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All questions of Friction for JEE Exam
Two blocks A and B masses 2m and m, respectively, are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in the figure. The magnitudes of acceleration of A and B, immediately after the string is cut, are respectively.
a)g, gb)g/2, gc)g, g/2d)g/2, g/2Correct answer is option 'B'. Can you explain this answer?
|
Riya Banerjee answered |
Before the string is cut the force applied by the spring is 3mg and tension in string is mg
Once the spring is cut, the tension comes to zero suddenly,
Thus acceleration of block b gets g downwards,
Whereas acceleration of a becomes g/2 upwards because the force from the spring is still the same at the moment the string is cut.
Once the spring is cut, the tension comes to zero suddenly,
Thus acceleration of block b gets g downwards,
Whereas acceleration of a becomes g/2 upwards because the force from the spring is still the same at the moment the string is cut.
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.
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 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
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
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 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
|
Suresh Reddy 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.
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.
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
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.
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:
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
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.
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 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
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
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
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
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.
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 are such that the sum of their magnitudes is 18 N and their resultant which has magnitude 12 N, is perpendicular to the smaller force. Then the magnitudes of the forces are- a)12 N, 6 N
- b)13 N, 5 N
- c) 10 N, 8 N
- d)16 N, 2 N
Correct answer is option 'B'. Can you explain this answer?
Two forces are such that the sum of their magnitudes is 18 N and their resultant which has magnitude 12 N, is perpendicular to the smaller force. Then the magnitudes of the forces are
a)
12 N, 6 N
b)
13 N, 5 N
c)
10 N, 8 N
d)
16 N, 2 N
|
Anshu Singh answered |
Solving Eqs. (i), (ii) and (iii), A = 5N, B = 13N
Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. a horizontal force F = 600 N is applied to the 20 kg mass so as to pull it. What is the tension in the string?- a)250 N
- b)300 N
- c)150 N
- d)200 N
Correct answer is option 'D'. Can you explain this answer?
Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. a horizontal force F = 600 N is applied to the 20 kg mass so as to pull it. What is the tension in the string?
a)
250 N
b)
300 N
c)
150 N
d)
200 N
|
Lavanya Menon answered |
For some instance assume both masses as one system, thus we get that
600 = 30a
Where a is the common acceleration of the system.
Now if we consider the 10kg block we get that
T = 10a
And a = 20m/s2
Thus we get T = 200N
600 = 30a
Where a is the common acceleration of the system.
Now if we consider the 10kg block we get that
T = 10a
And a = 20m/s2
Thus we get T = 200N
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
A monkey of mass 40 kg climbs on a rope which can stand a maximum tension of 600 N. What is the tension in the rope if the monkey climbs down with an acceleration of 4 ms−2- a)200 N
- b)300 N
- c)240 N
- d)280 N
Correct answer is option 'C'. Can you explain this answer?
A monkey of mass 40 kg climbs on a rope which can stand a maximum tension of 600 N. What is the tension in the rope if the monkey climbs down with an acceleration of 4 ms−2
a)
200 N
b)
300 N
c)
240 N
d)
280 N
|
|
Lavanya Menon 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 = 4 m/s2 downward
Using Newton’s second law of motion, we can write the equation of motion as:
mg T = ma
T = m (g- a)
= 40(10-4)
= 240 N
Acceleration due to gravity, g = 10 m/s
Maximum tension that the rope can bear, Tmax = 600 N
Acceleration of the monkey, a = 4 m/s2 downward
Using Newton’s second law of motion, we can write the equation of motion as:
mg T = ma
T = m (g- a)
= 40(10-4)
= 240 N
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
|
Arunaditya Das answered |
Impending motion is opposed by static friction..because it is the time when the object is going to slip in one direction...so that slipping of the object is opposed by static friction(max)...as soon as it starts moving friction becomes kinetic
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 body rests on an inclined plane and the angle of inclination is varied till the body just begins to slide down. The coefficient of friction is 
. What is the angle of inclination?- a)45o
- b)60o
- c)15o
- d)30o
Correct answer is option 'D'. Can you explain this answer?
