CBSE Previous Year Solved  Papers  Class 12 Physics Outside Delhi 2014 

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 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:




Question.1. Using the concept of force between two infinitely long parallel current carrying conductors, define one ampere of current.
Answer : The amount of current flowing in two parallel wire in same direction through two infinitely long parallel wire separated by one metre which produces an attractive force of 2 x 10-7 N/m, provided the wires, have negligible circular cross section and placed in vacuum, this flowing current can be defined as “One Ampere of Current.

Question.2. To which part of the electromagnetic spectrum does a wave of frequency 5 x 1019 Hz belong ?
Answer : The frequency 5 x 1019 Hz lies in the gamma region of the. electromagnetic spectrum.

Question.3. What is the force between two small charges of 2 x 10-7 C and 3 x 10-7 C placed 30 cm apart in air?
Answer : Taking all the units of physical quantities in S.I. units, we have,

Question.4. Define intensity of radiation on the basis of photon picture of light. Write its S.I. unit.
Answer : The intensity of radiation can be defined as the energy related with photons emitted from a unit surface area in unit time. Its S.I units is joule/metre2 second (J/m2s).

Question.5. The electric current flowing in a wire in the direction from B to A is decreasing. Find out the direction of the induced current in the metallic loop kept above the wire as shown.

Question.6. Why is it found experimentally difficult to detect neutrinos % in nuclear β -decay?
Answer : Neutrinos are difficult to detect experimentally in β decay because they do not have any charge with almost zero mass and also they do not interact easily with any matter.

Question.7. Why is the use of A.C. voltage preferred over D.C. voltage? Give two reasons.
Answer : The use of A.C. voltage is preferred over the use of D.C. voltage because of the following reasons :
(i) The energy losses while transmission of A.C. voltage are very less as compared to D.C. voltage.
(ii) A.C. voltage can be controlled as required by using a transformer (i. e., stepped up or stepped down).

Question.8. A biconvex lens made of a transparent material of refractive index 1.25 is immersed in water of refractive index 1.33. Will the lens behave as a converging or a diverging lens? Give reason.
Answer : In this case the biconvex lens will behave as a diverging lens because the refractive index of water (1.33) is more than that of the material (1.25) of the lens.

Question.9. Using Rutherford model of the atom, derive the expression for the total energy of the electron in hydrogen atom. What is the significance of total negative energy possessed by the electron?
Using Bohr’s postulates of the atomic model, derive the expression for radius of nth electron orbit. Hence obtain the expression for Bohr’s radius.
Answer : The magnitude of force according to Rutherford’s model of the atom is,
Here, negative sign indicates that the revolving electron is bound to the positive nucleus.
Suppose m be the mass of an electron and vnbe its speed in nth orbit of radius r. From Rutherford model, the centripetal force for revolution is produced by electrostatic attraction between electron and nucleus.

Question.10. A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined system to that stored initially in the single capacitor.
Answer : Let q be the charge on the charged capacitor. Energy stored in it is,

Question.11. Considering the case of a parallel plate capacitor being charged, show how one is required to generalize Ampere’s Circuital law to include the term due to displacement current.
The total current passing through any surface, of which the closed loop is the perimeter, is the sum of the conduction and displacement current.

Question.12. A cell of emf “E’ and internal resistance V is connected across a variable resistor ‘R’. Plot a graph showing variation of terminal voltage ‘V’ of the cell versus the current ‘I’. Using the plot, show how the emf of the cell and its internal resistance can be determined.

Question.13. Explain, with the help of a circuit diagram, the working of a p-n junction diode as a half-wave rectifier.
Answer : The circuit diagram for a p-n junction diode as a half wave rectifier is shown below :
Working : During the positive half cycle of the input a.c., the p-n junction is forward biased i.e., the forward current flows from p to n. In the forward biasing, the diode provides a very low resistance and allows the current to flow. Thus we get output across-load i.e. a.c. input will be obtained as d.c. output.
During the negative half cycle of the input a.c., the p-n junction is reversed biased i.e., the reverse current flows from n to p. In the reversed biasing, the diode provides a high resistance and hence a very small amount of current will flow through the diode which is of negligible amount. Thus no output is obtained across the load. The wave form of input and output is shown below :

Question.14. Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 1.0 x 10-7 m2 carrying a current of 1.5 A. Assume the density of conduction electrons to be 9 x 1028 m-3.

