All Exams >
Class 11 >
Physics Class 11 >
All Questions
All questions of Laws of Motion for Class 11 Exam
The dimensional formula of momentum is- a)[M L-1 T-1]
- b)[M L T-1]
- c)[ML-2T-1]
- d)[M L2 T-1]
Correct answer is option 'B'. Can you explain this answer?
The dimensional formula of momentum is
a)
[M L-1 T-1]
b)
[M L T-1]
c)
[ML-2T-1]
d)
[M L2 T-1]
|
Suresh Iyer answered |
Planck's Constant (h) = 6.626176 x 10-34 m2 kg/s
So, Unit of planck constant= m2 kg/s
So, Unit of planck constant= m2 kg/s
Dimensions =M L2 T −1 ________ (1)
Angular momentum l = mvr
Where, m-mass
v-velocity
r-radius
Dimensions of angular momentum = M L T −1 L = M L2 T −1 _______________ (2)
From (1) and (2).
From (1) and (2).
Planck's constant and angular momentum have the same dimensions.
For a body to be able to loop a vertical circle of radius R, the minimum velocity required at its lowest point is:- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
For a body to be able to loop a vertical circle of radius R, the minimum velocity required at its lowest point is:
a)
b)
c)
d)
|
|
Shreya Gupta answered |
For a body to be able to loop a vertical circle of radius R. the minimum velocity required at its lowest point is √5gR
How is inertia used when riding a bicycle?- a)Bicycles don’t use inertia.
- b)You can stop paddling and still continue rolling forward.
- c)You must paddle harder when going up hill.
- d)You must paddle slower when going up hill.
Correct answer is option 'B'. Can you explain this answer?
How is inertia used when riding a bicycle?
a)
Bicycles don’t use inertia.
b)
You can stop paddling and still continue rolling forward.
c)
You must paddle harder when going up hill.
d)
You must paddle slower when going up hill.
|
|
Suresh Iyer answered |
Due to inertia, even if we stop paddling of motion, we will continue to move forward up to a certain distance as friction will finally make us stop.
The dimensional formula for impulse is- a)[MLT-1]
- b)[ML2T-1]
- c)[M2LT]
- d)[ML-1T2]
Correct answer is option 'A'. Can you explain this answer?
The dimensional formula for impulse is
a)
[MLT-1]
b)
[ML2T-1]
c)
[M2LT]
d)
[ML-1T2]
|
Krishna Iyer answered |
We know that I = P, where P is momentum
As subtracting initial momentum from the final momentum won't affect its unit, we get unit if I is the same as that of P.
As subtracting initial momentum from the final momentum won't affect its unit, we get unit if I is the same as that of P.
Two masses are in the ratio 1:5. What is ratio of their inertia?- a)1:5
- b)5:1
- c)1:25
- d)25:1
Correct answer is option 'A'. Can you explain this answer?
Two masses are in the ratio 1:5. What is ratio of their inertia?
a)
1:5
b)
5:1
c)
1:25
d)
25:1
|
Sagar Goyal answered |
Force of inertia = ma
Let the masses be 1x and 5x
Force of inertia for 1st body= 1x * a
Force of inertia for 2nd = 5x * a
Ratio= x * a / 5x * a = 1:5
A block of 5 kg mass rests on a horizontal floor. The action of the block on the floor is- a)50 N vertically upward
- b)5 N vertically upward
- c)5 N vertically downward
- d)50 N vertically downward
Correct answer is option 'D'. Can you explain this answer?
A block of 5 kg mass rests on a horizontal floor. The action of the block on the floor is
a)
50 N vertically upward
b)
5 N vertically upward
c)
5 N vertically downward
d)
50 N vertically downward
|
|
Preeti Iyer answered |
Weight of the block, mg = 5kg x 10 m/s2 = 50 N.
According to Newton’s third law, the action of the block, that is the force exerted on the floor by the block is equal to 50 N in magnitude and is directly vertically downward.
A body of mass 2 kg is hung on a spring balance mounted vertically in a lift. If the lift moves up with an acceleration equal to the acceleration due to gravity, the reading on the spring balance will bea)5 kgb)8 kgc)7 kgd)4 kgCorrect answer is option 'D'. Can you explain this answer?
|
Geetika Shah answered |
As the lift moves upwards but the spring feels itself at rest hence we need to compensate the non inertial frame by adding an appropriate pseudo force to treat it as an inertial frame. Hence the pseudo force to be applied acts on every mass in the lift which is equal to mass x acceleration (=g) downwards.
Hence the tension in the spring would be 40N (20 due to weight and 20 pseudo). Thus the reading would be 4kg.
Hence the tension in the spring would be 40N (20 due to weight and 20 pseudo). Thus the reading would be 4kg.
