Medical College Admission Test (MCAT) which is accepted by medical schools in US, Australia, Canada and Carribean Islands for making admission decision.The test assesses candidates critical analysis and reasoning skills, biological and biochemical foundations of living systems, psychological, social and biological foundations of behaviour and chemical and physical foundations of biological systems.
In this collection there will be many samples to practice this test
It will be so important to Medical students who wants to prepare for this test
Explanation:
The gametophyte of a Pteridophyte such as fern is called as prothallus. It is a small, greenish heart-
shaped structure that produces male and female gametes. Thallus is the plant body of algae and
other lower organisms that is composed of filaments. The filamentous body of fungi is called as
hyphae. Conidia are the asexually produced fungal spores. Haustoria are the specialized absorbing
structures of parasitic plants.
Question: 13
Casuarina flowers exhibit
A. Porogamy
B. Chalazogamy
C. Apogamy
D. Mesogamy
E. Acrogamy
Correct Answer: B. Chalazogamy
Explanation:
Chalazogamy is a type of fertilization in which the pollen tube penetrates into the embryo sac through
the chalaza instead of entering through the micropyle (porogamy). Chalazogamy can be seen in
flowers of casuarina.
Question: 14
Which of the following mechanisms uses ATP molecule during transport across cell
membrane?
A. Diffusion
B. Passive transport
C. Osmosis
D. Active transport
E. All the above
Correct Answer: D. Active transport
Explanation:
Pumping of molecules or ions through a membrane against their concentration gradient is called
active transport. This requires a transmembrane protein called transporter and energy in the form of
ATP. In diffusion, passive transport and osmosis the movement of molecules occurs down its
concentration gradient. Osmosis and diffusion are types of passive transport.
Question: 15
Phenylketonuria is caused by a mutation in the gene coding for
A. Pyruvate carboxylase
B. Phenylalanine hydroxylase
C. Homogentisate dioxygenase
D. Glutaryl dehydrogenase
E. None of these
Correct Answer: B. Phenylalanine hydroxylase
Explanation:
Absence of phenylalanine hydroxylase enzyme causes phenylketonuria. Phenylalanine hydroxylase
(PAH) is a hepatic enzyme needed for phenylalanine metabolism. In the absence of PAH,
phenylalanine accumulates and is converted to phenyl pyruvate (also called as phenyl ketone).
Absence of homogentisate dioxygenase causes alkaptonuria.
Sample paper 4
Question: 1
Which of the following statements is false with respect to class Monocotyledoneae?
A. Grass is an example of class Monocotyledonae.
B. Embryo has one cotyledon
C. Monocotyledons have a single cotyledon
D. Vascular cambium usually absent
E. Monocotyledons generally have reticulate venation
Correct Answer: E. Monocotyledons generally have reticulate venation
Explanation:
Monocotyledons generally have parallel venation. In parallel venation, the veins are present parallel to
each other. Reticulate venation is a characteristic of dicot plant where the veins emerge from different
parts of the main vein.
Question: 2
Heart develops from
A. Lateral plate mesoderm
B. Intermediate mesoderm
C. Ectoderm
D. Endoderm
E. Paraxial mesoderm
Correct Answer: A. Lateral plate mesoderm
Explanation:
Lateral plate mesoderm is found at the periphery of the embryo. The circulatory system develops from
the lateral plate mesoderm. Intermediate mesoderm develops into kidneys and gonads. Cartilage,
skeletal muscles etc. develop from paraxial mesoderm.
Question: 3
Which of the following is/are the characteristics of euglenoids?
A. Autotrophic
B. Heterotrophic
C. Presence of pellicle
D. Presence of flagella
E. All the above
Correct Answer: E. All the above
Explanation:
Euglenoids are the protists which come under the group flagellates. They live in both fresh water and
marine environment. They contain pellicle, a protein rich membrane which helps them in flexibility.
They are mixotrophic which means that they are photosynthetic in the presence of sunlight and
heterotrophic when they lack sunlight.
Question: 4
Which of the following reactions block polyspermy?
