CBSE Sample Papers for Class 12 Physics Paper 2 are part of CBSE Sample Papers for Class 12 Physics. Here we have given CBSE Sample Papers for Class 12 Physics Paper 2.

## CBSE Sample Papers for Class 12 Physics Paper 2

Board |
CBSE |

Class |
XII |

Subject |
Physics |

Sample Paper Set |
Paper 2 |

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 2 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**

- All questions are compulsory. There are 26 questions in all.
- This question paper has five sections: Section A, Section B, Section C, Section D and Section E.
- Section A contains five questions of 1 mark each. Section B contains five questions of 2 marks each. Section C contains twelve questions of 3 marks each. Section D contains one value based question of 4 marks and Section E contains three questions of 5 marks each.
- There is no overall choice. However, an internal choice has been provided in 1 question of 2 marks, 1 question of 3 marks and all the 3 questions of 5 marks weightage. You have to attempt only 1 of the choices in such questions.
- You may use the following values of physical constants wherever necessary :

**Questions**

**SECTION : A**

Question 1.

Name the physical quantity whose S.I. units is JC^{-1} Is it a scalar or a vector quantity?

Question 2.

A beam of a-particles projected along + X-axis, experiences a force due to a magnetic field along + Y-axis. What is the direction of magnetic field?

Question 3.

Define self-inductance of a coil. Write its S.I units?

Question 4.

A converging lens is kept coaxially in contact with diverging lens – both the lenses being of equal focal lengths, what is the focal length of the combination?

Question 5.

Define ionization energy. What is its value for hydrogen atom?

**SECTION : B**

Question 6.

Two conducting wires X and Y of same diameter but different materials are joined in series across a battery. If the number density of electrons in X is twice that in Y, find the ratio of drift velocity of electrons in the two wires.

**OR**

A spherical conducting shell of inner radius r_{1} and outer radius r_{2} has a charge Q. A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surface of the shell?

Question 7.

State the underlying principle of working of a moving coil galvanometer. Write two reasons why a galvanometer cannot be used as such to measure current in a given circuit. Name any two factors on which the current sensitivity of a galvanometer depends.

Question 8.

In a meter bridge, the null point is found at a distance of l_{1} cm from A. If a resistance of X is connected in parallel with S, the null point occurs at l_{2} cm. Obtain a formula for X in terms of l_{1}l_{2 }and S.

Question 9.

Define the term modulation. Draw a block diagram of a simple modulator for obtaining AM signal.

Question 10.

In a series LCR circuit connected to an ac source of variable frequency and voltage v = v_{m} sin ωt, draw a plot showing the variation of current (I) with angular frequency (ω) for two different values of resistance R_{1}, and R_{2} (R_{1 }> R_{2}). Write the condition under which the phenomenon of resonance occurs. For which value of the resistance out of the two curves, a sharper resonance is produced? Define Q-factor of the circuit and give its significance.

**SECTION : C**

Question 11.

Draw a sketch of a plane electromagnetic wave propagating along the Z-direction. Depict clearly the directions of electric and magnetic fields varying sinusoidally with Z.

Question 12.

Show that the electric field at the surface of a charged conductor is given by

where σ is the surface charge density and \(\hat { n }\) is unit vector normal to the surface in the outward direction?

Question 13.

Two identical loops one of copper and the other of aluminium, are rotated with the same angular speed in the same magnetic field. Compare

- the induced emf and
- the current produced in the two coils, justify your answer.

Question 14.

An a-particle and a proton are accelerated from rest by the their de Broglie wavelengths.

Question 15.

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

Question 16.

Write Einstein photoelectric equation? State clearly the three salient features observe in photoelectric effect, which can be explained on the basis of above equation?

Question 17.

Draw a plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions which you can draw regarding the nature of nuclear forces.