If no resultant force acts on a body then the body will be in- a)rest
- b)motion
- c)earlier state (no change in state)
- d)none of the above
Correct answer is option 'C'. Can you explain this answer?
If no resultant force acts on a body then the body will be in
a)
rest
b)
motion
c)
earlier state (no change in state)
d)
none of the above
|
|
Preeti Iyer answered |
What do you think about Newton's 1st law, he was said that " every body have nature to maintain inertia of rest or motion untill there is not net force applied on that body". means body will be earlier state (no change in state) when net force applied on that body equals zero.
I know, you thought answer is option (a). but this is not true. for better understanding, Let's take. an example. a body moves with uniform velocity then, net force applied on body = 0 because acceleration of body is zero . but here you see body is not in rest . it is in motion. it is in earlier state . its state doesn't change.
I know, you thought answer is option (a). but this is not true. for better understanding, Let's take. an example. a body moves with uniform velocity then, net force applied on body = 0 because acceleration of body is zero . but here you see body is not in rest . it is in motion. it is in earlier state . its state doesn't change.
In the above questions what is the weight of the suspended block ?- a)
N - b)
N - c)
N - d)
N
Correct answer is option 'A'. Can you explain this answer?
In the above questions what is the weight of the suspended block ?
a)
Nb)
Nc)
Nd)
N|
|
Gaurav Kumar answered |
The question is incomplete and is too vague to be found
It should be removed so as to not cause confusion.
It should be removed so as to not cause confusion.
Which law is in control of a spacecraft that cruises through space at a constant speed without using any fuel?- a)Universal law of gravitation
- b)Newton’s third law
- c)Newton’s second law
- d)Newton’s first law
Correct answer is 'D'. Can you explain this answer?
Which law is in control of a spacecraft that cruises through space at a constant speed without using any fuel?
a)
Universal law of gravitation
b)
Newton’s third law
c)
Newton’s second law
d)
Newton’s first law
|
Mira Sharma answered |
According to Newton’s first law the spacecraft will continue moving with constant speed unless an external opposite force is applied on it.
Three forces 
act on an object of mass m = 2 kg. The acceleration of the object in m/s2 is:
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Three forces act on an object of mass m = 2 kg. The acceleration of the object in m/s2 is:
act on an object of mass m = 2 kg. The acceleration of the object in m/s2 is:a)
b)
c)
d)
|
|
Krishna Iyer answered |
Force vector follows the principle of superposition which says all the force vectors can be vectorially added if applied on one point to get the net force vector. Hence we get
F = F1 + F2 + F3
= (-2 + 3) i + (1 - 2) j
F = i - j = ma
Thus we get a = (i - j) /m
= (i - j) / 2
F = F1 + F2 + F3
= (-2 + 3) i + (1 - 2) j
F = i - j = ma
Thus we get a = (i - j) /m
= (i - j) / 2
Adjoining figure shows a force of 40 N acting at 30° to the horizontal on a body of mass 5 kg resting on a smooth horizontal surface. Assuming that the acceleration of free-fall is 10 ms_2, which of the following statements A, B, C, D, E is (are) correct?
[1] The horizontal force acting on the body is 20 N[2] The weight of the 5 kg mass acts vertically downwards[3] The net vertical force acting on the body is 30 N- a)1,2,3
- b)1,2
- c)2 only
- d)1 only
Correct answer is option 'C'. Can you explain this answer?
Adjoining figure shows a force of 40 N acting at 30° to the horizontal on a body of mass 5 kg resting on a smooth horizontal surface. Assuming that the acceleration of free-fall is 10 ms_2, which of the following statements A, B, C, D, E is (are) correct?
[1] The horizontal force acting on the body is 20 N
[2] The weight of the 5 kg mass acts vertically downwards
[3] The net vertical force acting on the body is 30 N
a)
1,2,3
b)
1,2
c)
2 only
d)
1 only
|
EduRev Humanities answered |
(C) 2 only
[2] The weight of the 5kg mass acts vertically downwards
The forces F1, F2, and F3 are acting on a particle of mass m, such that F2 and F3are mutually perpendicular and under the effect of F1, F2, and F3 , the particle remains stationary. What will be the acceleration of the particle, if the force F1 is removed?- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
The forces F1, F2, and F3 are acting on a particle of mass m, such that F2 and F3are mutually perpendicular and under the effect of F1, F2, and F3 , the particle remains stationary. What will be the acceleration of the particle, if the force F1 is removed?
a)
b)
c)
d)
|
|
Geetika Shah answered |
Concept: Forces.The particle is a stationary under the effect of forces F1, F2 and F3.
This shows that force F1 is equal and opposite to the resultant of forces F2 and F3.
Hence, if the force F1 is removed the particle will move under the action of the force -ve F1 and the acceleration will the particle will be,a =-F1/m.
This shows that force F1 is equal and opposite to the resultant of forces F2 and F3.
