CBSE Sample Papers for Class 12 Physics Paper 3 are part of CBSE Sample Papers for Class 12 Physics. Here we have given CBSE Sample Papers for Class 12 Physics Paper 3

## CBSE Sample Papers for Class 12 Physics Paper 3

Board |
CBSE |

Class |
XII |

Subject |
Physics |

Sample Paper Set |
Paper 3 |

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 3 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.

How does the fringe width, in Young’s double-slit experiment change when the distance of separation between the slits and screen is doubled?

Question 2.

Mention the two characteristic properties of the material suitable for making core of a transformer.

Question 3.

Mention the two characteristic properties of the material suitable for making core of a transformer.

Question 4.

How does the angular separation between fringes in single slit diffraction experiment change when the distance of separation between the slit and screen is doubled?

Question 5.

The speed of an electromagnetic wave in a material medium is given by v = \(\frac { 1 }{ \sqrt { \mu \varepsilon } } \),μ being

the permeability of the medium and s is its permittivity. How does its frequency change?

**SECTION-B**

Question 6.

A bar magnet is moved in the direction indicated by the arrow between two coils PQ and CD. Predict the directions of induced current in each coil.

A circular coil of closely wound N turns and radius r carries a current I. Write the expression for the following:

- the magnetic field at its centre.
- the magnetic moment of this coil.

Question 7.

For the same value of angle of incidence, the angles of refraction in three media A, B and C are 15°, 25° and 35° respectively. In which medium would the velocity of light be minimum?

Question 8.

A proton and an electron have same kinetic energy. Which one has greater de-Broglie wavelength and why? **OR**

Two wires of equal length, one of copper and the other of manganin have the same resistance. Which wire is thicker?

Question 9.

A charge ‘ q ’ is placed at the centre of a cube of side l. What is the electric flux passing through each face of the cube?

Question 10.

An electric dipole is held in a uniform electric field.

- Show that the net force acting on it is zero.
- The dipole is aligned parallel to the field. Find the work done in rotating it through the angle of 180°.

**SECTION-C**

Question 11.

State the underlying principle of a transformer. How is the large scale transmission of electric energy over long distances done with the use of transformers?

Question 12.

A capacitor of capacitance ‘C’ is being charged by connecting it across a d.c. source along with an ammeter. Will the ammeter show a momentary deflection during the process of charging? If so, how would you explain this momentary deflection and the resulting continuity of current in the circuit? Write the expression for the current inside the capacitor.

Question 13.

An object AB is kept in front of a concave mirror as shown in the figure.

(a) Complete the ray diagram showing the image formation of the object.

(b) How will the position and intensity of the image be affected if the lower half of the mirrors reflecting surface is painted black?

Question 14.

Draw a labelled ray diagram of a reflecting telescope. Mention its two advantages over the refracting telescope.

Question 15.

Describe briefly with the help of a circuit diagram, how the flow of current carriers in a p-n-p transistor is regulated with emitter-base junction forward biased and base-collector junction reverse biased.

Question 16.

A light bulb is rated 100 W for 220 V ac supply of 50 Hz. Calculate:

- The resistance of the bulb;
- The rms current through the bulb.

**OR**

An alternative voltage given by V = 140 sin 31 At is connected across a pure resistor of 50Ω. Find

- the frequency of the source.
- the rms current through the resistor.

Question 17.

A circular coil of ‘N’ turns and radius ‘R’ carries a current ‘I’. It is unwound and rewound to make another coil of radius ‘R/2’, current I remains the same. Calculate the ratio of the magnetic moments of the new coil and original coil.

Question 18.

Deduce the expression for the electrostatic energy stored in ( a capacitor of capacitance ‘C’ and having charge ‘Q’. How will the (i) energy stored and (ii) the electric field inside the capacitor be affected when it is completely filled with a dielectric material of dielectric constant ‘K’?

Question 19.

Calculate the value of the resistance R in the circuit shown in the figure so that the current in the circuit is 0.2 A. What would be the potential difference between points B and E?.

Question 20.

You are given three lenses L_{1}, L_{2} and L_{3} each of focal length 20 cm. An object is kept at 40 cm in front of L_{1}, as shown, The final real image is formed at the focus of L_{3}. Find the separations between L_{1}, L_{2} and L_{3}.

Question 21.

Define the terms (i) ‘cut-off voltage’ and (ii) ‘threshold frequency’ in relation to the phenomenon of photoelectric effect. Using Einstein’s photoelectric equation show how the cut-off voltage and threshold frequency for a given photosensitive material can be determined with the help of a suitable plot/graph.

Question 22.

A senes LCR circuit is connected to an ac source. Using the phasor diagram, derive the expression for the impedance of the circuit. Plot a graph to show the variation of current with frequency of the source, explaining the nature of its variation.

**SECTION-D**

Question 23.

Renu and her friend went to see an exhibition. There security guard standing on the entrance gate asked them to come through a metal detector gate. Her friend was scared of it. But Renu convinced her and explained the purpose and working of a metal detector.

- What values Renu possess?
- What is a metal detector and how it works?

**SECTION-E**

Question 24.

(a) In Young’s double slit experiment, derive the condition for (i) constructive interference and (ii) destructive interference at a point on the screen.

(b) A beam of light consisting of two wavelengths, 800 nm and 600 nm is used to obtain the interference fringes in a Young’s double slit experiment on a screen placed 1.4 m away. If the two slits are separated by 0.28 mm, calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide.

**OR**

(a) How does an unpolarized light incident on a polaroid get polarized? Describe briefly, with the help of a necessary diagram, the polarization of light by reflection from a transparent medium.

(b) Two polaroids ‘A’ and ‘B’ are kept in crossed position.

How should a third polaroid ‘C’ be placed between them so that the intensity of polarized light transmitted by polaroid B reduces to l/8th of the intensity of unpolarized light incident on A?

Question 25.

(a) Describe briefly, with the help of a diagram, the role of the two important processes involved in the formation of a p-n junction.

(b) Name the device which is used as a voltage regulator. Draw the necessary circuit diagram and explain its working.

**OR**

(a) Explain briefly the principle on which a transistor- amplifier works as an oscillator. Draw the necessary circuit diagram and explain its working.

(b) Identify the equivalent gate for the following circuit and write its truth table.

Question 26.

(a) Write the expression for the force \(\vec { F } \) , acting on a charged particle of charge ‘q’ moving with a velocity \(\vec { v } \) in the presence of both electric field E and magnetic field \(\vec { B } \) . Obtain the condition under which the particle moves undeflected through the fields.

(b) A rectangular loop of size l x b carrying a steady current I is placed in a uniform magnetic field \(\vec { B } \) . Prove that the torque \(\vec { \tau } \) acting on the loop is given by \(\vec { \tau } \) = \(\vec { m } \) x \(\vec { B } \) where \(\vec { m } \) is the magnetic moment of the loop.

**OR**

(a) Explain giving reasons, the basic difference in converting a galvanometer into (i) a voltmeter and (ii) an ammeter.

(b) Two long straight parallel conductors carrying steady currents I_{1}, and I_{2} are separated by a distance ‘d’. Explain briefly, with the help of suitable diagram, how the magnetic field due to one conductor acts on the other. Hence deduce the expression for the force acting between the two conductors. Mention the nature of this force.

**Answers **

**SECTION : A**

Answer 1.

The fringe width becomes double when the distance of separation between the slits and screen is doubled.

Answer 2.

Two characteristic properties of material:

- Low hysteresis loss.
- Low coercivity

Answer 3.

Two characteristic properties of material:

- Low hysteresis loss.
- Low coercivity

Answer 4.

We know angular separation is given as

Since 0 is independent of D i.e., the distance of separation between the screen and the slit, so when D is doubled, angular separation would remain same.

Answer 5.

The frequencies of electromagnetic waves have its inherent characteristics. When an electromagnetic wave travels from one medium to another, its wavelength changes but frequency remains unchanged.

**SECTION-B**

Answer 6.

According to Lenz’s law the polarity of the induced emf is such that it opposes the change in magnetic flux responsible for its production. Since North pole of bar magnet is receding away from the coil so the right end of the coil will develop South pole i.e., induced current as seen from the left end will be anticlock wise.

Again, since South pole is pushing towards the second coil so the left end of the coil will develop South pole in order to repel it and decrease the flux i.e., induced current as seen from the left end will be clockwise.

**OR**

Answer 6.

(i) Magnetic field at its centre is given by,

If current in the coil is anti-clockwise, then the direction of magnetic field is perpendicular to the plane of coil upward, but if the current in the coil is clockwise, then the direction of magnetic field will be perpendicular to the plane of coil downward.

(ii) Magnetic moment of this coil is

Answer 7.

As light travels from a rarer to denser medium it ‘ bends towards the normal as its speed decreases. So, if the bending is more, the speed of the light would be less in that medium, compared to other media. As the angle of refraction is measured with respect to the normal, the ray making the least angle of refraction would bend more and the speed of light would be minimum in that case. So, the correct option is medium A where refracting angle is 15°.

Answer 8.

Since de-Broglie wavelength X in terms of kinetic energy is given as

Where E_{k} is kinetic energy, m is mass of electron and h is the Planck’s constant.

Thus, for electron and proton with same kinetic energy, de-Broglie wavelength would depend on mass.

Hence, wavelength of electron is greater than wavelength of proton.

Where l is length and A is area of cross-section of material. Also p is the resistivity of the material.

Where ¡ is length and A is area of cross-section of material. Also p is the resistivity of the material.

Since the resistivity of alloy is greater than the resistivity of its constituents. We have resistivity of manganin (Cu + Mn + Ni) greater than resistivity of copper metal.

Answer 9.

By using Gauss’s Law,

Answer 10.

(i) Consider an electric dipole consisting of two equal and opposite point charges, -q at A and + q at B, separated by a small distance 2a.

Dipole moment of a dipole is given by,

(ii) Work done on dipole,W =ΔU =pE (cos θ_{1} – cos θ_{2})

W = pE (cos 0° – cos 180°)

W = 2pE.

**SECTION-C**

Answer 11.

**Transformer Principle:**

It is a device which converts high voltage AC into low voltage AC and vice-versa. It is based upon the principle of mutual induction. When alternating current is passed through a coil, an induced emf is set up in the neighbouring coil. Transformers are used for transmission of electrical energy over long distances. It step up the output voltage of power plant using step up transformer which reduce the current through cables and hence reduce resistive power loss. Then a step down transformer is used at consumer end to step down the voltage.

Answer 12.

Yes, the ammeter shows a momentary deflection during the process of charging because changing electric field produce displacement current between the plates of the capacitor. The resulting continuity of current in the circuit is because inside the capacitor, displacement current exist and in the wire conduction current flow.

Answer 13.

(i) Image formed will be inverted, between focus and centre of curvature and small in size.

(ii) If the lower half of the mirror’s reflecting surface is painted black, the position of image will be same but its intensity gets reduced.

Answer 14.

Reflecting Telescope:

**Advantages over refracting telescope:**

- It reduces the spherical aberration by using parabolic mirror and forms a clear focussed image.
- The resolving power of a large aperture mirror is high and hence minute details of distant stars can be obtained.

Answer 15.

In p-n-p transistor, the emitter base junction is always forward biased with voltage V_{EE} and the collector base junction is always reverse biased with voltage V_{CC}. The holes in the emitter are pushed into the base by the positive terminal of battery of voltage V_{EE}. Since, base is thin and lightly doped so only few holes combine with electrons in the base. Thus, base current I_{B} is small. Since V_{CC} is quite large almost 99% of holes coming from emitter are collected by collector. For each hole reaching the collector, an electron is released from the negative terminal of the collector base battery to neutralize the hole. For each hole consumed in collector, a bond breaks in emitter and electron is released that enters positive terminal of emitter-base battery. Thus, we can say the current is carried by holes inside the transistor and by electrons in external circuit.

Answer 16.

P= 100 W and V_{rms} =220 V

Answer 17.

The magnetic moment m of a current loop,

Let m_{1} and m_{2} be the magnetic moments of circular original coil of radius ‘R’ and new coil of radius ‘R/2‘.

Length of wire remains same. Thus,

Answer 18.

**Energy stored in a charged capacitor:**

The energy of a charged capacitor is measured by the total work done in charging the capacitor to a given potential. We know that, Capacitance, C= q/v

Where q is the charge on the plates and V is potential difference.

When an additional amount of charge dq is transferred from negative to positive plate, the small work done is given by

When dielectric material of dielectric constant ‘K’ is introduced inside the capacitor then

(i) V_{0}= E_{0}d …(1)

Where V_{0 }is the potential when there is vacuum between the plates of the capacitor and d is the separation between the plates of the capacitor.

When dielectric is introduced, potential difference is given by

Answer 19.

Answer 20.

But we have seen above that image by L_{1} is formed at 40 cm on the right of L_{1} which is at 20 cm left of L_{2} (focus of L_{2}). So, x_{1} = distance between L_{1} and L_{2} = (40 + 20) cm = 60 cm Again distance between L_{2} and L_{3} does not matter as the image by L_{2} is formed at infinity.

Hence, the distance between L_{2} and L_{3} can have any value.

Answer 21.

When light of suitable frequency is incident on a metal surface, electrons are ejected from the metal. This phenomenon is called the photoelectric effect.

(i) The cathode is illuminated with light of some fixed frequency v and fixed intensity. A small photoelectric current is observed due to few electrons that reach anode just because they have sufficiently large velocity of emission. If we make the potential of the anode negative with respect to cathode then the electrons emitted by cathode are repelled. Some electrons even go back to the cathode so that the current decreases. At a certain value of this negative potential, the current is completely stopped. The least value of this anode potential which just stops the photocurrent is called cut off potential or stopping potential.

(ii) For a given material, there is a certain minimum frequency that if the incident radiation has a frequency below this threshold, no photoelectric emission will take place, howsoever intense the radiation may be falling. According to Einstein’s photoelectric equation, maximum K.E. is given as

Where X is wavelength and v is corresponding frequency and Φ is work function. We expose a material to lights of various

frequencies and thus photoelectric current is observed and cut off potential needed to reduce this current to zero noted. A graph is plotted and that is straight line.

According to Einstein’s photoelectric equation

We can read the value of threshold frequency from graph. From equation (1), we can find the value of stopping potential (V_{0}).

Answer 22.

Let an alternating emf E = E_{0} sin ωt cot is applied to a series combination of inductor L, capacitor C and resistance R. Since all three of them are connected in series the current through them is same. But the voltage across each element has a different phase relation with current.

The potential difference V_{L}, V_{c} and V_{R} across L, C and R at any instant is given by V_{T }IX_{L}, V_{c} IX_{C} and V_{R}=IR Where I is the current at that instant.

X_{L} is inductive reactance and

X_{c} is capacitive reactance.

V_{R} is in phase with I, V_{L} leads I by 90° and V_{c} lags behind I by 90°

In the phasor diagram.

V_{L} and V_{c} are opposite to each other. If V_{L} > V_{c} then resultant (V_{L} – V_{c}) is represented by OD. OR represent the resultant of V_{R} and (V_{L} – V_{c}). It is equal to the applied emf E.

