CBSE Previous Year Solved Papers Class 12 Physics Delhi 2014

CBSE Previous Year Solved  Papers  Class 12 Physics Delhi 2014

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions:

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

SET I

Note : Except for the following questions, all the remaining questions have been asked in Previous Set.

Question.1. Define the term ‘Mobility’ of charge carriers in a conductor. Write its S.I. unit.
Answer : Mobility of charge carriers in a conductor is defined as the magnitude of their drift velocity per unit applied electric field.

Question.2. The carrier wave is given by C(t) = 2 sin (ωt) volt. The modulating signal is a square wave as shown. Find modulation index.

Question.3. For any charge configuration, equipotential surface through a point is normal to the electric field. Justify.
Answer: If the electric field were not normal to equipotential surface, it would have non-zero component along the surface. To move a charge against this component, work would have to be done. But no work is needed to move a test charge on an equipotential surface. Hence electric field must be normal to the equipotential surface at every point.

Question.4. Two spherical bobs, one metallic and the other of glass, of the same size are allowed to fall freely from the same height above the ground. Which of the two would reach earlier and why?
Answer : Glass bob will reach the ground earlier than the metallic bob. As the metallic bob falls, it intercepts earth’s magnetic field and induced currents are set up in it which oppose its downward motion. But no such currents are induced in the glass.

Question.5. Show variation of resistivity of copper as a function of temperature in a graph.
Answer: The variation of resistivity of copper with temperature is parabolic in nature. This is shown in the following graph :

Question.6. A convex lens is placed in contact with a plane mirror. A point object at a distance of 20 cm on the axis of this combination has its image coinciding with itself. What is the focal length of the lens ?
Answer : The figure shows a convex lens L placed in contact with a plane mirror M. P is the point object, kept in front of

this combination at a distance of 20 cm, from it. As the image coincides with itself, the rays from the object, after refraction from lens, should fall normally on the mirror M, so that they retrace their path. For this, the rays from P, after refraction from the lens must from a parallel beam perpendicular to M. For clarity, M has been shown at a small distance from L (in diagram). As the rays from P, form a parallel beam after refraction, P must be at the focus of the lens. Hence the focal length of the lens is 20 cm.

Question.7.

Answer: The Lorentz magnetic force is given by the following relation:

Question.8. The figure given below-shows the block diagram of a generalized communication system. Identify the element labeled ‘X’ and write its function.

Answer : The element labeled ‘X’ is called ‘channel’. The function of the channel is to connect the transmitter and the receiver. A channel may either be wireless or in the form of wires connecting the transmitter and the receiver.

Question.9. Out of the two magnetic materials, ‘A’ has relative permeability slightly greater than unity while ‘B’ has less than unity. Identify the nature of the materials ‘A’ and ‘B’. Will their susceptibilities be positive or negative ?

Question.10.

Question.11. For a single slit of width a, the first minimum of the interference pattern of a monochromatic light of wavelength k occurs at an angle of k/a. At the same angle of k/a, we get a maximum for two narrow slits separated by a distance a. Explain.

Question.12. Write the truth table for the combination of the gates shown. Name the gates used.

Identify the logic gates marked ‘P’ and ‘Q’ in the given circuit. Write the truth table for the combination.

Question.13. State Kirchhoff’s rules. Explain briefly how these rules are justified.
Answer : Kirchhoff s first Law—Junction Rule In an electrical circuit, the algebraic sum of the currents meeting at a junction is always zero.

Convention : The current flowing towards the junction is taken as positive.
The current flowing away from the junction is taken as negative.

This law is based on the law of conservation of charge. KirchhofFs Second Law — Loop rule
In a closed loop, the algebraic sum of the emf ‘s is equal to the algebraic sum of the products of the resistances and the currents flowing through them.

Question.14. A capacitor ‘C’ a variable resistor ‘R’ and a bulb ‘B’ are connected in series to the ac mains in circuit as shown. The bulb glows with some brightness. How will the glow of the bulb change if (i) a dielectric slab is introduced between the plates of the capacitor, keeping resistance R to be the same; (ii) the resistance R is increased keeping the same capacitance?

