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MEDICAL
Mega Test - III
{+1- Medical}
General Instructions:The question paper contains 90 objective multiple choice questions.
There are three parts in the question paper consisting of Biology (1 to 30),
Physics (31 to 60) and Chemistry (61 to 90).
Each right answer carries (4 marks) and wrong (–1marks)
The paper consists of 90 questions. The maximum marks are 360.
Maximum Time 3Hrs.
Give your response in the Answer Sheet provided with the Question Paper.
Note: Actual pattern of AIPMT consists of 45 questions each from
Zoology, Botany, Physics & Chemistry.
Which will be applicable in our next Mega Test.
Name: _______________________________Father Name:______________________________
Mobile: ______________________________School Name:______________________________
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Section – A {Biology}
1.
Which pair is mismatched ?
(a) Pedology–Soil Science
(b) Kphysiology–Study of processes and functions
of life
(c) Kinesiology–Study of fossils
(d) Limnology–Study of freshwater ecosystems
Ans. (c)
2.
In moss, spore on germination :
(a) Forms a protonema
(b) Forms a protonema on which moss plan comes out like a plant
(c) Produces antheridia and archegonia
(d) Directly forms a leafy gametophyte
Ans. (b)
3.
Which of the following pairs correctly represents the grouping spermatopyta
according to one of the schemes of classifying palnts ?
(a) Pinus, Cycas (b) Ginkgo, Pisum
(c) Acacia, Casuarina
(d) Rhizopus,
Triticum
Ans. (b)
4.
Mosses and ferns are found in moist and shady places because both :
(a) Require presence of water for fertilization
(b) Do no speed sunlight for photosynthesis
(c) Depend for their nutrition on micro-organisms which can survive only at low
temperature
(d) Cannot compete with sun-loving plants
Ans. (a)
5.
Consider the following four statements A, B, C and D and select the right option for two
correct statements :
A. In vexillary aestivation, the large posterior petal is called standard, two lateral ones
are wings and two small anterior petals are termed keel
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B. The floral formula for Liliaceae is
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o P3 3 A3 3G 3
C. In pea flower, the stamens are monoadelphous
D. The floral formula for Solanaceae is
The correct statements are :
(a) A and C
(b) A and B
(c) B and C
(d) C and D
Ans. (b)
6.
Function of pilli is :
(a) Attachment to other organisms
(b) Form conjugation tube
(c) Both (a) and (b)
(d) None of the above
Ans. (c)
7.
What low molecular weight proteins are produced by a host cell in response to viral
infection which protects a pre-infected cell from a viral infection ?
(a) Antibodies
(b) Phytotoxins
(c) Interferons
(d)
Phytoalexins
Ans. (c)
8.
At which stage of HIV infection does one usually show symptoms of AIDS :
(a) Within 15 days of sexual contact with an infected person
(b) When the infecting retrovirus enters host cells
(c) When viral DNA is produced by reverse transcriptase
(d) When HIV replicates rapidly in helper T-lymphocytes and damages large number
of these
Ans. (d)
9.
A pothecial cups are found in :
(a) Saccharomyces
(b) Ascobolus
(c) Aspergillus
(d) All of
these
Ans. (b)
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10. The annulus of sporangium of the fern assists in the dispersal of spores by ¨
(a) Collapsing in the strong wind
(b) Losing the air of its cells
(c) Losing water from its cells
(d) Taking up water from air
Ans. (c)
11. Polyembryony occurs commonly in :
(a) Gymnosperms
(b) Angiosperms
(c) Both (a) and (b)
(d) None of these
Ans. (a)
12. Some plants have roots and rhizome both underground structures, which
characteristic of rhizome would distinguish them from roots ?
(a) Rhizome are thinner than roots
(b) Rhizome are darker in colour than roots
(c) Rhizome have scale leaves with buds in the axils
(d) Rhizome are thicker than roots
Ans. (c)
13. The tendrils of pea plants are modified :
(a) Stipules
(b) Axillary buds
(c) Aerial roots
(d) Terminal leaflets
Ans. (d)
14. The edible part in Apple is :
(a) Mesocarp
(b) Cotyledons
(c) Thalamus
(d) Endocarp
Ans. (c)
15. Where two carpels of a flower give rise to two fruits, it can be a :
(a) Siliqua
(b) Silicula
(c) Etaerio of follicle
(d) Capsule
Ans. (c)
16. The fruit of tomato can be described as :
(a) A berry, since the entire ovary wall ripens into fleshy fruit tissue
(b) A capsule, since it is derived from a compound pistil
(c) Aggregate, since it is derived from a pentacarpellary pistil
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(d) A pome, as the entire fruit wall is fleshy and the fruit has many seeds
Ans. (a)
17. Mesophyll is undifferentiated into palisade and spongy parenchyma in :
(a) Dorsiventral leaf
(b) Isobilateral leaf
(c) Both (a) and (b)
(d) None of these
Ans. (b)
18. A timber merchant told his customer that log of wood which he was purchasing comes
from a 20 years old tree, he told so by inspecting the :
(a) Diameter of log
(b) Thickness of the heart wood
(c) Number of cork layers
(d) Growth rings
Ans. (d)
19. Match the names of the structures listed under column-I with the functions given
under column-II, choose the answer which gives the correct combination of the
alphabets of the two columns :
Column-I (Structure)
Column-II(Function)
A
B
C
D
p
q
r
s
t
Stomata
Bark
Cambium
Hydathode
(a) A = t, B = r, C = p, D = s
Protection of stem
Plant movement
Secondary growth
Transpiration
Guttation
(b) A = p, B = s, C = t, D = r
(c) A = q, B = s, C = p, D = r
(d) A = s, B = p, C = r, D = t
Ans. (d)
20. In plant cells, dictyosomes are involved in production of :
(a) Nuclear membrane material
(b) Cell wall material
(c) Vacuolar membrane material
(d) All of the above
Ans. (b)
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21. The catalytically active enzyme cofactor complex is called the :
