Tag: CBSE Sample Papers for Class 12

CBSE Previous Year Solved  Papers  Class 12 Physics Delhi 2015

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

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 μ_g lens < μ_m surroundings.
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.
Answer :

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 ?
Answer :

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 A^1/3, 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.
Answer :

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 ?
Answer:

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 X_C= X_L is inserted in series in the circuit. Justify your answer in each case.
Answer:

(i) When the number of turns in the inductor is reduced, its reactance X_L 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 X_L = ωL increases. This decreases the current in the circuit and the bulb glows dimmer.
(iii) With capacitor of reactance X_C=X_L, 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 µ₀ time 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.

CBSE Previous Year Solved  Papers  Class 12 Physics 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

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

Question.1. What are permanent magnets? Give one example.
Answer : The magnets which have high retentivity and high coercivity are known as permanent magnets. For example : Steel

Question.2. What is the geometrical shape of equipotential surface due to a single isolated charge?
Answer : The equipotential surfaces of an isolated charge are concentric spherical shells and the distance between the shells increase with the decrease in electric field and vice-versa.

Question.3. Which of the following waves can be polarized (i) Heat waves (ii) Sound waves? Give reason to support your answer.
Answer : Heat waves can be polarized as they are transverse waves whereas sound waves cannot be polarized as they are longitudinal waves.
Transverse waves can oscillate in the direction perpendicular to the direction of its transmission but longitudinal waves oscillate only along the direction of its transmission. So, longitudinal waves cannot be polarized.

Question.4. A capacitor has been charged by a dc source. What are the magnitude of conduction and displacement current, when it is fully charged?
Answer : Electric flux through plates of capacitor,

As voltage becomes constant when capacitor is fully charged.

Question.5. Write the relationship between angle of incidence ‘i’, angle of prism ‘A’ and angle of minimum deviations ∆m for a triangular prism.
Answer : The relation between the angle of incidence i, angle of prism A, and the angle of minimum deviation ∆m, for a triangular prism is given by i= (A+∆m)/2

Question.6. The given graph shows the variation of photo-electric current (I) versus applied voltage (V) for two different photosensitive materials and for two different intensities of the incident radiations. Identify the pairs of curves that correspond to different materials but same intensity of incident radiation.

Answer : Curves 1 and 2 correspond to similar materials while curves 3 and 4 represent different materials, since the value of stopping potential for 1, 2 and 3, 4 are the same. For the given frequency of the incident radiation, the stopping potential is independent of its intensity.
So, the pairs of curves (1 and 3) and (2 and 4) correspond to different materials but same intensity of incident radiation.

Question.7. A 10 V battery of negligible internal resistance is connected across a 200 V battery and a resistance of 38Ω as shown in the figure. Find the value of the current in circuit.

Answer : Since, the positive terminal of the batteries are connected together, so the equivalent emf of the batteries is given by
E = 200-10= 190 V
Hence, the current in the circuit is given by
I=E/R=190/38=5A

Question.8. The emf of a cell is always greater than its terminal voltage. Why? Give reason.
Answer : The emf of a cell is greater than its terminal voltage because there is some potential drop across the cell due to its small internal resistance.

Question.9. (a) Write the necessary conditions for the phenomenon of total internal reflection to occur.
(b) Write the relation between the refractive index and critical angle for a given pair of optical media.
Answer : (a) Necessary conditions for total internal reflection to occur are :
(i) The incident ray on the interface should travel in optically denser medium.
(ii) The angle of incidence should be greater than the critical angle for the given pair of optical media.

Where a and b are the rarer and denser media respectively. C is the critical angle for the given pair of optical media.

Question.10. State Lenzs law.
A metallic rod held horizontally along east-west direction, is allowed to fall under gravity. Will there be an emf induced at its ends? Justify your answer.
Answer : Lenz’s law states that the polarity of induced emf is such that it produces a current which opposes the change in magnetic flux that produces it.
Emf will be induced in the rod as there is change in magnetic flux.
When a metallic rod held horizontally along east-west direction, is allowed to fall freely under gravity i.e. fall from north to south, the magnetic flux changes and the emf is induced in it.

Question.11. A convex lens of focal length 25 m is placed co axially in contact with a concave lens of focal length 20 cm. Determine the power of the combination. Will the system be converging or diverging in nature?
Answer : We have focal length of convex lens,

The focal length of the combination = -1m = -100cm As the focal length is negative, the system will be diverging in nature.

Question.12. An ammeter of resistance 0.80 Ω can measure current up to 1.0 A.
(i) What must be the value of shunt resistance to enable the ammeter to measure current up to 5.0 A?
(ii) What is the combined resistance of the ammeter and the shunt?
Answer : We have, resistance of ammeter, R_A = 0.80 ohm
Maximum current across ammeter, I_A = 1.0 A.
So, voltage across ammeter, V= IR= 1 x 0.80 = 0.8 V
Let the value of shunt be x.

Question.13. In the given circuit diagram a voltmeter ‘V is connected across a lamp ‘L’. How would (i) the brightness of the lamp and (ii) voltmeter reading ‘V’ be affected, if the value of resistance ‘R’ is decreased? Justify your answer.

Answer : The given figure is Common Emitter (CE) configuration of an n-p-n transistor is shown in the figure. The input circuit is forward biased and collector circuit is reverse biased.
If resistance R decreases, forward biased in the input circuit will increase, thus the base current (I_b) will decrease and the emitter current (I_E) will increase. This will increase the collector current (I_C) as I_E = I_B + I_C.
When I_C increases which flows through the lamp, the voltage across the bulb will also increase making the lamp brighter and the voltmeter is-connected in parallel with the lamp, the reading in the voltmeter will also increase.

Question.14. (a) An EM wave is, travelling in a medium with a velocity \(v = v\hat { i } \)Draw a sketch showing the propagation of the EM wave, indicating the direction of the oscillating electric and magnetic fields.
(b) How are the magnitudes of the electric and magnetic fields related to velocity of the EM wave?

Question.15. Block diagram of a receiver is shown in the figure :
(a) Identify ‘X’ and ‘Y’
(b) Write their functions.

Answer : From the given block diagram of demodulator of a typical receiver, we can conclude the following things :
(a) X represents Intermediate Frequency (IF) stage while Y , represents an amplifier.
(b) At IF stage, the carrier frequency is transformed to a lower frequency then in this process, the modulated signal is detected. The function of amplifier is to amplify the detected signal which may not be strong enough to be made use of and hence is essential.