Answer : (i) As the dielectric slab is introduced between the f plates of the capacitor, its capacitance will increase. Hence, the potential drop across the capacitor will decrease (V= Q/C). As a result, the potential drop across the bulb will increase (since both are connected in series’). So, its brightness will increase.
(ii) As the resistance (R) is increased, the potential drop across the resistor will increase. As a result, the potential drop across the bulb will decrease (since both are connected in series). So, its brightness will decrease.

Question.15. State the underlying principle of a cyclotron. Write briefly how this machine is used to accelerate charged particles to high energies.
Answer : The underlying principle of a cyclotron is that an oscillating electric field can be used to accelerate a charge particle to high energy.
A cyclotron involves the use of an electric field to accelerate charge particles across the gap between the two D-shaped magnetic field regions. The magnetic field is perpendicular to the paths of the charged particles that makes them follow in circular paths with in the two D_s. An alternating voltage accelerates the charged particles each time they cross the D_s. The radius of each particles path increases with its speed. So, the accelerated particles spiral toward the outer wall of the cyclotron.

[The D_s are the semi-circular structures (D₁ and D₂) between which the charges move. The accelerating voltage is maintained across the opposite halves of the D_s.] Square wave electric fields are used to accelerate the charged particles in a cyclotron.

The accelerating electric field reverses just at the time the change particle finishes its half circle so that it gets accelerated across the gap between the D_s.
The particle gets accelerated again and again, and its velocity increases. Therefore, it attains high kinetic energy.
The positively charged ion adopts a circular path with a constant speed v, under the action of magnetic field B, which is perpendicular to the planes of D’s of radius r.
r=mv/qB

Question.16. An electric dipole of length 4 cm, when placed with its axis making an angle of 60° with a uniform electric field, experiences a torque of 4√3 Nm. Calculate the potential energy of the dipole, if it has charge ± 8 nC.

Question.17. A proton and a deuteron are accelerated through the same accelerating potential. Which one of the two has
(a) greater value of de-Broglie wavelength associated with it, and
(b) less momentum ?
Give reasons to justify your Answer.

Question.18. (i) Monochromatic light of frequency 6.0 x 10¹⁴ Hz is produced by a laser. The power emitted is 2.0 x 10^-3 W. Estimate the number of photons emitted per second on an average by the source.
(ii) Draw a plot showing the variation of photoelectric current versus the intensity of incident radiation on a given photosensitive

Question.19. A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. Up to which energy level the hydrogen atoms would be excited? Calculate the wavelength of the first member of Lyman and first member of Balmer series.

Question.20. When Sunita, a class XII student, came to know that her parents are planning to rent out the top floor of their house to a mobile company she protested. She tried hard to convince her patents that this move would be a health hazard. Ultimately her parents agreed :
(1) In what way can the setting up of transmission tower by a mobile company in a residential colony prove to be injurious to health ?
(2) By objecting to this move of her parents, what value did Sunita display?
(3) Estimate the range of e.m. waves which can be transmitted by an antenna of height 20 m. (Given radius of the earth = 6400 km)
Answer: (1) A transmitting tower makes use of electromagnetic waves such as microwaves, exposure to which can cause severe health hazards like, giddiness, headache, tumour and cancer. Also, the transmitting antenna operates on a very high power, so the risk of someone getting severely burnt in a residential area increases.
(2) By objecting to this move of her parents, Sunita has displayed awareness towards the health and environment of her society.
(3) Range of the transmitting antenna.

Question.21. A potentiometer wire of length 1 m has a resistance of 10Ω . It is connected to a 6 V battery in series with a resistance of 5 Ω . Determine the emf of the primary cell which gives a balance point at 40 cm.