Hence, if the force F1 is removed the particle will move under the action of the force -ve F1 and the acceleration will the particle will be,a =-F1/m.
A block of mass m is pushed by applying a force F at an angle θ with the horizontal surface. The normal force on the block is given as –- a)F = mg – F sin θ
- b)F = mg + F sin θ
- c)F = F sin θ
- d)F = mg
Correct answer is option 'A'. Can you explain this answer?
A block of mass m is pushed by applying a force F at an angle θ with the horizontal surface. The normal force on the block is given as –
a)
F = mg – F sin θ
b)
F = mg + F sin θ
c)
F = F sin θ
d)
F = mg
|
|
Krishna Iyer answered |
Both of them are vector quantities. And both of them can be easily simplified. If taken in the vector form then the task is even easier. Thus it is not necessary for the force or the couple to be vector only, even if the magnitude is taken, the simplification is done in the 2D.
A particle moves in a circle with a constant speed of 6m /s. Its centripetal acceleration is 120 m/s2, The radius of the circle is- a)20 cm
- b)200 cm
- c)300 cm
- d)30 cm
Correct answer is option 'D'. Can you explain this answer?
A particle moves in a circle with a constant speed of 6m /s. Its centripetal acceleration is 120 m/s2, The radius of the circle is
a)
20 cm
b)
200 cm
c)
300 cm
d)
30 cm
|
Naina Bansal answered |
a = v^2 /r
or, r = v^2 /a = 36 /120 = 0.3 m = 30 cm
A monkey of mass 40 kg climbs on a rope which can stand a maximum tension of 600 N. In which of the following cases will the rope break. When the monkey
(a) Climbs up with an acceleration of 6 ms−2.
- a)640 N
- b)632 N
- c)760 N
- d)740 N
Correct answer is option 'A'. Can you explain this answer?
A monkey of mass 40 kg climbs on a rope which can stand a maximum tension of 600 N. In which of the following cases will the rope break. When the monkey
(a) Climbs up with an acceleration of 6 ms−2.
(a) Climbs up with an acceleration of 6 ms−2.
a)
640 N
b)
632 N
c)
760 N
d)
740 N
|
|
Preeti Iyer answered |
Mass of the monkey, m = 40 kg
Acceleration due to gravity, g = 10 m/s
Maximum tension that the rope can bear, Tmax = 600 N
Acceleration of the monkey, a = 6 m/s2 upward
Using Newtons second law of motion, we can write the equation of motion as:
T mg = ma
T = m(g + a)
= 40 (10 + 6)
= 640 N
Since T > Tmax, the rope will break in this case.
Acceleration due to gravity, g = 10 m/s
Maximum tension that the rope can bear, Tmax = 600 N
Acceleration of the monkey, a = 6 m/s2 upward
Using Newtons second law of motion, we can write the equation of motion as:
T mg = ma
T = m(g + a)
= 40 (10 + 6)
= 640 N
Since T > Tmax, the rope will break in this case.
Can you explain the answer of this question below:A horizontal force of 100 N pulls two masses 5 kg and 10 kg tied to each other by a light string. What is the tension in the string if the force is applied on 10 kg mass?
- A:
30 N
- B:
23 N
- C:
43 N
- D:
33.3 N
The answer is d.
A horizontal force of 100 N pulls two masses 5 kg and 10 kg tied to each other by a light string. What is the tension in the string if the force is applied on 10 kg mass?
30 N
23 N
43 N
33.3 N
|
|
Geetika Shah answered |
At first considering both blocks as one system with only one external force F
We get common acceleration at right be a = 100/15 m/s2
Now considering 10 kg block
We get F - T = 10a
i.e. T = 100 - 10(100/15)
= 100 (1 - 2/3)
= 33.33 N
We get common acceleration at right be a = 100/15 m/s2
Now considering 10 kg block
We get F - T = 10a
i.e. T = 100 - 10(100/15)
= 100 (1 - 2/3)
= 33.33 N
Which of the following cannot be regarded as yet another kind of force?- a)centripetal force
- b)gravitational force
- c)electrostatic force
- d)magnetic force
Correct answer is option 'A'. Can you explain this answer?
Which of the following cannot be regarded as yet another kind of force?
a)
centripetal force
b)
gravitational force
c)
electrostatic force
d)
magnetic force
|
Lavanya Menon answered |
if an object is moving in a horizontal circle at constant speed, the centripetal force does not do any work and cannot alter the total mechanical energy of the object. For the reason, the kinetic energy and therefore the speed of the object will remain constant.