A. Capacitation
B. Spermatogenesis
C. Acrosome reaction
D. Cortical reaction
E. Fertilization
Correct Answer: D. Cortical reaction
Explanation:
Cortical reaction can be defined as the reaction of egg cell to fertilization which results in the change
of cell membrane. The cortical reaction is exocytosis of eggs’ cortical granules. The calcium signal
activates the egg to undergo cortical reaction. Cortical reaction prevents other sperms from entering
the egg.
Question: 5
Which of the following develops from ectoderm?
A. Liver
B. Pancreas
C. Kidney
D. Stomach
E. Brain
Correct Answer: E. Brain
Explanation:
Liver, pancreas and stomach develop from endoderm. Kidney develops from intermediate mesoderm.
Brain develops from the neural tube of the ectoderm. Ectoderm generates the outer layer of the
embryo. The ectoderm develops into the surface ectoderm, neural crest and neural tube.
Question: 6
To which of the following do ES cells belong?
A. Pluripotent
B. Oligopotent
C. Multipotent
D. Totipotent
E. None of the above
Correct Answer: A. Pluripotent
Explanation:
ES cells or embryonic stem cells can develop into any of the cells belonging to the three germ layers.
So they are pluripotent stem cells. ES cells will not develop extraembryonic cells. The differentiation
of ES cells to different cells depends on specific factors.
Question: 7
Presence of metameric segmentation is the characteristic of phylum
A. Protista
B. Porifera
C. Cnidaria
D. Annelida
E. Platyhelminthes
Correct Answer: D. Annelida
Explanation:
Metameric segmentation is the division of body into a number of similar segments. Metameric
segmentation is generally exhibited by phylum Annelida and phylum Arthropoda.
Question: 8
A process by which an egg undergoes a series of rapid cell divisions without cell growth or
gene expression is called
A. Apoptosis
B. Acrosome reaction
C. Cortical reaction
D. Cleavage
E. None of the above
Correct Answer: D. Cleavage
Explanation:
Cleavage is a series of rapid cell divisions without cell growth or gene expression. The organisms are
formed by growth and development of these cells. The different cells formed from cleavage are called
as blastomere. The compact mass of blastomere is called morula.
Question: 9
Which of the following is mixotropic?
A. Yeast
B. Dinoflagellates
C. Amoeba
D. Mushroom
E. Blue- green algae
Correct Answer: B. Dinoflagellates
Explanation:
A mixotroph is an organism that can utilize different forms of energy. Dinoflagellates are flagellated
mixotropic protists. They combine photosynthesis with ingestion of prey. Yeast, amoeba and
mushroom are heterotrophic. Blue- green algae is autotrophic.
Question: 10
Lung is a derivative of ________ germ layer.
A. Surface ectodermal
B. Neuroectodermal
C. Mesodermal
D. Endodermal
E. Both A and C
Correct Answer: D. Endodermal
Explanation:
During embryo development, the germ layers ectoderm, endoderm and mesoderm are differentiated
from blastula in a process called gastrulation. Later organogenesis takes place from those germ
layers. Lung is one of the foregut derivatives of endoderm.
Question: 11
The passage of materials between the nucleus and the cytosol takes place through
A. Vesicle
B. Nuclear pore
C. Nucleosome
D. Cajal body
E. Nuclear speckle
Correct Answer: B. Nuclear pore
Explanation:
Nuclear pores are small perforations on the nuclear envelope that allow ransportation of substances
across the nucleus and cytoplasm. Cajal body and nuclear speckle are small subnuclear organelles.
Nucleosome is a chromosomal structure containing histones in DNA. Vesicles present in cytoplasm
transport substances from one part of a cell to another part.
Question: 12
The state of a neuron which inhibits the generation of action potential is said to be
A. Depolarized
B. Repolarized
C. Hyperpolarized
D. Resting potential
E. A, B and D
Correct Answer: C. Hyperpolarized.
Explanation:
Resting potential is the state of a neuron which is not under any stimulus. The normal resting potential
of human nerves is -70 mV. When a neuron receives stimulus, the ce
ll’s interior will be depolarized
with influx of Na+ ions which will result in action potential. Immediately after the generation of action
potential the cell is hyperpolarized; that makes the cell more negative and keeps the neuron in a
refractory period so that the next action potential cannot generate. Next is the repolarization i.e.
neuron will come to its original resting state.
