QUESTIONS & SOLUTIONS OF

AIPMT-2010 (MAINS) TEST PAPER

Duration : 3 Hours Max. Marks : 480

Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.

IMPORTANT INSTRUCTIONS

1. The Test Booklet consists of one paper containing 120 objective type questions (four options with

single correct answer) from Physics, Chemistry and Biology (Botany & Zoology).

2. There are three parts in the question paper (Physics, Chemistry and Biology (Botany & Zoology)). The

distribution of marks subjectwise in each part is as under for each correct response.

3. Scoring and Negative Marking : Each question carries 4 marks. For each incorrect response

one mark will be deducted from the total score. No deduction from the total score will, however,
be made if no response is indicated for a question in the Answer Sheet. The candidates are advised not

to attempt such question in the Answer Sheet, if they are not sure of the correct response. More than

one answer indicated against a question will be deemed as incorrect response and will be
negatively marked.

Part A — PHYSICS (120 marks) – 30 Questions

Part B — CHEMISTRY (120 marks) – 30 Questions

Part C — BIOLOGY (240 marks) – 60 Questions

AIPMT (MAINS)-2010
PART- A (PHYSICS)
1. A current loop consists of two identical semicircular parts each of radius R, one lying in the x-y plane and the
other in x-z plane. If the current in the loop is i. The resultant magnetic field due to the two semicircular parts
at their common centre is

 0i  0i  0i  0i
(1) (2) (3) (4)
2 2R 2R 4R 2R
Ans. (1)
Sol. The loop mentioned in the question must look like one as shown in the figure.

1   0i 
Magnetic field at the centre due to semicircular loop lying in x-y plane, B xy    negative z direction
2  2R 
1   0i 
Similarly field due to loop in x–z plane, Bxz =   in negative y direction.
2  2R 
 Magnitude of resultant magnetic field,

2 2
 i  i  0i  0i
B= B 2xy  B 2xz   0    0  = 4R 2 
 4R   4R  2 2R

2. Two following figure shows a logic gate circuit with two inputs A and B and the output Y. the voltage wave
forms of A, B and Y are as given.

The logic gate is

(1) NOR gate (2) OR gate (3) AND gate (4) NAND gate
Ans. (4)

RESONANCE PAGE - 2
AIPMT (MAINS)-2010
Sol. It is clear from given logic circuit, that output Y is low when both the inputs are high, otherwise it is high. Thus
logic circuit is NAND gate.

A B Y
1 1 0
0 0 1
0 1 1
1 0 1

3. Two parallel metal plates having charges +Q and –Q face each other at a certain distance between them. If
the plates are now dipped in kerosene oil tank, the electric field between the plates will
(1) become zero (2) increase (3) decrease (4) remains same
Ans. (3)

Sol. Electric field between two parallel plates placed in vacuum is given by


E
0


In a medium of dielectric constant K, E' =  K For kerosene oil K > 1  E' = E
0

3R
4. The electric field at a distance from the centre of a charged conducting spherical shell of radius R is E.
2
R
The electric field at a distance from the centre of the sphere is
2
E E
(1) zero (2) E (3) (4)
2 3
Ans. (1)
Sol. Electric field insided charged conductor is always zero.

5. A student measures the distance traversed in free fall of abody, initially at rest, in a given time. He uses this
data to estimate g, the acceleration due to gravity. If the maximum percentage errors in measurement of the
distance and the time are e1 and e2 respectively, the percentage error in the estimation of g is
(1) e2 – e1 (2) e1 + 2e2 (3) e1 + e2 (4) e1 – 2e2
Ans. (2)
Sol. From the relation

1 2
h = ut + gt
2

1 2 2h
h= gt  g= ( body initially at rest)
2 t2
Taking natural logarithm on both sides, we get
ng = n h – 2n t

g h t
Differentiating, g  h  2 t

RESONANCE PAGE - 3
AIPMT (MAINS)-2010
For maximum permissible error,

 g   h   t 
or   100     100   2    100 
 g   h   t 

According to problem

h
 100 = e1 and t  100 = e2
h t

 g 
Therefore,  g  100  = e1 + 2e2
 

6. When monochromatic radiation of intensity  falls on a metal surface, the number of photoelectrons and their
maximum kinetic energy are N and T respectively. If the intensity of radiation is 2, the number of emitted
electrons and their maximum kinetic energy are respectively
(1) N and 2T (2) 2N and T (3) 2N and 2T (4) N and T
Ans. (2)
Sol. The number of photoelectrons ejected is directly proportional to the intensity of incident light. Maximum
kinetic energy is independent of intensity of incident light but depends upon the frequency of light. Hence
option (2) is correct.

7. The electric field of an electromagnetic wave in free space is given by E  10 cos(10 7 t  kx ) ĵ V/m, where t
and x are in seconds and metres respectively. It can be iferred that
(i) the wavelength  is 188.4 m. (ii) the wave number k is 0.33 rad/m
(iii) the wave amplitude is 10 V/m(iv) the wave is propagating along +x direction
Which one of the following pairs of statements is correct ?
(1) (iii) and (iv) (2) (i) and (ii) (3) (ii) and (iii) (4) (i) and (iii)
Ans. (4)
Sol. As given E = 10cos(107t + kx)
Comparing it with standard equation of e.m. wave,
E = E0cos(t + kx)
Amplitude E0 =10V/m and  = 107 rad/s

 2c 2  3  10 8
 c =  = or   = 188.4 m
2  10 7
Also

  10 7
c or k   = 0.033
k c 3  10 8
The wave is propagating along y direction.

8. The speed of light in media M1 and M2 are 1.5 × 108 m/s and 2.0 × 108 m/s respectively. A ray of light enters
from medium M1 to M2 at an incidence angle i. If the ray suffers total internal reflection, the value of i is

2 1  3 
(1) Equal to sin–1   (2) Equal to or less than sin  
3 5

1  3  1  2 
(3) Equal to or greater than sin   (4) Less than sin  
4 3
Ans. (3)

RESONANCE PAGE - 4
AIPMT (MAINS)-2010
Sol. Refractive index for medium M1 is

c 3  10 8
1   2
v 1 1.5  10 8

Refractive index for medium M2 is

c 3  108 3
2   8

v 2 2.0  10 2

For total internal reflection

sini  sinC
where i = angle of incidence
C = critical angle

2
But sinC = 
1

2 3 / 2 1  3 
sin i    i  sin  
1 2 4
9. A ray of light is incident on a 60° prism at the minimum deviation position. The angle of refraction at the first
face (i.e., incident face) of the prism is
(1) zero (2) 30° (3) 45° (4) 60°
Ans. (2)
Sol. Angle of prism, A = r1 + r2
For minimum deviation
r1 = r2 = r  A = 2r
Given, A = 60°

A 60
Hence, r =   30
2 2

10. For transistor action

(1) Base, emitter and collector regions should have similar size and doping concentrations.
(2) The base region must be very thin and lightly doped.
(3) The emitter-base junction is forward biased and base-collector junction is reverse biased
(4) Both the emitter-base junction as well as the base-collector junction are forward biased
(1) (4) and (1) (2) (1) and (2) (3) (2) and (3) (4) (3) and (4)
Ans. (3)

11. The additional kinetic energy to be provided to a satellite of mass m revolving around a planet of mass M, to
transfer it from a circular orbit of radius R1 to another of radius R2 (R2 > R1) is

 1 1   1 1   1 1  1  1 1 
(1) GmM 2  2  (2) GmM  (3) 2GmM  (4) GmM 
R  R  R  2 R
 1 R2 

 1 R2   1 R2   1 R2 
Ans. (4)

 GMm   GMm 
Sol. KE        
 2R 2   2R1 

GMm  1 1 
KE    
2  R1 R 2 

RESONANCE PAGE - 5
AIPMT (MAINS)-2010
12. The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is
(1) 60° (2) 15° (3) 45° (4) 60°
Ans. (1)
Sol. Let v be velocity of a projectile at maximum height H

v = ucos

u
According to given problem, v =
2

u 1
 = ucos  cos =
2 2
  = 60°

R
13. From a circular disc of radius R and mass 9M, a small disc of mass M and radius is removed concentrically..
3
The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing
through its centre is

40 4
(1) MR 2 (2) MR2 (3) 4MR2 (4) MR 2
9 9
Ans. (1)
Sol. Mass of the disc = 9M
Mass of removed portion of disc = M
The moment of inertia of the complete disc about an axis passing through its centre O and perpendicular to
its plane is

9
I1 = MR 2
2
Now, the moment of inertia of the disc with removed portion
2
1 R 1
I2 = M   MR 2
2 3 18

Therefore, moment of inertia of the remaining portion of disc about O is

I = I1 – I2

MR 2 MR 2 40MR 2
= 9  
2 18 9
14. A particle moves in x-y plane according to rule x = asint and y = acost. The particle follows
(1) an elliptical path (2) a circular path
(3) a parabolic path (4) a straight line path inclined equally to x and y-axes
Ans. (2)

RESONANCE PAGE - 6
AIPMT (MAINS)-2010

x
Sol. x = asint or  sin t ...(i)
a

y
y = acost or  cos t ...(ii)
a
Squaring and adding, we get

x2 y2
 1 ( cos2t + sin2t = 1)
a2 a2
or x2 + y2 = a2
This is the equation of a circle. Hence particle follows a circular path

15. A closely wound solenoid of 2000 turns and area of cross-section 1.5 × 10–4 m2 carries a current of 2.0 A. It
is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a
uniform magnetic field 5 × 10–2 tesla making an angle of 30° with the axis of the solenoid. The torque on the
solenoid will be
(1) 3 × 10–3 N m (2) 1.5 × 10–3 N m (3) 1.5 × 10–2 N m (4) 3 × 10–2 N m
Ans. (3)
Sol. Magnetic moment of the loop.
M = NIA = 2000 × 2 × 1.5 × 10–4 = 0.6 J/T
torque  = MBsin30°

1
= 0.6 × 5 × 10–2 ×
2
= 1.5 × 10–2 Nm

16. The decay constant of a radio isotope is . If A1 and A2 are its activities at times t1 and t2 respectively, the
number of nuclei which have decayed during the time (t1 – t2)
(1) A1t1 – A2t2 (2) A1 – A2 (3) (A1 – A2)/ (4) (A1 – A2)
Ans. (3)
Sol. A1 = N1 at time t1
A2 = N2 at time t2
Therefore, number of nuclei decayed during time interval (t1 – t2) is

[A1  A 2 ]
N1 – N2 =

17. A particle having a mass of 10–2 kg carries a charge  of 5 × 10 C. The particle
–8
 is given an initial horizontal
5
velocity of 10 ms in the presence of electric field E and magnetic field B . To keep the particle moving in a
–1

horizontal
 direction, it is necessary that 
(i) B should be perpendicular
 to the direction of velocity and E should be along the direction of velocity
(ii) Both B and E should be along the direction of velocity
(iii) Both
 B and E are mutually perpendicular and perpendicular
 to the direction of velocity..
(iv) B should be along the direction of velocity and E should be perpendicular to the direction of velocity
Which one of the following pairs of statements is possible
(1) (i) and (iii) (2) (iii) and (iv) (3) (ii) and (iii) (4) (ii) and (iv)
Ans. (3)

RESONANCE PAGE - 7
AIPMT (MAINS)-2010
18. The binding energy per nucleon in deuterium and helium nuclei are 1.1 MeV and 7.0 MeV, respectively.
When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is
(1) 23.6 MeV (2) 2.2 MeV (3) 28.0 MeV (4) 30.2 MeV
Ans. (1)
Sol. 1
H2 + 1H2  2He4 + E
The binding energy per nucleon of a deuteron = 1.1 MeV
 Total binding energy = 2 × 1.1 = 2.2 MeV
The binding energy per nucleon of a helium nuclei = 7 MeV
 Total binding energy = 4 × 7 = 28 MeV
 Hence, energy released
E = (28 – 2 × 2.2) = 23.6 MeV

19. The electron in the hydrogen atom jumps from excited state (n = 3) to its ground state (n = 1) and the
photons thus emitted irradiate a photosensitive material. If the work function of the material is 5.1 eV, the
13.6
stopping potential is estimated to be (the energy of the electron in nth state En =  2 eV )
n
(1) 5.1 V (2) 12.1 V (3) 17.2 V (4) 7V
Ans. (4)
Sol. Energy released when electron in the atom jumps from excited state (n = 3) to ground state (n = 1) is

 13.6   13.6   13 .6
E = h = E3 – E1 = 2
 2  = + 13.6 = 12.1 eV
3  1  9
Therefore, stopping potential
eV0 = h – 0
= 12.1 – 5.1 [ work function 0 = 5.1]
V0 = 7V

20. If cp and cv denote the specific heats (per unit mass) of an ideal gas of molecular weight M where R is the
molecular weight constant, then
(1) cp – cv = R/M2 (2) cp – cv = R (3) cp – cv = R/M (4) cp – cv = MR
Ans. (3)
Sol. Let CV and CP be molar specific heats of the ideal gas at constant volume and constant pressure, respectively,
then
CP = Mcp and CV = Mcv
 CP – CV = R  Mcp – Mcv = R
 cp – cv = R/M

21. A condenser of capacity C is charged to a potential difference of V1. The plates of the condenser are then
connected to an ideal inductor of inductance L. The current through the inductor when the potential difference
across the condenser reduces to V2 is

1 1
 C( V1  V2 )2  2 C( V12  V22 ) C( V12  V22 )  C( V12  V22 )  2
(1)   (2) (3) (4)  
 L 
 L L  L 

Ans. (4)
Sol. In case of oscillatory discharge of a capacitor through an inductor, charge at instant t is given by
q = q0cost

1
where =
LC

RESONANCE PAGE - 8
AIPMT (MAINS)-2010

q CV2 V2
 cost = q  CV  V ( q = CV) ...(i)
0 1 1

Current through the inductor

dq d( q0 cos t )
=  = –q0sint
dt dt

1
 = CV1 [1 – cos2t]1/2
LC

1/ 2 1/ 2
2
C   V2   C( V12  V22 ) 
= V1 1   
L   V1 



L
 (using (i))
   

22. The dependence of acceleration due to gravity g on the distance r from the centre of the earth, assumed to
be a sphere of radius R of uniform density is as shown in figures below. The correct figure is.

(1) (2) (3) (4)

Ans. (4)
Sol. The acceleration due to gravity at a depth d below surface of earth is

GM  d  d
g' =  1    g 1  
R2  R   R
g' = 0 at d = R
i.e., acceleration due to gravity is zero at the centre of earth .
Thus, the variation in value g with r is
for, r > R,

g gR 2
g'  
2
r2 1
 h  g' 
1   r2
 R

Here, R+h=r

 d  gr
For r < R, g' = g  1   
 R R
Here, R – d = r  g'  r
Therefore, the variation of g with distance from centre of the earth
will be as shown in the figure.

RESONANCE PAGE - 9
AIPMT (MAINS)-2010
23. A solid cylinder and a hollow cylinder, both of the same mass and same external diameter are relaeased from
the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach
the bottom first ?
(1) Both together only when angle of inclination of plane is 45°
(2) Both together
(3) Hollow cylinder
(4) Solid cylinder
Ans. (4)
Sol. time taken to reach the bottom of inclined plane

 K2 
2 1  2 
 R 
t=  
g sin 

Here,  is length of incline plane

R2
For solid cylinder K2 =
2
For hollow cylinder = K2 = R2
Hence, solid cylinder will reach the bottom first.

24. The thermo e.m.f. E in volts of a certain thermocouple is found to vary with temperature difference  in °C
between the two junctions according to the relation

2
E = 30 
15
The neutral temperature for the thermo-couple will be
(1) 450°C (2) 400°C (3) 225°C (4) 30°C
Ans. (3)

2
Sol. As given E = 30 –
15

dE
0
d

dE 2 
 30  0 or 30  2
d 15 15

15  30
= = 225°C
2
25. (i) Centre of gravity (C.G.) of a body is the point at which the weight of the body acts
(ii) Centre of mass coincides with the centre of gravity if the earth is assumed to have infinitely large radius
(iii) To evaluate the gravitational field intensity due to any body at an external point, the entire mass of the
body can be considered to be concentrated at its C.G.
(iv) The radius of gyration of any body rotating about an axis is the length of the perpendicular dropped from
the C.G. of the body to the axis.
Which one of the following pairs of statements is correct ?
(1) (iv) and (i) (2) (i) and (ii) (3) (ii) and (iii) (4) (iii) and (iv)
Ans. (1)

RESONANCE PAGE - 10
AIPMT (MAINS)-2010
26. The magnetic moment of a diamagnetic atom is
(1) much greater than one (2) one
(3) between zero and one (4) equal to zero
Ans. (4)
Sol. The magnetic momentum of a diamagnetic atom is equal to zero.

27. Two identical bar magnets are fixed with their centres at a distance d apart. a stationary charge Q is placed
at P in between the gap of the two magnets at a distance D from the centre O as shown in the figure

The force on the charge Q is

(1) Zero (2) directed along OP
(3) directed along PO (4) directed perpendicular to the plane of paper
Ans. (1)
Sol. Magnetic field due to bar magnets exerts force on moving charges only. Since the charge is at rest, zero
force acts on it.

28. A particle of mass M, starting from rest, undergoes uniform acceleration. If the speed acquired in time T is V,
the power delivered to the particle is

MV 2 1 MV 2 MV 2 1 MV 2
(1) (2) (3) (4)
T 2 T2 T2 2 T
Ans. (4)
Sol. Power delivered in time T is
P = F.V. = MaV

dV
or P = MV  PdT = MVdV
dT

MV 2 1 MV 2
 PT = or P =
2 2 T

29. A thin circular ring of mass M and radius r is rotating about its axis with constant angular velocity . Two
objects each of mass m are attached gently to the opposite ends of a diameter of the ring. The ring now
rotates with angular velocity given by

(M  2m) 2M (M  2m) M

(1) (2) (3) (4)
2m M  2m M M  2m
Ans. (4)
Sol. As no external toruqe is acting about the axis, angular momentum of system remains conserved.
11 = 22

11 Mr 2  M
 2   2

2 (M  2m)r (M  2m)

RESONANCE PAGE - 11
AIPMT (MAINS)-2010

1
30. A monoatomic gas at pressure P1 and volume V1 is compressed adiabatically to th of its original volume.
8
What is the final pressure of the gas
(1) 64P1 (2) P1 (3) 16P1 (4) 32P1
Ans. (4)

Sol. Ideal gas equation, for an adiabatic process is

PV = constant or P1V1 = P2V2

5
For monoatomic gas  =
3

5/3
V 
 P1V15 / 3  P2  1 
 8 
 P2 = P1 × (2)5 = 32P1

PART- B (CHEMISTRY)

31. Among the elements Ca, Mg, P and Cl, the order of increasing atomic radii is :
(1) Mg < Ca < Cl < P (2) Cl < P < Mg < Ca
(3) P < Cl < Ca < Mg (4) Ca < Mg < P < Cl
Ans. (2)
Sol. The atomic radii decreases on moving from left to right in a period, thus order of sizes for Cl, P and Mg is
Cl < P < Mg. Down the group size increases. Thus overall order is : Cl < P < Mg < Ca.

32. The reaction,

2A(g) + B(g) 3C(g) + D(g)
is begun with the concentrations of A and B both at an initial value of 1.00 M. When equilibrium is reached,
the concentration of D is measured and found to be 0.25 M. The value for the equilibrium constant for this
reaction is given by the expression :
(1) [(0.75)3 (0.25)] [(1.00)2 (1.00)] (2) [(0.75)3 (0.25)] [(0.50)2 (0.75)]
3 2
(3) [(0.75) (0.25)] [(0.50) (0.25)] (2) [(0.75)3 (0.25)] [(0.75)2 (0.25)]
Ans. (2)
Sol. 2A(g) + B(g) 3C(g) + D(g)
Initial moles : 1 1 0 0
Moles of eq. : 1 – (2 × 0.25) 1 – 0.25 3 × 0.25 0.25
= 0.5 = 0.75 = 0.75 = 0.75
Equilibrium constant,

[C]3 [D]
K=
[ A ]2 [B]

(0.75 )3 (0.25 )
 K=
(0.5)2 (0.75 )

RESONANCE PAGE - 12
AIPMT (MAINS)-2010
33. Which of the following expressions correctly represents the equivalent conductance at infinite dilution of

Al2(SO4)3. Given that  Al3+ and  SO 2  are the equivalent conductances at infinite dilution of the respective
4

ions?

(1) 2  Al3+ +3  SO 2  (2)  Al3+ +  SO 2 

4 4

1  1
(3) (  Al3+ + 3  SO 2  ) × 6 (4)  +  SO 2 
4 3 Al3+ 2 4

Ans. (2)
Sol. At infinite dilution, when dissociation is complete, each ion makes a definite contribution towards molar
conductance of the electrolyte irrespective of the nature of the other ion with which it is associated.

