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All questions of Newton’s Laws of Motion for EmSAT Achieve 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)
|
Ciel Knowledge answered |
For a body able to loop a vertical circle, the tension in the string must not be lesser than zero at the highest point. Hence when the tension at the highest point is zero we will obtain the minimum velocity. I.e.
At the highest point as T = 0, hence we get
Mg = Mv2/R
(comparing the centripetal acceleration with weight, where R is radius of vertical circle)
Hence we get the minimum velocity at the topmost point to be
Thus if we apply the work energy theorem upon the particle from its lowermost point to its top most point, we get net displacement as 2R and thus minimum velocity at lowermost point to be
At the highest point as T = 0, hence we get
Mg = Mv2/R
(comparing the centripetal acceleration with weight, where R is radius of vertical circle)
Hence we get the minimum velocity at the topmost point to be
Thus if we apply the work energy theorem upon the particle from its lowermost point to its top most point, we get net displacement as 2R and thus minimum velocity at lowermost point to be
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
|
Rajveer Kumar answered |
Given information:
- Mass of the object = 6 kg
- Three forces acting on the object:
- F1 = 20i + 30j N
- F2 = 8i - 50j N
- F3 = 2i + 2j N
To find: Acceleration of the object
Solution:
- We know that the net force acting on the object, F_net = F1 + F2 + F3
- Using vector addition, we can find the net force: F_net = (20+8+2)i + (30-50+2)j = 30i - 18j N
- Now, using Newton's second law of motion, F_net = m*a, where m is the mass of the object and a is the acceleration produced.
- Substituting the values, we get: 30i - 18j = 6*a
- Dividing both sides by 6, we get: a = (30/6)i - (18/6)j = 5i - 3j m/s^2
Therefore, the acceleration of the object is 5i - 3j m/s^2, which is option 'B'.
- Mass of the object = 6 kg
- Three forces acting on the object:
- F1 = 20i + 30j N
- F2 = 8i - 50j N
- F3 = 2i + 2j N
To find: Acceleration of the object
Solution:
- We know that the net force acting on the object, F_net = F1 + F2 + F3
- Using vector addition, we can find the net force: F_net = (20+8+2)i + (30-50+2)j = 30i - 18j N
- Now, using Newton's second law of motion, F_net = m*a, where m is the mass of the object and a is the acceleration produced.
- Substituting the values, we get: 30i - 18j = 6*a
- Dividing both sides by 6, we get: a = (30/6)i - (18/6)j = 5i - 3j m/s^2
Therefore, the acceleration of the object is 5i - 3j m/s^2, which is option 'B'.
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)
|
EduRev Humanities 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.
Find velocity of block 'B' at the instant shown in figure. 
- a)25 m/s
- b)20 m/s
- c) 22 m/s
- d)30 m/s
Correct answer is option 'A'. Can you explain this answer?
Find velocity of block 'B' at the instant shown in figure.
a)
25 m/s
b)
20 m/s
c)
22 m/s
d)
30 m/s
|
|
Rajesh Gupta answered |
Net Force = Force exerted by Cyclist - Frictional Force
Also, according to newton's second law
Fnet = m.a
250 N - Frictional Force = 30x4
∴ Frictional Force = 250 - 120 N
= 130 N
Also, according to newton's second law
Fnet = m.a
250 N - Frictional Force = 30x4
∴ Frictional Force = 250 - 120 N
= 130 N
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.
In the arrangement shown in figure, pulleys are massless and frictionless and threads are inextensible. The Block of mass m1 will remain at rest, if 
- a)
- b) m1 = m2 + m3
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
In the arrangement shown in figure, pulleys are massless and frictionless and threads are inextensible. The Block of mass m1 will remain at rest, if
a)
b)
m1 = m2 + m3
c)
d)
|
Nandini Iyer answered |
a = [(m3−m2 )/(m2+m3) ]g (m3 > m2)
T = [2m2m3g] / [m2+m3]
T′ = 2T = [4m2m3g] / [m2+m3]
m1g = 4m2m3gm / m2+m3
4/m1 = [1/m2] + [1/m3]
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
Two blocks are in contact on a frictionless table. One has mass m and the other 2m. A force F is applied on 2m as shown in the figure. Now the same force F is applied from the right on m. In the two cases respectively, the ratio force of contact between the two block will be :
- a) Same
- b)1 : 2
- c) 2 : 1
- d)1 : 3
Correct answer is option 'B'. Can you explain this answer?
