CBSE Sample Papers for Class 12 Physics Paper 5 are part of CBSE Sample Papers for Class 12 Physics. Here we have given CBSE Sample Papers for Class 12 Physics Paper 5

## CBSE Sample Papers for Class 12 Physics Paper 5

Board |
CBSE |

Class |
XII |

Subject |
Physics |

Sample Paper Set |
Paper 5 |

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 5 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.

Define electric dipole moment. Write its S.I. unit.

Question 2.

A cell of emf E’ and internal resistance r draws a current ‘I’. Write the relation between terminal voltage ‘V’ in terms of E, I, r.

Question 3.

How are side bands produced?

Question 4.

Graph showing the variation of current versus voltage for a material GaAs is shown in the figure, Identify the region of:

- negative resistance
- where Ohms law is obeyed.

Question 5.

A concave lens of refractive index 1.5 is immersed in a medium of refractive index 1.65. What is the nature of the lens ?

**SECTION-B**

Question 6.

Use Kirchhoff’s rules to obtain conditions for the balance condition in a Wheatstone bridge.

Question 7.

Show that the radius of the orbit in hydrogen atom varies as n. Where n is the principal quantum number of the atom.

Question 8.

Distinguish between ‘intrinsic’ semiconductors

Question 9.

Use the mirror equation to show that an object placed between ƒ and 2ƒ of a concave mirror produces a real image beyond 2ƒ.

**OR**

Find an expression for intensity of transmitted light when a Polaroid sheet is rotated between two crossed Polaroids. In which position of the Polaroid sheet will the transmitted intensity be maximum ?

Question 10.

A proton and an a-particle have the same de Broglie wavelength. Determine the ratio of (i) their accelerating potentials (ii) their speeds

**SECTION-C**

Question 11.

Write Einstein’s photoelectric equation and mention which important features in photoelectric effect can be explained with the help of this equation. The maximum kinetic energy of the photo electrons gets doubled when the wavelength of light incident on the surface changes from λ_{1} to λ_{2}. Derive the expressions for the threshold wavelength λ_{0} and work function for the metal surface.

Question 12.

In the study of Geiger-Marsdan experiment on scattering of a-particles by a thin foil of gold, draw the trajectory of α-particles in the Coulomb field of target nucleus. Explain briefly how one gets the information on the size of the nucleus from this study.

From the relation R = R_{0} A^{1/3}, where R_{0} is constant and A is the mass number of the nucleus, show that nuclear matter density is independent of A.

**OR**

Distinguish between nuclear fission and fusion. Show how in both these processes energy is released. Calculate the energy release in Me V in the deuterium-tritium fusion reaction:

Question 13.

A cell of emf ‘E’ and internal resistance ‘r’ is connected across a variable load resistor R. Draw the plots of the terminal voltage V versus (i) R and (ii) the current i. It is found that when R = 4 Ω, the current is 1 A when R is increased to 9 Ω, the current reduces to 0.5 A. Find the values of the emf E and internal resistance r.

Question 14.

Two capacitors of unknown capacitance C_{1} and C_{2} are connected first in series and then in parallel across a battery of 100 V. If the energy stored in the two combinations is 0.045 J and 0.25 J respectively, determine the value of C_{1} and C_{2}. Also calculate the charge on each capacitor in parallel combination.

Question 15.

Answer the following questions:

(a) In a double slit experiment using light of wavelength 600 nm, the angular width of the fringe formed on a distant screen is 0.1°. Find the spacing between the two slits. Light of wavelength 5000 Å propagating in air gets partly reflected from the surface of water. How will the wavelengths and frequencies of the reflected and refracted light be affected?

Question 16.

An inductor L of inductance X_{L} is connected in series with a bulb B and an ac source. How would brightness of the bulb change when

- number of turn in the inductor is reduced,
- an iron rod is inserted in the inductor and
- a capacitor of reactance X
_{c}= X_{L}is inserted in series in the circuit. Justify your answer in each case.

Question 17.