Question.22. (a) Draw a labeled ray diagram showing the formation of a final image by a compound microscope at least distance of distinct vision. ’
(b) The total magnification produced by a compound microscope is 20. The magnification produced by the eye piece is 5. The microscope is focussed on a certain object. The distance between the objective and eye-piece is observed to be 14 cm. If least distance of distinct vision is 20 cm, calculate the focal length of the objective and the eye-piece.
Answer: (a)

Question.23. (a) A mobile phone lies along the principal axis of a concave mirror. Show, with the help of a suitable diagram, the formation of its image. Explain why magnification is ‘ not uniform.
(b) Suppose the lower half of the concave mirrors reflecting surface is covered with an opaque material. What effect this will have on the image of the object? Explain.
Answer: (a)

The image of the mobile phone formed by the concave mirror is shown in the above figure. The part of the mobile phone that is at C will form an image of the same size only at C. In the figure, we can see that B’C = BC. The part of the mobile phone that lies between C and F will form enlarged image beyond C as shown in the figure. It can be observed that the magnification of each part of the mobile phone cannot be uniform on account of different locations. That is why the image formed is not uniform.
(b) As the laws of reflection are true for all points of the mirror, the height of the whole image will be produced. However, as the area of the reflecting surface has been reduced, the image intensity will be reduced. In other words, the image produced will be less bright.

Question.24. (a) Obtain the expression for the energy stored per unit volume in a charged parallel plate capacitor.
(b) The electric field inside a parallel plate capacitor is E. Find the amount of work done in moving a charge q over a closed rectangular loop abcda.

OR
(a) Derive the expression for the capacitance of a parallel plate capacitor having plate area A and plate separation d.
(b) Two charged spherical conductors of radii R₁ and R₂ when connected by a conducting wire acquire charge q₁ and q₂ respectively. Find the ratio of their surface charge densities in terms of their radii.
Answer: (a) Let us consider a parallel-plate capacitor of plate area A. If seperation between plates is d metre, capacitance C in given by

Question.25. (a) State Ampere’s Circuital law, expressing it in the integral form.
(b) Two long coaxial insulated solenoids, S₁ and S₂ of equal lengths are wound one over the other as shown in the figure. A steady current “I” flow through the inner solenoid S₁ to the other end B, which is connected to the outer solenoid S₂ through which the same current “I” flows in the opposite direction so as to come out at end A. If n₁ and n₂ are the number of turns per unit length, find the magnitude and direction of the net magnetic field at a point (i) inside on the axis and (ii) outside the combined system.

Answer: (a) Ampere’s Circuital law states that the circulation of the resultant magnetic field along a closed, plane curve is equal to po times the total current crossing the area bounded by the closed curve, provided the electric field inside the loop remains constant.

Question.26. Answer the following :
(a) Name the em waves which are suitable for radar systems
used in aircraft navigation. Write the range of frequency of these waves.
(b) If the earth did not have atmosphere, would its average surface temperature be higher or lower than what it is now? Explain.
(c) An em wave exerts pressure on the surface on which it is incident. Justify.
Answer: (a) Microwaves are suitable for radar systems used in aircrafts navigation. The range of frequency for these waves is 10⁹ Hz to 10¹² Hz.
(b) In the absence of atmosphere, there would be no greenhouse effect on the surface of the Earth. As a result, the temperature of the Earth would decrease rapidly, making it difficult for human survival.
(c) An em wave carries a linear momentum with it. The linear momentum carried by a portion of wave having energy U is given by p= UC
Thus, if the wave incident on a material surface is completely absorbed, it delivers energy U and momentum p = U C to the surface. If the wave is totally reflected, the momentum delivered is p = 2U C because the momentum of the wave changes from p to -p. Therefore, it follows that an em waves incident on a surface exert a force and hence a pressure on the surface.

Question.27.

