CBSE Sample Papers for Class 12 Physics Paper 1 are part of CBSE Sample Papers for Class 12 Physics. Here we have given CBSE Sample Papers for Class 12 Physics Paper 1.

## CBSE Sample Papers for Class 12 Physics Paper 1

Board |
CBSE |

Class |
XII |

Subject |
Physics |

Sample Paper Set |
Paper 1 |

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 1 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.

What is the electrostatic potential due to an electric dipole at an equatorial point?

Question 2.

Name the em waves used for studying crystal structure of solids. What is its frequency range?

Question 3.

An electron does not suffer any deflection while passing through a region of uniform magnetic field. What is direction of the magnetic field?

Question 4.

How would the angular separation of interference fringes in Young’s double slit experiment change when the distance between the slits and screen is doubled?

Question 5.

Two nuclei have mass numbers in the ratio 1 : 8. What is the ratio of their nuclear radii?

**SECTION : B**

Question 6.

Draw 3 equipotential surfaces corresponding to a field that uniformly increase in magnitude but remains constant along z-direction. How are these surfaces different from that of a constant electric field along z-direction?

Question 7.

Define electric flux. Write its S.I. unit. A charge q is enclosed by a spherical surface of radius R. If the radius is reduced to half, how would the electric flux through the surface change?

Question 8.

Define refractive index of a transparent medium. A ray of light passes through a triangular prism. Plot a graph showing the variation of the angle of deviation with the angle of incidence.

Question 9.

Calculate the current drawn from the battery in the given network.

Question 10.

Optical and radio telescopes are built on the ground while X-ray astronomy is possible only from satellites orbiting the Earth. Why?

**SECTION : C**

Question 11.

Define the term ‘linearly polarised light’. When does the intensity of transmitted light become maximum, when a polaroid sheet is rotated between two crossed polaroids?

Question 12.

A wire of 15 Cl resistance is gradually stretched to double its original length. It is then cut into two equal parts. These parts are then connected in parallel across a 3.0 volt battery. Find the current drawn from the battery.

Question 13.

- The mass of a nucleus in its ground state is always less than the total mass of its constituents – neutrons and protons. Explain.
- Plot of graph showing the variation of potential energy of a pair of nucleons as a function of their separation.

Question 14.

Write the function of

- Transducer and
- Repeater in the context of communication system.

^{OR}

Write two factors justifying the need of modulation for transmission of a signal.

Question 15.

A positive point charge (+ q) is kept in the vicinity of an uncharged conducting plate. Sketch electric field lines originating from the point on to the surface of the plate. Derive the expression for the electric field at the surface of a charged conductor.

Question 16.

- State the principle of working of a meter bridge.
- In a meter bridge, balance point is found at a distance l
_{1}as shown in the figure. When an unknown resistance X is connected in parallel with the resistance S, the balance point shifts to a distance l_{2}Find the expression for X in terms of l_{1}l_{2}and S.

Question 17.

- State Faraday’s law of electromagnetic induction.
- A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earths magnetic field at the location has a magnitude of 5 x 10
^{-4}T and the dip angle is 30°?

Question 18.

In Young’s double slit experiment, monochromatic light of wavelength 630 nm illuminates the pair of slits and produces an interference pattern in which two consecutive bright fringes are separated by 8.1 mm. Another source of monochromatic light produces the interference pattern in which the two consecutive bright fringes are separated by 7.2 mm. Find the wavelength of light from the second source. What is the effect on the interference fringes if the’ monochromatic source is replaced by a source of white light?

Question 19.

Draw a schematic arrangement of the Geiger-Marsden experiment. How did the scattering of α-particles by a thin foil of gold provide an important way to determine an upper limit on the size of the nucleus? Explain briefly.

Question 20.

Distinguish between sky wave and space wave propagation. Give a brief description with the help of suitable diagrams indicating how these waves are propagated.

Question 21.

- State the principle on which the working of an optical fiber is based.
- What are the necessary conditions for this phenomenon to occur?

Question 22.

Give a circuit diagram of a common emitter amplifier using an n-p-n transistor. Draw an input and out waveforms of the signal. Write the expression for its voltage gain.

**SECTION : D**

Question 23.