A body rests on an inclined plane and the angle of inclination is varied till the body just begins to slide down. The coefficient of friction is . What is the angle of inclination?
. What is the angle of inclination?a)
45o
b)
60o
c)
15o
d)
30o
|
Ambition Institute answered |
At the time when the block just starts to move, we get that net force acting upon it is 0, thus we get, f - mg.sin a = 0
Where f is friction force and a is angle of incline.
We also know that f = 1√3 x N
= 1√3 x mg.cos a
Thus we get 1√3 x mg.cos a = mg.sin a
Thus we get tan a = 1√3
And a = 30
Where f is friction force and a is angle of incline.
We also know that f = 1√3 x N
= 1√3 x mg.cos a
Thus we get 1√3 x mg.cos a = mg.sin a
Thus we get tan a = 1√3
And a = 30
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.
A rope of mass 5 kg is moving vertically in vertical position with an upwards force of 100 N acting at the upper end and a downwards force of 70 N acting at the lower end. The tension at midpoint of the rope is- a)100 N
- b)85 N
- c)75 N
- d)105 N
Correct answer is option 'B'. Can you explain this answer?
A rope of mass 5 kg is moving vertically in vertical position with an upwards force of 100 N acting at the upper end and a downwards force of 70 N acting at the lower end. The tension at midpoint of the rope is
a)
100 N
b)
85 N
c)
75 N
d)
105 N
|
Anjana Sharma answered |
When you are given one upward and one downward force on a rope of some mass, and you have to find the tension at the mid- point, then you can simply average the 2 forces. So, the correct answer is (100+70)/3 = 85N
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.
Three equal weights A, B, C of mass 2 kg each are hanging on a string passing over a fixed frictionless pulley as shown in the fig. The tension in the string connecting weights B and C is-
- a)Zero
- b) 13 Newton
- c)3.3 Newton
- d)19.6 Newton
Correct answer is option 'B'. Can you explain this answer?
Three equal weights A, B, C of mass 2 kg each are hanging on a string passing over a fixed frictionless pulley as shown in the fig. The tension in the string connecting weights B and C is-
a)
Zero
b)
13 Newton
c)
3.3 Newton
d)
19.6 Newton
|
|
Nandini Iyer answered |
For some time let us consider B and C to be one single block of mass 4kg. Now if we make the free body diagram for blocks A, B, C we get the net acceleration of the system by equation
4.g - 2.g = 6.a
Thus we get a = g/3
Now if we only make a F.B.D. of block C, we get
2.g - T = 2.a
Thus T = 2 (g - g/3)
= 4g/3
= 13.3 N
4.g - 2.g = 6.a
Thus we get a = g/3
Now if we only make a F.B.D. of block C, we get
2.g - T = 2.a
Thus T = 2 (g - g/3)
= 4g/3
= 13.3 N
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.
Three blocks A, B and C are suspended as shown in the figure. Mass of each blocks A and C is m. If system is in equilibrium and mass of B is M, then : 
- a) M = 2m (
- b) M<2m
- c) M > 2m
- d) M = m
Correct answer is option 'B'. Can you explain this answer?
Three blocks A, B and C are suspended as shown in the figure. Mass of each blocks A and C is m. If system is in equilibrium and mass of B is M, then :
a)
M = 2m (
b)
M<2m
c)
M > 2m
d)
M = m
|
|
Pooja Shah answered |
Given,
Mass of block A & B = m
Mass of block C = M
Let,
Mass of A = mA & Mass of B = mB
Tension in string is = T
At equilibrium, T = mAg = mBg = mg
Weight of block C is = Mg
Forces on block C,
2Tcosθ = Mgcosθ
⇒ Mg/2T = Mg/2mg
⇒ M/2m
If 0< θ <90° then 1> cosθ >0
1> M/2m >0
2m > M
Hence, 2m > M
A boy of mass 50 Kg running at 5 m/s jumps on to a 20Kg trolley travelling in the same direction at 1.5 m/s. What is the common velocity?- a)4m/s
- b)3 m/s
- c)3.5 m/s
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
A boy of mass 50 Kg running at 5 m/s jumps on to a 20Kg trolley travelling in the same direction at 1.5 m/s. What is the common velocity?