Question.16. Explain, with the help of a circuit diagram, the working of a photo diode. Write briefly how it is used to detect the optical signals.
OR
Mention the important consideration required while fabricating a p-n junction diode to be used as a Light Emitting Diode (LED). What should be the order of band gap of an LED if it is required to emit light in the visible range?
Answer : A junction diode made from light sensitive semi-conductor is called a photo diode.

An electrical device that is used to detect and convert light into an energy signal with the use of a photo detector is known as a photo diode. The light that falls on it controls the function of pn-junction. Suppose, the wavelength is such that the energy of a photon hc/λ is enough to break a valance
bond. There is an increase in number of charge carriers and hence the conductivity of the junction also increases. New hole-electron pairs are created when such light falls on the junction. If the junction is connected in a circuit, the intensity of the incident light controls the current in the circuit.
OR
1. The reverse breakdown voltage of LEDs are very low, which is around 5V. So enough care is to be taken while fabricating
a pn-junction diode such that the high reverse voltages do not occur across them.
2. There exist very small resistance to limit the current in LED. So, a resistor must be placed in series with the LED such that no damage is occurred to the LED.
The semiconductor used for fabrication of visible LEDs must at least have a band gap of 1.8 eV.

Question.17. Write three important factors which justify the need of modulating a message signal. Show diagrammatically how an amplitude modulated wave is obtained when a modulating signal is superimposed on a carrier wave.
Answer : Three important factors which justify the need of modulating a message signal:
(i) Size of antenna or aerial: For communication within the effective length of the antennas, the transmitting frequencies should be high, so modulation is required.
(ii) Effective power which is radiated by antenna : Since the power radiated from a linear antenna is inversely proportional to the square of the transmitting wavelength. As high powers are needed for good transmission, so higher frequency is required which can be achieved by modulation.
(iii) The interference of signals from different transmitters : To avoid the interference of the signals there is a need of high I frequency which can be achieved by the modulation.

Question.18. A capacitor of unknown capacitance is connected across a battery of Y volts. The charge stored in it is 360 μC. When potential across the capacitor is reduced by 120 V, the charge stored in it becomes 120 μC.
Calculate :
(i) The potential V and the unknown capacitance C.
(ii) What will be the charge stored in the capacitor, if the
voltage applied had increased by 120 V?
OR
A hollow cylindrical box of length 1 m and area of cross section 25 cm² is placed in a three-dimensional coordinate system as shown in the figure. The electric field in the region is given by over right arrow [/latex]{ E } =50x\hat { i } [/latex], where E is NC^-1 and x is in metres. Find
(i) Net flux through the cylinder.
(ii) Charge enclosed by the cylinder.

Question.19. (a) In a typical nuclear reaction, e.g.

although number of nucleons is conserved, yet energy is released. How? Explain.
(b) Show that nuclear density in a given nucleus is independent of mass number A.
Answer :
(a) In a nuclear reaction, the aggregate of the masses of the target nucleus (\(_{ 1 }^{ 2 }{ H }\)) and the bombarding particle may be greater or less than the aggregate of the masses of the product nucleus (\(_{ 3 }^{ 2 }{ He}\)) and the outgoing particle (\(_{ 0 }^{ 1 }{ n }\)) So from the law of conservation of mass-energy some energy (3.27 MeV) is evolved or involved in a nuclear reaction. This energy is called Q-value of the nuclear reaction.

Which shows that the density is independent of mass number A.

Question.20. (a) Why photoelectric effect cannot be explained on the basis of wave nature of light? Give reasons.
(b) Write the basic features of photon picture of electro-magnetic radiation on which Einstein’s photoelectric equation is based.
Answer : (a) Wave nature of radiation cannot explain the photoelectric effect because of:
(i) The immediate ejection of photo electrons.
(ii) The presence of threshold frequency for a metal surface.
(iii) The fact -that kinetic energy of the emitted electrons is independent of the intensity of light and depends upon its frequency.
Thus, the photoelectric effect cannot be explained on the basis of wave nature of light.
(b) Photon picture of electromagnetic radiation on which Einstein’s photoelectric equation is based on particle nature of light. Its basic features are :
(i) In interaction with matter, radiation behaves as if it is made up of particles called photons.

(iv) By increasing the intensity of light of given wavelength, there is only an increase in the number of photons per second crossing a given area, with each photon having the same energy. Thus, photon energy is independent of intensity of radiation.
(v) Photons are electrically neutral and are not deflected by electric and magnetic fields.
(vi) In a photon—particle collision (such as photon-electron collision), the total energy and total momentum are conserved. However, number of photons may not be observed.

Question.21. A metallic rod of length T is rotated with a frequency v with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius r, about an axis passing through the centre and perpendicular to the plane of the ring. A constant uniform magnetic field B parallel to the axis is present everywhere. Using Lorentz force, explain how emf is induced between the centre and the metallic ring and hence obtained the expression for it.
Answer : Suppose the length of the rod is greater than the radius of the circle and rod rotates anticlockwise and suppose the direction of electrons in the rod at any instant be along +y direction.
Suppose the direction of the magnetic field is along +z direction.

Thus, the direction of force on the electrons is along – x axis. So, the electrons will move towards the centre i.e., the fixed end of the rod. This movement of electrons will effect in current and thus it will generate an emf in the rod between the fixed end and the point touching the ring.

Question.22. Output characteristics of an n-p-n transistor in CE configuration is shown in the figure. Determine :
(i) dynamic output resistance
(ii) dc current gain and
(iii) at current gain at an operating point V_CE = 10 V, when I_B = 30 μA.

Question.23. Using Bohr’s postulates, obtain the expression for the total energy of the electron in the stationary states of the hydrogen atom. Hence draw the energy level diagram showing how the line spectra corresponding to Balmer series occur due to transition between energy levels.
Answer: According to Bohr’s postulates, in a hydrogen atom, a single electron revolves around a nucleus of charge +e. For an electron moving with a uniform speed in a circular orbit of a given radius, the centripetal force is provided by Coulomb force of attraction between the electron and the nucleus. The gravitational attraction may be neglected as the mass of electron and proton is very small. So,


When the electron in a hydrogen atom jumps from higher energy level to the lower energy level, the difference of energies of the two energy levels is emitted as a radiation of particular wavelength. It is called a spectral line.

Question.24. (a) In what way is diffraction from each slit related to the interference pattern in a double slit experiment.
(b) Two wavelengths of sodium light 590 nm and 596 nm are used, in turn to study the diffraction taking place at a single slit of aperture 2 x 10^-4 m. The distance between the slit and the screen is 1.5 m. Calculate the separation between the positions of the first maxima of the diffraction pattern obtained in the two cases.
Answer : (a) If the width of each slit is comparable to the wavelength of light used, the interference pattern thus obtained in the double-slit experiment is modified by diffraction from each of the two slits.