Question: 13
The word ‘arthropod’ means
A. Jointed feet
B. Jointed head
C. Presence of eight legs
D. Segmented body
E. Bilateral symmetry
Correct Answer: A. Jointed feet
Explanation:
The word arthropod in Greek means jointed feet. Presence of jointed feet is the characteristic feature
of arthropods.
Question: 14
Which of the following is included in aphotic zone?
A. Bathypelagic
B. Abyssopelgic
C. Mesopelagic
D. Hadopelagic
E. All of the above
Correct Answer: E. All of the above
Explanation:
All the above mentioned zones are devoid or have little light and are called aphotic zones. Aphotic
zone is the epipelagic zone. Very few organisms live in the hadopelagic (deepest) zone.
Question: 15
Engrailed proteins are
A. Tethering proteins
B. Transcription factors
C. Ubiquitins
D. Signaling molecules
E. Carrier proteins
Correct Answer: B. Transcription factors
Explanation:
Engrailed proteins are transcription factors which play a major role in brain development in many
species. They determine the midbrain/hindbrain border and aid in neuronal axon guidance.
Sample paper 1
Question 1
What is the dimensional formula of torque?
A. MLT
-2
B. MT
-2
C. ML
2
T
-2
D. MLT
-1
E. ML
3
T
-2
Correct Answer: C
Explanation:
Torque is the turning effect of force applied on a body. It is given as
t = r x F
Where
t is the torque vector;
r is the displacement vector;
F is the force vector;
Since dimensional formula of force is MLT
-2
and that of displacement is L hence, dimensional formula
of torque becomes
MLT
-2
* L
ML
2
T
-2
Therefore, C is the correct answer option.
Question 2
A car moving at 25 m/s suddenly stops by applying brakes on observing a stoplight. If the time
taken during the process of stopping is 1.5 s, what is the displacement of the car?
A. 17.55 m
B. 18.75 m
C. 16.25 m
D. 15.35 m
E. 17.75 m
Correct Answer: B
Explanation:
This problem can be solved by applying the kinematic equation of motion as shown below
v = u + at ------------ (1)
Where
v is the final velocity of the car;
u is the initial velocity of the car = 25 m/s;
a is the acceleration the car is undergoing;
t is the time under consideration = 1.5 s;
Since the car finally stops, hence v = 0 m/s. Putting the values in equation 1 we get
0 = 25 + a *1.5
Or, -25 = 1.5a --------------- (2)
Negative sign is coming because the car finally comes to rest and it is undergoing negative
acceleration. We can neglect the negative sign here. Therefore, equation 2 becomes
a = 25/1.5
Or, a = 16.67 m/s
2
Since, we know initial and final velocities and acceleration as well we can use kinematic equation
v
2
= u
2
- 2as -------------- (3)
to solve for displacement.
Where
S is the displacement of the car; We have taken the negative sign for acceleration here due to the fact
that acceleration is negative. Putting the values of v, u, a in equation 3 we get
0 = 25
2
- 2 × 16.67 × S
Or, 25
2
= 2 × 16.67 × S
Or, S = 25
2
/(2*16.67)
Or, S = 625/33.34
Or, S = 18.75m
Hence, B is the correct answer option.
Question 3
Net force acting on a body moving with constant velocity is zero. Which of the following is
incorrect?
A. Body will slow down
B. Body will move faster
C. Body will stop
D. Body will continue moving with same velocity
E. Body will accelerate
Correct Answer: D
Explanation:
A body moving with constant velocity or at rest is the condition of static equilibrium. For static
equilibrium, we need balanced force that means all the forces acting on the body balance each other
and in this way the net force acting on the body becomes zero. Therefore, only option D is correct
amongst the first four the options given above. Body cannot accelerate, as acceleration is time rate of
change of velocity. If velocity is remaining constant, the acceleration will become zero. Therefore, E is
an incorrect option.
Question 4
Two blocks of masses M and 4 M are moving under the effect of the same force, F on a
frictionless table as shown below. What would be acceleration of the smaller mass? (Assume
smaller mass is at rest with respect to the larger mass)
A. F/4M
B. F/M
C. F/5M
D. F/3M
E. F/2M
Correct Answer: C
Explanation:
Since the table is frictionless, hence both the bodies will continue moving under the common force F
and will have same acceleration. Since it is brought out in the question that the smaller block is not
moving over the larger block hence, both the blocks will move as a combined body.