a)
4m/s
b)
3 m/s
c)
3.5 m/s
d)
None of the above
|
|
Naina Sharma answered |
Since no external force, law of conservation of momentum can be applied
Initial Momentum
Final momentum
where v is the common velocity
Now we know that
Initial momentum = Final momentum
280 = 70v
V = 4m/s
Initial Momentum
Final momentum
where v is the common velocity
Now we know that
Initial momentum = Final momentum
280 = 70v
V = 4m/s
At a given instant, A is moving with velocity of 5 m/s upwards. What is velocity of B at the time 
- a)6 m/s
- b)15 m/s
- c) 0 m/s
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
At a given instant, A is moving with velocity of 5 m/s upwards. What is velocity of B at the time
a)
6 m/s
b)
15 m/s
c)
0 m/s
d)
None of these
|
|
Rajesh Gupta answered |
l1 = 2l2 + l3 = constant
dl1/dt + 2dl2/dt + dl3/dt = 0
−5 + 2(−5) + dl/dt = 0
dl3/dt = 15m/s
⇒ vB = 15m/s↓
dl1/dt + 2dl2/dt + dl3/dt = 0
−5 + 2(−5) + dl/dt = 0
dl3/dt = 15m/s
⇒ vB = 15m/s↓
Two fixed frictionless inclined plane making an angle 30° and 60° with the vertical are shown in the figure. Two block A and B are placed on the two planes. What is the relative vertical acceleration of A with respect to B ? [AIEEE 2010]- a)4.9 ms_2 in horizontal direction
- b)9.8 ms_2 in vertical direction
- c)Zero
- d)4.9 ms_2 in vertical direction
Correct answer is option 'D'. Can you explain this answer?
Two fixed frictionless inclined plane making an angle 30° and 60° with the vertical are shown in the figure. Two block A and B are placed on the two planes. What is the relative vertical acceleration of A with respect to B ? [AIEEE 2010]
a)
4.9 ms_2 in horizontal direction
b)
9.8 ms_2 in vertical direction
c)
Zero
d)
4.9 ms_2 in vertical direction
|
|
Raghav Bansal answered |
The forces acting on an object are shown in the fig. If the body moves horizontally at a constant speed of 5 m/s, then the values of the forces P and S are, respectively-
- a)0 N, 0 N
- b)300 N, 200 N
- c)300 N, 1000 N
- d)2000 N, 300 N
Correct answer is option 'C'. Can you explain this answer?
The forces acting on an object are shown in the fig. If the body moves horizontally at a constant speed of 5 m/s, then the values of the forces P and S are, respectively-
a)
0 N, 0 N
b)
300 N, 200 N
c)
300 N, 1000 N
d)
2000 N, 300 N
|
|
Naina Sharma answered |
As there is no net acceleration in horizontal or vertical direction, we can say that
P = 300N
And 2S = 2000N
Thus S = 1000N
P = 300N
And 2S = 2000N
Thus S = 1000N
A body of mass 8 kg is hanging another body of mass 12 kg. The combination is being pulled by a string with an acceleration of 2.2 m s-2. The tension T1 and T2 will be respectively : (use g = 9.8 m/s2) 
- a) 200 N, 80 N
- b)220 N, 90 N
- c) 240 N, 96 N
- d)260 N, 96 N
Correct answer is option 'C'. Can you explain this answer?
A body of mass 8 kg is hanging another body of mass 12 kg. The combination is being pulled by a string with an acceleration of 2.2 m s-2. The tension T1 and T2 will be respectively : (use g = 9.8 m/s2)
a)
200 N, 80 N
b)
220 N, 90 N
c)
240 N, 96 N
d)
260 N, 96 N
|
Top Rankers answered |
The tension T1 and T2 will be 240 N, 96 N.