Question.28. (a) (i) ‘Two independent mono-chromatic sources of light cannot produce a sustained interference pattern. Give reason.
(ii) Light wave each of amplitude ‘a’ and frequency ‘ω ’, emanating from two coherent light sources superpose at a point. If the displacements due to these waves is given by y₁ = a cos ωt and y₂ = a cos (ωt + ᶲ) where ᶲ is the phase difference between the two, obtain the expression for the resultant intensity at the point.
(b) In Youngs double slit experiment, using monochromatic light of Wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. Find out the intensity of light at a point where path difference is λ/3.
OR
(a) How does one demonstrate, using a suitable diagram, that unpolarised light when passed through a Polaroid gets polarized?
(b) A beam of unpolarised light is incident on a glass-air interface. Show, using a suitable ray diagram, that light reflected from the interface is totally polarised, when µ = tan i_B, where µ is the refractive index of glass with respect to air and i_B is the Brewster’s angle.
Answer : (a) (i) The condition for the sustained interference is that both the sources must be coherent (i.e. they must have the same wavelength and the same frequency, and they must have the same phase or constant phase difference).
Two sources are monochromatic if they have the same frequency and wavelength. Since they are independent, i.e. they have different phases with irregular difference, they are not coherent sources.

The intensity of light is directly proportional to the square of the amplitude of the wave. The intensity of light at point on the screen is given by:

OR

The phenomenon of restricting the vibration of light (electric vector) in a particular direction perpendicular to the direction of the wave propagation is called polarization of light.
When unpolarised light is passed through a Polaroid, only those vibrations of light pass through the crystal, which are parallel to the axis of the crystal (AB). All other vibrations are absorbed and that is why intensity of the emerging light is reduced.
The plane ABCD in which the vibrations of the polarised light are confined is called the plane of vibration. The plane KLMN that is perpendicular to the plane of vibration is defined as the plane of polarization.
(b) When unpolarised light is incident on the glass-air interference at Brewster angle iB, then reflected light is totally polarised. This is called Brewsters Law.
When light is incident at Brewster angle, the reflected component OB and the refracted component OC are mutually perpendicular to each other.
From the figure,

Question.29. (a) Describe a simple experiment (or activity) to show that the polarity of emf induced in a coil is always such that it tends to produce a current which opposes the change of magnetic flux that produces it.
(b)The current flowing through an inductor of self inductance L is continuously increasing. Plot a graph showing the variation.
(i) Magnetic flux versus the current
(ii) Induced emf versus dl l dt
(iii) Magnetic potential energy stored versus the current.
OR
(a) Draw a schematic sketch of an ac generator describing its basic elements. State briefly its working principle. Show a plot of variation of
(i) Magnetic flux and
(ii) Alternating emf versus time generated by a loop of wire rotating in a magnetic field.
(b) Why is choke coil needed in the use of fluorescent tubes with ac mains?
Answer: (a) Lenz law: According to Lenz s law, the polarity of the induced emf is such that it opposes a change in magnetic flux responsible for its production.

When the north pole of a bar magnet is pushed towards the coil, the amount of magnetic flux linked with the coil ,increase. Current is reduced in the coil from a direction such that it opposes the increase in magnetic flux. This is possible only when the current induced in the coil is in anti-clockwise
direction, with respect to an pb server. The magnetic moment  M associated with this induced emf has north polarity, towards the north pole of the approaching bar magnet. Similarly, when the north pole of the bar magnet is moved away from the coil, the magnetic flux linked with the coil
decreases. To counter this decrease in magnetic flux, current is induced in the coil in clockwise direction so that its south pole faces the receding north pole of the bat magnet. This would result in an attractive force which opposes the motion of the magnet and the corresponding decrease in magnetic flux.

OR

(a) Principle is “Based on the phenomenon of electromagnetic induction.
Construction :

Main parts of an ac generator :

• Armature : The rectangular coil ABCD
• Field Magnets Two pole pieces of a strong electromagnet
• Slip Rings. The ends of the coil ABCD are connected to two hollow metallic rings R₁ and R₂.
• Brushes : B₁ and B₂ are two flexible metal plates or carbon rods. They are fixed and are kept -is tight contact with R₁ and R₂, respectively.

Working : As the armature coil is rotated in the magnetic field, angle 6 between the field and the normal to the coil changes continuously. Therefore, magnetic flux linked with the coil changes and an emf is induced in the coil. According to Flemings right hand rule, current is induced from A to B in AB and from C to D in CD. In the external circuit, current flows from B₂ to B₁. To calculate the magnitude of emf induced :


The graph between alternating emf versus time is shown below:

(b) A choke coil is an electrical appliance used for controlling current in an a.c. circuit. Therefore, if we use a resistance R for the same purpose, a lot of energy would be wasted in the form of heat etc.