Question.15. Two monochromatic rays of light are incident normally on the face AB of an isosceles right-angled prism ABC. The refractive indices of the glass prism for the two rays ‘1’ and ‘2’ are respectively 1.35 and 1.45. Trace the path of these rays after entering through the prism.
Ray ‘1’ and ‘2’ will fall on the side AC at an angle of incidence (i) if 45°. Critical angle of ray T ’ is greater than i, so it will get refracted from the prism. Critical angle of ray ‘2’ is less than that of i, so it will undergo total internal reflection.

Question.16. Write the functions of the following in communication systems:
(i) Transducer
(ii) Repeater
Answer : (i) Transducer : A transducer is used to convert one form of energy to another form.
(ii) Repeater : A repeater is used to receive a signal from a transmitter, amplifies it and retransmits it to the receiver.

Question.17. Show diagrammatically the behaviour of magnetic field lines in the presence of (i) paramagnetic and (ii) diamagnetic substances. How does one explain this distinguishing feature.
Answer : (i) The behaviour of magnetic field lines in the presence of a paramagnetic substance is shown below :
(ii) The behaviour of magnetic field lines in the presence of a diamagnetic substance is shown below :
This distinguishing feature is because the relative permeability of the diamagnetic substance is less than 1, so the magnetic lines of force do not prefer passing through the substance and the relative permeability of a paramagnetic substance is greater than 1, so the magnetic lines of force prefer passing through the substance.

Question.18. Draw a circuit diagram n-p-n transistor amplifier, CE configuration. Under what condition does the transistor act as an amplifier?
Answer : The circuit diagram of an n-p-n transistor .amplifier in CE configuration is given below :
Condition : For a transistor to act as a amplifier, it must be operated close be to the centre of its active region.
The transistor behaves as an amplifier when the input circuit (emitter —base) is forward biased having low voltage VBB and the output circuit (collector-base) is reverse biased having high voltage Vcc.

Question.19. (a) Using the phenomenon of polarization, show how transverse nature of light can be demonstrated. (b) Two polaroids Pand P2 are placed with their pass axes perpendicular to each other. Unpolarised light of intensity IQ is incident on P1. A third Polaroid P is kept in between P and P2 such that its pass axis makes an angle of 30° with that of P1. Determine the intensity of light transmitted through P1, P2 and P.
Answer : (a) Suppose that an ordinary light is incident normally on a pair of crystals C and C2 . When the incident ray of light passes through crystal C, it gets plane polarised in the direction perpendicular-to the length of crystal. Now, we see that when the axes of two crystals are parallel, the intensity of the emerging light will be maximum. When the second crystal is placed perpendicular with respect to the first crystal, the intensity of light observed is zero. This is due to the electric field of the plane polarised light obtained from C can vibrate only in one direction. Hence, when the axis of the crystal is perpendicular to its direction of vibration of electric field, it gets blocked. This shows the transverse nature of light.
(b) As given in the question, the polaroids P and P2 are placed with their axes perpendicular to each other. Also, polaroid P. is placed at an angle of 30° with respect to P.

Question.20. Define the term ‘mutual inductance’ between the two coils. Obtain the expression for mutual inductance of a pair of long coaxial solenoids each of length l and radii r and r2 (r >> r2 ). Total number of turns in the two solenoids are N and N2 respectively.
Answer : The ratio of magnetic flux passing through one coil to the current passing through the other is known as coefficient of mutual inductance between the two coils.

Question.21. Answer the following:
(a) Why are the connections between the resistors in a meter bridge made of thick copper strips?
(b) Why is it generally preferred to obtain the balance point in the middle of the meter bride wire?
(c) Which material is used for the meter bridge wire and why?
A resistance of R Ω draws current from a potentiometer as shown in the figure. The potentiometer has a total resistance Ro Ω. A voltage V is supplied to the potentiometer. Derive an expression for the voltage across R when the sliding contact is in the middle of the potentiometer.
Answer : (a) The connection between the resistors in a meter bridge is made of thick copper strips because the resistivity of a copper wire is very low. As, the connections are thick, so the area becomes large and the resistance of the wires becomes almost negligible.
(b) It is preferred to obtain the balance point in the middle of the meter bridge wire because it improves the sensitivity of the meter bridge.
(c) Constant an nichrome/Manganine is used for meter bridge wire because its temperature coefficient of resistance is almost negligible due to which the resistance of the wire does not get affected on increasing temperature of the wire during flow of current.
While the slide is in the middle of the potentiometer only half of its resistance (Ro/2) will be between the points A and B. Hence, the total resistance between A and B, say. R1 will be given by the following expression :

