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Important Questions Class 11 Physics Chapter 7
Important Questions for CBSE Class 11 Physics Chapter 7 – Systems of Particles and Rotational Motion
Important Questions for Class 11 Physics Chapter 7 is a set of questions prepared by Extramarks’ subject matter experts that will help students revise the important concepts, exam patterns, and mark distribution related to Chapter 7 ‘Systems of Particles and Rotational Motion’ of Class 11 Physics. This questionnaire is comprehensive and set in accordance with the revised CBSE Syllabus.
Students can refer to Chapter 7 Class 11 Physics Important Questions to enhance their answers with accurate solutions to score better marks. These Important Questions contain detailed explanations of the concepts and terminologies which are easily available on the Extramarks website.
CBSE Class 11 Physics Chapter 7 Important Questions – Free Download
(add important questions)
Study Important Questions for Class 11 Physics Chapter 7 – System of Particles and Rotational Motion
A set of Class 11 Physics Chapter 7 Important Questions are given below. Students can also access the link provided to review additional questions.
Short Questions and Answers
[1 or 2 Mark]
Q1.) Prove that the cross product of two parallel vectors is zero.
Ans: AB=AB sinx
If A and B are parallel to each other,
Then, =0°
AB=0
Q2.) A rope of negligible mass is wound around a hollow cylinder of mass 3kg and radius 40cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30N? What is the linear acceleration of the rope? Consider that no slipping takes place.
Ans. The mass of the hollow cylinder is given as, m=3 kg
Radius of the hollow cylinder is given as, r=40cm=0.4m
Applied force on the given rope is given as, F=30N
The moment of inertia of the hollow cylinder about its geometric axis can be given as
I=mr2
I=30.42
I=0.48kgm2
Torque acting on the rope,
τ=Fr
τ=300.4
τ=12Nm
For angular acceleration , torque can also be given by the expression,
τ=I
=τI=120.48
=25 rad s2
Linear acceleration of the rope can be stated as =ra=0.425=10ms2.
Q3.) A child sits stationary at one end of a long trolley moving uniformly with a speed on a smooth horizontal floor. What will be the speed of the CM of the (trolley + child) system when the child gets up and runs about on the trolley?
Ans. The child runs about on the trolley with an acceleration that has no effect on the velocity of the centre of mass of the trolley. This occurs due to the force caused by the child’s internal motion.
There are no effects on the motion of the bodies on which they are acting when the motion is produced by internal forces. As no external force is involved in the (child + trolley) system, there will be no change in the speed of the centre of mass of the trolley due to the child’s motion.
Q4.) Determine which factors affect the moment of inertia of a body.
Ans: The factors that affect the moment of inertia of a body are as follows.
 Position of the axis of rotation
 Mass of the body
 Shape and size of the body
Q5.) A bullet of mass 10g and speed 500m/s is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0m wide and weighs 12kg. It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it. Hint: The moment of inertia of the door about the vertical axis at one end is ML23
Ans.
The mass of the bullet is given as, m=10g=10103kg
Velocity of the bullet is given as, =500m/s
Width of the door, L=1.0m
Radius of the door, r=12m
Mass of the door is given, M=12kg
Angular momentum transmitted by the bullet on the door:
=mr
=(100103)(500)12=2.5kgm2s1
Moment of inertia of the door can be given as
I=ML23
I=131212=4kgm2
But we have the relation,
=I
I=2.54=0.625rads1, which is the required angular speed.
Long Questions and Answers
[3 or 4 Marks]
Q1.) A circular ring of diameter 40cm and mass 1kg is rotating about an axis normal to its plane and passing through the centre with a frequency of 10 rotations per second. Calculate the angular momentum about its axis of rotation.
Ans. From the question we can write,
R=402cm=20cm=0.2m
M=1kg
=10 rotations/sec
Now, Moment of inertia can be calculated as,
M.I.=MR2=1(0.2)2=0.04kgm2
=2=210=20rad/s
Angular momentum can be measured as
L=0.0420
L=2.51kgm2/s
Q2.) Four particles of mass 1kg, 2kg, 3kg, and 4kg are placed at the four vertices A,B,C, and D of the square of side 1m. Find the position of the centre of mass of the particle.