**OR**

Draw a plot of the binding energy per nucleon as a function of mass number for a large number of nuclei, 2 ≤ A≤ 240. How do you explain the constancy of binding energy per nucleon in range 30<A< 170 using the property that nuclear force is short-ranged?

Question 18.

Which mode of propagation is used by short wave broadcast services having frequency range from a few MHz to 30 MHz? Explain diagrammatically how long distance communication can be achieved by this mode. Why there an upper limit to frequency of waves used in this mode?

Question 19.

Write any two factors on which internal resistance of a cell depends. The reading on a high resistance voltmeter, when a cell is connected across it, is 2.2 V When the terminals of the cell are also connected across to a resistance of 5 Ω as shown in the circuit, the voltmeter reading drops to 1.8 V. Find the internal resistance of the cell.

Question 20.

A network of four capacitors each of 12μF capacitance is connected to a 500 V supply as given in the figure. Determine

- equivalent capacitance of the network and
- charge on each capacitor.

Question 21.

- Draw a neat labelled ray diagram of an astronomical telescope in normal adjustment. Explain briefly its working.
- An astronomical telescope uses two lenses of powers 10 D and 1 D. What is its magnifying power in normal adjustment?

**OR**

- Draw a neat labelled ray diagram of a compound microscope. Explain briefly its working,
- Why must both the objective and the eye-piece of a compound microscope have short focal lengths?

Question 22.

- Write symbolically the β
^{–}decay process of \(^{ 32 }{ P }\). - Derive an expression for the average life of a radionuclide. Give its relationship with the half-life.

**SECTION : D**

Question 23.

Rajeev lives in an apartment of a group of housing society. From his apartment he daily observed that near his society on the road there was a sharp turn. The drivers used to face problems driving at this sharp turn and often met with an accident. Rajeev met with his RWA president and requested him to fix a convex lens at the sharp turn. And soon a convex mirror was fixed on this sharp turn.

- What values were exhibited by Rajeev?
- Why did Rajeev requested to fix a large convex mirror?

**SECTION : E**

Question 24.

An illuminated object and a screen are placed 90 cm apart. Determine the focal length and nature of the lens required to produce a clear image on the screen, twice the size of the object.

Question 25.

- With the help of a diagram, explain the principle and working of a moving coil galvanometer.
- What is the importance of a radial magnetic field and how is it produced?
- Why is it that while using a moving coil galvanometer as a voltmeter a high resistance’in series in required whereas in an ammeter a shunt is used?

**OR**

- Derive an expression for the force between two long parallel current carrying conductors.
- Use this expression to define S.I. unit of current.
- A long straight wire AB carries a current I. A proton P travels with a speed v, parallel to the wire, at a distance d from it in a direction opposite to the current as shown in the figure. What is the force experienced by the proton and what is its direction?.

Question 26.

(a) State Faraday’s law of electromagnetic induction.

Figure shows a rectangular conductor PQRS in which the conductor PQ is free to move in a uniform magnetic field B from x = 0 to x = b and is zero for x > b. Assume that only the arm PQ possesses resistance r. When the arm PQ is pulled outward from x = 0 to x = 2b and is then moved backward to x = 0 with constant speed v, obtain the expressions for the flux and the induced emf, the force necessary to pull the arm and the power dissipated as Joule heat. Sketch the variations of these quantities with distance 0 ≤ x ≤ 2b.

**OR**

(a) Draw a schematic diagram of a step-up transformer. Explain its working principle. Deduce the expression for the secondary to primary voltage in terms of the number of turns in the two coils. In an ideal transformer, how is this ratio related to the currents in the two coils? How is the transformer used in large scale transmission and distribution of electrical energy over long distances?

(b) Draw the circuit arrangement for studying the input and output characteristics of an n-p-n transistor in CE configuration. With the help of these characteristics define

- input resistance,
- current amplification factor.

**OR**

(b)

- Explain how a depletion region is formed in a junction diode.
- How is forward biasing different from reverse biasing in a p-n junction diode?
- Write the truth table for the logic circuit shown below and identify the logic operation by this circuit.

