 × 10–10,
respectively. The
equilibrium constant for the equilibrium
CN CH3COOH HCN CH3COO + ________ +
would be:
(1) 3.0 × 104 (2) 3.0 × 105
(3) 3.0 × 10–5 (4) 3.0 × 10–4
Sol. Answer (1)
– +
CH3COOH ________ CH3COO + H ...(i)
HCN ________ H++ CN– ...(ii)
(19)
For – –
CN + CH3COOH ________ HCN + CH3COO — K
K =
1
2
K
K =
–5
–10
1.5 10
4.5 10
.
. = 1 5
10
3
. = 3 × 104 (approx)
67. The values of ⊗H and ⊗S for the reaction, C(graphite)+CO2 (g) 2CO(g) are 170 kJ and 170 JK–1,
respectively.
This reaction will be spontaneous at:
(1) 510 K (2) 710 K
(3) 910 K (4) 1110 K
Sol. Answer (4)
⊗G = ⊗H – T. ⊗S
0 = 170 × 103 – T × 170
T =
170 103
170
.
= 1000 K (at eq.)
For spontaneous, T must be greater than 1000 K i.e. 1110 K
68. Half life period of a first-order reaction is 1386 seconds. The specific rate constant of the reaction is:
(1) 5.0 × 10–2 s–1 (2) 5.0 × 10–3 s–1
(3) 0.5 × 10–2 s–1 (4) 0.5 × 10–3 s–1
Sol. Answer (4)
K =
0.693
1386
= 0.5 × 10–3 s–1
69. In which of the following molecules/ions BF3, NO2,NH2 and H2O, the central atom is sp2 hybridized?
(1) BF3 and NO2 (2) NO2 and NH2
(3) NH2 and H2O (4) NO2 and H2O
Sol. Answer (1)
BF3 & –
NO2
70. Among the following which is the strongest oxidising agent?
(1) Cl2 (2) F2
(3) Br2 (4) I2
Sol. Answer (2)
F2 (since its °
ERed is highest)
71. According to MO theory which of the following lists ranks the nitrogen species in terms of increasing
bond
order?
(1) 2
N2 N2 N2 < < (2) 2
N2 N2 N2 < <
(3) 2
N2 N2 N2 < < (4) 2
N2 N2 N2 < <
(20)
Sol. Answer (3)
2– –
N2 <N2<N2
B.O. (2) < (2.5) < (3)
72. In the case of alkali metals, the covalent character decreases in the order
(1) MI > MBr > MCl > MF (2) MCI > MI > MBr > MF
(3) MF > MCl > MBr > MI (4) MF > MCl > MI > MBr
Sol. Answer (1)
MI > MBr > MCl > MF
73. Which of the following oxides is not expected to react with sodium hydroxide?
(1) BeO (2) B2O3
(3) CaO (4) SiO2
Sol. Answer (3)
CaO doesn't react with NaOH due to basic nature remaining are acidic. BeO is amphoteric.
74. Al2O3 is reduced by electrolysis at low potentials and high currents. If 4.0 × 104 amperes of current is
passed
through molten Al2O3 for 6 hours, what mass of aluminium is produced? (Assume 100% current
efficiency, At.
mass of Al = 27 g mol–1)
(1) 1.3 × 104 g (2) 9.0 × 103 g
(3) 8.1 × 104 g (4) 2.4 × 105 g
Sol. Answer (3)
w =
E
× i×t
96500
= 9 4
4 10 6 3600
96500
. . . . = 8.05 × 104 g
75. The stability of +1 oxidation state increases in the sequence:
(1) Ga < In < Al < Tl (2) Al < Ga < In < Tl
(3) Tl < In < Ga < Al (4) In < Tl < Ga < Al
Sol. Answer (2)
Al < Ga < In < Tl
On moving down in 13 group the stability of oxidation state +1, increases due to increase of inert pair
effect.
76. Copper crystallises in a face-centred cubic lattice with a unit cell length of 361 pm. What is the radius
of copper
atom in pm?
(1) 108 (2) 128
(3) 157 (4) 181
Sol. Answer (2)
For FCC
4r = a 2
a 361
r 127.65 128 pm
2 2 2 1.414
= = =
.
_
(21)
77. What is the dominant intermolecular force or bond that must be overcome in converting liquid CH3OH
to a gas?
