Please see below Class 12 Physics Sample Paper Term 1 Set F with solutions. We have provided Class 12 Physics Sample Papers with solutions designed by Physics teachers for Class 12 based on the latest examination pattern issued by CBSE. We have provided the following sample paper for Term 1 Class 12 Physics with answers. You will be able to understand the type of questions which can come in the upcoming exams.

CBSE Sample Paper for Class 12 Physics Term 1 Set F

SECTION – A

Question 1. If a LC circuit is considered analogous to a harmonically oscillating spring block system, which energy of the LC circuit would be analogous to potential energy and which one analogous to kinetic energy?
Answer.In LC circuit, magnetic energy is analogous to kinetic energy and electrical energy is analogous to potential energy.

Question 2. What happens to the width of depletion layer of a p-n junction when it is (i) forward biased, (ii) reverse biased ?
OR
Can the potential barrier across a p-n junction be measured by simply connecting a voltmeter across the junction?
Answer.(i) Forward biased : As forward voltage opposes the potential barrier, therefore effective barrier potential decreases. It makes the width of the depletion layer smaller.
(ii) Reverse biased : As reverse voltage supports the potential barrier, therefore effective barrier potential increases. It makes the width of the depletion layer larger.
OR
No, the voltmeter should have a very high resistance as compared to the resistance of p-n junction, which is nearly infinite.

Question 3. What is the rest mass of a photon?
Answer.Rest mass of a photon is zero.

Question 4. Is Huygen’s principle valid for longitudinal sound waves?
Answer.Yes, Huygen’s principle is valid for longitudinal as well as transverse waves and for all wave phenomena.

Question 5. Geologists claim that besides the main magnetic N-S poles, there are several local poles on the earth’s surface oriented in different directions. How is such a thing possible at all?
OR
If you made a map of magnetic field lines at Melbourne in Australia, would the lines seem to go into the ground or come out of the ground?
Answer.Localised magnetic dipoles can develop due to magnetised mineral deposits or movement of charged ions in atmosphere.
OR
Melbourne is closer to south pole, so north of the assumed magnet buried within earth lies inside, hence the field lines would seem to be coming out of the ground.

Question 6. Name the electromagnetic waves, which (i) maintain the Earth’s warmth and (ii) are used in aircraft navigation.
Answer.(i) Infra-red rays (ii) Microwaves.

Question 7. In the following diagram, is the junction diode forward biased or reverse biased?

Answer. oltage at p side is less than voltage at n side of the diode, so it is in reverse bias.

Question 8. Why are alloys used for making standard resistance coils?
OR
What are the factors on which the resistivity of a conductor depends ?
Answer. Alloys are used for making standard resistance coils because they have low value of temperature coefficient of resistance and high resistivity.
OR
Resistivity of a material, i.e., ρ = me/ne²𝜏 ,proportional to number density of electrons, and relaxation time.

Question 9. Both alternating current and direct current are measured in amperes. But how is the ampere defined for an alternating current?
Answer.Ampere for alternating current is defined in terms of Joule’s heating effect, which is independent of direction of current.

Question 10. Professor C.V. Raman surprised his student by suspending freely a tiny light ball in a transparent vacuum chamber by shining a laser beam on it. Which property of EM waves was he exhibiting? Give one more example of this property.
OR
Do electromagnetic waves carry energy and momentum?
For question numbers 11, 12, 13 and 14, two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and
(d) as given below.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is NOT the correct explanation of A.
(c) A is true but R is false.
(d) A is false and R is also false.
Answer.The property exhibited from the experiment is radiation pressure exerted by electromagnetic waves. Tails of comets are due to radiation pressure from sun.
OR
Yes, electromagnetic waves carry energy and momentum.

Question 11. Assertion (A) : When a magnetic dipole is placed in a non-uniform magnetic field, only a torque acts on the dipole.
Reason (R) : Force would act on dipole if magnetic field is uniform.

Question 12. Assertion (A) : Electric potential of earth is taken zero.
Reason (R) : No electric field exists on earth surface.

