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All questions of Rotational Motion for JEE Exam
The centre of mass of a body is locateda)outside the systemb)inside or outside the systemc)inside the systemd)at the centre of systemCorrect answer is option 'B'. Can you explain this answer?
|
Suresh Iyer answered |
The centre of mass of a body can lie within or outside the body.
Example
(i) Centre of mass of a uniform rod lies at its geometrical centre which lies within the rod
(ii) Centre of mass of a uniform ring lies at its geometrical centre which lies outside the ring.
Example
(i) Centre of mass of a uniform rod lies at its geometrical centre which lies within the rod
(ii) Centre of mass of a uniform ring lies at its geometrical centre which lies outside the ring.
Can you explain the answer of this question below:Two blocks of the masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block. The velocity of centre of mass is
- A:
10 m/s
- B:
5 m/s
- C:
15 m/s
- D:
20 m/s
The answer is a.
Two blocks of the masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block. The velocity of centre of mass is
10 m/s
5 m/s
15 m/s
20 m/s
|
Lavanya Menon answered |
Velocity of heavier block (v2) = 14m/s
Velocity of lighter block (v1) = 0m/s
Velocity of centre of mass,
Velocity of lighter block (v1) = 0m/s
Velocity of centre of mass,
A mass m is moving with a constant velocity along a line parallel to the x-axis, away from the origin. Its angular momentum with respect to the origin
- a)Is zero
- b)Remains constant
- c)Goes on increasing
- d)Goes on decreasing
Correct answer is option 'B'. Can you explain this answer?
A mass m is moving with a constant velocity along a line parallel to the x-axis, away from the origin. Its angular momentum with respect to the origin
a)
Is zero
b)
Remains constant
c)
Goes on increasing
d)
Goes on decreasing
|
Krishna Iyer answered |
Angular momentum (L) is defined as the distance of the object from a rotation axis multiplied by the linear momentum
L = mv×y
L = mv×y
As the particle moves, m; v; and y, all remain unchanged at any point of time
⇒ L = constant
⇒ L = constant
A person standing on a rotating platform with his hands lowered outstretches his arms. The angular momentum of the person- a)becomes zero
- b)decreases
- c)remains constant
- d)inreases
Correct answer is option 'C'. Can you explain this answer?
A person standing on a rotating platform with his hands lowered outstretches his arms. The angular momentum of the person
a)
becomes zero
b)
decreases
c)
remains constant
d)
inreases
|
|
Suresh Reddy answered |
Yes because there is absence of any external force or torque so angular momentum will remain constant
here outstretching the hands means internal forces are working.. so moment of inertia increases in this case and to make angular momentum constant when angular velocity decreases.
here outstretching the hands means internal forces are working.. so moment of inertia increases in this case and to make angular momentum constant when angular velocity decreases.
The moment of inertia of two spheres of equal masses is equal. If one of the spheres is solid of radius 
m and the other is a hollow sphere. What is the radius of the hollow sphere?- a)5 m
- b)√3 m
- c)3√3 m
- d)3 m
Correct answer is option 'C'. Can you explain this answer?
The moment of inertia of two spheres of equal masses is equal. If one of the spheres is solid of radius m and the other is a hollow sphere. What is the radius of the hollow sphere?
m and the other is a hollow sphere. What is the radius of the hollow sphere?a)
5 m
b)
√3 m
c)
3√3 m
d)
3 m
|
Sushil Kumar answered |
Moment of inertia of solid sphere Is= 2/5MR2
moment of inertia of hollow sphere Ih =2/3MR2
given mass of solid sphere =√45 kg.
Is=Ih
2MR2/5=2MR2/3
given their masses are equal 2 (√45)2/5= 2 R2/3
45/5=R2/3
9=R2/3
9×3=R2
27=R2
√27=R
√3×9=R
3√3 m=R.
moment of inertia of hollow sphere Ih =2/3MR2
given mass of solid sphere =√45 kg.
Is=Ih
2MR2/5=2MR2/3
given their masses are equal 2 (√45)2/5= 2 R2/3
45/5=R2/3
9=R2/3
9×3=R2
27=R2
√27=R
√3×9=R
3√3 m=R.
There are two objects of masses 1 kg and 2 kg located at (1, 2) and (-1, 3) respectively. The coordinates of the centre of mass are- a)( 2, -1 )
- b)( 8/3 ,-1/3 )
- c)( -1/3 , 8/3 )
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
There are two objects of masses 1 kg and 2 kg located at (1, 2) and (-1, 3) respectively. The coordinates of the centre of mass are
a)
( 2, -1 )
b)
( 8/3 ,-1/3 )
c)
( -1/3 , 8/3 )
d)
none of these
|
Riya Banerjee answered |
Body A has mass of 1kg and location (1,2)
Body B has mass of 2kg and location (-1,3)
Body B has mass of 2kg and location (-1,3)
mcxc = m1x1 + m2x2
(1+2) xc = (1 * 1) + (2 * -1)
xc = -1/3
(1+2) xc = (1 * 1) + (2 * -1)
xc = -1/3
Similarly,
mcyc = m1y1 + m2y2
(1 + 2) yc = (1 * 2) + (2 * 3)
yc= 8/3
mcyc = m1y1 + m2y2
(1 + 2) yc = (1 * 2) + (2 * 3)
yc= 8/3
Hence, the coordinates of the center of mass are (-1/3, 8/3).