At resonance, impedance of circuit becomes equal to R. Current becomes maximum and is equal to E/R

The graph of variation of peak current I_{max} with frequency is shown below:

With increase in frequency, current first increases and then decreases. At resonant frequency, the current amplitude is maximum.

**SECTION-D**

Answer 23.

- Sense of responsibility, leadership, general awareness.
- A metal detector is a LCR circuit tuned to resonance. When any person walks through the metal detector gate with any metal impedance of the circuit changes which is detected by electronic circuit and alarm sounded and security personnels become alert.

**SECTION-E**

Answer 24.

**(a) Condition of constructive and destructive interference :
**

In the given figure, S is a monochromatic source of light. S

_{1}and S

_{2}are two narrow pin holes equidistant from S and they act as coherent sources. Consider a point P on the screen XY placed parallel to S, and S

_{2}.

Let a

_{1}be the amplitude of the waves from S

_{1}and a

_{2}that from S

_{2}. Let Φ be the phase difference between the two waves reaching the point P. Let y

_{x}and y

_{2}be the displacements of the two waves, arriving at P.

Clearly the intensity of resultant wave at any point depends on the amplitude of individual waves and the phase difference between the waves at the point.

**Constructive Interference :**For maximum intensity at any point cos Φ = + 1

**Destructive Interference :**For minimum intensity at any

Clearly the maximum intensity is obtained in the region of superposition at those points where waves meet in the same phase or the phase difference between the waves is even multiple of % or path difference between them is the integral multiple of λ/2 and maximum intensity is (a

_{1}+ a

_{2})

^{2}which is greater than the sum intensities of individual waves by an amount 2a

_{1}a

_{2}.

**OR**

*(a)*Polaroid is made up of a special material work which blocks one of the two planes of vibration of an electromagnetic wave. Because of its chemical composition it allows only those vibrations of the electromagnetic wave which are parallel to its crystallographic axis. An ordinary beam of light on reflection from a transparent medium becomes partially polarised. The degree of polarization increases as the angle of incidence is increased. At a particular value of angle of incidence, the reflected

beam becomes completely polarised. This angle of incidence is called the polarizing angle (i

_{p}).

(b) By Malus law, the intensity of light emerging from the middle polaroid ‘C’ in given by

I

_{1}= I

_{0}cos

^{2}θ

This light (intensity Ij) falls on the polaroid ‘B’ whose polarisation axis makes an angle (90°- θ) with the polarisation axis of the

polaroid ‘C’ Therefore, the intensity of light emerging from ‘B’ is given by

Thus, polariod ‘C’, must be placed at an angle 45° with polaroid ‘B’.

Answer 25.

**(a) Two important process involved in the formation of a p-n junction are :**

**Diffusion and Drift:**

In p-type semiconductor, electrons are the majority carriers and holes are minority carriers. In the same way, in p-type semiconductor holes are majority and electrons are minority carriers. During the formation of p-n junction, due to concentration This space charge region is known as depletion region. An electric field directed from positive charge towards negative charge develops. Due to this field, electrons on p side of the junction move to this motion of charge carriers due to the electric field is called drift. Drift current is opposite in direction to the diffusion current. Initially, diffusion current is large and ‘drift current is small. Space- charge region on either side increases as the diffusion process continues. This increases the electric field and hence the drift current. This process continues until the diffusion current equals the drift current. Thus, a p-n junction is formed.

**(b) Zener diode is used as a voltage regulator:**

Voltage regulator converts an unregulated dc voltage into a constant regulated dc voltage using zener diode. The un regulated voltage is connected to the zener diode through a series resistance Ry such that the zener diode is reverse biased. If the input voltage side and holes on n side of the junction move to p side.

