CBSE Sample Papers for Class 12 Physics Paper 7

CBSE Sample Papers for Class 12 Physics Paper 7 are part of CBSE Sample Papers for Class 12 Physics. Here we have given CBSE Sample Papers for Class 12 Physics Paper 7

CBSE Sample Papers for Class 12 Physics Paper 7

Board	CBSE
Class	XII
Subject	Physics
Sample Paper Set	Paper 7
Category	CBSE Sample Papers

Students who are going to appear for CBSE Class 12 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 7 of Solved CBSE Sample Paper for Class 12 Physics is given below with free PDF download solutions.

Time Allowed: 3 hours
Maximum Marks: 80

General Instructions

SECTION C

Question 11.
(i) 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?
(ii) 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.

Answer :

• Calculation of phase difference between current and voltage Name of quantity which leads
• Calculation of value of ‘C’, is to be connected in parallel

Question 12.
Write the two processes that take place in the formation of a p-n junction. Explain with the help of a diagram, the formation of depletion region and barrier potential in a p-n junction.
Answer :

• Names of the two processes
• Diagram
• Explanation of formation of depletion region and Barrier Potential
  

Due to the diffusion of electrons and holes across the junction a region of (immobile) positive charge is created on the n-side and a region of (immobile) negative charge is created on the p-side, near the junction; this is called depletion region. Barrier potential is formed due to loss of electrons from n-region and gain of electrons by p-region. Its polarity is such that it opposes the movement of charge carriers across the junction.

Question 13.
(i) Obtain the expression for the cyclotron frequency.
(ii) A deuteron and a proton are accelerated by the cyclotron. Can both be accelerated with the same oscillator frequency? Give reason to justify your answer.
Answer :

• Derivation of the expression for cyclotron frequency
• Reason/justification for the correct answer
  
  (ii) No The mass of the two particles, i.e deuteron and proton, is different. Since (cyclotron) frequency depends inversely on the mass, they cannot be accelerated by the same oscillator frequency.

Question 14.
(i) How does one explain the emission of electrons from a photosensitive surface with the help
of Einstein’s photoelectric equation?
(ii) The work function of the following metals is given: Na = 2.75 eV, K = 2.3 eV, M0 = 4.17 eV and Ni = 5.15 eV. Which of these metals will not cause photoelectric emission for radiation of wavelength 3300 A from a laser source placed 1 m away from these metals? What happens if the laser source is brought nearer and placed 50 cm away?
Answer :

• Explanation of emission of electrons from the photosensitive surface
• Identification of metals/s which does/do not cause photoelectric effect
• Photoelectric emission
• Effect produced

(i) Einstein’s Photoelectric equation is
hv =Φ₀ + K_max
When a photon of energy ‘hv’ is incident on the metal, some part of this energy is utilized as work function to eject the electron and remaining energy appears as the kinetic energy of the emitted electron.

The work function of Mo and Ni is more than the energy of the incident photons; so photo-electric emission will not take place from these metals. Kinetic energy of photo electrons will not change, only photoelectric current will change.

Question 15.
A resistance of R draws current from a potentiometer. The potentiometer wire, AB, has 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.

Answer :
Derivation of expression of voltage across resistance R

Question 16.
Define the term ‘amplitude modulation’. Explain any two factors which justify the need for modulating a low frequency base-band signal.
Answer :

• Definition of amplitude modulation
• Explanation of two factors justifying the need of modulation

It is the process of superposition of information/message signal over a carrier wave in such a way that the amplitude of carrier wave is varied according to the information signal/message signal. Direct transmission, of the low frequency base band information signal, is not possible due to the following reasons;

1. Size of Antenna: For transmitting a signal, minimum height of antenna should be λ/4 with the help of modulation wavelength of signal decreases, hence height of antenna becomes manageable.
2. Effective power radiated by an antenna:
   Effective power radiated by an antenna varies inversely as λ², hence effective power radiated into the space, by the antenna, increases.
3. To avoid mixing up of signals from different transmitters. (Any two)

Question 17.
(i) Find equivalent capacitance between A and B in the combination given below. Each capacitor is of 2μF capacitance.