A right triangular plate ABC of mass m is free to rotate in the vertical plane about a fixed horizontal axis through A. It is supported by a string such that the side AB is horizontal. The reaction at the support A is : 
- a)
- b)
- c)
- d)mg
Correct answer is option 'B'. Can you explain this answer?
A right triangular plate ABC of mass m is free to rotate in the vertical plane about a fixed horizontal axis through A. It is supported by a string such that the side AB is horizontal. The reaction at the support A is :
a)
b)
c)
d)
mg
|
|
Lavanya Menon answered |
The distance of Centre Of Mass of the given right angled triangle is 2L/3 along BA and L/3 along AC from the point B.
Force of magnitude mg is acting downwards at its COM.
Moment balance around B gives:
mg(2L/3)−FA(L)=0
(Moment= × =rFsin(θ)=F(rsin(θ))=Fr⊥)
∴FA=2mg/3
Force of magnitude mg is acting downwards at its COM.
Moment balance around B gives:
mg(2L/3)−FA(L)=0
(Moment= × =rFsin(θ)=F(rsin(θ))=Fr⊥)
∴FA=2mg/3
A homogeneous cubical brick lies motionless on a rough inclined surface. The half of the brick which applies greater pressure on the plane is : 
- a) Left half
- b)Right half
- c)Both applies equal pressure
- d)The answer depend upon coefficient of friction
Correct answer is option 'A'. Can you explain this answer?
A homogeneous cubical brick lies motionless on a rough inclined surface. The half of the brick which applies greater pressure on the plane is :
a)
Left half
b)
Right half
c)
Both applies equal pressure
d)
The answer depend upon coefficient of friction
|
Mohit Rajpoot answered |
As weight of the body (mg) acts along the left side i.e. mg sinx acts along flow of object.
mgcosx acts perpendicular to flow of the object.
The pressure of right half comes on left half hence left half has maximum pressure.
mgcosx acts perpendicular to flow of the object.
The pressure of right half comes on left half hence left half has maximum pressure.
An engine develops a power of 360 kw, when rotating at 30 revolutions per second. The Torque required to deliver this power is- a)191.08 Nm
- b)19108 Nm
- c)1910.8 Nm
- d)19.108 Nm
Correct answer is option 'C'. Can you explain this answer?
An engine develops a power of 360 kw, when rotating at 30 revolutions per second. The Torque required to deliver this power is
a)
191.08 Nm
b)
19108 Nm
c)
1910.8 Nm
d)
19.108 Nm
|
Preeti Iyer answered |
The power delivered by the torque τ exerted on rotating body is given by
P=τω or τ=P/ω
Here P=360KW=360000 Watt
ω=30 x 2π rad/sec,
ω=60π rad/sec
now,
τ=360000 /60×3.14Nm
τ= 1910.8 Nm
P=τω or τ=P/ω
Here P=360KW=360000 Watt
ω=30 x 2π rad/sec,
ω=60π rad/sec
now,
τ=360000 /60×3.14Nm
τ= 1910.8 Nm
A rotating machine is maintained at a uniform angular speed of 150 rad s-1, with a torque of 100 Nm. The power is- a)15 kW
- b)1.5 kW
- c)1500 kW
- d)0.15 kW
Correct answer is option 'A'. Can you explain this answer?
A rotating machine is maintained at a uniform angular speed of 150 rad s-1, with a torque of 100 Nm. The power is
a)
15 kW
b)
1.5 kW
c)
1500 kW
d)
0.15 kW
|
|
Suresh Reddy answered |
Power = torque × angular velocity
P = 100 × 150
P = 15000 W
P = 15KW
P = 100 × 150
P = 15000 W
P = 15KW
If a shell at rest explodes then the centre of mass of the fragments
- a)remains at rest
- b)Moves along a parabolic path
- c)Moves along a straight line
- d)moves along an elliptical path
Correct answer is option 'A'. Can you explain this answer?
If a shell at rest explodes then the centre of mass of the fragments
a)
remains at rest
b)
Moves along a parabolic path
c)
Moves along a straight line
d)
moves along an elliptical path
|
Gaurav Kumar answered |
As a shell explosion is not governed by any external force, we get that no external force acts upon the shell, and hence it remains at rest.
The motion of a potter’s wheel is an example of
- a)rolling motion
- b)rotatory motion
- c)translatory motion
- d)precessional motion
Correct answer is option 'B'. Can you explain this answer?
The motion of a potter’s wheel is an example of
a)
rolling motion
b)
rotatory motion
c)
translatory motion
d)
precessional motion
|
Anjali Iyer answered |
Potter’s wheel is an example of rotary motion. Rotary motion is that kind of motion in which body of the mass moves along a circular path about an axis which remains fixed.
The centre of mass of a system of particles does not depend on
- a)masses of the particles
- b)forces on the particles
- c)relative distances between the particles
- d)position of the particles
Correct answer is option 'B'. Can you explain this answer?
The centre of mass of a system of particles does not depend on
a)
masses of the particles
b)
forces on the particles
c)
relative distances between the particles
d)
position of the particles
|
Neha Joshi answered |
The resultant of all forces, on any system of particles, is zero. Therefore, their centre of mass does not depend upon the forces acting on the particles.
