CBSE Previous Year Solved  Papers  Class 12 Physics Delhi 2012

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions:

  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 weightage. You have to attempt only one of the choices in such questions.
  5. You may use the following values of physical constants wherever necessary:


Question.1. When electrons drift in a metal from lower to higher potential, Does it mean that all the free electrons of the metal are moving in the same direction ?
Answer : No, when electric field is applied the electrons will have net drift from lower to higher field but locally electrons may Collide with ions and may change its direction of motion.

Question.2. The horizontal component of the earths magnetic field at a place is B and angle of dip is 60°. What is the value of vertical component of earth’s magnetic field at equator?
Answer : On the equator, the value of both angle of dip (8) and vertical component of earth’s magnetic field is zero. So, in t this case, Bv= 0.

Question.3. Show on a graph, the variation of resistivity with temperature for a typical semiconductor.
Answer: The following curve shows the variation of resistivity with temperature for a typical semiconductor.
This is because for a typical semiconductor,resistivity decreases rapidly with increasing temperature.

Question.4. Why should electrostatic field be zero inside a conductor?
Answer : Charge on conductor resides on its surface. So if we consider a Gaussian surface inside the conductor to find the electrostatic field,
Where, q = charge enclosed in Gaussian surface.
q= 0, inside the conductor, hence the electrostatic field inside the conductor is zero.

Question.5. Name the physical quantity which remains same for microwaves of wavelength 1 mm and UV radiations of 1600 Å in vacuum.
Answer : Both microwaves and UV rays are a part of the electromagnetic spectrum. Thus, the physical quantity that remains same for both types of radiation will be their speeds which is equal to c (3 x108m/s).

Question.6. Under what condition does a biconvex lens of glass having a certain refractive index act as a plane glass sheet when immersed in a liquid?
Answer : A biconvex lens twill act like a plane sheet of glass if it is immersed in a liquid having the same index of refraction
as itself.

Question.7.Predict the directions of induced currents in metal rings 1 and 2 lying in the same plane where current I in the wire is increasing steadily.

Question.8. State de-Broglie hypothesis.
Answer : de Broglie postulated that the material particles may exhibit wave aspect. Accordingly a moving material particle behaves as wave and the wavelength associated with material particle is

Question.9. A ray of light incident on an equilateral prism (μg = √3 ) moves parallel to the base line of the prism inside it. Find the angle of incidence for this ray.
Answer : It is given that the prism is equilateral in shape. So, all the angles are equal to 60°. Thus, the angle of

Question.10. Distinguish between‘Analog and Digital signals’,
Mention the functions of any two of the following used in communication system :

  1. Transducer
  2.  Repeater
  3. Transmitter
  4. Bandpass Filter

Answer : Analog Signal: It is continuous signal, which varies continuously with variable may be time or distance etc. E.g. Voice of human
Digital Signal : It is a type of signal which has only two values high or low. In digital high means 1 and low means 0.

  1.  Transducer : It is an electric device which coverts energy from one form to another form. E.g. microphone, which converts sound energy into electric energy and vice-versa.
  2.  Repeater : It is an electronic device used in transmission system to regenerate the signal. It picks up a signal amplifies it and transmits it to receiver.
  3.  Transmitter : Transmitter is an electronic device which is used to radiate electromagnetic waves. The purpose of the transmitter is to boost up the signal to be radiated to the required power level, so that it can travel long distances. The most familiar transmitters are mobile transmitter antennas, radio and T.V broadcasting antennas etc. ‘
  4.  Bandpass filter : It is an electronic filter, which pass the certain band (range) of frequency and reject rest of all.