A bomb of mass 16kg at rest, explodes into two pieces of masses 4kg and 12kg. After explosion, the velocity of the 12kg mass is 4m/s. What is the velocity of the 4kg piece?- a)-12 m/s
- b)12 m/s
- c)-4 m/s
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
A bomb of mass 16kg at rest, explodes into two pieces of masses 4kg and 12kg. After explosion, the velocity of the 12kg mass is 4m/s. What is the velocity of the 4kg piece?
a)
-12 m/s
b)
12 m/s
c)
-4 m/s
d)
none of these
|
Om Desai answered |
Simply by conserving the momentum of the system we get that,
0 (initial momentum) = 12 x 4 + 4 x v (final momentum)
Thus we get v = -12 m/s
0 (initial momentum) = 12 x 4 + 4 x v (final momentum)
Thus we get v = -12 m/s
A man of mass 70 kg stands on a weighing scale in a lift which is moving upwards with a uniform speed of 10 m s−1, what would be the reading on the scale?- a)105 kg
- b)75 kg
- c)70 kg
- d)35 kg
Correct answer is option 'C'. Can you explain this answer?
A man of mass 70 kg stands on a weighing scale in a lift which is moving upwards with a uniform speed of 10 m s−1, what would be the reading on the scale?
a)
105 kg
b)
75 kg
c)
70 kg
d)
35 kg
|
|
Pooja Shah answered |
Mass of the man, m = 70 kg
Acceleration, a = 0
Using Newton’s second law of motion, We can write the equation of motion as,
R – mg = ma
∴ R = mg = 70 × 10 = 700 N
∴ the weighing scale = 700 / g = 700 / 10 = 70 kg
Acceleration, a = 0
Using Newton’s second law of motion, We can write the equation of motion as,
R – mg = ma
∴ R = mg = 70 × 10 = 700 N
∴ the weighing scale = 700 / g = 700 / 10 = 70 kg
A block is placed on the table. What is the angle between the action of the block on the table and reaction of the table on the block?
- a)90o
- b)300
- c)0o
- d)180o
Correct answer is option 'D'. Can you explain this answer?
A block is placed on the table. What is the angle between the action of the block on the table and reaction of the table on the block?
a)
90o
b)
300
c)
0o
d)
180o
|
Hansa Sharma answered |
According to Newton's 3 rd law, every action and reaction have same magnitude and opposite direction. So the angle between them should be 180o.
What is the smallest radius of a circle at which a bicyclist can travel if his speed is 7 m/s and the coefficient of static friction between tyres and road is 0.25 - a)10 m
- b)20 m
- c)5 m
- d)15 m
Correct answer is option 'B'. Can you explain this answer?
What is the smallest radius of a circle at which a bicyclist can travel if his speed is 7 m/s and the coefficient of static friction between tyres and road is 0.25
a)
10 m
b)
20 m
c)
5 m
d)
15 m
|
Anjali Sharma answered |
Centripetal force = Frictional force
⇒ mv2/r=μmg
∴r=v2/μg
∴r=72/0.25×10
∴r=19.6 m≈20 m
⇒ mv2/r=μmg
∴r=v2/μg
∴r=72/0.25×10
∴r=19.6 m≈20 m
A block of mass 1kg is placed on the floor. The coefficient of static friction is 0.2. If a force of 1N is applied to the block, the force of friction is- a)0
- b)2 N
- c)1 N
- d)varies with the area of contact of the block
Correct answer is option 'C'. Can you explain this answer?
A block of mass 1kg is placed on the floor. The coefficient of static friction is 0.2. If a force of 1N is applied to the block, the force of friction is
a)
0
b)
2 N
c)
1 N
d)
varies with the area of contact of the block
|
Ayush Joshi answered |
The applied force is less than the maximum value of static friction which will be (0.2)(1kg)(10) = 2N. So the force of friction is equal to the applied force.
After the body starts moving, the friction involved with motion is- a)Static Friction
- b)Rolling Friction
- c)Sliding Friction
- d)Kinetic Friction
Correct answer is option 'D'. Can you explain this answer?
After the body starts moving, the friction involved with motion is
a)
Static Friction
b)
Rolling Friction
c)
Sliding Friction
d)
Kinetic Friction
|
|
Gaurav Kumar answered |
When the body is in rest it is under static friction but when it starts moving (neither rolling nor sliding), the static friction slowly chngs to kinetic friction as the coefficient of static friction start decreasing and that of kinetic friction starts increasing. In case it starts rolling motion then the friction is rolling friction & if it slides then sliding fiction.
Which of the following forces is not considered as a contact force in Mechanics?- a)Tensional force
- b)Gravitational force
- c)Viscous force
- d)Frictional force
Correct answer is option 'B'. Can you explain this answer?
