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**NCERT Solutions for Class 11 Physics Chapter 7 System Of Particles And Rotational Motion** are considered to be both highly important and challenging within the Class 11 NCERT syllabus. On this particular page of Careers360, you will find comprehensive class 11 physics chapter 7 exercise solutions, meticulously crafted by subject experts. These NCERT Solutions intricately elucidate each step in a simplified manner, making the understanding of this complex topic more accessible. The physics class 11 chapter 7 NCERT solutions encompass a total of thirty-three questions, spanning from 7.1 to 7.21 in the exercise questions section, and from 7.22 to 7.33 in the additional exercise section.

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- Significance of NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion:

ch 7 class 11 physics System of particles and rotational motion is part of Ncert solution for class 11 physics . The chapter starts with the concepts of the rigid body. In the solutions of NCERT Cass 11 physics chapter 7 System of Particles and Rotational Motion, a detailed explanation of answers is given by considering the bodies to be rigid. Ideally speaking a rigid body is a body with a perfectly definite and unaltering shape. The distance between all pairs of particles of such a body does not change. According to this definition, the bodies that we see in our real life are not rigid because they will deform when a force is applied. But when this deformation is negligible we consider the bodies as rigid.

The CBSE NCERT solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion are important as far as board and competitive exams are considered. The two main topics that we come across in NCERT are the moment of inertia and the centre of mass. There are questions and answers based on these topics in the NCERT solutions for Class 11 Physics Chapter 7 system of particles and rotational motion. NCERT solutions for class 11 physics chapter 7 pdf help students in self-study and score good marks in exams.

**NCERT Solutions for Class 11 Physics Chapter 7 pdf download**** **for free.

****According to the CBSE Syllabus for the academic year 2023-24, the chapter you previously referred to as Chapter 7, " System of particles and rotational motion," has been renumbered as Chapter 6.**

** Answer: **

For uniform mass density, the location of the centre of mass is the same as that of the geometric centre.

No, it is not necessary that the centre of mass lies inside the body. For example in the case of a ring the centre of mass outside the body.

** Answer: **

Let us assume that the centre of mass of the molecule is x cm away from the chlorine atom.

So, we can write

or

Here negative sign indicates that the centre of mass lies leftward of the chlorine atom.

** Answer: **

Since child and trolley are considered in a system thus speed of the centre of mass change only when an external force will act on the system. When the child starts to move on the trolley the force produced is the internal force of the system so this will produce ** no change ** in the velocity of the CM.

** Q7.4 ** Show that the area of the triangle contained between the vectors is one half of the magnitude of .

** Answer: **

Let a and b be two vectors having angle between them.

Consider MON,

or

or

= 2 (Area of MOK)

Therefore the area of MOK = One half of .

** Answer: **

A parallelepiped is shown in the figure given below:-

Volume is given by : = abc

We can write :

(The direction of is in the direction of vector a.)

Now,

This is equal to volume of parallelepiped.

** Q7.6 ** Find the components along the axes of the angular momentum of a particle, whose position vector is with components and momentum is with components , and . Show that if the particle moves only in the plane the angular momentum has only a z-component.

** Answer: **

Linear momentum of particle is given by :

And the angular momentum is :

When particle is confined to x-y plane then z = 0 and p _{ z } = 0.

When we put the value of z and p _{ z } in the equation of linear momentum then we observe that only the z component is non-zero.

** Answer: **

Assume two points (say A, B) separated by distance d.

So the angular momentum of a point about point A is given by :

And about point B :

Now assume a point between A and B as C which is at y distance from point B.

Now the angular momentum becomes :

Thus it can be seen that angular momentum is independent of the point about which it is measured.

** Answer: **

The FBD of the given bar is shown below :

Since the bar is in equilibrium, we can write :

or

or .....................................................(i)

For the rotational equilibrium :

(Use equation (i) to solve this equation)

or

Thus the center of gravity is at 0.72 m from the left.

** Answer: **

The FBD of the car is shown below :

We will use conditions of equilibrium here :

....................................(i)

For rotational equilibrium :

or

..............................(ii)

From (i) and (ii) we get :

and

Thus force exerted by the front wheel is = 3675 N

and force exerted by back wheel = 5145 N.