Anuj’s uncle, who was a scrap dealer, was getting weak day-by-day. His hair had also started falling. When Anuj inquire about it, he told that he had purchased a container from a lab. When Anuj saw that container, he found that it was filled with material emitting high dose of radiation. He immediately advised his uncle to be admitted in a hospital and start treatment,

- Which values are reflected by Anuj?
- Which possible reasons were responsible for deterioration of Anuj’s uncle’s health?

**SECTION : E**

Question 24.

Draw a schematic sketch of a cyclotron. Explain briefly how it works and how it is used to accelerate the charged particles.

- Show that time period of ions in a cyclotron is independent of both the speed and radius of circular path.
- What is resonance condition? How is it used to accelerate the charged particles?

**OR**

(a) Two straight long parallel conductors carry currents i_{1} and i_{2} in the same direction. Deduce the expression for the force per unit length between them.

(b) A rectangular current carrying loop EFGH is kept in a uniform magnetic field as shown in the figure.

- What is the direction of the magnetic moment of the current loop?
- When is the torque acting on the loop (a) maximum, (b) zero?

Question 25.

(a) What are eddy currents? Write their two applications.

(b) Figure shows a rectangular conducting loop PQRS in which arm RS of length 7’ is movable. The loop is kept in a uniform magnetic field ‘B’ directed downward perpendicular to the plane of the loop. The arm RS is moved with a uniform speed v.

**Deduce an expression for**

- The emf induced across the arm ‘RS’,
- The external force required to move the arm, and
- The power required to push RS.

**OR**

- State Lenz’s law. Give one example to illustrate this law. “The Lenz’s law is a consequence of the principle of conservation of energy”. Justify this statement.
- Deduce an expression for the mutual inductance of two long coaxial solenoids but having different radii and different number of turns.

Question 26.

(a)

- Draw a labelled ray diagram to show the formation of image in an astronomical telescope for a distant object,
- Write three distinct advantages of a reflecting type telescope.

(b) A convex lens of focal length 10 cm is placed coaxially 5 cm away from a concave lens of focal length 10 cm. If an object is placed 30 cm in front of the convex lens, find the position of the final image formed by the combined system.

**OR**

- With the help of a suitable ray diagram, derive the mirror formula for a concave mirror.
- The near point of a hypermetropic person is 50 cm from the eye. What is the power of the lens required to enable the person to read clearly a book held at 25 cm from the eye?

**Answers **

**SECTION : A
**

Answer 1.

The electric potential due to an electric dipole at an equatorial point is zero.

Answer 2.

X-rays are used to study crystal structure of solids. The frequency range is 1 nm to 10^{-3}

Answer 3.

The magnetic field should be in the direction in which the electron moves. The force on electron is given by F = \(\vec { q } \) ( \(\vec { v } \) x \(\vec { B } \)) = 0. Here force is zero so \(\vec { v } \) and \(\vec { B } \) must be parallel to each other.

Answer 4.

Fringe width is given by :

β = \(\frac { \lambda D }{ d } \). When D doubles, β_{1} = \(\frac { 2\lambda D }{ d } \) = 2β Fringe width also doubles.

Answer 5.

The radii of the nucleus is given by R = R_{0}A^{1/3} which gives

So, the nuclear radii are in the ratio of 1 : 2.

**SECTION : B**

Answer 6.

For constant electric field in z-axis equipotential surfaces will be plane parallel to xy-planes.

If field that increases in magnitude, the equipotential surfaces will be planes parallel to XY plane but, as the field increases such planes will get closer.

Answer 7.

Electric flux is the total number of electric field lines crossing an area. Its SI units in Nm /C. The electric flux through a spherical surface of radius R for a charge q enclosed by the surface remains the same even if radius is reduced to half.

Answer 8.

**Refractive Index :**

The ratio of velocity of light in vacuum to the velocity of light in medium is called absolute refractive index of the medium.

**Graph :**

The plot of angle of deviation versus angle of incidence for a triangular prism is shown below :

Answer 9.

Given circuit can be redrawn as It is a balanced Wheatstone’s bridge as no current flows through the section BC. The resistance of R = 5 Ω in BC is therefore, ineffective. (R_{1} and R_{5} in series) and then (R_{4} and R_{3} in series). R_{15} and R_{34} are in parallel with each other.

**Total resistance of the combination is found as follows :**

R _{1}_{5 }= 3 Ω**
**R

_{34}= 6 Ω

1/R = 1/3 + 1/6

R = 2 Ω

Current drawn from the battery.

Answer 10.

X-rays from celestial objects cannot reach the earth’s surface. So for X-ray astronomy to be possible the satellite has to be present in space. However, visible light and lower wavelength light emitted by celestial objects reach the surface of the earth. So these objects can be observed with optical and radio telescopes.

**SECTION : C**

Answer 11.

‘Linearly polarized light’ is an electromagnetic wave, in which the vibrations of electric field are restricted to a single plane.

The intensity of transmitted light is given by I = I_{m} cos^{2} θ The intensity of transmitted light is maximum, when θ = 0° or 180°, or the polarizing axis of the two polaroids is parallel to each other.

Answer 12.

When any resistor is stretched to double its original length, the new resistance becomes four times of original resistance.

Given, R = 15 Ω and V = 3.0 volt

New resistance = 4 R = 4 x 15 = 60 Ω

Resistance for each part = 30 Ω

So, R_{1} = 30 Ω and R_{2} = 30 Ω

Effective Resistance in parallel combination is

Answer 13.

(a) The mass of a nucleus in its ground state is always less than the total mass of its constituents because some mass is converted into energy, in accordance with the equation E = mc^{2}.This difference in mass is called the mass defect and the energy corresponding to the mass defect is the binding energy. This is the energy that has to be supplied to the nucleus to break it up into its constituents.

(b) The graph is

Answer 14.

- Transducer converts energy from one form to another either at the input or at the output. For example, sound signals are converted to electrical signals so that they can be transmitted through the communication channel.
- Repeater stations receive the signal, amplify it and then transmit it. So they are a combination of a receiver and transmitter. This helps to increase the range of transmission of signals.

**OR**

**Two factors justifying the need of modulation for transmission of signal are :**

- Modulation helps to increase the frequency of the signal. This helps to transmit it over larger distances. This is because the power radiated by an antenna is proportional to 1/λ
^{2 }The power radiated by the antenna increase for high frequencies. - The size of the antenna required is proportional to (wavelength/4). So if waves of large wavelength are transmitted then the size of the antenna required is impractical. By increasing the frequency of the signal by modulation, its wavelength is decreased. So now shorter and more practical size antennas can be built.

Answer 15.

**Representation of electric field :**

The electric field due to a positive charge (+q) is represented as

Electric field due to a point charge: Consider a point charge +q placed at the origin O of the coordinate frame. Let q_{0} be any test charge placed at P.

According to Coulomb’s law, force on the test charge q_{0} due to charge q is given by

The magnitude of the electric field at point P is given by

The electric field due to a point charge is spherically symmetric about the charge.

Answer 16.

(i) A meter bridge is like a Wheatstone bridge. The jockey is moved along the wire and the resistances of the arms are varied, until the balance condition is reached.

**(ii) When resistance R and S are connected :**

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

Answer 17.

- Faraday’s law of electromagnetic induction states that the induced emf is proportional to the rate of change of magnetic flux.
- The induced emf will be given by E = Blv where the magnetic field B is perpendicular to the length l. In the question the earth’s magnetic field is given and the angle of dip is 30°, so the magnetic field perpendicular to the direction of the plane is B sin 30°.
- Hence induced emf

Answer 18.

When the monochromatic source is replaced by a source of white light, the fringe width is changed.

Answer 19.

Diagram of the Geiger Marsden experiment.

The alpha particles that are incident head-on with the gold nucleus experience a very large force of repulsion and undergo maximum deflection. Equating the kinetic energy of the incident alpha particle with the potential energy of the alpha particle and gold nucleus, the sum of the approximate radius of the gold nucleus and alpha particles, can be found.

Answer 20.

The sky waves are reflected from the ionosphere and received by a receiver. Space waves penetrate the ionosphere and are intercepted by a satellite. They can also be used for line of sight communication. This mode of propagation is used by short-wave broadcast service. The space waves are the radiowaves of very high frequency (i.e. between 54 MHz to 4.2 MHz).