There are two circular iron discs A and B having masses in the ratio 1:2 and diameter in the ratio 2:1. The ratio of their moment of inertia is- a)4:1
- b)1:3
- c)2:1
- d)8:1
Correct answer is option 'C'. Can you explain this answer?
There are two circular iron discs A and B having masses in the ratio 1:2 and diameter in the ratio 2:1. The ratio of their moment of inertia is
a)
4:1
b)
1:3
c)
2:1
d)
8:1
|
|
Preeti Iyer answered |
Given,
Mass of A=1,
Mass of B=2.
diameter if A=2,
diameter if B=1.
radius (r) of A=d/2=2/2=1.
radius (r) of B=d/2=1/2.
we know ,
moment of inertia of disc=MR2/2.
moment of inertia (I)of A/moment of inertia (I)of B=MR2/2/MR2/2.
(I) of A/(I) of B=1×12/2/2×(1/2)2/2.
=1×1/2/2×(1/4)/2.
=1/2/(1/2)/2.
=1/2/1/4.
=4/2.
=2/1.
Mass of A=1,
Mass of B=2.
diameter if A=2,
diameter if B=1.
radius (r) of A=d/2=2/2=1.
radius (r) of B=d/2=1/2.
we know ,
moment of inertia of disc=MR2/2.
moment of inertia (I)of A/moment of inertia (I)of B=MR2/2/MR2/2.
(I) of A/(I) of B=1×12/2/2×(1/2)2/2.
=1×1/2/2×(1/4)/2.
=1/2/(1/2)/2.
=1/2/1/4.
=4/2.
=2/1.
There are some passengers inside a stationary railway compartment. The centre of masses of the compartment itself(without the passengers) is C1, while the centre of mass of the compartment plus passengers’ system is C2. if the passengers moves about inside the compartment
- a)both C1 and C2 will move with respect to the ground
- b)neither C1 nor C2 will move with respect to the ground
- c)C1 will move but C2 will be stationary with respect to the ground
- d)C2 will move but C1 will be stationary with respect to the ground
Correct answer is option 'C'. Can you explain this answer?
There are some passengers inside a stationary railway compartment. The centre of masses of the compartment itself(without the passengers) is C1, while the centre of mass of the compartment plus passengers’ system is C2. if the passengers moves about inside the compartment
a)
both C1 and C2 will move with respect to the ground
b)
neither C1 nor C2 will move with respect to the ground
c)
C1 will move but C2 will be stationary with respect to the ground
d)
C2 will move but C1 will be stationary with respect to the ground
|
|
Lavanya Menon answered |
When net Fexternal=0, then the centre of mass of the system remains at rest.
Thus if the passenger move inside the compartment which donot require any external force, so the centre of mass of the "passenger + compartment" system must remain at rest and hence C2 will be fixed w.r.t ground.
Also due to the movement of the passenger, the position of centre of mass of the passengers only will change, thus C1 will have to move in such a way that C2 may remain fixed w.r.t ground.
If a man of mass M jumps to the ground from height h and his centre of mass moves a distance x in the time taken by him to hit the ground, the average force acting on him is (assuming his retardation to be constant during his impact with the ground)- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
If a man of mass M jumps to the ground from height h and his centre of mass moves a distance x in the time taken by him to hit the ground, the average force acting on him is (assuming his retardation to be constant during his impact with the ground)
a)
b)
c)
d)
|
|
Neha Joshi answered |
As the center of mass moves through a distance x, the average force F acting on the man is calculated from :
Work done = Change in potential energy
⇒ Fx = Mgh
⇒ F =
Work done = Change in potential energy
⇒ Fx = Mgh
⇒ F =
The mass of an electron is 
kg. It revolves around the nucleus of an atom in a circular orbit of radius 4.5 Å, with a speed of 
m/s. The angular momentum of electron is- a)2.24 x 10-4 kgm2s-1
- b)3.24 x 10-34 kgm2s-1
- c)4.24 x10-34 kgm2s-1
- d)2.24 x 10-34 kgm2s-1
Correct answer is option 'B'. Can you explain this answer?
The mass of an electron is kg. It revolves around the nucleus of an atom in a circular orbit of radius 4.5 Å, with a speed of m/s. The angular momentum of electron is
kg. It revolves around the nucleus of an atom in a circular orbit of radius 4.5 Å, with a speed of m/s. The angular momentum of electron isa)
2.24 x 10-4 kgm2s-1
b)
3.24 x 10-34 kgm2s-1
c)
4.24 x10-34 kgm2s-1
d)
2.24 x 10-34 kgm2s-1
|
|
Krishna Iyer answered |
M=9x10-31Kg,r=4.5x10-10
V=8x105m/s
L=mvr
=9x10-31 x 4.5 x 10-10 x 8 x 105 = 3.24 x 10-34 kgm2s-1
V=8x105m/s
L=mvr
=9x10-31 x 4.5 x 10-10 x 8 x 105 = 3.24 x 10-34 kgm2s-1
Can you explain the answer of this question below:The angular position of a particle (in radians), along a circle of radius 0.8 m is given by the function in time (seconds) by 
. The linear velocity of the particle
- A:
1.84 m/s
- B:
1.80 m/s
- C:
1.82 m/s
- D:
1.88 m/s
The answer is a.