**Answers **

**SECTION : A
**

Answer 1.

The SI unit of electrostatic potential is JC^{-1}. It is a scalar quantity.

Answer 2.

The direction of magnetic field is perpendicular to x and y-axis i.e. along z axis.

Answer 3.

It is the phenomenon of production of induced e.m.f. in a coil when a changing current passes through it. S.I unit is Henry (H).

Answer 4.

Let f_{1} is the focal length of convex lens and f_{2} is the focal length of concave lens. Then their equivalent focal length F would be:

Answer 5.

Ionisation energy is the required energy to knock an electron out of the atom, that is energy required to take an electron from its ground state to the outermost orbit (n = ∞). Ionisation energy of hydrogen E =E_{∞} – E_{1} = 0 – (- 13.6 eV) = 13.6 eV

**SECTION : B**

Answer 6.

The current I through the wire is related with drift velocity as :

I = nAev_{d
}Since two wires X and Y of same diameter but different materials are joined in series.

**OR**

The charge (+ q) at the centre induces charge – q on the inner surface of the + q on the outer surface of the shell

(a) (i) Surface charge density on the inner surface

(ii) Surface charge density on the outer surface

(b) For x > r_{2
}

Answer 7.

Torque acts on a current carrying coil suspended in magnetic field. (τ = NIAB sin θ) Two reasons.

- Galvanometer is a very sensitive device, it gives a full-scale deflection for a current of the order of μA.
- For measuring currents, the galvanometer has to be connected in series, and it has a finite resistance, this will change the value of the current in the circuit.

**Two factors :**

The current sensitivity of a moving coil galvanometer can be increased by

- increasing the number of turns,
- increasing cross-section area of the loop

Answer 8.

**When resistance R and S are connected :**

Then balanced point is found at distance l_{1} from the zero

Answer 9.

In order to carry the audio signal message to large distances, it is superimposed on a high frequency carrier wave. The process of superimposing audio signal over a high frequency carrier wave is called modulation.

Answer 10.

Figure shows the variation of i_{m} with co in a LCR series circuit for two values of Resistance R_{1} and R_{2} (R_{1} > R_{2}),

The condition for resonance in the LCR circuit is,

We can observe that the current amplitude is maximum at the resonant frequency ω_{0}. Since i_{m} = V_{m}/R at resonance, the current amplitude for case R_{2} is sharper to that for case R_{1}. Quality factor or simply the Q-factor of a resonant LCR circuit is defined as the ratio of voltage drop across the capacitor (or inductor) to that of applied voltage. It is given by

The Q factor determines the sharpness of the resonance curve and if the resonance is less sharp, the maximum current decreases and also the circuit is close to the resonance for a larger range of frequencies and the regulation of the circuit will not be good. So, less sharp the resonance, less is the selectivity of the circuit while higher is the Q, sharper is the resonance curve and lesser will be the loss in energy of the circuit.

When X_{L} = X_{c} or V_{L} = V_{c}, the LCR circuit is said to be in resonance condition.

**SECTION : C**

Answer 11.

E along x-axis, B along y-axis or vice-versa.

Answer 12.

Let σ the surface charge density at point P and ds be the area of cross-section According to the Gauss law

Answer 13.

Since induced emf of a coil rotating in magnetic field B with angular velocity ω is given by And induced current

E_{0} = NBAω

And induced current I_{0} = ε_{0} / R

- The ratio of induced emf = 1 : 1
- The ratio of induced current = 1 : 1

Since the induced emf does not depend on nature of material of coil.

Answer 14.

The kinetic energies of proton (charge e) and a particle (charge 2e) accelerated through V volt are :

K_{p} = 1 eV and K_{α} = 2 eV

Thus

**Thus de-Broglie wavelength of a particle and proton are :**

Where m_{α} and m_{p} are respectively masses of α – particle and proton. Thus

Answer 15.

