CBSE Previous Year Solved Papers Class 12 Physics Outside Delhi 2013

CBSE Previous Year Solved  Papers  Class 12 Physics Outside Delhi 2013

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

Question.1. Two charges of magnitudes -2Q and +Q are located at point (a, 0) and (4a, 0) respectively. What is the electric flux due to these charges through a sphere of radius ‘3a’ with its centre at the origin?

Question.2. How does the mutual inductance of a pairs coils change when
(i) distance between the coils is increased and
(ii) number of turns in the coils is increased?

Question.3. The graph shown in the figure represents a plot of current versus voltage for given semiconductor. Identify the region, if any, over which the semiconductor has a negative resistance.

Answer : Resistance of a material can be found out by the slope of the V-I curve. Part BC of the curve shows the negative resistance as with the increase in voltage, current decreases.

Question.4. Two identical cells, each of emf E, having negligible internal resistance, are connected in parallel with each other across an external resistance R, What is the current through this resistance?

Answer: The cells are arranged as shown in the circuit diagram. As the internal resistance is negligible, so total resistance of the circuit = R.
So, current through the resistance, I = E/R .

Question.5. The motion of copper plate is damped when it is allowed
to oscillate between the two poles of a magnet. What is the cause of this damping?
Answer : As the copper plates oscillate in the magnetic field between the two poles of the magnet, there is a continuous change of magnetic flux linked with the copper plate. Due to this, eddy currents are set up in the copper plate which try to oppose the motion of the pendulum according to the Lenz’s law and finally bring it to rest.

Question.6. Define the activity of a given radioactive substance. Write its S.I. unit.
Answer: Activity of a radioactive substance is defined as number of radioactive disintegrations taking place in one second in the sample.
S.I. unit of activity is Becquerel (Bq).
1 Becquerel = 1 Bq = 1 decay per second.

Question.7. Welders wear special goggles or face masks with glass windows to protect their eyes from electromagnetic radiations. Name the radiations and write the range of their frequency.
Answer: Welders wear special goggles with glass windows to protect the eyes from ultraviolet rays (UV rays). The range of frequency of UV rays is 4 x 10^-7 m (400 nm) to 6 x 10^-10 m (0.6 nm).

Question.8. Write the expression for the de-Broglie wavelength associated with a charged particle having charge ‘q and mass ‘m’, when it is accelerated by a potential V.
Answer : Let v be the velocity gained by the given charge particle when it is accelerated through a potential difference of ‘V’ volts kinetic energy of the particle = 1/2 mv²
Kinetic energy of the particle = Work done on the particle by electric field.

Question.9. Draw a typical output characteristics of an n-p-n transistor in CE configuration. Show how these characteristics can be used to determine output resistance.
Answer : Output characteristics is the plot between collector- emitter voltage (V_CE ) and the collector current (I_C ) at different constant values of base current (I_B).

Output resistance is defined as the ratio of the change in collector-emitter voltage (∆V_CE) to the change in collector current (∆I_C) at a constant base current (I_B).
Initially, with the increase in V_CE the collector current increases almost linearly, this is because the junction is not reverse biased. When the supply is more than required to reverse bias the base-collector junction, I_C increases very little with V_CE.

The reciprocal of slope of the linear part of the curve gives the value of output resistance i.e.,

Question.10. A parallel beam of light of 500 nm bills on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Calculate the width of the slit.

Question.11. A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness dl 2, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.
Answer : Initially when there is vacuum between the two plates, the capacitance of the two parallel plate is

Question.12. A capacitor, made of two parallel plates each of plate area A and separation d, being charged by an external ac source. How the displacement current inside the capacitor is same as the current charging the capacitor.


Thus, the displacement current inside the capacitor is the same as the current charging the capacitor.

