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NCERT Exemplar Class 11 Physics Solutions Chapter 7 - System of Particles and Rotational Motion will give a broad conceptual clearance about particles and their rotational motion. NCERT Class 11 physics solutions defines center of mass and moment of inertia. The experts have prepared the NCERT exemplar class 11 Physics chapter 7 solutions so that the students will never face understanding issues. Newton’s laws play a role here in the chapter, thus using the NCERT exemplar class 11 Physics solutions chapter 7 will make the understanding process easier for students. Overall, the class 11 Physics NCERT exemplar solutions chapter 7 teaches about the motion of extended bodies.

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This Story also Contains

- NCERT Exemplar Class 11 Physics Solutions Chapter 7: MCQI
- NCERT exemplar class 11 physics solutions chapter 7: MCQII
- NCERT exemplar class 11 physics solutions chapter 7: very short answer
- NCERT exemplar class 11 physics solutions chapter 7: short answer
- NCERT exemplar class 11 physics solutions chapter 7: long answer
- NCERT Exemplar Class 11 Physics Solutions Chapter 7 - Main subtopics
- What will students learn in NCERT Exemplar Class 11 Physics Solutions Chapter 7?
- NCERT Exemplar Class 11 Physics Solutions Chapter-Wise
- NCERT Exemplar Class 11th Solutions
- Important topics to cover from NCERT exemplar class 11 Physics solutions chapter 7

Question:7.1

For which of the following does the centre of mass lie outside the body?

(a) A pencil

(b) A shotput

(c) A dice

(d) A bangle

Answer:

The answer is the option (d) A Bangle.The centre of mass of a pencil, a shot put a bangle and a dice all lie at their centre. Since pencil, shot put and dice are solid objects their centre of mass lies within the body. For a bangle, the centre of mass is at its centre, but since it is hollow from inside, it lies outside the bangle.

Question:7.2

Which of the following points is the likely position of the center of mass of the system shown in Fig. 7.1?

(a) A

(b) B

(c) C

(d) D

Answer:

The answer is the option (c) CThe volume of the hollow sphere is occupied equally by the air and sand. Since the pressure exerted by the air region is less than that of sand, the mass of sand is greater than the air. The centre of mass shifts towards a heavier portion which in this case is sand. B is exactly at the line dividing air and sand, so it is incorrect. D is at the farthest end and divides the sand area very unequally, so it also cancels out. C is almost at the centre with regard to the mass.

Question:7.4

When a disc rotates with uniform angular velocity, which of the following is not true?

(a) The sense of rotation remains same.

(b) The orientation of the axis of rotation remains same.

(c) The speed of rotation is non-zero and remains same.

(d) The angular acceleration is non-zero and remains same.

Answer:

The answer is the option (d)Angular acceleration is given by

When is constant, is equal to zero. Thus

Question:7.5

A uniform square plate has a small piece Q of an irregular shape removed and glued to the centre of the plate leaving a hole behind (Fig. 7.3). The moment of inertia about the z-axis is then

(a) increased

(b) decreased

(c) the same

(d) changed in unpredicted manner.

Answer:

The answer is the option (b) The formula for Moment of Inertia is given byWhen the piece Q is removed and glued to the centre of the plate, the mass increases around the axis of rotation. However, we can see from the formula that greater the distance from the axis of rotation higher is the moment of Inertia. Hence B

Question:7.6

In problem 7.5. the CM of the plate is now in the following quadrant of x-y plane

a) I

b) II

c) III

d) IV

Answer:

The answer is the option (c) III. The centre of mass will shift towards the side opposite to Q along the line passing through the axis of rotation. Thus, the new centre of mass will be in quadrant III. Hence, C is the answer.Question:7.7

Answer:

The correct answer is the optionThe density of non-uniform rod is given by

So if b = 0, the density of the rod would be uniform and its centre of mass would lie at the middle point of the length of the rod, i.e.) 0.5

Putting b = 0 for all the options, we get 0.5 only for A.

Hence A is the correct option.

Question:7.9

Choose the correct alternatives:

(a) For a general rotational motion, angular momentum L and angular velocity need not be parallel.

(b) For a rotational motion about a fixed axis, angular momentum L and angular velocity are always parallel.

(c) For a general translational motion , momentum p and velocity v are always parallel.

(d) For a general translational motion, acceleration a and velocity v are always parallel.