Hence  Al (SO ) =  Al3+ +  SO 2 

2 43 4

34. The pressure exerted by 6.0 g of methane gas in a 0.03 m3 vessel at 129° C is (Atomic masses) : C = 12.01,
H = 1.01 and R = 8.314 J K–1 mol–1) :
(1) 215216 Pa (2) 13409 Pa (3) 41648 Pa (4) 31684 Pa
Ans. (3)
Sol. Given, mass of CH4, w = 6 g
Volume of CH4, V = 0.03 m3
T = 129°C = 129 + 273 = 402 K
R = 8.3114 J K–1 mol–1
Molecular mass of CH4, M = 12.01 + 4 × 1.01 = 16.05
w
PV = nRT = RT
M

w RT 6 8.314  402
 P= = 
M V 16 .05 0.03
= 41647.7 Pa  41648 Pa

35. Match List I (Equations) with List II (Types of processes) and select the correct option.
List I List II
Equations Types of process
(a) Kp > Q (i) Non-spontaneous
(b) G° < RT In Q (ii) Equilibrium
(c) Kp = Q (iii) Spontaneous and endothermic
H
(d) T> (iv) Spontaneous
S
(1) a - (i), b - (ii), c - (iii), d - (iv) (2) a - (iii), b - (iv), c - (ii), d - (i)
(3) a - (iv), (b -(i), c - (ii), d - (iii) (4) a - (ii), b - (i), c - (iv), d - (iii)
Ans. (3)
Sol. When Kp > Q, rate of forward reaction > rate of backward reaction.
 Reaction is sontaneous.
When G° < RT ln Q, G° is positive, reverse reaction is feasible, thus reaction is non-spontaneous.
When Kp = Q, rate of forward reaction = rate of backward reaction
 Reaction is in equilibrium.
When TS > H, G will be negative only when H = + ve.

RESONANCE PAGE - 13
AIPMT (MAINS)-2010
36. Among the following four compounds :
(i) Phenol (ii) Methyl phenol (iii) Meta-nitrophenol (iv) Para-nitrophenol
The acidity order is :
(1) (iv) > (iii) > (i) > (ii) (2) (iii) > (iv) > (i) > (ii) (3) (i) > (iv) > (iii) > (ii) (4) (ii) > (i) > (iii) > (iv)
Ans. (1)
Sol. In phenols, the presence of electron releasing groups decreases the acidity, whereas presence of electron
withdrawing groups increase the acidity, compared to phenol. Among the meta and para-nitrophenols, the
latter is more acidic as the present of – NO2 group at para position stabilises the phenoxide ion to a greater
extent than when it is present at meta position. Thus, correct order of acidity is :
Para-nitrophenol > meta-nitrophenol > phenol > methyl phenol
(iv) (iii) (i) (ii)
37. Among the following which one has the highest cation to anion size ratio?
(1) CsI (2) CsF (3) LiF (4) NaF
Ans. (2)
Sol. The order of size of given cations is
Li+ < Na+ < Cs+
and the order of size of given anions is
I– > F–
Thus, when the cation is large and anion is smallest, the cation to anion size ratio is maximum.
Hence, cation to anion size ratio is maximum for CsF.

38. Three moles of an ideal gas expanded spontaneously into vacuum. The work done will be :
(1) Infinite (2) 3 Joules (3) 9 Joules (4) Zero
Ans. (4)
Sol. Since the ideal gas expands spontaneously into vacuum, Pext = 0, hence work done is also zero.

39. Which of the following species is not electrophilic in nature?

  
(1) Cl (2) BH3 (3) H3O (4) N O2
Ans. (3)
Sol. Electrophiles are electron deficient species. Among the given, H3O has lone pair of electrons for donation,
thus it is not electron deficient and hence, does not behave like an electrophile.

40. A 0.66 kg ball is moving with a speed of 100 m/s. The associated wavelength will be : (h = 6.6 × 10–34 Js)
(1) 6.6 × 10–32 m (2) 6.6 × 10–34 m (3) 1.0 × 10–35 m (4) 1.0 × 10–32 m
Ans. (3)
h
Sol According to de-Broglie equation,  =
mv
Given, h = 6.6 × 10–34 J s
m = 0.66 kg
v = 100 m s–1
6.6  10 34
 = = 1 × 10–35 m
0.66  100
41. Consider the following relations for emf of an electrochemical cell :
(i) EMF of cell = (Oxidation potential of anode) – (Reduction potential of cathode)
(ii) EMF of cell = (Oxidation potential of anode) + (Reduction potential of cathode)
(iii) EMF of cell = (Reduction potential of anode) + (Reduction potential of cathode)
(iv) EMF of cell = (Oxidation potential of anode) – (Oxidation potential of cathode)
Which of the above relations are correct
(1) (iii) and (i) (2) (i) and (ii) (3) (iii) and (iv) (4) (ii) and (iv)
Ans. (4)

RESONANCE PAGE - 14
AIPMT (MAINS)-2010
Sol. EMF of a cell = Reduction potential of cathode – Reduction potential of anode
= Reduction potential of cathode + Oxidation potential of anode
= Oxidation potential of anode – Oxidation potential of cathode.

42. In which of the following molecules, the central atom does not have sp3 hybridization?
(1) CH4 (2) SF4 (3) BF–4 (4) NH+4
Ans. (2)
Sol. For neutral molecules,
No. of electron pairs = No. of atoms bonded to it + 1/2 [Gp. no of central atom – Valency of central atom]
1
 For CH4, no. of e– pairs = 4  [ 4  4] = 4 (sp3 hybridisation)
2

1
For SF4, no. of e– pairs = 4  [6  4] = 5 (sp3 d hybridisation)
2
For ions,
No. of electron pairs = No. of atoms bonded to it + 1/2 [Gp. no. of central atom – Valency of central atom ±
No. of electrons]
1
 For BF4–, no. e– pairs = 4  [3  4  1] = 4 (sp3 hybridisation)
2

1
For NH+4, no. of e– pairs = 4  [5  4  1] = 4 (sp3 hybridisation)
2

43. For vaporization of water at 1 atmospheric pressure, the values of H and S are 40.63 kJ mol–1 and 108.8
JK–1 mol–1, respectively. The temperature when Gibb's energy change (G) for this transformation will be
zero, is :
(1) 273.4 K (2) 393.4 K (3) 373.4 K (4) 293.4 K
Ans. (3)
Sol. According to Gibb's equation, G = H – TS when G = 0, H = TS
Given, H = 40.63 kJ mol–1 = 40.63 × 103 J mol–1
S = 108.8 J K–1 mol–1

H 40.63  103
 T= = = 373.43 K.
S 108.8

44. Match List I (substances) with List II (processes) employed in the manufacture of the substances and select
the correct option :
List I List II
(Substances) (Processes)
(a) Sulphuric acid (i) Haber's process
(b) Steel (ii) Bessemer's process
(c) Sodium hydroxide (iii) Leblanc process
(d) Ammonia (iv) Contact process
(1) a - (i), b- (iv), c- (ii), d- (iii) (2) a- (i), b- (ii), c- (iii), d- (iv)
(3) a - (iv), b - (iii), c - (ii), d - (i) (4) a- (iv), b - (ii), c - (iii), d - (i)
Ans. (4)
45. When glycerol is treated with excess of HI, it produces :
(1) 2-iodopropane (2) allyl iodide (3) propene (4) glycerol triiodide
Ans. (1)

RESONANCE PAGE - 15
AIPMT (MAINS)-2010

Sol.

