NCERT Solutions for Class 12 Physics

NCERT Solutions for Class 12 Physics in PDF format is available for free download. CBSE Solutions Apps as well as NCERT Solutions and their answers, solutions of additional exercises, intext questions, back exercises questions with assignments from popular books like S L Arora, Concepts of Physics by H C Verma, Pradeep’s fundamental physics, A B C Physics, Arihant publications books, Full Marks question bank etc.




Class 12:Physics
Contents:NCERT Solutions for 2019-2020

Table of Contents

NCERT Solutions for Class 12 Physics

For the regular preparation for CBSE, IIT – JEE Mains and Advance, NEET, BITSAT, GGSIPU use latest books available in the market. A few questions related to these books are given along with each chapter named as Competitive Exams.

Chapter 1: Electric Charges and Fields
Chapter 2: Electrostatic Potential and Capacitance
Chapter 3: Current Electricity
Chapter 4: Moving Charges and Magnetism
Chapter 5: Magnetism and Matter
Chapter 6: Electromagnetic Induction
Chapter 7: Alternating Current
Chapter 8: Electromagnetic Waves
Chapter 9: Ray Optics and Optical Instruments
Chapter 10: Wave Optics
Chapter 11: Dual Nature of Radiation and Matter
Chapter 12: Atoms
Chapter 13: Nuclei
Chapter 14: Semiconductor Electronics: Materials, Devises and Simple Circuits
Chapter 15: Communication Systems



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Must focus on the following

Chapter 1: Electric Charges and Fields

  1. Does the charge given to a metallic sphere depend on whether it is hollow or solid? Give reason for your answer. [Delhi 2017]

Chapter 2: Electrostatic Potential and Capacitance

  1. A 12 pF capacitor is connected to a 50 v battery. How much electrostatic energy is stored in the capacitor? If another capacitor of 6 pf is connected in series with it with the same battery connected across the combination, find the charge stored and potential difference across each capacitor. [Delhi 2017]
  2. Derive the expression for the electric potential due to an electric dipole at a point on its axial line. Depict the equipotential surfaces due to an electric dipole. [Delhi 2017]



Chapter 3: Current Electricity

  1. Derive an expression for drift velocity of electrons in a conductor. Hence deduce Ohm’s law. A wire whose cross sectional area is increasing linearly from its one end to the other, is connected across a battery of V volts. Which of the following quantities remain constant in the wire? (a) drift velocity (b) current density (c) electric current (d) electric field. Justify your answer. [Delhi 2017]
  2. State the two Kirchhoff’s laws. Explain briefly how these rules are justified. The current is drawn from a cell of emf E and internal resistance r connected to the network of resistors each of resistance r as shown in the figure. Obtain the expression for (i) the current draw from the cell and (ii) the power consumed in the network. [Delhi 2017] NCERT Solutions Class 12 Physics PDF
  3. A resistance of R draws current from a potentiometer. The potentiometer wire, AB, has a total resistance of R0. A voltage V is supplied to the potentiometer. Derive an expression for the voltage across R when the sliding contact is in the middle of potentiometer wire. [Delhi 2017] Physics 12 Chapter 3 - image 1

Chapter 4: Moving Charges and Magnetism

  1. Describe the working principle of a moving coil galvanometer. Why is it necessary to use (i) a radial magnetic field and (ii) a cylindrical soft iron core in a galvanometer? Write the expression for current sensitivity of the galvanometer. Can a galvanometer as such be used for measuring the current? Explain. [Delhi 2017]
  2. An electron of mass m revolves around a nucleus of charge +Ze. Show that it behaves like a tiny magnetic dipole. Hence prove that the magnetic moment associated with it is expressed as µ = (e/2m)L, where L is the orbital angular momentum of the electron. Give the significance of negative sign. [Delhi 2017]
Chapter 5: Magnetism and Matter
  1. At a place, the horizontal component of Earth’s magnetic field is B and angle of dip is 60. What is the value of horizontal component of the Earth’s magnetic field at equator? [Delhi 2017]