Two blocks are in contact on a frictionless table. One has mass m and the other 2m. A force F is applied on 2m as shown in the figure. Now the same force F is applied from the right on m. In the two cases respectively, the ratio force of contact between the two block will be :
a)
Same
b)
1 : 2
c)
2 : 1
d)
1 : 3
|
Kamna Science Academy answered |
- If F is applied force then acceleration of the system is F/2m + m = F/3m
- Now when we apply the force from the left, the force applied on the block m is F/3m = F/3. This will be the force in the contact.
- When we apply the force from the right from on the block will be 2Fm/3 = 2F/3, this will be the force on the contact then.
So the ratio is F/3 : 2F/3 = 1 : 3
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
Three block are connected as shown, on a horizontal frictionless table and pulled to the right with a force T3 = 60 N. If m1 = 10 kg, m2 = 20 kg and m3 = 30 kg, the tension T2 is-
- a)10 N
- b)20 N
- c)30 N
- d)60 N
Correct answer is option 'C'. Can you explain this answer?
Three block are connected as shown, on a horizontal frictionless table and pulled to the right with a force T3 = 60 N. If m1 = 10 kg, m2 = 20 kg and m3 = 30 kg, the tension T2 is-
a)
10 N
b)
20 N
c)
30 N
d)
60 N
|
Sushil Kumar answered |
Let a be the acceleration of the system.
T1 = M1a .....(1)
T2 − T1 = M2a ....(2)
F − T2 = M3a ......(3)
Adding (1), (2) and (3) we get
(M1 + M2 + M3)a = F
or (10+20+30)a = 60
⇒ a = 1m/s2
Now , T2 = (M1+M2)a
⇒ (10+20)(1) = 30N
T1 = M1a .....(1)
T2 − T1 = M2a ....(2)
F − T2 = M3a ......(3)
Adding (1), (2) and (3) we get
(M1 + M2 + M3)a = F
or (10+20+30)a = 60
⇒ a = 1m/s2
Now , T2 = (M1+M2)a
⇒ (10+20)(1) = 30N
Find out the reading of the weighing machine in the following cases. 
- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Find out the reading of the weighing machine in the following cases.
a)
b)
c)
d)
|
Hansa Sharma answered |
N = mgcosФ
= 2 x 10 x cos30
= 2 x 10 x √3 / 2
= 10√3 N
= 2 x 10 x cos30
= 2 x 10 x √3 / 2
= 10√3 N
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.
A particle moves in the xy plane under the action of a force F such that the value of its linear momentum (P) at any time t is, Px = 2 cost, Py = 2 sint. The angle q between P and F at that time t will be -- a)0º
- b)30º
- c)90º
- d)180º
Correct answer is option 'C'. Can you explain this answer?
A particle moves in the xy plane under the action of a force F such that the value of its linear momentum (P) at any time t is, Px = 2 cost, Py = 2 sint. The angle q between P and F at that time t will be -
a)
0º
b)
30º
c)
90º
d)
180º
|
Infinity Academy answered |
Fx = dpx / dt = - 2sint
Fy = dpy / dt = 2cost
So angle between F and P will be 90º because we see that their dot product is zero.
Fy = dpy / dt = 2cost
So angle between F and P will be 90º because we see that their dot product is zero.
In the arrangement shown in fig. the ends P and Q of an unstretchable string move downwards with uniform speed U. Pulleys A and B are fixed. Mass M moves upwards with a speed. 
