SPED TEST 4
Type: Custom Test Questions: 180 Total Marks: 720 Difficulty: Easy, Medium
Topics covered
Physics
Newton's Laws of Motion - Aristotle’s Fallacy, The Law of Inertia , Newton’s Law of Motion, Law of Conservation of
Linear Momentum, Equilibrium of a Particle , Constrain Motion, Applications of Newton’s Laws, Friction, Circular Motion,
Miscellaneous Problems in Newton’s Laws of Motion
Chemistry
Chemical Bonding - Cause of Chemical Combination , Lewis Dot Structures of Elements, Molecules, and Polyatomic
Ions, Kossel-Lewis Theory or Electronic Theory of Valency, Octet Rule and Its Limitations, Ionic Bonding, Covalent Bond,
Comparision between Ionic and Covalent Bond (or Compounds) , Coordinate Bond or Dative Bond, Formal Charge,
Bond Parameters, Resonance Structures, Polarity of a Covalent Bond, Polarisation and Fajans' Rules, VSEPR Theory,
Valence Bond Theory, VBT, Hybridisation, Molecular Orbital Theory (MOT), Comparision of Valence Bond Theory and
Molecular Orbital Theory, Hydrogen Bonding, Metallic Bonding, Miscellaneous questions on chemical bonding
Biology
Structural Organisation in Animals - Introduction, Animal Tissue, Organ and Organ System, Earthworm, Cockroach,
Frog
1. A horizontal force 10N is applied to a block A as 2. Read the Statement − A and Statement − B
shown in figure. The mass of blocks are 2 kg and carefully to mark the correct option given below:
3 kg respectively. The blocks slide over a frictionless
The impulse Impulse is given by the
surface. The force exerted by block on block is : experienced by an integral of force over
object is equal to the the time during which
change in momentum the force acts.
of the object.
(1) Both Statement A and Statement B are true.
(1) Zero (2) Statement A is true, but Statement B is false.
(3) Statement A is false, but Statement B is true.
(2) 4N
(4) Both Statement A and Statement B are false.
(3) 6N
(4) 10N
,3. A body of mass 2 kg has an initial velocity of 6. The acceleration of m and m are a and a then
1 2 1 2
3 ms
−1
along OE and it is subjected to a force of
4N in a direction perpendicular to OE . The
distance of body from O after 4s will be.
(1) 12 m
(2) 20 m
(3) 8 m (1) a1 = a2
(4) 48 m (2) a 1 = 2a 2
(3) 2a 1 = a 2
4. A man weighing 60 kg is in a lift moving down with
(4) a1 ≤ a2
an acceleration of 1.8 ms
−1
. The force exerted by
the floor on him is 7. A car is moving along a straight horizontal road
(1) 588 N with a speed v . If the coefficient of friction
0
(2) 480 N
between the tyres and the road is μ, the shortest
distance in which the car can be stopped is
(3) zero
2
(1)
v
0
(4) 696 N 2 μg
(2)
v0
μg
5. Two masses M and M are accelerated uniformly
1 2 2
(3)
v0
( )
on a frictionless surface as shown in figure. The μg
ratio of the tensions
T1
is (4) v0
μ
T2
8. Which one of the following statements is not true
about Newton’s second law of motion F→ = ma→?
(1) The second law of motion is consistent with the
(1)
M1
first law.
M2
(2) The second law of motion is a vector law.
M2
(2) (3) The second law of motion is applicable to a
M1
(M 1 + M 2 ) single point particle.
(3)
M2 (4) The second law of motion is not applicable for a
(4)
M1
system of particles.
(M 1 + M 2 )
,9. Read the Statement − A and Statement − B 12. The coefficient of friction between the tyres and
carefully to mark the correct option given below: road is 0.4. The minimum distance covered before
attaining a speed of 8 ms
−1
starting from rest is
The concept of The concept of
nearly g = 10 ms −2
acceleration is acceleration has not
necessary to describe been used to determine (1) 8.0 m
uniform motion. what governs the (2) 4.0 m
motion of bodies.
(3) 10.0 m
(1) Both Statement A and Statement B are true. (4) 16.0 m
(2) Statement A is true, but Statement B is false.