Modulation of a signal is necessary to transmit a signal in the audio frequency range over a long distance due to

- Size of antenna
- Effective power radiated by antenna.

Answer 16.

**Einstein photoelectric equation :**

K_{max} = 1/2 mv^{2} = hv – Φ_{0
}Where Φ_{0} is work function. If the energy of the photon absorbed by the electron is less than the work function Φ_{0} of the metal, then the electron will not be emitted. Therefore, for the given metal, the thershold, frequency of light be v_{0}, then an amount of energy hv_{0} of the photon of light will be spent in ejecting the electron out of the metal, that is, it will be equal to the work function W.

W = hv_{0}

Thus, E_{k} = hv – hv_{0
}

**This equation is called ‘Einstein Photoelectric equation’. Salient features :**

- The stopping potential and hence the maximum kinetic energy of emitted electrons varies linearly with the frequency of incident radiation.
- There exists a minimum cut-off frequency v
_{0}, for which the stopping potential is zero. - Photoelectric emission is instantaneous.

Answer 17.

Plot of potential energy of a pair of nucleons as a function of their separation:

**Conclusions :**

- The nuclear force is much stronger than the Coulomb force acting between the charges or g too gravitational forces.
- The nuclear force between two nucleons falls rapidly to zero as their distance is more than few fermi.
- For a separation greater than r
_{o}the force is attractive and for separations less than r_{o}force is strongly repulsive.

**OR**

Constancy of binding energy per nucleon in range 30<A<170a consequence of the fact that nuclear force is short-ranged. It will be under the influence of only some of its neighbours.

Answer 18.

Sky wave propagation can be used by short wave broadcast services having frequency f and from a MHz to 30 MHz.In this mode sky waves are directed towards the sky and are reflected by the ionosphere toward the desired station on the earth.

Sky wave propagation is restricted to frequencies upto 30 MHz because ionosphere cannot reflect electro-magnetic waves having frequencies greater than 30 MHz.

Answer 19.

The internal resistance of a cell depends on nature of electrolyte as well as the concentration of the electrolyte.Let r be the internal resistance of the cell

When the terminals of the cell are connected to a resistance R_{1} = 5 Ω then

Answer 20.

(a) The capacitors C_{1} C_{2} and C_{3} are in series

Now C_{s} and C_{4} are in parallel combination.

Therefore the equivalent capacitance,

C_{eq }= C_{s }+ C_{4 }= 4 + 12 = 16 μF

**(b) Charge on each capacitor :
**Charge on capacitor C

_{4 }Q

_{4 }= C

_{4}V = 12 x 500 μC = 6000 μC = 6 x 10

^{-3 }C

Charge on capacitors C

_{1}, C

_{2}and C

_{3 }Q

_{123}

^{=}4 μF x 500 V = 2 x 10

^{-3 }

Answer 21.

**(i) Working :**

Telescope has an objective and eyepiece. The objective has a large focal length and much larger aperture than the eyepiece. Light from a distant object enters the objective and a real image is formed in the tube at its second focal point. The eyepiece magnifies this image producing a final inverted image at infinity.

**(ii) Calculation of magnifying power :
**Here, Power of eyepiece = 10 D

∴ Focal length, f

_{e }= 100/10 = 10 cm

and power of objective = 1 D

Focal length,f

_{0}= 100/1 = 100 cm

Hence, Magnifying power in normal adjustment :

Negative sign shows that final image is inverted.

**OR**

(i) Labelled ray diagram of compound microscope

The objective forms a real, inverted, magnified image of the object. This serves as the object for the second lens, the eyepiece, which functions essentially like a simple microscope or magnifier, produces the final image, which is enlarged and virtual.

(ii) To achieve a large magnification of a small object; both the objective and eyepiece should have small focal lengths.

Answer 22.