State the principle of working of a galvanometer. A galvanometer of resistance G is converted into a voltmeter to measure up to V volts by connecting a resistance R_{1} in series with the coil. If a resistance R_{2} is connected in series with it, then it can measure up to V/2 volts. Find the resistance, in terms of R_{1} and R_{2}, required to be connected to convert it into a voltmeter that can read up to 2 V. Also find the resistance G of the galvanometer in terms of R_{1} and R_{2}.

Question 18.

With what considerations in view, a photo diode is fabricated? State its working with the help of a suitable diagram. Even though the current in the forward bias is known to be more than in the reverse bias, yet the photo diode works in reverse bias. What is the reason?

Question 19.

Draw a circuit diagram of a transistor amplifier in CE configuration. Define the terms:

- Input resistance and
- Current amplification factor. How are these determined using typical input and output characteristics?

Question 20.

Name the parts of the electromagnetic spectrum which is

(a) suitable for radar systems used in aircraft navigation

(b) used to treat muscular strain

(c) used as a diagnostic tool in medicine

Question 21.

- A giant refracting telescope has an objective lens of focal length 15 m. If an eye-piece of focal length 1.0 cm is used. What is the angular magnification of the telescope?
- If this telescope is used to view the moon. What is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 x 10
^{8}m and the radius of lunar orbit is 3.8 x 10^{8}m.

Question 22.

- Why is communication line of sight mode limited to frequencies above 40 MHz?
- A transmitting antenna at the top of a tower has a height 32 m and the height of the receiving antenna is 50 m. What is the maximum distance between them for satisfactory communication in line of sight mode?

**SECTION-D**

Question 23.

A group of students while coming from the school noticed a box marked “Danger H.T. 2200 V” at a substation in the main street. They did not understand the utility of such a high voltage, while they argued the supply was only 220 V. They asked their teacher this question the next day. The teacher thought it to be an important question and therefore, explained to the whole class.

Answer the following questions:

- What device is used to bring the high voltage down to low voltage of a.c. current and what is the principle of its working?
- Is it possible to use this device for bringing down the high dc voltage to the low voltage? Explain.
- Write the values displayed by the students and the teacher.

**SECTION-E**

Question 24.

(a) An electric dipole of dipole moment \(\vec { p } \) consists of point charges \(\underrightarrow { + } \) q and – q separated by a distance 2a apart. Deduce the expression for the electric field \(\vec { E } \) due to the dipole at a distance x from the centre of the dipole on its axial line in terms of the dipole moment \(\vec { p } \). Hence show that in the limit

(b) Given the electric field in the region E =2 *x̂* l, find the net electric flux through the cube and the charge enclosed by it.

**OR**

(a) Explain, using suitable diagrams, the difference in the behaviour of a (i) conductor and (ii) dielectric in the presence of external electric field. Define the terms polarization of a dielectric and write its relation with susceptibility.

(b) A thin metallic spherical shell of radius r carries a charge Q on its surface. A point charge Q/2 is placed at its centre C and another charge + 2Q is placed outside the shell at a distance x from the centre as shown in the figure. Find (i) the force on the charge at the centre of shell and at the point A, (ii) the electric flux through the shell.

Question 25.

(a) State Ampere’s Circuital law. Use this law to obtain the expression for the magnetic field inside an air cored toroid of average radius, having ‘n’ turns per unit length and carrying a steady current I.

(b) An observer to the left of a solenoid of N turns each of cross section area A observes that a steady current I in / it flows in the clockwise direction. Depict the magnetic f field lines due to the solenoid specifying its polarity and N ; show that it acts as a bar magnet of magnetic momentum

M = NIA.

**OR**

(a) Define mutual inductance and write its S.I. units.

(b) Derive an expression for the mutual inductance of two long co-axial solenoids of same length wound one over the other.

(a) In an experiment two coils c_{1} and c_{2} are placed close to each other. Find out the expression for the emf induced in the coil c_{1} due to a change in the current through the coil c_{2}

Question 26.

(a) A point object ‘O’ is kept in a medium of refractive index n_{1} in front of a convex spherical surface of radius of curvature R which separate the second medium of refractive index n_{2} from the first one as shown in the figure. Draw the ray diagram showing the image formation and deduce the relationship between the object distance and the image distance in terms of n_{1},n_{2} and R.