- a)2 U cos q
- b) U cos q
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
In the arrangement shown in fig. the ends P and Q of an unstretchable string move downwards with uniform speed U. Pulleys A and B are fixed. Mass M moves upwards with a speed.
a)
2 U cos q
b)
U cos q
c)
d)
|
|
Naina Sharma answered |
Thus option D is correct
Free body diagram of a situation is shown below. The net force is known, however, the magnitude of few of the forces is not known. The magnitude of the unknown forces C and D will be:
- a)C = 40 N, D = 80 N
- b)C = 40 N, D = 100 N
- c)C = 40 N, D = 80 N
- d)C = 40 N, D = 60 N
Correct answer is option 'B'. Can you explain this answer?
Free body diagram of a situation is shown below. The net force is known, however, the magnitude of few of the forces is not known. The magnitude of the unknown forces C and D will be:
a)
C = 40 N, D = 80 N
b)
C = 40 N, D = 100 N
c)
C = 40 N, D = 80 N
d)
C = 40 N, D = 60 N
|
|
Naina Bansal answered |
Force on A’ and B’ is 40 N and 60 N respectively. Net force of the situation is 40 N upward. So to balance the force on A’ a force of 40 N has to apply on C’. Similarly a force of 100 N is also be applied to get the net force of 40 N in upward direction.
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
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
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|
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.
Three blocks connected together lie on a horizontal frictionless table and pulled to the right with a force of F = 100 N. If m1 = 10 kg, m2 = 20 kg, m3 = 30kg. Then the tension T1 and T2 will be
- a)T1 = 10 N, T2 = 50 N
- b)T1 = 16.66 N, T2 = 50 N
- c)T1 = 10 N, T2 = 20 N
- d)T1 = 10 N, T2 = 40 N
Correct answer is option 'B'. Can you explain this answer?
Three blocks connected together lie on a horizontal frictionless table and pulled to the right with a force of F = 100 N. If m1 = 10 kg, m2 = 20 kg, m3 = 30kg. Then the tension T1 and T2 will be
a)
T1 = 10 N, T2 = 50 N
b)
T1 = 16.66 N, T2 = 50 N
c)
T1 = 10 N, T2 = 20 N
d)
T1 = 10 N, T2 = 40 N
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Gaurav Kumar answered |
Considering the three block system, we have
100 = 60.a
where a is the common acceleration of all the blocks
a = 5/3 m/s2
Considering mass 1 we get,
T1 = 10 a
T1 = 50/3 N
Considering mass 1 and 2 we get,
T2 = 30 a
T2 = 50N
100 = 60.a
where a is the common acceleration of all the blocks
a = 5/3 m/s2
Considering mass 1 we get,
T1 = 10 a
T1 = 50/3 N
Considering mass 1 and 2 we get,
T2 = 30 a
T2 = 50N
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 block is placed on a rough floor and a horizontal force F is applied on it. The force of friction f by the floor on the block is measured for different values of F and a graph is plotted between them-- a) The graph is a straight line of slope 45°
- b)The graph is straight line parallel to the F axis
- c) The graph is a straight line of slope 45º for smal lF and a straight line parallel to the F-axis for large F.
- d) There is small kink on the graph
Correct answer is option 'D'. Can you explain this answer?
A block is placed on a rough floor and a horizontal force F is applied on it. The force of friction f by the floor on the block is measured for different values of F and a graph is plotted between them-
a)
The graph is a straight line of slope 45°
b)
The graph is straight line parallel to the F axis
c)
The graph is a straight line of slope 45º for smal lF and a straight line parallel to the F-axis for large F.
d)
There is small kink on the graph
|
|
Suresh Reddy answered |
For a small value of the applied force the force of friction f increases linearly upto the limiting friction. When it crosses the maximum point of static friction then the friction force due to kinetic friction does not change any more.
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
|
|
Arka Desai answered |
Given data:
Weight of man (W) = 70 kg
Acceleration of lift (a) = 5 m/s²
Acceleration due to gravity (g) = 10 m/s²
To find: Reading on the weighing scale
Explanation:
When the lift is moving upwards with an acceleration of 5 m/s², the man inside the lift will experience an apparent weight due to the net force acting on him. The apparent weight of the man will be greater than his actual weight when the lift is accelerating upwards, and will be less than his actual weight when the lift is decelerating or moving downwards.
The net force acting on the man can be calculated using Newton's second law of motion:
Net force (F) = ma, where m is the mass of the man and a is the acceleration of the lift.