Question.30. (a) State briefly the processes involved in the formation of p-n junction explaining clearly how the depletion region is formed.
(b) Using the necessary circuit diagram, show how the V-I characteristics of a p-n junction are obtained in
(i) Forward biasing
(ii) Reverse biasing
How are these characteristics made use of in rectification?
OR
(a) Differentiate between three segments of a transistor on the bias of their size and level of doping.
(b) How is a transistor biased to be in active state?
(c) With the help of necessary circuit diagram, describe briefly how n-p-n transistor in CE configuration amplifies a small sinusoidal input voltage. Write the expression for the ac current gain.
Answer : (a) As we know that n-type semi-conductor has more concentration of electrons than that of a hole and p-type semi-conductor has more concentration of holes than an electron. Due to the difference in concentration of charge carriers in the two regions of p-n junction, the holes diffuse from p-side to n-side and electrons diffuse from n-side to p-side.
When an electron diffuses from n to p, it leaves behind an ionized donor on n-side. The ionised donor (+ve charge) is immobile as it is bonded by the surrounding atoms. Therefore, a layer of positive charge is developed on the «-side of the junction. Similarly, a layer of negative charge is developed on the p-side.

Hence, a space-charge region is formed on both side of the junction, which has immobile ions and is devoid of any f charge carrier, called as depletion layer or depletion region.
(b) (i) p-n junction diode under forward bias

A p-n junction diode is said to be forward biased if the positive terminal of the external battery B is connected to p-side and the negative terminal to the n-side of p-n junction.
The applied voltage of battery mostly drops across the depletion region and the voltage drop across the p-side and n-side of the p-n junction in negligible small. The resistance of depletion region is very high as it has no free change carriers.
Electron in n-region moves towards the p-n junction and holes in the p-region move towards the junction. The width of the depletion layer decreases and hence, it offers less resistance. Diffusion of majority carriers takes place across the junction. This leads to the forward current.
The V-I characteristics of p-n junction is forward bias is shown below:

(ii) The p-n junction under reverse bias Positive terminal of battery is connected to n-side and negative terminal to p-side.
Reverse bias supports the potential barrier. Therefore, the barrier height increases and the width of depletion region also increases. Due to the majority carriers, there is no conduction across the junction. A few minority carriers cross the junction after being accelerated by high reverse bias voltage.

This constitutes a current that flows in opposite direction, which is called reverse current.
The V-I characteristics of p-n junction diode in reverse bias is shown on previous page :

p-n junction diode Is used as a half-wave rectifier. Its working is based on the fact that the resistance of p-n junction becomes low when forward biased and becomes high when reverse biased. These characteristics of diode is used in rectification.
OR
(a) Emitter (E) : It is the left hand side thick layer of the transistor, which is heavily doped.
Base (B) : It is the central thin layer of the transistor, which is lightly doped.

Collector (C) : It is the right hand side thick layer of the transistor, which is moderately doped.
(b) There are two conditions for a transistor to be into an active region.
(1)The input circuit should be forward biased by using a low voltage battery.
(2) The out put circuit should be reverse biased by using a high voltage battery.
(c) n-p-n transistor as an amplifier :
The operating point is fixed in the middle of its active region.

SET II

Note : Except for the following questions, all the remaining questions have been asked in Privious Sets.

Question.1. Define the term ‘electrical conductivity’ of a metallic wire. Write its S.I. unit.
Answer : The electric conductivity of a metallic wire is defined as the ratio of the current density to the electric field it creates. Electrical conductivity.

Question.2. The carrier wave is represented by C (t) = 5 sin (10 πt) volt . A modulating signal is a square wave as shown. Determine modulation index.
Answer: Modulation index p is the ratio of the amplitude of the modulating signal to the amplitude of the carrier wave.

Question.10. An electric dipole of length 2 cm, when placed with its axis making an angle of 60° with a uniform electric field, experiences a torque of 8√3 Nm. Calculate the potential energy of the dipole, if it has a charge of ±4 nC.