Question.13. Explain the term ‘drift velocity of electrons in a conductor. Hence obtain the expression for the current through a conductor in terms of ‘drift velocity’.
OR
Describe briefly, with the help of a circuit diagram, how a potentiometer is used to determine the internal resistance of a cell.
Answer : Drift velocity of electrons in a conductor : Metals contain a large number of free electrons. These electrons are in continuous random motion. Due to the random motion, the free electrons collide with positive metal ions with high frequency and undergo change in direction at each collision. So the average velocity for the electrons in a conductor is zero.
Now, when this conductor is connected to a source of emf an electric field is established in the conductor,
such that E=V/L
Where, V = potential difference across the conductor and L = length of the conductor.
The electric field exerts an electrostatic force ‘-Ee on each free electron in the conductor.
The acceleration of each electron is given by

Where, e = electric charge on the electron and m = mass of electron
The negative sign indicates that the force and hence the acceleration is in a direction opposite to the direction of the electric field. Due to this acceleration, the electrons attain ‘ a velocity in addition to thermal velocity in the direction opposite to that of electric field. The average velocity of all the free electrons in the conductor is called the drift velocity of free electrons of the conductor.


OR
Measurement of internal resistance of a cell using potentiometer is shown in figure. The cell of emf, E is connected across a resistance box (R) through key K₂.

Question.14. A convex lens of focal length f₁ is kept in contact with a concave lens of focal length f₂. Find the focal length of the combination.
Answer : For convex lens, focal length, f=f₁ and for concave lens, the focal, f= -f₂
The equivalent focal length of a combination of convex lens and concave lens given as :

Question.15. In the block diagram of a simple modulator for obtaining an AM signal, shown in the figure, identify the boxes A and B. Write their functions.

Answer: In the block diagram of modulator. A is Square Law Device and B is Band pass filter.
Band pass filter rejects low and high frequencies and allows a band of frequencies to pass through.
Square Law Device is a non linear device. It produces a non linear output of message and carrier signals. The output of square law device is

Question.16. In the circuit shown in the figure, identify the equivalent gate of the circuit and make its truth table.

Question.17.

Question.18. A rectangular conductor LMNO is placed in a uniform magnetic field of 0.5T. The field is directed perpendicular to the plane of the conductor. When the arm MN of length of 20 cm is moved towards left with a velocity of 10 ms-1, calculate the emf induced in the arm. Given the resistance of the arm to be 5 Ω (assuming that other arms are of negligible resistance) find the value of the current in the arm.

OR
A wheel with 8 metallic spokes each 50 cm long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of the Earth’s magnetic field. The Earth’s magnetic field at the place is 0.4G and the angle of dip is 60°. Calculate the era/-induced between the axle and the rim of the wheel. How will the value of emf be affected if the number of spokes were increased?
Answer : Let ON be x at some instant.

The value of emf induced is independent of the number of spokes as the emf s across the spokes are in parallel. So, the emf will be unaffected with the increase in spokes.

Question.19. Define the current sensitivity of a galvanometer. Write S.I. unit. Figure shows two circuits each having a galvanometer and a battery of 3 V
When the galvanometers in each arrangement do not show any deflection, obtain the ratio R₁/R₂ .

Answer : Current sensitivity of a galvanometer is defined as the deflection in galvanometer per unit current. Its SI unit is radians/ampere.
For balanced Wheatstone bridge, there will be no deflection in the galvanometer.
4/R₁ =6/9
R₁=(4 × 9)/6 =6Ω

For the equivalent circuit, when the Wheatstone bridge is balanced, there will be no deflection in the galvanometer.

Question.20. A wire AB is carrying a steady current 12 A and is lying on the table. Another wire CD carrying 5 A is held directly above AB at a height of 1 mm. Find the mass per unit length of the wire CD so that it remains suspended at its position when left free. Give the direction of the current flowing in CD with respect to that in AB. [Take the value of g = 10 ms^-2]
Answer : Force per unit length between the current carrying wires is given as :

Question.21. Draw V-I characteristics of a p-n junction diode. Answer the following questions, giving reasons :
(i) Why is the current under reverse bias almost independent of the potential up to a critical voltage?
(ii) Why does the reverse current show a sudden increase at the critical voltage?
Name any semiconductor device which operates under the reverse bias in the breakdown region.
Answer : V-I-characteristics of p-n junction diode :

(i) Under the reverse bias condition, the holes of p-side are attracted towards the negative terminal of the battery and the electrons of the n-side are attracted towards the positive terminal of the battery. This increases the depletion layer and the potential barrier. However the minority charge carriers are drifted across the junction producing a small current. At any temperature the number of minority carriers is constant and very small so there is the small current at any applied potential. This is the reason for the current under reverse bias to be almost independent of applied potential. At the critical voltage, avalanche break down takes place which results in a sudden flow of large current.
(ii) At the critical voltage, the holes in the p-side and conduction electrons in the n-side are accelerated due to the reverse-bias voltage. These minority carriers acquire sufficient kinetic energy from the electric field and collide with a valence electron. Thus the bond is finally broken and the valence electrons move into the conduction band resulting in enormous flow of electrons and thus formation of hole-electron pairs. Thus there is a sudden increase in the current in the critical voltage.
Zener diode is a semiconductor device which operates under the reverse bias in the break down region.