Question 13. Assertion (A) : UV radiation causes photo dissociation of ozone into O₂ and O, thus causing damage to the stratospheric ozone layer.
Reason (R) : Ozone hole is resulting in global warming and climate change.

Question 14. Assertion (A) : An emf is induced in a closed loop where magnetic flux is varied.
Reason (R) : Line integral of induced electric field around the closed loop is non-zero.

SECTION – B

Question 15. According to Bohr’s theory, electrons of an atom revolve around the nucleus on certain orbits, or electron shells. Each orbit has its specific energy level, which is expressed as a negative value. This is because the electrons on the orbit are “captured” by the nucleus via electrostatic forces, and impedes the freedom of the electron. The orbits closer to the nucleus have lower energy levels because they interact more with the nucleus, and vice versa.

(i) An excited state electron drops down to the ground state and emits a photon of yellow light. If instead the electron drops down to a lower excited state, then the light emitted would be
(a) red
(b) blue
(c) ultraviolet
(d) green

(ii) Which of the following transitions in hydrogen atom will have the longest wavelength?
(a) n = 4 to n = 1
(b) n = 5 to n = 4
(c) n = 7 to n = 1
(d) n = 2 to n = 1

(iii) The electron in a hydrogen atom is in the first excited state, when the electron acquires an additional 2.86 eV of energy. What is the quantum number, n, of the state into which the electron moves?
(a) 2
(b) 3
(c) 4
(d) 5

(iv) Calculate the energy required to excite a hydrogen atom by causing an electronic transition from the energy level with n = 1 to the level with n = 4.
(a) 2.043 × 10⁻¹⁸ J
(b) 1.632 × 10⁻²⁹ J
(c) 2.192 × 10⁵ J
(d) 2.254 × 10⁻¹⁸ J

(v) In the hydrogen atom, the transition that gives radiation in the visible region is
(a) from n > 1 to n = 1
(b) from n > 1 to n = 1
(c) from n > 3 to n = 1
(d) from n > 2 to n = 2

Question 16. According to Coulomb’s Law, force F₀ between two electrostatic charges q₁, q₂ separated in air/vacuum by a distance r is F₀ =q₁.q ₂ / 4πε₀r²

(i) If the medium between two charges is air, then the value of constant k in SI units will be
(a) 5 × 10⁹ N m² C^–2
(b) 7 × 10⁹ N m² C^–2
(c) 8 × 10⁹ N m² C^–2
(d) 9 × 10⁹ N m² C^–2

(ii) Ratio of force of repulsion between two electrons and two protons separated by same distance in air is
(a) 1 : 1
(b) me : mp
(c) mp : me
(d) none of these.

(iii) Two charges +3μC and –5μC are held in air at unit distance. The ratio of the force exerted by one on the other is
(a) 3 : 5
(b) 5 : 3
(c) 1 : 1
(d) 15 : 1

(iv) Two charges +8μC and –6μC held certain distance apart in air attract each other with a force of 12 N. A charge of +4μC is added to each. The force between the new charges held the same distance apart in air would be
(a) 12 N
(b) 2 N
(c) 6 N
(d) 24 N

(v) A force of 2.25 N acts on a charge of 15 × 10^–4 C. The intensity of electric field at the point is
(a) 150 N C^–1
(b) 15 N C^–1
(c) 1500 N C^–1
(d) 1.5 N C^–1

SECTION – C

Question 17. Prove that the average energy density of the oscillating electric field is equal to that of the oscillating magnetic field.
Answer.In an electromagnetic wave, both E and B fields vary sinusoidally in space and time. The average energy density u of an e.m. wave can be obtained by replacing E and B by their rms value

Question 18. (a) An electrostatic field line is a continuous curve, i.e., a field line cannot have sudden breaks. Why?
(b) Explain why two field lines never cross each other at any point.
Answer.
(a) An electrostatic field line is a continuous curve, because it represents the actual path of a unit positive charge, which experiences a continuous force.
It cannot have sudden breaks because the moving test charge never jumps from one position to the other.
(b) No two electric field lines of force can intersect each other because at the point of intersection, we can draw two tangents to the two lines of force. This would mean two directions of electric field intensity at the same point which is not possible. Hence, no two electric lines of force can cross each other.

Question 19. Find the total energy stored in the capacitors in the given network.

Answer.

Question 20. Can the instantaneous power output of an ac source ever be negative? Can the average power output be negative?
Answer.Instantaneous power output can be negative when E and I are out of phase in an ac circuit as P = EI. Average power over a cycle cannot be negative as P_av = E_rms I_rms cosΦ and cos f ≥ 0.

Question 21. Light from bulb falls on a wooden table, but no photoelectrons are emitted. Why ?
OR
Draw a graph showing the variation of stopping potential with frequency of incident radiation. What does the slope of the line with frequency axis indicate ?
Answer.No photoelectrons are emitted, because frequency of light is less than the threshold frequency of wooden table.
OR
The V₀-u graph is a straight line as shown in the figure.

Comparing the above relation with the equation of straight line, y = mx + C, we see the slope of V₀ – u graph is h/e.

Question 22. A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60°, and one of the fields has a magnitude of 1.2 × 10^–2 T. If the dipole comes to stable equilibrium at an angle of 15° with this field, what is the magnitude of the other field?
Answer.The magnetic dipole experiences torque due to both the fields and is in equilibrium.

MB₁ sin15° = MB₂ sin45°
1.2 × 10^–2 × 0.26 = B₂ (0.71)
or B₂ = 0.44 × 10^–2 T.

Question 23. (a) State the condition for total internal reflection.
(b) Calculate the speed of light in the medium whose critical angle is 45°.
Answer.(a) Conditions for total internal reflection are
(i) light must travel from denser to rarer medium.
(ii) angle of incidence must be greater than critical angle (C).

Question 24. With what considerations in view is a photodiode fabricated ?
OR
Assuming that the two diodes D₁ and D₂ used in the electric circuit shown in the figure are ideal, find out the value of the current flowing through 1 W resistor.

Answer.   A photodiode is fabricated by allowing light to fall on a diode through a transparent window. It is fabricated such that the generation of e-h pairs take place near the depletion region.
OR
According to the question, R = 2 + 1 = 3Ω

Question 25. Yellow light (λ = 6000 Å) illuminates a single slit of width 1 × 10^–4 m. Calculate the distance between two dark lines on either side of the central maximum, when the diffraction pattern is viewed on a screen kept 1.5 m away from the slit.
OR
A ray of light travelling through rarer medium is incident at very small angle i on a glass slab and after refraction its velocity is reduced by 20%. Find the angle of deviation.
Answer.Here a = 1 × 10^–4 m, D = 1.5 m
l = 6000 Å = 6000 × 10^–10 m
The distance between the two dark bands on each side of central band is equal to width of the central bright band, i.e.,

SECTION – D

Question 26. Draw the energy band diagram of (i) n-type, and (ii) p-type semiconductors at temperatures T > 0 K.
In the case of n-type Si-semiconductor, the donor energy level is slightly below the bottom of conduction band whereas in p-type semiconductor, the acceptor energy level is slightly above the top of valence band.Explain, giving examples, what role do these energy levels play in conduction and valence bands.
Answer.The required energy band diagrams are given here.

In n-type extrinsic semiconductors, the number of free electrons in conduction band is much more than the number of holes in valence band. The donor energy level lies just below the conduction band.
In p-type extrinsic semiconductor, the number of holes in valence band is much more than the number of free electrons in conduction band. The acceptor energy level lies just above the valence band.

Question 27. In Young’s double slit experiment, explain with reason in each case, how the interference pattern changes, when
(i) width of the slit is doubled
(ii) separation between the slits is increased and
(iii) screen is moved away from the plane of slits.