The situation above is shown below.
As both the bodies are moving together, hence, their combined mass becomes
M
total
= M + 4M
Or, M
total
= 5M
Common force acting on the bodies is F. Therefore, by Newton’s second law of motion we have
F = ma ------------- (1)
Where
F is the force applied;
m is the mass of the body;
a is the acceleration of both the blocks;
Mass of the body here is M
total
= 5M. Putting the values in equation 1 we get
F = 5M * a
Or, F/5M = a
Hence, with this acceleration both the smaller and the larger mass will move under the effect of
common force, F.
Therefore, C is the correct answer option.
Question 5
What would be the moment of inertia of a thin-walled fluorescent lamp having radius of 3 cm
and mass of 50 grams?
A. 4.5 × 10
-5
kgm
2
B. 3.5 × 10
-5
kgm
2
C. 3.5 × 10
-6
kgm
2
D. 4.5 × 10
-6
kgm
2
E. 4 × 10
6
kgm
2
Correct Answer: A
Explanation:
Moment of inertia of thin-walled fluorescent lamp is given as
I = MR
2
---------------- (1)
Where
I is the moment of inertia;
M is the mass of the lamp = 50 g;
R is the radius of the lamp = 3 cm;
We have been given mass in grams so, we need to convert it into kg.
1 g = 10
-3
kg
50g = 50 * 10
-3
kg
50g = 0.05 kg
Also, radius is given in cm so we need to convert it into m.
1 cm = 10
-2
m
3 cm = 3 * 10
-2
m
3 cm = 0.03 m
Putting the values of M and R in equation 1 we get
I = 0.05 * 0.03
2
Or, I = 4.5 * 10
-5
kg m
2
Therefore, A is the correct answer option.
Question 6
Consider the figure given below. B is the nodal plane or ground. A body of mass M is kept on a
horizontal plane A at a certain height above the plane B. Plane C is at same height from the
ground (B) as that of plane A. Which of the following statements is correct?
Potential energy of the body is non-zero with respect to the plane A
Potential energy of the body is non-zero with respect to the plane C
Potential energy of the body is non-zero with respect to the plane B
A. 1
B. 2
C. 3
D. Both 1 and 2
E. Both 2 and 3
Correct Answer: C
Explanation:
Potential energy is the energy stored in a body by virtue of its height from the reference plane.
Potential energy is given as
PE = mgh --------------- (1)
Where
m is the mass of the body;
g is the acceleration due to gravity;
h is the height of the body above the reference plane;
The reference plane plays a vital role in defining the potential energy of a body. Most often reference
plane is taken as the ground and hence plane B is the reference plane here. We will discuss potential
energy of the body for all the planes individually.
For plane A, the height of the body is zero as it is kept on the same plane A. Hence, putting the value
of h in equation 1 we get
PE = 0
Therefore, potential energy of the body is zero with respect to the plane A. Thus, A is an incorrect
option.
Both the planes A and C are at same height above the reference plane. Therefore, potential energy of
the body with respect to the plane C is same as that in respect of A. Hence, B is incorrect as well.
Since the plane A is at some height from the reference plane, h will have a non-zero value. Therefore,
potential energy of the body with respect to the ground plane will have some non-zero value. Thus, C
is the correct answer option. In the view of the above, D and E are also incorrect options.
Question 7
Displacement of a particle in SHM is given as
x(t) = 23 sin(314t) cm
What is the amplitude and frequency of oscillation of the particle?
A. 2.6 cm, 40 Hz
B. 2.3 cm, 40 Hz
C. 2.3 cm, 50 Hz
D. 2.6 cm, 50 Hz
E. 3.14 cm, 314 Hz
Correct Answer: C
Explanation:
General expression for displacement of a body in SHM is given as
X(t)
= A sin(ωt) ----------- (1)
Where
X(t) is the displacement of the particle;
A is the amplitude of the oscillation;
ω is the angular frequency;
Displacement of a particle as given in the question is
X(t) = 23 sin(314t) cm ------------- (2)
Comparing equation 1 with 2 we get
A = 2.3 cm and ω = 314 rad/s
Angular frequency is related to frequency as ω = 2Πf ----------- (3)
Where
f is the frequency of the oscillation;
Putting the value of in equation 3 we get
314 = 2Πf
Or, 314/(2*Π) = f
Or, 314/(2*3.14) = f
(Π = 3.14)
Or, f = 50 Hz
Hence, C is the correct answer option.