Question.22. A convex lens of focal length 20 cm is placed co-axially with a convex mirror of radius of curvature 20 cm. The two are kept at 15 cm apart from each other. A point object lies 60 cm in front of the convex lens. Draw a ray diagram to show the formation of the image by the combination. Determine the nature and position of the image formed.
Answer : Let us first locate the image of the point object S formed by the convex lens.
Here : u = -60 cm and f= 20 cm
From the lens formula, we have :
The image I1 is formed behind the mirror and acts as a virtual source for the mirror. The convex mirror forms the image H, whose distance from the mirror can be determined as :

Question.23. A voltage V = V sin ωt is applied to a series LCR circuit. Derive the expression for the average power dissipated over a cycle. Under what condition is (i) no power dissipated even though the current flows through the circuit, (ii) maximum power dissipated in the circuit ?
The average power over a cycle is average of the two terms on the R.H.S. of the above equation. The second term is time dependent, so, its average is zero.

Question.24. Write any two distinguishing features between conductors, semiconductors and insulators on the basis of energy band diagrams.
Answer: Conductors :
(i) In case of conductors, the valence band is completely filled and the conduction band can have two cases-either it is partially filled with an extremely small energy gap between the valence and conduction bands or it is empty, with the two bands overlapping each other as shown below :
(ii) Even when a small current is applied, conductors can conduct electricity.
Insulators :
(i) In case of insulators, the energy gap between the conduction and valence bands is very large. And the conduction band is practically empty.
(ii) When an electric field is applied to such kind of material, the electrons find hard to receive such a large amount of energy to attain the conduction band. Thus, the conduction band remains to be empty. That is why no current flows through insulators.
Semiconductors :
(i) In case of semiconductor, the energy band structure of semiconductors is similar to insulators, but in this case, the size of forbidden energy gap is quite smaller than that of the insulators.
(ii) When an electric field is applied to a semiconductor, the electrons in the valence band find it relatively easier to jump to the conduction band. So, the conductivity of semiconductors lies between the conductivity of conductors and insulators.

Question.25. For the past some time, Aarti had been observing some erratic body movement, unsteadiness and lack of coordination in the activities of her sister Radha, who also used,to complain of severe headache occasionally. Aarti suggested to her parents to get a (a) What, according to you, are the values displayed by Aarti?
(b) How can radioisotopes help a doctor to diagnose brain tumour?
Answer : (a) Aarti has showed awareness and responsibility towards her sister.
(b) A little amount radio-isotope like radioiodine is inserted into the body along with organic dyes which are absorbed strongly by the tumour tissue than the normal tissues. By detecting the emitted radiation, the radiologist get information about the size and location of the tumour.

Question.26. Write two basic modes of communication. Explain the process of amplitude modulation. Draw a schematic sketch showing how amplitude modulated signal is obtained by superposing a modulating signal over a sinusoidal carrier wave.
Answer : The two basic modes of communication are :
(1) Point-to-point communication.
(2) Broadcast communication
Amplitude Modulation: Amplitude modulation is obtained by changing the amplitude of the carrier waves according to the amplitude of the modulating wave.
Amplitude modulated signal is obtained by superposing a modulating signal over a sinusoidal carrier wave.

Question.27. An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de-Broglie wavelength associated with the electrons. Taking other factors, such as numerical aperture etc. to be same, how does the resolving power of an electron microscope Compare with what of an optical microscope which uses yellow light ?
Answer : The de-Broglie wavelength of the electrons is given by:
This formula shows that to enhance resolution, we have to use minimized wavelength and media with large indices of refraction.
For an electron microscope, p is equal to 1 (vacuum).
For an electron microscope, the electrons are accelerated through a 60,000 V potential difference.
As, λ is very little (roughly 10-5 times smaller) for electron microscope than an optical microscope which use yellow light of wavelength (5700 A to 5900 A). So, the resolving power of electron microscope is about 105 greater than that of optical microscope.

