CBSE Previous Year Solved Papers Class 12 Physics Delhi 2016
Time allowed : 3 hours Maximum Marks: 70
- All questions are compulsory. There are 26
questions in all.
- This question paper has five sections : Section A, Section B, Section C, Section D and Section E.
- 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.
- 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.
- You may use the following values of physical constants wherever necessary:
Question.1. A point charge +Q is placed at point O as show in the figure. Is the potential difference VA — VB positive, negative or zero ?
Question.2. How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased ?
Answer : According to Gauss’s law
Flux depends only on the charge enclosed.
Hence, the electric flux remains constant.
Question.3. Write the underlying principle of a moving coil galvanometer.
Answer : When a current carrying coil is placed in magnetic field then it experiences a torque.
Question.4. Why are microwaves considered suitable for radar systems ,used in aircraft navigation ?
Answer : Microwaves of frequency 1 GHz to 300 GHz bounces from even the smallest aircraft so that they are suitable to avoid getting bombed. Microwaves can penetrate through clouds also.
Question.5. Define ‘quality factor ‘ of resonance in series LCR circuit. What is its SI unit ?
Answer : The Q factor of series resonance circuit is defined as the ratio of the voltage developed across the inductor or capacitor at resonance to the impressed voltage, which is the voltage across R.
It is dimensionless hence, it has no units.
Question.6. Explain the terms (i) Attenuation and (ii) Demodulation used in Communication System.
Answer: (i) Attenuation: The loss of strength of a signal while . propagating through a medium is known as attenuation.
(ii) Demodulation : The process of retrieval of information from the carrier wave at the receiver is termed demodulation. This is the reverse process of modulation.
Question.8. A nucleus with mass number A = 240 and BE/A = 7.6 MeV breaks into two fragments each of A = 120 with BE/A = 8.5 MeV Calculated the released energy.
Answer : Gain in binding energy for nucleon is about 0.9 MeV.
Question.9. Two cell of emfs 1.5 V and 2.0 V having internal resistances 0.2 Ω and 0.3 Ω respectively are connected in parallel. Calculate the emf and internal resistance of the equivalent cell.
Question.10. State Brewster’s law.
The value of Brewster angle for a transparent medium is different for light of different colour. Give reason.
Answer : Brewster’s law : The law states that the tangent of the polarising angle of incidence of a transparent medium is equal to its refractive index. The light incident at this angle. When reflects back is perfectly polarised.
M = tan ip
The refractive index of a material depends on the colour or wavelength of light. As the polarising angle depends on refractive index (M = tan ip), so it also depends on wavelength of light.
SECTION – C
Question.11. A charge is distributed uniformly over a ring of radius ‘a’. Obtain an expression for the electric intensity E at a point on the axis of the ring. Hence show that for points at large distances from the ring, it behaves like a point charge.
Answer : Suppose we have a ring of radius a that carries a uniformly distributed positive charge q.
(i) The axial components dE cos θ and
(ii) The perpendicular component dE sin θ.
Since the perpendicular component of any two diametrically opposite elements are equal and opposite, they cancel out in pairs. Only the axial components will add up to produce the resultant field.
This expression is similar to electric field due to point charge.
Question.12. Write three characteristic features in photoelectric effect which cannot be explained on the basis of wave theory of light, but can be explained only using Einstein’s equation.
Answer: (i) Existence of threshold frequency: According to wave theory, there should not exist any threshold frequency but Einstein’s theory explains the existence of Threshold frequency.
(ii) Dependence of kinetic energy on frequency of incident light: According to wave theory, the maximum kinetic energy of emitted electrons should depend on intensity of incident light and not on frequency whereas Einstein’s equation explains that it depends on frequency and not on intensity.
(iii) Instantaneous emission of electrons : According to wave theory there should be time lag between emission of electrons and incident of light whereas Einstein’s equation explains why there is no time lag between incident of light and emission of electrons.
Question.13. (a) Write the expression for the magnetic force acting on a charged particle moving with velocity v in the presence of magnetic field B.
(b) A neutron, an electron and an alpha particle moving with equal velocities, enter a uniform magnetic field going into the plane of the paper as shown. Trace their paths in the field and justify your
Question.14. (i) Define mutual inductance.
(ii) A pair of adjacent coils has a mutual inductance of 1.3 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil ?
Answer : (i) Mutual induction is the phenomenon of production of induced emf in one coil due to change of current in the neighbouring coil. The coil in which the current changes is called primary coil and the coil in which emf is induced is called the secondary coil.
Question.15. Two parallel plate capacitors X and Y have the same area of plates and same separation between them. X has air between the plates while Y contains a dielectric medium of εr= 4.
(i) Calculate capacitance of each capacitor if equivalent capacitance of the combination is 4 µF.
(ii) Calculate the potential difference between the plates of X and Y.
(iii) Estimate the ratio of electrostatic energy stored in X and Y
Question.16. Two long straight parallel conductors carry steady current I1 and I2 separated by a distance d. If the currents are flowing in the same direction, show how the magnetic field set up in one produces an attractive force on the other. Obtain the expression for this force. Hence define one ampere.
Question.17. How are e.m. waves produced by oscillating charges ?
Draw a sketch of linearly polarized e.m. waves propagating in the Z-direction. Indicate the directions of the oscillating electric and magnetic fields.
Answer: These waves are constituted by varying or oscillating electric and magnetic fields. The electric and magnetic fields are perpendicular to each other and are also perpendicular to the direction of propagation of the wave E is the envelope of electric intensity vector and B is the envelope of magnetic intensity vector.