(ii) If a dc source of 7 V is connected across AB, how much charge is drawn from the source and what is the energy stored in the network?
Answer:

• Calculation of equivalent capacitance
  
• Calculation of charge and energy stored
  

Question 18.
(i) Derive the expression for electric field at a point on the equatorial line of an electric dipole.
(ii) Depict the orientation of the dipole in (i) stable, (ii) unstable equilibrium in a uniform elec­tric field.
Answer :

• Derivation of expression of electric field on the equatorial line of the dipole
• Depiction of orientation for stable and unstable equilibrium

Question 19.
(i) A radioactive nucleus ‘A’ undergoes a series of decays as given below.

The mass number and atomic number of A₂ are 176 and 71 respectively. Determine the mass and atomic numbers of A₄ and A.
(ii) Write the basic nuclear processes underlying β⁺ and β^–decays.
Answer :

• Determining the mass and atomic number of A₄ and A
• Basic nuclear processes of β⁺ and β^– decays.

(i) A₄: Mass Number : 172
Atomic Number: 69
(ii) A : Mass Number : 180
Atomic Number: 72

[Alternatively: Give full credit if student considers β⁺ decays and find atomic and mass numbers accordingly]
[Note: Give full credit of this part, if student writes the processes as conversion of proton into neutron for β⁺decay and neutron into proton for β^– decay.]

Question 20.
(i) 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.

(ii) Find the angle of incidence at face AB so that the emergent ray grazes along the face AC.
Answer :
(i) Calculation of speed of light
(ii) Calculation of angle of incidence at face AB

Question 21.
For a CE-transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 2 V. 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.
Answer :
Calculation of collector current I, base current I and input signal voltage V

[Note: give full credit if student calculates the required quantities by any other alternative method]

Question 22.
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.
OR
(a) Define the term ‘self-inductance’ and write its S.I. unit.
(b) 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 turns per unit length, when a current I is set up in the outer solenoid S .
Answer :

• Working and Principle of moving coil galvanometer
• Necessity of (i) radial magnetic field (ii) cylindrical soft iron core
• Expression for current sensitivity
• Explanation of use of Galvanometer to measure current

When a coil, carrying current, and free to rotate about a fixed axis, is placed in a uniform magnetic field, it experiences a torque (Which is balanced by a restoring torque of suspension).
(i) To have deflection proportional to current / to maximize the deflecting torque acting on the current carrying coil.
(ii) To make magnetic field radial / to increase the strength of magnetic field.
Expression for current sensitivity

where θ is the deflection of the coil No The galvanometer, can only detect current but cannot measure them as it is not calibrated. The galvanometer coil is likely to be damaged by currents in the (mA/A) range
OR

• Definition of self inductance and its SI unit
• Derivation of expression for mutual inductance

Self inductance of a coil equals, the magnitude of the magnetic flux, linked with it, when a unit current flows through it.
Alternatively Self inductance, of a coil, equals the magnitude of the emf induced in it, when the current in the coil, is changing at a unit rate.
SI unit: henry / (weber/ampere) / (ohm second.)

[Note: If the student derives the correct expression, without giving the diagram of two coaxial coils, full credit can be given]

SECTION D

Question 23.
Mrs. Rashmi Singh broke her reading glasses. When she went to the shopkeeper to order new specs, 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 were thicker than the earlier ones. She asked this question to the shopkeeper but he could not offer satisfactory explanation for this. At home, Mrs. Singh raised the same question to her daughter Anuja who explain why plastic lenses were thicker.
(a) Write two qualities displayed each by Anuja and her mother.
(b) How do you explain this fact using lens maker’s formula?
Answer :

• Two qualities each of Anuj a and her mother
• Explanation, using lens maker’s formula

(a) Anuja: Scientific temperament, co-operative, knowledgeable (any two)
Mother: Inquisitive, Scientific temper/keen to leam/has no airs (any two) (or any other two similar values)

As the refractive index of plastic material is less than that of glass material therefore, for the same power (= 1/ƒ), the radius of curvature of plastic material is small. Therefore, plastic lens is thicker. Alternatively, If student just writes that plastic has a different refractive index than glass, award one mark for this part.