46. Some statements about heavy water are given below :

(i) Heavy water is used as a moderator in nuclear reactors.
(ii) Heavy water is more associated than ordinary water.
(iii) Heavy water is more effective solvent than ordinary water.
Which of the above statements are correct?
(1) (i) and (ii) (2) (i), (ii) and (iii) (3) (ii) and (iii) (4) (i) and (iii)
Ans. (1)
Sol. Heavy water is used for slowing down the speed of neutrons in nuclear reactors, hence used as moderators.
Boiling point of heavy water is greater (374.42 K) than that of ordinary water (373 K), hence heaver water is
more associated. Dielectric constant of ordinary water is greater than that of heavy water, hence ordinary
water is a better solvent.

47. The compound A on heating gives a colourless gas and a residue that is dissolved in water to obtain B.
Excess of CO2 is bubbled through aqueous solution of B, C is formed which is recovered in the solid form.
Solid C on gentle heating gives back A. The compound A is :
(1) CaCO3 (2) Na2CO3 (3) K2CO3 (4) CaSO4.2H2O
Ans. (1)
Sol. The reactions can be summarised as follows
A 
 colourless gas + residue
excess CO 2
Residue + H2O  B   C  A


This is possible only when A is CaCO3. The reactions are as follows :
CaCO3 
 CO2 + CaO
(A) (colourless gas) (residue)

CO 2
CaO + H2O  Ca(OH)2    Ca(HCO )
( excess ) 3 2  CaCO3


(B) (C) (A)

RESONANCE PAGE - 16
AIPMT (MAINS)-2010
48. Match the compounds given in List I with their characteristic reactions given in List II. Select the correct
option.
List I List II
(Compounds) (Reactions)
(a) CH3(CH2)3NH2 (i) Alkaline hydrolysis
(b) CH3C  CH (ii) With KOH (alcohol) and CHCl3 produces bad smell
(c) CH3CH2COOCH3 (iii) Gives white ppt. with ammoniacal AgNO3
(d) CH3CH (OH)CH3 (iv) With Lucas reagent cloudiness appears after 5 minutes
(1) a-(ii), b - (i), c - (iv), d- (iii) (2) a - (iii), b - (ii), c - (i), d - (iv)
(3) a - (ii), b - (iii), c - (i), d - (iv) (4) a - (iv), b - (ii), c - (iii), d - (i)
Ans. (3)
KOH(alc.)
Sol. CH3CH2CH2CH2NH2   CH3CH2CH2CH2NC + 3 KCl + 3 H2O
CHCl3

1° Amine Bad smell

CH3C  CH + Amm. AgNO3  CH3C  C Ag  + HNO3

White ppt.

Heat
CH3CH2COOCH3 + NaOH  CH3CH2COONa + CH3OH

49. Which one of the following compounds will be most readily dehydrated?

(1) (2) (3) (4)

Ans. (3)
Sol. (3) : The ease of dehydration of the given compounds can be explained on the basis of the stability of the
carbocation formed. In case of options (1), (2) and (4), a secondary carbocation is formed but the presence
of an electron withdrawing > C = O group adjacent to the positively charged carbon, intensifies the charge
and hence destabilies the species.

(a) (b) (c)

However, in case of option (c), a secondary carbocation is formed, but the electron withdrawing > C = O
group is present away, as a farther result, the effect of this group is diminished and hence the carbocation is
relatively more stable.

(c)
(more stable)
RESONANCE PAGE - 17
AIPMT (MAINS)-2010
50. The rate of reaction
2 NO + Cl2  2 NOCl is given by the rate equation rate = k[NO]2 [Cl]2. The value the rate of constant can
be increased by :
(1) increasing the temperature (2) increasing the concentration of NO
(3) increasing the concentration of the Cl2 (4) doing all of above
Ans. (1)
Sol. Rate constant is independent of the initial concentration of the reactants. It has a constant value at fixed
temperature. Hence the value of rate constant can be increased by increasing the temperature.

51. Which of the following complexes is not expected to exhibit isomerism?

(1) [Ni (NH3)4 (H2O)2]2+ (2) [Pt (NH3)2Cl2] (3) [Ni(NH3)2Cl2] (4) [Ni(en)3]2+
Ans. (3)

Sol. [Ni(NH3)2Cl2] has tetrahedral geometry and thus, deos not exhibit isomerism due to presence of symmetry
elements.

52. Which of the following conformers for ethylene glycol is most stable?

(1) (2) (3) (4)

Ans. (4)

Sol. The conformation (4) is most stable because of intermolecular H - bonding.

53. The IUPAC name of the compound CH3CH == CHC  CH is :

(1) Pent-4-yn-2-ene (2) Pent-3-en-1-yne (3) Pent-2-en-4-yne (4) Pent-1-yn-3-ene
Ans. (2)
5 4 3 2 1
Sol. CH3 CH  CHC  CH
Pent-3-en-1-yne
Fact : If a molecule contains both carbon-carbon double or triple bonds, the two are treated at par in seeking
the lowest number combination. However, if the sum of numbers turns out to be the same starting from either
of the carbon chain, then lowest number is given to the C == C double bond.

54. Which of the following oxidation states is the most common among the lanthanoids?
(1) 4 (2) 2 (3) 5 (4) 3
Ans. (4)
Sol. The common stable oxidation state of all the lanthanoids is + 3. The oxidation state of + 2 and + 4 are also
exhibited by some of the elements. These oxidation states are only stable in those cases where stable 4f°,
4f7 or 4f14 configuration are achieved.

55. How many bridging oxygen atoms are present in P4O10?

(1) 6 (2) 4 (3) 2 (4) 5
Ans. (1)

RESONANCE PAGE - 18
AIPMT (MAINS)-2010
Sol. P4O10 has the following structure :

Thus, the number of bridging oxygen atoms = 6.

56. Some of the properties of the two species, NO–3 and H3O+ are described below. Which one of them is
correct?
(1) Dissimilar in hybridization for the central atom with different structures
(2) Isostructural with same hybridization for the central atom
(3) Isostructural with different hybridization for the central atom
(4) Similar in hybridiation for the central atom with different structures
Ans. (1)
–
Sol. In NO 3,
1
H= [5  0  0  1]  3
2
Thus, in NO–3, the central atom is sp2 hybridised and it has trigonal planar geometry.

In H3O+,
1
H= [ 6  3  1  0]  4
2
Thus, O is sp3 hybridised in H3O+ and H3O+ has pyramidal geometry due to the presence of one lone pair
of electrons.

