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.
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.

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2.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

\( \frac{q}{4 \pi \epsilon_0 l^2} \) pointing along MA
\( \frac{q}{\pi \epsilon_0 l^2} \) pointing along AM
\( \frac{q}{2 \pi \epsilon_0 l^2} \) pointing along AM
Zero

View Solution

3.
Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

attract with a force \( \frac{F}{2} \)
repel with a force \( \frac{F}{2} \)
repel with a force \( F \)
attract with a force \( F \)

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4.
Two thin lenses of focal length \( f_1 \) and \( f_2 \) are placed in contact with each other coaxially. Prove that the focal length \( f \) of the combination is given by \[ f = \frac{f_1 f_2}{f_1 + f_2}. \]

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5.
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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6.
Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \( \mathbf{B} \) at a point \( (1 \, \text{m}, 1 \, \text{m}, 0) \).

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.
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.

2.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

3.
Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

4.
Two thin lenses of focal length \( f_1 \) and \( f_2 \) are placed in contact with each other coaxially. Prove that the focal length \( f \) of the combination is given by \[ f = \frac{f_1 f_2}{f_1 + f_2}. \]

6.
Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \( \mathbf{B} \) at a point \( (1 \, \text{m}, 1 \, \text{m}, 0) \).

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.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.

2.The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

3.Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

4.Two thin lenses of focal length \( f_1 \) and \( f_2 \) are placed in contact with each other coaxially. Prove that the focal length \( f \) of the combination is given by \[ f = \frac{f_1 f_2}{f_1 + f_2}. \]

6.Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \( \mathbf{B} \) at a point \( (1 \, \text{m}, 1 \, \text{m}, 0) \).

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
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.

2.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

3.
Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

4.
Two thin lenses of focal length \( f_1 \) and \( f_2 \) are placed in contact with each other coaxially. Prove that the focal length \( f \) of the combination is given by \[ f = \frac{f_1 f_2}{f_1 + f_2}. \]

6.
Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \( \mathbf{B} \) at a point \( (1 \, \text{m}, 1 \, \text{m}, 0) \).