Question.25. In a series LCR circuit connected to an ac source of variable frequency and voltage v = v_m sin ωt, draw a plot showing the variation of current (I) with angular frequency (ω) for two different values of resistance R₁ and R₂ (R₁ > R₂). Write the condition under which the phenomenon of resonance occurs. For which value of the resistance out of the two curves, a sharper resonance is produced? Define Q-factor of the circuit and give its significance.


We can observe that the current amplitude is maximum at the resonant frequency ω₀. Since i_m = V_m/R at resonance, the current amplitude for case R₂ is sharper to that for case R₁.
Quality factor or simply the Q-factor of a resonant LCR circuit is defined as the ratio of voltage drop across the capacitor (or inductor) to that of applied voltage.

The Q factor determines the sharpness of the resonance curve and if the resonance is less sharp, the maximum current decreases and also the circuit is close to the resonance for a larger range Aoo of frequencies and the regulation of the circuit will not be good. So, less sharp the resonance, less is the selectivity of the circuit while higher is the Q, sharper is the resonance curve and lesser will be the loss in energy of the circuit.
When X_L = X_C or V_L = V_C, the LCR circuit is said to be in resonance condition.

Question.26. While travelling back to his residence in the car, Dr. Pathak was caught up in a thunderstorm. It became very dark. He stopped driving the car and waited for thunderstorm to stop. Suddenly he noticed a child walking alone on the road. He asked the boy to come inside the car till the thunderstorm stopped. Dr. Pathak dropped the boy at his residence. The boy insisted that Dr. Pathak should meet his parents. The parents expressed their gratitude to Dr. Pathak for his concern for safety of the child.
Answer the following questions based on the above information :
(a) Why is it safer to sit inside a car during a thunderstorm?
(b) Which two values are displayed by Dr. Pathak in his action?
(c) Which values are reflected in parents’ response to Dr. Pathak?
(d) Give an example of similar action on your part in the part from everyday life.
Answer: (a) It is safer to be inside a car during thunderstorm because the car acts like a Faraday cage.
(b) Awareness and humanity
(c) Gratitude and obliged
(d) Once I came across to a situation where a puppy was struck in the middle of a busy road during rain and was not able to cross due to heavy flow, so I quickly rushed and helped him.

Question.27. (a) Draw a ray diagram showing the image formation by a compound microscope. Hence obtain expression for total magnification when the image is formed at infinity.
(b) Distinguish between myopia and hypermetropia. Show diagrammatically how these defects can be corrected.
OR
(a) State Huygens principle. Using this principle draw a diagram to show how a plane wave front incident at the interface of the two media gets refracted when it propagates from a rarer to a denser medium. Hence verify Snell’s law of refraction.
(b) When monochromatic light travels from a rarer to a denser medium, explain the following, giving reasons :
(i) Is the frequency of reflected and refracted light same as the frequency of incident light?
(ii) Does the decrease in speed imply a reduction in the energy carried by light wave?
Answer : (a) A compound microscope consists of two convex lenses parallel separated by some distance. The lens nearer to the object is called the objective. The lens through which the final image is viewed is called the eyepiece.
Magnifying power, when final image is at infinity : The magnification produced by the compound microscope is the product of the magnifications produced by the eyepiece and objective.


(b) (i) Nearsightedness or Myopia s A person suffering from myopia can see only nearer objects clearly, but cannot see the objects beyond a certain distance clearly.
Myopic eye:

Correction : To correct the eye from this defect, a concave lens of appropriate focal length is positioned close to the eye so that the parallel ray of light from an object at infinity after refraction through the lens appears to come from the far point P’ of the myopic eye.

(ii) Farsightedness Or Hypermetropia : A person suffering from hypermetropia can see distant objects clearly, but cannot see nearer objects.

Correction : To correct this defect, a convex lens of suitable focal length is positioned close to the eye so that the rays of light from an object placed at the point N after refraction through the lens appear to come from the near point N’ of the hypermetropic eye.

OR
(a) Hyugens Principle:

• Each point on the primary wave front acts as a source of secondary wavelets,- transferring out disturbance in all directions in the same way as the original source of light does.
• The new position of the wave front at any instant is the envelope of the secondary wavelets at that instant. Refraction on the basis of wave theory
• Consider any point Q on the incident wave front.
• Suppose when disturbance from point P on incident wave
  front reaches point P’ on the refracted wave front, the disturbance from point Q reaches Q’ on the refracting surface XY.
  

Since P’ A’ represents the refracted wave front, the time taken by light to travel from a point on incident wave front to the corresponding point on refracted wave front should always be the same. Now, time taken by light to go from Q to Q’ will be

The rays from different points on the incident wave front will take the same time to reach the corresponding points on the refracted wave front i.e. given equation (iv) is independent of AK. It will happen so, if

This is the Snell’s law for refraction of light.
(b) (i) The frequency of reflected and refracted light remains constant as the frequency of incident light because frequency only depends on the source of light.
(ii) Since the frequency remains same, hence there is no reduction in energy.

Question.28. (a). State the working principle of a potentiometer. With the help of the circuit diagram, explain how a potentiometer is used to compare the emfs of two primary cells. Obtain the required expression used for comparing the emfs.
(b) Write two possible causes for one sided deflection in a potentiometer experiment.
OR
(a) State Kirchhoff’s rules for an electric network. Using Kirchhoff’s rules, obtain the balance condition in terms of the resistance of four arms of Wheatstone bridge.
(b) In the meter bridge experimental set up, shown in the figure, the null point ‘D’ is obtained at a distance of 40 cm from end A of the meter bridge wire. If a resistance of 10 Ω is connected in series with R1, null point is obtained at AD = 60 cm. Calculate the values of R₁ and R₂.
Answer : (a) Working principle of Potentiometer : When a constant current is passed through a wire of uniform area of cross-section, the potential drop across any portion of the wire is directly proportional to the length of that portion.
Applications of Potentiometer for comparing emfs of two cells : The following figure stows an application of the potentiometer to compare the emf of two cells of emf E₁ and E₂
E₁, E₂ are the emf of the two cells.
1, 2, 3 form a two way key.
When 1 and 3 are connected, E1 is connected to the galvanometer (G).
Jokey is moved to N1, which is at a distance Li from A, to find the balancing length.

E₁/E₂ =l₁/l₂
Thus, we can compare the emfs of any two sources. Generally, one of the cells is chosen as a standard cell whose emf is known to a high degree of accuracy. The emf of the other cell is then calculated from equation (3).
(b) (i) The emf of the cell connected in main circuit may not be more than the emf of the primary cells whose emfs are to be compared.
(ii) The positive ends of all cells are not connected to the same end of the wire.
OR
(a) Kirchhoff’s First Law – Junction Rule : The algebraic sum of the currents meeting at a point in an electrical circuit is always zero.