The angular position of a particle (in radians), along a circle of radius 0.8 m is given by the function in time (seconds) by . The linear velocity of the particle
1.84 m/s
1.80 m/s
1.82 m/s
1.88 m/s
|
Knowledge Hub answered |
The answer is option A
So, as we know,
After differentiating angular velocity with respect to t,
Linear velocity=2t+2.3=ω
Now,
Velocity=r x ω(where r is 0.8m)
=0.8x2.3
=1.84m/s
So, as we know,
After differentiating angular velocity with respect to t,
Linear velocity=2t+2.3=ω
Now,
Velocity=r x ω(where r is 0.8m)
=0.8x2.3
=1.84m/s
When external forces acting on a body are zero, then its centre of mass- a)remains stationary
- b)moves with uniform velocity
- c)either remains stationary or moves with uniform velocity
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
When external forces acting on a body are zero, then its centre of mass
a)
remains stationary
b)
moves with uniform velocity
c)
either remains stationary or moves with uniform velocity
d)
none of these
|
|
Naina Sharma answered |
When force acting upon the body results zero, the resulting acceleration due to net force applied is also zero, and hence by the law of inertia the motion of the body either at rest or constant velocity wont change.
Two rings have their moment of inertia in the ratio 2:1 and their diameters are in the ratio 2:1. The ratio of their masses will be:- a)1:2
- b)2:1
- c)1:4
- d)1:1
Correct answer is option 'A'. Can you explain this answer?
Two rings have their moment of inertia in the ratio 2:1 and their diameters are in the ratio 2:1. The ratio of their masses will be:
a)
1:2
b)
2:1
c)
1:4
d)
1:1
|
|
Suresh Reddy answered |
We know that MI of a ring is mr2
Where m is mass of the ring and r is its radius
When we have ratio of I = 2:1
And ratio of r = 2:1
We get ratio of r2 = 4:1
Thus to make this ratio 2:1 , that ratio of masses must be 1:2
Where m is mass of the ring and r is its radius
When we have ratio of I = 2:1
And ratio of r = 2:1
We get ratio of r2 = 4:1
Thus to make this ratio 2:1 , that ratio of masses must be 1:2
A solid sphere and a hollow sphere of the same mass have the same moments of inertia about their respective diameters, the ratio of their radii is- a)(5)1/2 : (3)1/2
- b) (3)1/2 : (5)1/2
- c)3 : 2
- d)2 : 3
Correct answer is option 'A'. Can you explain this answer?
A solid sphere and a hollow sphere of the same mass have the same moments of inertia about their respective diameters, the ratio of their radii is
a)
(5)1/2 : (3)1/2
b)
(3)1/2 : (5)1/2
c)
3 : 2
d)
2 : 3
|
|
Gaurav Kumar answered |
We know moment of inertia of solid sphere Is=2/5msRs2 and
moment of inertia of hollow sphere IH=2/3mHRH2 As per question Is=IH
Now,
2/5msRs2=2/3mHRH2
as the masses are equal the ratio of their radii will be
Rs2 /RH2 =2/3/2/5=√5/3=(5)1/2: (3)1/2
moment of inertia of hollow sphere IH=2/3mHRH2 As per question Is=IH
Now,
2/5msRs2=2/3mHRH2
as the masses are equal the ratio of their radii will be
Rs2 /RH2 =2/3/2/5=√5/3=(5)1/2: (3)1/2
If three balls of same radius are placed touching each other on a horizontal surface such that they will form an equilateral triangle, when their centers are joined. What will be the position of the centre of mass of the system?- a)At the point of intersection of the medians.
- b)At the line joining the centers of any two balls.
- c)At the centre of one of the ball.
- d)At the horizontal surface.
Correct answer is option 'A'. Can you explain this answer?
If three balls of same radius are placed touching each other on a horizontal surface such that they will form an equilateral triangle, when their centers are joined. What will be the position of the centre of mass of the system?
a)
At the point of intersection of the medians.
b)
At the line joining the centers of any two balls.
c)
At the centre of one of the ball.
d)
At the horizontal surface.
|
|
Neha Joshi answered |
The COM will be at the intersection point of the medians as that is the most symmetric point of this symmetric system.
The M.I. of a disc about its diameter is 2 units. Its M.I. about axis through a point on its rim and in the plane of the disc is- a)4 unit
- b)6 unit
- c)8 unit
- d)10 unit
Correct answer is option 'D'. Can you explain this answer?
The M.I. of a disc about its diameter is 2 units. Its M.I. about axis through a point on its rim and in the plane of the disc is
a)
4 unit
b)
6 unit
c)
8 unit
d)
10 unit
|
|
Krishna Iyer answered |
We know that for a disc of mass m and radius r
MI of a disc about its diameter = mr2/4 = 2
And also MI about a point on its rim = mr2/4 + mr2
= 5mr2/4
= 5 x 2 = 10
MI of a disc about its diameter = mr2/4 = 2
And also MI about a point on its rim = mr2/4 + mr2
= 5mr2/4
= 5 x 2 = 10
A 2 m rod of mass 1 kg rotates at an angular speed of 15 rad/s about its ends. Then K.E. is- a)150 J
- b)1500 J
- c)100 J
- d)1.5 J
Correct answer is option 'A'. Can you explain this answer?
A 2 m rod of mass 1 kg rotates at an angular speed of 15 rad/s about its ends. Then K.E. is
a)
150 J
b)
1500 J
c)
100 J
d)
1.5 J
|
Princi Gupta answered |
We use 1/2 I omega square in which we take I is equal to 1/3 ML square and got the answer ...