(3) Statement A is false, but Statement B is true. 13. Which of the following graph depicts spring
(4) Both Statement A and Statement B are false. constant k versus length l of the spring correctly?
10. At time t = 0 , a force F = αt , where t is time in
(1)
seconds, is applied to a body of mass 1 kg , resting
on a smooth horizontal plane. If the direction of the (2)
force makes an angle of 45 with the horizontal,
∘
(3)
then the velocity of the body at the moment of its
breaking off the plane is (4)
100
(1) m/s 14. A particle of mass ‘m’ and initially at rest is acted by
α
50√ 2 a force F = at . Newtons best representation of
(2) m/s
α Force- Displacement graph is:
50α
(3) m/s
√2
50 (1)
(4) m/s
α
(2)
11. Read the Statement − A and Statement − B
(3)
carefully to mark the correct option given below:
(4)
The centripetal force The centripetal force is
needed for circular always greater than the
motion can be provided force of gravity.
by the tension in a
string.
(1) Both Statement A and Statement B are true.
(2) Statement A is true, but Statement B is false.
(3) Statement A is false, but Statement B is true.
(4) Both Statement A and Statement B are false.
, 15. The coefficient of static friction, μ between a block
s
17. A box of mass m is in equilibrium under the
A of mass 2 kg and the table as shown in the figure application of three forces as shown below. If the
is 0.2 . What would be the maximum mass value of magnitude of F is 10 N , what is the magnitude of
1
block B so that, the two blocks do not move? The F3 ?
string and the pulley are assumed to be smooth
and massless (Take, g = 10 ms ). −2
(1) 2.0 kg
(2) 4.0 kg
(3) 0.2 g
(4) 0.4 kg
(1) 5N
16. A bead A can slide freely along a smooth rod bent (2) 15 N
in the form of a half circle of Radius R. The system (3) 20 N
is set in rotation with a constant angular velocity ω
(4) 30 N
about a vertical axis ’. Find the angle
OO θ
corresponding to steady position of the bead.
(1)
2
−1 Rω
cos ( )
g
(2) −1 g
cos ( 2
)
Rω
(3) −1 g
sin ( 2
)
Rω
(4)
2
−1 Rω
sin ( )
g
Type: Custom Test Questions: 180 Total Marks: 720 Difficulty: Easy, Medium
Topics covered
Physics
Newton's Laws of Motion - Aristotle’s Fallacy, The Law of Inertia , Newton’s Law of Motion, Law of Conservation of
Linear Momentum, Equilibrium of a Particle , Constrain Motion, Applications of Newton’s Laws, Friction, Circular Motion,
Miscellaneous Problems in Newton’s Laws of Motion
Chemistry
Chemical Bonding - Cause of Chemical Combination , Lewis Dot Structures of Elements, Molecules, and Polyatomic
Ions, Kossel-Lewis Theory or Electronic Theory of Valency, Octet Rule and Its Limitations, Ionic Bonding, Covalent Bond,
Comparision between Ionic and Covalent Bond (or Compounds) , Coordinate Bond or Dative Bond, Formal Charge,
Bond Parameters, Resonance Structures, Polarity of a Covalent Bond, Polarisation and Fajans' Rules, VSEPR Theory,
Valence Bond Theory, VBT, Hybridisation, Molecular Orbital Theory (MOT), Comparision of Valence Bond Theory and
Molecular Orbital Theory, Hydrogen Bonding, Metallic Bonding, Miscellaneous questions on chemical bonding
Biology
Structural Organisation in Animals - Introduction, Animal Tissue, Organ and Organ System, Earthworm, Cockroach,
Frog
1. A horizontal force 10N is applied to a block A as 2. Read the Statement − A and Statement − B
shown in figure. The mass of blocks are 2 kg and carefully to mark the correct option given below:
3 kg respectively. The blocks slide over a frictionless
The impulse Impulse is given by the
surface. The force exerted by block on block is : experienced by an integral of force over
object is equal to the the time during which
change in momentum the force acts.
of the object.
(1) Both Statement A and Statement B are true.
(1) Zero (2) Statement A is true, but Statement B is false.
(3) Statement A is false, but Statement B is true.
(2) 4N
(4) Both Statement A and Statement B are false.
