NCERT Solutions for Class 11 Physics Chapter 7
Physics essentially deals with what matter constitutes, its motion and behaviour, and energy and force. While this subject can seem daunting to many students, regular practice and thorough knowledge of basic concepts right from Class 11 can help in studying Physics in a much easier way.
Physics class 11 Chapter 7 can be a tough subject for many students. Hence, getting help from experts becomes necessary while preparing for board exams. Moreover, having a strong base and grip on the topics is very important for students who want to pursue physics or engineering or subjects in any other related field in college. It will help them to understand new topics better and have an edge over the other students.
This is why students must refer to Class 11 Physics Chapter 7 NCERT Solutions that are prepared by the teachers, who are experts in teaching these subjects for years. The main motive of the solutions is to make conceptual understanding of the subject easier for the students and make it seem less daunting for them.
NCERT Solutions for Class 11 Physics Chapter 7 (System of Particles and Rotational Motion)
The chapter will cover the motion of the centre of mass of a system of particles and the usefulness of this concept in understanding the motion of extended bodies. It explains the theoretical and mathematical parts of the concept. You will learn about the unchanging shape of an ideal rigid body in the NCERT solutions Class 11 Physics rotational motion.
The chapter will focus on the angular velocity relation with linear velocity, torque and angular momentum. It includes the kind of motion a rigid body can have and how the centre of mass of a system of particles proceeds.
With the right resources, students can easily crack this chapter. At Extramarks, the solutions for Physics Chapter 7 class 11 have been prepared by the teachers and subject matter experts, who make sure that the answers are easy-to-understand and learn.
NCERT Solutions for Class 11 Physics Chapter 7 – (Free Download)
Physics is a lovely subject but can also be quite difficult to master. It forms the base for many scientific subjects that students take later in life. Class 11 Physics introduces students to topics like thermodynamics, laws of motion, etc. These topics are extremely important for the students to prepare from the exam point of view. They have to prepare these subjects well as they are further elaborated upon in higher classes and college. Many times, these topics are also asked in competitive exams.
It is essential that you have a good grasp of a subject like Physics as it includes plenty of formulae and fundamental concepts that form the base for your CBSE syllabus. We provide the CBSE Class 11 Physics NCERT solutions which are planned out to aid students to perform better by making the subject easier for each student. Having the solutions handy will also allow you to revise before the exam as all the syllabus is merged into one place for the students to quickly go through.
NCERT Solution for Class 11 Physics Chapter 7
Class 11 is an important class for students as it prepares you for Class 12 and board exams. The marks scored in Class 11 also are important in getting admission into a good college. For this, you need to have a strong grip on the basic concepts of Physics. This can only be achieved with regular practice, in-depth knowledge, and a problem-solving approach to any difficult topics.
The Class 11 Physics NCERT solutions will help you understand the various difficult topics and master the subject. The solutions contain all the chapters laid out and answered by our panel of expert teachers. With many years of expertise, they mark out important questions and give you tips on how to solve them. With step-by-step solutions, you can also clear any doubts you may have had.
Chapter 7 System of Particles and Rotational Motion
In the usual case scenario, the body is considered to be inflexible wherein the distances between different particles of the body do not change even after the acting of force on it. Only rotational motion is possible in the case of a rigid body fixed at one point or along a line. It is highly possible for this kind of body to either have pure translation or a combination of translation and rotation. One should have an understanding of the external forces of the body to determine the motion of the centre of mass of a system without knowledge of the internal forces of the system. Newton’s Second Law for finite-sized bodies (or systems of particles) is based on Newton’s Second Law and also Newton’s Third Law for particles.
Key Features of NCERT Solutions for Class 11 Physics Chapter 7
NCERT solutions physics class 11 chapter 11 will help students to solve questions that they might get stuck at. The solutions are explained and narrated by experienced teachers to make it easier for students to practise and revise.