A particle of mass 1 kg is fired from the origin of the co-ordinate axis-making angle 45 with the horizontal. After time t its position vector is 
and velocity 
. What is the angular momentum of the particle at that instant?- a)-25
- b)-7
- c)25
- d)7
Correct answer is option 'A'. Can you explain this answer?
A particle of mass 1 kg is fired from the origin of the co-ordinate axis-making angle 45 with the horizontal. After time t its position vector is and velocity . What is the angular momentum of the particle at that instant?
and velocity . What is the angular momentum of the particle at that instant?a)
-25
b)
-7
c)
25
d)
7
|
|
Anaya Patel answered |
A uniform circular disc of radius 50 cm at rest is free to turn about an axis having perpendicular to its plane and passes through its centre. This situation can be shown by the figure given below:
Therefore,
Therefore,
Angular acceleration = α = 2 rad/s2
Angular speed, ω = αt = 4 rad/s
Centripetal acceleration, ac = ω2r
42 x 0.5
=16 x 0.5
= 8m/s2
Linear acceleration at the end of 2 s is,
at = αt = 2 x 0.5 = 1 m/s2
Therefore, the net acceleration at the end of 2.0 sec is given by
If the net force acting on the system of particles is zero, then which of the following may vary
- a)moment of inertia
- b)Kinetic energy of the system
- c)velocity of centre of mass
- d)position of centre of mass
Correct answer is option 'D'. Can you explain this answer?
If the net force acting on the system of particles is zero, then which of the following may vary
a)
moment of inertia
b)
Kinetic energy of the system
c)
velocity of centre of mass
d)
position of centre of mass
|
|
Suresh Kumar answered |
Correct answer is B . because when a=0 then v=constant hence K.E. will be constant and position of com will vary
A CD accelerates uniformly from rest to 200 spins per minute in 8 seconds. If the rate of change of speed is constant, determine the instantaneous acceleration in rad/s2 .- a)200.2 π rad/s2
- b)25 rad/ s2
- c)20.9 rad/ s2
- d)2.6 rad/ s2
Correct answer is option 'D'. Can you explain this answer?
A CD accelerates uniformly from rest to 200 spins per minute in 8 seconds. If the rate of change of speed is constant, determine the instantaneous acceleration in rad/s2 .
a)
200.2 π rad/s2
b)
25 rad/ s2
c)
20.9 rad/ s2
d)
2.6 rad/ s2
|
Varun Kapoor answered |
The correct option is A
2.6 rad/ s2
A child sits stationary at one end of long trolley moving uniformly with speed v on a smooth horizontal floor. If the child gets up and runs about on the trolley in the forward direction with speed u. The centre of mass of the system (child + trolley) will move with speed- a)v
- b)zero
- c)u + v
- d)v/u
Correct answer is option 'A'. Can you explain this answer?
A child sits stationary at one end of long trolley moving uniformly with speed v on a smooth horizontal floor. If the child gets up and runs about on the trolley in the forward direction with speed u. The centre of mass of the system (child + trolley) will move with speed
a)
v
b)
zero
c)
u + v
d)
v/u
|
|
Preeti Iyer answered |
The child is running arbitrarily on a trolley moving with velocity v. However, the running of the child will produce no effect on the velocity of the centre of mass of the trolley. This is because the force due to the boy’s motion is purely internal. Internal forces produce no effect on the motion of the bodies on which they act. Since no external force is involved in the boy–trolley system, the boy’s motion will produce no change in the velocity of the centre of mass of the trolley.
A rigid body is one
- a)the sum of distances of all particles from the axis remains constant
- b)in which the distance between all pairs of particles remains fixed
- c)whose centre of mass follows a parabolic path
- d)that deforms and comes back to its original shape after getting deformed
Correct answer is option 'B'. Can you explain this answer?
A rigid body is one
a)
the sum of distances of all particles from the axis remains constant
b)
in which the distance between all pairs of particles remains fixed
c)
whose centre of mass follows a parabolic path
d)
that deforms and comes back to its original shape after getting deformed
|
|
Krishna Iyer answered |
A body is said to be a rigid body if the body remains in its original shape even under the influence of external force. We can also say that if distance between two points of the body does not change with time regardless of external forces exerted on it, then the body is said to be a rigid body.
When a shell was following a parabolic path in the air, it explodes somewhere in its flight. The centre of mass of fragments will continue to move in- a)any direction
- b)horizontal direction
- c)same parabolic path
- d)vertical direction
Correct answer is option 'C'. Can you explain this answer?
When a shell was following a parabolic path in the air, it explodes somewhere in its flight. The centre of mass of fragments will continue to move in
a)
any direction
b)
horizontal direction
c)
same parabolic path
d)
vertical direction
|
|
Riya Banerjee answered |
The internal forces have no effect on the trajectory of the center of mass, and the forces due to explosion are the internal forces. So the center of mass will follow the same parabolic path even after the explosion.
A mixer grinder rotates clockwise, its angular velocity will be :- a)zero
- b)negative
- c)uniform but not zero
- d)positive
Correct answer is option 'B'. Can you explain this answer?
