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

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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 mv 2

  • 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.

CBSE CLASS XII Related Questions

  • 1.
    Consider a cylindrical conductor of length \( l \) and area of cross-section \( A \). Current \( I \) is maintained in the conductor and electrons drift with velocity \( \vec{v}_d \, (|\vec{v}_d| = \frac{eE}{m} \tau) \), where symbols have their usual meanings. Show that the conductivity of the material of the conductor is given by \[ \sigma = \frac{n e^2 \tau}{m}. \]


      • 2.

        The electric potential (V ) and electric field (⃗ E) are closely related concepts in electrostatics. The electric field is a vector quantity that represents the

          • Production of AC is economical.
          • AC can be easily and efficiently converted from one voltage to another.
          • AC can be transmitted economically over long distances.
          • AC is less dangerous.

        • 3.
          Assertion : In Young’s double-slit experiment, the fringe width for dark and bright fringes is the same. Reason (R): Fringe width is given by \( \beta = \frac{\lambda D}{d} \), where symbols have their usual meanings.

            • 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.

          • 4.
            The energy of an electron in an orbit in hydrogen atom is \( -3.4 \, \text{eV} \). Its angular momentum in the orbit will be:

              • \( \dfrac{3h}{2\pi} \)
              • \( \dfrac{2h}{\pi} \)
              • \( \dfrac{h}{\pi} \)
              • \( \dfrac{h}{2\pi} \)

            • 5.
              The figure represents the variation of the electric potential \( V \) at a point in a region of space as a function of its position along the x-axis. A charged particle will experience the maximum force at:

                • P
                • Q
                • R
                • S

              • 6.
                A circular coil of 100 turns and radius \( \left(\frac{10}{\sqrt{\pi}}\right) \, \text{cm}\) carrying current of \( 5.0 \, \text{A} \) is suspended vertically in a uniform horizontal magnetic field of \( 2.0 \, \text{T} \). The field makes an angle \( 30^\circ \) with the normal to the coil. Calculate:
                the magnetic dipole moment of the coil, and
                the magnitude of the counter torque that must be applied to prevent the coil from turning.

                  CBSE CLASS XII Previous Year Papers

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