NCERT Solutions For Class 11 Physics Chapter 7: System of Particles and Rotational Motion

NCERT Solutions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion covers all the concepts discussed in the Class 11 Physics Chapter 7. The combination of rotational motion and the translational motion of a rigid body is known as rolling motion. According to the law of conservation of angular momentum, if there is no external couple acting, the total angular momentum of a rigid body or a system of particles is conserved.

Class 11 Physics Chapter 7 System of Particles and Rotational Motion has a weightage of 17 marks along with Unit 4 Work, Energy, and Power and Unit 6 Gravitation. The Class 11 Physics Chapter 7 discusses the concepts of Torque, Angular Momentum, and Rotational Kinetic Energy.

Download PDF: NCERT Solutions for Class 11 Physics Chapter 7

NCERT Solutions for Class 11 Physics Chapter 7

Class 11 Physics Chapter 7 – Concepts Covered

Centre of Mass: For a system of particles, the centre of mass is the balancing point where the entire mass of the system is concentrated, for consideration of its translational motion.

If there are 2 particles with mass m1 and m2 with position vectors \(\overrightarrow{r_1}\ and\ \overrightarrow{r_2}\), then the position vector of centre of mass is given as:

\(\overrightarrow{r_{cm}} = {{m_1}\overrightarrow{r_{1}} + {m_2}\overrightarrow{r_{2}} \over m_1 + m_2}\)

The cross product of two vectors \(\overrightarrow{A}\) and \(\overrightarrow{B}\) is another vector \(\overrightarrow{C}\), which has a magnitude equal to the product of the magnitudes of 2 vectors and the sine of the smaller angle \(\theta\) between them.

\(\overrightarrow{A} \times \overrightarrow{B} = \overrightarrow{C} = ABsin\theta \hat{c}\)

Torque or moment of force is the product of the magnitude of the force acting on a particle and the perpendicular distance of the application of this force from the axis of rotation of the particle.

\(Torque = Force \times perpendicular\ distance\)

The angular momentum about an axis of rotation is a vector quantity, with a magnitude equal to the product of the magnitude of momentum and the perpendicular distance of the line of action of momentum from the axis of rotation. Its direction is perpendicular to the plane that contains the momentum and the perpendicular distance.

\(\overrightarrow{L} = \overrightarrow{r} \times \overrightarrow{p} \)

Torque and angular momentum are correlated to each other.

\(\tau = {\overrightarrow{dL} \over dt}\)

Class 11 Physics Chapter 7 Related Guides:

Unit of Moment of Inertia	Angular Speed	Radial Acceleration
Relation Between Torque and Moment of Inertia	Angular Speed Formula	Angular Acceleration
Moment of Inertia	Rolling Motion	Unit of Moment of Inertia
Angular displacement	Angular speed	Radial Acceleration
Radius of Gyration	Rigid Bodies	Tangential Velocity Formula

Class 11 Physics Study Guides:

NCERT Solutions for Class 11 Physics	Difference between in Physics	Class 11 Physics Notes
Physics Formula	Physics Constants	SI Units in Physics
Derivations in Physics	Relation between articles in Physics	Physics Study Notes

CBSE CLASS XII Related Questions

1.
The ends of six wires, each of resistance R (= 10 \(\Omega\)) are joined as shown in the figure. The points A and B of the arrangement are connected in a circuit. Find the value of the effective resistance offered by it to the circuit.

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2.
A vertically held bar magnet is dropped along the axis of a copper ring having a cut as shown in the diagram. The acceleration of the falling magnet is:

zero
less than \( g \)
\( g \)
greater than \( g \)

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3.
Answer the following giving reason:
(a) All the photoelectrons do not eject with the same kinetic energy when monochromatic light is incident on a metal surface.
(b) The saturation current in case (a) is different for different intensity.
(c) If one goes on increasing the wavelength of light incident on a metal sur face, keeping its intensity constant, emission of photoelectrons stops at a certain wavelength for this metal.

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4.
Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).

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5.
A system of two conductors is placed in air and they have net charge of \( +80 \, \mu C \) and \( -80 \, \mu C \) which causes a potential difference of 16 V between them.
(1) Find the capacitance of the system.
(2) If the air between the capacitor is replaced by a dielectric medium of dielectric constant 3, what will be the potential difference between the two conductors?
(3) If the charges on two conductors are changed to +160µC and −160µC, will the capacitance of the system change? Give reason for your answer.

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6.
Write the mathematical forms of three postulates of Bohr’s theory of the hydrogen atom. Using them prove that, for an electron revolving in the \( n \)-th orbit,
(a) the radius of the orbit is proportional to \( n^2 \), and
(b) the total energy of the atom is proportional to \( \frac{1}{n^2} \).

View Solution

NCERT Solutions For Class 11 Physics Chapter 7: System of Particles and Rotational Motion

NCERT Solutions for Class 11 Physics Chapter 7

Class 11 Physics Chapter 7 – Concepts Covered

CBSE CLASS XII Related Questions

1.
The ends of six wires, each of resistance R (= 10 \(\Omega\)) are joined as shown in the figure. The points A and B of the arrangement are connected in a circuit. Find the value of the effective resistance offered by it to the circuit.

2.
A vertically held bar magnet is dropped along the axis of a copper ring having a cut as shown in the diagram. The acceleration of the falling magnet is:

4.
Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).

Similar Physics Concepts

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NCERT Solutions For Class 11 Physics Chapter 7: System of Particles and Rotational Motion

NCERT Solutions for Class 11 Physics Chapter 7

Class 11 Physics Chapter 7 – Concepts Covered

CBSE CLASS XII Related Questions

1.The ends of six wires, each of resistance R (= 10 \(\Omega\)) are joined as shown in the figure. The points A and B of the arrangement are connected in a circuit. Find the value of the effective resistance offered by it to the circuit.

2.A vertically held bar magnet is dropped along the axis of a copper ring having a cut as shown in the diagram. The acceleration of the falling magnet is:

4.Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
The ends of six wires, each of resistance R (= 10 \(\Omega\)) are joined as shown in the figure. The points A and B of the arrangement are connected in a circuit. Find the value of the effective resistance offered by it to the circuit.

2.
A vertically held bar magnet is dropped along the axis of a copper ring having a cut as shown in the diagram. The acceleration of the falling magnet is:

4.
Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).