A mixer grinder rotates clockwise, its angular velocity will be :
a)
zero
b)
negative
c)
uniform but not zero
d)
positive
|
Ashish Roy answered |
**Explanation:**
A mixer grinder is a device that is used for grinding and mixing various ingredients. It consists of a motor and a set of blades that rotate at high speeds to perform the grinding and mixing tasks. When the mixer grinder is turned on, the motor starts rotating the blades in a clockwise direction.
**Angular Velocity:**
Angular velocity is a measure of how quickly an object rotates or moves around a central point. It is defined as the rate of change of angular displacement with respect to time. The direction of the angular velocity is determined by the direction of rotation. In the case of a mixer grinder rotating clockwise, the angular velocity will be negative.
**Direction of Angular Velocity:**
The direction of angular velocity is determined by the right-hand rule. According to the right-hand rule, if the fingers of the right hand curl in the direction of rotation, the thumb will point in the direction of the angular velocity vector. In the case of a mixer grinder rotating clockwise, the fingers of the right hand curl in the clockwise direction, and the thumb points in the opposite direction, which is counterclockwise or negative.
**Significance of Negative Angular Velocity:**
A negative angular velocity indicates that the object is rotating in the opposite direction compared to the conventional positive direction. In the case of a mixer grinder, a negative angular velocity means that the blades are rotating counterclockwise when viewed from above. This counterclockwise rotation is necessary for the blades to effectively grind and mix the ingredients.
**Conclusion:**
In conclusion, a mixer grinder rotates clockwise, which means its angular velocity will be negative. The negative angular velocity indicates that the blades are rotating counterclockwise when viewed from above, allowing them to efficiently perform the grinding and mixing tasks.
A mixer grinder is a device that is used for grinding and mixing various ingredients. It consists of a motor and a set of blades that rotate at high speeds to perform the grinding and mixing tasks. When the mixer grinder is turned on, the motor starts rotating the blades in a clockwise direction.
**Angular Velocity:**
Angular velocity is a measure of how quickly an object rotates or moves around a central point. It is defined as the rate of change of angular displacement with respect to time. The direction of the angular velocity is determined by the direction of rotation. In the case of a mixer grinder rotating clockwise, the angular velocity will be negative.
**Direction of Angular Velocity:**
The direction of angular velocity is determined by the right-hand rule. According to the right-hand rule, if the fingers of the right hand curl in the direction of rotation, the thumb will point in the direction of the angular velocity vector. In the case of a mixer grinder rotating clockwise, the fingers of the right hand curl in the clockwise direction, and the thumb points in the opposite direction, which is counterclockwise or negative.
**Significance of Negative Angular Velocity:**
A negative angular velocity indicates that the object is rotating in the opposite direction compared to the conventional positive direction. In the case of a mixer grinder, a negative angular velocity means that the blades are rotating counterclockwise when viewed from above. This counterclockwise rotation is necessary for the blades to effectively grind and mix the ingredients.
**Conclusion:**
In conclusion, a mixer grinder rotates clockwise, which means its angular velocity will be negative. The negative angular velocity indicates that the blades are rotating counterclockwise when viewed from above, allowing them to efficiently perform the grinding and mixing tasks.
Moment of inertia of a solid sphere of mass M and radius R is ______- a)MR2
- b)2/5MR2
- c)2/3MR2
- d)1/2MR2
Correct answer is option 'B'. Can you explain this answer?
Moment of inertia of a solid sphere of mass M and radius R is ______
a)
MR2
b)
2/5MR2
c)
2/3MR2
d)
1/2MR2
|
|
Sarita Pandey answered |
Moment of inertia of a solid sphere of mass M and radius R = 2/5 MR^2
The moment of inertia and rotational kinetic energy of a fly wheel are 20kg-m2 and 1000 joule respectively. Its angular frequency per minute would be -- a)600/π
- b)25/π2
- c)5/π
- d)300/π
Correct answer is option 'D'. Can you explain this answer?
The moment of inertia and rotational kinetic energy of a fly wheel are 20kg-m2 and 1000 joule respectively. Its angular frequency per minute would be -
a)
600/π
b)
25/π2
c)
5/π
d)
300/π
|
|
Riya Banerjee answered |
Rotational K.E. = ½ × moment of inertia × square of angular velocity = ½Iw2
Can you explain the answer of this question below:An isolated particle of mass m is moving in a horizontal plane (x,y) along the x axis at a certain height above the ground. It suddenly explodes into two fragments of masses m/4 and 3m/4. An instant later, the smaller fragments is at y = +15 cm. The larger fragment at this instant is at
- A:
y = +5 cm
- B:
y = +20 cm
- C:
y = -20 cm
- D:
y = -5 cm
The answer is d.
An isolated particle of mass m is moving in a horizontal plane (x,y) along the x axis at a certain height above the ground. It suddenly explodes into two fragments of masses m/4 and 3m/4. An instant later, the smaller fragments is at y = +15 cm. The larger fragment at this instant is at
y = +5 cm
y = +20 cm
y = -20 cm
y = -5 cm
|
|
Suresh Iyer answered |
As the particle is exploded only due to its internal energy.
⇒ net external force during this process is 0 i.e. center mass will not change.
⇒ net external force during this process is 0 i.e. center mass will not change.
Let the particle while the explosion was above the origin of the coordinate system i.e. just before explosion xcm =0 and ycm =0
After the explosion, the Centre of mass will be at xcm =0 and ycm =0
Since smaller fragment has fallen on the y-axis.