(a) Coenzyme
(b) Holoenzyhme
(c) Apoenzyme
(d) None of these
Ans. (b)
22. The different membrane systems of the cell which have a close structural and
functional relationship are:
(a) Golgi complex, ER and lysosomes
(b) Golgi complex, cytochrome and
lysosome
(c) Polysomes, nucleus and ribosomes
(d) None of these
Ans. (a)
23. Mitochondria are believed to be bacterial endosymbionts of cells because :
(a) They have their own nucleic acid
(b) Their membrane resembles those of
bacterium
(c) These do not arise de novo
(d) They have all of the above attributes
Ans. (d)
24. How many equational divisions are necessary in a cell of onion root tip to form 128
cells?
(a) 64
(b) 128
(c) 7
(d) none of these
Ans. (c)
25. At which stage of meiosis, the bivalents become far apart, chromosomes achieve
maximum condensation and nuclear membrane as well as nucleolus disappear ?
(a) Zygotene
(b) Diplotene
(c) Pachytene
(d) Diakinesis
Ans. (d)
26. The number of chromosome pairs at diakinesis in a meiocyte is 12, the chromosome
number expected in each nucleus after first and second cycle of division could
respectively be :
(a) 12 and 6
(b) 12 and 12
(c) 6 and 12
(d) 24 and 12
Ans. (b)
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27. Select the matched ones
A.
S phase
DNA replication
B.
Zygotene
Synapsis
C.
Diplotene
Crossing over
D.
Meiosis
Both haploid and diploid cells
E.
Gap 2 phase
Quiescent stage
(a)
A and B only
(b) C and D only
(c)
C and E only
(d) A, C and E only
Ans. (a)
In the questions given below, two statements-an Assertion (A) and a Reason (R) are
given. Give the appropriate response as :
(a) If both A and R are true and R is the correct explanation of A.
(b) If both A and R are true and R is not correct explanation of A.
(c) If A is true but R is false.
(d) If both A and R are false.
28. A : In collateral vascular bundles, phloem is situated towards the inner side.
R : In monocot stem, the cambium is present.
Ans. (d)
29. A:Eubacteria are found in three basic shapes : rod, round or spherical and spiral.
R : The Cocci and Bacilli may from clusters or chains of a length typical of a particular
bacterium.
Ans. (b)
30. A : Meiosis leads to recombinations of genes.
R: Crossing over occurs in pachytene sub-stage of prophase I of meiosis.
Ans. (a)
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Section B {Physics}
31. The dimensional formula of latent heat is :
(a) [M0L2T–2]
(b) [ML2T–2]
(c) [MLT–2]
(d) [ML2T–1]
Sol : (a)
32. The dimensional formula of
e4
(where
2 3
m
m
c
G
0 p e
0
= permittivity, c = speed of light, mp =
mass of proton, me = mass of electron, G = gravitational constant) is :
(a) [M]
(b) [T]
(c) [AT]
(d) [MA4T–2]
Sol : (b)
Coulomb’s electrostatic force F =
e2
Dimension of
2
0r
4
0
1 e2
4 0 r2
[Force F]
Newton’s gravitational force F =
Dimensional of
q1q2
r2
1
Gmpme
r2
Gm1m2
r2
= [force F]
2
Dimensions of
F2r2
=
Fmec3
Fr2
mec3
e4
2
2 3
0m p m e c G
e2
r2
2
0r
Gmpme
mec3
2
r
[MLT 2 ][L2 ]
[T]
[M][LT 1 ]
33. A boy performs an experiment in which he uses a simple pendulum to find the value of g
4 2l
using the formula g = 2 , where l
T
1m. Error in measurement of length is l , human
error in time measurement is 0.15s and least count of stop watch T , for what value of
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amplitude (A), l and T , error in calculation of g is minimum :
(a) A = 1.4 m,
l = 0.5m
(c) A = 0.5 m,
l = 0.5 m,
T = 0.1 s
(b) A = 0.4 m,
T = 0.1 s
(d) A = 0.05 m,
l = 0.1 m,
l = 0.1 m,
T = 0.05 s
T = 0.05 s
Sol : (d)
4 2l
The formula g = 2 is valid for SHM in which A should have leave value
T
g
g
l
l
2 T
T
g will be minimum when
l and
T have least values and it should also have a least
value of A.
34. A body starts from rest and moves with constant acceleration. The ratio of distance
covered by the body in nth second to that covered in n seconds is :
n2
(c)
2n 1
2n 1
(b)
n2
(a) 1 : n
(d)
2n 1
2n2
Sol : (c)
Distance traversed in n sec, s1
1 2
gn
2
Distance traversed in nth sec,
s2
1
g 2n 1
2
s1
s2
n2
.
2n 1
35. A stone is thrown upwards from the top of a tower with some initial speed and it
reaches the ground in t1 seconds. Now it is allowed to fall with the same initial speed
downward and it reaches the ground in t2 seconds. In how much time will the stone
reach the ground if it is allowed to fall freely under gravity from the same place ?
(a)
t1 t2
2
(b)
t1 t2
2
(c)
t 12 t 22
(d)
t 1t 2
Sol : (d)
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When stone is thrown up, u is negative.
So, h
1 2
gt 1
2
ut 1
..... 1
When stone is thrown downward
h
1 2
gt 2
2
ut 2
.. 2
When stone is released,
h
1 2
gt 3
2
..(3)
Multiplying (1) by t2 and (2) by t1
ht 2
ht 1
ut 1t 2
ut 1t 2
1 2
gt 1t 2
2
... 4
1 2
gt 2t 1
2
... 5
Adding, we get
h t1 t2
1
gt 1gt 2 t 1 t 2
2
i.e.,
1
gt 1t 2
2
h
... 6
Comparing (3) and (6), we get get
t 32
t 1t 2
t3
t 1t 2
36. A car starting from rest, accelerates at the rate ‘f’ through a distance s, then continues at
constant speed for time t and then decelerates at the rate
f
to come to rest. If the total
2
distance traversed is 15 s, then
(a) s
1 2
ft
2
(b) s
1 2
ft
4
(c) s
1 2
ft
6
(d) s
1 2
ft
72
Sol: (d)
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the velocity time graph for the given situation canbe drawn as below. Magnitude of
slope of OA = f
and slope of BC
v
f
2
f
t2
2
ft 1
t2
2t 1
In graph area of AOD gives distance,
s
1 2
ft 1
2
…(i)
Area of rectangle ABED gives distance travelled in time t.
s2
ft 1 t
Distance travelled in time t2,
1f
2t 1
22
s3
2
Thus, s1 s2 s3 15s
s
or
or
ft 1 t ft 12 15s
s
ft 1 t 2s 15s
s
ft 1 t 12s
1 2
ft 1
2
..(ii)
From Eqs. (i) and (ii), we have
12s
s
ft 1 t
1
ft t
2 1 1
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or
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t
6
t1
From Eq. (i), we get
s
1
f t1
2
s
1 t
f
2 6
2
2
1 2
ft
72
 