57. The following two reactions are known

Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g) ; H = – 26.8 kJ

FeO(s) + CO(g)Fe(s) + CO2(g) ; H = – 16.5 kJ

The value of H for the following reaction
Fe2O3(s) + CO(g)  2FeO(s) + CO2(g) is :
(1) + 10.3 kJ (2) – 43.3 kJ (3) – 10.3 kJ (4) + 6.2 kJ
Ans. (4)

RESONANCE PAGE - 19
AIPMT (MAINS)-2010
Sol. Given
(I) Fe2O3(s) + 3CO(g)  3Fe(s) + 3CO2(g) ; H = – 26.8 kJ
(II) FeO(s) + CO(g)  Fe(s) + CO2(g) ; H = – 16.5 kJ
On multiplying equation (II) with 2, we get
(III) 2FeO(s) + 2CO(g)  2Fe(s) + 2CO2(g) ; H = – 33 kJ
On substracting equation (IIII) from, I we get
Fe2O3(s) + CO(g)  2FeO(s) + CO2(g) ; H = – 26.8 – (– 33) = + 6.2 kJ

58. Following compounds are given

(i) CH3CH2OH (ii) CH3COCH3 (iii) (iv) CH3OH

Which of the above compound (s), on being warmed with iodine solution and NaOH, will give iodoform?
(1) (i), (iii) and (iv) (2) Only (ii) (3) (i), (ii) and (iii) (4) (i) and (ii)
Ans. (3)

Sol. Compounds having either group or CH3CHOH – group, give iodoform when warmed with I2 and

NaOH.
Thus,

give iodoform when warmed with I2 and NaOH, (Remember, NaOI oxidises CH3CH2OH to CH3CHO, thus it
gives positive iodoform test.)

59. Fructose reduces Tollen's reagent due to :

(1) Asymmetric carbons
(2) Primary alcoholic groups
(3) Secondary alcoholic group
(4) Enolisation of fructose followed by conversion to aldehyde by base
Ans. (4)
Sol. In aqueous solution, fructose is enolised and then converted into aldehyde in basic medium. All aldehydes
generally reduce Tollen's reagent, thus fructose also reduces Tollen's reagent.

60. In the following reaction,

1.Mg Ether
C6H5CH2Br    
 X, the product 'X' is :
2.H3O 

(1) C6H5CH2OCH2C6H5 (2) C6H5CH2OH

(3) C6H5CH3 (4) C6H5CH2CH2C6H5
Ans. (3)

Sol.

Cornea is the transparent anterior portion of the outer coat of the vertebrate eye covering the iris and the
pupil. It is avascular as it lacks lymphatic and blood vessels.

RESONANCE PAGE - 20
AIPMT (MAINS)-2010
PART- C (BIOLOGY)
61. Which one of the following is manoecious
(1) Date palm (2) Marchantia (3) Cyas (4) Pinus
Ans. (4)

62. Jaundice is disorder of

(1) Circulatory system (2) Excretory system
(3) Skin and eyes (4) Digestive system
Ans. (4)

63. Given below are four statements (A-D) each with one or two blanks. Select the option which correctly fills up
the blanks in two.
Statements :
(A) Wings of butterfly and birds look alike and are the results of ..... (i)....., evolution
(B) Miller showed that CH4, H2, NH3 and .... (i) ..., when exposed to electric discharge in a flask resulted in
formation of ...... (ii).....
(C) Vermiform appendix is a ...... (i)..... organ and an ...... (ii).... evidence of evolution.
(D) According to Darwin evolution took place due to ...... (i).... and .......(ii)..... of the fittest.
Options :
(1) (C) – (i) vestigial, (ii) anatomical, (D) (i) mutations (ii) multiplication
(2) (D) – (i) small variation, (ii) survival, (A) – (i) convergent
(3) (A) – (i) convergent, (B) – (i) oxygen, (ii) nucleosides
(4) (B) – (i) water vapour, (ii) amino acids, (C) – (i) rudimentary (ii) anatomical

Ans. (2)

64. In eukaryotic cell transcription, RNA splicaing and RNA capping take place inside the
(1) ER (2) Ribosomes (3) Nucleus (4) Dictyosomes
Ans. (3)

65. Select the correct matching of a hormone, its source and function.
Horm one Source Function
Regulates growth of m am mary
1 Prolactin Posterior pituitary
glands and milk form ation in females
2 Vasopression Posterior pituitary Increases loss of water through urine
Increases heart beat, rate of
3 Norepinephrine Adrenal m edulla
respiration and alertness
Beta-cells of lslets
4 Glucagon Stim ulates glycogenolysis
of langerhans
Ans. (3)

66. A cross in which an organism showing a dominant phenotype crosed with the recessive parent in order to
know its genotype is called
(1) Dihybrid cross (2) Monohybrid cross (3) Back cross (4) Test cross
Ans. (4)
67. A person suffering from a disease caused by Plasmodium, experinces recurring chill and fever at the time
when
(1) The microgametocytes and megagametocytes are being destoryed by the WBCs.
(2) The sporozoites released from RBCs are being rapidly killed and broken down inside spleen.

RESONANCE PAGE - 21
AIPMT (MAINS)-2010
(3) The trophyzoitges reach maximum growth and give out certain toxins
(4) The parasite after its rapid multiplicaiton inside RBCs ruptures them, releasing the stage to enter fresh
RBCs.
Ans. (3)

68. Select the correct combination of the statements (a-d) regarding the characteristics of certain organisms.
(a) Methanogens are archaebacteria which produce methane in marshy areas
(b) Notoc is a filamentous blue-green alga which fixes atmospheric nitrogen
(c) Chemosynthetic autotrophic bacteria synthesize cellulose form glucose
(d) Mycoplasma lack a cell wall and can survive wihout oxygen.
The correct statemetns are
(1) (a), (b), (d) (2) (b), (c) (3) (a), (b), (c) (4) (b), (c), (d)
Ans. (1)

69. Read the following four statements, A, B, C and D and select the right option having both correct statements.
Statement :
(A) Z scheme of light reaction takes place in presence of PSI only
(B) Only PSI is functional in cyclic photophosphorylaiton.
(C) Cyclic photophosphorylation results into synthesis of ATP and NADPH2
(D) Stroma lamellae lack PSII as well as NADP.
Options :
(1) C and D (2) B and D (3) A and B (4) B and C
Ans. (2)
70. Which one of the following lechniques is safest for the detection of cancers
(1) Histopathological studies (2) Magnetic resonance imaging (MRI)
(3) Radiography (X-ray) (4) Computed tomography (CT)
Ans. (2)
71. Select the answer with correct matching of the structure, its location and function.

Structure Loca tion Function

Near the place Rods and cones are
1 Blind spot W here optic nerve present but inactive
leaves the eye here

Equalizes air pressure

Anterior part of
2 Eustachian tube on either sides of
internal ear
tympanic membrane

Controls respiration
3 Cerebellum Mid brain
and ganieie
Controls Body
temperature, urge for
4 Hypothalamus Fore brain
eating and drinking

Ans. (4)

RESONANCE PAGE - 22
AIPMT (MAINS)-2010
72. Identify the components labelled A, B, C and D in the diagram below from the list (i) to (viii) given with.