** Answer: **

We know that moment of inertia of a sphere about diameter is :

Using parallel axes theorem we can find MI about the tangent.

Moment of inertia of a sphere about tangent :

** Answer: **

We know that moment of inertia of a disc about its diameter is :

Using perpendicular axes theorem we can write :

Moment of inertia of disc about its centre:-

Using parallel axes theorem we can find the required MI :

Moment of inertia about an axis normal to the disc and passing through a point on its edge is given by :

** Answer: **

The moment of inertia of hollow cylinder is given by

And the moment of inertia of solid cylinder is given by :

We know that :

Let the torque for hollow cylinder be and for solid cylinder let it be .

According to the question :

So we can write the ratio of the angular acceleration of both the objects.

Now for angular velocity,

Clearly, the angular velocity of the solid sphere is more than the angular velocity of the hollow sphere. (As the angular acceleration of solid sphere is greater).

** Answer: **

Firstly we will calculate moment of inertia of the solid cyliner :

So the kinetic energy is given by :

And the angular momentum is given by :

** Answer: **

We are given with the initial angular speed and the relation between the moment of inertia of both the cases.

Here we can use conservation of angular momentum as no external force is acting the system.

So we can write :

** Answer: **

The final and initial velocities are given below :

and

Taking the ratio of both we get,

or

or

Thus the final energy is 2.5 times the initial energy.

The increase in energy is due to the internal energy of the boy.

** Answer: **

The moment of inertia is given by :

or

And the torque is given by :

Also,

So

And the linear acceleration is

** Answer: **

The relation between power and torque is given by :

or

Hence the required power is 36 KW.

** Answer: **

Let the mass per unit area of the disc be .

So total mass is

Mass of the smaller disc is given by :

Now since the disc is removed, we can assume that part to have negative mass with respect to the initial condition.

So, the centre of mass of the disc is given by the formula :

or

Hence the centre of mass is shifted leftward from point O.

** Q7.17 ** A metre stick is balanced on a knife edge at its centre. When two coins, each of mass are put one on top of the other at the mark, the stick is found to be balanced at . What is the mass of the metre stick?

** Answer: **

The centre of mass of meter stick is at 50 cm.

Let the mass of meter stick be m.

Now according to the situation given in the question, we will use the rotational equilibrium condition at the centre point of the meter stick.

or

Thus the mass of the meter stick is 66 g.

** Q7.18 (a) ** A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination.

(a) Will it reach the bottom with the same speed in each case?

** Answer: **

Let the height of the plane is h and mass of the sphere is m.

Let the velocity at the bottom point of incline be v.

So the total energy is given by :

Using the law of conservation of energy :

For solid sphere moment of inertia is :

So,

Put v = wr and solve the above equation.

We obtain :

or

Thus the sphere will reach the bottom at the same speed since it doesn't depend upon the angle of inclination.

(b) Will it take longer to roll down one plane than the other?

** Answer: **

Let us assume the angle of inclination to be and .

And the acceleration of the sphere will be and respectively.

Thus (since )

Since their acceleration are different so the time taken to roll down the inclined plane will be different in both the cases.

** Answer: **

The equation of motion gives :

So

Since we know that

So

Hence the inclined plane having a smaller angle of inclination will take more time.

** Answer: **

Total energy of hoop is given by :

Also the moment inertia of hoop is given by :

We get,

And v = rw

So the work required is : _{}

Answer:

We are given the moment of inertia and the velocity of the molecule.

Let the mass of oxygen molecule be m.

So the mass of each oxygen atom is given by :

Moment of inertia is :

or

or

We are given that :

or

or

or

(a) How far will the cylinder go up the plane?

** Answer: **

The rotational energy is converted into the translational energy.(Law of conservation of energy)

Since the moment of inertia for the cylinder is :

Putting the value of MI and v = wr in the above equation, we get :

or

or

Now using the geometry of the cylinder we can write :

or

or

Thus cylinder will travel up to 3.82 m up the incline.

** Q7.21 (b) ** A solid cylinder rolls up an inclined plane of angle of inclination . At the bottom of the inclined plane the centre of mass of the cylinder has a speed of .

b) How long will it take to return to the bottom?