37. If a1 and a2 are non-collinear unit vectors and if a1 a2
3 ; then the value of
 
 
2a1 5a2 . 3a1 a2 is
(a)
41
2
(b)
Sol : (c)


a
a2
1 or a1
11
2
Squaring 1 1 2 cos

2a1
1
2


5 a 2 . 3 a1
11
2
(d) –
41
2
a2 1
Given a12 a22 2a1 a2 cos
or
(c) –
3
3
cos

a2
 
 
= 6 a1 . a1 2a1 . a2
= 6 2 cos
= 1 13
 
15 a2 . a1
15 cos
1
2
 
5a2 . a2
5 1 13 cos
11
2
38. A boat which has a speed a 13 km/h in still water crosses a river of width 1 km along
the shortest possible path in 12 minutes. The velocity of river water in km/h is.
(a) 12
(b) 10
(c) 8
(d) 6
Sol : (a)
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Shortest distance = AD = 1km
Time = 12 /60 =
Speed =
AE
1
hr
5
1
=5 km
1 /5
169 25
145
12
39. A shooter aims his rifle at an angle of 300 with the horizontal to hit an object at A but the
bullet hits at point B, y0 below A. If the initial velocity of the bullet is 600 m/s; the value
of y0 is (g = 10 m/s2)
(a) 1.85 m
(b) 18.5 m
(c) 185 m
(d) 1850 m
Sol : (a)
Time t taken by bullet to reach point P
t
x
ux
1000
4 cos 300
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1000
=
600
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10
sec
3 3
3
2
Vertical deflection of bullet in this time,
y0
1 2
gt
2
1
=
10
2
=
10
3 3
2
1
100
10
2
27
500
= 18.5 m
27
40. A projectile can have same range R for the angles of projections. If t1 and t2 be the times
of flight in the two cases, then the product of the two times of flights is proportional to
(a)
1
2
(b) R
(c) R2
(d)
1
R
Sol : (b)
R
u2 sin 2
g
For the same
t1
2 usin
g
1
2u sin
, t2
g
t 1t 2
t 1t 2
u cos
2
900
2u sin 90
g
4 u sin
u cos
g2
2u cos
g
u cos
2 2 u sin
.
g
g
2R
g
R
41. A person with his hands in his pockets is skating on ice at the rate of 10 m/s and
describes a circle of radius 50 m. What is his inclination of the vertical?
(a) tan–1 (1/2)
(b) tan–1 (1/5)
(c) tan–1(3/5)
(d) tan–1 (1/10)
Sol : (b)
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2
10
1
50 50 5
v2
rg
tan
tan
1
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1 /5
42. The roadway bridge over a canal is in the form of an arc of a circle of radius 20 m. What
is the minimum speed with which a car can cross the bridge without leaving contact
with the surface at the highest point ? (g = 9.8 m/s2)
(a)
10
m/s
7
(c) 14 2 m/s
(b) 14 m/s
(d) 14 5 m/s
Sol : (b)
V
rg
20 9.8
14 m / s
43. A block A can slide on a frictionless incline of angle
and length l inside an elevator
moving up with uniform acceleration a0 (see figure). Find the time taken by the block to
slide down the length l of the incline if it is released from the top of the incline ?
(a) t
2l
(g ao )sin
(c) t
2l
(g ao )sin
(b) t
(d) ) t
3l
(g ao )sin
3l
(g a o )cos
Sol : (a)
The block m is in non-internal frame. Therefore it experiences a fictitious force ma0 in
downward direction. The forces acting on the block are
(i) Weight mg vertically downward.
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(ii) fictitious force ma0 vertically downward
(iii) normal reaction N.
As block slides downward, the forces action along the slope downward are
F = mg sin
+ ma0 sin
If a is downward acceleration of block, then equation of motion of block along the
incline is
ma = mg sin
+ ma0 sin
a = (g + a0) sin
…(1)
The block remains in equilibrium normal to incline; the forces doing so are
N
mg cos
ma0 cos
…(2)
As block starts from rest and traverses a distance l with uniform acceleration a, we have
,
l
t
ut
1 2
at
2
0
1 2
at
2
2l
a
Using (1), we get
t
2l
g a0 sin
44. Two blocks of mass 2m and m are connected as shown in figure. Now the string
between the blocks is suddenly broken. The accelerations of blocks A and B respectively
at that instant are
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(a) g and g
(b) g and
g
2
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(c)
g
and g
2
(d)
g
g
and
2
2
Sol : (c)
After breaking bogy m is moving with free fall
mg ma
a g
when the system is in equilibrium, the spring force = 3mg. when the staring is cut, the
net force on block A = 3 mg – 2mg = mg
a
mg
2m
g /2
45. A player caught a cricket ball of mass 150 g moving at the rate of 20 m/s. If the catching
process is completed in 0.1 s, the force of blow exerted by the ball on the hand of the
player is equal to
(a) 150 N
(b) 3 N
(c) 30 N
(d) 300 N
Sol : (c)
F
p
t
mu
=
t
M V
t
150 10
0.1
3
20
= 30N
46. Two blocks A and B of equal masses are released from an inclined plane of inclination
450 at t = 0. Both the blocks are initially at rest. The coefficient of kinetic friction
between the block A and the inclined plane is 0.2, while it is 0.3 for block B. initially the
block A is 2 m behind the block B , Both the blocks will come in a line after A has
travelled a distance “K m “Down the plane. Then K is equal to (Take g = 10 m/s2)
(a) 8 2
(b) 6 2
(c) 8 3
(d) 6 3
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Sol : (a)
The forces action on block A are shown in fig.
New downward force on block A
= MA g sin 450
k
MA g cos 450
Acceleration of blocks A down the plane is
MA a A
1
2
MAg
g
1
2
aA
k
MAg
1
2
10
1 0.2
2
k
4 2 m / s2
Similarly acceleration of block B down the plane is
g
1
2
aB
10
1 0.3
2
k
7 2
m / s2
2
The front faces of A and B will come to the line, when
sA
sB
2
1 2 1 2
aAt
a Bt
2
2
2
1
2
7 2 2
t
2
4 2 t2
2t 2
sA
1
2
7 2
t 1
4
1
aA t2
2
t
2
2 second
1
4 2 2
2
2
8 2 m.
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47. A body is placed on a plane. The coefficient of friction between the body and the plane is
. The plane is gradually titled up. If
is the inclination of the plane, then the frictional
force on the body is
(a) constant throughout
(b) constant up to
tan
(c) increases up to
tan
(d) decreases up to
1
tan
and decreases after that
1
1
and constant after that
and constant after that
Sol : (c)
48. A block is placed on a rough surface. The coefficient of friction between the block and
the surface is
1
. What is the maximum value of force so that the block shown in
2 3
the arrangement, does not move?
(a) 10 N
(b) 12 N
(c) 15 N
(d) 20 N
Sol : None of these

K yi xj (where K is a positive constant)acts on a particle moving in the
49. A force F
X-Y plane. Starting from the origin, the particle is taken along the positive X-axis to the
point (a, 0) and then parallel to the Y-axis to the point (a, a). The total work done by the

force F on the particle is :
(a) 2Ka2
(b) 2Ka2
(c) Ka2
(d) Ka2
Sol : (c)
Along X-axis, y = 0
 
W1 F. s
K x j . dx i 0
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Parallel to y-axis (x = 0)
 