Answer:

The correct answer is the option (a) and (c).For general translation motion , where m is a scalar quantity. So, the direction of p is the same as that of v. Hence; they are always parallel.

In case of general rotation motion where the axis of rotation is not symmetric, L is not parallel to the angular velocity.

Question:7.10

Figure 7.4 shows two identical particles 1 and 2, each of mass m, moving in opposite directions with same speed v along parallel lines.

At a particular instant, and are their respective position vectors drawn from point A which is in the plane of the parallel lines . Choose the correct options :

(a) Angular momentum of particle 1 about A is

(b) Angular momentum of particle 2 about A is

(c) Total angular momentum of the system about A is

(d) Total angular momentum of the system about A is

represents a unit vector coming out of the page.

represents a unit vector going into the page.

Answer:

The correct answer is the option (a) and (d)Using the direction of L is determined as perpendicular to the plane of r and p by Right-hand thumb rule.

(out of the page)

(into the page)

Hence a is correct and b wrong.

For total angular momentum

(out of page) (into page)

[ into the page]

Question:7.11

The net external torque on a system of particle about an axis is zero. Which of the following are compatible with it?

a) the forces may be acting radially from a point on the axis

b) the forces may be acting on the axis of rotation

c) the forces may be acting parallel to the axis of rotation

d) the torque caused by some forces may be equal and opposite to that caused by other forces

Answer:

The correct answer is the option (a, b, c, d) Torque is given byOr

where is the angle between both the vectors and is the unit vector perpendicular to the plane of and

- When the force is acting radially in the direction of ,

- When the forces are acting on the axis of rotation then and thus
- The component of forces in the plane of and when they are parallel to the axis of rotation is which is equal to 0. Hence
- When torques are equal in magnitude but opposite in direction, the net resultant is 0.

Question:7.12

Figure 7.5 shows a lamina in x-y plane. Two axes z and z′ pass perpendicular to its plane. A force F acts in the plane of lamina at point P as shown. Which of the following are true? (The point P is closer to z′-axis than the z-axis.)

(a) Torque caused by F about z axis is along

(b) Torque caused by F about z' axis is along

(c) Torque caused by F about z axis is greater in magnitude than that about z axis.

(d) Total torque is given be

Answer:

The correct answer is the option (b) and (c)By right-hand thumb rule, in the formula the direction of is perpendicular to the plane of and

So a is incorrect

So b is correct

is only valid if and are along the same axis, which is not true in this case. Hence d is incorrect.

Question:7.13

With reference to Fig. 7.6 of a cube of edge a and mass m, state whether the following are true or false. (O is the centre of the cube.)

(a) The moment of inertia of cube about z-axis is

(b) The moment of inertia of cube about z' is

(c) The moment of inertia of cube about z'' is

(d)

Answer:

The correct answer is the option (a), (b), and (d).By perpendicular axis theorem

So a is correct

Z and z' axis are parallel, and the distance between them can be calculated as

By parallel axis theorem

So b is correct

Since Z-axis and z" axis are not parallel to each other, so parallel theorem is not applicable here, rendering option c wrong

Since Z axis passes through the centre of the cube, so x and y-axis are symmetric

Hence, d is correct

Question:7.14

Answer:

An object is called small if it is the vertical height or geometric centre lies remarkably close to the earth surface. According to this concept, a building and a pond are smaller objects with their geometrical centres remarkably close to the earth surface. Whereas mountain and lake are extended objects as their geometrical centres are above and below the surface of the earth, respectively.Question:7.15

Answer:

Moment of Inertia is given by where r is the distance of the mass from the axis of rotation. In solid sphere entire mass is distributed from centre to the radius of the sphere whereas in a hollow sphere whole mass is concentrated at the periphery of the sphere. Thus in hollow sphere average value of becomes larger and hence results in a greater moment of Inertia.Question:7.17

Uniform cube of mass m and side a is placed in a frictionless horizontal surface. A vertical force F is applied to the edge as shown in the figure.

Match the following

a) | i) cube will move up |

b) | ii) cube will not exhibit motion |

c) | iii) cube will begin to rotate and slip at A |

d) | iv)normal reaction effectively at from A, no motion |

Answer:

(a) – (ii), (b) – (iii), (c) – (i), (d) – (iv)Let be the moment of force due to F at A and be the moment of force due to mg at A

is anti-clockwise whereas is clockwise

and

Cube will rotate only if

If normal reaction force acts effectively at from A

implies

So due to block will not move

Question:7.18

A uniform sphere of mass m and radius R is placed on a rough horizontal surface. The sphere is struck horizontally at a height h from the floor.