Question.28. Draw a labelled diagram of Van de Graff generator. State its working principle to show how by introducing a small charged sphere into a larger sphere, a large amount of charge can be transferred to the outer sphere. State the use of this machine and also point out its limitations.
(b) Consider two hollow concentric spheres, S and S2 , enclosing charges 2Q and 4Q respectively as shown in the figure, (i) Find out the ratio of the electric flux through them, (ii) How will the electric flux through the sphere Si change if a medium of dielectric constant ‘ε’ ‘ε’ is introduced in the space inside Sin place of air? Deduce the necessary expression.
Answer : Van de Graff generator is a device used for building up high potential differences of the order of a few million volts. Principle : The working of Vande Graff Graff generator is based on the following two electrostatic phenomena.
(i) Discharging action at sharp points (corona discharge i.e., electric discharge takes place in air or gases readily at the pointed ends of conductors.
(ii) If a charged conductor is brought into internal contact with a hollow conductor, all of its charge transfers to the hollow conductor, howsoever high the potential of the latter may be.
Construction : It has a big spherical conducting shell (S) kept over insulating pillars. A long narrow insulating belt is wound around 2 pulleys’ P and P2 . Band B2 are 2 metal combs with sharp points. Bis known as spray comb and B2 collecting comb.
Working : The spray comb provides positive potential by high tension source. The positive charge is sprayed on belt. As belt moves and touches the sphere, a negative charge is induced on the sharp ends of collecting comb B2 and similar positive charge is induced on the further end of B2 . This positive charge moves immediately to the outer surface of S. because of discharging action of sharp points of B2 , the positive charge on the belt is neutralized. The uncharged belt moves downwards and collects the positive charge from B, which is then collected by B2 . This process is repeated and the positive charge on S goes on accumulating. In this way, voltage differences of as much as 6 or 8 million volts (with respect to the ground) can be created.
Use : Van de Graff generator used to create high potential differences that are used to accelerate charged particles such as electrons, protons, ions, etc. used for nuclear breakdown.
Limitations :
1. It is a series of combination that allows only one way for moving charge.
2. It can accelerate only the charged particles and not the uncharged particles.

Question.29. (a) In Young’s double slit experiment, describe briefly how bright and dark fringes are obtained on the screen kept in front of a double slit. Hence obtain the expression for the fringe width.
(b) The ratio of the intensities at minima to the maxima in the Youngs double slit experiment is 9 : 25. Find the ratio of the widths of the two slits.
(a) Describe briefly how a diffraction pattern is obtained on a screen due to a single narrow slit illuminated by a monochromatic source of light. Hence obtain the conditions for the angular width of secondary maxima and secondary minima.
(b) Two wavelengths of sodium light of 590 nm and 596 nm are used in turn to study the diffraction taking place at a single slit of aperture 2 x 10-16 m. The distance between the slit and the screen is 1.5 m. Calculate the separation between the positions of first maxima of the diffraction pattern obtained in the two cases.
Answer : (a) In Youngs double slit experiment, the wave fronts from the two illuminated slits superpose on the screen. This results in formation of alternate bright and dark fringes because of constructive and destructive interference, respectively. The intensity of light is maximum at the centre C of the screen and it is called central maxima.

Let S and S2 be two slits separated by a distance d. GG’ is the screen at a distance D from the slits S1 and S2 . Both the slits are equidistant from point C. The intensity of light will be maximum at this point due to the path difference of the waves reaching this point will be zero.
At point P, the path difference between the rays coming from the slits S and S2 is S2 P — S1P.



(a) The phenomenon of bending of light round the sharp corners of an obstacle and spreading into the regions of the geometrical shadow is called diffraction.
Consider a parallel light beam from a lens is incident on slit AB. As diffraction happens, the pattern is focussed on screen XY with help of lens L2 . We will get a diffraction pattern that is a central maximum at the centre O flanked by a number of dark and bright fringes known as secondary maxima and minima.
Central Maximum: Each point on the plane wave front AB sends secondary wavelets in all directions: The waves from points equidistant from the centre C kept on the upper and lower half reach point O with zero path difference and so, reinforce each other, making maximum intensity at point O.
Positions and Widths of Secondary Maxima and Minima
Consider a point P on screen at which wavelets moving in a direction making angle θ with CO are brought to focus by the lens. The wavelets from points A and B will have a path difference similar to BN :