a
W2 F. s
K
yi aj . dy j
0
= Ka y
a
Ka2
0
50. A wind powered generator converts wind energy into electrical energy. Assume that the
generator converts a fixed fraction of wind energy intercepted by its blades into
electrical energy. For wind speed v, the electrical power will be proportional to
(a) v
(b) v2
(c) v3
(d) v4
Sol : (c)
Power P
= v2
dp
v
dt
Fv
v2
dm
dt
d
dv
(volume × density) = v 2p
dt
dt
= v2p A v (where A = cross-sectional area)
v3
51. A particle is acted on by a force of constant magnitude, which is always perpendicular to
the velocity of the particle; the motion of the particle takes place in a plane. It follows
that
(a) its velocity is constant
(b) its acceleration is constant
(c) its kinetic energy is constant
(d) it moves in a straight line
Sol : (c)
Change in K.E. = Work done by external force
= Fs cos 900
0
52. An elastic string of unscratched length L and force constant K is stretched by a small
length x. If is further stretched by another small length y. The work done in second
stretching is
(a)
1 2
Ky
2
(b)
1
K x2 y 2
2
(c)
1
K x y2
2
(d)
1
Ky 2x y
2
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Sol : (d)
1 2
Kx
2
Initial elastic energy U1
1
K x y
2
Final elastic U2
W
=
U U2 U1
2
1
K x y
2
1
K x2 y 2 2xy x2
2
2
1 2
Kx
2
1
y y 2x .
2
53. A ball impinges directly on another ball at rest and is brought to rest by the impact. If
half initial kinetic energy is lost in the impact, the coefficient of restitution is
(a)
1
4
(b)
1
3
(c)
3
4
(d)
1
2
Sol : (d)
v1
0
v2
e u1 u2 (Newton’s experimental formula)
v2
eu
v2 eu1
Also
…(1)
1 1
m1u12
2 2
1 p2
2 2m1
p2
2m2
m2
Also
2m1
m1u1
m2v2
m1u1
2m1v2
v2
1
m2v22
2
u1
2
Comparing (1) and (2), e =
..(2)
1
2
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54. Two blocks of masses 10 kg and 1 kg are connected by a spring of negligible mass and
placed on frictionless horizontal surfaces. An impulse gives a speed of 14 ms–1 to the
heavier block in the direction of lighter block. Then the velocity of the centre of mass is
(a) 30 ms–1
(b) 25 ms–1
(c) 10 ms–1
(d) 5 ms–1
Sol : (a,b,c,d)
There is no option in the given question (in order to minimize fluke and guessing
probability )
55. A uniform disc of radius R is spanned about its axis to an angular velocity
0
and then
carefully placed with its plane on a rough horizontal surface. Find the time for the disc
to stop rotating. Given the coefficient of friction between the disc and the surface =
.
Assume that the disc exerts pressure on the horizontal surface uniformly.
(a) t
4
3
R
g
0
(b) t
3
4
R
g
0
(c) t
1
2
g
R
0
(d) None of these
Sol : (b)
Initial kinetic energy of disc
1
I
2
2
0
.
Final kinetic energy of disc = 0. The kinetic energy is dissipated in doing work against
frictional couple. The value of this couple is different for different parts of disc. The disc
may be considered to be formed a large number of thin concentric rings. Consider one
such ring of radius x and thickness dx. If
Mass of ring, m
is the mass per unit area, then
2 x dx
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Frictional couple on this ring
d
mg x
2 x dx
gx
gx2dx
=2
2
g
R
0
x
2
dx 2
R3
g
3
m
R2
As
Frictional couple
R3 2
=
g
3 3
m
R2
2
mg R
mR 2
2
For disc I
Angular acceleration
2
3
I
mgR
mR 2
2
=
4 g
3 R
From relation
0
at,
0
we have
3
4
R
g
0
0
4 g
t
3 R
t
56. A small sphere rolls down without slipping from the top of a track in a vertical plane.
The track has an elevated section and a horizontal part. The horizontal part is 1.0 metre
above the ground level and the top of the track is 2.4 metre above the ground. Find the
distance on the ground with respect to the point B (which is vertically below the end of
the track as shown in Figure) where the sphere lands.
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(a) 3 m
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(b) 5m
(c) 2m
(d) 7m
Sol : (c)
The loss in potential energy = gain in total kinetic energy
Mg H h
1
1
. Mv 2
.I
2
2
For a sphere I
Mg H h
=
2
2
. MR 2 and
5
1
1
mv2
2
2
v
R
2
MR 2
5
v2
R2
7
mv 2
10
i.e., Speed of particle at A
10 H h g
7
v
Here H = 2.4m, h = 1m, g = 9.8 m/s2
10
v
2.4 1
7
9.8
19.6 m / s2
This is horizontal velocity at point A. The vertical component of velocity at point A is
zero.
Time taken to fall a distance 1 m is
t
2h
g
2 1
9.8
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Distance of the position of landing the ball on ground from B
s
vt
19.6
19.6 2
9.8
=
2
9.8
2m
57. A thin circular ring of mass M and radius R is rotating about its axis with a constant
angular velocity
. Two objects each of the mass m are attached gently to the opposite
ends of a diameter of the ring. The wheel now rotates with angular velocity.
(a)
M
M m
(b)
M 2m
M 2m
(c)
M
M 2m
(d)
M 2m
M
Sol : (c)
As
I1
0, J
I2
1
MR2
2
I
constan t
2
MR2 2mR2
2
M
.
M 2m
58. Consider a body consisting of two identical balls, each of mass M, connected by a light
rigid rod. If an impulse J = Mv is imparted to the body at one of its ends, what should be
its angular velocity ?
(a)
v
L
(b)
2v
L
(c)
v
3L
(d)