Question.11. A cell of emf E and internal resistance r is connected to two external resistance Ri and R2 and a perfect ammeter. The current in the circuit is measured in four different situations:
(i) without any external resistance in the circuit
(ii) with resistance Ri only
(iii) withR1 and Rin series combination
(iv) with R1 and R2 in parallel combination
The currents measured in the four cases are 0.42 A, 1.05 A, 1.4 A and 4.2 A, but not necessarily in the order. Identify the currents corresponding to the four cases mentioned above.
Answer : The current relating to corresponding situations is as follows :

Question.12. The susceptibility of a-magnetic material is -2.6 x 10-5. Identify the type of magnetic material and state its two properties.
Answer : Diamagnetic materials have negative susceptibility. So the given magnetic material is diamagnetic.
Two properties of diamagnetic material:

  1. They do not obey Curie’s law.
  2. They are feebly repelled by a magnet.

Question.13. Two identical circular wires P and Q each of radius R and carrying current ‘I’ are kept in perpendicular planes such that they have a common centre as shown in the figure. Find the magnitude and direction of the net magnetic field at the common centre of the two cods.
Answer: Magnetic field produced by the two coils at their common centre having currents I1 and I2, radius a1 and a2, number of turns N1, N2, are given by :

Question.14. When an ideal capacitor is charged by a dc battery, no current flows. However, when an ac source is used the current flows continuously. How does one explain this based on the concept of displacement current?
Answer : When an ideal capacitor is charged by dc battery, charge flows till the capacitor gets fully charged.
When an ac source is connected then conduction current

Question.15. Draw a plot showing the variation of (i) electric field (E) and (ii) electric potential (V) with distance r due to a point charge Q.
Answer : We know, that for a point charge Q

Question.16. Define self-inductance of a coil. Show that magnetic energy required to build up the current I in a coil of self inductance L is given by 1/2 L I2.
Answer : Self inductance is the inherent inductance of a circuit, given by the ratio of the electromotive force produced in the circuit by self-induction to the rate of change of current producing it. It is also called coefficient of self-induction.

Question.17. The current in the forward bias is known to be more (∼ mA) than the current in the reverse bias (∼ μA). What is the reason, then, to operate the photodiode in reverse bias ?
Answer : The current in the forward bias is due to majority carriers where as current in the reverse bias is due to minority carriers. So current in forward bias is more (~mA) than current in reverse bias (~μA).
On illumination of photodiodes with light, the fractional change in the majority carriers would be much less than that’ in -minority carriers. It implies fractional change due to light on minority carrier dominated reverse bias current is more easily measurable
than fractional change in forward bias current. f So photodiodes are operated in reverse bias condition.

Question.18.A metallic rod of ‘L’ length is rotated with angular frequency of ‘o’ with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius L about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field B parallel to the axis is present everywhere. Deduce the expression for the emf between the centre and the metallic ring.

Question.19. The figure shows a series LCR circuit with L = 75 H, C = 80 μF, R = 40Ω connected to a variable frequency 240 V source, calculate
(i) the angular frequency of the source which drives the circuit at resonance,
(ii)the current at the resonating frequency,
(iii)the rms potential drop across the inductor at resonance.

Question.20. A rectangular loop of wire of size 4 cm x 10 cm carries a steady current of 2 A. A straight long wire carrying 5 A current is kept near the loop as shown. If the loop and the wife are coplanar, find
(i) the torque acting on the loop and
(ii)the magnitude and direction of the force on the loop due to the current carrying wire.

Question.21. (a) Using Bohr’s second postulate of quantization of orbital angular momentum show that the circumference of the electron in the nth orbital state in hydrogen atom is n times the de Broglie wavelength associated with it.
(b) The electron in hydrogen atom is. initially in the third excited state. What is the maximum number of spectral lines which can be emitted when it finally moves to the ground state?
Answer : (a) According to Bohr’s ‘second postulate of quantization, the electron can revolve around the nucleus only in those circular orbits in which the angular momentum of the electron is integral multiple of where h is Planck’s

Question.22. In the figure a long uniform potentiometer wire AB is having .a constant potential gradient along its length. The null points for the two primary cells of emfs ε1 and ε2 connected in the manner shown are obtained at a distance of 120 cm and 300 cm from the end A. Find (i) ε1/ε2 and (ii) position of null point for the cell ε1.
How is the sensitivity of a potentiometer increased?
Using Kirchhoffs rules determine the value of unknown resistance R in the circuit so that no current flows through 4 Ω resistance. Also find the potential between A and D.


