‘s
Pioneer’s
Aspire Scholarship Exam
11th (Medical)
Solution
Examination Centre: Pioneer Education, Sector – 40-D
General Instructions:The question paper contains 90 objective multiple choice questions.
There are three Sections in the question paper consisting of Section – A BIOLOGY (1 to 30),
Section – B PHYSICS (31 to 60) and Section – C CHEMISTRY (61 to 90).
Each right answer carries 4 marks and wrong –1mark
The maximum marks are 360.
Maximum Time is 3Hrs.
Give your response in the Answer Sheet provided with the Question Paper.
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Section – A {Biology}
1.
A sudden increase in CO2 concentration around a leaf will cause:
(a) Wider opening of stomata
(b) Increase in transpiration
(c) Closure of stomata
(d) Decrease in transpiration due to closure of stomata
Ans. (d)
2.
Hydathodes are found in:
(a) Stem
(b) Leaves
(c) Roots
(d) All of these
(c) Xerophytes
(d) Halophytes
Ans. (b)
3.
Osmotic pressure is highest in
(a) Hydrophytes
(b) Mesophytes
Ans. (d)
4.
A person passes much urine and drinks much water but his blood glucose level is normal. This
condition may be the result of:
(a) a reduction in insulin secretion from pancreas
(b) a reduction in vasopressin secretion from posterior pituitary
(c) a fall in the glucose concentration in urine
(d) an increase in secretion of glucagon
Ans. (b)
5.
Both adrenaline and cortisol are secreted in response to stress. Which of the following statements is
also true for both of these hormones? :
(a) They act to increase blood glucose
(b) They are secreted by the adrenal cortex
(c) Their secretion is stimulated by adrenocorticotropin
(d) They are secreted into blood within seconds of the onset of stress.
Ans. (a)
6.
Thigmotropism is shown by :
(a) Lamina
(b) Tendrils
(c) Root apex
(d) Thorns
Ans. (b)
7.
Parthenocarpic fruits can be produced by application of auxin :
(a) IAA
(b) IBA
(c) NAA
(d) All of these
Ans. (d)
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8.
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Immediately after fat absorption, monoglycerides move into :
(a) lymph vessels
(b) blood vessels
(c) villi
(d) chylomicron
Ans. (a)
9.
Main role of minor elements is to act as :
(a) Constituent of hormones
(b) Binder of cell structure
(c) Cofactor of enzymes
(d) Constituent of amino acids
Ans. (c)
10. Plants require sulphur for :
(a) ATP synthesis
(b) Protein synthesis
(c) Glucose synthesis
(d) DNA replication
Ans. (b)
11.
Zinc is required for:
(a) Stomatal opening
(b) Stomatal closure
(c) Biosynthesis of lAA
(d) Oxidation of carbohydrates
Ans. (c)
12. Pons connects :
(a) Brain with spinal cord
(b) Cerebrum with cerebellum
(c) Two-lobes of cerebellum
(d) Two cerebral hemispheres
Ans. (c)
13. Mammalian brain is characterized from the brain of other animals through :
(a) Cerebellum
(b) Cerebrum
(c) Corpus callosum
(d) Optic lobes
Ans. (c)
14. In sugarcane, CO2 is fixed in malic acid with the help of enzyme :
(a) PEP carboxylase
(b) RuBP carboxylase
(c) Ribulose phosphate kinase
(d) Fructose phosphatase
Ans. (a)
15. Enzyme required in early CO2 fixation in C4 plant:
(a) RuBP carboxylase
(b) RuBP oxygenase
(c) PEP carboxylase
(d) PGA dehydrogenase
Ans. (c)
16. In photosynthesis photolysis of water is used in :
(a) reduction of NADP+
(b) oxidation of NADP
(c) oxidation of FAD
(d) none of these
Ans. (a)
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17. Blood maintains homeostasis in body by :
(a) Replenishment of O2 and nutrients, elimination of wastes.
(b) Replenishment of O2 and elimination of CO2
(c) Maintenance of sugar level, conversion of amino acids into urea and destruction of worn out RBCs
(d) Maintenance of ion concentration in blood and body fluids by eliminating nitrogenous wastes
Ans. (a)
18. In ventricular systole, oxygenated blood is pumped into:
(a) Pulmonary artery and deoxygenated into artery
(b) Aorta and deoxygenated into pulmonary vein
(c) Pulmonary vein and deoxygenated into pulmonary artery.