The space waves can travel through atmosphere from transmitter antenna to- receiver antenna either directly or after reflection from ground in the earth’s troposphere region. That is why space wave propagation is also known as tropospherical propagation. The space waves travel in straight line from transmitting antenna to receiving antenna. Therefore, the space waves are used for line of sight communication such as television broadcast, microwave link and satellite communication.

Answer 21.

(i) The phenomenon of reflection of light when the light travelling in a denser medium strike the interface separating the denser medium and the rarer medium at an angle greater than the critical angle is called total internal reflection.

**(ii) Conditions :**

- Light must travel from a denser medium to rarer medium.
- The angle of incidence in the denser medium must be greater than the critical angle for the two media in contact.

Answer 22.

The diagram of a common emitter amplifier using an n-p-n transistor.

The output waveform is amplified but reversed in phase by 180 degrees.

**Voltage gain :**

It is the ratio of small change in output voltage (ΔV_{CE}) to the small change in input voltage (ΔV_{BE}).

Voltage gain,

**SECTION : D**

Answer 23.

- Concern for elders’ health, awareness, presence of mind.
- Radioactive radiation emitted by the material in the container were affecting the functioning of body organs of his uncle. Hair loss is also due to the radiation.

**SECTION : E**

Answer 24.

Charged particles are introduced between the dees. An alternating voltage applied between the dees accelerates, these particles by an electric field. A magnetic field that is perpendicular to the plane of the dees exerts a force on the particles that is given by F = q (v x B). This cause the particles to follow a circular trajectory. As they reach the dees the polarity is reversed and the particles are once more accelerated. This continues and highly energetic beams of charged particles are obtained. Magnetic ^ field is perpendicular to dees.

F = q(v x B)

(i) The particle traverses a distance 2πr in one revolution. If its time-period be T, then

Hence, the time period is independent of speed and radius of circular path.

**(ii) Resonance condition :**

When the angular frequency of the rotating charged particle and the angular frequency of the alternating voltage applied across the dees of the cyclotron are same. The charged particles are only accelerated by the electric field. The magnetic field only keeps it moving along a circular track. The acceleration happens when the charged particle crosses the gap between the two dees. At this instant the field between the dees has to be reversed so that the electric field can accelerate the charged particle.

**OR**

(a) Force on first wire due to current in second wire

This will also be equal to force on second wire due to first wire.

(b) (i) The direction of the magnetic moment of the current loop will be perpendicular to the plane of paper downward.

(ii)

- When axis of the loop is perpendicular to the magnetic field (that is, the plane of the loop is parallel to the magnetic field \(\vec { B } \) ), θ = 90°, the torque becomes maximum, τ = niAB sin θ = mB sin θ
- When axis of the loop is parallel to the magnetic field (that is, the plane of the loop is parallel to the magnetic field \(\vec { B } \) ), θ = 0, the torque becomes zero

τ = niAB sin θ

τ = mB sin θ = 0

Answer 25.

(a) When a bulk piece of conductor is subjected to changing magnetic flux, the induced current developed in it is called eddy current.

**Applications of eddy currents :**

- Magnetic brakes in trains.
- Electromagnetic damping.
- Induction furnaces.
- Electric power meter.

(b) (i) The emf induced across the arm is Blv On account of the presence of the magnetic field, there will be a force on the arm RS. This force I(l x B), is directed outwards in the direction opposite to the velocity of the arm RS. The magnitude of this force is,

(ii) On account of the presence of the magnetic field, there will be a force on the arm RS. This force I (l x B), is directed outwards in the direction opposite to the velocity of the arm RS. The magnitude of this force is,

The arm RS is being pushed with a constant speed v, power required to do this is,

P = Fv

**OR**

**(a) Lenz’s Law :**

An induced electromotive force (emf) always gives rise to a current whose magnetic field opposes the original change in magnetic flux.

Lenz’s law is shown with the negative sign in Faraday’s law of induction :

Lenz’s law states that the current induced in a circuit due to a change or a motion in a magnetic field is so directed as to oppose the change in flux or to exert a mechanical force opposing the motion.