Chapter 6: Electromagnetic Induction

  1. A long straight current carrying wire passed normally through the centre of circular loop. If the current through the wire increases, will there be an increase induced emf in the loop? Justify. [Delhi 2017]
  2. Define the term ‘Self Inductance’ and write its S.I. unit. Obtain the expression for the mutual inductance of two long co-axial solenoids S1 and S2 wound one over the other, each of length L and radii r1 and r2 and n1 and n2 number of terns per unit length, when a current I is set up in the outer solenoid S2. [Delhi 2017]
  3. Draw a labelled diagram of AC generator. Derive the expression for the instantaneous value of the emf induced in the coil. A circular coil of cross-sectional area 200 sq. cm and 20 terns is rotated about the vertical diameter with angular speed of 50 rad/s in a uniform magnetic field of magnitude 3.0 × 10^-2 T. Calculate the maximum value of the current in the coil. [Delhi 2017]




Chapter 7: Alternating Current

  • Draw a labelled diagram of a step-up transformer. Obtain the ratio of secondary to primary voltage in terms of number of turns and currents in the two coils. A power transmission line feeds input power at 2200 V to a step-down transformer with its primary windings having 3000 turns. Find the number of turns in the secondary to get the power output at 220 V. [Delhi 2017]
  • Find the value of the phase difference between the current and the voltage in the series LCR circuit shown below. Which one leads in phase: current or voltage? Without making any other change, find the value of the additional capacitor C1, to be connected in parallel with the capacitor C, in order to make the power factor of the circuit unity. [Delhi 2017] Physics 12 image 7-1

Chapter 8: Electromagnetic Waves

  1. How is the speed of em-waves in vacuum determined by the electric and magnetic fields? [Delhi 2017]
  2. How does Ampere-Maxwell law explain the flow of current through a capacitor when it is being charged by a battery? Write the expression for the displacement current in terms of the rate of change of electric flux. [Delhi 2017]

Chapter 9: Ray Optics and Optical Instruments

  1. A ray of light incident on face AB of an equilateral glass prism, shows minimum deviation of 30. Calculate the speed of light through the prism. Find the angle of incident at face AB so that the emergent ray grazes along the face AC. [Delhi 2017] NCERT Solutions Class 12 Physics
  2. Mrs. Rashmi Singh broke her reading glasses. When she went to the shopkeeper to order new spects, he suggested that she should get spectacles with plastic lenses instead of glass lenses. On getting the new spectacles, she found that the new ones could not offer satisfactory explanation for this. At home, Mrs. Singh raised the same question to her daughter Anuja who explained why plastic lenses were thicker. (a) Write the two qualities displayed each by Anuja and her mother. (b) How do you explain this fact using lens maker’s formula? [Delhi 2017]

Chapter 10: Wave Optics

  1. Why should the objective of a telescope have large focal length and large aperture? Justify your answer. [Delhi 2017]
  2. Distinguish between unpolarised light and linearly polarised light. How does one get linearly polarised light with the help of a Polaroid? A narrow beam of unpolarised light of intensity I0 is incident on a Polaroid P1. The light transmitted by it is then incident on a second Polaroid P2 with its pass axis making angle of 60 relative to the pass axis of P1. Find the intensity of the light transmitted by P2. [Delhi 2017]
  3. Explain two features to distinguish between the interference patterns in Young’s double slit experiment with the diffraction pattern obtained due to a single slit. A monochromatic light of wavelength 500 nm is incident normally on a single slit of width 0.2 mm to produce a diffraction pattern. Find the angular width of the central maximum obtained on the screen. Estimate the number of fringes obtained in Young’s double slit experiment with fringe width 0.5 mm, which can be accommodated within the region of total angular spread of the central maximum due to single slit. [Delhi 2017]

