## CBSE Previous Year Solved  Papers  Class 12 Physics Delhi 2015

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:

### SET I

SECTION-A

Question.1.Define capacitor reactance. Write its S.I. units.
Answer : Capacitor reactance is the resistance offered by a capacitor to the flow of a.c. It is given by

Question.2. What is the electric flux through a cube of side 1 cm which encloses an electric dipole ?
Answer : The electric flux through a cube of side 1 cm which encloses an electric dipole will be zero, as net charge enclosed by a cube is zero.

Question.3. A concave lens of refractive index 1.5 is immersed in a medium of refractive index 1.65. What is the nature of the lens ?
Answer : Since μg lens < μsurroundings.
It behaves like a converging lens.

Question. 4. How are side bands produced ?
Answer : Side bands are produced during the process of modulation. During modulation the audio frequency modulating signal wave is superimposed on a high frequency wave called carrier wave. Any form of modulation produces , frequencies that are the sum and difference of the carrier and modulating frequencies. These frequencies are called as side bands.

Question.5. Graph showing the variation of current versus voltage for a material GaAs is shown in the figure, Identify the region of:
(i) negative resistance
(ii) where Ohms law is obeyed.

Answer : (i) DE is the region of negative resistance because the slope of curve in this part is negative.
(ii) BC is the region where Ohms law is obeyed because in this part, the current varies linearly with the voltage.

SECTION – B

Question.6. A proton and an α -particle have the same deBroglie wavelength. Determine the ratio of (i) their accelerating potentials (ii) their speeds.
Answer : (i) The deBroglie wavelength of a particle is given by

Question.7. Show that the radius of the orbit in hydrogen atom varies as n2. Where n is the principal quantum number of the atom.

Question.8. Distinguish between ‘intrinsic’ semiconductors.

Question.9. Use the mirror equation to show that an object placed between f and 2 f of a concave mirror produces a real image beyond 2f.
OR
Find an expression for intensity of transmitted light when a Polaroid sheet is rotated between two crossed Polaroids. In which position of the Polaroid sheet will the transmitted intensity be maximum ?

Question.10. Use Kirchhoff’s rules to obtain conditions for the balance condition in a Wheatstone bridge.
Answer : Let us consider a Wheatstone bridge arrangement as shown below:
Wheatstone bridge is a special bridge type circuit which consists of four resistances, a galvanometer and a battery. It is used to determine unknown resistance.
In figure four resistance P, Q, R and S are connected in the form of four arms of a quadrilateral. Let the current given by battery in the balanced position be I. This current on reaching point A is divided into two parts I1 and I2. As there is no current in galvanometer in balanced state, therefore, current in resistances P and Q is Ii and in resistances R and S it is I2.

SECTION-C

Question.11. Name the parts of the electromagnetic spectrum which is
(a) suitable for radar systems used in aircraft navigation
(b) used to treat muscular strain
(c) used as a diagnostic tool in medicine
Write in brief, how these waves can be produced.
Answer : (a) Microwaves are suitable for radar systems that are used in aircraft navigation.
These rays are produced by special vacuum tubes, namely -klystrons, magnetrons and Gunn diodes.
(b) Infrared waves are used to treat muscular strain.
These rays are produced by hot bodies and molecules.
(c) X-rays are used as a diagnostic tool in medicine.
These rays are produced when high energy electrons are stopped suddenly on a metal of high atomic number.

Question.12. (i) A giant refracting telescope has an objective lens of focal length 15 m. If an eye-piece of focal length 1.0 cm is used. What is the angular magnification of the telescope ? (ii) If this telescope is used to view the moon. What is the diameter of the image of the moon formed by the objective lens ? The diameter of the moon is 3.48 x 108 m and the radius of lunar orbit is 3.8 x 108 cm.

Question.13. Write Einstein’s photoelectric equation and mention which important features in photoelectric effect can be explained with the help of this equation.
The maximum kinetic energy of the photo electrons gets doubled when the wavelength of light incident on the surface changes from λ1 to λ2. Derive the expressions for the threshold wavelength λ0 and work function for the metal surface.