Match the following

a) | i) sphere rolls without slipping with a constant velocity and no loss of energy |

b) | ii) sphere spins clockwise, loses energy by friction |

c) | iii) sphere spins anti-clockwise, loses energy by friction |

d) | iv) sphere has only a translational motion, loses energy by friction |

Answer:

(a)-(iii), (c)-(ii), (d) -(i), (b)-(iv)The sphere rolls without slipping when

Let the velocity of the sphere after applying F be v

Then by the law of conservation of angular momentum

Therefore, the sphere rolls without slipping with a constant velocity and no loss of energy. Thus (d) -(i)

Torque due to force

If , and thus

In this case, the sphere will only have a translation motion and slip against the force of friction. Thus (b)-(iv)

For clockwise rotation of sphere

Or , Thus (c) - (ii)

For anti-clockwise rotation

Thus (a) - (iii)

Question:7.19

Answer:

The vector sum of all torques due to forces at a point is 0. Let us assume that be the torque about a point PThen (as per question)

Now torque about any other point say A will be given by

Since a and are not equal to zero. Thus the sum of all torques about any arbitrary point is not 0 necessarily.

Question:7.20

Answer:

The wheel is a rigid elastic body. When a wheel is in uniform motion about the axis passing through its centre and perpendicular to the plane of the wheel, every particle of the wheel is also in a circular motion about the above axis. The centripetal acceleration acting on each particle is directed towards the axis of rotation due to elastic forces which are in pairs.To set a half wheel into uniform motion about an axis passing through the centre of mass of wheel and perpendicular to its plane an external torque is required. This is because in case of a half-wheel, the distribution of mass of half wheel is not symmetric about the axis of the wheel and thus the direction of angular momentum does not coincide with angular velocity.

Question:7.21

Answer:

In the direction of is perpendicular to the plane of and . So a force can produce torque only along the axis in the direction normal to force.The weight of the door acts along –y-axis. The door is in X-Y plane. So it can rotate the door in the axis along the z-axis and not along the y-axis.

Question:7.22

Answer:

The centre of mass of a regular polygon with n sides lies on its geometric centre. If mass m is placed at all the n vertices, then the C.O.M is again at the geometric centre. Let be the position vector of the COM and of the vacant vertex. Then(when mass is placed at nth vertex also)

The negative sign depicts that the C.O.M lies on the opposite side of the nth vertex.

Question:7.23

Find the centre of mass of a uniform

a) half-disc

b) quarter-disc

Answer:

Let the mass of half-disc be MThen

mass per unit area m

Let us assume that the half-disc is divided into many semi-circular strips with radius from 0 to R

Surface area of a semicircular strip

Area

Mass of strip (dm) =

Let (x,y) be the co-ordinates of c.m. of this strip (x,y)

So the centre of mass of the circular half-disc

Mass per unit area of quarter disc centre of mass of a uniform quarter disc=

For a half disc along y-axis com will be at y = (Using symmetry)

So, centre of mass of quarter disc =

Question:7.24

Answer:

(a) The law of conservation of angular momentum is applicable to this situation because there is no net external torque acting on the system. Gravitational and its normal reaction cancel out each other thus resulting in a 0 net torque.(a) According to the law of conservation of angular momentum

(c) Final Kinetic Energy = (rotational +translational) Kinetic Energy

In the absence of translation energy

As discs don't have any translational motion

The negative sign in the answer shows that final kinetic energy is less than the initial since energy is lost during friction between moving surfaces of the disc.

Question:7.25

Answer:

(a) Before coming in contact with the table, the disc was undergoing only rotational motion about its axis that passes through the centre. The velocity of C.O.M =0 since the point on the axis is considered at rest(b)When placed in contact with the table, the force of friction reduces the linear velocity of a point on the rim.

(c) The linear velocity causes a change in momentum (action force). By Newton’s third law of motion wherein, every object has an equal and opposite reaction a reaction force is applied on the disc due to which it moves in the direction of reaction force, so Center of mass acquires a linear velocity.

(d) Force of friction

(e) The C.O.M acquires a velocity due to reaction force due to rotation. Therefore

at the start when just comes in contact with table is

Angular retardation produced by frictional torque =

As R and F are perpendicular,

(for linear velocity)

(for rotational motion)

Thus frictional force help in pure rolling motion without slipping.