** Answer: **

The velocity of cylinder is given by :

or

We know that for cylinder :

Thus

Required time is :

or

Hence required time is 0.764(2) = 1.53 s.

** Answer: **

The FBD of the figure is shown below :

Consider triangle ADH,

(AD and DH can be found using geometrical analysis.)

Now we will use the equilibrium conditions :

(i) For translational equilibrium :

......................................(i)

(ii) For rotational equilibrium :

or

or ............................................................(ii)

Using (i) and (ii) we get :

and

Now calculate moment about point A :

Solve the equation :

(a) What is his new angular speed? (Neglect friction.)

** Answer: **

Moment of inertia when hands are stretched :

So the moment of inertia of system (initial ) = 7.6 + 8.1 = 15.7 Kg m ^{ 2 } .

Now, the moment of inertia when hands are folded :

Thus net final moment of inertia is : = 7.6 + 0.4 = 8 Kg m ^{ 2 } .

Using ** conversation of angular momentum ** we can write :

or

(b) Is kinetic energy conserved in the process? If not, from where does the change come about?

** Answer: **

No, the kinetic is not constant. The kinetic energy increases with decrease in moment of inertia. The work done by man in folding and stretching hands is responsible for this result.

** Answer: **

The imparted angular momentum is given by :

Putting all the given values in the above equation we get :

Now, the moment of inertia of door is :

or

Also,

or

(a) What is the angular speed of the two-disc system?

** Answer: **

Let the moment of inertia of disc I and disc II be I _{ 1 } and I _{ 2 } respectively.

Similarly, the angular speed of disc I and disc II be w _{ 1 } and w _{ 2 } respectively.

So the angular momentum can be written as :

and

Thus the total initial angular momentum is :

Now when the two discs are combined the angular momentum is :

Using conservation of angular momentum :

Thus angular velocity is :

(b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? Take .

** Answer: **

The initial kinetic energy is written as :

or

Now the final kinetic energy is :

Put the value of final angular velocity from part (a).

We need to find :

Solve the above equation, we get :

or

(Since none of the quantity can be negative.)

Thus initial energy is greater than the final energy. (due to frictional force).

** Q7.26 (a) ** Prove the theorem of perpendicular axes.

(Hint: Square of the distance of a point in the plane from an axis through the origin and perpendicular to the plane is ).

** Answer: **

Consider the figure given below:

The moment of inertia about x axis is given by :

And the moment of inertia about y-axis is :

Now about z-axis :

or

** Answer: **

Consider the figure given below :

The moment of inertia about RS axis :-

Now the moment of inertia about QP axis :-

or

or

Thus

Hence proved.

using dynamical consideration (i.e. by consideration of forces and torques). Note is the radius of gyration of the body about its symmetry axis, and is the radius of the body. The body starts from rest at the top of the plane.

** Answer: **

Consider the given situation :

The total energy when the object is at the top (potential energy) = mgh.

Energy when the object is at the bottom of the plane :

Put and , we get :

By the law of conservation of energy we can write :

or

.

** Q7.28 ** A disc rotating about its axis with angular speed is placed lightly (without any translational push) on a perfectly frictionless table. The radius of the disc is . What are the linear velocities of the points , and on the disc shown in Fig. 7.41? Will the disc roll in the direction indicated?

** Answer: **

Let the angular speed of the disc is .

So the linear velocity can be written as

(a) Point A:-

The magnitude of linear velocity is and it is tangentially rightward.

(b) Point B:-

The magnitude linear velocity is and its direction is tangentially leftward.

(c) Point C:-

The magnitude linear velocity is and its direction is rightward.

The disc cannot roll as the table is frictionless.

** Q7.29 (a) ** Explain why friction is necessary to make the disc in Fig. 7.41 roll in the direction indicated

(a) Give the direction of frictional force at , and the sense of frictional torque, before perfect rolling begins.

** Answer: **

Since the velocity at point B is tangentially leftward so the frictional force will act in the rightward direction.

The sense of frictional torque is perpendicular (outward) to the plane of the disc.