(d) Aorta and deoxygenated into pulmonary artery
Ans. (d)
19. Nucleated biconvex RBCs are found in:
(a) rat
(b) human
(c) dog
(d) frog
Ans. (d)
20. The last electron acceptor of ETC during oxidative phosphorylation is:
(a) Cyt b
(b) Cyt a3
(c) H2
(d) CO2
(c) Chloroplast
(d) Nucleus
(c) Oblique
(d) Normal
(c) arachnida
(d) earthworm
Ans. (b)
21. Site for glycolysis or EMP is:
(a) Cytoplasm
(b) Mitochondria
Ans. (a)
22. During expiration diaphragm becomes:
(a) Flattened
(b) Dome-shaped
Ans. (b)
23. Book lungs are respiratory organs of :
(a) mollusca
(b) mammals
Ans. (c)
24. One of the following is not a function of bones :
(a) place for muscle attachment
(b) protection of vital organs
(c) secretion of hormones for calcium regulation in blood and bones
(d) production of blood corpuscles
Ans. (c)
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25. Total number of bones in the hind limb of man is:
(a) 21
(b) 14
(c) 30
(d) 24
Ans. (b)
26. Haemodialysis is done in the condition when person is suffering from :
(a)Diabetes
(b) Uremia
(c) Anaemia
(d) Goitre
Ans. (c)
27. Renin is released from:
(a) Juxtaglomerular apparatus
(b) Cortical nephron
(c) Collecting duct
(d) Pelvis
Ans. (a)
28. Glucose is taken back from glomerular filtrate through:
(a) active transport
(b) passive transport
(c) osmosis
(d) diffusion
Ans. (a)
29. Emulsification is the function of:
(a) bile
(b) lipases
(c) esterases
(d) proteases
(c) cellulose
(d) sucrose
Ans. (a)
30. Glucose is stored in liver as :
(a) starch
(b) glycogen
Ans. (b)
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Section – B {Physics}
31. The volume V of water passing any point of a uniform tube during t seconds is related to the
cross- sectional area A of the tube and velocity u of water by the relation V
A u t which one of the
following will be tube?
(a)
(b)
(c)
(d)
Ans. (b)
Solution:
The dimensions of the two sides of proportionality are
L3 L2α (LT 1 )β Tγ
L2α βT γ
β
Equating the powers of dimensions on both sides, we have
2α β 3
γ β 0
which give β
γ and α
1
3 β , i.e. α
2
β
γ.
Thus, the correct choice is (b).
32. When a wave traverses a medium, the displacement of a particle located at x at time is given by
y asin bt cx where a, b and c are constants of the wave. The dimensions of b are the same as those
by
(a) wave velocity
(b) amplitude
(c) wavelength
(d) wave frequency
Ans. (d)
Solution:
Since the argument of a sine function (or any trigonometric function) must be dimensionless, bt and cx
are dimensionless. Since bt is dimensionless, the dimensions of b = dimensions of 1/t = T–1, which are
the dimensions of frequency. Hence, the correct choice is (d).
33. If velocity (V), force (F) and energy (E) are taken as fundamental units, then dimensional formula for
mass will be
(a) V 2F0E 1
(b) V0FE2
(c) VF 2E0
(d) V 2F0E
Ans. (d)
Solution:
Let m
V αFbEc
[M1 ] [LT 1 ]a [MLT 2 ]b [ML2T 2 ]c
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Equation the powers of M, L and T are solving, we get a = – 2, b = 0 an c = 1. Hence, the dimensional
formula of m = [V–2 F0 E] which is choice (d).
34. The acceleration a of a particle moving with an initial velocity u varies with distance x as a = k x
where k is a constant. The distance covered by the particle when its velocity becomes 3u is given by
2/3
3u2
(a)
k
(b) 3ku
2
4/3
6u2
(c)
k
2/3
u2
(d)
3k
2/3
Ans. (c)
Solution:
dν
dt
a
ν dν
dν dx
dx dt
adx
ν
dν
dx
k x dx
Integrating
3u
νdν
k
x dx
u
ν2
2
3u
3/2
kx
3/ 2
u
1
9u2 u2
2
4u2
x
2k 3/2
x
3
2k 3/2
x
3
6u2
k
2/3
35. The velocity v of a particle moving in a straight line varies with distance x as v = k x where k is a
positive constant. The distance x varies with time t as
(a) x
t
(b) x t
(c) x t 3/2
(d) x t 2
Ans. (d)
Solution:
ν k x
Acceleration
=ν
a
dν
dt
dν
dx
dx
dt
dν
dx
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d
k x
dx
=k x
k
x
2
= k x1/2
dν
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1/2
k2
2
k2
dt
2
Integrating
k2
dt
2
d ν
k2
t
2
ν
dx
dt
k2
t
2
dx
k2
t dt
2
Intergating
k2
t dt
2
dx
or
x
k 2t 2
2
x
t2
36. A police jeep moving at a constant speed v on a straight road is chasing a thief riding on motor-cycle.
The thief starts from the rest when the police jeep is at a distance x away and accelerates at a constant
rate a. The police will be able to catch the thief if
(a) v
(b) v
ax
2ax
(c) v 2 ax
(d) v
ax
2
Ans. (c)
Solution:
Suppose the thief is caught at a time t after the starting of the motorcycle. The distance travelled by the
motorcycle in this time t is
S
1 2
at
2
(i)
During this time, the jeep must travel a distance
S + x = νt
(ii)
Using (i) and (ii)
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1 2
at x νt
2
at 2 2νt 2x 0
The roots of this quadratic equation are
ν2 2ax
a
v
t
Since t must be real and positive, we must have
ν2
2ax
ν
2ax
37. The driver of a train moving at a speed v1 sights another train at a distance d ahead of him, moving in
the same direction with a slower speed v2. He immediately applies brakes to achieve a constant
retardation a. There will be no collision if d is greater than
(a)
v1 v 2
2
(b)
a
v 12 v 22
a
(c)
v1 v 2
2
(d)
2a
v 12 v 22
2a
Ans. (c)
Solution :
The two trains will not collide if the initial relative velocity u
ν1 ν2 reduces to zero relative
velocity ν 0 in a minimum distance S = dmin under a retardation (–a). Using ν2
0
ν1 ν2
2
ν1 ν2
dmin
u2 2a S, we have
2 a dmin
2
2a
So, the correct choice is (c).