Kirchhoff’s Second Law – Loop Rule : In a closed loop, the algebraic sum of the emfs is equal to the algebraic sum of the products of the resistance and current flowing through them.

Wheatstone Bridge is an arrangement of four resistances as shown in the following figure.

This is the required balanced condition of Wheatstone Bridge, (b) Considering both the situations and writing them in the form of equations Let R’ be the resistance per unit length of the potential meter wire,

Question.29. (a) Derive the expression for the torque on a rectangular current carrying loop suspended in a uniform magnetic field. 
(b) A proton and a deuteron having equal momenta enter in a region of a uniform magnetic field at right angle to the direction of a the field. Depict their trajectories in the field.
OR
(a) A small compass needle of magnetic moment ‘tn is free to turn about an axis perpendicular to the direction of uniform magnetic field ‘B’. The moment of inertia of the needle about the axis is T. The needle is slightly disturbed from its stable position and then released. Prove that it executes simple harmonic motion. Hence deduce the expression for its time period.
(b) A compass needle, free to turn in a vertical plane orients itself with its axis vertical at a certain place on the earth. Find out the values of (i) horizontal component of earth s magnetic field and (ii) angle of dip at the place.
Answer: (a) Consider a rectangular loop-ABCD carrying current I.

Case I : The rectangular loop is placed such that the uniform magnetic field B is in the plane of loop.
No force is exerted by the magnetic field on the arms AD and BC. Magnetic field exerts a force F₁ on arm AB.


If there are V such turns the torque will be nlAB
Where, b —> Breadth of the rectangular coil
a —> Length of the rectangular coil
A = ab —> Area of the coil .
Case II: Plane of the loop is not along the magnetic field, but makes angle with it.

SET II

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

Question.1. A cell of emf‘E’ and internal distance V draws a current
‘I’. Write the relation between terminal voltage ‘V’ in terms of E, I, r.
Answer : When the current I draws from a cell of emf E and internal resistance r, then the terminal voltage is V = E – I_r

Question.2. Which of the following substances are diamagnetic?
Bi, Al, Na, Cu, Ca and Ni
Answer : Bi and Cu are diamagnetic substances.

Question.3. A heating element is marked 210 V, 630 W. What is the value of current drawn by the element when connected to a 210 V  dc source?
Answer: In dc source, P = VI
Given that P = 630 Wand V = 210 V
So, I=P/V =630/210 =3 A

Question.10. An ammeter of resistance 1 Ω can measure current up to 1.0 A. (i) What must be the value of the shunt resistance to enable the ammeter to measure up to 5.0 A ? (ii) What is the combination resistance of the ammeter and the shunt?

Question.14. A convex lens of focal length 20 cm is placed co axially in contact with a concave lens of focal length 25 cm. Determine the power of the combination. Will the system be converging or diverging in nature?

The focal length of the combination = 1 m = 100 cm As the focal length is positive, the system will be converging in nature.

Question.15. Using Bohr’s postulates, obtain the expressions for (i) kinetic energy and (ii) potential energy of the electron in stationary state of hydrogen atom.
Draw the energy level diagram showing how the transitions between energy levels result in the appearance of Lyman series.
Answer: According to Bohr’s postulates, in a hydrogen atom, a single electron revolves around a nucleus of charge +ke. For an electron moving with a uniform speed in a circular orbit or a given radius, the centripetal force is provided by Coulomb force of attraction between the electron and the nucleus. The gravitational attraction may be neglected as the mass of electron and proton is very small.

Question.22. Figure shows a rectangular loop conducting PQRS in which the arm PQ is free to move. A uniform magnetic field acts in the direction perpendicular to the plane of the loop. Arm PQ is moved with a velocity v towards the arm RrS Assuming that the arms QR, RS and SP have negligible resistances and the moving arm PQ has the resistance r, obtain the expression for (i) the current in the loop (ii) the force and (iii) the power required to move the arm PQ.

Question.23. Distinguish between ‘sky waves’ and ‘space waves’ modes of propagation in communication system.
(a) Why is sky wave mode propagation restricted to frequencies up to 40 MHz?
(b) Give two examples where space wave mode of propagation is used.
Answer : Sky wave : Sky waves are the AM radio waves, which are received after being reflected from the ionosphere. The propagation of radio wave signals from one point to another via reflection from ionosphere, is known as sky wave propagation. The sky wave propagation is an important consequence of the total internal reflection of radio waves. As we go higher in the ionosphere, there is an increase in the free electron density. Consequendy there is a decresise of refractive index. Thus, as a radio wave travels up in the ionosphere, it finds itself travelling from denser to rarer medium. It continuously bends away from its path till it suffers total internal reflection to reach back the Earth.
Space waves : Space waves are the waves which are used for satellite communication and line of sight path. The waves have frequencies up to 40 MHz provides essential communication and limited the line of sight paths.
(a) The e.m. waves of frequencies greater than 40 MHz penetrate the ionosphere and escape so, the sky wave propagation is restricted to the frequencies up to 40 MHz.
(b) In television broadcast and satellite communication, the space wave mode of propagation is used.

SET III

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

Question.6. A 5 V battery of negligible internal resistance is connected across a 200V battery and a resistance of 39 Ω as shown in the figure. Find the value of the current.

Question.7. Which of the following substances are para-magnetic? Bi, Al, Cu, Ca, Pb, Ni
Answer : Paramagnetic substances are Aluminium (Al) and Calcium (Ca).

Question.9. An ammeter of resistance 0.6 Q can measure current up to 1.0 A. Calculate (i) The shunt resistance required to enable the ammeter to measure current up to 5.0 A (ii) The combined resistance of the ammeter and the shunt.

Question.15. A convex lens of focal length 30 cm is placed coaxially in contact with a concave lens? of focal length 40 cm. Determine the power of the combination. Will the system be converging or diverging in naituire?

The focal length of combination = 1.2 m = 120 cm
As the focal length is positive the system will be converging in nature.

Question.18. A capacitor of unknown capacitance is connected across a battery of v volts. The charge stored in it is 300 μc. When potential across the capacitor is reduced by 100 V, the charge stored in it becomes 100 μC. Calculate the potential V and , the unknown capacitance. What will be the charge stored in the capacitor if the voltage applied had increased by 100 V?
OR
A hollow cylindrical box of length 0.5 m and area of cross-section 20 cm is placed in a three-dimensional coordinate system as shown in the figure. The electric field in the region is given by E = 20 x_i, where E is NC^-1 and x is in meters. Find
(i) Net flux through the cylinder
(ii) Charge enclosed in the cylinder.