Let positon of larger fragment be y.
Let positon of larger fragment be y.
m * ycm = (m/4 * 15) + (3m/4 * y)
⇒ (m/4 * 15) + (3m/4 * y) = 0
⇒ y = - 5 cm
⇒ (m/4 * 15) + (3m/4 * y) = 0
⇒ y = - 5 cm
A small cylinder rolling with a velocity v along a horizontal surface encounters a smooth inclined surface. The height ‘h’ up to which the cylinder will ascend is- a)3v2/4g
- b)v2/2g
- c)3v2/2g
- d)v2/4g
Correct answer is option 'B'. Can you explain this answer?
A small cylinder rolling with a velocity v along a horizontal surface encounters a smooth inclined surface. The height ‘h’ up to which the cylinder will ascend is
a)
3v2/4g
b)
v2/2g
c)
3v2/2g
d)
v2/4g
|
|
Shreya Gupta answered |
A body can roll along a surface only if the surface is rough. The body will roll up to the foot of the inclined smooth surface. It will continue to spin with the angular speed it has acquired, and will slide up to a certain height, maintaining its spin motion throughout the smooth surface. Its translational kinetic energy alone is responsible for its upward motion along the smooth incline so that the height up to which it will rise is given by
A rigid body is rotating about an axis. Different particles are at different distances from the axis. Which of the following is true?- a)Different particles possess the same angular acceleration
- b)Different particles possess the same angular velocity
- c)Different particles possess the same linear acceleration
- d)Different particles possess the same linear velocity.
Correct answer is option 'B'. Can you explain this answer?
A rigid body is rotating about an axis. Different particles are at different distances from the axis. Which of the following is true?
a)
Different particles possess the same angular acceleration
b)
Different particles possess the same angular velocity
c)
Different particles possess the same linear acceleration
d)
Different particles possess the same linear velocity.
|
|
Suresh Reddy answered |
In the fixed axis rotation we see that every point on the body has two components of velocity, one in the radial direction and one in the tangential direction. The resultant of these velocities is not the same for any two points lying in the plane of the body.
Any two points on the radial line have the radial acceleration directed towards the center of equal magnitude and the tangential acceleration of equal magnitude as well. Thus option B is correct.
All the particles lying on the curved surface of a cylinder whose axis coincides with the axis of rotation have the same speed but different velocities.
Any two points on the radial line have the radial acceleration directed towards the center of equal magnitude and the tangential acceleration of equal magnitude as well. Thus option B is correct.
All the particles lying on the curved surface of a cylinder whose axis coincides with the axis of rotation have the same speed but different velocities.
An automobile engine develops 100H.P. when rotating at a speed of 1800 rad/min. The torque it delivers is- a)3.33 W-s
- b)200W-s
- c)248.7 W-s
- d)2487 W-s
Correct answer is option 'D'. Can you explain this answer?
An automobile engine develops 100H.P. when rotating at a speed of 1800 rad/min. The torque it delivers is
a)
3.33 W-s
b)
200W-s
c)
248.7 W-s
d)
2487 W-s
|
Manish Aggarwal answered |
100 HP = 74570 W or 74.57 KW Now, P = 2*π*N*T/60 where, P is the power (in W), N is the operating speed of the engine (in r.p.m.) and T is the Torque (in N.m). Therefore, 74570 = 2*π*1800*T/60 i.e. T = 395.606 N.m
Calculate the M.I. of a thin uniform ring about an axis tangent to the ring and in a plane of the ring, if its M.I. about an axis passing through the centre and perpendicular to plane is 4 kg m2.- a)12 kg m2
- b)3 kg m2
- c)6 kg m2
- d)9 kg m2
Correct answer is option 'C'. Can you explain this answer?
Calculate the M.I. of a thin uniform ring about an axis tangent to the ring and in a plane of the ring, if its M.I. about an axis passing through the centre and perpendicular to plane is 4 kg m2.
a)
12 kg m2
b)
3 kg m2
c)
6 kg m2
d)
9 kg m2
|
|
Naina Sharma answered |
At tangent I =MR2/2 + MR2 --- (1)
MR2/2 = 4, MR2 = 8 then substitute in (1) you will get I = 12, this is for tangent perpendicular to plane then divide by 2 you will get tangent along the plane
MR2/2 = 4, MR2 = 8 then substitute in (1) you will get I = 12, this is for tangent perpendicular to plane then divide by 2 you will get tangent along the plane
The angular velocity of a second hand of a clock is :- a)π/ 30 rad/s
- b)π/5 rad/s
- c)2π rad/s
- d)π /3 rad/s
Correct answer is option 'A'. Can you explain this answer?
The angular velocity of a second hand of a clock is :
a)
π/ 30 rad/s
b)
π/5 rad/s
c)
2π rad/s
d)
π /3 rad/s
|
|
Krishna Iyer answered |
Time taken to complete one rotation is 60 seconds.
One rotation involves 360 degree = 2π
Formula of angular velocity = theta/time taken
Hence, omega = 2π/60
= π/30 rad/s
One rotation involves 360 degree = 2π
Formula of angular velocity = theta/time taken
Hence, omega = 2π/60
= π/30 rad/s
An earth satellite is moving around the earth in a circular orbit. In such case, what is conserved?- a)force
- b)velocity
- c)angular momentum
- d)linear momentum
Correct answer is option 'C'. Can you explain this answer?