The above equations explains the following results.
(1) If v < v0, then the maximum kinetic energy is negative, which is impossible. Hence, photoelectric emission does not take place for the incident radiation below the threshold frequency. Thus, the photoelectric emission can take place if v>v0.
(2) The maximum kinetic energy of emitted photoelectrons is directly proportional to the frequency of the incident radiation. This means that maximum kinetic energy of photo electron depends only on the frequency of incident light.

Question.14. In the study of Geiger-Marsdon experiment on scattering of a-particles by a thin foil of gold, draw the trajectory of a-particles in the Coulomb field of target nucleus. Explain briefly how one gets the information on the size of the nucleus from this study.
From the relation R = R0 A1/3, where R0 is constant and A is the mass number of the nucleus, show that nuclear matter density is independent of A.
OR
Distinguish between nuclear fission and fusion. Show how in both these processes energy is released. Calculate the energy release in MeV in the deuterium-tritium fusion reaction :

From this experiment, the following is observed :
(1) Most of the alpha particles pass straight through the gold foil. It means that they do not suffer any collision with golf atoms.
(2) About one alpha particle in every 8000 alpha particles deflects by more than 90°.
As most of the alpha particles go undeflected and only a few get deflected, this shows that most of the space in an atom is empty and at the centre of the atom, there exists a nucleus. By the number of the alpha particles deflected, the information regarding size of the nucleus can be known.
If m is the average mass of a nucleon and R is the nuclear radius, then mass of nucleus = mA, where A is the mass number of the element.

Question.15. Draw a block diagram of a detector for AM signal and show, using necessary processes and the wave forms, how the original message signal is detected from the input AM wave.

When a message is received, it gets attenuated through the channel therefore, receiving antenna is to be followed by an amplifier and a detector. The camera frequency is usually changed to a lower frequency in an Intermediate Frequency (IF) stage. The detected signal may not be strong enough to be made use of and hence is required to be amplified.
In order to obtain the original message signal m(t) of angular frequency a simple method is used which is shown below in the form of a block diagram.

When the received modulated signal is passed through a rectifier, an envelope signal is produced. This envelope signal is the message signal. In order to retrieve the message, the signal is passed through an envelope detector.

Question.16. A cell of emf ‘E’ and internal resistance ‘r’ is connected across a variable load resistor R. Draw the plots of the terminal voltage V versus (i) R and (ii) the current i. It is found that when R = 4Ω, the current is 1A when R is increased to 9Ω, the current reduces to 0.5 A. Find the values of the emf E and internal resistance r.

Question.17. Two capacitors of unknown capacitance C1 and C2 are connected first in series and then in parallel across a battery of 100 V. If the energy stored in the two combinations is 0.045 J and 0.25 J respectively, determine the value of C1 and C2. Also calculate the charge on each capacitor in parallel combination.

Question.18. State the principle of working of a galvanometer. A galvanometer of resistance G is converted into a voltmeter to measure up to V volts by connecting a resistance R1 in series with the coil. If a resistance R2 is connected in series with it, then it can measure up to V/2 volts. Find the resistance, in terms of R1 and R2, required to be connected to convert it into a voltmeter that can read up to 2 V. Also find the resistance G of the galvanometer in terms of R1 and R2.
Answer : Principle : When a current-carrying coil is placed in a magnetic field, it experiences a torque. From the measurement of the deflection of the coil, the strength of the current can be computed. A high resistance is connected in series with the galvanometer to convert it into voltmeter. The value of the resistance is given by

Question.19. With what considerations in view, a photo diode is fabricated ? State its working with the help of a suitable diagram.Even though the current in the forward bias is known to be more than in the reverse bias, yet the photo diode works in reverse bias. What is the reason ?
Answer: A photo diode is used to observe the change in current with change in the light intensity under reverse bias condition.
In fabrication of photo diode, material chosen should have band gap —1.5 eV or lower so that solar conversion efficiency is better. This is the reason to choose Si or GaAs material.
Working : It is a p-n junction fabricated with a transparent window to allow light photons to fall on it. These photons generate electron hole pairs upon absorption. If the junction is reverse biased using an electrical circuit, these electron hole pair move in opposite directions so as to produce current in the circuit. This current is very small and is detected by the micro ammeter placed in the circuit.