Question.19. (a) Write two characteristic features distinguishing the diffraction pattern from the interference fringes obtained in Young’s double slit experiment.
(b) Two wavelength of sodium light 590 nm and 596 nm are used, in turn, to study the diffraction taking place due to a single slit of aperture 1 x 10^-4 m. The distance between the slit and the screen is 1.8 m. Calculate the separation between the positions of the first maxima of the diffraction pattern obtained in the two cases.
Answer: (a) Two characteristic features distinguishing the diffraction pattern from the interference fringes obtained in Young’s double slit experiment are :
(i) The interference fringes may or may not be of the same width whereas the fringes of diffraction pattern are always of varying width.
(ii) In interference the bright fringes are of same intensity whereas in diffraction patten the intensity falls as we go to successive maxima away from the centre, on either side.
(b) Wavelength of the light beam, λ₁ = 590 nm = 5.9 x 10^-7 m
Wavelength of another light beam, λ₂ = 596 nm = 5.96 x 10^-7 m
Distance of the slits from the screen = D = 1.8 m
Distance between the two slits =1 x 10^-4 m
For the first secondary maxima,

Question.21. (a) In a nuclear reaction

though the number of nucleons is conserved on both sides of the reaction, y et the energy is released. How? Explain.
(b) Draw a plot of potential energy between a pair of nucleons as a function of their separation. Mark the regions where potential energy is (i) positive and (ii) negative.
Answer : (a) ’In a nuclear reaction, the sum of the masses of the target nucleus \(_{ 2 }^{ 3 }{ He }\) may be greater or less the sum of the masses of tine product nucleus \(_{ 4 }^{ 2 }{ He }\) and the \(_{ 1 }^{ 1 }{ H }\) . So from the law of conservation of mass energy some energy (12.86 MeV) is evolved in nuclear reaction. This energy is called Q-value of the nuclear reaction. The binding energy of the nuclear reaction. The binding energy of the nucleus on the left side is not equal to the right side. The difference in the binding energies on two sides appearance energy released or absorbed in the nuclear reaction.
(b) Th e potential energy is minimum at ro : For distance larger than ro the negative potential energy goes on decreasing and for the distances less than ro the negative potential energy decrease to zero and then becomes positive and increases abruptly. Thus, A t o B is the positive potential energy region and B to C is the negative potential energy region.

Question.25. (b) What is the significance of negative sign in the expression for the energy?
(c) Draw the energy level diagram showing how the line spectra corresponding to Paschen series occur due to transition between energy levels.

Must Refer:

SAIL Pivot Point Calculator

CBSE Previous Year Solved  Papers  Class 12 Physics Delhi 2014

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

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

Question.1. Define the term ‘Mobility’ of charge carriers in a conductor. Write its S.I. unit.
Answer : Mobility of charge carriers in a conductor is defined as the magnitude of their drift velocity per unit applied electric field.

Question.2. The carrier wave is given by C(t) = 2 sin (ωt) volt. The modulating signal is a square wave as shown. Find modulation index.

Question.3. For any charge configuration, equipotential surface through a point is normal to the electric field. Justify.
Answer: If the electric field were not normal to equipotential surface, it would have non-zero component along the surface. To move a charge against this component, work would have to be done. But no work is needed to move a test charge on an equipotential surface. Hence electric field must be normal to the equipotential surface at every point.

Question.4. Two spherical bobs, one metallic and the other of glass, of the same size are allowed to fall freely from the same height above the ground. Which of the two would reach earlier and why?
Answer : Glass bob will reach the ground earlier than the metallic bob. As the metallic bob falls, it intercepts earth’s magnetic field and induced currents are set up in it which oppose its downward motion. But no such currents are induced in the glass.

Question.5. Show variation of resistivity of copper as a function of temperature in a graph.
Answer: The variation of resistivity of copper with temperature is parabolic in nature. This is shown in the following graph :

Question.6. A convex lens is placed in contact with a plane mirror. A point object at a distance of 20 cm on the axis of this combination has its image coinciding with itself. What is the focal length of the lens ?
Answer : The figure shows a convex lens L placed in contact with a plane mirror M. P is the point object, kept in front of

this combination at a distance of 20 cm, from it. As the image coincides with itself, the rays from the object, after refraction from lens, should fall normally on the mirror M, so that they retrace their path. For this, the rays from P, after refraction from the lens must from a parallel beam perpendicular to M. For clarity, M has been shown at a small distance from L (in diagram). As the rays from P, form a parallel beam after refraction, P must be at the focus of the lens. Hence the focal length of the lens is 20 cm.

Question.7.

Answer: The Lorentz magnetic force is given by the following relation:

Question.8. The figure given below-shows the block diagram of a generalized communication system. Identify the element labeled ‘X’ and write its function.

Answer : The element labeled ‘X’ is called ‘channel’. The function of the channel is to connect the transmitter and the receiver. A channel may either be wireless or in the form of wires connecting the transmitter and the receiver.

Question.9. Out of the two magnetic materials, ‘A’ has relative permeability slightly greater than unity while ‘B’ has less than unity. Identify the nature of the materials ‘A’ and ‘B’. Will their susceptibilities be positive or negative ?

Question.10.

Question.11. For a single slit of width a, the first minimum of the interference pattern of a monochromatic light of wavelength k occurs at an angle of k/a. At the same angle of k/a, we get a maximum for two narrow slits separated by a distance a. Explain.

Question.12. Write the truth table for the combination of the gates shown. Name the gates used.

Identify the logic gates marked ‘P’ and ‘Q’ in the given circuit. Write the truth table for the combination.

Question.13. State Kirchhoff’s rules. Explain briefly how these rules are justified.
Answer : Kirchhoff s first Law—Junction Rule In an electrical circuit, the algebraic sum of the currents meeting at a junction is always zero.

Convention : The current flowing towards the junction is taken as positive.
The current flowing away from the junction is taken as negative.

This law is based on the law of conservation of charge. KirchhofFs Second Law — Loop rule
In a closed loop, the algebraic sum of the emf ‘s is equal to the algebraic sum of the products of the resistances and the currents flowing through them.

Question.14. A capacitor ‘C’ a variable resistor ‘R’ and a bulb ‘B’ are connected in series to the ac mains in circuit as shown. The bulb glows with some brightness. How will the glow of the bulb change if (i) a dielectric slab is introduced between the plates of the capacitor, keeping resistance R to be the same; (ii) the resistance R is increased keeping the same capacitance?