OR
(a) Use Huygens geometrical construction to show how a plane wavefront at t = 0 propagates and produces a wavefront at a later time.
(b) When monochromatic light is incident on a surface separating two media, the reflected and refracted light both have the same frequency. Explain why?
Answer.Fringe width, β = Dλ/d
(i) When the width of the slit is doubled; the intensity of interfering waves increases and fringes become brighter.

(ii) As β ∝ 1/d, therefore when separation between the slits is increased the fringe width decreases, i.e., fringes come closer.
(iii) b ∝ D, therefore when screen is moved away from the plane of the slits, the fringe width increases, i.e., fringes become farther away.
OR
(a) Consider a spherical or plane wavefront moving towards right. Let AB be its position at any instant of time. The region on its left has received the wave while region on the right is undisturbed.

Huygens geometrical construction for the propagation of (a) spherical, (b) plane wavefront.
According to Huygens principle, each point on AB becomes a source of secondary disturbance, which takes with the same speed c. To find the new wavefront after time t, we draw spheres of radii ct, from each point on AB.
The forward envelope or the tangential surface CD of the secondary wavelets gives the new wavefront after time t.
The lines aa′, bb′, cc′, etc., are perpendicular to both AB and CD. Along these lines, the energy flows from AB to CD. So these lines represent the rays. Rays are always normal to wavefronts.
(b) Reflection and refraction arise through interaction of incident light with atomic constituents of matter which vibrate with the same frequency as that of the incident light. Hence frequency remains unchanged.

Question 28. State Gauss’s law in electrostatics. A cube with each side ‘a’ is kept in an electric field is given by E = Cxi,
(as is shown in the figure) where C is a positive dimensional constant. Find out

(i) the electric flux through the cube (ii) the net charge inside the cube.
Answer.Gauss’s law in electrostatics states that the total electric flux through a closed surface enclosing a charge is equal to 1/ε₀ times the magnitude of that charge.

Question 29. A rectangular loop, and a circular loop having the same area, are moved out of a uniform magnetic field region, to a field free region, with a constant velocity v . Would the induced emf remain constant in the two loops as they move out of the field region? Justify your answer.

OR
Consider an infinitely long wire carrying a current I(t), with dI/dt =λ (constant). Find the current produced in the rectangular loop of wire ABCD placed next to the wire at distance x0 if its resistance is R.

Answer.Magnitude of induced emf is directly proportional to the rate of area moving out of the field, for a constant
magnetic field, ε = − dΦ/dt = − B .dA/dt
For the rectangular coil, the rate of area moving out of the field remains same while it is not so for the circular coil. Therefore, the induced emf for the rectangular coil remains constant.
OR

Question 30. Two wires made of tinned copper having identical cross-section (= 10^–6 m²) and lengths 10 cm and 15 cm are to be used as fuses. Show that the fuses will melt at the same value of current in each case.
Answer.temperature of the wire rises to a certain steady temperature q if the heat produced per second by the current just becomes equal to the rate of loss of heat by radiation from it.Heat produced per second by the current

where l is the length, r is radius and r is the specific resistance of a wire.
Let H = heat lost per second per unit surface area of the wire.Neglecting the loss of heat from the end faces of the wire.
Heat lost per second by the wire = H × 2prl At steady state temperature,

From equation (i) we conclude that the rate of loss of heat (H), which in turn depends upon the temperature of wire, is independent of length of the wire. Hence the fuses of two wires of same values of r and r but of different lengths will melt for the same value of current in each case.

SECTION – E

Question 31. (a) Draw a ray diagram to show the formation of the image of an object placed on the axis of a convex refracting surface of radius of curvature ‘R’, separating the two media of refractive indices ‘n₁’ and ‘n₂’ (n₂ > n₁). Use this diagram to deduce the relation n₂/v − n₁/u =  (n₂ − n₁)= R, where u and v represent respectively the distance of the object and the image formed.
(b) A convex lens of focal length f₁ is kept in contact with a concave lens of focal length f₂. Find the focal length of the combination.
OR
(a) Draw a ray diagram showing the image formation by an astronomical telescope when the final image is formed at infinity.
(b) (i) A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. Find the magnifying power of the telescope for viewing distant objects when the telescope is in normal adjustment and the final image is formed at the least distance of distinct vision.
(ii) Also find the separation between the objective lens and the eyepiece in normal adjustment.
Answer.(a) Refraction at convex spherical surface When object is in rarer medium and image formed is real.