The total force acting on the man is the sum of his actual weight and the net force:
Total force (T) = Weight of man (W) + Net force (F)
The reading on the weighing scale will be equal to the total force acting on the man.
Calculation:
Weight of man (W) = 70 kg × 10 m/s² = 700 N
Net force (F) = ma = 70 kg × 5 m/s² = 350 N
Total force (T) = W + F = 700 N + 350 N = 1050 N
Therefore, the reading on the weighing scale will be 1050 N.
Answer: a) 1050 N
Weight of man (W) = 70 kg
Acceleration of lift (a) = 5 m/s²
Acceleration due to gravity (g) = 10 m/s²
To find: Reading on the weighing scale
Explanation:
When the lift is moving upwards with an acceleration of 5 m/s², the man inside the lift will experience an apparent weight due to the net force acting on him. The apparent weight of the man will be greater than his actual weight when the lift is accelerating upwards, and will be less than his actual weight when the lift is decelerating or moving downwards.
The net force acting on the man can be calculated using Newton's second law of motion:
Net force (F) = ma, where m is the mass of the man and a is the acceleration of the lift.
The total force acting on the man is the sum of his actual weight and the net force:
Total force (T) = Weight of man (W) + Net force (F)
The reading on the weighing scale will be equal to the total force acting on the man.
Calculation:
Weight of man (W) = 70 kg × 10 m/s² = 700 N
Net force (F) = ma = 70 kg × 5 m/s² = 350 N
Total force (T) = W + F = 700 N + 350 N = 1050 N
Therefore, the reading on the weighing scale will be 1050 N.
Answer: a) 1050 N
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:
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.
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 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 mass M is suspended by a rope from a rigid support at A as shown in figure. Another rope is tied at the end B, and it is pulled horizontally with a force F. If the rope AB makes an angle q with the vertical in equilibrium, then the tension in the string AB is : 
- a)F sin q
- b)F/sin q
- c)F cos q
- d)F/cos q
Correct answer is option 'B'. Can you explain this answer?
A mass M is suspended by a rope from a rigid support at A as shown in figure. Another rope is tied at the end B, and it is pulled horizontally with a force F. If the rope AB makes an angle q with the vertical in equilibrium, then the tension in the string AB is :
a)
F sin q
b)
F/sin q
c)
F cos q
d)
F/cos q
|
|
Neha Sharma answered |
As the block is still at the equilibrium, we get
T.cos q = mg
T.sin q = F
Thus we T = F / sin q
T.cos q = mg
T.sin q = F
Thus we T = F / sin q
A mass of 100 kg is resting on a rough inclined plane of 60o. If the coefficient of friction is 0.5, then the least force acting parallel to the plane to keep the mass in equilibrium is- a)605 N
- b)100 N
- c)500 N
- d)603.7 N
Correct answer is option 'D'. Can you explain this answer?
A mass of 100 kg is resting on a rough inclined plane of 60o. If the coefficient of friction is 0.5, then the least force acting parallel to the plane to keep the mass in equilibrium is
a)
605 N
b)
100 N
c)
500 N
d)
603.7 N
|
|
Geetika Shah answered |
The force parallel to the plane but downwards, W = mg.sin 60 = 980 x √3/2 = 848.7N
Maximum friction force acting, f = mg.cos 60 x 0.5 = 980 x ½ x 0.5 = 245N
Thus the minimum extra force required, let say F = 848.7 - 245 = 603.3 N
Maximum friction force acting, f = mg.cos 60 x 0.5 = 980 x ½ x 0.5 = 245N
Thus the minimum extra force required, let say F = 848.7 - 245 = 603.3 N
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.
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
A particle of mass 50 gram moves on a straight line. The variation of speed with time is shown in figure. find the force acting on the particle at t = 2, 4 and 6 seconds. 
- a)0.25 N along motion, zero, 0.25 opposite to motion
- b)0.25 N along motion, zero, 0.25 along to motion
- c)0.25 N opposite motion, zero, 0.25 along to motion
- d)0.25 N opposite motion, zero, 0.25 opposite to motion
Correct answer is option 'A'. Can you explain this answer?