A photo diode is preferably operated in reverse bias condition. Consider an n- type semiconductor. Its majority carrier (electron) density is much larger than the minority hole density i.e. n < < p. When illuminated with light, both types of carriers increase equally in number

That is, the fractional increase in majority carries is much less than the fractional increase in minority carriers. Consequently, the fractional change due to the photo-effects on the minority carrier dominated reverse bias current is more easily measurable than the fractional change in the majority carrier dominated forward bias current. Hence, photo diodes are preferable used in the reverse bias condition for measuring light intensity.

Question.20. Draw a circuit diagram of a transistor amplifier in CE configuration.Define the terms :
(i) Input resistance and (ii) Current amplification factor. How are these determined using typical input and output characteristics ?

Question.21. Answer the following questions :
(a) In a double slit experiment using light of wavelength 600 nm, the angular width of the fringe formed on a distant screen is 0.1°. Find the spacing between the two slits. Light of wavelength 5000 A propagating in air gets partly reflected from the surface of water. How will the wavelengths and frequencies of the reflected and refracted light be affected ?

Question.22. An inductor L of inductance XL is connected in series with a bulb B and an ac source. How would brightness of the bulb change when (i) number of turn in the inductor is reduced, (ii) an iron rod is inserted in the inductor and (iii) a capacitor of reactance XC= XL is inserted in series in the circuit. Justify your answer in each case.

(i) When the number of turns in the inductor is reduced, its reactance XL decreases. The current in the circuit increases and hence brightness of the bulb increases.
(ii) When an iron rod is inserted in the inductor, the self inductance increases. Consequently, the inductive reactance XL = ωL increases. This decreases the current in the circuit and the bulb glows dimmer.
(iii) With capacitor of reactance XC=XL, the impedance

becomes minimum, the current in circuit becomes maximum. Hence the bulb glows with maximum brightness.

SECTION – D

Question.23. A group of students while coming from the school noticed a box marked “Danger H.T. 2200 V” at a substation in the main street. They did not understand the utility of a such a high voltage, while they argued the supply was only 220 V. They asked their teacher this question the next day. The . teacher thought it to be an important question and therefore, explained to the whole class.
Answer the following questions :
(i) What device is used to bring the high voltage down to low voltage of a.c. current and what is the principle of its working ?
(ii) Is it possible to use this device for bringing down the high dc voltage to the low voltage ? Explain.
(iii) Write the values displayed by the students and the teacher.
Answer : (i) The device that is used to bring high voltage down to low voltage of an a.c. current is a transformer. It works on the principle of mutual induction of two windings or circuits. When current in one circuit changes, emf is induced in the neighbouring circuit.
(ii) The transformer cannot convert d.c. voltages because it works on the principle of mutual induction. When the current linked with the primary coil changes the magnetic flux linked with the secondary coil also changes. This change in flux induces emf in the secondary coil. If we apply a direct current to the primary coil the current will remain constant. Thus, there is no mutual induction and hence no emf is induced.
(iii) The value of gaining knowledge and curiosity about learning new things is being displayed by the students. The value of providing good education and undertaking the doubts of students has been displayed by the teacher.

SECTION-E

Question.24. (a) State Ampere’s Circuital law. Use this law to obtain the expression for the magnetic field inside an air cored toroid of average radius, having ‘n’ turns per unit length and carrying a steady current I.
(b) An observer to the left of a solenoid of N turns each of cross section area A observes that a steady current I in it flows in the clockwise direction. Depict the magnetic field lines due to the solenoid specifying its polarity and show that it acts as a bar magnet of magnetic momentum M = NIA.