  1.  What characteristic property of nuclear force explains the constancy of binding energy per nucleon (BE/A) in the range of mass number ‘A’ lying 30 < A < 170?
  2. Show that the density of nucleus over a wide range of nuclei is constant-independent of mass number A.


  1.  The constancy of BE/A over most of the range is saturation property of nuclear force, ‘
    In heavy nuclei: nuclear size > range of nuclear force.
    So, a nuclear sense approximately a constant number of neighbours and thus, the nuclear BE/A levels off at high ‘A’.
    This is saturation of the nuclear force.
  2.  To find the density of nucleus of an atom, we have an atom with mass number let say A and let mass of the nucleus of the atom of the mass number A be mcbse-previous-year-solved-papers-class-12-physics-outside-delhi-2012-32

This expression is independent of mass number A and is constant.

Question.24.Write any two factors which justify the need for modulating a signal. Draw a diagram showing an amplitude modulated wave by superposing a modulating signal over a sinusoidal carrier wave.
Answer : Factors needed for modulating a signal:

  1.  To send the signal over large distance for communication.
  2. Practical size of antenna.

Question.25.Write Einsteins photoelectric equation. State clearly how this equation is obtained using the photon picture of electromagnetic radiation.
Write the three salient features observed in photoelectric effect which can be explained using this equation.
Answer : Einstein’s photoelectric equation,
According to Planck’s quantum theory, light radiations consist of small packets of energy.
Einstein postulated that a photon of energy hv is absorbed by the electron of the metal surface, then the energy equal to øis used to liberate electron from the surface and rest of the energy hv —ø becomes the kinetic energy of the electron.

Sailent features observed in photoelectric effect:

  •  The stopping potential and hence the maximum kinetic energy of emitted electrons varies linearly with the frequency of incident radiation. –
  •  There exists a minimum cut — off frequency V0, for which the stopping potential is zero.
  • Photoelectric emission is instantaneous.

Question.26.(a) Why are coherent sources necessary to produce a sustained interference pattern?
(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 isλ, is K units. Find out the intensity of light at a point where path difference is λ/3.
Answer : (a) Coherent sources have constant phase difference between them i.e., phase difference does not change with time. Hence, the intensity distribution on the screen remains constant and sustained.
(b) We know

Question.27.Use Huygens principle to explain the formation of diffraction pattern due to a single slit illuminated by a monochromatic source of light.
When the width of the slit is made double the original width, how would this affect the size and intensity of the central diffraction band?
Answer : Consider a parallel beam of monochromatic light is incident normally in a slit of width b as shown in figure. According to Huygens principle every point of slit acts as a source of secondary wavelets spreading in all directions. Screen is placed at a larger distance.
Consider a particular point P on the screen receives waves from all the secondary sources. All these waves start from different point of the slit and interference at point P to give resultant intensity.
Point P0 is a bisector plane of the slit. At P0, all waves are travelling equal optical path. So all wavelets are in phase thus interface constructively with each other and maximum, intensity is observed. As we move from P0, the wave arrives with different phases and intensity is changed. Intensity at point P is given by
Where, D = Distance between screen and slit,
λ = Wavelength of the light,
b = size of slit
So with the increase in size of slit the width of central maxima decrease. Hence, double the size of the slit would result in half the width of the central maxima.