v
4L
Sol : (a)
Angular momentum L
Mv
L
2
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L
M
2
Moment of inertia I
L
I
MvL / 2
ML2 / 2
2
L
M
2
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2
ML2
2
V
L
59. A cord is wound round the circumference of a wheel of radius r. The axis of the wheel is
horizontal and its moment of inertia about this axis is I. A weight ‘mg’ is attached to the
end of the cord and is allowed to fall from rest. The angular velocity of the wheel, when
the weight has fallen through a distance h is
2gh
(a)
I mr2
1/2
2mgh
(b)
I mr2
1/2
2mgh
(c)
I 2mr2
1/2
(d) 2 gh
1/2
Sol : (b)
According to conservation of energy
mgh
1
1
mv 2
I
2
2
2
For rolling of wheel v
mgh
1
m r2
2
1
mgh
mr2 I
2
r
2
2
1
I
2
2
2mgh
I mr2
1/2
60. A disc of radius R has a mass 9 m. A hole of radius
R
is cut from it as shown in fig. The
3
moment of inertia of the remaining part about an axis passing through centre O of the
disc and perpendicular to the plane of disc is
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(b) 4mR 2
(a) 8mR2
(c)
40
mR 2
9
(d)
37
mR 2
9
Sol : (b)
R
3
R2
Mass of cut disc
Mass of total disc
Mass of cut disc =
M.I. of full disc, I1
9m
9
2
1
9
m
1
9m R2
2
M.I. of removed disc, I2
1
R
m
2
3
2
2R
m
3
2
1
mR2
2
M.I. of remaining disc
= I1 I2
1
1
9m R 2
mR 2 4mR 2
2
2
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Section – C {Chemistry}
61. If 6.3 g of NaHCO3 are added to 15.0 g CH3COOH solution, the residue is found to weigh
18.0. What is the mass of CO2 released in the reaction?
(a) 4.5 g
(b) 3.3 g
(c) 2.6 g
(d) 2.8 g
Sol:(b)
According to law of conservation of mass,
mass of reactant = mass of products
6.3 + 15.0 = 18.0 + x
or
x = 21.3 – 18.0 = 3.3g
62. Consider a titration of potassium dichromate solution with acidified Mohr’s salt solution
using diphenylamine as indicator. The number of moles of Mohr’s salt required per
mole of dichromate is
(a) 3
(b) 4
(c) 5
(d) 6
Sol :(d)
6Fe2+ + Cr2O 72 + 14 H+
6
Cr2 O72
6 Fe3+ 2Cr3+ + 7H2O
Cr3
x factor 6
2
Mohr’s salt FeSO4 . (NH4)2SO4 . 6H2O
Oxidation Fe2+
Fe3+
x factor = 1
Mole ratio is reverse of x-factor ratio
Thus, one mole of dichromate required 6 moles of Mohr’s salt
63. The maximum number of molecule is present in
(a) 15 L H2 gas at STP
(b) 5 L of N2 gas at STP
(c) 0.5 g of H2 gas
(d) 10 g of O2 gas
Sol: (a)
In 15 L of H2 gas at STP,
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The number of molecules =
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6.023 1023
15 4.033 1023
22.4
In 5 L of N2 gas at STP,
6.023 1023
the no. of molecules =
5 1.344 1023
22.4
In 0.5 g of H2 gas at STP,
6.023 1023 0.5
the no. of molecules =
1.505 1023
2
In 10 g of O2 gas at STP,
6.023 1023 10
the no. of molecules =
32
1.882 1023
Hence, maximum molecules are present in 15 L of H2 at STP
64. Which one of the following sets of ions represents a collection of isoelectronic species ?
(a) K+, Cl–, Ca2+, Sc3+ (b) Ba2+, Sr2+, K+, S2–
(c) N3–, O2–, F–, S2–
(d) Li+, Na+, Mg2+,
Ca2+
Sol: (a)
Isoelectronic means having same no. of electrons K+, Cl–, Ca2+, Sc3+(All are having 18
electrons)
65. The energy of second Bohr orbit of the hydrogen atom is –328 kJ mol–1; hence the
energy of fourth Bohr orbit would be
(a) –41 kJ mol–1
(b) –1312 kJ mol–1
(c) –164 kJ mol–1
(d) –82 kJ mol–1
Sol: (d)
The energy of second Bohr orbit of hydrogen atom (E2) is –328 kJ mol–1 because
E2
En
1312
kJmol
22
1312
kJmol
n2
1
1
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If n = 4
E4
1312
kJmol
42
82 kJmol 1
1
66. The orbital angular momentum of an electron revolving in a p-orbital is
(a) zero
(b)
h
2
(c)
h
2
(d)
h
2 2
Sol: (b)
Orbital angular momentum = l l 1
h
2
For p-orbital, l =1
Orbital angular momentum
11 1
h
2
2h
2
h
2
67. At the same conditions of pressure, volume and temperature, work done is maximum
for which gas if all gases have equal masses ?
(a) NH3
(b) N2
(c) Cl2
(d) H2S
Sol: (a)
When p, V and are same and mass is also same, work done only depends upon
molecular mass.
W
1
where M molecular mass,
M
Among the given gases, NH3 has lowest molecular mass so work done is maximum for it.
68. The most probable velocity (in cm/s) of hydrogen molecules at 270C will be
(a) 19.3
104
(b) 17.8
104
(c) 24.93
109
(d) 17.8
108
Sol: (a,b,c,d)
69. Which one of the following arrangements represents the correct order of electron gain
enthalpy (with negative sign) of the given atomic species ?
(a) Cl < F < S < O
(b) O < S < F < Cl
(c) S < O < Cl < F
(d) F < Cl < O < S
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Sol: (b)
The correct order of electron gain enthalpy is
O < S <F <Cl
Element
O
Electron gain enthalpy (in eV) 1.48
S
F
Cl
2.07
3.45
3.61
70. The electron affinity values (in kg mol–1) of three halogens X, Y and Z are respectively–
349, –333 and –325. Then X, Y and Z respectively, are
(a) F2, Cl2 and Br2 (b) Cl2, F2 and Br2
(c) Cl2, Br2 and F2
(d) Br2, Cl2 and F2
Sol: (b)
The electron affinity ( in kJ/mol)
Flourine = 332.6
Chlorine =348.5
Bromine =324.7
Iodine =295.5