(a)

**(b) Derivation of average life :
**Hence the average life period of a radioactive element is 1.44 times the half life period of the element.

**SECTION : C**

Answer 23.

(i) Social, responsible to society, helping nature, knowledge of science and caring for others.

(ii) Convex mirrors are fixed on sharp turns of the road because convex mirrors diverge the image and due to divergence, we can know about the vehicle coming from opposite side of the turn.

(iii) A polaroid consists of long chain molecules aligned in a particular direction. The electric vectors (associated with the propagating light wave) along the direction of the aligned molecules get absorbed. Thus, if an unpolarised light wave is incident on such a polaroid, then the light wave will get linearly polarised with the electric vector oscillating along a direction perpendicular to the aligned molecules.

When the polaroid is rotated in the path of plane polarized light, its intensity will vary from maximum to minimum, let I_{0} be intensity of polarized light after passing through first polarizer P_{1}. Then intensity of polarized light after passing through second polarizer P_{2} is given by

I = (I_{0} cos^{2} θ)

Expression for the intensity transmitted through second polaroid

I = (I_{0} cos^{2} θ) cos^{2} (90° – θ) = I_{0} (cos θ sin θ)^{2} = I_{0} sin^{2} 2θ/4

where I_{0} is the intensity of the polarized light after passing through the first polaroid.

- Intensity will be maximum when θ = 45°
- Minimum when θ = 0°

**SECTION : E**

Answer 24.

**Nature of lens :**

Convex/Converging.

Answer 25.

**(a)
Moving coil galvanometer :**

It is a device used for the detection and measurement of small electric current.

**Construction :**

It consists of a coil having a large number of turns of insulated copper wire wound around metallic frame. A hair spring is attached to lower end of coil, the other end is attached to scale through the pointer.

**Principle :**

Galvanometer action is based upon the principle that when electric current flows in a coil placed in a magnetic field, a deflecting torque acts upon the coil whose magnitude depends upon the strength of the current.

**Working :**

The magnetic torque tends to rotate the coil. A spring provides a counter torque that balances the magnetic torque; resulting in a steady angular deflection. The deflection is indicated on the scale by a pointer attached to the spring (Sp). If I is current flowing through coil.

B = magnetic field supposed to be uniform and parallel to coil

A = area of coil

Deflecting torque acting on the coil is

τ = NI AB sin 90° = NIBA [ v sin 90° = 1]

Due to deflecting torque, the coil rotates and suspension wire gets twisted.

**(b) Importance and production of radial magnetic field :**

In a radial magnetic field, magnetic torque remains maximum for all positions of the coils. It is produced due to cylindrical pole pieces and soft iron core.

**Reason:**

Voltmeter :

Voltmeter :

This ensures that a very low current passes through the voltmeter and hence does not change (much) the original potential difference to be measured.

**Ammeter :**

This ensures that the total resistance of the circuit does not change much and the current flowing remains (almost) as its original value.

(c) A galvanometer can be converted into a voltmeter by connecting a high resistance in series with galvanometer to draw a very small current. A galvanometer can be converted into an ammeter by connecting a low resistance shunt in parallel with galvanometer to draw a very large current.

**OR**

(a) Two long parallel conductors ‘a’ and ‘b’ are separated by a distance d and carry (parallel) currents l_{a} and I_{b} respectively. The conductor V produces, the same magnetic field B_{a }at all points along the conductor ‘b’ perpendicular to the plane of the page directed downwards.

Y_{ba} is the force on a segment L of A’ due to ‘a’. The magnitude of this force is given by

Force per unit length, of conductor ‘b’ is, therefore,

By Fleming left hand rule, this force acts on CD towards AB force. This force will be attractive.

**(b) Definition of ampere :**

The ampere is the value of that steady current which, when maintained in each of the two very long, straight, parallel conductors of negligible crosssection, and placed one metre apart in vacuum, would produce on each of these conductors a force equal to 2 x 10^{-7} newton per metre of length.