Question.28. Explain the principle of a device that can build up high voltage of the order of a few million volts.
Draw a schematic diagram and explain the working of this device.
Is there any restriction on the upper limit of the high voltage set up in this machine? Explain.
(a) Define electric flux. Write its S.l. units.
(b) Using Gauss’s law, prove that the electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance from it.
(c) How is the field directed if (i) the sheet is positively charged, (ii) negatively charged?
Answer: Van de Graff generator is the device used for building up high voltages of the order of a few million volts.
Such high voltages are used to accelerate charged particles such as electrons, protons, ions, etc.
It is based on the principle that charge given to a hallow conductor is transferred to outer surface and is distributed uniformly over it.
It consists of large spherical conducting shell (S) supported over the insulating pillars. A long narrow belt of insulating material is wound around two pulleys P1 and P2,B1 and Bare two sharply pointed metal combs. B1 is called the spray comb and Bis called the collecting comb.
Working: The spray comb is given a positive potential by high tension source. The positive charge gets sprayed on the belt. As the belt moves and reaches the sphere, a negative charge is induced on the sharp ends of collecting comb B2 and an equal positive charge is induced on the farther end of B2 . This positive charge shifts immediately to the outer surface of S. Due to discharging action of sharp points of B2 , the positive charge on the belt is neutralized. The uncharged belt returns down and collects the positive charge from B1 , which in turn is collected by B2. This is repeated. Thus, the positive charge of S goes on accumulating. In this way, potential differences of as much as 6 or 8 million volts (with respect to the ground) can be built up. The main limiting factor on the value of high potential is the radii If the electric field just outside the sphere is sufficient for dielectric breakdown, of air, no more charge can be transferred to it.
For a conducting sphere,
Electric field just outside sphere

Question.29. Define magnifying power of a telescope. Write its expression. A small telescope has an objective lens of focal length 130 cm and an eye-piece of focal length 5 cm. If this telescope is used to view a 100 m high tower 3 km away, find the height of the final image when it is formed 25 cm away from the eye-piece.
How is the working of a telescope different from that of a microscope?
The focal lengths of the objective and eyepiece of a microscope are 1.25 cm and 5 cm respectively. Find the position of the object relative to the objective in order to obtain an angular magnification of 30 in normal adjustment.
Answer : Magnifying power of a telescope is defined as the ratio of the angle subtended at the eye by the image formed at the least distance of distinct vision to the angle subtended at the eye by the object lying at infinity, when seen directly.
The formula for magnifying power is,
A microscope is used to look into smaller objects like structure of cells etc. On the other hand, a telescope is used to see larger objects that are very far away like stars, planets etc.
Telescope mainly focuses on collecting the light into the objective lens, which should thus be large, where the microscope already has a focus and the rest is blurred around it. There is a big difference in their magnification factors.
For telescope the angular magnification is given by

Question.30. Draw a simple circuit of a CE transistor amplifier. Explain its working. Show that the voltage gain, Av, of the amplifier is give by cbse-previous-year-solved-papers-class-12-physics-outside-delhi-2012-48 is the current gain; ri RL is the load resistance and r, is the input resistance of the transistor. What is the significance of the negative sign in the expression for the voltage gain?
(a) Draw the circuit diagram of a full wave rectifier using p-n junction diode.
Explain is working and show the output, input waveforms.
(b) Show the output waveforms (Y) for the following inputs A and B of
(i) OR gate (ii) NAND gate
Answer : Circuit diagram of CE transistor amplifier :
Working : If a small sinusoidal voltage is applied to the input of a CE configuration, the base current and collector current will also -have sinusoidal variations. Because the collector current drives the load, a large sinusoidal voltage Vo-will be observed at the output. The expression for voltage gain of the transistor in CE configuration is :
Working: When the diode rectifies the whole of the AC wave, it is called full wave rectifier.
The figure shows the arrangement for using diode as full wave rectifier. The alternating input signal is fed to the primary P1 P2 of a transformer. The output signal appears across the load resistance RL.
During the positive half of the input signal, suppose P1 and P2 are negative and positive respectively. This would mean that S1 and S2 are positive and negative respectively. Therefore, the diode D1is forward biased and D2 is reverse biased. The flow of current in the load resistance AL is from A to B.
During the negative half of the input signal, S1 and S2 are negative and positive respectively. Therefore, the diode D1is reverse biased and D2 is forward biased. The flow of current in the load resistance RL is from A to B.


Note : Except for the following questions, all the remaining . questions have been asked in Set-I 2012.

Question.1. Why must electrostatic field be normal to the surface at every point of a charged conductor?
Answer : The electrostatic field must be normal to the surface , of the conductor of every point because if the electric field is not normal to the surface of the charged conductor, there will be a component of the electric field along the surface of the conductor, which would exert a force on the charges at the surface. Due to this charge starts flowing which is not possible.