(b) When the image formed above acts as a virtual object for a concave spherical surface separating the medium n_{2} from n_{1} (n_{2}> n_{1}) draw this ray diagram and write the similar (similar to (a)) relation. Hence obtain the expression for the Lens Makers formula.

**Answers **

**SECTION-A**

Answer 1.

Electric dipole moment: Dipole moment is a measure of strength of electric dipole. It is vector quantity whose magnitude is equal to product of magnitude of charge and the distance between them.

SI unit of dipole moment is coulomb-metre (C-m).

Answer 2.

When the current I draws from a cell of emf E and internal resistance r, then the terminal voltage is V = E – Ir.

Answer 3.

Side bands are produced during the process of modulation. During modulation the audio frequency modulating signal wave is superimposed on a high frequency wave called carrier wave. Any form of modulation produces, frequencies that are the sum and difference of the carrier and modulating frequencies. These frequencies are called as side bands.

bands

Lower side band frequency =ƒ_{c} – ƒ_{m
}Upper side band frequency = ƒ_{c} + ƒ_{m }where

ƒ_{c}→_{ }Carrier wave frequency

ƒ_{m }→Modulating wave frequency

Answer 4.

- DE is the region of negative resistance because the slope of curve in this part is negative.
- BC is the region where Ohms law is obeyed because in this part, the current varies linearly with the voltage.

Answer 5.

Since n_{g} lens < μ_{m} surroundings. It behaves like a converging lens.

**SECTION-B**

Answer 6.

Let us consider a Wheatstone bridge arrangement as shown: Wheatstone bridge is a special bridge type circuit which consists of four resistances, a galvanometer and a battery. It is used to determine unknown resistance.

In figure four resistance P, Q, R and S are connected in the form of four arms of a quadrilateral. Let the current given by battery in the balanced position be I. This current on reaching point A is divided into two parts I_{1} and I_{2}. As there is no current in galvanometer in balanced state, therefore, current in resistances P and Q is I_{1} and in resistances R and S it is I_{2}.

Applying Kirchhoff’s first law at point A

I – I_{1} – 1_{2}=0 or I = I_{1}+1_{2} …(1)

Applying Kirchhoff’s second law to closed mesh ABDA

– I_{1}P + I_{2}R = 0 or I_{1}P = I_{2}R …(2)

Applying Kirchhoff s second law to mesh BCDB

This is condition of balance Wheatstone’s bridge.

Answer 7.

According to the Bohr’s theory of hydrogen atom,

the angular momentum of a revolving electron is given by

Here,

m = Mass of the electron v= Velocity of the electron

r = Radius of the orbit n h= Planck’s constant

n = Principal quantum number of the atom

If an electron of mass m and velocity v is moving in a circular orbit of radius r, then the centripetal force required is given by

F = \(\frac { { mv }^{ 2 } }{ r } \)

Also, if the charge on the nucleus is Ze, then the force of electrostatic attraction between the nucleus and the electron will provide the necessary centripetal force.

Answer 8.

Intrinsic semiconductor |
Extrinsic semiconductors |

1. It is pure semi-conducting material with no impurity atoms added to it. | It is prepared by doping a small quantity of impurity atoms to the pure semiconductor. |

2. The number of free electrons in the conduction band and the number of holes in valence band is exactly equal. | The number of free electrons and holes is never equal. There is an excess of electrons in n-type semiconductors and an excess of holes in p-type semiconductors |

3. Its electrical conductivity is a function of temperature alone | Its electrical conductivity depends upon the temperature and the amount of impurity added in them. |

Answer 9.

Mirror equation is \(\frac { 1 }{ f } =\frac { 1 }{ u } +\frac { 1 }{ v } \)

Where u is distance of object from the mirror, v is the distance of image from the mirror and ƒ is the focal length of the mirror.

For a concave mirror/is negative i.e.,

f< 0, v < 0 .