SECTION E

Question 24.
(a) Draw a labelled diagram of AC generator. Derive the expression for the instantaneous value of the emf induced in the coil.
(b) A circular coil of cross-sectional area 200 cm² and 20 turns is rotated about the vertical diameter with angular speed of 50 rad s ‘ in a uniform magnetic field of magnitude 3.0 x 10^-2 T. Calculate the maximum value of the current in the coil.
OR
(a) 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.
(b) 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.
Answer :
(a) Labelled diagram of AC generator Expression for instantaneous value of induced emf.
(b ) Calculation of maximum value of current

[Deduct 1/2 mark, If diagram is not labelled] When the coil is rotated with constant angular speed co, the angle 0 between the magnetic field and area vector of the coil, at instant, t is given by θ = ω t,
Therefore, magnetic flux, (Φ_B), at this instant, is

[Note 1: It the student calculates the value of the maximum induced emf and says that “since R is not given the value of maximum induced current cannot be calculated”, the 1/2 mark, for the last part, of the question, can be given.]
[Note 2 : The direction of magnetic field has not been given. If the student takes this direction along the axis of rotation and hence obtains the value of induced emf and, therefore, maximum current, as zero, award full marks for this part.]
OR

• Labelled diagram of a step up transformer
  Derivation of ratio of secondary and primary voltage
• Calculation of number of turns in the secondary

[Note: Deduct 14 mark, if labelling is not done]
(a) When AC voltage is applied to primary coil the resulting current produces an alternating magnetic flux, which links the secondary coil. The induced emf, in the secondary coil, having N, turns, is

Question 25.
(a) Distinguish between unpolarized light and linearly polarized light. How does one get line­arly polarised light with the help of a polaroid?
(b) 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.
OR
(a) Explain two features to distinguish between the interference pattern in Young’s double slit experiment with the diffraction pattern obtained due to a single slit.
(b) 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.
Answer :

• Distinction between unpolarised and linearly polarized light Obtaining linearly polarized Light
• Calculation of intensity of light

(a) In an unpolarised light, the oscillations, of the electric field, are in random directions, in planes perpendicular to the direction of propagation. For a polarized light, the oscillations are aligned along one particular direction.
Alternatively Polarized light can be distinguished, from unpolarised light when it is allowed to pass through a Polaroid. Polaroid light does can show change in its intensity remains same in case of unpolarised light.  When unpolarised light wave is incident on a polaroid, then the electric vectors along the direction of its aligned molecules, get absorbed; the electric vector, oscillating along a di-rection perpendicular to the aligned molecules, pass through. This light is called linearly polarized light.
(b) According to Malus’ Law:

OR

• Explanation of two features (distinguishing between interference pattern and diffraction pattern.)
• Calculation of angular width of central maxima Estimation of number of fringes

(a)

Interference Pattern	Diffraction pattern
1.fringes are of equal width.	1. Width of central maxima is twice the width of higher order bands.
2. Intensity of all bright bands is equal.	2. Intensity goes on decreasing for higher order of diffraction bands.

[Note: Also accept any other two correct distinguishing features.]
(b) Angular width of central maximum


[Award the last 1/2 mark if the student writes the answer as 2 (taking d = a), or just attempts to do these calculation.]

Question 26.
(i) Derive an expression for drift velocity of electrons in a conductor. Hence deduce Ohm’s law.
(ii) A wire whose cross-sectional area is increasing linearly from its one end to the other, is connected across a battery of volts. Which of the following quantities remain constant in the wire?
(a) drift speed
(b) current density
(c) electric current
(d) electric field
Justify your answer.
OR
(i) State the two Kirchhoff’s laws. Explain briefly how these rules are justified.
(ii) 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.

Answer :

• Derivation of the expression for drift velocity Deduction of Ohm’s law
• Name of quantity and justification

(i) Let an electric field E be applied the conductor. Acceleration of each electron is

Let the conductor contain n electrons per unit volume. The average value of time between their successive collision, is the relaxation time, ‘r’.


(ii) Electric current well remain constant in the wire. All other quantities, depend on the cross sectional area of the wire.
OR

• Statement of Kirchhoff’s laws Justification
•  Calculation of (i) current drawn and; (ii) Power consumed

(i) Junction Rule: At any Junction, the sum of currents, entering the junction, is equal to the sum of currents leaving the junction.
Loop Rule: The Algebraic sum, of changes in potential, around any closed loop involving resistors and cells, in the loop is zero.
∑ (ΔV) = 0
Justification: The first law is in accord with the law of conservation of charge.
The Seconds law is in accord with the law of conservation of energy.
(ii) Equivalent resistance of the loop

[Note: Award the last 1 \(\frac { 1 }{ 2 } \) marks for this part, if the calculations, for these parts, are done by using (any other) value of equivalent resistance obtained by the student.]

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