Question.22. Draw a labelled ray diagram of a refracting telescope. Define its magnifying power and write the expression for it.
Write two important limitations of a refracting telescope over a reflecting type telescope.
Answer: Refracting telescope :

Magnifying power: The magnifying power is the ratio of the angle a subtended at the eye by the final image to the angle P which the object subtends at the lens or the eye.

Limitations of refracting telescope over reflecting type telescope:
(i) Refracting telescope suffers from chromatic aberration as it uses large sized lenses.
(ii) The requirement of big lenses tend to be very heavy and therefore difficult to make and support by their edges.

Question.23. Write Einsteins photoelectric equation and point out any two characteristic properties of photons on which this equation is based.
Briefly explain the three observed features which can be explained by this equation.
Answer : Einsteins photoelectric effect equation :

Question.24. Name the type of waves which are used for line of sight (LOS) communication. What is the range of their frequencies?
A transmitting antenna at the top of a tower has a height, of 20 m and the height of the receiving antenna is 45 m. Calculate the maximum distance between them for satisfactory communication in LOS mode. (Radius of the Earth = 6.4 x 10⁶m)

Question.25. (a) What is linearly polarized light. Describe briefly using a diagram how sunlight is polarised.
(b) Unpolarised light is incident on a Polaroid. How would the intensity of transmitted light change when the Polaroid is rotated?
Answer : (a) Natural light is unpolarised i. e., the electric vector takes all possible directions in the transverse plane, rapidly and randomly, during a measurement. A polarizer transmits only one component. This resulting light is called linear or plane polarized.
The incident sunlight is unpolarised. The dot and double arrows show the polarization in the perpendicular and in the plane of the figure. Under the influence of the electric field of the incident wave, the electrons in the molecules of the atmosphere acquire components of motion in both these directions. An observer looking at 90° to the direction of the sun, the charges accelerating parallel to the double arrows do not radiate energy towards this observer since their acceration has no transverse component. The radiation scattered by the molecule is therefore, represented by dots. It is linearly polarized perpendicular to the plane of the figure.

Thus, the intensity of the transmitted light remains uncharged when the polaroid is rotated.

Question.26. One day Chetaris mother developed a severe stomach ache all of a sudden. She was rushed to the doctor who suggested for an immediate endoscopy test and gave an estimate of expenditure for the same. Chetan immediately contacted his class teacher and shared the information with her. The class , teacher arranged for the money and rushed to the hospital. On realizing that Chetan belonged to a below average income group family, even the doctor offered concession for the test fee. The test was conducted successfully.
Answer the following questions based on the above information:
(a) Which principle in optics is made use of in endoscopy?
(b) Briefly explain the values reflected in the action taken by the teacher.
(c) In what way do you appreciate the response of the doctor on the given situation?
Answer : (a) Endoscopy is based on total internal reflection principle. It has tubes which are made up of optical fibres and are used for transmitting and receiving electrical signals, which are converted into light by suitable transducer.
(b) Humanity and charity.
(c) Doctor gave monetary help to Chetan by understanding his poor financial condition.

Question.27. (a) Using Biot-Savart’s law, derive the expression for the magnetic field in the vector form at a point on the axis of a circular current loop.
(b) What does a toroid consist  of Find out the expression for the magnetic field inside a toroid for N turns of the coil having the average radius r and carrying a current I. Show that the magnetic field in the open space inside and exterior to the toroid is zero.
OR
(a) Draw a schematic sketch of a cyclotron. Explain clearly the role of crossed electric and magnetic field in accelerating the charge. Hence derive the expression for the kinetic energy acquired by the particles.
(b) An α —particle and a photon, are released from the centre of the cyclotron and made to accelerate.
(i) Can both be accelerated at the same cyclotron frequency? Give reason to justify your answer.
(ii) When they are accelerated in turn, which of the two will have higher velocity at the exit slit of the dees?
Answer : (a) Magnetic field on the axis of a circular loop
I —> current
R —> Radii
X —> Axis
x —> Distance of OP .
dl —> Conducting element of the loop



(b) Toroid is a hollow circular ring on which a large number of turns of a wire are closely wound. Figure shows a sectional view of the toroid. The direction of the magnetic field inside is clockwise as per the right-hand thumb rule for circular loops. Three circular Amperian loops 1, 2 and 3 are shown by dashed lines.