Chapter 11: Dual Nature of Radiation and Matter

  1. In the study of photoelectric effect the graph between the stopping potential V and frequency v of the incident radiation on two different metals P and Q is shown below:
    NCERT Solutions Class 12 Physics 2
    (i) Which one of the two metals has higher threshold frequency?
    (ii) Determine the work function of the metal which has greater value.
    (iii) Find the maximum kinetic energy of electron emitted by light of frequency 8 × 10^14 Hz for this metal. [Delhi 2017]

Chapter 12: Atoms

  1. Find the wavelength of the electron orbiting in the first exited state in hydrogen atom. [Delhi 2017]
  2. Define the distance of closest approach. An α-particle of kinetic energy ‘K’ is bombarded on a thin gold foil. The distance of the closest approach is ‘r’. What will be the distance of closest approach for an α-particle of double the kinetic energy? [Delhi 2017]
  3. Write two important limitations of Rutherford nuclear model of the atom. [Delhi 2017]

Chapter 13: Nuclei

  • A radioactive nucleus ‘A’ undergoes a series of decays as given below:
    NCERT Solutions Class 12
    The mass number and atomic number of A2 are 176 and 71 respectively. Determine the mass and atomic number of A4 and A. Write the basic nuclear processes underlying β+ and β- decays. [Delhi 2017]

Chapter 14: Semiconductor Electronics: Materials, Devises Simple Circuits

  1. Name the junction diode whose I-V characteristics are drawn below:
    NCERT Solutions Class 12 Physics 4
    For a CE-transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 2V. Given the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1 kΩ. [Delhi 2017]
  2. Zener diode is fabricated by heavily doping both p- and n- sides of the junction. Explain, why? Briefly explain the use of Zener diode as a dc voltage regulator with the help of a circuit diagram. [Delhi 2017]

Chapter 15: Communication System

  • Distinguish between a transducer and a repeater. [Delhi 2017]
  • Define the term ‘amplitude modulation’. Explain any two factors which justify the need for modulating a low frequency base-band signal. [Delhi 2017]

Table of Contents

When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.

Rubbing produces charges of equal magnitude but of opposite nature on the two bodies because charges are created in pairs. This phenomenon of charging is called charging by friction. The net charge on the system of two rubbed bodies is zero. This
is because equal amount of opposite charges annihilate each other. When a glass rod is rubbed with a silk cloth, opposite nature charges appear on both the bodies. This phenomenon is in consistence with the law of conservation of energy. A similar
phenomenon is observed with many other pairs of bodies.

Three capacitors of capacitances 2pF, 3pF and 4pF are connected in parallel. What is the total capacitance of the combination?

Capacitances of the given capacitors: C1 = 2 pF, C2 = 3 pF and C3 = 4 pF
For the parallel combination of the capacitors, equivalent capacitor is given by Ceq the algebraic sum,
Therefore Ceq = C1 + C2 + C3 = 2 + 3 + 4 = 9 pF
Therefore, total capacitance of the combination is 9 pF.

If Coulomb’s law involved 1/r^3 dependence (instead of 1/r^2), would Gauss’s law be still true?

Gauss’s law will not be true, if Coulomb’s law involved 1/r^3 dependence, instead of1/r^2, on r.

Is Ohm’s law universally applicable for all conducting elements? If not, give examples of elements which do not obey Ohm’s law.

No, Ohm’s law is not universally applicable for all conducting elements. Vacuum diode semi-conductor is a non-ohmic conductor. Ohm’s law is not valid for it.

A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60º with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.

Number of turns on the circular coil, n = 30
Radius of the coil, r = 8.0 cm = 0.08 m
Area of the coil πr^2=π(0.08)^2=0.0201 m^2
Current flowing in the coil, I = 6.0 A
Magnetic field strength, B = 1 T
Angle between the field lines and normal with the coil surface, θ = 60°
The coil experiences a torque in the magnetic field.
Hence, it turns.
The counter torque applied to prevent the coil from turning is given by the relation,
τ = n IBA sin θ
= 30 × 6 × 1 × 0.0201 × sin60°
= 3.133 N m

A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10−4 m2, carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane. What is the magnetic moment associated with the solenoid?