Question.6. Predict the direction of induced current in a metal ring when the ring is moved towards a straight conductor with constant speed v. The conductor is carrying current I in the direction shown in the figure.
Answer: Clockwise

Question.10. Derive the expression for the self inductance of a long solenoid of cross-sectional area A and length l, having n turns per unit length.
Answer : Self-Inductance of a long Solenoid : The magnetic field B at any point inside such a solenoid is constant and given by,

Question.16. Two identical circular loops, P and Q, each of radius r and carrying currents I and 21 respectively are lying in parallel planes such that they have a common axis. The direction of
current in both the loops is clockwise as seen from O which is equidistance from the both loops. Find the magnitude of the net magnetic field at point O.

Question.20. A series LCR circuit with L = 4.0, C = 100 μF, R = 60 Ω connected to a variable frequency 240 V source as shown in figure calculate:
(i) the angular frequency of the source which drives the circuit at resonance,
(ii)the current at the resonating frequency,
(iii)the rms potential drop across the inductor at resonance.

Question.22. A rectangular loop of wire of size 2 cm x 5 cm carries a steady current of 1 A. A straight long wire carrying 4 A current is kept near the loop as shown in the figure. If the loop and the wire are coplanar, find (i) the torque acting on the loop and (ii) the magnitude and direction of the force on the loop due to the current carrying wire.

Question.27. Name the three different modes of propagation of electromagnetic waves. Explain, using a proper diagram the mode ofpropagationusedintheftequencyrangeabove40MHz.
Answer : Three different modes of propagation of electromagnetic waves are :

  1.  Ground (surface) wave propagation.
  2.  Sky waves propagation
  3. Space wave propagation (LOS communication)
    For frequencies about 40 MHz, space wave propagation is being used as the ionosphere will not reflect the signals and ground transmission is not possible. In space wave the transmitter and receiver must be on the line of sight (LOS) together.
    Such wave propagation used for Television broadcast and Satellite Communication
    In this, propagation the uplink and downlink frequencies are kept different to avoid the mixing up of the Signals.


Note : Except for the following questions, all the remaining questions have been asked in Set-I and Set-II 2012.

Question.6. Why is electrostatic potential constant throughout the volume of the conductor and has the same value (as inside) on its surface?
Answer : We know that the electric field inside a conductor is zero, so E = 0. This is the reason why electrostatic potential is

Question.8. Predict the direction of induced current in metal rings 1 and 2 when current I in the wire is steadily decreasing?
Answer : Using Lenz’s law we can predict the direction of induced current in the ring. Induce current oppose the cause of increase of magnetic flux in moving towards the conductor.
In metal Ring 1 induced current will be in is clockwise In metal Ring 2 induced current will be in is anticlockwise

Question.9. The relative magnetic permeability of a magnetic material is 800. Identify the nature of magnetic material and state its own properties.
Answer : A magnetic material having relative permeability of 800 would be classified as a ferromagnet. A few examples of such materials include iron and nickel.
Its two properties are :

  1.  All ferromagnetic materials become paramagnetic when heated to a temperature above the Curie temperatue (Tc).
  2.  These materials show a strong attraction towards magnetic fields and have a’tendency to become magnets themselves.

Question.16. Two identical circular loops, P and Q, each of radius r and carrying equal currents are kept in the parallel planes having a common axis passing through O. The direction of current in P is clockwise and. in Q is anti-clockwise as seen from O which is equidistant from the loops P and Q. Find the magnitude of the net magnetic field at O.
Now, as the current flowing in loop P is clockwise by using right hand thumbs rule the direction of the magnetic field will be towards left and as the current in loop Q is clockwise then the direction of magnetic field is towards left. So the net magnetic field at point O will be the sum of the magnetic fields due to loops P and Q.
Also, as the fields produced are at an equal distance to O, Bp = BQ, So, net field

Question.23.A rectangular loop of wire of size 2.5 cm x 4 cm carries a steady current of 1 A. A straight wire carrying 2 A current is kept near the loop as shown. If the loop and the wire are coplanar, find the (i) torque acting on the loop and (ii) the magnitude and direction of the force on the loop due to the current carrying wire.