By symmetry, the magnetic field should be tangential to each of them and constant in magnitude for a given loop.
Let the magnetic field inside the toroid be B. We shall now consider the magnetic field at S. Once again we employ :

Open space exterior to the toroid :
Each turn of current carrying, wire is cut twice by the loop
3. Thus, the current coming out of the plane of the paper is cancelled exactly by the current going into it. Thus,
I= 0, and B = 0
OR
(a) Schematic sketch of cyclotron : The combination of crossed electric and magnetic field is used to increase the energy of the charged particle. Cyclotron uses the fact that the frequency of revolution – of the charged particle in a magnetic field is independent of its energy. Inside the dees particle is shielded from the electric field and magnetic field acts on the particle and makes it to go round in a circular path inside a dee. Every time the particle moves from one dee to the other it comes under the influence of electric field which ensures to increase the energy of the particles as the sign of the electric field changed alternately. The increased in energy, increases the radius of the circular path so the accelerated particle moves in a spiral path.

Question.28. (a) Define electric dipole moment. It is a scalar or a vector? Derive the expression for the electric field of a dipole at a point on the equatorial plane of the dipole.
(b) Draw the equipotential surfaces due to an electric dipole. Locate the points where the potential due to the dipole is zero.
OR
Using Gauss’ law deduce the expression for the electric field due to a uniformly charged spherical conducting shell of radius R at a point (i) outside and (ii) inside the shell. Plot a graph showing variation of electric field as a function of r > R and r < R. (r being the distance from the centre of the shell)
Answer : (a) Electric dipole moment : The strength of the electric dipole is measured by the quantity of electric dipole moment. Its magnitude is equal to the product of the magnitude of either charge and the distance between the two charges.



(b) Equipotential surface due to electric dipole : The potential due to the dipole is zero at the line bisecting the dipole length.

Answer : Electric field due to a uniformly charged thin spherical shell:
(i) When point P lies outside the spherical shell: Suppose that we have to calculate electric field at the point P at a distance r (r > R) from its centre. Draw the Gaussian surface through point P so as to enclose the charged spherical shell. The Gaussian surface is a spherical shell of radius r and centre O.

Question.29. Using Bohr’s postulates, derive the’ expression for the frequency of radiation emitted when electron in hydrogen atom undergoes transition from higher energy state (quantum number n_i) to the lower state, (n_f)
When electron in hydrogen atom jumps from energy state n_i = 4 to n_f= 3, 2, 1, identify the spectral series to which the emission lines belong.
OR
(a) Draw the plot of binding energy per nucleon (BE/A) as a function of mass number. Write two important conclusions that can be drawn regarding the nature of nuclear source.
(b) Use this graph to explain the release of energy in both the processes of nuclear fusion and fission.
(c) Write the basic nuclear process of neutron undergoing β-decay. Why is the detection of neutrinos found very difficult?
Answer : In the hydrogen atom, Radius of electron orbit,



OR
(a) Plot of binding energy per nucleon as the function of mass number A is given as below :
Following are the two conclusion that can be drawn regarding the nature of the nuclear force :
(i) The force is attractive and strong enough to produce a binding energy of few MeV per nucleon.
(ii) The constancy of the binding energy in the range 30 <A<170 is a consequence of the fact that the nuclear force is short range force.

(b) Nuclear fission: A very heavy nucleus (say A = 240) has lower binding energy per nucleon as compared to the nucleus with A = 120. Thus, if the heavier nucleus breaks to the lighter nucleus with high binding energy per nucleon, nucleons are tightly bound. This implies that energy will be released in the process which justifies the energy release in fission reaction.
Nuclear fusion : When two light nuclei (A< 10) are combined to form a heavier nuclei, the binding energy of the fused heavier nuclei is more than the binding energy per nucleon of the lighter nuclei. Thus the final system is more tightly bound than the initial system. Again the energy will be released in fusion reaction.
(c) The basic nuclear process of neutron undergoing β-decay is given as :

Neutrinos are massless and charge less particles. Nutrinos interact very weakly with matter that it because very difficult to defect then that’s why the detection of neutrinos is found very difficult.