Question 32. (a) What is the relationship between the current and the magnetic moment of a current carrying circular loop? Use the expression to derive the relation between the magnetic moment of an electron moving in a circle and its related angular momentum.
(b) A muon is a particle that has the same charge as an electron but is 200 times heavier than it. If we had an atom in which the muon revolves around a proton instead of an electron, what would be the magnetic moment of the muon in the ground state of such an atom?
OR
(a) Magnetic force is always normal to the velocity of a charge and therefore does no work. An iron nail held near a magnet, when released, increases its kinetic energy as it moves to cling to the magnet. What agency is responsible for this increase in kinetic energy if not the magnetic field?
(b) If magnetic monopoles existed, how would Gauss’s law of magnetism be modified?
(c) Does a bar magnet exert a torque on itself due to its own field? Does one element of a current carrying wire exert a force on another element of the same wire?
(d) Magnetic field arises due to charges in motion. Can a system have magnetic moment even though its net charge is zero?
Answer.(a) Magnetic moment of current carrying loop is M = IA.
For an electron revolving in a circular orbit of radius r with speed v, effective current is

OR
(a) An iron nail is made up of a large number of atoms, in which so many electronic charges are in motion. All these charges in motion experience a magnetic force when held near a magnet. The magnetic forces do not change speed of the charges, but they do change their velocity. The velocity of centre of mass may increase at the expense of nail’s internal energy. Thus internal energy of the nail is responsible for increase in kinetic energy of the nail, as a whole.
(b) According to Gauss’s law in magnetism, magnetic flux over any closed surface is always zero.

If monopoles existed, the magnetic flux would no longer be zero, but equal to m0 times the pole strength enclosed by the surface.
(c) No, there is no force or torque on an element due to the field produced by that element itself. But there is a force (or torque) on an element of the same wire.
However, for the special case of a straight wire, this force is zero.
(d) Yes, a system can have magnetic moment even if its net charge is zero. For example, every atom of paramagnetic and ferromagnetic materials has a magnetic moment, though every atom is electrically neutral. Again, a neutron has no charge, but it does have some magnetic moment.

Question 33. It is proposed to use the nuclear reaction:

₁H² + ₁H² → ₂He⁴
in a nuclear reactor of 200 MW rating. If the energy from the above reaction is used with a 25% efficiency in reaction, how many grams of deuterium fuel will be needed per day? The masses of ₁H² and ₂He⁴ nuclei are 2.0141 a.m.u. and 4.0026 a.m.u. respectively?
OR
(a) Using Bohr’s postulates for hydrogen atom, show that the total energy (E) of the electron in the stationary states can be expressed as the sum of kinetic energy (K) and potential energy (U), where K = –2U. Hence deduce the expression for the total energy in the nth energy level of hydrogen atom.
(b) Using the postulates of Bohr’s model of hydrogen atom, obtain an expression for the frequency of radiation emitted when atom make a transition from the higher energy state with quantum number ni to the lower energy state with quantum number n_f (n_f < n_i).
Answer. The fusion reaction used in the reactor is

OR
(a) According to Bohr’s postulates for hydrogen atom, electron revolves in a circular orbit around the heavy positively charged nucleus. These are the stationary (orbits) states of the atom.
For a particular orbit, electron moves there, so it has kinetic energy.
Also, there is potential energy due to charge on electron and heavy positively charged nucleus.
Hence, total energy (E) of atom is sum of kinetic energy (K) and potential energy (U).
i.e., E = K + U
Let us assume that the nucleus has positive charge Ze.An electron moving with a constant speed v along a circle of radius r with centre at the nucleus.
Force acting on electron due to nucleus is given by