Question.15. A proton and an alpha particle are accelerated through the same potential. Which one of the two has (i) greater value of de-Broglie wavelength associated with it and (ii) less kinetic energy. Give reasons to justify your answer.
Answer: (i) de-Broglie wavelength of a particle depends upon its mass and charge for same accelerating potential such that

Question.16.

Question.20. A 12.9 eV beam of electronic is used to bombard gaseous hydrogen at room temperature. Up to which energy level the hydrogen atoms would be excited? Calculate the wavelength of the first member of Paschen series and first member of Balmer series.

Question.22. Answer the following:
(a) Name the em waves which are used for the treatment of certain forms of cancer. Write their frequency range.
(b) Thin ozone layer on top of stratosphere is crucial for human survival. Why?
(c) Why is the amount of the momentum transferred by the em waves incident on the surface so small?
Answer: (a) X-ray, Gamma (y) rays are used for the treatment of certain forms of cancer. Their frequency range is 10¹⁸m to 10²²m.
(b) The thin ozone layer on top of stratosphere absorb most of the harmful ultraviolet rays coming from the Sun towards the Earth. They include UVA, UVB and UVC radiations, which can destroy the life system on the Earth. Hence, this layer is crucial for human survival.
(c) Momentum transferred = Energy Speed of light = hvc = 10^-22 (for v – 10²⁰ Hz)
Thus, the amount of the momentum transferred by the em waves incident on the surface is very small.

Question.24. A potentiometer wire of length 1.0 m has a resistance of 15 Ω. It is connected.to a 5V battery in series with a resistance of 5 Ω. Determine the emf of the primary cell which gives a balance point at 60 cm.

SET III

Note : Except for the following questions, all the remaining questions have been asked in Set—I and Privious Set.

Question.1. Define the term ‘drift velocity’ of charge carriers in a conductor and write its relationship with the current flowing through it.
Answer : The net speed achieved by an electron due to a current carrying conductor is called as drift velocity.
The average velocity acquired by the tree electrons along the length of a metallic conductor under a potential differnce applied across the conductor in called drift velocity of the electrons.

Here :
I is the current following through the conductor.
n is the number density of an electron.
A is the area of the conductor.
e is the charge of the electron.

Question.2. The carrier, wave of a signal is given by C(t) = 3 sin (8πt) volt. The modulating signal is a square wave as shown. Find its modulation index.

Question.4. Plot a graph showing variation of current versus voltage for the material Ga.
Answer : Current—Voltage characteristics graph for Ga :

Question.9. An electric dipole of length 2 cm, when placed with its axis making an angle of 60° with a uniform electric field, experiences a torque of 6 √3 Nm. Calculate the potential energy of the dipole, if it has a charge of ± 2 nC.

Question.12. A deuteron and an alpha particle are accelerated with the same accelerating potential. Which one of the two has
(1) greater value of de-Broglie wavelength, associated with it and
(2) less kinetic energy? Explain.
Answer: (1) de-Broglie wavelength of a particle is dependent on its mass and charge for same accelerating potential, such that

Charge of a deuteron is less as compared to an alpha particle. So, deuteron will have less of K.E.

Question.15.

Question.20. A 12.3 eV electron beam is used to bombard gaseous hydrogen at room temperature. Up to which energy level the hydrogen atoms would be excited? Calculate the wavelength of the second member of Lyman series and second member of Balmer series.

24. Answer the following :
(a) Name the em waves which are used for the treatment of certain forms of cancer. Write their frequency range.
(b) Welders wear special glass goggles while working. Why? Explain.
(c) Why are infrared waves often called as heat waves? Give their one application.
Answer:
(a) Gamma rays are used for the treatment of certain forms of cancer. There frequency range.
(b) Welders wear special glass googles while working so that they can protect their eyes from harmful electromagnetic radiation.
(c) Infrared waves are often called as heat waves because they induce resonance in molecules and increase internal energy in a substance.
Infrared waves are used in burglar alarms, security lights and remote controls for television and DVD players.