38. Two bodies are projected horizontally in opposite directions with velocities u1 and u2 from the top of a
building. If their velocities become perpendicular to each other after falling through a distance h, then
(a) h
2
u1u2
g
(b) h
2u1u2
g
(c) h
2u1u2
2g
(d) h
u1u2
2g
Ans. (d)
Solution:
Let u1 be along the positive x-axis and u2 along the negative x-axis. Their velocities at any time t are
ν1
u1 i
gt j
ν2
u2 i
gt j
Let t be the time at which ν1 becomes perpendicular to ν2 , then
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ν1 . ν2
u1 i
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0
gt j .
u1u2
gt
t
1
u1u2
g
h
1 2
gt
2
2
u2 i
gt j
0
0
1/2
u1u2
1
g
2
g2
u1u2
2g
39. A body is projected with a speed u at an angle
velocity vector makes an angle
(a)
ucos
cos
(b)
with the horizontal. The speed of the body when its
with vertical is
ucos
sin
(c)
usin
cos
(d)
usin
sin
Ans. (b)
Solution:
Let the speed of the body be ν when its velocity vector makes an angle β with the vertical fig.
Horizontal component of u is u cos α and the horizontal component of ν is ν sin β . Since the horizontal
component of velocity remains constant,
u cos α
ν
ν sin β
ucos α
, which is choice (b)
sin β
40. A body is projected with kinetic energy K at an angle of 600 with the vertical. The kinetic energy of the
body when it is at the highest point on the trajectory is
(a) zero
(b)
3K
4
(c)
K
2
(d)
K
4
Ans. (b)
Solution:
Angle with the horizontal is θ 900 600 = 300. At the highest point, the horizontal component of
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velocity is u cos θ and the vertical component is zero.
Given
K'
1
mu2 K. At the highest point,
2
1
m u cos θ
2
= K cos2 300
2
1 2
mu cos2 θ
2
3K
4
41. A force F is applied horizontal to block A of mass m1 which is in contact with a block B of mass m2, as
shown in fig. If the surface are frictionless, the force exerted by A on B is given by
(a)
m1
F
m2
(b)
m2
F
m1
(c)
m1 F
(m2 m2 )
(d)
m2F
m2 m1
Ans. (d)
Solution:
The contact force exerted by A on B is F2 =
m2F
, which is choice (d).
m1 m2
42. A block of mass 10 kg is placed at a distance of 5 m from the rear end of a long trolley as shown in fig..
The coefficient of friction between the block and the surface below is 0.2. Starting from rest, the trolley
is given a uniform acceleration of 3 ms–2. At what distance from the starting point will the block fall off
the trolley ? Take g = 10 ms–2.
(a) 15 m
(b) 20 m
(c) 25 m
(d) 30 m
Ans. (a)
Solution:
Since the block is placed on the trolley, the acceleration of the block = acceleration of the trolley a = 3
ms–2. Therefore, the force acting on the block is
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F = ma = 10
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3 = 30 N
The weight mg of the block is balanced by the normal reaction R. As the trolley accelerates in the
forward direction, it exerts a reaction force F = 30 N on the block in the backward direction, as shown
in the figure. The force of friction will oppose this force and will act in a direction opposite to that of F.
The force of limiting friction f is given by
μ
or
f
R
f
mg
f μmg
0.2 10 10 20N
Thus, the block is acted upon by two forces-force F = 30N towards the right and frictional force f = 20 N
towards the left see fig. The net force on the block towards the right, i.e. towards the rear end of the
trolley is
F’ = F – f = 30 – 20 = 10 N.
Due to this force, the block experiences an acceleration towards the rear end which is given by
a'
F'
m
10
1 ms
10
2
Let t be the time taken for the block to fall from the rear end of the trolley. Clearly, the block has to
travel a distance S’ = 5 m to fall off the trolley. Since the trolley starts from rest, initial velocity u = 0.
Now t can be obtained from the relation
s ut
1 2
at
2
Putting s = 5m, u = 0 and a = a’ = 1 ms–2, we get t = 10 s.
The distance covered by the trolley in time t = 10 s is ( u 0)
s' ut
1 2
1
at 0
3 10 15m
2
2
Hence, the correct choice is (a).
43. A given object takes n times as much time to side down a 450rough incline as it takes to side down a
perfectly smooth 450 incline. The coefficient of kinetic friction between the object and the incline given
by
(a)
k
1 / 1 n2
(b)
k
1 1/ n2
(c)
k
1/ 1 n2
(d)
1 1 / n2
Ans. (b)
Solution :
The square of the time of slide is inversely proportional to the acceleration. The acceleration in the two
cases are
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g
a1 g sin 450
2
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and a2
= g sin 450 μk gcos 450
=
g
2
1 μk
or
t 22
t 12
n2
μk
1
a1
a2
1
1 μk
1
.
n2
Hence, the correct choice is (b).