For a real object (on the left of mirror)

Real image is formed beyond 2ƒ. This implies that v is negative and greater than 2ƒ This means that the image lies beyond 2ƒ and it is real.

OR

Let us consider two crossed polarisers P, and P_{2} with a polaroid sheet P_{3} placed between them. Let I_{0} be the intensity of polarised light after passing through the first polarise P_{1} If θ is the angle between the axes of P_{1} and P_{2}, then the intensity of the polarized light after passing through P_{3} will be I = I_{0} cos^{2} θ.

As P_{1} and P_{2} are crossed the angle between the axes of P_{1} and P_{2} = 90°.

∴ Angle between the axes of P_{2} and P_{3} = (90° – 0)

Answer 10.

(i) The deBroglie wavelength of a particle is given by

where, V is the accelerating potential of the particle It is

given that

(ii) We can also write de Broglie wavelength as λ = h/mv where h is Planck’s constant, m is mass of the particle and v is speed of the particle.

**SECTION-C**

Answer 11.

Einstein’s photoelectric equations is given by

K_{ma}x ^{=} Maximum kinetic energy of the photoelectron

^{v}max ^{=} Maximum velocity of the emitted photoelectron

m = Mass of the photoelectron

v = Frequency of the light radiation

_{Φ0} = Work function

h = Planck’s constant

If v_{0} is the threshold frequency, then the work function can be written as

The above equations explains the following results.

- If v < v
_{0 }then the maximum kinetic energy is negative, which is impossible. Hence, photoelectric emission does not take place for the incident radiation below the threshold frequency. Thus, the photoelectric emission can take place if v > v_{0}. - The maximum kinetic energy of emitted photelectrons is directly proportional to the frequency of the incident radiation. This means that maximum kinetic energy of photo electron depends only on the frequency of incident light.

According to the photoelectric equation,

Let the maximum K.E. for the wavelength of the incident λ_{1} be k_{1
}

Answer 12.

From this experiment, the following is observed:

- Most of the alpha particles pass straight through the gold foil. It means that they do not suffer any collision with gold atoms.
- About one alpha particle in every 8000 alpha particles deflects by more than 90°.

As most of the alpha particles go undeflected and only a few get deflected, this shows that most of the space in an atom is empty and at the centre of the atom, there exists a nucleus. By the number of the alpha particles deflected, the information regarding size of the nucleus can be known. If m is the average mass of a nucleon and R is the nuclear radius, then mass of nucleus = mA, where A is the mass number of the element.

This shows that the nuclear density is independent of A.

OR

**Nuclear fission:** Nuclear fission is a disintegration process, in which a heavier nucleus gets split up into two lighter nuclei, with the release of a large amount of energy.

**Nuclear fusion:** When two or more light nuclei combine to from a heavy stable nuclide, part of mass disappears and is converted into energy. This phenomenon is called nuclear fusion.

Answer 13.

Answer 14.

When the capacitors are connected in parallel.

Answer 15.

(a) Angular width (θ) of fringe in double-slit experiment is given by θ = λ/d

Where, d = Spacing between the slits

Given, Wavelength of light λ = 600 nm = 600 x 10^{-9} m

Angular width of fringe, θ = 0.1° = 0.0018 rad

The refracted wave will have same frequency, only wavelength will change.

The velocity of light in water is given by

As light ray in travelling from rare (air) medium to denser medium, its speed will decrease. Hence wavelength (λ) will also decrease.

Answer 16.

(i) When the number of turns in the inductor is reduced, its reactance X_{L} The current in the circuit increases and hence brightness of the bulb increases.

(ii) When an iron rod is inserted in the inductor, the self inductance (L) increases. Consequently, the inductive reactance X_{L} = coL increases. This decreases the current in the circuit and the bulb glows dimmer.

(iii) With capacitor of reactance X_{c} = X_{L}, the impedance Z = \(\sqrt { { R }^{ 2 }+{ \left( { X }_{ L }-{ X }_{ C } \right) }^{ 2 } } \) R = becomes minimum, the current in circuit becomes maximum. Hence the bulb glows with maximum brightness

Answer 17.

