NCERT Solutions for class 11 Physics Chapter 6: Work, Energy, and Power

NCERT Solutions for Class 11 Physics Chapter 6 Work, Energy, and Power deals with the important concepts of the chapter. Work is referred to as the process of energy that is transferred to an object's motion by applying force. It is generally represented as the product of displacement and force. Power is the amount of energy that is transferred in a unit of time.

Unit 4 Chapter 6 Work, Energy, and Power along with Unit 5 and Unit 6 have a weightage of 17 marks in the Class 11 Physics Examination. The NCERT Solutions for Class 11 Physics Chapter 6 covers Conservative Force, Conservation Of Mechanical Energy, and Relation Between eV And Joule.

Download PDF: NCERT Solutions for Class 11 Physics Chapter 6

NCERT Solutions for Class 11 Physics Chapter 6

Class 11 Physics Chapter 6 – Concepts Covered

Work is said to be done when a force that is applied to the body displaces it through a certain distance in the direction of the applied force.

It is measured as the product of the force and the distance moved in the direction of the force applied, i.e., W = F.S

If an object undergoes a displacement ‘S’ along a straight line while it is acted on by a force F that makes an angle 0 with S. The work done would then be the product of the magnitude of displacement and the component of force in the direction of displacement.

\(W = FScos \theta = \overrightarrow {F}.\overrightarrow {S}\)

The energy possessed by a body is its capacity to do work. Energy is measured in the unit Joule.

Mechanical energy can be classified as Kinetic energy and Potential energy.

Kinetic Energy is the energy possessed by a body by virtue of its motion. For an object having mass m and velocity v, the kinetic energy is given by:

Kinetic Energy or KE = ½ mv²

Potential Energy is the energy of a body by virtue of its position or condition.

Potential energy is of two types – gravitational and elastic potential energy.

Class 11 Physics Chapter 6 Related Guides:

Heat	Principle of Conservation of Energy	Elastic and Inelastic Collisions
Pulley	Unit of Power	Types of Energy
Elastic Collision	Unit of Work	Derivation of Kinetic Energy
Pulley System	Types of Springs	Derivation of Work Energy Theorem
Linear Momentum Formula	Mechanical Advantage Formula	Difference between Kinetic and Potential Energy

Class 11 Physics Study Guides:

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

CBSE CLASS XII Related Questions

1.
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

2.
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 \( F \)

View Solution

3.
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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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

View Solution

5.
Write any two features of nuclear forces.

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6.
Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

\((P_1+P_2)\)
\(\dfrac{P_1+P_2}{2}\)
\(\dfrac{P_1P_2}{2(P_1+P_2)}\)
\(\dfrac{P_1P_2}{4(P_1+P_2)}\)

View Solution

NCERT Solutions for class 11 Physics Chapter 6: Work, Energy, and Power

NCERT Solutions for Class 11 Physics Chapter 6

Class 11 Physics Chapter 6 – Concepts Covered

CBSE CLASS XII Related Questions

1.
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.

2.
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:

3.
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.

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.
Write any two features of nuclear forces.

6.
Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

Similar Physics Concepts

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NCERT Solutions for class 11 Physics Chapter 6: Work, Energy, and Power

NCERT Solutions for Class 11 Physics Chapter 6

Class 11 Physics Chapter 6 – Concepts Covered

CBSE CLASS XII Related Questions

1.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.

2.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:

3.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.

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. Write any two features of nuclear forces.

6.Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
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.

2.
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:

3.
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.

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.
Write any two features of nuclear forces.

6.
Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –