## CBSE Previous Year Solved  Papers  Class 12 Physics Outside Delhi 2010

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:

### SET I

Question.1. Name the physical quantity whose S.I. units is JC-1. Is it a
scalar or a vector quantity?
Answer : The SI unit of electrostatic potential is JC-1. It is a scalar quantity.

Question.2. A beam of a-particles projected along + X-axis, experiences a force due to a magnetic field along +y-axis. What is the direction of magnetic field?

Answer: The direction of magnetic field is perpendicular to x andy axis i.e. along z axis.

Question.3. Define self-inductance of a coil. Write its S.I units?
Answer : It is the phenomenon of production of induced e.m.f. in a coil when a changing current passes through it. S.I unit .is Henry (H).

Question.4. A converging lens is kept coaxially in contact with diverging
lens — both the lenses being of equal focal lengths, what is the focal length of the combination?
Answer : Let f1 is the focal length of convex lens and  f2 is the focal length of concave lens. Then their equivalent focal length F would be:

Question.5. Define ionization energy. What is its value for hydrogen atom?
Answer: Ionisation energy is the required energy to knock an electron out of the atom, that is energy required to take an electron from its ground state to the outermost orbit (n = ∞).
Ionisation energy of hydroge n = E– E1
= 0 – (-13.6 eV) = 13.6 eV

Question.6. Two conducting wires X and Y of same diameter but different materials are joined in series across a battery. If the number density of electrons in X is twice that in Y, find the ratio of drift velocity of electrons in the two wires.
Answer : The current I through the wire is related with drift velocity as :
I = nAevd
Since two wires X and Y of same diameter but different materials are joined in series.

Question.7. X-ray is the part of electromagnetic spectrum whose wavelength lies in the range of 10-10 m. Give its one use.
Answer : The wavelength range of 10-10m lies in x-rays. Its uses are : X-rays are used as a diagnostic tool in medicine and as a treatment for certain forms of cancer.

Question.8. When light travels from a rarer to a denser medium, the speed decreases. Does this decrease in speed imply a decrease in the energy carried by the light wave? Justify your answer?
Answer: No, since energy E = hv, where h is. Planck’s constant and u is frequency that does not change when light goes from one medium to another. Hence energy does not change.

Question.10. A spherical conducting shell of inner radius r1 and outer radius r2  has a charge ‘Q’, a charge‘q’ is placed at centre ofdie shell.
(a) What is the surface charge density on (i) inner surface (ii) outer surface of the shell?
(b) Wire the expression for the electric field at a point x > r2, from the centre of the shell.
Answer : The charge (+ q) at the centre induces charge – q on the inner surface of the shell and charge + q on the outer surface of the shell

Question.11.Draw a sketch of a plane electromagnetic wave propagating along the Z-direction. Depict clearly the directions of electric and magnetic fields varying sinusoidally with z
Answer : E along x-axis, B along y-axis.or vice-versa X

Question.12. Show that the electric field at the surface of a charged conductor is given by.

where a is the surface charge density and n is unit vector normal to the surface in the outward direction?

Question.13.Two identical loops one of copper and the other of aluminum, are rotated with the same angular speed in the same magnetic field. Compare (i) the induced emf and (ii) the current produced in the two coils, justify your answer.

Question.14. An a-particle and a proton are accelerated from rest by the same potential. Find the ratio of their de Broglie wavelengths?

Question.15. Write two factors justifying the need of modulating a signal. A carrier wave of peak voltage 12 V is used to transmit a message signal. What should be peak voltage of the modulating signal in order to have a modulation index of 75%.
Answer : Modulation of a signal is necessary to transmit a signal in the audio frequency range over a long distance due to
(i) Size of antenna
(ii) Effective power radiated by antenna.

Question.16 Write Einstein photoelectric equation? State clearly the three salient features observe in photoelectric effect, which can be explained on basis of above equation?
Answer : Einstein photoelectric equation :

Where ø0 is work function. If the energy of the photon absorbed by the electron is less than the work function ø0 of the metal, then the electron will not be emitted. Therefore, for the given metal, the thershold, frequency of light be v0, then an amount of energy hv0 of the photon of light will be spent in ejecting the electron out of the metal, that is, it will be equal to the work function W. Thus

This equation is called ‘Einstein Photoelectron equation. Salient features :
1. The stopping potential and hence the maximum kinetic energy of emitted electrons varies linearly with the * frequency of incident radiation.
2. There exists a minimum cut-off frequency v0, for which the stopping potential is zero.
3. Photoelectric emission is instantaneous.

Question.17.Draw a plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions which you can draw regarding the nature of nuclear forces.
OR
Draw a plot of the binding energy per nucleon as a function of mass number for a large number of nuclei, 2 ≤ A≤ 240. How do you explain the constancy of binding energy per nucleon in range 30 < A < 170 using the property that nuclear force is short-ranged?
Answer : Plot of potential energy of a pair of nucleons as a fuction of their seperation:

Conclusions:
(i) The nuclear force is much stronger than the Coulomb force acting between the charges or gravitational forces between
(ii) The nuclear force between two nucleons falls rapidly to zero as their distance is more than few fermi.
(iii) For a separation greater than r0 the force is attractive and for separations less than ro force is strongly repulsive.

Constancy of binding energy per nucleon in range 30 < A < 170 a consequence of the fact that nuclear force is short-ranged. It will be under the influence of only some of its neighbours.

Question.18. (i) Identify the logic gates marked P and Q in the given logic circuit.

(ii) Write down output at X for the inputs A = 0, B = 0 and A = 1, B = 1.
Answer : (i) The logic gate P is NAND gate and logic gate Q is OR gate.
(ii) Let output of logic gate P is Y.

Question.19. Which mode of propagation is used by short wave broadcast services having frequency range from a few MHz to 30 MHz? Explain diagrammatically how long distance communication can be achieved by this mode. Why there an upper limit to frequency of waves used in this mode?
Answer : Sky wave propagation can be used by short wave broadcast services having frequency r and from a MHz to 30 MHz.

In this mode sky waves are directed towards the sky and are reflected by the ionosphere toward the desired station on the earth.
Sky wave propagation is restricted to frequencies upto 30 MHz because ionosphere cannot reflect electro-magnetic waves having frequencies greater than 30 MHz.

Question.20.Write any two factors on which internal resistance of a cell depends. The reading on a high resistance voltmeter, when a cell is connected across it, is 2.2 V When the terminals of the cell are also connected accross to a resistance of 5 Ω as shown in the circuit, the voltmeter reading drops to 1.8 V. Find the internal resistance of the cell.

Answer: The internal resistance of a cell depends on nature of electrolyte as well as the concentration of the electrolyte.
Let r be the internal resistance of the cell

Question.21. A network of four capacitors each of 12 μF capacitance is connected to a 500 V supply as given in the figure. Determine (a) equivalent capacitance of the network and
(b) charge on each capacitor.

Question.22. (i) Draw a neat labelled ray diagram of an astronomical telescope in normal adjustment. Explain briefly its working.
(ii) An astronomical telescope uses two lenses of powers 10 D and 1 D. What is its magnifying power in normal adjustment?
OR
(i) Draw a neat labelled ray diagram of a compound microscope. Explain briefly its working.
(ii)Why must both the objective and the eye-piece of a compound microscope have short focal lengths?

Working : Telescope has an objective and eyepiece. The objective has a large focal length and much larger aperture than the eyepiece. Light from a distant object enters the objective and a real image is formed in the tube at its second focal point. The eyepiece magnifies this image producing a final inverted image at infinity.
(ii) Calculation of magnifying power :

The objective forms a real, inverted, magnified image of the object. This serves as the object for the second lens, the eyepiece, which functions essentially like a simple microscope or magnifier, produces the final image, which is enlarged and virtual.
(ii) To achieve a large magnification of a small object; both the * objective and eyepiece should have small focal lengths.

Question.23. In Young’s double slit experiment, the two slits 0.15 mm apart are illuminated by monochromatic light of wavelength  450 nm. The screen is 1.0 m away from the slits.
(a) Find the distance of the second (i) bright fringe, (ii) dark fringe from the central maximum.
(b) How will the fringe pattern change if the screen is moved
away from the slits?

Question.24. State Kirchhoff’s rules. Use these rules to write the expressions for the currents I1, I2 and I3 in the circuit diagram shown.

Answer: (a) Junction rule: At any junction, the sum of the currents entering the junction is equal to the sum of currents leaving the junction.
∑I = 0
(b) Loop rule : The algebraic sum of changes in potential, around any closed loop, involving resistors and cells in the loop, is zero.

Question.25.(a) Write symbolically the
(b) Derive an expression for the average life of a radionuclide. Give its relationship with the half-life.

(b) Derivation of average life :

Hence the average life period of a radioactive element is 1.44 times the half life period of the element.