**Explanation of conservation of energy :**

When a magnet is moved near a current carrying coil, the direction of the induced current opposes the motion of the magnet. When the north pole of the magnet is moved towards the coil, the induced current flows in a direction so that near face of the coil acts as a magnetic north pole. The repulsion between two poles opposes the motion of the magnet towards the coil. Similarly, when the north pole of the magnet moves away from the coil, the direction of the induced current is such as to make the near face of the coil a south pole. The attraction between the two poles opposes the motion of the magnet away from the coil. In either case, therefore, work has to be done in moving the magnet. This mechanical work appears as electrical energy in the coil. Thus Lenz’s law is in accordance with the principle of conservation of energy.

**(b) Mutual Inductance of two long coaxial solenoids :**

Consider the following fig. which shows two long co-axial solenoids each of length l. Let the radius of the inner solenoid S_{1} be r_{1} and the number of turns per unit length be n_{1}.

The corresponding quantities for the outer solenoid S_{2} are r_{2} and n_{2} respectively. Let N_{1} and N_{2}, be the total number of turns of coils S_{1} and S_{2} respectively. When a current I_{2} is set up through S_{2}, a magnetic flux Φ_{1 }is setup in it. The flux linkage with solenoid S_{1 }is

_{ }N_{1}Φ_{1}_{ } = M_{12}I_{2} …………..(1)

Where M_{12} is called the mutual inductance of solenoid S_{1} with respect to solenoid S_{2}. The magnetic field due to the current I_{2} in S_{2} is μ_{0}n_{2}I_{2}. The resulting flux linkage with coil S_{1} is

N_{1}Φ_{1 }= (n_{1}l) (πr_{1}^{2}) (μ_{0}n_{2}I_{2})

= μ_{0}n_{1}n_{2 }πr_{1}^{2}lI_{2}_{ }………..(2)

Where n_{12}l is the total number of turns in solenoid S_{1} From equation (1) and (2), we get

M_{12}I_{2} = μ_{0}n_{1}n_{2 }πr_{1}^{2}lI_{2} _{ }

_{OR}

M_{12} = μ_{0}n_{1}n_{2 }πr_{1}^{2}l …………..(3)

Now consider the reverse case, a current I_{1} is passed through the solenoid S_{1} and the flux linkage with coil S_{2} is

_{ }N_{2}Φ_{2} = M_{21}I_{2} …………….(4)

Where M_{21} is called the mutual inductance of solenoid S_{2} with respect to solenoid S_{1}The flux due to the current in Sj can be assumed to be confined solely inside S_{1}. Since the solenoids are very long. Thus, flux linkage with solenoid S_{2} is

_{ }N_{2}Φ_{2 } = μ_{0}n_{1}n_{2 }πr_{1}^{2}lI_{2}_{ } ………………(5)

Where n_{2}l is the total number of turns S_{2}. From equation (4) and (5), we get

M_{21} = μ_{0}n_{1}n_{2 }πr_{1}^{2}l ……………………(6)

From eq. (3) and (6), we get

M_{12} = M_{21} = μ_{0}n_{1}n_{2 }πr_{1}^{2}l

Therefore,

M = μ_{0}n_{1}n_{2 }πr_{1}^{2}l .

Answer 26.

(a) (i) Diagram of telescope

(ii) Advantage of reflecting type telescope over a refracting type telescope are :

- Lenses suffer from chromatic aberrations that are not there in mirrors.
- Lenses also have spherical aberration; a parabolic mirror will be free of spherical aberration.
- It is easier to support large mirrors as the back surface is non-reflecting, but a lens needs support around its rim.

(b)

For convex lens f = +10 cm, u = – 30 cm

image formed at I will serve as a virtual object for concave lens :

u = 10cm, f = – 10 cm

**OR**

(a)

**Mirror formula for concave Mirror :**

When image formed is real ΔABC and Δ A’B’C are similar

(b) The near point of the eye of this person is at 50 cm. The required lens must have a focal length such that the virtual image of the book placed at 25 cm is formed at 50 cm. Then image can be focused by the eye.

Thus, for the lens we have u = – 25 cm and v = – 50 cm putting these values in the lens formula

That is to serve the purpose a convex lens of focal length 50 cm is required. Further, the power of the lens is

We hope the CBSE Sample Papers for Class 12 Physics Paper 1 help you. If you have any query regarding CBSE Sample Papers for Class 12 Physics Paper 1, drop a comment below and we will get back to you at the earliest.