Which of the following forces is not considered as a contact force in Mechanics?
a)
Tensional force
b)
Gravitational force
c)
Viscous force
d)
Frictional force
|
|
Pooja Mehta answered |
Gravitational force
The force exerted by the earth on a body is called gravitational force. Actually this force exists between any two bodies in the universe.This force is always of attraction. e.g. When a body is dropped from a height it moves in downward direction towards the Earth with increasing speed (with constant acceleration). This constant acceleration by which all bodies fall down is called acceleration due to gravity. Its value is 9.8 m/s' (approx 10 m/s' )on the surface of the earth. e.g. i) A fruit from tree falls down;ii) Water falls down on a ground from a tap.iii) We feel the weight of bucket full of water holding in our hand.
A lift is moving down with the acceleration 3 m/s2. A ball is released 1.7 m above the the lift floor. How long will it take to hit the lift floor.- a)0.5s
- b)1.7s
- c)7.1s
- d)0.71s
Correct answer is option 'D'. Can you explain this answer?
A lift is moving down with the acceleration 3 m/s2. A ball is released 1.7 m above the the lift floor. How long will it take to hit the lift floor.
a)
0.5s
b)
1.7s
c)
7.1s
d)
0.71s
|
|
Mira Sharma answered |
Relative acceleration is,
a = 9.8 – 3 = 6.8 m/s2
Now, we have
S = ut + (1/2)at2
Here, u = 0 and S = 1.7 m
Therefore, t = 0.71 s
A man weighing 100kgf carries a load of 10kgf on his head. He jumps from tower with that load. What will be the weight of load experienced by the man.- a)0
- b)110 kgf
- c)10 kgf
- d)slightly more than 10kgf
Correct answer is option 'A'. Can you explain this answer?
A man weighing 100kgf carries a load of 10kgf on his head. He jumps from tower with that load. What will be the weight of load experienced by the man.
a)
0
b)
110 kgf
c)
10 kgf
d)
slightly more than 10kgf
|
Infinity Academy answered |
When an object falls freely, it experiences weightlessness. This is because the object and the load on it are both accelerating towards the ground at the same rate due to gravity. Therefore, the object and the load on it will have the same weight as they would have if they were stationary on the ground.
In this case, the man is carrying a load of 10 kgf on his head and jumps from a tower. As he falls freely, both the man and the load on his head will experience weightlessness. Therefore, the weight of the load experienced by the man will be zero.
Newton’s third law states that when two bodies interact.- a)they exert forces on each other that at each instant are equal in magnitude and opposite in direction
- b)they exert forces on each other that at each instant are equal in magnitude and same in direction
- c)they exert forces on each other that at some instants are equal in magnitude and opposite in direction
- d)they exert forces on each other that at some instants are equal in magnitude and same in direction
Correct answer is option 'A'. Can you explain this answer?
Newton’s third law states that when two bodies interact.
a)
they exert forces on each other that at each instant are equal in magnitude and opposite in direction
b)
they exert forces on each other that at each instant are equal in magnitude and same in direction
c)
they exert forces on each other that at some instants are equal in magnitude and opposite in direction
d)
they exert forces on each other that at some instants are equal in magnitude and same in direction
|
|
Krishna Iyer answered |
- The third law states that all forces between two objects exist in equal magnitude and opposite direction.
- If one object A exerts a force FA on a second object B, then B simultaneously exerts a force FB on A, and the two forces are equal in magnitude and opposite in direction, FA = −FB
- Newton's third Law:
Force that produces an acceleration of 1 ms−2 in a body of mass of 1 kg is called- a)slow newton
- b)zero newton
- c)one newton
- d)two newton
Correct answer is option 'C'. Can you explain this answer?
Force that produces an acceleration of 1 ms−2 in a body of mass of 1 kg is called
a)
slow newton
b)
zero newton
c)
one newton
d)
two newton
|
Riya Banerjee answered |
We know that F = ma, and for m = 1kg and a = 1m/s2
We get F = 1N which is one newton
We get F = 1N which is one newton
A man weighs 70 kg. He stands on a weighing scale in a lift which is moving upwards with an acceleration of 5ms2.What would be the reading on the scale? (g=10 ms2)- a)1050 N
- b)1200 N
- c)220 N
- d)1000 N
Correct answer is option 'A'. Can you explain this answer?
A man weighs 70 kg. He stands on a weighing scale in a lift which is moving upwards with an acceleration of 5ms2.What would be the reading on the scale? (g=10 ms2)
a)
1050 N
b)
1200 N
c)
220 N
d)
1000 N
|
|
Krishna Iyer answered |
As the moving elevator is a non inertial frame hence newton's laws can’t be applied directly to it. So to apply Newton's laws we need to add a pseudo force to the man's body equal to mass times the acceleration of lift in the opposite direction to that of acceleration. Thus the balancing normal force is equal to the weight of the man + mass times the acceleration which is,
Reading = Normal force = 700 + 70 x 5
= 700 + 350
= 1050 N
Reading = Normal force = 700 + 70 x 5
= 700 + 350
= 1050 N
Centripetal force always acts at 90 degrees to the velocity, and away from the centre of the circle.- a)true
- b)cannot predict
- c)false
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
Centripetal force always acts at 90 degrees to the velocity, and away from the centre of the circle.
a)
true
b)
cannot predict
c)
false
d)
none of these
|
|
Anjana Sharma answered |
1. The centripetal force is always towards the centre of the circle.
2. The force always acts at 90 degrees to the direction of the movement.
A mass of 6 kg is suspended by a rope of length 2 m from a ceiling. A force of 50 N in the horizontal direction is applied at the mid-point of the rope, as shown. What is the angle the rope makes with the vertical in equilibrium?