44. Two blocks A and B are connected to each other by a string and a spring of a force constant k, as shown
in Fig., The string passes over a frictionless pulley as shown. The block B slides over the horizontal top
surface of a stationary block C and the block A slides along the vertical side of C both the same uniform
speed. The coefficient of friction between the surfaces of the block B and C is
. If the mass of block A
is m, what is the mass of block B?
(a)
m
(b)
m
(c)
m
(d) m
Ans. (b)
Solution:
Since the blocks slide at the same uniform speed, no net force acts on them. If M is the mass of block B,
then the tension in the string is T = μ M g. Also T = mg. Equating the two, we get μ M m or M
m
,
μ
which is choice (b).
45. A smooth inclined plane of angle of inclination 300 is placed on the floor of a compartment of a train
moving with a constant acceleration a. When a block is placed on the inclined plane, it does not slide
down or up the plane. The acceleration a must be
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(a) g
(b)
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g
2
(c)
g
(d)
2
g
3
Ans. (d)
Solution:
Refer to fig. The component of acceleration vector a along the plane is a cos θ . The component of
acceleration due to gravity g along the plane is g sin θ . The block will stay at rest if a cos θ = g sin θ
or a = g tan θ
Now θ 300. Therefore, a = g tan 300
g
3
. Hence the correct choice is (d).
46. A body is sliding down a rough inclined plane of angle of inclination
friction varies with distance x as
x
for which the coefficient of
kx , where k is a constant. Here x is the distance moved by the
body down the plane. The net force on the body will be zero at a distance x0 given by
(a)
tan
k
(b) k tan
(c)
cot
k
(d) k cot
Ans. (a)
Solution:
The net downward force on the body at a distance x is
f x
mg sin θ μmg cos θ
= mg sin θ μ cos θ
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= mg sin θ kx cos θ
f x
sin θ
0 at a value of x = x0 given by
kx0 cos 0
which gives
x0
tanθ
k
Thus, the correct choice is (a).
47. The displacement x of a particle of mass m moving in a straight line varies with time t as x = kt3/2 under
the action of a force F where k is a constant. The work done by the force is proportional to
(a) t
(c) t 3/2
(b)t
(d)t2
Ans. (b)
x kt 3/2
Velocity ν
dx
dt
dν
dt
Acceleration a
Force
F ma
Work done
3 1/2
kt
2
3
k
2
3km
t
4
W
(i)
1
t
2
1/2
1/2
3k
t
4
1/2
(ii)
(iii)
Fdx
1
3
From Eq (i) dx = k t 2 dt
2
(iv)
Using (ii) and (iv) in (iii) we have
=
W
3km
t
4
9k2m
dt
8
9k2m
t
8
Hence W
1/2
1
3 2
kt dt
2
t.
48. A uniform chain of mass M and Length L has a part l of its length hanging over the edge of the table. If
the friction between the chain and the edge is neglected, the word done to pull the length l on the table
is
(a)
Mgl2
L
(b)
Mgl2
2L
(c)
MgL2
l
(d)
MgL2
2l
Ans. (b)
Solution:
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Mass per unit length =
F mg
M
Ml
. Mass of part of length l is m =
. Force (weight) of this part is
L
L
Mlg
L
l
l
Mg
W Fdl
ldl
L 0
0
Work done
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Mgl2
2L
49. One end of an elastic spring of natural length L and spring constant k is fixed to a wall and the other
end is attached to a block of mass m lying on a horizontal frictionless table (fig.,). The block is moved to
a position A so that the spring is compressed to half its natural length and then the released. What is
the velocity of the block when it reaches position B which is at a distance
(a)
L k
4 m
(b)
L 2k
2 m
(c)
3L
from the wall
4
L 3k
4 m
(d)
L k
2 m
Ans. (c)
Solution:
At position A, the compression of the spring is
L
3L
. At position B, the compression is L
2
4
L
.
4
Therefore, loss potential energy as the block moves from A to B is given by
1 L
k
2 2
2
1 L
k
2 4
2
3kL2
32
From conservation of energy, loss in P.E. = gain in K.E., i.e.
3kL2
32
ν
1 2
mν
2
L 3k
4 m
50. A car of mass m starts from rest at time t =0 and is driven on straight horizontal road by the engine
which exerts a constant force F. If friction is negligible, the car acquires kinetic energy E at time t and
develops a power P. Which of the following is correct?
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t2
(b) E
(a) E t
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(c) P
t 3/2
(d) P
t2
Ans. (b, c)
Solution:
Since F is constant, acceleration a =
ν
u at 0 at at
F
is constant. At time t, the velocity of the car is
m
Ft
.
m
Kinetic energy E at time
t
1 2
mν
2
1
Ft
m
2
m
Since F is constant, E
Power P at time t = Fν
2
Ft
m
2
t2.
t2.
Ft
F
m
F2
t
m
Thus P t . So the correct choices are (b) and (c).
51. Two identical balls marked 2 and 3, in contact with each other and at rest on a horizontal frictionless
table, are hit head-on by another identical ball marked 1 moving initially with speed v as shown in fig..
What is observed, if collision is elastic?
(a) Ball 1 comes to rest and balls 2 and 3 roll out with speed
v
each.
2
(b) Balls 1 and 2 come to rest and ball 3 rolls out with speed v each.