Answer : (i) As the dielectric slab is introduced between the f plates of the capacitor, its capacitance will increase. Hence, the potential drop across the capacitor will decrease (V= Q/C). As a result, the potential drop across the bulb will increase (since both are connected in series’). So, its brightness will increase.
(ii) As the resistance (R) is increased, the potential drop across the resistor will increase. As a result, the potential drop across the bulb will decrease (since both are connected in series). So, its brightness will decrease.

Question.15. State the underlying principle of a cyclotron. Write briefly how this machine is used to accelerate charged particles to high energies.
Answer : The underlying principle of a cyclotron is that an oscillating electric field can be used to accelerate a charge particle to high energy.
A cyclotron involves the use of 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 with in the two D_s. An alternating voltage accelerates the charged particles each time they cross the D_s. The radius of each particles path increases with its speed. So, the accelerated particles spiral toward the outer wall of the cyclotron.

[The D_s are the semi-circular structures (D₁ and D₂) between which the charges move. The accelerating voltage is maintained across the opposite halves of the D_s.] Square wave electric fields are used to accelerate the charged particles in a cyclotron.

The accelerating electric field reverses just at the time the change particle finishes its half circle so that it gets accelerated across the gap between the D_s.
The particle gets accelerated again and again, and its velocity increases. Therefore, it attains high kinetic energy.
The positively charged ion adopts a circular path with a constant speed v, under the action of magnetic field B, which is perpendicular to the planes of D’s of radius r.
r=mv/qB

Question.16. An electric dipole of length 4 cm, when placed with its axis making an angle of 60° with a uniform electric field, experiences a torque of 4√3 Nm. Calculate the potential energy of the dipole, if it has charge ± 8 nC.

Question.17. A proton and a deuteron are accelerated through the same accelerating potential. Which one of the two has
(a) greater value of de-Broglie wavelength associated with it, and
(b) less momentum ?
Give reasons to justify your Answer.

Question.18. (i) Monochromatic light of frequency 6.0 x 10¹⁴ Hz is produced by a laser. The power emitted is 2.0 x 10^-3 W. Estimate the number of photons emitted per second on an average by the source.
(ii) Draw a plot showing the variation of photoelectric current versus the intensity of incident radiation on a given photosensitive

Question.19. A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. Up to which energy level the hydrogen atoms would be excited? Calculate the wavelength of the first member of Lyman and first member of Balmer series.

Question.20. When Sunita, a class XII student, came to know that her parents are planning to rent out the top floor of their house to a mobile company she protested. She tried hard to convince her patents that this move would be a health hazard. Ultimately her parents agreed :
(1) In what way can the setting up of transmission tower by a mobile company in a residential colony prove to be injurious to health ?
(2) By objecting to this move of her parents, what value did Sunita display?
(3) Estimate the range of e.m. waves which can be transmitted by an antenna of height 20 m. (Given radius of the earth = 6400 km)
Answer: (1) A transmitting tower makes use of electromagnetic waves such as microwaves, exposure to which can cause severe health hazards like, giddiness, headache, tumour and cancer. Also, the transmitting antenna operates on a very high power, so the risk of someone getting severely burnt in a residential area increases.
(2) By objecting to this move of her parents, Sunita has displayed awareness towards the health and environment of her society.
(3) Range of the transmitting antenna.

Question.21. A potentiometer wire of length 1 m has a resistance of 10Ω . It is connected to a 6 V battery in series with a resistance of 5 Ω . Determine the emf of the primary cell which gives a balance point at 40 cm.

Question.22. (a) Draw a labeled ray diagram showing the formation of a final image by a compound microscope at least distance of distinct vision. ’
(b) The total magnification produced by a compound microscope is 20. The magnification produced by the eye piece is 5. The microscope is focussed on a certain object. The distance between the objective and eye-piece is observed to be 14 cm. If least distance of distinct vision is 20 cm, calculate the focal length of the objective and the eye-piece.
Answer: (a)

Question.23. (a) A mobile phone lies along the principal axis of a concave mirror. Show, with the help of a suitable diagram, the formation of its image. Explain why magnification is ‘ not uniform.
(b) Suppose the lower half of the concave mirrors reflecting surface is covered with an opaque material. What effect this will have on the image of the object? Explain.
Answer: (a)

The image of the mobile phone formed by the concave mirror is shown in the above figure. The part of the mobile phone that is at C will form an image of the same size only at C. In the figure, we can see that B’C = BC. The part of the mobile phone that lies between C and F will form enlarged image beyond C as shown in the figure. It can be observed that the magnification of each part of the mobile phone cannot be uniform on account of different locations. That is why the image formed is not uniform.
(b) As the laws of reflection are true for all points of the mirror, the height of the whole image will be produced. However, as the area of the reflecting surface has been reduced, the image intensity will be reduced. In other words, the image produced will be less bright.

Question.24. (a) Obtain the expression for the energy stored per unit volume in a charged parallel plate capacitor.
(b) The electric field inside a parallel plate capacitor is E. Find the amount of work done in moving a charge q over a closed rectangular loop abcda.

OR
(a) Derive the expression for the capacitance of a parallel plate capacitor having plate area A and plate separation d.
(b) Two charged spherical conductors of radii R₁ and R₂ when connected by a conducting wire acquire charge q₁ and q₂ respectively. Find the ratio of their surface charge densities in terms of their radii.
Answer: (a) Let us consider a parallel-plate capacitor of plate area A. If seperation between plates is d metre, capacitance C in given by

Question.25. (a) State Ampere’s Circuital law, expressing it in the integral form.
(b) Two long coaxial insulated solenoids, S₁ and S₂ of equal lengths are wound one over the other as shown in the figure. A steady current “I” flow through the inner solenoid S₁ to the other end B, which is connected to the outer solenoid S₂ through which the same current “I” flows in the opposite direction so as to come out at end A. If n₁ and n₂ are the number of turns per unit length, find the magnitude and direction of the net magnetic field at a point (i) inside on the axis and (ii) outside the combined system.

Answer: (a) Ampere’s Circuital law states that the circulation of the resultant magnetic field along a closed, plane curve is equal to po times the total current crossing the area bounded by the closed curve, provided the electric field inside the loop remains constant.