Principle: When a current-carrying coil is placed in a magnetic field, it experiences a torque. From the measurement of the deflection of the coil, the strength of the current can be computed. A high resistance is connected in series with the galvanometer to convert it into voltmeter. The value of the resistance is given by

Answer 18.

A photo diode is used to observe the change in current with change in the light intensity under reverse bias condition. In fabrication of photo diode, material chosen should have band gap -1.5 eV or lower so that solar conversion efficiency is better. This is the reason to choose Si or GaAs material.

**Working:** It is a p-n junction fabricated with a transparent window to allow light photons to fall on it. These photons generate electron hole pairs upon absorption. If the junction is reverse biased using an electrical circuit, these electron hole pair move in opposite directions so as to produce current in the circuit. This current is very small and is detected by the micro ammeter placed in the circuit.

A photo diode is preferably operated in reverse bias condition. Consider an n-type semiconductor. Its majority carrier (electron) density is much larger than the minority hole density i.e. n>>p. When illuminated with light, both types of carriers increase equally in number

That is, the fractional increase in majority carriers is much less than the fractional increase in minority carriers. Consequently, the fractional change due to the photo-effects on the minority carrier dominated reverse bias current is more easily measurable than the fractional change in the majority carrier dominated forward bias current. Hence, photo diodes are preferable used in the reverse bias condition for measuring light intensity.

Answer 19.

Input resistance may be defined as the ratio of small change in the base-emitter voltage (ΔV_{BE}) to the resulting change in base current (I_{B}) at constant collector-emitter voltage (V_{CF}).

To find the input resistance mark a point P on the input characteristic. Now, draw a tangent at point P. The reciprocal of slope of AB will give the input resistance.

(ii) Current amplification factor

defined as the ratio of change in collector current (ΔI_{C}) to the change in base current (ΔI_{B})

Answer 20.

(a) Microwaves are suitable for radar systems that are used in aircraft navigation. These rays are produced by special vacuum tubes, namely — klystrons, magnetrons and Gunn diodes.

(b) Infrared waves are used to treat muscular strain. These rays are produced by hot bodies and molecules.

(c) X-rays are used as a diagnostic tool in medicine. These rays are produced when high energy electrons are stopped suddenly on a metal of high atomic number.

Answer 21.

Answer 22.

At frequencies above 40 MHz, communication is essentially limited to line-of-sight paths. At these frequencies, the antennas are relatively smaller and can be placed at height of many wavelengths above the ground. Because of the line- of-sight nature of propagation, direct waves get blocked at some point by the curvature of earth.

**SECTION-D**

Answer 23.

(i) The device that is used to bring high voltage down to low voltage of an a.c. current is a transformer. It works on the principle of mutual induction of two windings or circuits. When current in one circuit changes, emf is induced in the neighbouring circuit.

(ii) The transformer cannot convert d.c. voltages because it works on the principle of mutual induction. When the current linked with the primary coil changes the magnetic flux linked

with the secondary coil also changes. This change in flux induces emf in the secondary coil. If we apply a direct current to the primary coil the current will remain constant. Thus, there is no mutual induction and hence no emf is induced.

(iii) The value of gaining knowledge and curiosity about learning new things is being displayed by the students. The value of providing good education and undertaking the doubts of students has been displayed by the teacher.

**SECTION-E**

Answer 24.

(a) Electric field at a point on the axial line

(b) Since, the electric field is parallel to the faces parallel to xy and xz planes, the electric flux through them is zero.

OR

(a) (i) Conductor:

E_{0} → external field

E_{m} → internal field created by the redistribution of electrons inside the metal.

When a conductor like a metal is subjected to external electric field, the electrons experience a force in the opposite direction collecting on the left hand side. A positive charge is therefore induced on the right hand side.

This creates an opposite electric field (E_{m}) that balances out (E_{0}).

∴ The net electric field inside the conductor becomes zero.

(ii) Dielectric: When external electric field is applied, dipoles are created (in case of non-polar dielectrics). The placement of dipoles is as shown in the given figure. An internal electric field is created which reduces the external electric field. Polarisation of dielectric (P) is defined as the dipole moment per unit volume of the polarised dielectric.