(1) London dispersion force (2) Hydrogen bonding
(3) Dipole-dipole interaction (4) Covalent bonds
Sol. Answer (2)
Hydrogen bonding
78. Which of the following complex ions is expected to absorb visible light?
(1) [Zn (NH3)6]2+ (2) [Sc (H2O)3(NH3)3]3+
(3) [Ti (en)2(NH3)2]4+ (4) [Cr (NH3)6]3+
(At. number Zn = 30, Sc = 21, Ti = 22, Cr = 24)
Sol. Answer (4)
[Cr (NH3)6]3+ (due to presence of unpaired 'd' electron d-d transition is possible)
79. Out of 2 3
TiF6 , COF6 Cu2Cl2 and 2
NiCl4 (Z of Ti = 22, CO = 27, Cu = 29, Ni = 28) the colourless species
are:
(1) 3
COF6 and 2
NiCl4 (2) 2
TiF6 and 3
COF6
(3) Cu2Cl2 and 2
NiCl4 (4) 2
TiF6 and Cu2Cl2
Sol. Answer (4)
2
TiF6 and Cu2Cl2 (Due to absence of unpaired e–)
80. Which of the following does not show optical isomerism?
(1) [CO (en)3]3+ (2) [CO (en)2Cl2]+
(3) [CO (NH3)3Cl3]0 (4) [CO(en)Cl2(NH3)2]+
(en = ethylenediamine)
Sol. Answer (3)
[Co(NH3)3Cl3]0
81. Which one of the elements with the following outer orbital configurations may exhibit the largest
number of
oxidations states?
(1) 3d24s2 (2) 3d34s2
(3) 3d54s1 (4) 3d54s2
Sol. Answer (4)
3d54s2
82. Which of the following molecules acts as a Lewis acid?
(1) (CH3)3 N (2) (CH3)3 B
(3) (CH3)2 O (4) (CH3)3 P
Sol. Answer (2)
(CH3)3 B
(22)
83. Amongst the elements with following electronic configurations, which one of them may have the
highest
ionization energy?
(1) Ne[3s23p1] (2) Ne[3s23p3]
(3) Ne[3s23p2] (4) Ar[3d104s24p3]
Sol. Answer (2)
3p3-configuration is more stable than 4p3-configuration.
84. The straight chain polymer is formed by
(1) Hydrolysis of (CH3)2 SiCl2 followed by condensation polymerisation
(2) Hydrolysis of (CH3)3 SiCl followed by condensation polymerisation
(3) Hydrolysis of CH3 SiCl3 followed by condensation polymerisation
(4) Hydrolysis of (CH3)4 Si by addition polymerisation
Sol. Answer (1)
Silicone formation.
85. The IUPAC name of the compound having the formula CH α C—CH = CH2 is
(1) 1-butene-3-yne (2) 3-butene-1-yne
(3) 1-butyn-3-ene (4) but-1-yne-3-ene
Sol. Answer (1)
1-butene-3-yne.
86. Which of the following compounds will exhibit cis-trans (geometrical) isomerism?
(1) 2 Butenol (2) 2-Butene
(3) Butanol (4) 2-Butyne
Sol. Answer (2)
2-butene.
87. H2COH.CH2OH on heating with periodic acid gives :
(1) 2 C O
H
H
(2) 2CO2
(3) 2HCOOH (4)
CHO
CHO
Sol. Answer (1)
CH OH 2
CH OH 2
HIO4 2HCHO
88. Consider the following reaction, ethanol ↓↓PB↓r3 ↓ X↓↓alc↓. KO↓H↓ Y 2 4
2
(i) H SO room temperature
(ii) H O, heat ↓↓↓↓↓↓↓↓↓↓↓ Z ;
The product Z is
(1) CH3CH2OH (2) CH2 = CH2
(3) CH3CH2—O—CH2—CH3 (4) CH3—CH2—O—SO3H
(23)
Sol. Answer (1)
PBr3 alc. KOH H2SO4, 25 C
C2H5OH C2H5Br CH2 CH2 CH3CH2OH ↓↓↓↓ ↓↓↓↓↓ = ↓↓↓↓↓° ↓
89. Benzene reacts with CH3Cl in the presence of anhydrous AlCl3 to form
(1) Xylene (2) Toluene
(3) Chlorobenzene (4) Benzylchloride
Sol. Answer (2)
CH3
+ CH Cl 3
Anhyd.
AlCl3
90. Nitrobenzene can be prepared from benzene by using a mixture of conc. HNO3 and conc. H2SO4. In
the
mixture, nitric acid acts as a/an
(1) Catalyst (2) Reducing agent
(3) Acid (4) Base
Sol. Answer (4)
Base.