Chlorine has the highest electron affinity value so the correct order of electron affinity
Cl > F2 > Br2
71. The first ionisation energy of oxygen is less than that of nitrogen. Which of the following
is the correct reason for this observation ?
(a) Lesser effective nuclear charge of oxygen than nitrogen
(b) Lesser atomic size of oxygen than nitrogen
(c) Greater interelectron repulsion between two electrons in the same p-orbital counter
balances the increase in effect nuclear charge on moving from nitrogen to oxygen
(d) Greater effective nuclear charge of oxygen than nitrogen
Sol: (c)
The electronic configuration of nitrogen is
7
N 1s2 ,2s2 ,2p3
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due to the presence of half filled p-orbitals a large amount of energy is required to
remove an electron from nitrogen
Hence, first ionization energy of nitrogen is greater than that of oxygen
The electronic configuration of oxygen is
8
O 1s2 ,2p2 ,2p4
The other reason for the greater IP of nitrogen is that in oxygen there is a greater inter
electronic repulsion b/w electron present in the same p-orbital which counter balance
the increase in effective nuclear charge from nitrogen to oxygen
72. Which substance has the greatest ionic character ?
(a) Cl2O
(b) NCl3
(c) PbCl3
(d) BaCl2
Sol: (d)
(i) Covalent character
1
sizeof cation
size of anion
(according to Fajan’s rule)
(ii) Lower the covalent character higher will be the ionic character
Cl2O, contains O2– , NCl3 contains N3–, PbCl2 contains Pb2+ and BaCl2 contains Ba2+
Hence, order of covalent character is NCl3 > Cl2O > PbCl2 > BaCl2
BaCl2 has the greatest ionic character
73. The sequence that correctly describes the relative bond strength pertaining to oxygen
molecule and its cation or anions is
(a) O22
O2
(c) O2
O2 O22
O2 O2
O2
O2
O22
O2 O2
O22
(b) O2 O2
(d) O2
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Sol: (d)
Bond order of oxygen molecule(O2) = 2
Bond order of oxygen molecule ion (O2+) =2.5
Bond order of super oxide ion(O2 ) 1.5
Bond order of peroxide ion(O22 ) 1
Hence, the order of bond strength is as: O2
74. How many
and
(a) 14 , 8
O2 O2
O22
-bonds are there in ?
(b) 18 , 8
(c) 19 , 4
(d) 14 , 2
Sol: (c)
The first bond b / w any two atom is
and rest are
draw detailed structure of
compound and find the total bond in it.
19 bons and
bonds.
75. Which one of the following compounds has the smallest bond angle in the its molecule ?
(a) SO2
(b) OH2
(c) SH2
(d) NH3
Sol: (c)
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Molecule
SO2
OH2
SH2
NH3
Hybridization
Repulsion
Sp2
Lp – lp, bp –bp
Sp3
Lp – lp, bp – lp
bp – bp
Sp2
–do–
3
Sp
Lp – bp, bp –bp
76. Which is the correct statement about
(i)
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and
Bond angle
1190
104.50
900
1070
molecular orbitals? Statements are
bonding orbitals are ungerade
(ii)
antibonding orbitals are ungerade
(iii)
antibonding orbitals are gerade
(a) (i) only
(b) (ii) and (iii) only
(c) (iii) only
(d) (ii) only
Sol: (a)
According to molecular orbital theory,
bonding orbital are ungerade
77. The molecular shapes of SF4, CF4 and XeF4 are
(a) different with 1, 0 and 2 lone pairs of electrons on the central atom, respectively
(b) different with 0, 1 and 2 long pairs of electrons on the central atom, respectively
(c) the same with 1, 1 and 1 lone pair of electrons on the central atoms, respectively
(d) the same with 2, 0 and 1 lone pairs of electrons on the central atom, respectively
Sol:(a)
Molecular Structure Hybridisation
Lone pair
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SF4
Sp3d
One
CF4
Sp3
Zero
XeF4
Sp3d2
Two
78. The number of electrons in the valence shell of sulphur in SF6 is
(a) 12
(b) 10
(c) 8
(d) 11
Sol:(a)
S has 6 electrons in its valence shell and it shares 6 electrons with 6 flourine atoms
in SF6, S has 12 electrons
79. In which reaction there will be increase in entropy ?
(a) Na s
H2O l
NaOH l
(c) H2 g
1
O2 g
2
H2O l
1
H2 g
2
(b) Ag aq
(d) Cu2 aq
Cl aq
4NH3 g
AgCl s
[Cu NH3 3 ]2 aq
Sol: (a)
1
H g
In reaction (a) Na(s) reacts with H2O(l) to give NaOH(l) and 2 2
.
Since, the entropy of a gas is higher then that liquid and the entropy of liquid is higher
than that of solid, therefore, the change in entropy for this reaction increases.
80. The enthalpy of formation of NH3 is –46 kJ mol–1. The enthalpy change for the reaction
2NH3 g
N2 g
3H2 g is
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(a) + 184 kJ
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(b) + 23 kJ
(c) + 92 kJ
(d) + 46 kJ
Sol: (c)
2NH3 g
Hr
N2 g
3H2 g
2 enthalpy of formation of NH3
2
81. C s
46
92kJ
CO2 g ; H
O2 g
2CO g
94 kcal
H = – 135.2 kcal
2CO2 g ;
O2 g
The heat of formation of CO(g) is
(a) – 26.4 kcal
(b) 41.2 kcal
(c) 26.4 kcal
(d) 292.2 kcal
Sol: (b)
The heat of formation of CO is calculated by using Hess’s law. According to it, the total
heat changes occurring during a chemical reaction are independent of path.
2CO g
O2 g
i Co2 g
ii
2CO2 g
CO2 g
C s
O2 g
H
CO2 g
H
135.2kcal
135.2
kcal
2
H
94kcal
Add Eqs. (i) and (ii)
C s
1
O2 g
2
CO g ;
H 41.2kcal