(c) Balls 1, 2 and 3 roll out with speed
v
each.
3
(d) Balls 1, 2 and 3 come to rest.
Ans. (b)
Solution:
The system consists of three identical balls marked 1, 2 and 3. Let m be the mass of each ball. Before
the collision,
KE of the system = KE of 1 + KE of 2 + KE of 3
=
1 2
1 2
mν 0 0
mν
2
2
Case (a) After the collision,
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KE of the system = KE of 1 + KE of 2 + KE of 3
=0
=
2
1
ν
m
2
2
1
ν
m
2
2
2
1 2
mν
4
Case (b)
KE of the system = KE of 1 + KE of 2 + KE of 3
=0 0
1
mν2
2
1 2
mν
2
Case (c)
KE of the system = KE of 1 + KE of 2 + KE of 3
=
1
ν
m
2
3
=
1 2
mν
6
2
1
ν
m
2
3
2
1
ν
m
2
3
2
Case (d)
KE of the system = 0
Now, in an elastic collision, the kinetic energy of the system remains unchanged. Hence choice (b) is
the only possible result of the collision.
52. A body of mass m = 1 kg is dropped from a height h = 40 cm on a horizontal platform fixed to one end
of an elastic spring, the other being fixed to a base, as shown in fig.. As a result the spring is
compressed by an amount x = 10 cm. What is the force constant of the spring. Taken g = 10 ms–2.
(a) 600 Nm–1
(b) 800 Nm–1
(c) 1000Nm–1
(d) 1200 Nm–1
Ans. (c)
Solution:
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Since the platform is depressed by an amount x, the total work done on the spring is mg (h + x). This
work is stored in the spring in the form of potential energy
1 2
kx . Equating the two, we have
2
1 2
kx mg h x
2
or
k
2mg h x
x2
Given, h = 0.4 m, x = 0.1 m, m = 1 kg and g = 10 ms–2. Substituting these values, we get k = 1000 Nm–1.
Hence the correct choice is (c).
53. A particle, which is constrained to move along the x-axis, is subjected to a force in the same direction
which varies with the distance x of the particle from the origin as F(x) = –kx + ax3. Here k and a are
positive constants. For x 0 , the functional form of the potential energy U(x) of the particle is (see fig.)
Ans. (d)
Solution:
The potential energy of the particle is given by
U
or
U k
kx ax3 dx
Fdx
x2
2
a
x4
4
x2
2k ax2
4
(i)
From Eq. (i) it follows that U = 0 at two values of x which are x = 0 and x =
2k / a . Hence graphs (b)
and (c) are not possible. Also U is maximum or minimum at a value of x given by
d kx2
0
dx 2
dU
= 0, i.e.
dx
ax 4
4
= kx – ax3 = x(k – ax2)
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x
or
k /a .
At this value of x, U is maximum if
d2U
Now
dx2
At x =
d2U
dx2
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d
kx ax3
dx
d2 U
0,
dx2
k 3ax2 .
k /a ,
k 3a
k
a
2k ,
k 3k
which is negative.
Hence U is maximum at x =
k /a .
Hence graph (a) is also not possible. Also U is negative for x
2k / a . Therefore, the correct graph is
(d).
54. Two particles of masses m and 4m have linear momenta in the ratio 2 : 1. What is the ratio of their
kinetic energies?
(a) 2
(b) 2
(c) 4
(d) 16
Ans. (d)
Solution:
Given p1
m1 ν1
m 1 ν2
m1 ν1
2p and p2
m2 ν2
p , so that
2
The ratio of their kinetic energies is
KE
KE
1
2
1
m ν2
2 1 1
1
m2ν22
2
m2
But
KE
KE
1
2
2
m12ν12 m2
.
m22ν22 m1
4 m1 and
m1 ν1
m2 ν2
2. Therefore,
4 16
2
Hence the correct choice is (d).
55. Three particles of the same mass lie in the x – y plane. The (x, y) coordinates of their positions are
(1, 1), (2, 2) and (3, 3) respectively. The (x, y) coordinates of the centre of mass are
(a) (1, 2)
(b) (2, 2)
(c) (4, 2)
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(d) (6, 6)
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Ans. (b)
Solution:
The x and y co-ordinates of the centre of mass are
m1 x1 m2x2 m3x3
m1 m2 m3
x
=
1
x1 x2 x3
2
=
1
1 2 3
3
and
y
m1
m2
m3
2
1
y1 y 2 y3
3
1
1 2 3
3
2.
Hence the correct choice is (b).