- a)60o
- b)40o
- c)50o
- d)30o
Correct answer is option 'B'. Can you explain this answer?
A mass of 6 kg is suspended by a rope of length 2 m from a ceiling. A force of 50 N in the horizontal direction is applied at the mid-point of the rope, as shown. What is the angle the rope makes with the vertical in equilibrium?
a)
60o
b)
40o
c)
50o
d)
30o
|
Sushil Kumar answered |
Making the free body diagram of the body we get
T1.cosθ = T2 = mg = 60
T1.sinθ = 50N
By dividing the above two equations we get
tanθ = 5/6
Thus we get θ = 40°
T1.cosθ = T2 = mg = 60
T1.sinθ = 50N
By dividing the above two equations we get
tanθ = 5/6
Thus we get θ = 40°
A constant force acting on a body of mass 3 kg changes its speed from 2 m/s to 3.5 m/s in 10 second. If the direction of motion of the body remains unchanged, what is the magnitude and direction of the force?- a)0.45 N in the direction opposite to motion.
- b)2.45 N in the direction of motion.
- c)0.45 N in the direction of motion.
- d)1.45 N in the direction opposite to motion.
Correct answer is 'C'. Can you explain this answer?
A constant force acting on a body of mass 3 kg changes its speed from 2 m/s to 3.5 m/s in 10 second. If the direction of motion of the body remains unchanged, what is the magnitude and direction of the force?
a)
0.45 N in the direction opposite to motion.
b)
2.45 N in the direction of motion.
c)
0.45 N in the direction of motion.
d)
1.45 N in the direction opposite to motion.
|
|
Nandini Patel answered |
Mass of the body, m = 3 kg
Initial speed of the body, u = 2 m/s
Final speed of the body, v = 3.5 m/s
Time, t = 10 s
Using the first equation of motion, the acceleration (a) produced in the body can be calculated as:
v = u + at
∴ a = (v – u) / t
= (3.5 – 2) / 10 = 0.15 ms^-2
As per Newton’s second law of motion, force is given as:
F = ma
= 3 x 0.15 = 0.45 N
Since the application of force does not change the direction of the body, the net force acting on the body is in the direction of its motion.
A block of mass 2 kg is placed on the floor. The coefficient of static friction is 0.4. If a force of 2.8 N is applied on the block parallel to floor, the force of friction between the block and floor is (Take g = 10 m/s2) - a)2.8 N
- b)8 N
- c)2 N
- d)zero
Correct answer is option 'A'. Can you explain this answer?
A block of mass 2 kg is placed on the floor. The coefficient of static friction is 0.4. If a force of 2.8 N is applied on the block parallel to floor, the force of friction between the block and floor is (Take g = 10 m/s2)
a)
2.8 N
b)
8 N
c)
2 N
d)
zero
|
|
Geetika Shah answered |
Weight of the block is 20N thus maximum friction that can act is 0.4 x 20 = 8N
But as the external force acting, F = 2.8N < 8N,
We get f = F = 2.8N
But as the external force acting, F = 2.8N < 8N,
We get f = F = 2.8N
A particle moves in a circle of radius 0.30m with a constant speed of 6m/s. Its centripetal acceleration is- a)20 m/s2
- b)zero
- c)12 m/s2
- d)120 m/s2
Correct answer is option 'D'. Can you explain this answer?
A particle moves in a circle of radius 0.30m with a constant speed of 6m/s. Its centripetal acceleration is
a)
20 m/s2
b)
zero
c)
12 m/s2
d)
120 m/s2
|
Ciel Knowledge answered |
We know that centripetal acceleration, a = v x v /r
Here v = 6m/s and r = 0.3m
Thus a = 36 / 0.3 m/s2
= 120 m/s2
Here v = 6m/s and r = 0.3m
Thus a = 36 / 0.3 m/s2
= 120 m/s2
Velocity-time graph of the particle is- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Velocity-time graph of the particle is
a)
b)
c)
d)
|
|
Gaurav Kumar answered |
In the velocity time graph, the slope value should give the value of acceleration.
The component of contact force normal to the surfaces in contact is called- a)Tension
- b)Friction
- c)Gravitational component
- d)Normal reaction
Correct answer is 'D'. Can you explain this answer?