82. The value of log10 K for a reaction A 
 B is
(Given :
r
Ho298 k
(a) 5
54.07 kJ mol 1 ,
(b) 10
o
r 298 K
S
10 KJ–1 mol–1 and R = 8.314 JK–1 mol–1, )
(c) 95
(d) 100
Sol: (b)

A 
B
G0
H0 T S0
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G0
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2.303RT log k
H0 T S0
2.303RT
log k
54.07 103 298 10
10
2.303 8.314 298
83. For a reaction H
3 kJ , S
10 J / K beyond which temperature, this reaction
will be spontaneous ?
(a) 300 K
(b) 200 K
(c) 273 K
(d) 373 K
Sol: (a)
G
H T. S
For the spontaneous reaction the G must be negative
H
S
3kJ 3000J
10J / K
If T =300 K
G 3000 300 10 0
If T = 200 K
G 3000 200 10 1000J
If T = 273 K
G 3000 273 10 170J
If T = 373 K
G 3000 373 10
730J
Hence, beyond 300 K temperature the reaction will be spontaneous.
84. Ka for HCN is 5 10–10 at 250C. For maintaining a constant pH = 9, the volume of 5 M
KCN solution required to be added to 10 mL of 2 M HCN solution is
(a) 2 mL
(b) 4 mL
(c) 8.2 mL
(d) 6.4 mL
Sol: (a)
According to Henderson’s equation
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pH
9
10
log 5 10
log 5 10
10
log
log
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5 V 10 2
/
V 10 V 10
V
4
9 = –0.6989 + 10 + log V – log 4
9 = –0.6989 + 10 + logV –0.6020
log V = 9–8.699 = 0.301
V = antilog 0.301 = 2
On solving, V = 2mL

