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.
A charged particle is moving in a circular path with velocity \( V \) in a uniform magnetic field \( \vec{B} \). It is made to pass through a sheet of lead and as a consequence, it looses one half of its kinetic energy without change in its direction. How will (1) the radius of its path change? (2) its time period of revolution change?

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2.
In a circular loop of radius \( R \), current \( I \) enters at point \( A \) and exits at point \( B \), as shown in the figure. The value of the magnetic field at the centre \( O \) of the loop is:

\( \dfrac{\mu_0 I}{R} \)
zero
\( \dfrac{\mu_0 I}{2R} \)
\( \dfrac{\mu_0 I}{4R} \)

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3.
The ratio of the number of turns of the primary to the secondary coils in an ideal transformer is 20:1. If 240 V AC is applied from a source to the primary coil of the transformer and a 6.0 \( \Omega \) resistor is connected across the output terminals, then the current drawn by the transformer from the source will be:

4.0 A
3.8 A
0.97 A
0.10 A

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4.
A conductor of length \( l \) is connected across an ideal cell of emf E. Keeping the cell connected, the length of the conductor is increased to \( 2l \) by gradually stretching it. If R and \( R' \) are initial and final values of resistance and \( v_d \) and \( v_d' \) are initial and final values of drift velocity, find the relation between:
\( R' \) and \( R \)
\( R' = 4R \)

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5.
A small spherical shell \( S_1 \) has point charges \( q_1 = -3 \, \mu C \), \( q_2 = -2 \, \mu C \) and \( q_3 = 9 \, \mu C \) inside it. This shell is enclosed by another big spherical shell \( S_2 \). A point charge \( Q \) is placed in between the two surfaces \( S_1 \) and \( S_2 \). If the electric flux through the surface \( S_2 \) is four times the flux through surface \( S_1 \), find charge \( Q \).

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6.
A 1 cm straight segment of a conductor carrying 1 A current in \( x \)-direction lies symmetrically at the origin of Cartesian coordinate system. The magnetic field due to this segment at point (1m, 1m, 0) is:

\( 1.0 \times 10^{-9} \, \text{T} \)
\( -1.0 \times 10^{-9} \, \text{T} \)
\( \frac{5.0}{\sqrt{2}} \times 10^{-10} \, \text{T} \)
\( -\frac{5.0}{\sqrt{2}} \times 10^{-10} \, \text{T} \)

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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

2.
In a circular loop of radius \( R \), current \( I \) enters at point \( A \) and exits at point \( B \), as shown in the figure. The value of the magnetic field at the centre \( O \) of the loop is:

6.
A 1 cm straight segment of a conductor carrying 1 A current in \( x \)-direction lies symmetrically at the origin of Cartesian coordinate system. The magnetic field due to this segment at point (1m, 1m, 0) is:

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

2.In a circular loop of radius \( R \), current \( I \) enters at point \( A \) and exits at point \( B \), as shown in the figure. The value of the magnetic field at the centre \( O \) of the loop is:

6.A 1 cm straight segment of a conductor carrying 1 A current in \( x \)-direction lies symmetrically at the origin of Cartesian coordinate system. The magnetic field due to this segment at point (1m, 1m, 0) is:

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

2.
In a circular loop of radius \( R \), current \( I \) enters at point \( A \) and exits at point \( B \), as shown in the figure. The value of the magnetic field at the centre \( O \) of the loop is:

6.
A 1 cm straight segment of a conductor carrying 1 A current in \( x \)-direction lies symmetrically at the origin of Cartesian coordinate system. The magnetic field due to this segment at point (1m, 1m, 0) is: