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A rubber ball changes shape when squeezed, and a metal rod can bend when pressure is applied. However, when the deformation is very small and doesn't affect the overall behaviour of the object, we often treat it as a rigid body for simplicity in calculations.
System of particles and rotational motion is an important but tricky topic in Class 11 Physics. Here are the detailed solutions for Chapter 6, created by experts. These solutions explain each step clearly, making the topic easier to understand. The NCERT solutions for Class 11 Physics cover all 17 questions, from 6.1 to 6.17, in the exercise section.
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For both board examinations and competitive exams, the System of particles and rotational motion class 11 solution is crucial.
With several questions and answers based on these ideas, the chapter concentrates on important subjects like the centre of mass and the moment of inertia. These solutions will help students do well on tests and are especially helpful for self-study.
Download Solution PDF Access Class 11 Physics Chapter 6 Exercise solutions
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.
Q6.4 Show that the area of the triangle contained between the vectors
Answer:
Let a and b be two vectors having
Consider
or
or
= 2 (Area of
Therefore the area of
Answer:
A parallelepiped is shown in the figure given below:-
Volume is given by : = abc
We can write :
Now,
This is equal to volume of parallelepiped.
Q6.6 Find the components along the
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
For the rotational equilibrium :
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 :
For rotational equilibrium :
or
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:
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
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 :
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
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.
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
Chapter 6 of the NCERT Physics Solutions includes a total of 17 questions. The solutions are explained in a step-by-step manner, making them easy to follow. Students can download the solutions in PDF format for easy access and revision. It's recommended that students first attempt the problems on their own and then refer to the provided solutions to assess their understanding and pinpoint areas for improvement.
Get a better grasp of Physics with our class 11 physics chapter 6 solutions. These solutions include important formulas (listed below), easy-to-follow diagrams, and a handy eBook link. Improve your understanding of the System of Particles and Rotational Motion and perform well in the Chapter 6 physics problems.
- Motion of COM
Average angular velocity =∆θ/∆t
Instantaneous angular velocity, ω=dθ/dt
Download- Formula Sheet for Physics Class 11: Chapterwise
Thorough Understanding: The Class 11 Physics Chapter 6 (System of Particles and Rotational Motion) NCERT solutions are designed to give students a deep and complete understanding of the chapter, covering all critical concepts.
In summary, the NCERT solutions for Class 11 Physics help students effectively understand and solve problems in the chapter, strengthening their knowledge and enhancing their performance in exams.
Also Check-
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 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
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.
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