(3) 6N
(4) 10N
,3. A body of mass 2 kg has an initial velocity of 6. The acceleration of m and m are a and a then
1 2 1 2
3 ms
−1
along OE and it is subjected to a force of
4N in a direction perpendicular to OE . The
distance of body from O after 4s will be.
(1) 12 m
(2) 20 m
(3) 8 m (1) a1 = a2
(4) 48 m (2) a 1 = 2a 2
(3) 2a 1 = a 2
4. A man weighing 60 kg is in a lift moving down with
(4) a1 ≤ a2
an acceleration of 1.8 ms
−1
. The force exerted by
the floor on him is 7. A car is moving along a straight horizontal road
(1) 588 N with a speed v . If the coefficient of friction
0
(2) 480 N
between the tyres and the road is μ, the shortest
distance in which the car can be stopped is
(3) zero
2
(1)
v
0
(4) 696 N 2 μg
(2)
v0
μg
5. Two masses M and M are accelerated uniformly
1 2 2
(3)
v0
( )
on a frictionless surface as shown in figure. The μg
ratio of the tensions
T1
is (4) v0
μ
T2
8. Which one of the following statements is not true
about Newton’s second law of motion F→ = ma→?
(1) The second law of motion is consistent with the
(1)
M1
first law.
M2
(2) The second law of motion is a vector law.
M2
(2) (3) The second law of motion is applicable to a
M1
(M 1 + M 2 ) single point particle.
(3)
M2 (4) The second law of motion is not applicable for a
(4)
M1
system of particles.
(M 1 + M 2 )
,9. Read the Statement − A and Statement − B 12. The coefficient of friction between the tyres and
carefully to mark the correct option given below: road is 0.4. The minimum distance covered before
attaining a speed of 8 ms
−1
starting from rest is
The concept of The concept of
nearly g = 10 ms −2
acceleration is acceleration has not
necessary to describe been used to determine (1) 8.0 m
uniform motion. what governs the (2) 4.0 m
motion of bodies.
(3) 10.0 m
(1) Both Statement A and Statement B are true. (4) 16.0 m
(2) Statement A is true, but Statement B is false.
(3) Statement A is false, but Statement B is true. 13. Which of the following graph depicts spring
(4) Both Statement A and Statement B are false. constant k versus length l of the spring correctly?
10. At time t = 0 , a force F = αt , where t is time in
(1)
seconds, is applied to a body of mass 1 kg , resting
on a smooth horizontal plane. If the direction of the (2)
force makes an angle of 45 with the horizontal,
∘
(3)
then the velocity of the body at the moment of its
breaking off the plane is (4)
100
(1) m/s 14. A particle of mass ‘m’ and initially at rest is acted by
α
50√ 2 a force F = at . Newtons best representation of
(2) m/s
α Force- Displacement graph is:
50α
(3) m/s
√2
50 (1)
(4) m/s
α
(2)
11. Read the Statement − A and Statement − B
(3)
carefully to mark the correct option given below:
(4)
The centripetal force The centripetal force is
needed for circular always greater than the
motion can be provided force of gravity.
by the tension in a
string.
(1) Both Statement A and Statement B are true.
(2) Statement A is true, but Statement B is false.
(3) Statement A is false, but Statement B is true.
(4) Both Statement A and Statement B are false.
, 15. The coefficient of static friction, μ between a block
s
17. A box of mass m is in equilibrium under the
A of mass 2 kg and the table as shown in the figure application of three forces as shown below. If the
is 0.2 . What would be the maximum mass value of magnitude of F is 10 N , what is the magnitude of
1
block B so that, the two blocks do not move? The F3 ?
string and the pulley are assumed to be smooth
and massless (Take, g = 10 ms ). −2
(1) 2.0 kg
(2) 4.0 kg
(3) 0.2 g
(4) 0.4 kg
(1) 5N
16. A bead A can slide freely along a smooth rod bent (2) 15 N
in the form of a half circle of Radius R. The system (3) 20 N
is set in rotation with a constant angular velocity ω
(4) 30 N
about a vertical axis ’. Find the angle
OO θ
corresponding to steady position of the bead.
(1)
2
−1 Rω
cos ( )
g
(2) −1 g
cos ( 2
)
Rω
(3) −1 g
sin ( 2
)
Rω
(4)
2
−1 Rω
sin ( )
g