An earth satellite is moving around the earth in a circular orbit. In such case, what is conserved?
a)
force
b)
velocity
c)
angular momentum
d)
linear momentum
|
|
Geetika Shah answered |
An earth satellite is moving in a circular motion due to change in the direction of velocity. Its angular momentum about centre is conserved as there is no external force acting on the system.
A weightless rod is acted on by upward parallel forces of 2N and 4N ends A and B respectively. The total length of the rod AB = 3m. To keep the rod in equilibrium a force of 6N should act in the following manner :- a)Downwards at any point between A and B.
- b) Downwards at mid point of AB.
- c)Downwards at a point C such that AC = 1m.
- d) Downwards at a point D such that BD = 1m.
Correct answer is option 'D'. Can you explain this answer?
A weightless rod is acted on by upward parallel forces of 2N and 4N ends A and B respectively. The total length of the rod AB = 3m. To keep the rod in equilibrium a force of 6N should act in the following manner :
a)
Downwards at any point between A and B.
b)
Downwards at mid point of AB.
c)
Downwards at a point C such that AC = 1m.
d)
Downwards at a point D such that BD = 1m.
|
|
Riya Banerjee answered |
A boy is playing with a tire of radius 0.5m. He accelerates it from 5rpm to 25 rpm in 15 seconds. The linear acceleration of tire is- a)70 m/s2
- b)7 m/s2
- c)10 m/s2
- d)0.07 m/s2
Correct answer is option 'D'. Can you explain this answer?
A boy is playing with a tire of radius 0.5m. He accelerates it from 5rpm to 25 rpm in 15 seconds. The linear acceleration of tire is
a)
70 m/s2
b)
7 m/s2
c)
10 m/s2
d)
0.07 m/s2
|
|
Naina Sharma answered |
r=0.5m, t=15s
n1=5rpm=5/60 rps
n2=25rpm=25/60rps
… ω1=2π(5) rad/s
… ω2=2π(25) rad/s
As,
a=r∝ [∝=ω2- ω1/t]
so, a=0.5[2π(n2-n1)/t]
a=0.5x6.28x20/60x15
a=6.28x2/60x3
a=6.28/90
a=0.697 m/s2
a≈0.7 m/s2
n1=5rpm=5/60 rps
n2=25rpm=25/60rps
… ω1=2π(5) rad/s
… ω2=2π(25) rad/s
As,
a=r∝ [∝=ω2- ω1/t]
so, a=0.5[2π(n2-n1)/t]
a=0.5x6.28x20/60x15
a=6.28x2/60x3
a=6.28/90
a=0.697 m/s2
a≈0.7 m/s2
A disc of radius 2m and mass 200kg is acted upon by a torque 100N-m. Its angular acceleration would be- a) 1 rad/sec2
- b)0.25 rad/sec2
- c)0.5 rad/sec2
- d)2 rad/sec2.
Correct answer is option 'B'. Can you explain this answer?
A disc of radius 2m and mass 200kg is acted upon by a torque 100N-m. Its angular acceleration would be
a)
1 rad/sec2
b)
0.25 rad/sec2
c)
0.5 rad/sec2
d)
2 rad/sec2.
|
|
Neha Joshi answered |
F = ma
a = Wr
So, F = mwr
a = wr = acceleration
m = 200kg
F = 100 n-m, r = 2m, w = to be calculated
100 = 200 x w2
w = 0.25 rad/sec2
a = Wr
So, F = mwr
a = wr = acceleration
m = 200kg
F = 100 n-m, r = 2m, w = to be calculated
100 = 200 x w2
w = 0.25 rad/sec2
In an equilateral triangle of length 6 cm , three masses m1= 40g , m2= 60g and m3= 60g are located at the vertices. The moment of inertia of the system about an axis along the altitude of the triangle passing through m1 is
- a)1080 g cm2
- b)1020 g cm2
- c)1000 g cm2
- d)1100 g cm2
Correct answer is option 'A'. Can you explain this answer?
In an equilateral triangle of length 6 cm , three masses m1= 40g , m2= 60g and m3= 60g are located at the vertices. The moment of inertia of the system about an axis along the altitude of the triangle passing through m1 is
a)
1080 g cm2
b)
1020 g cm2
c)
1000 g cm2
d)
1100 g cm2
|
|
Naina Sharma answered |
Axis is along AD so Moment of inertia of first mass is zero and 2nd and 3rd masses are at distance of 6/2 from axis, so by using simply I=mr2 we get, I=60×36/4+60×36/4=1080 g cm2
The earth rotates about an axis passing through its north and south poles with a period of one day. If it is struck by meteorites, then- a)moment of inertia decreases and angular velocity increases
- b)moment of inertia decreases and angular velocity decreases
- c)moment of inertia increases and angular velocity decreases
- d)moment of inertia increases and angular velocity increases
Correct answer is option 'C'. Can you explain this answer?
The earth rotates about an axis passing through its north and south poles with a period of one day. If it is struck by meteorites, then
a)
moment of inertia decreases and angular velocity increases
b)
moment of inertia decreases and angular velocity decreases
c)
moment of inertia increases and angular velocity decreases
d)
moment of inertia increases and angular velocity increases
|
|
Neha Joshi answered |
Because angular momentum remains constant as the moment of inertia is increasing here so the angular velocity will decrease.
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