Question 8
Which of the following do not require any medium to propagate?
A. Sound waves
B. Seismic waves
C. Tides
D. Light waves
E. Waves on the surface of water
Correct Answer: D
Explanation:
Sound waves can only propagate in the presence of a medium like solid, liquid or gas. Sound waves
cannot propagate in the absence of a medium. Therefore, A is an incorrect option. Seismic waves are
generated due to the fault in tectonic plates of the Earth. Seismic waves originate from a point and
travel through the Earth’s layers. Hence, without the presence of Earth’s layers seismic waves cannot
propagate. Hence, B is an incorrect option. Tides and waves on the surface of water originate at a
point of disturbance and progress along all directions. Without any medium surrounding the point
where the tides or surface waves on water originate, the waves will not propagate. The medium for
tides and surface waves on water is essentially liquid. Therefore, C and E are incorrect options as
well. Light waves do not require any medium to propagate. For example, light waves coming from the
Sun travel most of the distance through the outer space that is devoid of any medium. Therefore, D is
the correct answer option
Question 9
A cylinder of height 2.5 m is filled completely with water. A hole is made at the bottom of the
cylinder in such a way that water is coming out of it. What is the velocity of water coming out
of the cylinder?
A. 6.4 m/s
B. 4 m/s
C. 9.8 m/s
D. 2.5 m/s
E. 7 m/s
Correct Answer: E
Explanation:
Velocity of water coming out of a cylinder is given as
v
2
= 2gh ----------------- (1)
Where
v is the velocity of the water coming out of the cylinder;
h is the height to which water is filled= 2.5 m;
g is the acceleration due to gravity = 9.8 m/s2;
Putting the values of h and g in equation 1 we get
v
2
= 2 * 9.8 * 2.5
Or, v
2
= 49
Or, v = 7 m/s
Hence, E is the correct answer option.
Question 10
An object A of mass 40 kg at 60
0
C has internal energy of 4000 J. If it is kept in contact with
another object B of mass 40 kg at 65
0
C having internal energy 4500 J, then which of the
following statements is true regarding heat transfer? (Assume boundary of contact can allow
heat transfer to take place)
A. Heat will flow from A to B
B. Heat will flow from B to A
C. No heat will flow between the objects A and B
D. Temperature of A decreases
E. Temperature of B increases
Correct Answer: B
Heat always flows from a hot body to a cooler body. In other words, heat flows from a body at a
higher temperature to a body at a lower temperature. If an object A, at 60
0
C is kept in contact with
another object B, at 65
0
C, then this means that object A is at lower temperature than that of object B.
Hence, heat will flow from B to A. Also, as the heat is transferred from B to A, temperature of B will
reduce but temperature of A will increase to reach thermal equilibrium. Thus, B is the correct answer
option and A, C, D and E are incorrect answer options.
Question 11
How many electronic charges comprise a coulomb of charge?
A. 6 × 10
18
B. 10
18
C. 2 × 10
18
D. 2 × 10
19
E. 5 × 10
15
Correct Answer: A
Explanation:
Charge is given as
Q = ne --------------- (1)
Where
Q is one coulomb of charge;
n is the number of electronic charges;
e is the charge on electron = 1.6 * 10
-19
C;
Putting the values in equation 1 we get
1 = n * 1.6 * 10
-19
Or, n = 6.25 * 10
18
Since n is a number therefore number of electronic charges that make one coulomb of charge is 6 *
10
18
. Therefore, A is the correct answer option
Question 12
Three batteries each of 25 V having zero internal resistance are connected in series with a
resistance of 100 Ω. What would be the current in the circuit?
A. 4 A
B. 0.75 A
C. 6.5 A
D. 0.33 A
E. 0.25 A
Correct Answer: B
Explanation:
Since the batteries are connected in series, equivalent potential of batteries can be found by adding
the individual potential. Let equivalent potential be V
eq
. Therefore, V
eq
= 25 + 25 + 25
Or, V
eq
= 3 * 25 V
Or, V
eq
= 75V
Since internal resistance of the batteries is zero, only resistance to which the equivalent potential is
connected is 100 Ω. Hence, current flowing through the circuit can be found by Ohm’s law. Ohm’s law
is given as
V = IR -------------- (1)
Where
V is the potential of the cell = 75 V;
I is the current flowing through the circuit;
R is the equivalent r
esistance of the circuit = 100 Ω
Putting the values of V and R in equation 1 we get
75 = I * 100
Or, I = 75/100
Or, I = 0.75 A
Therefore, B is the correct answer option.
Question 13
Keeping the current same in a circular wire, radius is getting doubled. What will the new value
of magnetic moment be, if the initial magnetic moment is 3 × 10
-5
Am
2
?
A. 3 × 10
-5
Am
2
B. 0.75 × 10
-5
Am
2
C. 4 × 10
-5
Am
2
D. 3.2 × 10
-4
Am
2
E. 1.2 × 10
-4
Am
2
Correct Answer: E
Explanation:
Magnetic moment is given as
M = IA ------------ (1)
Where
M is the magnetic moment;
I is the current flowing in the wire;
A is the area of the circular loop;
Since the wire is of circular shape hence, area of the circular loop is given as
A = πr
2
-------------- (2)
Where
r is the radius of the wire;
Let A
1
and r
1
be the area and radius of the initial wire. Therefore, expressing equation 2 in terms of
A
1
and r
1
we get
A
1
= πr
1
2
----------- (3)
Similarly, let A
2
and r
2
be the area and radius of the final wire. Therefore, expressing equation 2 in
terms of A
2
and r
2
we get
A
2
= πr
2
2
------------ (4)
Since it is given in the question radius of the final wire is twice that of the initial wire hence, r
2
= 2r
1
Using the value of r
2
in equation 4 we get
A
2
= π( 2r
1
)
2
Or, A
2
= 4πr
1
2
------------- (5)
Comparing equation 5 with 3 we find that final area can be expressed in terms of initial area and is
given as
A
2
= 4πr
1
2
A
2
= 4A
1
------------------ (6)
Expressing initial and final magnetic moment in equation 1 in terms of initial area and final area we
get
M
1
= IA
1
---------------- (7)
And, M
2
= IA
2
---------- (8)
Both initial and final current in the wire is same. Dividing equation 8 by equation 7 we get
M
2
/ M
1
= (IA
2
)/(IA
1
) ---------------- (9)
Cancelling out the common term, in 9 we get
M
2
/ M
1
= A
2
/A
1
--------------- (10)
Using equation 6 in 10 we get
M
2
/ M
1
= 4A
2
/A
1
Or, M
2
/ M
1
= 4
Therefore, M
2
= 4 M
1
-------------- (11)
Putting the value of M
1
in equation 11 we get M
2
= 4 * 3 * 10
-5
Or, M
2
= 1.2 * 10
-4
Am
2
Hence, E is the correct answer option.
Question 14
Consider the figure given below.
What are the values of Θ
1
and Θ
2
for the figure if a ray of light is incident at from air to a
medium of refractive index 1.5?
A. sin
-1
(1/3.7)
B. cos
-1
(1/3)
C. tan
-1
(1/3.7)
D. cot
-1
(1/3)
E. sin
-1
(1/3)
Correct Answer: E
Explanation:
In the figure given below, the brown arrow represents the incident ray, the green arrow represents the
transmitted or the reflected ray, and the orange arrow represents the refracted ray. OA represents the
normal to the interface between the air and the medium.
Since the incident light is travelling from one medium to another, some part of it will be reflected and
some will be refracted. As we know from the property of reflection that, angle of incidence is same as
angle of reflection, the green arrow, which is the reflected light will make the same angle with the
normal as made by the incident beam with the normal. Therefore,
θ
1
= 30
o
θ
2
can be found by applying Snell’s law. According to Snell’s law we have
µ
1
sin θ
i
= µ
2
+ sin θ
r
---------------- (1)
Where
µ
1
is the refractive index of medium 1 (here air = 1);
θ
i
is the angle of incidence = 30
o
;
µ
2
is the refractive index of medium 2 = 1.5;
θ
r
is the angle of refraction = θ
2
;
Putting these values in equation 1 we get
Or, 1 * sin 30
o
= 1.5 * sin θ
r
Or, 1/2 = 1.5 * sin θ
r
Or, 1/(2*1.5) = sin θ
r
Or, 1/3 = sin θ
r
Or, θ
r
= sin
-1
(1/3)
As θ
r
is same as θ
2
. The angle of refraction is sin
-1
(1/3).