Answer 25.

(a) Amperes circuital law in electro magnetism is analogous to Gauss law in electrostatics. This law states that “The line integral of resultant magnetic field along a closed plane curve is equal to μ_{0} time the total current crossing the area bounded by the closed curve provided the electric field inside the loop remains constant. Thus § B.dl = μ_{0 }I_{enc} where μ_{0} is permeability of free space and I_{enc} is the net current enclosed by the loop.

A toroid is a hollow circular ring on which a large number of turns of a wire are closely wound. Consider an air-cored toroid (as shown below) with centre O.

Given:

r = Average radius of the toroid

I = Current through the solenoid .

n = Number of turns per unit length

To determine the magnetic field inside the toroid, we consider three amperian loops (loop 1, loop 2 and loop 3) as show in the figure here.

According to Ampere’s Circuital law, we have § \(\vec { B } \)

.\(\vec { dl } \) = μ_{0 }I_{enc }= μ_{0} I_{enc} (Total current)

Total current for loop 1 is zero because no current is passing through this loop

So, for loop \(\vec { B } \).\(\vec { dl } \) = 0 = μ_{0} (Total current)

**For loop 3:**

According to Ampere’s Circuital law, we have j> B .dl = p_{0} (Total current)

Total current for loop 3 is zero because net current coming out of this loop is equal to the net current going inside the loop.

**For loop 2:**

The total current flowing through the toroid is NI, where N is the total number of turns

This is the expression for magnetic field inside air-cored toroid.

(b) Given that the current flows in the clockwise direction for an observer on the left side of the solenoid. This means that left face of the solenoid acts as south pole and right face acts as north pole. Inside a bar magnet the magnetic field lines are directed from south to north. Therefore, the magnetic field lines are directed from left to right in the solenoid. Magnetic moment of single current carrying loop is given by

m = IA

where I = Current flowing through the loop, A = Area of the loop So, Magnetic moment of the whole solenoid is given by M = Nm = N(IA)

OR

(a) Mutual inductance is the property of two coils by the virtue of which each opposes any change in the value of current flowing through the other by developing an induced emf. The SI unit of mutual inductance is Henry and its symbol is H.

(b) Consider two long solenoids S_{1} and S_{2} of same length l such that solenoid S_{2} surrounds solenoid S_{1} completely.

Let:

n_{1} = Number of turns per unit length of S,

n_{2} = Number of turns per unit length of S_{2
}

When current is passed through solenoid S_{1} an emf is induced in solenoid S_{2}. Magnetic field produced inside solenoid S_{1} on passing current through it is given by

B_{1} — μ_{0}n_{1}I_{1}

Magnetic flux linked with each turn of solenoid S_{2} will be equal to B_{1} times the area of cross-section of solenoid S_{1
}Magnetic flux linked with each turn of the solenoid Φ_{21} = B_{1}A Therefore, total magnetic flux linked with the solenoid S_{2} is given by

Similarly the mutual inductance between the two solenoids when current is passed through solenoid S_{2} and induced emf is produced in solenoid S, is given by

M_{12} = μ_{0 }N_{1 }n_{2}A

where N, is total number of turns wound over primary coil.

Answer 26.

(a) Let a spherical surface separate a rarer medium of refractive index n_{1 }from second medium of refractive index n_{2} Let C be the centre of curvature and R = MC be the radius of the surface. Consider a point object O lying on the principal axis of the surface. Let a ray starting from O incident normally on the surface along OM and pass straight. Let another ray of light incident on NM along ON and refract along NI.

From N, draw MN perpendicular to OI.” The figure shows the geometry of formation of image I of an object O and the principal axis of a spherical surface with centre of curvature C and radius of curvature R. Let us make the following assumptions:

(i) The aperture of the surface is small as compared to the other distance involved.

(ii) NM will be taken as nearly equal to the length of the perpendicular from the point N on the principal axis.

(b) Now the image I acts as virtual object for the second surface that will form real image at l As refraction takes place from denser to rarer medium

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