This increases, the current through R_{?} and zener diode also increases. This increases the voltage drop across R_{s} without any change in voltage drop across zener diode. This is because in the breakdown region, zener voltage remain constant even though the current through zener diode changes. Similarly, if the input voltage decreases, the current through R_{y} and zener diode decrease. The voltage drop across R_{s}, decrease without any change in the voltage across the zener diode. Thus any change in input voltage results the change in voltage drop across R_{y }without any change in voltage across the zener diode. Thus, zener diode acts as a voltage regulator.

OR

**(a) Transistor amplifier as an oscillator: **

In an oscillator, the output at a desired frequency is obtained without applying any external input voltage. The common emitter n-p-n transistor as an oscillator is shown in the following figure. A variable capacitor C of suitable range is connected in parallel to coil T_{2} to give the variation in frequency.

**Oscillator action: **

As in an amplifier, the base- emitter junction is forward biased while the base collector junction is reverse biased. When the switch S is put on, charge of collector flows in the coil T_{2}. The inductive coupling between coil T_{2} and T, cause a current to flow in the emitter circuit i.e. feedback from input to output. As a result of positive feedback, the collector current reaches at maximum. When there is no further feedback from T_{2} to T_{p} the emitter current begins to fall and collector current decreases. Therefore, the transistor has reverted back to its original state. The whole process now repeats itself. The resonance frequency (f) of the oscillator is given by:

The tank of tuned circuit is connected in the oscillator side. Hence, it is known as tuned collector oscillator.

Answer 26.

(a) Force \(\vec { F } \) acting on a charge ‘q’ moving with velocity \(\vec { v } \) in the presence of both electric field \(\vec { E } \) and magnetic field \(\vec { B } \) .

Consider a region in which magnetic field, electric field and velocity of charge particle are perpendicular to each other.

To move charge particle undeflected the net force acting on the particle must be zero i.e., the electric force must be equal and opposite to the magnetic force.

The direction of electric and magnetic forces are in opposite direction. Their magnitudes are in such a way that they cancel out each other to give net force zero and so the charge particle does not deflect.

(b) When an electric current flows in closed loop of wire, placed in a uniform magnetic field, the magnetic forces produce a torque which tends to rotate the loop so that area of the loop is perpendicular to the direction of the magnetic field.

Consider a rectangular coil PQRS placed in an external magnetic field as shown in diagram (a). Let T be the current flowing through the coil. Each part of the coil experiences one Lorentz. Each part of the forces is \(\vec { { F }_{ 1 } } ,\vec { { F }_{ 1 } } ,\vec { { F }_{ 3 } } \) as shown. The

\(\vec { { F }_{ 4 } } \) and \(\vec { { F }_{ 5 } } \) are equal in magnitude but act in opposite directions along the same straight line. Elence they cancel out.

**OR**

(a) (i) In converting a galvanometer into a voltmeter, a very high suitable resistance is connected is series to its coil. So that galvanometer gives full scale deflection.

(ii)In converting a galvanometer into an ammeter, a very small suitable resistance is connected in parallel to its coil. The remaining pair of the current i.e. (I – I_{g} ) flows through the resistance.

(b) Assumption: Current flows in the same direction. Using Right hand thumb rule, the direction of the magnetic field at point P due to current I_{2} is perpendicular to the plane of paper and inwards.

Similarly, at point Q on X_{2}Y_{2}, the direction of magnetic field due to current I_{1} is perpendicularly outward. Using Flemings left hand rule we can find the direction of forces \(\vec { { F }_{ 12 } } \) and \(\vec { { F }_{ 21 } } \) which are in opposite directions thus, By Amperes circuital law, we have,

From above we get the magnitude of forces \(\vec { { F }_{ 12 } } \) and \(\vec { { F }_{ 21 } } \) are equal but in opposite direction.

So, \(\vec { { F }_{ 12 } } \) = — \(\vec { { F }_{ 21 } } \)

Therefore, two parallel straight current carrying conductors in the same direction attract each other. Similarly, we can prove if two parallel straight conductors carry currents in opposite direction, they repel each other with the same magnitude as equation (1).

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