Question.26. How does an unpolarised light get polarised when passed through a Polaroid? Two polaroids are set in crossed positions. A third Polaroid is placed between the two making an angle θ with the pass axis of the first Polaroid. Write the expression for the intensity of light transmitted from the second Polaroid. In what orientations will be transmitted intensity be (i) minimum and (ii) maximum?
Answer. A polaroid consists of long chain molecules aligned in a particular direction. The electric vectors (associated with the propagating light wave) along the direction of the aligned molecules get absorbed. Thus, if an unpolarised light wave is incident on such a polaroid, then the light wave will get linearly polarised with the electric vector oscillating along a direction perpendicular to the aligned molecules.
When the polaroid is rotated in the path of plane polarized light, its intensity will vary from maximum to minimum, let Io be intensity of polarized light after passing through first polarizer P1. Then intensity of polarized light after passing through second polarizer P2 is given by
I = (I0 cos2 θ)
Expression for the intensity transmitted through second polaroid
I = (I0 cos2 θ) cos2 (90° – θ) = I0 (cos θ sin θ)2 = I0sin2 2θ/4
where I0 is the intensity of the polarized light after passing through the first polaroid.
(i) Intensity will be maximum when θ= 45°
(ii) Minimum when θ=0°

Question.27. An illuminated object and a screen are placed 90 cm apart.
Determine the focal length and nature of the lens required to produce a clear image on the screen, twice the size of the object.
Answer : For real image, m = v/u =2
v = 2u

Question.28. (a) With the help of a diagram, explain the principle and working of a moving coil galvanometer.
(b) What is the importance of a radial magnetic field and how is it produced?
(c) Why is it that while using a moving coil galvanometer as a voltmeter a high resistance’in series in required whereas in an ammeter a shunt is used?
OR
(a) Derive an expression for the force between two long parallel current carrying conductors.
(b) Use this expression to define S.I. unit of current.
(c) A long straight wire AB carries a current I. A proton P travels with a speed v, parallel to the wire, at a distance d from it in a direction opposite to the current as shown in the figure. What is the force experienced by the proton and what is its direction?

Answer : (a) Moving coil galvanometer : It is a device used for the detection and measurement of small electric current.

Construction : It consists of a coil having a large number of turns of insulated copper wire wound around metallic frame.
A hair spring is attached to lower end of coil, the other end is attached to scale through the pointer.
Principle : Galvanometer action is based upon the principle that when electric current flows in a coil placed in a magnetic field, a deflecting torque acts upon the coil whose magnitude depends upon the strength of the current.
Working : The magnetic torque tends to rotate the coil. A spring provides a counter torque that balances the magnetic torque; resulting in a steady angular deflection. The deflection is indicated on the scale by a pointer attached to the spring.
If I is current flowing through coil.
B = magnetic field supposed to be uniform and parallel to coil
A = area of coil
Deflecting torque acting on the coil is
τ = NI AB sin 90° = NIBA             ………. [sin 90° = 1 ]
Due to deflecting torque, the coil rotates and suspension wire gets twisted.
(b) Importance and production of radial magnetic field :
In a radial magnetic field, magnetic torque remains maximum for all positions of the coils. It is produced due to cylindrical pole pieces and soft iron core.
Reason:
Voltmeter : This ensures that a very low current passes through the voltmeter and hence does not change (much) the original potential difference to be measured.
Ammeter : This ensures that the total resistance of the circuit
does not change much and the current flowing remains (almost) as its original value.
(c) A galvanometer can be converted into a voltmeter by connecting a high resistance in series with galvanometer to
draw a very small current. A galvanometer can be converted into an ammeter by connecting a low resistance shunt in parallel with galvanometerto draw a very large current.

Two long parallel conductors ‘a’ and ‘b’ are separated by a distance d and carry (parallel) currents  Ia and  Ib, respectively. The conductor ‘a’produces, the same magnetic field Ba at all points along the conductor ‘b’ perpendicular to the plane of the page directed downwards.

By Fleming left hand rule, this force acts on CD towards AB
force. This force will be attractive.
(b) Definition of ampere : The ampere is the value of that teady current which, when maintained in each of the two very long, straight, parallel conductors of negligible cross-section, and placed on meter apart in vacuum, would produce on each of these conductors a force equal to 2 x 10-7 newton per meter of length.
(c) Magnetic field due to the straight wire AB at a perpendicular , distance r from it.

The direction of force on proton acts in plane of paper towards right. ,

Question.29.State Faraday’s law of electromagnetic induction.
Figure shows a rectangular conductor PQRS in which the conductor PQ is free to move in a uniform magnetic field B
from x = 0 to x = b and is zero for x > b. Assume that only the arm PQ possesses resistance r. When the arm PQ is pulled outward from x = 0 to x = 2b and is then moved backward to x = 0 with constant speed v, obtain the expressions for the flux and the induced emf, the force necessary to pull the arm and the power dissipated as Joule heat. Sketch the variations of these quantities with distance 0≤ x ≤ 2b

OR
Draw a schematic diagram of a step-up transformer. Explain its working principle. Deduce the expression for the secondary to primary voltage in terms of the number of turns in the two coils. In an ideal transformer, how is this ratio related to the currents in the two coils? How is the transformer used in large scale transmission and distribution of electrical energy over long distances?
Answer: Part I: Faradays law of induction : The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit.