91. Which of the following reactions is an example of nucleophilic substitution reaction?
(1) RX + Mg ↓ RMgX (2) RX + KOH ↓ ROH + KX
(3) 2RX + 2Na ↓ R—R + 2NaX (4) RX + H2 ↓ RH + HX
Sol. Answer (2)
RX + KOH ↓ ROH + KX
92. Which one of the following is employed as a tranquilizer?
(1) Chlorpheninamine (2) Equanil
(3) Naproxen (4) Tetracycline
Sol. Answer (2)
Equanil.
93. Structures of some common polymers are given. Which one is not corretly presented?
(1) Nylon 66
—[NH(CH2)6NHCO(CH2)4 — CO—]2
(2) Teflon
—( CF2—CF2—)n
(3) Neoprene
—CH2—C CH—CH2—CH2—
|
Cl n
=



(4) Terylene
COOCH — CH — O —) 2 2 n —( OC
(24)
Sol. Answer (3)
CH2—C CH—CH2
— | —
Cl
ϒ = ⁄
′ ∞
′≤ ∞ƒ
[Note : The answer will also be (1) because the structure of nylon-66 given is dimer structure not
polymer structure.]
94. Predict the product : NHCH3 + NaNO2 + HCl Product
(1) N
OH
CH3 (2)
N N
CH3
O
(3) N NO2
CH3
(4) NO
NHCH3 NHCH3
+
NO
Sol. Answer (2)
NHCH3 + NaNO + HCl 2 N N
CH3
O
95. Propionic acid with Br2 | P yields a dibromo product. Its structure would be
(1) C COOH
Br
Br
CH3 (2) CH2Br—CHBr—COOH
(3) C CH COOH 2
Br
Br
H (4) CH2Br—CH2—COBr
Sol. Answer (1)
Br2 /P
3 2 HVZ reaction 3
Br
|
CH —CH —COOH CH — C—COOH
|
Br
↓↓↓↓↓↓
(25)
96. Trichloroacetaldehyde, CCl3CHO reacts with chlorobenzene in presence of sulphuric acid and
produces
(1) Cl CH Cl
CCl3
(2) Cl C Cl
CH2Cl
Cl
(3) Cl C Cl
H
Cl
(4) Cl C Cl
Cl
OH
Sol. Answer (1)
CCl CHO + 2 Cl 3
H2SO4 CCl CH 3
Cl
Cl
D.D.T.
97. Consider the following reaction :
3 4
3 2
CH Cl Alkaline KMnO
Zn dust Anhydrous AlCl H O/H
Phenol X Y Z + ↓↓↓↓ ↓↓↓↓↓↓↓ ↓↓↓↓↓↓↓ , the product Z is
(1) Benzene (2) Toluene
(3) Benzaldehyde (4) Benzoic acid
Sol. Answer (4)
OH
CH3Cl
AlCl3
Zn
Dust
CH3
(i) Alkaline
(ii) H /H O +
2
COOH
98. The state of hybridization of C2, C3, C5 and C6 of the hydrocarbon,
CH C 3 CH CH CH C CH
CH3
CH3
CH3
7 6 5 4 3 2 1
is in the following sequence
(1) sp, sp2, sp3 and sp2 (2) sp, sp3, sp2 and sp3
(3) sp3, sp2, sp2 and sp (4) sp, sp2, sp2 and sp3
Sol. Answer (2)
sp, sp3, sp2 and sp3
(26)
99. The segment of DNA which acts as the instrumental manual for the synthesis of the protein is
(1) Nucleoside (2) Nucleotide
(3) Ribose (4) Gene
Sol. Answer (4)
Gene.
100. Which of the following hormones contains iodine?
(1) Thyroxine (2) Insulin
(3) Testosterone (4) Adrenaline
Sol. Answer (1)
Thyroxine
101. Which one of the following has haplontic life cycle?
(1) Wheat (2) Funaria
(3) Polytrichum (4) Ustilago
Sol. Answer (4)
Wheat – diplontic life cycle.
Funaria and Polytrichum – Haplodiplontic life cycle.
102. T.O. Diener discovered a
(1) Bacteriophage (2) Free infectious RNA
(3) Free infectious DNA (4) Infectious protein
Sol. Answer (2)
He discovered viroids having ssRNA and no capsid.
103. Mannitol is the stored food in
(1) Gracillaria (2) Chara
(3) Porphyra (4) Fucus
Sol. Answer (4)
Reserve food of brown algae (Fucus) is Laminarin and Mannitol.
104. Which one of the following is a vascular cryptogram?
(1) Cedrus (2) Equisetum
(3) Ginkgo (4) Marchantia
Sol. Answer (2)
Equisetum is a pteridophyte i.e., vascular cryptogam.
105. Phylogenetic system of classification is based on
(1) Floral characters (2) Evolutionary relationships
(3) Morphological features (4) Chemical constituents
Sol. Answer (2)
Phylogeny deals with evolutionary interrelations.
(27)
106. Which one of the following groups of animals is bilaterally symmetrical and triploblastic?
(1) Sponges (2) Coelenterates (Cnidarians)
(3) Aschelminthes (round worms) (4) Ctenophores
Sol. Answer (3)
Aschelminthes are bilaterally symmetrical and triploblastic with pseudocoelom. Sponges are generally
asymmetrical and some are radially symmetrical and diploblastic. Coelenterates and Ctenophores are
radially
symmetrical and diploblastic.
107. Peripatus is a connecting link between
(1) Coelenterata and Porifera (2) Ctenophora and Platyhelminthis
(3) Mollusca and Echinodermata (4) Annelida and Arthropoda
Sol. Answer (4)
Peripatus is a connecting link between annelida and arthropoda. It has annelidian character like presence
of
nephridia and unjointed legs and like arthopods it respires by trachea.
108. Which one of the following pairs of animals comprises ‘jawless fishes’?
(1) Guppies and hag fishes (2) Lampreys and eels
(3) Mackerals and Rohu (4) Lampreys and hag fishes
Sol. Answer (4)
'Jawless' fishes belong to the class Cyclostomata. The example of this class are Petromyzon (lamprey)
and
Myxine (hagfish).
109. If a live earthworm is pricked with a needle on its outer surface without damaging its gut, the fluid
that comes
out is
(1) Slimy mucus (2) Excretory fluid
(3) Coelomic fluid (4) Haemolymph
Sol. Answer (3)
Annelids are bilaterally symmetrical, triploblastic with a true coelom. The coelom is filled with coelomic
fluid
which acts as hydraulic skeleton and even helps in locomotion.
110. Plasmodesmata are
(1) Connections between adjacent cells
(2) Lignified cemented layers between cells
(3) Locomotary structures
(4) Membranes connecting the nucleus with plasmalemma
Sol. Answer (1)
These are protoplasmic strands for communication and transport between adjacent cells.
111. Stroma in the chloroplasts of higher plant contains
(1) Chlorophyll (2) Light-independent reaction enzymes
(3) Light-dependent reaction enzymes (4) Ribosomes
Sol. Answer (2)
Dark reaction is light independent reaction and is an enzymatic process.
(28)
112. Synapsis occurs between
(1) Two homologous chromosomes (2) A male and a female gamete
(3) mRNA and ribosomes (4) Spindle fibres and centromere
Sol. Answer (1)
Synapsis (Bivalent formation) stands for pairing of homologous chromosomes.
113. Middle lamella is composed mainly of
(1) Phosphoglycerides (2) Hemicellulose
(3) Muramic acid (4) Calcium pectate
Sol. Answer (4)
Middle lamella is composed of Ca-pectate (mainly) and Mg-pectate.
114. Cytoskeleton is made up of
(1) Proteinaceous filaments (2) Calcium carbonate granules
(3) Callose deposits (4) Cellulosic microfibrils
Sol. Answer (1)
Microtubule – Tubulin protein
Microfilament – Actin protein
Intermediate filament – Acidic proteins
115. The cell junctions called tight, adhering and gap junctions are found in
(1) Neural tissue (2) Muscular tissue
(3) Connective tissue (4) Epithelial tissue
Sol. Answer (4)
The various cells in a tissue are held together by ECF (extracellular fluid), it is made up of glycoproteins
and
acts as a binder. In the epithetial tissue the cells are in close contact with each other with little or no
extracellular fluid. So, the cells in epithelial tissue are held together by cell junctions.
116. The kind of tissue that forms the supportive structure in our pinna (external ears) is also found in
(1) Tip of the nose (2) Vertebrae
(3) Nails (4) Ear ossicles
Sol. Answer (1)
Ear pinna has elastic cartilage, which is also present in the tip of nose; but the nasal septum has hyaline
cartilarge.
117. The epithelial tissue present on the inner surface of bronchioles and fallopian tubes is
(1) Squamous (2) Cuboidal
(3) Glandular (4) Ciliated
Sol. Answer (4)
The epithelial tissue present in the bronchioles and fallopian tubes is ciliated epithelium. Ciliated
epithelium
is present on the surfaces which involve movement of particles, mucous or cells.
(29)
118. Given below is a schematic break-up of the phases/stages of cell cycle
Which one of the following is the correct indication of the stage/phase in the cell cycle?