The component of contact force normal to the surfaces in contact is called
a)
Tension
b)
Friction
c)
Gravitational component
d)
Normal reaction
|
|
Shreya Gupta answered |
The component of contact force normal to the surfaces in contact is called normal reaction. The component parallel to the surfaces in contact is called friction.
Which of the conditions shows that the particles are in equilibrium state for two forces F1 and F2 acting on the body?- a)F1= F2
- b)F1= 2F2
- c)F1 = -F2
- d)2F1=F2
Correct answer is option 'C'. Can you explain this answer?
Which of the conditions shows that the particles are in equilibrium state for two forces F1 and F2 acting on the body?
a)
F1= F2
b)
F1= 2F2
c)
F1 = -F2
d)
2F1=F2
|
|
Raghav Bansal answered |
At equilibrium condition net force on body is 0
F1 + F2 = 0
F1 = -F2
F1 + F2 = 0
F1 = -F2
If second law is applied to a rigid body- a)the acceleration is that of the centre of mass
- b)the acceleration is the average of all particles in the body
- c)the acceleration is that of any particle in the body
- d)none of the above
Correct answer is option 'A'. Can you explain this answer?
If second law is applied to a rigid body
a)
the acceleration is that of the centre of mass
b)
the acceleration is the average of all particles in the body
c)
the acceleration is that of any particle in the body
d)
none of the above
|
|
Geetika Shah answered |
The net external force on the rigid body is always equal to the total mass times the translational acceleration (i.e., Newton's second law holds for the translational motion, even when the net external torque is nonzero, and/or the body rotates).
Impending motion of a body is opposed by- a)sliding friction
- b)rolling friction
- c)static friction
- d)kinetic friction
Correct answer is option 'C'. Can you explain this answer?
Impending motion of a body is opposed by
a)
sliding friction
b)
rolling friction
c)
static friction
d)
kinetic friction
|
|
Rajat Kapoor answered |
When the motion is not impending, b) Impending motion is the moment where the body is on the verge of slipping. Static friction force reaches the max value. For a given pair of mating surfaces, . c) Motion The body starts moving in the direction of the applied force.
When a wheel rolls on a level road, the direction of frictional force at the point of contact of wheel and ground is:
- a)along the tangent to the wheel
- b)forward direction
- c)along the centre of the wheel
- d)backward direction
Correct answer is option 'B'. Can you explain this answer?
When a wheel rolls on a level road, the direction of frictional force at the point of contact of wheel and ground is:
a)
along the tangent to the wheel
b)
forward direction
c)
along the centre of the wheel
d)
backward direction
|
Advait Ghoshal answered |
Frictional force is the opposing force which plays between two surfaces and it destroys the relative motion between them. Frictional force is a non-conservative force. The force produced by two surfaces that contact and slide against each other, that force is called the frictional force. These forces are affected by the nature of the surface and amount of force acting on them.
In case of a bicycle, the front wheel of the bicycle is connected to a rod passing through its centre. The force acting on the wheel about its central axis by the force coming from the rest of the bicycle is zero. Front wheel obtains linear velocity by pedalling but it cannot rotate it.
Wheel or ball can also be rolled by pushing on it. The frictional force prevents the wheel from sliding forward at the point of contact. Here, the frictional force prevents the wheel from sliding forward and it is in the opposite direction.
So, in the case of the wheel, the point P which is in contact with the ground tries to go backward due to rotation. Frictional force will oppose this motion. Hence it will move forward.
Hence the direction of frictional force at the point P of the wheel is in forward direction.
Note: Frictional force opposes the motion. Here static friction holds a wheel or a ball on the surface. Frictional force is equal and opposite in direction to the applied force parallel to the contacting surfaces. The resistance due to the rolling body on a surface is called rolling friction. Torque is a force that acts on a body that is undergoing rotation.
Two forces F1 and F2 are acting on a particle. The particle will remain at rest if two forces are- a)Opposite
- b)Equal and Opposite
- c)Unequal and in same direction
- d)Equal
Correct answer is option 'B'. Can you explain this answer?
Two forces F1 and F2 are acting on a particle. The particle will remain at rest if two forces are
a)
Opposite
b)
Equal and Opposite
c)
Unequal and in same direction
d)
Equal
|
|
Ameya Choudhury answered |
For an object at rest to be at rest, no net acceleration must act upon it, which implies no net force. Thus the two forces need to nullify each other, which is only possible if both are equal but opposite.
A block of mass 2kg rests on a plane inclined at an angle of 30o with the horizontal. The coefficient of friction between the block and the surface is 0.7. The frictional force acting on the block is
- a)9.8N
- b)0.7×9.8×3–√N
- c)12.5 N
- d)0.7×9.8N
Correct answer is option 'A'. Can you explain this answer?