** Q7.29 (b) ** Explain why friction is necessary to make the disc in Fig. 7.41 roll in the direction indicated.

(b) What is the force of friction after perfect rolling begins?

** Answer: **

Perfect rolling will occur when the velocity of the bottom point (B) will be zero. Thus the frictional force acting will be zero.

** Answer: **

Friction is the cause for motion here.

So using Newton's law of motion we can write :

or

Now by the equation of motion, we can write :

or

The torque is given by :

or

or

or

Now using the equation of rotational motion we can write :

or

Condition for rolling is

So we can write :

For ring the moment of inertia is :

So we have :

or

Now in case of the disc, the moment of inertia is :

Thus

or

Hence disc will start rolling first.

(a) How much is the force of friction acting on the cylinder?

** Answer: **

Consider the following figure :

Moment of inertia of cylinder is :

Thus acceleration is given by :

or

Now using Newton's law of motion :

or

or

or

Hence frictional force is 16.3 N.

(b) What is the work done against friction during rolling ?

** Answer: **

We know that the bottommost point of body (which is in contact with surface) is at rest during rolling. Thus work done against the frictional force is zero.

(c) If the inclination of the plane is increased, at what value of does the cylinder begin to skid, and not roll perfectly?

** Answer: **

In case of rolling without any skidding is given by :

Thus

or

or

** Q7.32 (a) ** Read each statement below carefully, and state, with reasons, if it is true or false;

(a) During rolling, the force of friction acts in the same direction as the direction of motion of the of the body.

** Answer: **

** False ** . Friction also opposes the relative motion between the contacted surfaces. In the case of rolling, the cm is moving in backward direction thus the frictional force is directed in the forward direction.

** Q7.32 (b) ** Read each statement below carefully, and state, with reasons, if it is true or false;

(b) The instantaneous speed of the point of contact during rolling is zero.

** Answer: **

** True. ** This is because the translational speed is balanced by rotational speed.

** Q7.32 (c) ** Read each statement below carefully, and state, with reasons, if it is true or false;

(c) The instantaneous acceleration of the point of contact during rolling is zero.

** Answer: **

** False. ** The value of acceleration at contact has some value, it is not zero as the frictional force is zero but the force applied will give some acceleration.

** Q7.32 (d) ** Read each statement below carefully, and state, with reasons, if it is true or false;

(d) For perfect rolling motion, work done against friction is zero.

** Answer: **

True. As the frictional force at the bottommost point is zero, so the work done against it is also zero.

** Q7.32 (e) ** Read each statement below carefully, and state, with reasons, if it is true or false;

(e) A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.

** Answer: **

** True. ** Since it is a frictionless plane so frictional force is zero thus torque is not generated. This results in slipping not rolling.

(a) Show

where is the momentum of the ith particle (of mass )and Note is the velocity

of the ith particle relative to the centre of mass.Also, prove using the definition of the centre of mass

** Answer: **

The momentum of i _{ th } particle is given by :

The velocity of the centre of mass is V.

Then the velocity of ith particle with respect to the center of mass will be :

Now multiply the mass of the particle to both the sides, we get :

or (Here p' _{ i } is the momentum of ith particle with respect to center of mass.)

or

Now consider :

But as per the definition of centre of mass, we know that :

Thus

(b) show where is the total kinetic energy of the system of particles, is the total kinetic energy of the

system when the particle velocities are taken with respect to the centre of mass and is the kinetic energy of the translation of the system as a whole (i.e. of the centre of mass motion of the system). The result has been used in Sec. 7.14.

** Answer: **

From the first part we can write :

or

Squaring both sides (in vector form taking dot products with itself), we get :

or

or

Hence

(c) Show

where

is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember ; rest of the notation is the standard notation used in the chapter. Note and can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.

** Answer: **

The position vector of the ith particle (with respect to the centre of mass) is given by :

or

From the first case we can write :

Taking cross product with position vector we get ;

or

or

** 7.33 (d) ** Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass :

(d) Show

Further, show that

where is the sum of all external torques acting on the system about the centre of mass.

(Hint: Use the definition of centre of mass and third law of motion. Assume the

internal forces between any two particles act along the line joining the particles.)