Question.26. Answer the following :
(a) Name the em waves which are suitable for radar systems
used in aircraft navigation. Write the range of frequency of these waves.
(b) If the earth did not have atmosphere, would its average surface temperature be higher or lower than what it is now? Explain.
(c) An em wave exerts pressure on the surface on which it is incident. Justify.
Answer: (a) Microwaves are suitable for radar systems used in aircrafts navigation. The range of frequency for these waves is 10⁹ Hz to 10¹² Hz.
(b) In the absence of atmosphere, there would be no greenhouse effect on the surface of the Earth. As a result, the temperature of the Earth would decrease rapidly, making it difficult for human survival.
(c) An em wave carries a linear momentum with it. The linear momentum carried by a portion of wave having energy U is given by p= UC
Thus, if the wave incident on a material surface is completely absorbed, it delivers energy U and momentum p = U C to the surface. If the wave is totally reflected, the momentum delivered is p = 2U C because the momentum of the wave changes from p to -p. Therefore, it follows that an em waves incident on a surface exert a force and hence a pressure on the surface.

Question.27.

Question.28. (a) (i) ‘Two independent mono-chromatic sources of light cannot produce a sustained interference pattern. Give reason.
(ii) Light wave each of amplitude ‘a’ and frequency ‘ω ’, emanating from two coherent light sources superpose at a point. If the displacements due to these waves is given by y₁ = a cos ωt and y₂ = a cos (ωt + ᶲ) where ᶲ is the phase difference between the two, obtain the expression for the resultant intensity at the point.
(b) In Youngs double slit experiment, using monochromatic light of Wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. Find out the intensity of light at a point where path difference is λ/3.
OR
(a) How does one demonstrate, using a suitable diagram, that unpolarised light when passed through a Polaroid gets polarized?
(b) A beam of unpolarised light is incident on a glass-air interface. Show, using a suitable ray diagram, that light reflected from the interface is totally polarised, when µ = tan i_B, where µ is the refractive index of glass with respect to air and i_B is the Brewster’s angle.
Answer : (a) (i) The condition for the sustained interference is that both the sources must be coherent (i.e. they must have the same wavelength and the same frequency, and they must have the same phase or constant phase difference).
Two sources are monochromatic if they have the same frequency and wavelength. Since they are independent, i.e. they have different phases with irregular difference, they are not coherent sources.

The intensity of light is directly proportional to the square of the amplitude of the wave. The intensity of light at point on the screen is given by:

OR

The phenomenon of restricting the vibration of light (electric vector) in a particular direction perpendicular to the direction of the wave propagation is called polarization of light.
When unpolarised light is passed through a Polaroid, only those vibrations of light pass through the crystal, which are parallel to the axis of the crystal (AB). All other vibrations are absorbed and that is why intensity of the emerging light is reduced.
The plane ABCD in which the vibrations of the polarised light are confined is called the plane of vibration. The plane KLMN that is perpendicular to the plane of vibration is defined as the plane of polarization.
(b) When unpolarised light is incident on the glass-air interference at Brewster angle iB, then reflected light is totally polarised. This is called Brewsters Law.
When light is incident at Brewster angle, the reflected component OB and the refracted component OC are mutually perpendicular to each other.
From the figure,

Question.29. (a) Describe a simple experiment (or activity) to show that the polarity of emf induced in a coil is always such that it tends to produce a current which opposes the change of magnetic flux that produces it.
(b)The current flowing through an inductor of self inductance L is continuously increasing. Plot a graph showing the variation.
(i) Magnetic flux versus the current
(ii) Induced emf versus dl l dt
(iii) Magnetic potential energy stored versus the current.
OR
(a) Draw a schematic sketch of an ac generator describing its basic elements. State briefly its working principle. Show a plot of variation of
(i) Magnetic flux and
(ii) Alternating emf versus time generated by a loop of wire rotating in a magnetic field.
(b) Why is choke coil needed in the use of fluorescent tubes with ac mains?
Answer: (a) Lenz law: According to Lenz s law, the polarity of the induced emf is such that it opposes a change in magnetic flux responsible for its production.

When the north pole of a bar magnet is pushed towards the coil, the amount of magnetic flux linked with the coil ,increase. Current is reduced in the coil from a direction such that it opposes the increase in magnetic flux. This is possible only when the current induced in the coil is in anti-clockwise
direction, with respect to an pb server. The magnetic moment  M associated with this induced emf has north polarity, towards the north pole of the approaching bar magnet. Similarly, when the north pole of the bar magnet is moved away from the coil, the magnetic flux linked with the coil
decreases. To counter this decrease in magnetic flux, current is induced in the coil in clockwise direction so that its south pole faces the receding north pole of the bat magnet. This would result in an attractive force which opposes the motion of the magnet and the corresponding decrease in magnetic flux.

OR

(a) Principle is “Based on the phenomenon of electromagnetic induction.
Construction :

Main parts of an ac generator :

• Armature : The rectangular coil ABCD
• Field Magnets Two pole pieces of a strong electromagnet
• Slip Rings. The ends of the coil ABCD are connected to two hollow metallic rings R₁ and R₂.
• Brushes : B₁ and B₂ are two flexible metal plates or carbon rods. They are fixed and are kept -is tight contact with R₁ and R₂, respectively.

Working : As the armature coil is rotated in the magnetic field, angle 6 between the field and the normal to the coil changes continuously. Therefore, magnetic flux linked with the coil changes and an emf is induced in the coil. According to Flemings right hand rule, current is induced from A to B in AB and from C to D in CD. In the external circuit, current flows from B₂ to B₁. To calculate the magnitude of emf induced :


The graph between alternating emf versus time is shown below:

(b) A choke coil is an electrical appliance used for controlling current in an a.c. circuit. Therefore, if we use a resistance R for the same purpose, a lot of energy would be wasted in the form of heat etc.

Question.30. (a) State briefly the processes involved in the formation of p-n junction explaining clearly how the depletion region is formed.
(b) Using the necessary circuit diagram, show how the V-I characteristics of a p-n junction are obtained in
(i) Forward biasing
(ii) Reverse biasing
How are these characteristics made use of in rectification?
OR
(a) Differentiate between three segments of a transistor on the bias of their size and level of doping.
(b) How is a transistor biased to be in active state?
(c) With the help of necessary circuit diagram, describe briefly how n-p-n transistor in CE configuration amplifies a small sinusoidal input voltage. Write the expression for the ac current gain.
Answer : (a) As we know that n-type semi-conductor has more concentration of electrons than that of a hole and p-type semi-conductor has more concentration of holes than an electron. Due to the difference in concentration of charge carriers in the two regions of p-n junction, the holes diffuse from p-side to n-side and electrons diffuse from n-side to p-side.
When an electron diffuses from n to p, it leaves behind an ionized donor on n-side. The ionised donor (+ve charge) is immobile as it is bonded by the surrounding atoms. Therefore, a layer of positive charge is developed on the «-side of the junction. Similarly, a layer of negative charge is developed on the p-side.

Hence, a space-charge region is formed on both side of the junction, which has immobile ions and is devoid of any f charge carrier, called as depletion layer or depletion region.
(b) (i) p-n junction diode under forward bias

A p-n junction diode is said to be forward biased if the positive terminal of the external battery B is connected to p-side and the negative terminal to the n-side of p-n junction.
The applied voltage of battery mostly drops across the depletion region and the voltage drop across the p-side and n-side of the p-n junction in negligible small. The resistance of depletion region is very high as it has no free change carriers.
Electron in n-region moves towards the p-n junction and holes in the p-region move towards the junction. The width of the depletion layer decreases and hence, it offers less resistance. Diffusion of majority carriers takes place across the junction. This leads to the forward current.
The V-I characteristics of p-n junction is forward bias is shown below:

(ii) The p-n junction under reverse bias Positive terminal of battery is connected to n-side and negative terminal to p-side.
Reverse bias supports the potential barrier. Therefore, the barrier height increases and the width of depletion region also increases. Due to the majority carriers, there is no conduction across the junction. A few minority carriers cross the junction after being accelerated by high reverse bias voltage.

This constitutes a current that flows in opposite direction, which is called reverse current.
The V-I characteristics of p-n junction diode in reverse bias is shown on previous page :

p-n junction diode Is used as a half-wave rectifier. Its working is based on the fact that the resistance of p-n junction becomes low when forward biased and becomes high when reverse biased. These characteristics of diode is used in rectification.
OR
(a) Emitter (E) : It is the left hand side thick layer of the transistor, which is heavily doped.
Base (B) : It is the central thin layer of the transistor, which is lightly doped.

Collector (C) : It is the right hand side thick layer of the transistor, which is moderately doped.
(b) There are two conditions for a transistor to be into an active region.
(1)The input circuit should be forward biased by using a low voltage battery.
(2) The out put circuit should be reverse biased by using a high voltage battery.
(c) n-p-n transistor as an amplifier :
The operating point is fixed in the middle of its active region.

SET II

Note : Except for the following questions, all the remaining questions have been asked in Privious Sets.

Question.1. Define the term ‘electrical conductivity’ of a metallic wire. Write its S.I. unit.
Answer : The electric conductivity of a metallic wire is defined as the ratio of the current density to the electric field it creates. Electrical conductivity.

Question.2. The carrier wave is represented by C (t) = 5 sin (10 πt) volt . A modulating signal is a square wave as shown. Determine modulation index.
Answer: Modulation index p is the ratio of the amplitude of the modulating signal to the amplitude of the carrier wave.

Question.10. An electric dipole of length 2 cm, when placed with its axis making an angle of 60° with a uniform electric field, experiences a torque of 8√3 Nm. Calculate the potential energy of the dipole, if it has a charge of ±4 nC.

Question.15. A proton and an alpha particle are accelerated through the same potential. Which one of the two has (i) greater value of de-Broglie wavelength associated with it and (ii) less kinetic energy. Give reasons to justify your answer.
Answer: (i) de-Broglie wavelength of a particle depends upon its mass and charge for same accelerating potential such that

Question.16.

Question.20. A 12.9 eV beam of electronic is used to bombard gaseous hydrogen at room temperature. Up to which energy level the hydrogen atoms would be excited? Calculate the wavelength of the first member of Paschen series and first member of Balmer series.

Question.22. Answer the following:
(a) Name the em waves which are used for the treatment of certain forms of cancer. Write their frequency range.
(b) Thin ozone layer on top of stratosphere is crucial for human survival. Why?
(c) Why is the amount of the momentum transferred by the em waves incident on the surface so small?
Answer: (a) X-ray, Gamma (y) rays are used for the treatment of certain forms of cancer. Their frequency range is 10¹⁸m to 10²²m.
(b) The thin ozone layer on top of stratosphere absorb most of the harmful ultraviolet rays coming from the Sun towards the Earth. They include UVA, UVB and UVC radiations, which can destroy the life system on the Earth. Hence, this layer is crucial for human survival.
(c) Momentum transferred = Energy Speed of light = hvc = 10^-22 (for v – 10²⁰ Hz)
Thus, the amount of the momentum transferred by the em waves incident on the surface is very small.

Question.24. A potentiometer wire of length 1.0 m has a resistance of 15 Ω. It is connected.to a 5V battery in series with a resistance of 5 Ω. Determine the emf of the primary cell which gives a balance point at 60 cm.

SET III

Note : Except for the following questions, all the remaining questions have been asked in Set—I and Privious Set.

Question.1. Define the term ‘drift velocity’ of charge carriers in a conductor and write its relationship with the current flowing through it.
Answer : The net speed achieved by an electron due to a current carrying conductor is called as drift velocity.
The average velocity acquired by the tree electrons along the length of a metallic conductor under a potential differnce applied across the conductor in called drift velocity of the electrons.

Here :
I is the current following through the conductor.
n is the number density of an electron.
A is the area of the conductor.
e is the charge of the electron.

Question.2. The carrier, wave of a signal is given by C(t) = 3 sin (8πt) volt. The modulating signal is a square wave as shown. Find its modulation index.

Question.4. Plot a graph showing variation of current versus voltage for the material Ga.
Answer : Current—Voltage characteristics graph for Ga :

Question.9. An electric dipole of length 2 cm, when placed with its axis making an angle of 60° with a uniform electric field, experiences a torque of 6 √3 Nm. Calculate the potential energy of the dipole, if it has a charge of ± 2 nC.

Question.12. A deuteron and an alpha particle are accelerated with the same accelerating potential. Which one of the two has
(1) greater value of de-Broglie wavelength, associated with it and
(2) less kinetic energy? Explain.
Answer: (1) de-Broglie wavelength of a particle is dependent on its mass and charge for same accelerating potential, such that

Charge of a deuteron is less as compared to an alpha particle. So, deuteron will have less of K.E.

Question.15.

Question.20. A 12.3 eV electron beam is used to bombard gaseous hydrogen at room temperature. Up to which energy level the hydrogen atoms would be excited? Calculate the wavelength of the second member of Lyman series and second member of Balmer series.

24. Answer the following :
(a) Name the em waves which are used for the treatment of certain forms of cancer. Write their frequency range.
(b) Welders wear special glass goggles while working. Why? Explain.
(c) Why are infrared waves often called as heat waves? Give their one application.
Answer:
(a) Gamma rays are used for the treatment of certain forms of cancer. There frequency range.
(b) Welders wear special glass googles while working so that they can protect their eyes from harmful electromagnetic radiation.
(c) Infrared waves are often called as heat waves because they induce resonance in molecules and increase internal energy in a substance.
Infrared waves are used in burglar alarms, security lights and remote controls for television and DVD players.