SET II

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

Question.4. Two charges of magnitudes -3Q and +2Q are located at points (a, 0) and (4a, 0) respectively. What is the electric flux due to these charges through a sphere of radius ‘5d with its centre at the origin?
Answer : According to Gauss’ theorem, the electric flux through a closed surface enclosing a charge is equal to 1/ε₀ times the magnitude of the charge enclosed.

Question.7. A light metal disc on the top of an electromagnet is thrown up as the current is switched on. Why? Give reason.
Answer : A light metal disc on the top of an electromagnet is thrown up as the current is switched on because when the current flows through the electromagnet, the magnetic flux through the disc increases which leads in setting up an eddy current in the disc in the same direction of the electromagnetic current: So the upper surface of electromagnet and the lower surface of the disc acquire same polarity. Since body with same polarity repels, so the disc is thrown up.

Question.9. In the circuit shown in the figure, identify the equivalent gate of the circuit and make its.truth table.

Question.13. A parallel beam of light of 600 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1.2 m away. It is observed that the first minimum is at a distance of 3 mm from the centre of the screen. Calculate the width of the slit.

Question.19. A wire AB is carrying a steady current of 10 A and is lying on the table. Another wire CD carrying 6 A is held directly above AB at a height of 2 mm. Find the mass per unit length of the wire CD so that it remains suspended at its position when left free. Give the direction of the current flowing in CD with respect to that in AB. [Take the value of g = 10 ms^-2]
Answer : Force per unit length between the current carrying wires is given as :

Question.23. Name the type of waves which are used for line of sight (LOS) communication. What is the range of their frequencies?
A transmitting antenna at the top of a tower has a height of 45 m and the height of the receiving antenna is 80 m. Calculate the maximum distance between them for satisfactory communication is LOS mode. (Radius of the Earth = 6.4 x 10⁶ m)
Answer : Space wave are used for the line of sight (LOS) communication.
The range of their frequencies 4 40 MHz and above.
We have,
Height of transmitting antenna, b_T= 45 m
Height of receiving antenna, b_R = 80 m

Question.24.

SET III

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

Question.1. Write the expression for the de Broglie wavelength associated with a charged particle having charge ‘q and mass m, when it is accelerated by a potential V.
Answer: The charged particle has a mass m and charge q. The kinetic energy of the particle is equal to the work done on it by the electric field.

Question.3. Two charges of magnitudes + 4Q and -Q are located at points (a, 0) and (-3a, 0) respectively. What is the electric flux due to these charges through a sphere of radius ‘”la with its centre at the origin’.
Answer : Gauss’s theorem states that the electric flux through a closed surface enclosing a charge is equal to 1/ε₀ times the magnitude of the charge enclosed.

Question.9. In the circuit shown in the figure, identify the equivalent gate of the circuit and make its truth table.

Question.11. A parallel beam of light of 450 nm Ms on a narrow slit and the resulting diffraction pattern is observed on a screen 1.5 m away. It is observed that the first minimum is at a distance of 3 mm from the centre of the screen. Calculate the width of the slit.

Question.17. A wire AB is carrying a steady current of 6 A and is lying on the table. Another wire CD carrying 4 A is held directly above AB at a height of 1 mm. Find the mass per unit length of the wire CD so that it remains suspended at its position when left free. Give the direction of the current flowing in CD with respect to that in AB. [Take the value of g = 10 ms^-2]
Answer : Force per unit length between the current carrying wife is given as :

Question.18. A light bulb is rated at 125 W for a 250 V a.c. supply. Calculate the resistance of the bulb.

Question.25. Name the type of waves which are used for line of sight (LOS) communication. What is the range of their frequencies?
A transmitting antenna at the top of a power has a height of 45 m and the receiving antenna is on the ground. Calculate the maximum distance between them for satisfactory communication in LOS mode. (Radius of the Earth = 6.4 x 10⁶m)
Answer : Space wave are used for the line of sight (LOS) communication.
The range of their frequencies is 40 MHz and above. We have height of transmitting antenna,