Number of turns on the solenoid, n = 2000
Area of cross-section of the solenoid, A = 1.6 × 10^−4 m^2
Current in the solenoid, I = 4 A
The magnetic moment along the axis of the solenoid is calculated as:
M = nAI
= 2000 × 1.6 × 10−4 × 4
= 1.28 Am^2

Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point. (This fact is analogous to the static electric field lines being normal to the surface of a conductor at every point.) Why?

The permeability of a ferromagnetic material is not less than one. It is always greater than one. Hence, magnetic field lines are always nearly normal to the surface of such materials at every point.

The peak voltage of an ac supply is 300 V. What is the rms voltage?

Peak voltage of the ac supply, Vo = 300

Vrms voltage is given as:
V = Vo/√2
= 300/√2
= 212.1 V

The small ozone layer on top of the stratosphere is crucial for human survival. Why? (e) If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now?

In the absence of an atmosphere, there would be no greenhouse effect on the surface of the Earth. As a result, the temperature of the Earth would decrease rapidly, making it chilly and difficult for human survival.

Does short-sightedness (myopia) or long-sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of accommodation? If not, what might cause these defects of vision?

A myopic or hypermetropic person can also possess the normal ability of accommodation of the eye-lens. Myopia occurs when the eye-balls get elongated from front to back. Hypermetropia occurs when the eye-balls get shortened. When the eye-lens loses its ability of accommodation, the defect is called presbyopia.

When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?

When we place our eyes too close to the eyepiece of a compound microscope, we are unable to collect much refracted light. As a result, the field of view decreases substantially. Hence, the clarity of the image gets blurred.
The best position of the eye for viewing through a compound microscope is at the eye-ring attached to the eyepiece. The precise location of the eye depends on the separation between the objective lens and the eyepiece.

In what way is diffraction from each slit related to the interference pattern in a double-slit experiment?

The interference pattern in a double-slit experiment is modulated by diffraction from each slit. The pattern is the result of the interference of the diffracted wave from each slit.

As you have learnt in the text, the principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffraction and interference patterns. What is the justification of this principle?

The principle of linear superposition of wave displacement is essential to our understanding of intensity distributions and interference patterns. This is because superposition follows from the linear character of a differential equation that governs wave motion. If y1 and y2 are the solutions of the second order wave equation, then any linear combination of y1 and y2 will also be the solution of the wave equation.

Why should gases be insulators at ordinary pressures and start conducting at very low pressures?

At atmospheric pressure, the ions of gases have no chance of reaching their respective electrons because of collision and recombination with other gas molecules. Hence, gases are insulators at atmospheric pressure. At low pressures, ions have a chance of reaching their respective electrodes and constitute a current. Hence, they conduct electricity at these pressures.

Is the probability of backward scattering (i.e., scattering of α-particles at angles greater than 90°) predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?

Much less.
The probability of scattering of α-particles at angles greater than 90° predicted by Thomson’s model is much less than that predicted by Rutherford’s model.

Why are elemental dopants for Silicon or Germanium usually chosen from group XIII or group XV?

The size of dopant atoms should be such as not to distort the pure semiconductor lattice structure and yet easily contribute a charge carrier on forming co-valent bonds with Si or Ge.

A carrier wave of peak voltage 12 V is used to transmit a message signal. What should be the peak voltage of the modulating signal in order to have a modulation index of 75%?

Amplitude of the carrier wave, Ac = 12 V
Modulation index, m = 75% = 0.75
Amplitude of the modulating wave = Am Using the relation for modulation index:
m = A_m/A_c
⇒ A_m = m A_m
= 0.75 × 12
= 9 V

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