Ans. From the given data we can infer,
m1=1kg at x1,y1=(0,0)
m2=2kg at x2,y2=(1,0)
m3=3kg at x3,y3=(1,1)
m4=4kg at x4,y4=(0,1)
Now,
Xcm=m1x1+ m2x2+ m3x3+ m4x4, m1+ m2+ m3+ m4=0.5m
Ycm=m1y1+ m2y2+ m3y3+ m4y4m1+ m2+ m3+ m4=0.7m
Therefore, the centre of mass is located at 0.5m,0.7m
Q3.) A hoop of radius 2m weighs 100kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20m/s. How much work has to be done to stop it?
Ans. Given that,
Radius of the hoop is given as, r=2m
Mass of the hoop is, m=100 kg
Velocity of the hoop is,
=20cm/s=0.2m/s
Total energy of the hoop = Translational KE + Rotational KE
Er=12m2+12I2
Moment of inertia of the hoop about its centre can be given as I=mr2
Er=12m2+12(mr2)2
But we have the formula,
=r
E1=12m2+12(mr2)2
E1=12m2+12mr22
The work needed to be done for halting the hoop is same to the total energy of the hoop.
Hence, the required work to be done can be given as,
W=m2=1000.22=4J
Q4.) State whether the statements mentioned below are true or false, along with suitable reasons;
(ⅰ). The instantaneous acceleration of the point of contact during rolling is zero.
Ans: False
During the rolling of a body, its instantaneous acceleration is not equivalent to zero and has a certain value.
(ⅱ). A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.
Ans: True.
A frictional force will work between the body and the surface, resulting in the rolling of the body. The torque provided by this frictional force is needed for rolling. In the absence of frictional force, the body slides from the inclined plane due to its own weight.
(ⅲ) During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body.
Ans: False
Frictional force works in the opposite direction of motion of the centre of mass of a body and the motion of the centre of mass is in the backward direction. Thus, the frictional force works in the forward direction.
(ⅳ). The instantaneous speed of the point of contact during rolling is zero.
Ans: True
The rotation of a body around an axis that goes across the point of contact of the body with the ground is rolling. Thus, the instantaneous speed of the point of contact while rolling is zero.
(ⅴ). For perfect rolling motion, the work done against friction is zero.
Ans: True
The frictional force that works at the lowermost point turns to zero when the rolling motion starts. Hence, the work done against friction is zero as well.
Very Long Questions and Answers
[5 or 6 Marks]
Q1.)
(a). A cat is able to land on its feet after a fall. State a reason to justify this statement.
Ans: A cat stretches its tail when it lands on the ground, leading to an increase in the moment of inertia.
Also, I=constant
The increase in moment of inertia results in a small change in angular speed. Hence, the cat lands on its feet.
(b). If the angular momentum moment of inertia is decreased, will its rotation also be conserved? Explain.
Ans: The moment of inertia of a system decreases from I to I’.
Then the angular speed will increase from to ‘.
We can write,
I=I”
Since, I=constant
‘=II’
K.E. of rotation of the system can be given as
K.E.=12I”2
K.E.=12I’II’2
K.E.=12 I22I’
Therefore, because I'< I the K.E. of the system will increase. Hence, the kinetic energy will not be conserved.
Q2.) Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?
Ans. Assuming that m is the mass and r is the radius of the hollow cylinder as well as the solid sphere.
The moment of inertia of the hollow cylinder about its standard axis can be expressed as,
I1=mr2
The moment of inertia of the solid sphere about an axis that passes through its centre can be presented as,
I2=25mr2
The formula for torque in terms of angular acceleration and moment of inertia can be expressed as,
τ=I
Where,
τ= Torque
= Angular acceleration
I= Moment of inertia
For the hollow cylinder the expression can be given as,
τ1=I11
For the solid sphere the expression can be given as,
τ2=I22
As an equal amount of torque is applied to both bodies it can be stated as,
12=I1I2=mr225mr2=25
2>1 ……….(1)
Using the relation =0+t
Where,
= Angular acceleration
t= Time of rotation
0= Initial angular velocity
= Final angular velocity
For equal 0 and t, we have:
= ………(2)
From equations (1) and (2), we can conclude:
2>1
Therefore, the angular velocity of the solid sphere will be greater than that of the hollow cylinder.
Important Questions For Class 11 Physics Chapter 7
Class 11 Physics Chapter 7 Important Questions
Class Physics 11 Chapter 7 Important Questions is a set of questionnaires prepared by Extramarks for Class 11 students. These Important Questions can be accessed from the Extramarks website as per students’ convenience.
Class 11 Physics Rotational Motion Important Questions
Students of Class 11 must learn the basic concepts of this chapter as it is important to know the basics to understand the complex concepts of the subject in Class 12. These Important Questions include the following.