56. A particle is moving in a circle of radius r under the action of a force F = αr2 which is directed towards
centre of the circle. Total mechanical energy (Kinetic energy + potential energy) of the particle is
(take potential energy = 0 for r = 0) :
(a)
1 3
αr
2
(b)
5 3
αr
6
(c)
4 3
αr
3
(d) αr3
Ans. (b)
Solution:
dU F.dr
r
αr3dr
U
0
mv 2
r
αr2
3
αr2
m2v2 mαr3
1 3
αr
2
2m KE
αr3 αr2
Total E
3
2
5 3
αr
3
57. A vector A is rotated by a small angle
θ radians
θ
1 to get a new vector B . In that case B
is :
(a) A
θ
(b) B θ
A
(c) A 1
θ2
2
(d) 0
Ans. (a)
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A
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Solution :
Arc length = Radius × Angle
B
A
A
θ
58. A uniform thin rod AB of length L has linear mass density μ x
the CM of the rod lies at a distance of
(a) a = 2b
a
bx
, where x is measure from A. If
L
7
L from A, then a and b are related as :
12
(b) 2a = b
(c) a = b
(d) 3a = 2b
Ans. (b)
Solution:
L
x cm
bx2
(ax
)dx
L
0
L
(a
0
7L
12
bx
)dx
L
a b
2 3
b
a
2
b =2a
59. Diameter of a steel ball is measured using a vernier calipers which has divisions of 0.1 cm on its main
side (MS) and 10 divisions of its vernier scale (VS) match 9 divisions on the main scale. Three such
measurements for a ball are given as :
S. No.
MS (cm)
VS divisions
1
0.5
8
2
0.5
4
3
0.5
6
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If the zero error is – 0.03 cm, then mean corrected diameter is :
(a) 0.53 cm
(b) 0.56 cm
(c) 0.59 cm
(d) 0.52 cm
Ans. (c)
Solution:
Least count = 0.01 cm
d1 = 0.5 + 8 × 0.01 + 0.03 = 0.61 cm
d2 = 0.5 + 4 × 0.01 + 0.03 = 0.57 cm
d3 = 0.5 + 6 × 0.01 + 0.03 = 0.59 cm
Mean diameter =
0.61 0.57 0.59
0.59 cm
2
60. A block of mass 0.1 kg is connected to a spring of unknown spring constant k. It is compressed to a
distance x from rest. After approaching half the distance
x
from equilibrium position, it hits another
2
block and comes to rest momentarily, while the other block moves with a velocity 3ms–1, the total
initial energy of the spring is
(a) 0.3 J
(b) 0.6 J
(c) 1.5 J
(d) 0.8 J
Ans. (b)
Solution :
Apply momentum conservation
0.1u + m(0) = 0.1(0) + m(3)
1
0.1u2
2
1
m 3
2
2
Solving u = 3
1 2
kx
2
1 x
K
2 2
2
1
0.1 32
2
3 2
Kx 0.9
4
1 2
Kx 0.6 J
2
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Section – C {Chemistry}
61. A 5 M solution of H2SO4 is diluted from 1 litre to 10 litres. What is the normality of the solution?
(a) 0.25 N
(b) 1 N
(c) 2N
(d) 7 N
Ans. (b)
Solution:
M1V1= M2V2
5 1 M2 10
M2 0.5
Normality = 0.5
2 = 1N.
62. If 0.5 mol of BaCl2 is mixed with 0.2 mol of Na3(PO4) the maximum number of Ba3(PO4)2 that can be
formed is
(a) 0.7
(b) 0.5
(c) 0.3
(d) 0.1
Ans. (d)
Solution:
3BaCl2 + 2Na3PO4
3 mole
Ba3(PO4)2 + 6NaCl
2 mole
1 mole
BaCl2 and Na3PO4 react in the molar ratio 3 : 2 respectively. The given amount is in the ratio 5 : 2, hence
Na3PO4 is limiting reagent.
Ba3(PO4)2 obtained = (1/2)
0.2 = 0.1 mole
63. At a given temperature, density of gas x is twice as that of has y and molar mass of x is one third of gas
y. The ratio of their pressure will be
(a)
Px
Py
1
4
(b)
Px
Py
(c)
4
Px
Py
6
(d)
Px
Py
1
6
Ans. (c)
Solution:
We know, PM = dRT
So, PxMx = dyRT
Dividing one by other
Px Mx
Py My
Px
Py
dx RT
d y RT
given dx
Mx
2dy and
1
My
3
6.
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64. Which of the following plots is incorrect?
T= constant
V
(a)
P= constant
(b)
V
P
V= constant
(c)P
T
(d)
PV
T
T
Ans. (a)
Solution:
V
1
at constant temperature.
P
65. Which of the following statements is not true?
(a) The pressure of a gas is due to collision of the gas molecular with the wall of the container (b) The
molecular velocity of any gas is proportional to the square root of the absolute temperature.
(c) The rate of diffusion of a gas is directly proportional to the density of the gas at constant pressure.
(d) Kinetic energy of an ideal has is directly proportional to the absolute temperature.
Ans. (c)
Solution:
Rate of diffusion of a gas is inversely proportional to the density of the gas at constant pressure.
66. The van der Waals equation reduces itself to the ideal gas equation at
(a) High pressure and low temperature
(b) Low pressure and low temperature
(c) Low pressure and high temperature
(d) High pressure alone.
Ans. (c)
Solution:
At low pressure and high temperature, the real gases tend to behave ideally.
67. The radius of which of the following orbits is same as that of the first Bohr’s orbit of hydrogen atom?
hydrogen atom is
(a) He+ (n=2)
(b) Li2+ (n=2)
(c) Li2+ (n=3)
(d) Be3+ (n=2)
Ans. (d)
Solution :
Applying, rn
As given,
r0 n2
where r0 is the radius of 1st Bohr orbit or ground state of H-atom
Z
r0 n2
Z
r0
i.e. n2 = Z which is possible only for Be3+
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For Be3+,
n = 2, Z = 4
n2 = Z = 4.