Question.30. (a) Deduce an expression for the frequency of revolution of a charged particle in a magnetic field and show that it is independent of velocity or energy of the particle. (b) Draw a schematic sketch of a cyclotron. Explain, giving the essential details of its construction, how it is used to accelerate the charged particles.
(a) Draw a labelled diagram of a moving coil galvanometer. Describe briefly its principle and working.
(b) Answer the following :
(i) Why is it necessary to introduce a cylindrical soft iron core inside the coil of a galvanometer.
(ii) Increasing the current sensitivity of a galvanometer may not necessarily increase its voltage sensitivity. Explain, giving reason.
Therefore, the frequency of the revolution of the charged particle is independent of the velocity or the energy of the particle.
(b) The working principle of a cyclotron is that a charge particle can be accelerated to high energy by an oscillating electric field. A cyclotron uses 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 within the two Ds. Each time the charged particles cross the Ds, it is accelerated by an alternating voltage. As its speed increases the radius of path of each particle also increases. So, the accelerated particles move in a spiral path to the other wall of the cyclotron.
Square wave electric fields are used to accelerate the charged particles in a cyclotron.
At the time the charge particle finishes its half circle, the accelerating electric field reverse so that it gets accelerated across the gap between the Ds.
The particle gets accelerated again and again, and its velocity increases. Therefore, a high kinetic energy is achieved.
(a) Moving coil galvanometer : It is a device used for detecting and measuring small electric current.
Principle : The working is based upon the principle, when a current carrying coil suspended in a magnetic field experiences a torque.
Construction : It is consists of a coil with a large number of turns of insulated copper wire wounded on a metallic frame.
The coil is suspended by means of a phosphor bronze strip and a horse shoe magnet NS surrounds it. The lower end of the coil is attached with a hair spring. The scale of the pointer is attached to the other end of the spring.
Working : When current is passed, the couple acts on it. Since the plane remains parallel to the magnetic field in all position of the coil, the force on the vertical arms always remains perpendicular to the place of the coil.
(b) (i) When the cylindrical soft iron core placed inside the coil of a galvanometer, the magnetic field gets stronger and radial in the space between it and pole pieces, such that whatever the position of the rotation of the coil, the magnetic field is always parallel to its plane.
(ii) Current Sensitivity : It is defined as the small deflection produced in the galvanometer when a unit current is passed through it.


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

Question.1. A conducting loop is held above a current carrying wire ‘PQ’ as shown in the figure. Depict the direction of the current induced in the loop when the current in the wire PQ is constantly increasing.

Question.4. Why do the electrostatic field lines not form closed loops ?
Answer : Electrostatic field lines do not form closed loops because electric field is conservative in nature.

Question.5. A biconvex lens made of a transparent material of refractive index 1.5 is immersed in water of refractive index 1.33. Will the lens behave as a converging or a diverging lens? Give reason.
Answer: The refractive index of material of lens (1.5) is greater than the refractive index of water (1.33). So, it will behave as a converging lens.

Question.7. To which part of the electromagnetic spectrum does a wave of frequency 3 x 1013 Hz belong ?
Answer : The frequency 3 x 1013 Hz belongs to infrared, region of electromagnetic spectrum.

Question.9. Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 2.5 x 10-17m2 carrying a current of 1.8. A. Assume the density of conduction electrons to be 9 x 1028m-3.

Question.13. Write the functions of the following in communication systems:
(i) Transmitter
(ii) Modulator
Answer: (i) Transmitter: A transmitter converts the message signal from the source of information into a form suitable for transmission through a channel.
(ii) Modulator: Modulator : is a device which changes some characteristics like amplitude/frequency/phase angle of a high frequency carrier wave in accordance with the instantaneous value of the low frequency message signal through the appropriate superposition.

Question.21. (a) Show, with the help of a diagram, how unpolarised sunlight gets polarised due to scattering.
Explanation : The dots in Diagram show Vibrations Perpendicularte to the plane of paper and double arrow show vibrations in the plane of paper. The elctrons in the molecule define to vibrate in both of these directions. The electrons vibrating parallel to the double arrows cannot send energy toward an observer looking at 90° to the direction of sun because their acceleration has no transverse component. The light scattered by the molecules in this direction has only dots. It is polarised perpendicular to the plane of paper.
(b) Two polaroids P1 and P2 are placed with their pass axes perpendicular to each other. Unpolarised light of intensity Io is incident on P1. A third Polaroid P3 is kept in between P1 and P2 such that its pass axis makes an angle of 45° with that of P1. Determine the intensity of light transmitted through P1, Pand P3.
Answer : (a) Unpolarized light scattering from air molecules shakes their electrons perpendicular to the direction of the original ray. The scattered light therefore, has a polarization perpendicular to the original direction and none parallel to the original direction.
(b) As given in the question, the polaroids P1 and P2 are placed with their pass axes perpendicular to each other. Also, P3 is placed at an angle of 45° with respect of P1.
Now, we have :
Intensity of light after falling on P1, I’ = I0 /2