A particle of mass 50 gram moves on a straight line. The variation of speed with time is shown in figure. find the force acting on the particle at t = 2, 4 and 6 seconds.
a)
0.25 N along motion, zero, 0.25 opposite to motion
b)
0.25 N along motion, zero, 0.25 along to motion
c)
0.25 N opposite motion, zero, 0.25 along to motion
d)
0.25 N opposite motion, zero, 0.25 opposite to motion
|
|
Geetika Shah answered |
- To know the force acting on the particle, we need to know the acceleration of the particle. In the velocity-time graph the slope of the curve gives the instantaneous acceleration. Let us find it for the given instant.
- At t=2 seconds
The graph is a straight line with positive slope. It means the particle has a constant acceleration with magnitude =15/3 =5 m/s.
So the force acting on it = mass x acceleration = 0.05 kg x 5 m/s
= 0.25 N, it acts along the motion because it is positive. - At t=4 seconds
The graph is horizontal to time axis meaning thereby the velocity is constant and no acceleration. It can be understood in this way too that the slope which represents acceleration is zero. Since there is no acceleration at t=4 s. so there is no force acting on the particle at this instant, Force= zero. - At t=6 seconds
The graph shows that velocity is uniformly decreasing with the time and the acceleration which is represented by the slope is negative. From the graph value of acceleration = -15/3 =-5 m/s
Force = mass x acceleration = 0.05 x -5 N =-0.25 N - So the force acting on the particle is 0.25 N and negative sign shows that its direction is opposite to the motion.
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).
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.
The pulley arrangements shown in figure are identical the mass of the rope being negligible. In case I, the mass m is lifted by attaching a mass 2m to the other end of the rope. In case II, the mass m is lifted by pulling the other end of the rope with cosntant downward force F = 2mg, where g is acceleration due to gravity. The acceleration of mass in case I is 
- a)Zero
- b)More than that in case II
- c) Less than that in case
- d)Equal to that in case II
Correct answer is option 'C'. Can you explain this answer?
The pulley arrangements shown in figure are identical the mass of the rope being negligible. In case I, the mass m is lifted by attaching a mass 2m to the other end of the rope. In case II, the mass m is lifted by pulling the other end of the rope with cosntant downward force F = 2mg, where g is acceleration due to gravity. The acceleration of mass in case I is
a)
Zero
b)
More than that in case II
c)
Less than that in case
d)
Equal to that in case II
|
|
Pooja Shah answered |
When we make the free body diagram of two mass system in case 1, we get the equation
2mg - mg = (2m + m) a
Thus we get a = g/3
But in the case 2 the equations is as follows
2mg - mg = ma
Thus we get a - g here.
2mg - mg = (2m + m) a
Thus we get a = g/3
But in the case 2 the equations is as follows
2mg - mg = ma
Thus we get a - g here.
If a block moving up at an inclined plane with θ =30° with a velocity 5 m/s , stops after 0.5 sec , then coefficient of friction is:- a)0.5
- b)1.25
- c)0.6
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
If a block moving up at an inclined plane with θ =30° with a velocity 5 m/s , stops after 0.5 sec , then coefficient of friction is:
a)
0.5
b)
1.25
c)
0.6
d)
none of these
|
|
Naina Sharma answered |
The initial velocity is 5 m/s and the block stops after ½ sec,
Hence the net negative acceleration is 10 m/s2
The net force active downwards parallel the plane is mg sin 30° + kmg cos 30°
Hence we get a = g ( sin 30° + kcos 30°) = 10
Thus we get sin 30° + kcos 30° = 1
k = sec 30° - tan 30°
= 2/1.73 - 1/1.73
= 1/1.73
= 0.6
Hence the net negative acceleration is 10 m/s2
The net force active downwards parallel the plane is mg sin 30° + kmg cos 30°
Hence we get a = g ( sin 30° + kcos 30°) = 10
Thus we get sin 30° + kcos 30° = 1
k = sec 30° - tan 30°
= 2/1.73 - 1/1.73
= 1/1.73
= 0.6
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