68. The energy of hydrogen atom in its ground state is -13.6 eV. The energy of the level corresponding to
the quantum number n=5 is
(a) –0.54 eV
(b) –5.40 eV
(c) –0.58 eV
(d) –2.72 eV
Ans. (a)
Solution :
En
E0
for H-atom
n2
13.6
52
Es
0.54eV.
69. The wavelength associated with a golf ball weighting 200 g and moving at a speed of 5 m/h is of the
order
(a) 10–10 m
(b) 10–20 m
(c) 10–30 m
(d) 10–40 m
Ans. (c)
Solution :
λ
h
mv
m = 200 g = 0.2 kg,
λ
v = 5 m/h =
6.6 10 34 3600
2.37 10
0.2 5
30
5
m/s
3600
m.
70. The electrons identified by quantum number n and l,
(i) n = 4, l = 1
(ii) n = 4, l = 0
(iii) n = 3, l = 2
(iv) n = 3, l = 1
can be placed in order of increasing energy, from the lowest to highest, as
(a) (iv) < (ii) < (iii) < (i)
(b) (ii) < (iv) < (i) < (iii)
(c) (i) < (iii) < (ii) < (iv)
(d) (iii) < (i) < (iv) < (ii)
Ans. (a)
Solution :
By applying (n + l) rule, energies of different electrons are in the order (iv) < (ii) < (iii) < (i).
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71. Uncertainty in position and momentum are equal, Uncertainty in velocity is
(a)
h
(b)
h
2
(c)
1 h
2m
(d) none of these
Ans. (c)
Solution:
h
but
4π
x. m v
m v
h
4π
2
x
m v (given).
v2
h
,
4πm2
1 h
.
2m π
v
72. Which of the following represents the given mode of hybridization sp2 – sp2 –sp – sp from left to right?
(a) H2C=CH–C N
(b) HC C – C H CH
2
(d) CH2
(c) H2C=C=C=CH2
Ans. (a)
Solution:
sp2
sp2
sp
sp
H2C CH C N.
73. Which of the following are isoelectronic and isostructural?
NO3 ,CO32 ,ClO3 ,SO3
(a) NO3 ,CO32
(c) ClO3 ,CO32
(b) SO3 ,NO3
(d) CO32 ,SO3
Ans. (a)
Solution:
Number of electrons in NO3 = 7 + 3 + 8
Number of electrons in CO32 = 6 + 3
Number of electrons in ClO3 = 17 + 3
Number of electrons in SO3 = 16 + 3
1 = 32
8 + 2 = 32
8 + 1 = 42
8 = 40
Thus, NO3 and CO32 are isoelectronic.
O
O-
O
C
N
O
NO3
O
O
CO32
Thus, NO3 and CO32 are isostructural also.
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74. According to molecular orbital theory which of the following statements about the magnetic character
and bond order is correct regarding O2 .
(a) Paramagnetic and bond order < O2
(b) Paramagnetic and bond order > O2
(c) Diamagnetic and bond order < O2
(d) Diamagnetic and bond order > O2
Ans. (b)
Solution :
O2 15 electrons :
KK σ2s2σ *2s2σ2p2z π2p2x
π2p2y π *2p1x
π *2p0y
one unpaired electron, hence paramagnetic.
B.O. =
1
Nb Na
2
8 3
2.5
2
but B.O. of O2 = 2, Thus B.O.O
2
B.O.O2
75. Which of the following reaction corresponds to the definition of enthalpy of formation?
(a) H2 g
1
O
2 2g
H2O l
(b) H2 g
1
O
2 2s
(c) H2 g
1
O
2 2l
H2O l
(d) H2 g
O2 g
H2O l
2H2O l
Ans. (a)
Solution:
Heat of formation is the heat change when 1 mole of a substance is formed from its constituent
elements in gaseous state.
76. Calculate H of the reaction,
H
H
C
Cl (g)
C(g) + 2H(g) + 2Cl(g)
Cl
The average bond energies of C–H bond and C – Cl bond are 416 kJ and 325 kJ mol–1 respectively.
(a) 1482 kJ
(b) –1482 kJ
(c) 1282 kJ
(d) None of these
Ans. (a)
Solution:
We know that
H
B.E. Reactant
B.E.
Product
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= [2 B.E. C
2 B.E. C
H
Cl
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]
= 2 416 2 325 1482 kJ.
77. Consider the reaction, N2(g) + 3H2(g)
2NH3(g) carried out at constant temperature and pressure. If H
and U are enthalpy and internal energy change for the reaction, which of the following expressions is
true?
(a) H 0
(b) H
U
(c) H
U
(d) H
U
Ans. (c)
Solution:
N2 g
3H2 g
H
U
ng
2
2NH3 g
ngRT
1 3
H
2
Hence H
U 2RT.
78. Kp/Kc for the reaction CO g
(a) RT
1
O
2 2g
(b) (RT)1/2
U.
CO2 g
(c) Kp
(d)
1
RT
Ans. (d)
Solution:
Kp
K c RT
For CO g
ng
Kp
1
ng
1
O
2 2g
1
1
2
K c RT
Κp
1
Kc
RT
CO2 g
1
2
1/2
.