** Answer: **

Since we know that :

Differentiating the equation with respect to time, we obtain :

or

or

Now using Newton's law of motion we can write :

Thus

Chapter 7 of NCERT Physics Solutions consists of a total of thirty-three questions. Questions 22-33 belong to the additional exercise section. The solutions provided in the NCERT Solutions are explained step by step and are easy to understand. Students can also download the solutions in PDF format and use them according to their convenience. It is always advisable to attempt the problems on their own first and then refer to the solutions provided to check their understanding and identify areas of improvement.

Chapter 1 | |

Chapter 2 | |

Chapter 3 | |

Chapter 4 | |

Chapter 5 | |

Chapter 6 | |

Chapter 7 | System of Particles and Rotational motion |

Chapter 8 | |

Chapter 9 | |

Chapter 10 | |

Chapter 11 | |

Chapter 12 | |

Chapter 13 | |

Chapter 14 | |

Chapter 15 |

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Download EBookGet a grip on Physics with our helpful class 11 physics chapter 7 exercise solutions. They come with important formulas (provided below), easy-to-understand diagrams, and a useful eBook link. Boost your understanding of the System of Particles and Rotational Motion and do great in chapter 7 physics class 11 numericals.

#### Center of mass

#### Motion of COM

#### Angular Velocity & Linear Velocity

Average angular velocity =∆θ/∆t

Instantaneous angular velocity, ω=dθ/dt

#### Moment of Inertia of few useful configurations

Careers360 official website offers PDF format of System of particles and rotational motion ncert solutions from the NCERT textbook. rotational motion class 11 ncert solutions are well-crafted, taking into account your understanding and proficiency, and can aid you in achieving high scores on exams.

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These ncert solutions for class 11 physics chapter 7 pdf can help students score good marks in board exams as well as competitive exams like JEE or NEET.

ncert solutions for class 11 physics chapter 7 are an excellent resource for exam preparation. Students can download the PDF version of these solutions quickly.

Overall, the system of particles class 11 NCERT solutions can help students master the entire chapter with ease.

- To understand the NCERT solutions for Class 11 Physics chapter 7 System of Particles and Rotational Motion knowledge of previous chapters of Class 11 Physics is a prerequisite.
- The NCERT solutions for class 11 help to understand how to apply concepts studied in numerical problems.
- For exams like NEET and JEE Mains, two or more questions are expected from the chapter System of Particles and Rotational Motion.

**NCERT Physics Exemplar Solutions Class 11 For All The Chapters:**

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- NCERT Books Class 11 Physics
- NCERT Syllabus Class 11 Physics
- NCERT Books Class 11
- NCERT Syllabus Class 11

1. What is the weighage for the Class 11 NCERT chapter 7 for NEET exam

For NEET exam around 5% of questions are asked from the chapter system of particles and rotational motion. NEET questions can be practiced refering the previous papers of NEET. For more number of questions refer to NCERT exemplar questions for Rotational Motion.

2. How many questions can be expected from the chapter system of particles and rotational motion for JEE Main exam

2 questions can be expected from System of Particles and Rotational motion for JEE Main exam. Oscillation and Waves is one of the important topics for exams like KVPY, NSEP and JEE Main

3. How to solve difficult questions related to Chapter 7 of NCERT Solutions system of particles and rotational motion?

To answer difficult questions regarding Chapter 7 of NCERT Solutions for Class 11 Physics, it's essential to carefully read the question, review relevant concepts, break down the question, apply concepts, check your answer, and seek help if needed.

4. Which chapter is system of particles and rotational motion Class 11?

The System of Particles and Rotational Motion is the 7th chapter in the Class 11 Physics textbook of NCERT Solutions.

5. how to convert rev min to rad s ?

To convert revolutions per minute (rev/min) to radians per second (rad/s), you can use the following formula:

angular velocity (ω) in rad/s = (2π × n)/60

where n is the angular velocity in revolutions per minute.

So, to convert rev/min to rad/s, multiply the angular velocity in revolutions per minute by 2π/60.

For example, if the angular velocity is given as 100 rev/min, then we can calculate it in radians per second as:

ω = (2π × 100)/60 = 10.47 rad/s

Therefore, the angular velocity in radians per second is 10.47 rad/s.

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