- The solution set covers all the necessary topics covered in the NCERT class 11 textbook to ensure students gain enough knowledge about everything.
- The solutions are written in easy language to ensure clarity so that you can score better in the exams.
- Time management will be one of the best benefits for students if they refer to NCERT Solutions.
- It becomes convenient for students to revise the solutions before appearing for the exams.
Q.1 Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?
The centre of mass (C.M.) of a body is a point where the mass of a body is supposed to be concentrated. A sphere, cylinder, ring and a cube are symmetrical objects and have their centre of mass at their geometric centres. C.O.M. of a body need not necessarily lie inside the body, for example; hollow sphere and a ring.
Q.2 In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10-10 m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.
Q.3 A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?
Internal forces do not affect the motion of the bodies on which they act. No external force is not involved in the system, so the child’s motion will not change the velocity of the centre of mass of the system.
Q.4 Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b.
Q.5 Show that a. (b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors, a, b and c.
Q.7 Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the vector angular momentum of the two particle system is the same whatever be the point about which the angular momentum is taken.
Q.8 A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.7.39. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long.
Calculate the distance d of the centre of gravity of the bar from its left end.
Q.9 A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.
Q.10 (a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2MR2/5, where M is the mass of the sphere and R is the radius of the sphere.
(b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be MR2/4, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.
Q.11 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?
If m be the mass of the hollow cylinder and r is the radius of the solid sphere.
Q.12 A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s–1. The radius of the cylinder is 0.25 m. 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?
Q.13 (a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction.
(b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?
Q.14 A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope? Assume that there is no slipping.
Q.15 To maintain a rotor at a uniform angular speed of 200 rad s–1, an engine needs to transmit a torque of 180 Nm. What is the power required by the engine?
(Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient.
Q.16 From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.
Q.17 A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?
Q.18 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?
(b) Will it take longer to roll down one plane than the other?
(c) If so, which one and why?
Q.19 A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20 cm/s. How much work has to be done to stop it?
Q.20 The oxygen molecule has a mass of 5.30 × 10–26 kg and a moment of inertia of 1.94×10–46 kg m2 about an axis through its centre perpendicular to the lines joining the two atoms.
Suppose the mean speed of such a molecule in a gas is 500 ms-1 and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.
Q.21 A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s.
(a) How far will the cylinder go up the plane?
(b) How long will it take to return to the bottom?
Q.22 As shown in Fig.7.40, the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE, 0.5 m is tied half way up. A weight 40 kg is suspended from a point F, 1.2 m from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. (Take g = 9.8 ms-2)(Hint: Consider the equilibrium of each side of the ladder separately.)
The situation of the problem can be shown in the given figure.
Q.23 A man stands on a rotating platform, with his arms stretched horizontally holding a 5 kg weight in each hand. The angular speed of the platform is 30 revolutions per minute. The man then brings his arms close to his body with the distance of each weight from the axis changing from 90 cm to 20cm. The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg m2.
(a) What is his new angular speed? (Neglect friction.)
(b) Is kinetic energy conserved in the process? If not, from where does the change come about?
Q.24 A bullet of mass 10 g and speed 500 ms-1 is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0 m wide and weighs 12 kg. 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 ML2/3.)
Q.25 Two discs of moments of inertia I1 and I2 about their respective axes (normal to the disc and passing through the centre), and rotating with angular speeds ω1 and ω2 are brought into contact face to face with their axes of rotation coincident.
(a) What is the angular speed of the two-disc system?
(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 ω1 ≠ ω2.
Q.26 (a) Prove the theorem of perpendicular axes. (Hint: Square of the distance of a point (x, y) in the x–y plane from an axis through the origin perpendicular to the plane is x2 + y2).
(b) Prove the theorem of parallel axes. (Hint: If the centre of mass is chosen to be the origin).
Q.28 A disc rotating about its axis with angular speed ωo is placed lightly (without any translational push) on a perfectly frictionless table. The radius of the disc is R. What are the linear velocities of the points A, B and C on the disc shown in Fig. 7.41? Will the disc roll in the direction indicated?