Question:7.27

Answer:

The direction of

and at point of contact is tangentially upward. Frictional force acts due to difference in the two velocities.

is force on 1 due to 2 and acts upward and acts downward

(b) External forces acting on system are and which are equal and opposite in direction

external Torque = (anti clockwise)

As Velocity of Drum 2 is twice,

(e) When velocities of drum 1 and drum 2 become equal, no force of friction will act and hence

Question:7.28

Answer:

Area of square = Area of Rectangle

Thus

(b)

(c)

Question:7.29

Answer:

Let us assume a to be the linear acceleration and as angular acceleration.

(In case of linear motion )

;

This is the maximum force applied on the disc to roll on surface without slipping.

Also, check NCERT solutions for Class 11 other subjects

NCERT Exemplar Class 11 Physics Solutions Chapter 7 PDF download is to be available. But till then, students can use the webpage download function in their browser. It will prove to be helpful for them.

NCERT exemplar solutions for class 11 Physics chapter 7 System of Particles and Rotational, is exclusively prepared only by experts. These solutions help students to prepare for competitive exams like JEE Main and NEET.

- NCERT Solutions for Class 11 Maths
- NCERT Solutions for Class 11 Chemistry
- NCERT Solutions for Class 11 Biology

- NCERT Notes Class 11 Maths
- NCERT Notes for Class 11 Physics
- NCERT Notes for Class 11 Chemistry
- NCERT Notes for Class 11 Biology

Class 11 Physics NCERT Exemplar solutions Chapter 7 System of Particles and Rotational Motion include the following topics:

7.1 Introduction

7.2 Centre of mass

7.3 Motion of centre of mass

7.4 Linear momentum of a system of particles

7.5 Vector product of two Vectors

7.6 Angular velocity and its relation with linear velocity

7.7 Torque and angular momentum

7.8 Equilibrium of a rigid body

7.9 Moment of inertia

7.10 Theorems of perpendicular and parallel axes

7.11 Kinematics of rotational motion about a fixed axis

7.12 Dynamics of rotational motion about a fixed axis

7.13 Angular momentum in the case of rotation about a fixed axis

7.14 Rolling motion

In the chapter, students will learn about the rigid body, how they do not change place even after exerting pressure over them. The concept of the fixed axis, i.e. fixed positioning of the rigid body, is what the students will learn. Newton’s second and third law of motion over particles is explained with many practical examples. In the System of Particles and Rotational Motion, Terms such as vector, mass centering, angular momentum, kinematic and dynamic are used. Class 11 Physics NCERT Exemplar solutions Chapter 7 that are so well prepared by the experts, that it will create better learning and knowledge grasping practice for students.

- NCERT Exemplar for Class 11 Physics
- NCERT Exemplar for Class 11 Maths
- NCERT Exemplar for Class 11 Biology
- NCERT Exemplar for Class 11 Chemistry

It is important that students cover some key topics from this NCERT exemplar class 11 Physics chapter 7 solutions for exams. Here are a few of them:

· Newton’s second and third law of motion- relating to particle size and rigid body is one of the most important topics to learn here.

· A rigid body is the distances between different particles of the body do not change, despite force acting on them. A rigid body if fixed at a single place or in a line, will have only one rotational movement. Nevertheless, if not in a single fixed positioning, there could be pure or combination of transitional movements.

· NCERT exemplar class 11 Physics solutions chapter 7 can be referred to find the motion centre of mass, external body knowledge is required when the one for the internal body is missing.

· The total torque and total external force are two independent subjects and can exist with one being absent.

· The chapter is filled with many terms and concept. It is said to be an important one. Class 11 Physics NCERT Exemplar solutions Chapter 7, will help solve all of these issues.

Chapter 1 | Physical world |

Chapter 2 | Units and Measurement |

Chapter 3 | Motion in a straight line |

Chapter 4 | Motion in a Plane |

Chapter 5 | Laws of Motion |

Chapter 6 | Work, Energy and Power |

Chapter 7 | System of Particles and Rotational motion |

Chapter 8 | Gravitation |

Chapter 9 | Mechanical Properties of Solids |

Chapter 10 | Mechanical Properties of Fluids |

Chapter 11 | Thermal Properties of Matter |

Chapter 12 | Thermodynamics |

Chapter 13 | Kinetic Theory |

Chapter 14 | Oscillations |

Chapter 15 | Waves |

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Application Date:01 June,2024 - 15 June,2024

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