Mass of the body = m1 = 12kg (Given)
Mass of the body = m2 = 8kg (Given)
Acceleration of the string - 2.2m/s² (Given)
Considering positive direction upwards -
Thus,
T1− (m1+m2)g = (m1+m2)a
T1 = (m1+m2)g + (m1+m2)a
T1= (12+8)(2.2)+(12+8)(9.8)
T1 = 44 + 196
= 240 N
T2− (m2)g=(m2)a
T2=(m2)g+(m2)a
T2= (8)(2.2)+(8)(9.8)
T2 = 17.6 + 78.4
= 96 N
Mass of the body = m1 = 12kg (Given)
Mass of the body = m2 = 8kg (Given)
Acceleration of the string - 2.2m/s² (Given)
Considering positive direction upwards -
Thus,
T1− (m1+m2)g = (m1+m2)a
T1 = (m1+m2)g + (m1+m2)a
T1= (12+8)(2.2)+(12+8)(9.8)
T1 = 44 + 196
= 240 N
T2− (m2)g=(m2)a
T2=(m2)g+(m2)a
T2= (8)(2.2)+(8)(9.8)
T2 = 17.6 + 78.4
= 96 N
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.
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.
A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m while applying the force and the ball goes upto 2 m height further, find the magnitude of the force. Consider g = 10 m/s2. [AIEEE 2006]- a) 4 N
- b)16 N
- c) 20 N
- d)22 N
Correct answer is option 'D'. Can you explain this answer?
A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m while applying the force and the ball goes upto 2 m height further, find the magnitude of the force. Consider g = 10 m/s2.
[AIEEE 2006]
a)
4 N
b)
16 N
c)
20 N
d)
22 N
|
|
Suresh Reddy answered |
The situation is shown in figure. At initial time, the ball is at P, then under the action of a force (exerted by hand) from P to A and then from A to B let acceleration of ball during PA is a ms-2 [assumed to be constant] in upward direction and velocity of ball at A is v m/s.
Then for PA, v2 = 02 +2a × 0.2
Then for PA, v2 = 02 +2a × 0.2
For AB, 0 = v2 - 2×g×2 => v2 = 2g × 2
From above equations,
a = 10g =100ms-2
Then for PA, FBD of ball is F - mg = ma
[F is the force exted by hand on ball] => F = m(g+a) = 0.2(11g)
= 22 N
Mass of the object is quantitative measure of its inertia stated law is newton's- a)first law
- b)second law
- c)third law
- d)fourth law
Correct answer is option 'A'. Can you explain this answer?
Mass of the object is quantitative measure of its inertia stated law is newton's
a)
first law
b)
second law
c)
third law
d)
fourth law
|
|
Raghav Bansal answered |
Mass of the object is quantitative measure of its inertia stated law is Newton's First Law.
A man getting down a running bus, falls forward because-- a)Due to inertia of rest, road is left behind and man reaches forward
- b)Due to inertia of motion upper part of body continues to be in motion in forward direction while feet come to rest as soon as they touch the road
- c)He leans forward as a matter of habbit
- d)Of the combined effect of all the three factors stated in (A), (B) and (C)
Correct answer is option 'B'. Can you explain this answer?
A man getting down a running bus, falls forward because-
a)
Due to inertia of rest, road is left behind and man reaches forward
b)
Due to inertia of motion upper part of body continues to be in motion in forward direction while feet come to rest as soon as they touch the road
c)
He leans forward as a matter of habbit
d)
Of the combined effect of all the three factors stated in (A), (B) and (C)
|
|
Preeti Khanna answered |
The explanation is the question itself. Inertia is a property of which it resists its change of state of rest or state of motion, so as soon as the person jumps the lower part immediately comes to rest by sharing contact with the ground the upper body due to inertia of motion resists its change in state of motion.
A block of mass m is connected to another block of mass M by a spring (massless) of spring constant k. The block are kept on a smooth horizontal plane. Initially the blocks are at rest and the spring is unstretched. Then a constant force F starts acting on the block of mass M to pull it. Find the force of the block of mass m.- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
A block of mass m is connected to another block of mass M by a spring (massless) of spring constant k. The block are kept on a smooth horizontal plane. Initially the blocks are at rest and the spring is unstretched. Then a constant force F starts acting on the block of mass M to pull it. Find the force of the block of mass m.
a)
b)
c)
d)
|
Rohit Jain answered |
Drawing free body-diagrams for m & M,
we get T = ma and F – T = Ma where T is force due to spring
⇒ F – ma = Ma or,, F = Ma + ma
⇒ F – ma = Ma or,, F = Ma + ma
Now, force acting on the block of mass m is
Two men of unequal masses hold on to the two sections of a light rope passing over a smooth light pulley. Which of the following are possible? 
