NCERT Solutions For Class 11 Physics Chapter 5: Laws of Motion

NCERT Solutions for Class 11 Physics Chapter 5 Laws of Motion are provided in the article below. The branch of physics where we study the motion of a body by considering the cause which is the force which generates the motion is called Dynamics. Newton’s three laws of motion make us understand how objects act when standing still, while moving or when forces behave upon them.

Class 11 Physics Chapter 5 Laws of Motion belongs to Unit 3 which has a weightage of 23 marks along with Unit 2 Kinematics and Unit 3. The Class 11 Physics Chapter 5 NCERT Solutions deals with the concepts of force and Laws of Motion, Non-contact force, Inertia and Mass.

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NCERT Solutions for Class 11 Physics Chapter 5

NCERT Solutions

Class 11 Physics Chapter 5 – Concepts Covered

Newton’s First Law of Motion is referred to as the “Law of Inertia”. The law defines inertia, an inertial frame of reference, and force.

A body will remain at rest or continue to be in motion with a uniform velocity unless an external force is applied to it.

Newton’s Second Law of Motion: When an external force is applied to a body with constant mass, the force produces an acceleration. The acceleration that is produced is directly proportional to the force and then inversely proportional to the mass of the body.

\(\overrightarrow {F} = K {dP \over dt} = Km \overrightarrow a\)

Newton’s Third Law of Motion: To every action, there is an equal and opposite reaction force.

When body A exerts a force on some other body B, then B exerts an equal and opposite force on A.

Linear momentum of a body is the product of the mass and velocity of the body.

Linear Momentum = mass x velocity

Impulse is the product of force and the small-time interval for which the force acts.

\(Impulse= \int Fdt\)

Class 11 Physics Chapter 5 Related Guides –

Banking of Roads	Types of Friction	Rolling Friction
Sliding Friction	Pseudo Force	Tension Force
Kinetic Friction	Motion in Physics	Derivation of Centripetal Acceleration
Coefficient of Static Friction	Aristotle Fallacy	Static Friction
Impending Motion	Interaction of Forces	Difference between momentum and inertia

Class 11 Physics Study Guides:

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

CBSE CLASS XII Related Questions

1.
Assertion (A) : All atoms have a net magnetic moment. Reason (R) : A current loop does not always behave as a magnetic dipole.

Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
Assertion (A) is true, but Reason (R) is false.
Both Assertion (A) and Reason (R) are false.

View Solution

2.
Two parallel plate capacitors X and Y are connected in series to a 6 V battery. They have the same plate area and same plate separation but capacitor X has air between its plates, whereas capacitor Y contains a material of dielectric constant 4. Calculate the capacitances of X and Y, if the equivalent capacitance of the combination of X and Y is \( 4 \, \mu\text{F} \). Calculate the potential difference across the plates of X and Y.

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3.
Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
Assertion (A) is true, but Reason (R) is false.
Both Assertion (A) and Reason (R) are false.

View Solution

4.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

5 V
3.5 V
2.5 V
Zero V

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5.
Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

(i) and (iii) only
(iii) and (iv) only
(i), (iii) and (iv) only
All of them

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6.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

View Solution

NCERT Solutions For Class 11 Physics Chapter 5: Laws of Motion

NCERT Solutions for Class 11 Physics Chapter 5

Class 11 Physics Chapter 5 – Concepts Covered

CBSE CLASS XII Related Questions

1.
Assertion (A) : All atoms have a net magnetic moment. Reason (R) : A current loop does not always behave as a magnetic dipole.

3.
Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

4.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

5.
Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

6.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

Similar Physics Concepts

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NCERT Solutions For Class 11 Physics Chapter 5: Laws of Motion

NCERT Solutions for Class 11 Physics Chapter 5

Class 11 Physics Chapter 5 – Concepts Covered

CBSE CLASS XII Related Questions

1.Assertion (A) : All atoms have a net magnetic moment. Reason (R) : A current loop does not always behave as a magnetic dipole.

3.Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

4.A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

5.Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

6.Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Assertion (A) : All atoms have a net magnetic moment. Reason (R) : A current loop does not always behave as a magnetic dipole.

3.
Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

4.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

5.
Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

6.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.