Components :
(i) Cristae of mitochondria
(ii) Inner membrane of mitochondria
(iii) Cytoplasm
(iv) Smooth endoplasmic reticulum
(v) Rough endoplasmic reticulum
(vi) Mitochondrial matrix
(vii) Cell vacuole
(viii) Nucleus
The correct components are
A B C D
(1) (v) (i) (iii) (ii)
(2) (v) (iv) (viii) (iii)
(3) (i) (iv) (viii) (vi)
(4) (vi) (v) (iv) (vii)
Ans. (2)

73. Signals from fully developed foetus and placenta ultimately lead to parturition which requires the release of
(1) Ralaxin from placenta (2) Estrogen from placenta
(3) Oxytocin from maternal pituitary (4) Oxytocin from foetal pituitary
Ans. (3)

74. Consider the following four statements A, B, C and D select the right option for two correct statements.
Statements :
(A) In vexillary aestivation, the large posteior petal is called-standard, two lateral ones are wings and two
small anteior petals are termed keel.

(B) The floral formula for Litiaceae is P3+3 A3+3 G3

(C) In pea flower the stamens are monadelphous

(D) The floral formula for Solanaceae is K(5)C(3) G(2)

The Correct statements are

(1) (C) and (D) (2) (A) and (C) (3) (A) and (B) (4) (B) and (C)
Ans. (1)

RESONANCE PAGE - 23
AIPMT (MAINS)-2010
75. The most apparent change during the evloutionary history of Homo sapiens is traced in
(1) Remarkable increase in the brains size (2) Loss of body hair
(3) Walking upright (4) Shortening of the jaws
Ans. (1)

76. Transport of food material in higher plants takes place through

(1) Sieve elements (2) Companion cells (3) Transfusion cells (4) Tracheids
Ans. (1)

77. Crocodile and Penguin are similar to Whale and Dogfish in which one of the following fetures
(1) Have gill slits at some stage
(2) Possess a solid single stranded central nerovous system
(3) Lay eggs and guard them till they hatch
(4) Possess bony skeleton.
Ans. (1)

78. The Indian Rhinoceros is a natural inhabitant of which one of the Indian states
(1) Assam (2) Uttarakhand (3) Uttar Pradesh (4) Himachal Pradesh
Ans. (1)

79. In Antirrhinum two plants with pink flowers were hybridized. The F1 plants produced red, pink and white
flowers in the proportion of 1 red, 2 pink and 1 white. What could be the genotype of the two plants used for
hybridization? Red flower colour is determined by RR, and white by rr genes.
(1) rr (2) rrr (3) RR (4) Rr
Ans. (4)

80. Fastest distribution of some injectible material / medicine and with no risk of any kind can be achieved by
injecting it into the
(1) Lymph vessels (2) Muscles (3) Arteries (4) Veins
Ans. (4)

81. Which one of the following is most appropriately defined

(1) Parasite is an organism which always lives inside the body of other organism and may kill it.
(2) Host is an organism which provides food to another organism.
(3) Amensalism is a relationship in which one species is benefited where as the other is unaffected.
(4) Predator is an organism that catches and kills other organism for for food.
Ans. (4)

82. Kranz anatomy is one of the characteristics of the leaves of

(1) Mustard (2) Potato (3) Wheat (4) Sugarcane
Ans. (4)

83. The fruit fly Drosophila melanogaster was found to be very suitbable for experimental verification of chromo-
somal theory of inheritance by Morgan and his colleagues because
(1) It completes life cycle in about two weeks
(2) It reproduces parthenogenetically
(3) A single mating produces two young flies
(4) Smaller female is easily recognisable from larger male
Ans. (1)

RESONANCE PAGE - 24
AIPMT (MAINS)-2010
84. Leguminous plants are able to fix atmospheric nitrogen through the process of symbiotic nitrogen fixation.
Which one of the following statements is not correct during this process of nitrogen fixation
(1) Nitrogenase is insensitive to oxygen
(2) Leghaemoglobin scavenges oxygen and is pinkish in colour
(3) Nodules act as sites nitrogen fixation
(4) The enzyme nitrogenase catalyses the conversion of atmospheric N2 to NH3
Ans. (1)

85. Which one of the following can not be used for preparation of vaccines against plague
(1) Heat-Killed suspensions of virulent bacteria
(2) Formalin-inactivated suspensions of virulent bacteria
(3) Avirulent live baeteria
(4) Synthetic capsular polysaccharide material
Ans. (1)

86. One of the commonly used plant growth hormone in tea plantations is
(1) Indole-3-acetic acid (2) Ethylene (3) Abscisic acid (4) Zeatin
Ans. (1)

87. The haemoglobin content per 100 ml of blood of a normal healthy adult is
(1) 12 - 16 g (2) 5 - 11 g (3) 25 -30 g (4) 17 - 20 g
Ans. (1)

88. Black (stem) rust of wheat is caused by

(1) Xanthomonas oryzae (2) Alternaria solani (3) Ustilago nuda (4) Puccinia graminis
Ans. (4)

89. In human female the blastocyst

(1) Gets implanted in endometrium by the trophoblast cells
(2) Forms placenta even before implantation
(3) Gets implanted into uterus 3 days after ovulation
(4) gets nutrition from uterine endometrial secretion only after implantation
Ans. (1)

90. Which of the following are used in gene cloning

(1) Plamids (2) Nucleoids (3) Lomasomes (4) Mesosomes
Ans. (1)

91. ABO blood grouping is controlled by gene I which has three alleles and show co-dominance. There are six
genotypes. How many phenotypes in all are possible
(1) Five (2) Six (3) Three (4) Four

92. Given below are four statements (a-d) regarding human blood circulatory system
(a) Arteries are thick-walled and have narrow lumen as compared to veins
(b) Angina is acute chest pain when the blood ciruclation of the brain is reduced
(c) Persons with blood group AB can donate blood to any person with any blood group under ABO system
(d) Calcium ions play a very important role in blood clotting
Which two of the above statements are correct
(1) (c) and (d) (2) (a) and (d) (3) (a) and (b) (4) (b) and (c)
Ans. (2)

RESONANCE PAGE - 25
AIPMT (MAINS)-2010
93. Vegetative propagation in Pistia occurs by
(1) Sucker (2) Stolen (3) Offset (4) Runner
Ans. (3)

94. Which one of the following pairs of structure is correctly matched with their correct description
Structure s De scription
1 Premolars and molars – 20 in all and 3-rooted
2 Tibia and fibula – Both form parts of knee joint
3 Cartilage and cornea – No blood supply but do require
oxygen for respiratory need
4 Shoulder joint and elbow joint – Ball and socket type of joint
Ans. (3)

95. Secretions from which one of the following are rich in fructose, calcium and some enzymes
(1) Salivary glands (2) Male accessory glands
(3) Liver (4) Pancreas
Ans. (2)

96. When domestic sewage mixes with river water

(1) The river water is still suitable for drinking as impurities only about 0.1%
(2) Small animals like rats will die after drinking river water
(3) The increased microbial activity releases micro-nutrients such as iron
(4) The increased microbial activity uses up dissolved oxygen
Ans. (4)

97. Three of the following statements about enzymes are correct and one is wrong which one is wrong
(1) Most enzymes are proteins but some are lipids
(2) Enzymes require optimum pH for maximal activity
(3) Enzymes are denatured at high temperatures but in certain exceptional organisms they are effective even
at tempreatures 80°- 90°C
(4) Enzymes are highly specific
Ans. (1)

98. Root development is promoted by

(1) Ethylene (2) Abscisic acid (3) Auxin (4) Gibberellin
Ans. (3)