Q.29 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 B, and the sense of frictional torque, before perfect rolling begins.
(b) What is the force of friction after perfect rolling begins?
In order to roll a disc, we need a torque tangential to the disc; provided only by frictional force in this case.
(a) Frictional force acts in the direction opposite to the direction of velocity at point B. The linear velocity at point B acts along tangentially leftward. Thus, force of friction will act tangentially rightward and the sense of frictional torque will be in normal direction to the plane of the disc and outwards before the start.
(b) Since force of friction acts opposite velocity at point B, perfect rolling will start when the velocity at that point becomes zero. The force of friction will also become zero at this stage.
Q.30 A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10 π rad s-1. Which of the two will start to roll earlier? The co-efficient of kinetic friction is μk = 0.2.
Q.31 A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30°. The coefficient of static friction µs = 0.25.
(a) How much is the force of friction acting on the cylinder?
(b) What is the work done against friction during rolling?
(c) If the inclination θ of the plane is increased, at what value of θ does the cylinder begin to skid, and not roll perfectly?
Q.33 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 CM of the body.
(b) The instantaneous speed of the point of contact during rolling is zero.
(c) The instantaneous acceleration of the point of contact during rolling is zero.
(d) For perfect rolling motion, work done against friction is zero.
(e) A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.
(a)The given statement is falseas the force of friction and motion of CM are in opposite directions.
(b) The given statement is trueas rolling can be considered to be rotation of a body about an axis passing through the point of contact of the body with the ground. Thus, instantaneous speed of the point of contact of the body during rolling is zero.
(c) The given statement is falseas a rortating body has an instantaneous acceleration not equal to zero.
(d) The given statement is true. On perfect rolling frictional force becomes zero and therefore the work done against it also becomes zero.
(e) The given statement true. The required torque for rolling comes from the tangential force generated by friction between the body and the surface. Without friction the body will simply slip from the inclined plane.
FAQs (Frequently Asked Questions)
1. What are the learning students will get in the chapter 7 systems of particles and rotational motion of NCERT solutions for class 11 physics?
The student will get an understanding of the external forces of the body to determine the motion of the centre of mass of a system without knowledge of the internal forces of the system. Newton’s Second Law for finite-sized bodies (or systems of particles) is based on Newton’s Second Law and also Newton’s Third Law for particles. The total torque on a system is independent of the origin if the total external force is zero.
Students can master these concepts with the NCERT solutions class 11 physics chapter 7 provided by Extramarks.
2. Why should I download the Extramarks solutions for class 11 physics chapter 7?
The language used in the NCERT solutions is very easy to understand and simple for students to grasp difficult concepts easily and score high marks in the first term exams. The solutions will help the students in CBSE exams and other competitive exams like JEE Main, Jee Advanced etc. It will play a vital role in clearing all the doubts students have. By repeated practice, students can gain confidence in their abilities and secure good marks. Many times, these topics are also asked in competitive exams. It is extremely important that the students are thorough and well prepared with Physics for Class 11.
3. Will these help students answer difficult questions related to chapter 7 of NCERT solutions for class 11 physics?
Yes, these solution sets will help students solve complicated questions more easily. With proper focus on learning the fundamental concepts and top priority given to core topics of the chapter, the students can surely score good marks. This will also help students in coming up with new ideas to answer questions based on their understanding.
Apart from these, our focus is to provide all the study materials to students in the simplest of forms hence, we also have NCERT solutions class 1, NCERT solutions class 2, NCERT solutions class 3, NCERT solutions class 4, NCERT solutions class 5, NCERT solutions class 6, NCERT solutions class 7, NCERT solutions class 8, NCERT solutions class 9, NCERT solutions class 9, NCERT solutions class 10, NCERT solutions class 11 and NCERT solutions class 12.