(1) A-Cytokinesis (2) B-Metaphase
(3) C-Karyokinesis (4) D-Synthetic phase
Sol. Answer (4)
It represents S-phase that occurs between G1 and G2 phase.
119. What is not true for genetic code?
(1) It is unambiguous
(2) A codon in mRNA is read in a non-contiguous fashion
(3) It is nearly universal
(4) It is degenerate
Sol. Answer (2)
All three bases are to be read continuously to code for an amino acid.
120. Removal of introns and joining the exons in a defined order in a transcription unit is called
(1) Capping (2) Splicing
(3) Tailing (4) Transformation
Sol. Answer (2)
Splicing is eukaryotic feature to remove introns (non coding sequences).
121. Semiconservative replication of DNA was first demonstrated in
(1) Salmonella typhimurium (2) Drosophila melanogaster
(3) Escherichia coli (4) Streptococcus pneumoniae
Sol. Answer (3)
Meselson and Stahl conducted this experiment on E.coli.
122. Whose experiments cracked the DNA and discovered unequivocally that a genetic code is a
‘‘triplet’’?
(1) Beadle and Tatum (2) Nirenberg and Mathaei
(3) Hershey and Chase (4) Morgan and Sturtevant
Sol. Answer (2)
Triplet code was first deciphered by Nirenberg and Matthaei using homopolymer of ‘Poly U’ nucleotides.
123. Point mutation involves
(1) Deletion (2) Insertion
(3) Change in single base pair (4) Duplication
Sol. Answer (3)
True gene mutations that may occur as transition or transversion.
(30)
124. In the case of peppered moth (Biston betularia) the black-coloured form became dominant over the
light-coloured
form in England during industrial revolution. This is an example of
(1) Inheritance of darker colour character acquired due to the darker environment
(2) Natural selection whereby the darker forms were selected
(3) Appearance of the darker coloured individuals due to very poor sunlight
(4) Protective mimicry
Sol. Answer (2)
Natural selection selects alleles which makes an individual live and reproduce successfully. After
industrial
revolution the soot was deposisted on tree trunks, so the black/melanic moth could easily camouflage and
they were naturally selected.
125. Sickle cell anemia is
(1) Characterized by elongated sickle like RBCs with a nucleus
(2) An autosomal linked dominant trait
(3) Caused by substitution of valine by glutamic acid in the beta globin chain of haemoglobin
(4) Caused by a change in a single base pair of DNA
Sol. Answer (4)
Sickle cell anaemia is due to a single base change in the ®-chain of haemoglobin. Glutamic acid at
position
number six is replaced by valine.
126. Study the pedigree chart given below
What does it show?
(1) Inheritance of a recessive sex-linked disease like haemophilia
(2) Inheritance of a sex-linked inborn error of metabolism like phenylketonuria
(3) Inheritance of a condition like phenylketonuria as an autosomal recessive trait
(4) The pedigree chart is wrong as this is not possible
Sol. Answer (3)
Aa Aa
Aa aa Aa Aa aa Aa
aa Aa Aa
Phenylketonuria is an autosomal recessive trait.
(31)
127. The most popularly known blood grouping is the ABO grouping. It is named ABO and not ABC
because ‘‘O’’
in it refers to having
(1) No antigens A and B on RBCs
(2) Other antigens besides A and B on RBCs
(3) Overdominance of this type on the genes for A and B types
(4) One antibody only–either anti–A or anti–B on the RBCs
Sol. Answer (1)
In the ABO blood group system the blood group 'O' has no antigen on RBC but both the antibodies a and
b
are present in blood plasma.
128. Select the incorrect statement from the following
(1) Baldness is a sex-limited trait
(2) Linkage is an exception to the principle of independent assortment in heredity
(3) Galactosemia is an inborn error of metabolism
(4) Small population size results in random genetic drift in a population
Sol. Answer (1)
Baldness is a sex influenced trait.
129. Cotyledons and testa respectively are edible parts in
(1) Cashew nut and litchi
(2) Groundnut and pomegranate
(3) Walnut and tamarind
(4) French bean and coconut
Sol. Answer (2)
Fleshy testa is edible for pomegranate and cotyledons are edible in groundnut.
130. An example of a seed with endosperm, perisperm and caruncle is
(1) Castor (2) Cotton
(3) Coffee (4) Lily
Sol. Answer (1)
Ricinus (Castor) has all three structures, where caruncle is an overgrowth of outer integument at
micropyle
showing hygroscopic nature. Perisperm is persistent nucellus in seed.


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