99. In which one of the following organisms its excretory organs are correctly stated
(1) Frog – Kidneys, skin and buccal epithelium
(2) Humans – Kidneys, sebaceous glands and tear glands
(3) Earthworm – Pharyngeal, integumentary and septal nephridia
(4) Cockroach – Malpighian tubules and enteric caeca
Ans. (3)

100. If for some reason the pariental cells of the gut epithelium become partially nonfunctional, what is likely to
happen.
(1) Proteins will not be adequately hydrolysed by pepsin into proteoses and peptones
(2) The pancreatic enzymes and specially the trypsin and lipase will not work efficiently

RESONANCE PAGE - 26
AIPMT (MAINS)-2010
(3) The pH of stomach will fall abruptly
(4) Steapsin will be more effective
Ans. (1)

101. Which one of the following is now being commercially produced by biotechnological procedures
(1) Insulin (2) Nicotine (3) Morphine (4) Quinine
Ans. (1)

102. The figure given below shows the conversion of a substrate into produced by an enzyme. In which one of the
four option (1-4) the components of reaction labelled as A, B, C and D are identitied correctly.

Options :
A B C D
Activation energy Activation energy
1 Transition state Potential energy
with enzyme without enzym e

Activation energy
2 Potential energy Transition state Activation energy without enzyme
with enzym e

Activation energy
3 Transition state Potential energy Activation energy with enzyme
without enzym e

Activation energy
4 Potential energy Transition state Activation energy without enzyme
with enzym e

Ans. (3)

103. An eleborate network of filamentous proteinaceous structures present in the cytoplasm which helps in the
maintenance of cell shape is called.
(1) Cytoskeleton (2) Thylakoid
(3) Endosplasmic Reticulum (4) Plasmalemma
Ans. (1)

104. Given below is the diagram of a stomatal apparatus. In which of the folloiwng all the four parts Labelled as A,
B, C and D are correctly identified.

RESONANCE PAGE - 27
AIPMT (MAINS)-2010

A B C D
1 Epidermal cell Subsidiary cell Stomatal aperture Guard cell
2 Subsidiary cell Epiderm al cell Guard cell Stomatal aperture
3 Guard cell Stomatal aperture Subsidiary cell Epidermal cell
4 Epidermal cell Guard cell Stomatal aperture Subsidiary cell

Ans. (1)

105. Which one of the following statements about the particular entity is true.
(1) DNA consists of a core of eight histones
(2) Centromere is found in animal cells, which produces aster during cell division
(3) The gene for producing insulin is present in every body cell
(4) Nucleosome is formed of nucleotides
Ans. (3)

106. Study the pathway given below

In which of the following options correct words for all the three blanks A, B and C are indicated.

A B C
1 Carboxylation Decarboxylation Reduction
2 Decarboxylaiton Reduction Regeneration
3 Fixation Transamination Regeneration
4 Fixation Decrboxylaiton Regeneration

Ans. (4)

107. Which one of the following is a xerophytic plant in which the stem is modified into a flat, green and succulent
structure
(1) Acacia (2) Opuntia (3) Casuarina (4) Hydrilla
Ans. (2)

RESONANCE PAGE - 28
AIPMT (MAINS)-2010
108. Examine the figures A, B, C and D. In which one of the four options all the items, A, B, C and D are correct

Options:

Opn. A B C D
1 Funaria Adinatum Salvinia Riccia
2 Chara Marchantia Fucus Pinus
3 Equisetum Ginkgo Selaginella Lycopodium
4 Selaginella Equisetum Salvinia Ginkgo

Ans. (4)

109. The lac operon consists of

(1) Three regulatory genes and three structural genes
(2) Four regulatory genes only
(3) One regulatory gene and three structural genes
(4) Two regulatory genes and two structural genes
Ans. (3)

110. Given below is the diagram of a bacteriophage. In which one of the options all the four parts A, B, C and D are
correct.

RESONANCE PAGE - 29
AIPMT (MAINS)-2010
Options :
A B C D
1 Collar Tail fibres Head Sheath
2 Tail fibres Head Sheath Collar
3 Sheath Collar Head Tail fibres
4 Head Sheath Collar Tail fibres
Ans. (4)

111. Which one of the following is the the correct description of a certain part of a normal human skeleton
(1) Glenoid cavity is a depression to which the thigh bone articulates
(2) Parietal bone and the temporal bone of the skull are joined by fibrous joint
(3) First vertebra is axis which articulates with the occipital condyles
(4) The 9th and 10th pairs of ribs are called the floating ribs
Ans. (2)

112. An example of endomycorrhiza is

(1) Rhizobium (2) Nostoc (3) Glomus (4) Agaricus
Ans. (3)

113. Aestivation of petals in the flower of cotton is correctly shown is

(1) (2) (3) (4)

Ans. (1)

114. Examine the figures (A-D) given below and select the right option out of 1-4, in which all the four structures
A, B, C and D are identified correctly.
Structures :

Options :

RESONANCE PAGE - 30
AIPMT (MAINS)-2010
A B C D
Megaspore
1 Sucker Seta Gemma cup
mother cell
2 Rhizome Sporangiophore Polar cell Globule
3 Runner Archegoniophore Synergid Antheridium
4 Offset Antheridiophore Antipodals Oogonium
Ans. (4)

115. Which of the following representations shows the pyramid of number in a forest ecosystem

(1) C (2) D (3) A (4) B

Ans. (4)

116. Study the pedigree chart of a certain family given below and select the correct conclusion which can be
drawn for the character.

(1) The male parent is homozygous dominant

(2) The female parent is heterozygous
(3) The parents could not have had a normal daughter for this character
(4) The trait under study could not be colour-blindness.
Ans. (4)
117. The correct floral formula of soyabean is
(1) % K(5)C1+2+(2) A1 + (9) G1 (2) % K(5)C1+(2)+2 A(9) + 1 G 1

(3) % K(5)C1+(2)+2 A(9) + 1 G1 (4) % K(5)C1+2+(2) A(9) + 1 G1

Ans. (1)
118. The 3'-5' phosphodiester linkages inside a polynucleotide chain serve to join
(1) One nitrogenous base with pentose sugar (2) One DNA strand with the other DNA strand
(3) One nucleoside with another nucleoside (4) One nucleotide with another nucleotide
Ans. (4)

119. In genetic engineering, a DNA segment (gene) of interest, is transferred to the host cell through a vector.
Consider the following four agents (A-D) in this regard and select the correct option about which one or more
of these can be used as a vector/vectors:
(A) a bacterium (B) plasmid (C) plasmodium (D) bacteriophage
Options :
(1) (B) and (D) only (2) (A), (B) and (D) only (3) (A) only (4) (A) and (C) only
Ans. (1)

RESONANCE PAGE - 31
AIPMT (MAINS)-2010
120. Study the cycle shown below and select the option which gives correct words for all the four blanks A, B, C
and D.

Options :

A B C D
1 Denitrification Nitrificaiton Plants Anim als
2 Nitrification Am m onification Anim als Plants
3 Denitrification Am m onification Plants Anim als
4 Nitrification Denitrification Anim als Plants

Ans. (3)

RESONANCE PAGE - 32