Question.22.Define the term self-inductance of a solenoid. Obtain the expression for the magnetic energy stored in an inductor of self-inductance L to build up a current I through it.
Answer : The ratio of magnetic flux through the solenoid to the current passing through it is called self-inductance of a solenoid. It is given by
Energy stored in an inductor : When a current grows through an inductor, a back e.m.f. is set up which opposes the growth of current. So work needs to be done against back e.m.f. (e) in building up the current. This work done is stored as magnetic potential energy.
Let I be the current through the inductor L at any instant t. The current rises at the rate dl/dt.
So the induced e.m.f. is

Question.24. A convex lens of focal length 20 cm is placed co axially with a concave mirror of focal length 10 cm at a distance of 50 cm apart from each other. A beam of light coming parallel to the principal axis is incident on the convex lens. Find the position of the final image formed by this combination. Draw the ray diagram showing the formation of the image.
Answer : The beam incident on lens L is parallel to principal axis. Hence the lens forms an image I1 at its focus, i. e., at a distance OI1 (= 20 cm) from the lens.
The image I1 is formed in front of the mirror and hence, acts as a real source for the mirror. The concave mirror forms the image I2, whose distance from the mirror can be calculated as;


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

Question.1. A conducting loop is held below a current carrying wire PQ as shown. Predict the direction of the induced current in the loop when the current in the wire is constantly increasing.

Question.2. The graph shows variation of stopping potential V0 versus frequency of incident radiation v for two photosensitive metals A and B. Which of the two metals has higher threshold frequency and why ?
Answer : Metal A has higher threshold energy because from the graph it is clear that the minimum frequency required to start photo emission is more in A than that of B.

Question.5. Why do the electric field lines never cross each other ?
Answer : At any point, if electric field lines cross each other it means at that point there are two directions of electric field, which is impossible.

Question.6. To which part of the electromagnetic spectrum does a wave of frequency 5 x 1011 Hz belong ?
Answer : A wave of frequency 5 x 1011 Hz will belong to the microwaves of electromagnetic spectrum.

Question.10. Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 2.5 x 10-17 m2 carrying a current of 2.7 A. Assume the density of conduction electrons to be 9 x 1018m-3.

Question.18. Write the functions of the following in communication systems:
(i) Receiver
(ii) Demodulator
(i) Receiver : The receiver functions on the received signal collected from transmitted signal at the channel output and reproduce the original message signal for sending.
(ii) Demodulator : Demodulator is a device which recovers the original information signal from the modulated wave at the receiver end.

Question.19. A convex lens of focal length 20 cm is placed co axially with a convex mirror of radius of curvature 20 cm. The two are kept at 15 cm from each other. A point object placed 40 cm in front of the convex lens. Find the position of the image formed by this combination. Draw a ray diagram to show the formation.
The positive sign describes that the image is. formed to the right of the lens.
The image I1 is formed behind the mirror and thus acts as a virtual source for the mirror. The convex mirror forms the image I2, whose distance from the mirror is given by :

Question.25. (a) A rod of length l is moved horizontally with a uniform velocity V in a direction perpendicular to its length through a region in which a uniform magnetic field is acting vertically downward. Derive the expression for the emf induced across the ends of the rod.
(b) How does one understand this motional emf by invoking the Lorentz force acting on the free charge carriers of the conductor? Explain.
After this, resultant force on the free electrons of the wire PQ becomes zero. The potential difference between the ends Q and P is given by, ‘
V = El = vBl
Thus, the potential difference is maintained by the magnetic force on the moving free electron and hence, produces an emf, e = Bvl
(b) Lorentz force acting on a charge q which is moving with a speed v in a (normal) uniform magnetic field B, is Bqv.
All the charges will experience the same force.

Question.26. (a) Show, giving via suitable diagram, how unpolarized light can be polarised by reflection.
(b) Two polaroids P1 and P2 are placed with their pass axes perpendicular to each  other. Unpolarised light of intensity Io is incident on P1. A third Polaroid P3 is kept in between P1 and P2 such that its pass axis makes an angle of 60° with that of P1. Determine the intensity of light transmitted through P1, Pand P3.
Answer: (a) On reflection from a transparent medium a normal light beam becomes partially polarised. As the angle of incidence gets higher, the degree of polarization gets higher.
The reflected light beam becomes fully polarised at a certain  value. This angle of incidence is known as polarizing angle (p). At the interface of a refracting medium when light is incident at polarizing angle, the refractive index of the medium is similar to the tangent of the polarizing angle.
(b) As given, the Polaroid P1 and P2 are placed with their axes perpendicular to each other and Polaroid P3 placed at 60° with respect to P1.