Where ∅B is flux linked with one turn of the coil. The negative sign indicates the direction of 6 and hence the direction of current in a closed loop. If the circuit is closed a current
I=∈/R is set up in it.
Part II : Refer to following figure. The arm PQ of the rectangular conductor is moved from x = 0 outwards. The uniform magnetic field is perpendicular to the plane and extends from x = 0 to x = b and is zero the situation when the arm PQ possesses substantial resistance r. Consider the situation when arm PQis pulled outwards from x=0 to x=2b and is then moved back to x = 0 with constant speed v.

OR
Step up Transformer:
Principle : When an alternating voltage is applied to the primary, the resulting current produces an alternating magnetic flux which links the secondary and induces an emf in it (mutual induction). The value of this emf depends on the number of turns in the secondary.

secondary windings. The core laminations are joined in the form of strips; in between the strips you can see that there are some narrow gaps right through the cross-section of the core. These staggered joints are said to be ‘imbricated’. Working : Both the coils have high mutual inductance. A mutual electro-motive force is induced in the transformer from the alternating flux that is setup in the laminated core, due to the coil that is connected to a source of alternating voltage. Most of the alternating flux developed by this coil is linked with the other coil and thus produces the mutual induced electro-motive force. Thus so produced electro-motive force can be explained with the help of Faradays laws of Electromagnetic Induction as
If the secondary coil circuit is closed, a current flow in it and thus electrical energy is transferred magnetically from the first to the second coil.
The alternating current supply is given to the first coil and hence it can be called as the primary winding. The energy is drawn out from the second coil and thus can be called as the secondary winding.

Large scale transmission : The large scale transmission and distribution of electrical energy over long distances is done with the use of transformers. The voltage output of the generator is  stepped-up (so that current is reduced and consequendy, the I2R loss is cut down). It is then transmitted over long distances to an area sub-station near the consumers. There the voltage is stepped down. It is further stepped down at distributing sub-stations and utility poles before a power supply of 240 V reaches our homes.

Question.30. (a) Draw the circuit arrangement for studying the input and output characteristics of an n-p-ti transistor in CE configuration. With the help of these characteristics define
(i) input resistance, (ii) current amplification factor. ‘
(b) Describe briefly with the help of a circuit diagram how an n-p-n transistor is used to produce self-sustained oscillations.

(b) Circuit diagram :

Working : In an oscillator, we get an ac output without any external input signal. Hence, the output in an oscillator is self-sustained. To attain this, an amplifier is taken. A portion of the output power is returned back (feedback) to the input in phase with the starting power (this process is termed positive feedback).

### SET II

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

Question.1.Find the ratio of energies of photons produced due to transition of an electron of hydrogen atom form its
(i) second permitted energy level to the first level, and
(ii)the highest permitted energy level to the first permitted level.

Question.4. A beam of electrons projected along + x-axis, experiences a force due to a magnetic field along the +y-axis. WTiat is the direction of the magnetic field?

Answer : The direction of magnetic field is along z-axis.

Question.12. A rectangular loop and a circular loop are moving out of a uniform magnetic field to a field-free region with a constant velocity V as shown in the figure. Explain in which loop do you expect the induced emf to be constant during the passage out of the field region. The magnetic field is normal to the loops.

Answer : The induced emf is expected to be constant only in the area of rectangular loop. In the case of circular loop, the rate of change of area of the loop during its passage out of one field region is not constant. Hence induced emf will vary accordingly.

Question.19. A network of four capacitors each of 15 pF capacitance is connected to a 500 V supply as shown in the figure. Determine (a) equivalent capacitance of the network and (b) charge on each capacitor.

Question.20. Write any two factors on which internal resistance of a cell depends. The reading on a high resistance voltmeter, when a cell is connected across it, is 2.0 V, when the terminals of the cell are also connected to a resistance of 3 Ω as shown in the circuit, the voltmeter reading drops to 1.5 V Find the internal resistance of the cell.

Question.22.In Youngs double slit experiment, the two slits 0.12 mm apart are illuminated by monochromatic light of wavelength 420 nm. The screen is 1.0 m away from the slits.
(a) Find the distance of the second (i) bright fringe, (ii) dark fringe from the central maximum.
(b) How will the fringe pattern change if the screen is moved
away from the slits?

the screen is moved away from the slits fringes become further apart as D increases.

Question.23. State KirchhofFs rules. Apply Kirchhoff’s rules to the loops ACBPA and ACBQA to write the expressions for the currents I1, I2 and I3 in the network.

Question.27. The image obtained with a convex lens is erect and its length is four times the length of the object. If the focal length of the lens is 20 cm, calculate the object and image distances.

### SET III

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

Question.1. A beam of protons projected along +x-axis, experience a force due to a magnetic field along the —y-axis. What is the direction of the magnetic field?

Answer. The direction of megnetic field is along z-axis.

Question.8.The ground state energy of hydrogen atom is -13.6 eV. What are the kinetic and potential energies of electron in this state ?