OR
(a) Define mutual inductance and write its S.I. units.
(b) Derive an expression for the mutual inductance of two long co-axial solenoids of same length wound one over the other.
(c) In an experiment two coils c1 and c2 are placed close to each other. Find out the expression for the emf induced in the coil c1 due to a change in the current through the coil c2.
Answer : (a) Amperes circuital law in electro magnetism is analogous to Gauss law in electrostatics. This law states that “The line integral of resultant magnetic field along a closed plane curve is equal to µtime the total current crossing the area bounded by the closed curve provided the electric field

Given :
r = Average radius of the toroid
I = Current through the solenoid .
n = Number of turns per unit length
To determine the magnetic field inside the toroid, we consider three amperian loops (loop 1, loop 2 and loop 3) as show in the figure below.

This is the expression for magnetic field inside air-cored toroid.
(b) Given that the current flows in the clockwise direction for an observer on the left side of the solenoid. This means that left face of the solenoid acts as south pole and right face acts as north pole. Inside a bar magnet the magnetic field lines are directed from south to north. Therefore, the magnetic field lines are directed from left to right in the solenoid.
Magnetic moment of single current carrying loop is given by
m = LA
where
I = Current flowing through the loop A = Area of the loop
So, Magnetic moment of the whole solenoid is given by
M = Nw’ = N(IA)
OR
(a) Mutual inductance is the property of two coils by the virtue of which each opposes any change in the value of current flowing through the other by developing an induced emf. The SI unit of mutual inductance is henry and its symbol is H.
(b) Consider two long solenoids S1 and S2 of same length l such that solenoid S2 surrounds solenoid S1 completely.

Question.25. (a) A point object ‘O’ is kept in a medium of refractive index n in front of a convex spherical surface of radius of curvature R which separate the second medium of refractive index n% from the first one as shown in the figure.
Draw the ray diagram showing the image formation and deduce the relationship between the object distance and the image distance in terms of n, n1 and R.

(b) When the image formed above acts as a virtual object for a concave spherical surface separating the medium n2 from n1 (n2 > n1) draw this ray diagram and write the similar (similar to (a)) relation. Hence obtain the expression for the Lens Makers formula.
Answer : (a) Let a spherical surface separate a rarer medium of refractive index n1 from second medium of refractive index n2. Let C be the centre of curvature and R = MC be the radius of the surface.

Consider a point object O lying on the principle axis of the surface. Let a ray starting from O incident normally on the surface along OM and pass straight. Let another ray of light incident on NM along ON and refract along NI.
From M, draw MN perpendicular to OI. ”
The above figure shows the geometry of formation of image I of an object O and the principal axis of a spherical surface with centre of curvature C. and radius of curvature R.
Let us make the following assumptions :
(i) The aperture of the surface is small as compared to the other distance involved.
(ii) Nil will be taken as nearly equal to the length of the perpendicular from the point N on the principal axis.

Question.26.

OR
(a) Explain, using suitable diagrams, the difference in the behaviour of a (i) conductor and (ii) dielectric in the presence of external electric field. Define the terms polarization of a dielectric and write its relation with susceptibility.
(b) A thin metallic spherical shell of radius carries a charge Q on its surface. A point charge Q/2 is placed at its centre C and another charge +2Q is placed outside the shell at a distance x from the centre as shown in the figure. Find (i) the force on the charge at the centre of shell and at the point A, (ii) the electric flux through the shell.

When external electric field is applied, dipoles are created (in case of non-polar dielectrics). The placement of dipoles is as shown in the given figure. An internal electric field is created which reduces the external electric field.
Polarization of dielectric (P) is defined as the dipole moment per unit volume of the polarized dielectric.

## One thought on “CBSE Previous Year Solved Papers Class 12 Physics Delhi 2015”

1. NITISH SINGh says:

Thanks

Thanks