- a)The lighter man is stationary while the heavier man slides with some acceleration
- b)The heavier man is stationary while the lighter man climbs with some acceleration
- c)The two men slide with the same acceleration in the same direction
- d)The two men move with accelerations of the same magnitude in opposite directions
Correct answer is option 'A,B,D'. Can you explain this answer?
Two men of unequal masses hold on to the two sections of a light rope passing over a smooth light pulley. Which of the following are possible?
a)
The lighter man is stationary while the heavier man slides with some acceleration
b)
The heavier man is stationary while the lighter man climbs with some acceleration
c)
The two men slide with the same acceleration in the same direction
d)
The two men move with accelerations of the same magnitude in opposite directions
|
Neha Joshi answered |
Here m2>m1
If lighter man is stationary,
Vertical force on m1
T−m1g=0
⇒T=m1g
Vertical force on m2
m2a=T−m2g=m1g−m2g
a= m1/ m2 g−g
since m1/ m2<1, so in this case, a is negative. thus the heavier man of mass m2 is sliding down.
If heavier man is stationary,
Vertical force on m2=T−m2g=0
⇒T=m2g
Vertical force on m1=m1a=T−m1g=m2g−m1g
a= m2/ m1 g−g
since m2>m1 so in this case, a is positive . Thus the lighter man will climb up with some acceleration.
From the above two cases we can say that they will move with the same acceleration over the pulley but in the opposite direction.
Ans:(A),(B),(D)
For ordinary terrestrial experiments, which of the following observers below are inertial.- a)A child revolving in a "giant wheel".
- b)A driver in a sports car moving with a constant high speed of 200 km/h on a straight road.
- c)The pilot of an aeroplane which is taking off.
- d)A cyclist negotiating a sharp turn.
Correct answer is option 'B'. Can you explain this answer?
For ordinary terrestrial experiments, which of the following observers below are inertial.
a)
A child revolving in a "giant wheel".
b)
A driver in a sports car moving with a constant high speed of 200 km/h on a straight road.
c)
The pilot of an aeroplane which is taking off.
d)
A cyclist negotiating a sharp turn.
|
|
Naina Sharma answered |
For an inertial frame the acceleration of the object should be = 0
Static friction force- a)increases until the body starts to move
- b)decreases before moving and increases later
- c)remains constant
- d)is a strong interaction force
Correct answer is option 'A'. Can you explain this answer?
Static friction force
a)
increases until the body starts to move
b)
decreases before moving and increases later
c)
remains constant
d)
is a strong interaction force
|
Priyanka Roy answered |
Explanation:It is the maximum friction, where a body just starts to move over the surface. for example. A large block of mass m is placed on a horizontal table. Apply a small force. the block does not move due to the friction force which balances the applied force. Now Gradually increase the applied force untill it start moving. This maximum applied force acting in opposite direction is called as Static Friction.
The velocity of end `A' of rigid rod placed between two smooth vertical walls moves with velocity `u' along vertical direction. Find out the velocity of end `B' of that rod, rod always remains in constant with the vertical walls. 
- a)u tan 2q
- b)u cot q
- c)u tan q
- d)2u tan q
Correct answer is option 'C'. Can you explain this answer?
The velocity of end `A' of rigid rod placed between two smooth vertical walls moves with velocity `u' along vertical direction. Find out the velocity of end `B' of that rod, rod always remains in constant with the vertical walls.
a)
u tan 2q
b)
u cot q
c)
u tan q
d)
2u tan q
|
|
Anjana Sharma answered |
Along the length of the rod the velocities should be same or the rod will break
and the velocity of B is always horizontal (net)
Velocity of A is vertical (net)
now take the components along the length and equate them
usinθ=vcosθ
utanθ=v
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