 Detailed explanation of the concept of torque
 Evaluating the motion of rigid bodies
 Determining the body’s density based on the type of distribution of particles
 Study of conservation laws
 Rotational motion visualisation, etc.
Conclusion
Class 11 Physics Chapter 7 Important Questions helps students cover all the important topics while revising the chapter. It ensures that students have accurate solutions for each question in this chapter.
Q.1 A solid cylinder of mass 20 kg rotates on its axis with angular speed 100 rad s^{1}.The radius of the cylinder is 0.25m.What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?
Marks:4
Ans
Rotational kinetic energy of the cylinder
Q.2
Two particles of masses 1 gm and 2gm are at distance 1 m apart.Find their centre of mass.
Marks:4
Ans
Q.3 Read the assertion and reason carefully to mark the correct option out of the options given below.
Assertion: According to Newton?s second law of motion, If the net force on the system of particles is zero, then the linear momentum of the system is constant.
Reason: The linear momentum of the system is constant because the internal force acting on the particle is zero and the centre of mass is at rest.
AAssertion is true but reason is false.
BAssertion and reason both are false.
CBoth assertion and reason are true and the reason is the correct explanation of the assertion.
DBoth assertion and reason are true but reason is not the correct explanation of the assertion.
Marks:1
Ans
Assertion is true but reason is false.
As per Newton’s second law of motion, the net external force acting on the system is the rate of change of linear momentum. So, the net external force acting on all individual particles is zero, which implies that the momentum of the system is conserved but it does not indicate that centre of mass is at rest and internal forces;on particles within an object is zero.
The centre of mass of the system can move with constant velocity even if its acceleration is zero. Acceleration of the centre of mass is only due to external forces acting on a system of particles and internal forces have no contribution in the acceleration of the system.
Q.4 Angular momentum and aerial velocity of a body of mass m are related as
AL = 2m × areal velocity
BL = 2? × areal velocity
Careal velocity = 2m × L
D2m = L × areal velocity
Marks:1
Ans
L = 2m Ã— areal velocity
Angular momentum, L = 2m × areal velocity
Q.5
\begin{array}{l}\text{A particle starts rotating from rest according to formula}\\ \text{? =}\frac{{\text{3t}}^{\text{3}}}{\text{20}}\text{\u2013}\frac{{\text{t}}^{\text{2}}}{\text{3}}\\ \text{The angular velocity and angular acceleration at t = 5 s will be}\end{array}
A9.72 rad/s and 3.83 rad/s^{2}
B7.92 rad/s and 3.83 rad/s^{2}
C7.92 rad/s and 2.83 rad/s^{2}
D5.92 rad/s and 3.83 rad/
Marks:1
Ans
\begin{array}{l}\text{As, ? =}\frac{{\text{3t}}^{\text{3}}}{\text{20}}\text{\u2013}\frac{{\text{t}}^{\text{2}}}{\text{3}}\\ ?\text{? =}\frac{\text{d?}}{\text{dt}}\\ \text{=}\frac{{\text{9t}}^{\text{2}}}{\text{20}}\text{\u2013}\frac{\text{2t}}{\text{3}}\\ \text{=}\frac{\text{9 \xd7}{\left(\text{5 s}\right)}^{\text{2}}}{\text{20}}\text{\u2013}\frac{\text{2 \xd7}\left(\text{5 s}\right)}{\text{3}}\text{}\\ \text{= 7}{\text{.92 rads}}^{\text{1}}\\ \text{Now, ? =}\frac{\text{d?}}{\text{dt}}\\ \text{=}\frac{\text{18t}}{\text{20}}\text{\u2013}\frac{\text{2}}{\text{3}}\\ \text{=}\frac{\text{18 \xd7 5 s}}{\text{20}}\text{\u2013}\frac{\text{2}}{\text{3}}\\ \text{= 3}{\text{.83 rads}}^{\text{2}}\end{array}Please register to view this section
FAQs (Frequently Asked Questions)
1. What is a flywheel?
The rotational motion is produced by machines like the steam engine and the automobile engine, which contain a disc with a big moment of inertia, known as a flywheel.
2. How is the velocity of the centre of mass of a system of particles represented?
The velocity of the centre of mass of a system of particles is represented by
V=PM,
where P = linear momentum of the system.
3. How Important Questions Class 11 Physics Chapter 7 is useful for students?
This questionnaire consists of comprehensive questions with concise answers. Students can refer to Important Questions Class 11 Physics Chapter 7 by Extramarks before the exams to have a clear understanding of the whole chapter and do the last round of quick revision.