A block of mass 2kg rests on a plane inclined at an angle of 30o with the horizontal. The coefficient of friction between the block and the surface is 0.7. The frictional force acting on the block is
a)
9.8N
b)
0.7×9.8×3–√N
c)
12.5 N
d)
0.7×9.8N
|
|
Rohan Singh answered |
Since the frictional force is self adjusting, the weight component acting down the inclined plane is mgsin?, which comes out to be 2 x 10 sin 30 = 10 N. So the frictional force balancing this downward force will also be 10 N acting up the plane.
Which law says that every force is accompanied by an equal and opposite force?- a)Newton’s first law of motion
- b)Newton’s third law of motion
- c)law of inertia
- d)Newton’s second law of motion
Correct answer is option 'B'. Can you explain this answer?
Which law says that every force is accompanied by an equal and opposite force?
a)
Newton’s first law of motion
b)
Newton’s third law of motion
c)
law of inertia
d)
Newton’s second law of motion
|
|
Raghav Bansal answered |
"Every action has an equal and opposite reaction” -Newton's 3rd Law
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m s−1How long does the body take to stop?- a)1.7.0 s
- b)6.0 s
- c)5.0 s
- d)none
Correct answer is option 'B'. Can you explain this answer?
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m s−1How long does the body take to stop?
a)
1.7.0 s
b)
6.0 s
c)
5.0 s
d)
none
|
|
Lavanya Menon answered |
According to Newton's second Law
Force = mass x accleration
F = m.a
= m.v/t
t = m.v/F
= 20x15/50
= 6 sec
Force = mass x accleration
F = m.a
= m.v/t
t = m.v/F
= 20x15/50
= 6 sec
In a tug-of-war contest, two men pull on a horizontal rope from opposite sides. The winner will be the man who- a)Exerts greater force on the rope
- b) Exerts greater force on the ground
- c)Exerts a force on the rope which is greater than the tension in the rope
- d) Makes a smaller angle with the vertical
Correct answer is option 'B'. Can you explain this answer?
In a tug-of-war contest, two men pull on a horizontal rope from opposite sides. The winner will be the man who
a)
Exerts greater force on the rope
b)
Exerts greater force on the ground
c)
Exerts a force on the rope which is greater than the tension in the rope
d)
Makes a smaller angle with the vertical
|
|
Naina Sharma answered |
The greater the force exerted on the ground, the greater is the reaction and therefore the greater pull will be given to the rope.
A man pulls a block heavier than himself with a light horizontal rope. The coefficient of friction is the same between the man and the ground, and between the block and the ground. 
- a)The block will not move unless the man also moves
- b)The man can move even when the block is stationary
- c)If both move, the acceleration of the man is greater than the acceleration of the block
- d)None of the above assertions is correct
Correct answer is option 'A,B,C'. Can you explain this answer?
A man pulls a block heavier than himself with a light horizontal rope. The coefficient of friction is the same between the man and the ground, and between the block and the ground.
a)
The block will not move unless the man also moves
b)
The man can move even when the block is stationary
c)
If both move, the acceleration of the man is greater than the acceleration of the block
d)
None of the above assertions is correct
|
|
Krishna Iyer answered |
The friction force between the block and ground is more as compared to friction force between man and ground.
such that unless man doesn't move the block will not be moved.
The block of mass say M is heavier than the man of mass say m. The surface is rough with friction coefficient say μ. So when the man applies the force on the block the force cannot exceed the frictional force μmg without moving as μMg>μmg. Now if he starts moving (i.e. the force applied is increased and now the friction between him and the surface is not holding him stationary) there is a possibility that the block may move. Now as there is no other force acting on the system and as the man is lighter than block so he would have greater acceleration than the block when both move.
such that unless man doesn't move the block will not be moved.
The block of mass say M is heavier than the man of mass say m. The surface is rough with friction coefficient say μ. So when the man applies the force on the block the force cannot exceed the frictional force μmg without moving as μMg>μmg. Now if he starts moving (i.e. the force applied is increased and now the friction between him and the surface is not holding him stationary) there is a possibility that the block may move. Now as there is no other force acting on the system and as the man is lighter than block so he would have greater acceleration than the block when both move.
Chapter doubts & questions for Laws of Motion - Physics Class 11 2025 is part of Class 11 exam preparation. The chapters have been prepared according to the Class 11 exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Class 11 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
Chapter doubts & questions of Laws of Motion - Physics Class 11 in English & Hindi are available as part of Class 11 exam.
Download more important topics, notes, lectures and mock test series for Class 11 Exam by signing up for free.
Physics Class 11
135 videos|438 docs|100 tests
|
|
© EduRev
|
Education Revolution
|
|
Signup on EduRev and stay on top of your study goals
10M+ students crushing their study goals daily