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79. When pressure is applied to the equilibrium system
Ice
Water
which of the following phenomenon will happen?
(a) More ice will be formed
(b) Water will evaporate
(c) More water will be formed
(d) Equilibrium will not be formed.
Ans. (c)
Solution:
Increase of pressure favours the melting of ice which is associated with decrease of volume.
80. The conjugate base of OH– is
(a) O2
(b) H2O
(c) O–
(d) O2–
Ans. (d)
Solution:
O2
OH
acid
conjugate
base
H .
81. pH of solution produced when an aqueous solution of pH = 6 is mixed with an equal volume of an
aqueous solution of pH = 3 is about
(a) 3.3
(b) 4.3
(c) 4.0
(d) 4.5
Ans. (a)
Solution:
[H+] in solution having pH of 6 = 10–6 M
[H+] in solution having pH of 3 = 10–3 M
After mixing total [H+] =
pH
log [H ]
10
6
10
3
2
5.005 10
4
log (5.005 10 4 )
= 4 log 5.005 3.3
82. The solubility of A2X3 is ‘s’ mol dm–3. Its solubility product is
(a) 6 s4
(b) 64 s4
(c) 36 s5
(d) 108 s5
Ans. (d)
Solution:
Let A2X3
2A3 3X 2
Ksp [A3 ]2[X2 ]3
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Since solubility = y mol dm–3
K sp
2y
2
3y
2
108 y 5 .
83. In the standardization of Na2S2O3 using K2Cr2O7 by iodomentry, the equivalent weight of K2Cr2O7 is
(a) (Molecular weight)/2
(b) (Molecular weight)/6
(c) (Molecular weight)/3
(d) Same as molecular weight
Ans. (b)
Solution:
K2Cr2O7 acts as an oxidising agent in presence of dil. H2SO4.
Cr2O72
14H
2Cr3
6e
7H2O
Equivalent weight of K2Cr2O7 =
84. The reaction, 3ClO
aq
ClO3
Molecular weight
Number of electrons gained
aq
2Cl
aq
M
.
6
is an example of
(a) Oxidation reaction
(b) Reduction reaction
(c) Disproportion reaction
(d) Decomposition reaction
Ans. (c)
Solution :
1
3ClO
5
C lO3 2Cl
1
In the given reaction chlorine is getting oxidised ClO
5
1
ClO3 as well as reduced ClO
Cl
simultaneously. Therefore it is a disproportionation reaction.
85. The maximum number of hydrogen bonds in which water molecule can participate is
(a) 1
(b) 2
(c) 3
(d) 4
Ans. (d)
Solution :
A water molecule can participate in 4 hydrogen bond formation. One each by two H-atoms and one
each by two long pairs present on O-atom.
H-bond
H
H
O
H
H
O
H
H
O
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86. Which of the following statements is incorrect?
(a) H2O2 is a pale blue viscous liquid
(b) H2O2 can act as an oxidizing as well as a reducing agent.
(c) In H2O2 the two hydroxyl groups lie on the same plane.
(d) H2O2 has an ‘open-book’ structure.
Ans. (c)
Solution:
H2O2 has non-planar structure. Two hydroxyl groups lie in different planes.
87. The ionic mobility of alkali metal ions in aqueous solution in maximum for
(a) K
(b) Rb
(c) Li
(d) Na
Ans. (b)
Solution:
The alkali metal ion undergoes hydration in the aqueous solution. The degree of hydration, decreases
with ionic size as we go down the group. Since the mobility of ions is inversely proportional to the size
of the their hydrated ions, hence the increasing order of ionic mobility is
Li+ < Na+ < K+ < Rb+.
88. Thermal stability of alkaline earth metal carbonates decreases in the order
(a) BaCO3 > SrCO3 > CaCO3 > MgCO3
(b) BaCO3 > SrCO3 > MgCO3 > CaCO3
(c) CaCO3 > SrCO3 > MgCO3 >BaCO3
(d) MgCO3 > CaCO3 > SrCO3 > BaCO3
Ans. (a)
Solution:
Thermal stability of alkaline earth metal carbonates increases down the group. Thus order of stability
should be :
BaCO3 > SrCO3 > CaCO3 > MgCO3
89. Amongst LiCl, RbCl, BeCl2 and MgCl2 the compounds with greater and the least ionic character,
respectively are
(a) LiCl and RbCl
(b) RbCl and BeCl2
(c) RbCl and MgCl2
(d) MgCl2 and BeCl2
Ans. (b)
Solution:
Smaller cation with large magnitude of charge (Be2+) leads to more covalent character i.e., less ionic
character and large cation with small magnitude of charge (Rb+) leads to less covalent character i.e.,
more ionic character according to Fajan rules. Thus RbCl is most ionic and BeCl2 is least ionic.
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Pioneer Education The Best Way To Success
IIT – JEE /AIPMT/NTSE/Olympiad Classes
IUPAC name of
OH
CH3
(a) 3-Methylhexan-2-enol
(b) 4- Methylhexan-2-enol
(c) 4-Methyhpentan-3-enol
(d) 4-Methylpentan-2-enol
Ans. (d)
Solution :
3
5
4
1
OH
4-Methylpent-2-en-1-ol
2
CH3
–Good Luck
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