Hence, E is the correct answer option.
Question 15
Which of the following gets deflected by electric field?
A. Gamma rays
B. Infrared rays
C. Beta rays
D. Neutrons
E. X-ray
Correct Answer: C
Explanation:
Electric field can be negative or positive in nature. If the electric field is created by negative charges
then the electric field is negative and if the electric field is created by positive charges then the electric
field is positive. Depending on the type of the charged particle passing through the region of electric
field, deflection can be either attractive or repulsive. Since gamma rays have no charge they will move
further into the electric field without being deflected. Hence, A is an incorrect option. Infrared rays
consist of uncharged particles and therefore, there would not be any deflection from the normal path.
Thus, B is an incorrect option. Beta rays are equivalent to electrons i.e. beta rays have the same
charge as that on electron. Hence, when beta rays pass through an electric field they will either get
attracted or repelled depending on the nature of the electric field. Thus, C is the correct answer
option. Neutrons are neutral like gamma rays or infrared rays. Hence, they will not show any
deflection while passing through the electric field. Therefore, D is an incorrect option. X-rays consists
of neutral particles. Due to this, when they pass through a region of electric field, they undergo no
deflection. Thus, E is an incorrect option as well.
Sample paper 2
Question 1
Force acting on an object is given as
F = Xx
-1
(Where x is the distance by which the body is displaced due to the applied force)
What would be the dimensional formula for X?
A.
MLT
-2
B.
MLT
-1
C.
M
-1
LT
-1
D.
M
-1
L
2
T
-1
E.
ML
2
T
-2
Correct Answer: E
Explanation:
From dimensional analysis we know that dimensional formula of the left hand side of an expression
should be the same as the dimensional formula of the individual quantities on the right hand side.
Force has the SI unit of kgms
-2
. Therefore, the dimensional formula of force becomes MLT
-2
.
Dimensional formula of x is L. Hence, dimensional formula of x
-1
becomes L
-1
. As said earlier, Xx
-
1
should have the same dimensional formula as F. Therefore, it can be written as
MLT
-2
= X * L
-1
X = ML
2
T
-2
Therefore, E is the correct answer option.
Question 2
Consider the figure and the statements given below.
(v is velocity and t is time)
The graph is for a body that is accelerating
The graph is for a body that is falling freely
The graph is for a body that has constant velocity
Which of the following statement(s) is true in accordance to the given figure?
A.
2
B.
1
C.
3
D.
Both 1 and 2
E.
Both 2 and 3
Correct Answer: A
Explanation:
The graph given in the question is between velocity and time. Slope of the velocity time graph is
linear here. Therefore, velocity increases with each instant but change in velocity per unit time is
constant throughout the travel. Thus, 3 is an incorrect statement and so is option C. Acceleration of
a body is defined as the rate of change of velocity per unit time. As said earlier, for a linear slope
between velocity and time, change in velocity is constant per unit time and thus, acceleration is
constant for the entire travel. Therefore, statement 1 isincorrect and so is option B. A body falls
freely under gravity only. A free falling body always experiences uniform acceleration throughout
the travel. Thus, we can say that acceleration is constant for a free fall and the graph essentially
represents the motion of free fall. Thus, statement 2 is correct and so A is the correct answer option.
In view of the above discussion, options D and E are incorrect as well.
Question 3
If a block of wood weighs 1000 N, what would be its mass?
A.
102.04 kg
B.
10.24 kg
C.
1020.44 kg
D.
1.24 kg
E.
110.24 kg
Correct Answer: A
Explanation:
Weight of a body is given as
W = mg -------------- (1)
Putting the value of W and g in equation 1 we get
1000 = m * 9.8
m = 102.04 kg
Therefore, A is the correct answer option.