(c) Magnetic field due to the straight wire AB at a perpendicular, distance r from it.

Therefore, force on proton moving with velocity V perpendicular to B, is

The direction of force on proton acts in plane of paper towards right.

Answer 26.

(a) Faraday’s law of induction : The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit.

Where Φ_{B} is flux linked with one turn of the coil. The negative sign indicates the direction of e and hence the direction of current in a closed loop. If the circuit is closed a current

I = ε / R is set up in it.

Refer to above figure. The arm PQ of the rectangular conductor is moved from x = 0 outwards. The uniform magnetic field is perpendicular to the plane and extends from x = 0 to x = b and is zero the situation when the arm PQ possesses substantial resistance r. Consider the situation when arm PQ is pulled outwards from x = 0 to x = 2b and is then moved back to x = 0 with constant speed v.

Let us consider forward motion from x = 0 to x = 2b, the flux Φ_{B} linked with the circuit SPQR is

Φ_{B }= Blx 0< x < b

= Blb 0< x < b

When the induced emf is non-zero the current I is (in magnitude)

The joule heating loss is

**(a) Step up Transformer :**

**Principle :**

When an alternating voltage is applied to the primary, the resulting current produces an alternating magnetic flux which links the secondary and induces an emf in it (mutual induction). The value of this emf depends on the number of turns in the secondary.

**Secondary Windings :**

The core laminations are joined in the form of strips; in between the strips you can see that there are some narrow gaps right through the cross-section of the core. These staggered joints are said to be ‘imbricated’.

**Working :**

Both the coiLs have high mutual inductance. A mutual electro-motive force is induced in the transformer from the alternating flux that is setup in the laminated core, due to the coil that is connected to a source of alternating voltage. Most of the alternating flux developed by this coil is linked with the other coil and thus produces the mutual induced electro-motive force. Thus so produced electro-motive force can be explained with the help of Faraday’s laws of Electromagnetic Induction as – It’ the secondary coil circuit is closed, a current flow In it and thus electrical energy is transferred magnetically from the first to the second coil.

The alternating current supply is given to the first coil and hence it can be called as the primary winding. The energy is drawn out from the second coil and thus can be called as the secondary winding.

**Large scale transmission :**

The large scale transmIssIon and distribution of electrical energy over long distances is done with the use of transformers. The voltage output of the generator is stepped-up (so that current is reduced and consequently, the l2R loss is cut down). h is then transmitted over long distances to an area sub-station near the consumers. There the voltage is stepped down. Ii is further stepped down at distributing sub-stations and utility poles before a power supply of 220 V reaches our homes.

(b)

(i) Input resistance :

This is defined as the ratio of change ïn base emitter voltage (ΔV_{BE}) to the resulting change in base current (ΔI_{B}) at constant collector-emitter voltage (ΔV_{CE}).

**(ii) Current ampliflcation factor :**

This is defined as the ratio of the change in collector current to the change in base current at a constant collector-emitter voltage (V_{CE}) when the transistor is in active state.

**OR**

(b) (i) Depletion region is formed in a junction diode ‘because depletion region is created around thep-n junction which is devoid of free charge carriers and has immobile ions. It is created due to diffusion of majority carriers across the junction when p – n junction is formed.

**(ii) Forward Biasing :**

The positive terminal of the external battery is connected to p-side and negative terminal of battery to n-side of p-n junction. The forward bias voltage oppose the potential barrier. Due to this, the potential barrier is reduced and hence the depletion becomes thin.

**Reverse biasing :**

The negative terminal of the external battery is connected top-side and positive terminal of battery to n-side of p-n junction. The reverse bias voltage supports the potential barrier. Due to this, the potential barrier is increased. The resistance of p-n junction becomes more

(iii)

Clearly Y = A + B

Hence. the given circuit perform the function of an OR gate.

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