85. Ammonium carbonate decomposes as: NH2COONH4 s 
 2NH3 g
For the reaction, Kp = 2.9
CO2 g
10–5 atm3. If we start with 1 mole of the compound, the total
pressure at equilibrium would be
(a) 0.0766 atm
(b) 0.0582 atm
(c) 0.388 atm
(d) 0.0194 atm
Sol: (b)

NH2COONH4(g) 
 2NH3(g) + CO2(g)
At. Eq. if partial pressure of CO2= p
then that of NH3 = 2p
Kp
PNH3 PCO2
2p
2
p 4p3
2.9 10 5 or p3 0.725 10
or p 1.935 10
5
2
Hence, total pressure (p)=3×0.0194=5.81×10–2 = 0.0581 atm
86. For a concentrated solution of weak electrolyte AxBy of concentration ‘C’ the degree of
dissociation ' ' is given as
(a)
(c)
Keq /C x y
Keq / Cx
y 1
xxy y
1/ x y
(b)
K eqC / xy
(d)
Keq /Cxy
Sol: (c)
The weak electrolyte AxBy dissociate as follow
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y

A xBy 
 A
y
x
C
C1
where,
Bx
0
0
C
y
x
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C
deg ree of association
C = concentration
K eq
x
x
Ay
y
Bx
A x By
x
C
y
y
C
C 1
x x .Cx .
x
.y y .Cy .
C
x x .y y .
x y
.Cx
y
1
1
y 1
K eq
x y
x x .y y .Cx
y 1
1
x y
K eq
x x .y y .Cx
y 1
87. What volume of M/10 NaOH is to be added in 50 mL, M/10 acetic acid solution to get a
buffer solution having highest buffer capacity ?
(a) 50 mL
(b) 25 mL
(c) 10 mL
(d) 40 mL
Sol: (b)
Buffer capacity of a buffer solution is maximum when the concentration of the weak
acid and its salt or weak base and its salt are equal, i.e., for higher buffer capacity pH =
pK a .
For this salt [salt] = [acids]
Thus, 25 mL
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88. The number of hydroxyl ions in 10 cm3 of 0.2 M HCl solution is
(a) 5
10–14
(b) 3
109
(c) 3
1012
(d) 15
10–12
Sol: (c)
[H+] = 0.2 M (given, HCl= 0.2 m)
[H+][OH–] = 10–14
[OH–]
10 14
=
0.2
5 10
14
M
Concentration of hydroxyl ions in 0.2 M HCl
= 5 10
14
M
The number of moles of hydroxyl ions
5 10
14
1000
The number of OH– ion in 10 cm3
5 10
14
1000 6.023 1023
10
3 1012
89. According to hard and soft acid-base principle, a hard acid
(a) has low charge density
(b) shows preference for soft bases
(c) shows preference for donor atoms of low electronegativity
(d) is not polarisable
Sol: (d)
A hard acid has high charge density, shows preference for hard bases, i.e., for atoms of
high Electronegativity. Moreover, its outer electrons are not easily distorted, i.e., it nonpolarisable
90. The pKa of acetylsalicylic acid (aspirin) is 3.5. The pH of gastric juice in human stomach
is about 2-3 and the pH in the small intestine is about 8. Aspirin will be
(a) unionised in the small intestine and in the stomach
(b) completely ionised in the small intestine and in the stomach
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(c) ionised in the stomach and almost unionised in the small intestine
(d) ionised in the small intestine and almost unionised in the stomach
Sol: (d)
Aspirin is a weak acid. Due to common ion effect, it is unionized in acid
medium(stomach) but completely ionized alkaline medium (intestine).
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