Correction in Amperes Circuital law (Modified Ampere’s law) : Maxwell removed the problem of current continuity and inconsistency observed in Ampere’s Circuital law by introducing the concept of displacement current, Displacement current arises due to change in electric flux
Conduction current is because of How of charges but displacement current is not because of How of charges but because of change in electric flux.
Question.18. (a) Explain any two factors which justify the need of modulating a low frequency signal.
(b) Write two advantages of frequency modulation over amplitude modulation.
Question.19. (i) Write the functions of three segments of a transistor.
(ii) Draw the circuit diagram for studying the input and output characteristics of n-p-n transistor in common emitter configuration. Using the circuit, explain how input, output characteristics are obtained.
Answer : (i) Three segments of transistor are :
(i) Emitter (ii) Base
Emitter : It is of moderate size and heavily doped, it supplies a large number of majority carriers which flow through the transistor.
Base : It is very thin and lightly doped and it separates emitter and collector region of transistor and controls the flow of charge carriers.
Collector : This segment is moderately doped and larger in size as compared to emitter. It collects a major portion of majority carriers supplied by the emitter.
Question.20. (a) Calculate the distance of an object of height h from a concave mirror of radius of curvature 20 cm, so as to obtain a real image of magnification 2. Find the location of image also,
(b) Using mirror formula, explain why does a convex mirror always produce a virtual image.
Question.21. (i) State Bohr’s quantization condition for defining stationary orbits. How does de-Broglie hypothesis explain the stationary orbits ?
(ii) Find the relation between the three wavelengths λ1, λ2 and λ3 from the energy level diagram shown below.
Question.22. Draw a schematic ray diagram of reflecting telescope showing how rays coming from a distant object are received at the eyepiece. Write its two important advantages over a refracting telescope.
Answer : Reflecting Telescope : The reflecting telescope makes us of a concave mirror as objective. The rays of light coming from distant object are incident on the objective (parabolic reflective). After reflection the rays of light meet at a point where another convex mirror is placed. This mirror focusses light inside the telescope tube. The final image is seen through the eye piece. The images produced by the reflecting telescope is very bright and its resolving power is high.
(i) The resolving power (the ability to observe two object distinctly) is high, due to the large diameter of the objective.
(ii) There is no chromatic aberration as the objective is a mirror.
SECTION – D
Question.23. Meeta’s father was driving her to the school. At the traffic signal she noticed that each traffic light was made of many tiny lights instead of a single bulb. When Meeta asked this question to her father, he explained the reason for this.
Answer the following questions based on above information:
(i) What were the values displayed by Meeta and her father ?
(ii) What answer did Meeta’s father give ?
(iii) What are the tiny lights in traffic signals called and how do these operate ?
Answer : (i) Awareness for energy conservation, power saving and knowledge about traffic rules.
(ii) Meeta’s father said that these are LED light which consume less power and high reliability.
(iii) The tiny lights in traffic signals are Light Emitting Diode. These are operated by connecting the P-N junction diode in forward biased condition.
SECTION – E
Question.24. (i) An a. c. source of voltage V = V0 sin tor is connected to a series combination of L, C and R. Use the phasor diagram to obtain expressions for impedance of the circuit and phase angle between voltage and current. Find the condition when current will be in phase with the voltage. What is the circuit in this condition called ? (ii) In a series LR circuit XL = R and power factor of the circuit is P1. When capacitor with capacitance C such that XL= XC is put in series, the power factor becomes P2. Calculate P1/P2.
(i) Write the function of a transformer. State is principle of working with the help of a diagram. Mention various energy losses in this device.
(ii) The primary coil of an ideal step up transformer has 100
turns and transformation ratio is also 100. The input voltage and power are respectively 220V and 1100 W Calculate:
(a) number of turns in secondary.
(b) current in primary.
(c) voltage across secondary.
(d) current in secondary.
(e) power in secondary.
Answer : (i) Let a, series LCR circuit is connected to an ac source V (Fig). We take the voltage of the source to be V = V0 sin ωt.
(i) A transformer is an electrical device for converting an alternating current at low voltage into that at high voltage or vice-versa.
- If it increases the input ac voltage, it is called step up transformer.
- If it decreases the input ac voltage, it is called step down transformer.
Principle : It works on the principle of mutual induction i.e., When a changing current is passed through one of the tow inductively coupled coils, an induced emf is set up in the other coil.
Working Theory : As the AC flows through the primary, it generate an alternating Magnetic flux in the core which passes through the secondary coil.
Let N1 = No. of turns in primary coils
N2 = No. of turns in secondary coils
This changing flux set up an induced emf in the secondary,/ also a self induced emf in the primary.
If there is no leakage of magnetic flux, then flux linked with each turn of the primary will be equal to that linked with each of the secondary. According to Faraday’s law of induction.
- Copper loss : Some energy is lost due to the heating of copper wires used in the primary and secondary winding’s. This power less (P = 12R) can be minimised by using thick copper wires of low resistance.
- Eddy current loss : The alternating magnetic flux induces eddy current in the iron core which leads to some energy loss in the form of heat. This loss can be reduced by using laminated iron core.
- Hysteresis loss : The alternating current carries the iron core through cycles of magnetisation and demagnetisation. Work done in each of these cycles and is lost as heat. This is called hysteresis loss and can be magnetised by using c.ore material having narrow hysteresis loop.
- Flux leakage: The magnetic flux produced by the primary may not fully pass through the secondary. Some of the flux may leak into air. This loss can be minimised by winding the primary and secondary coils over one another.
Question.25. (i)In Young’s double slit experiment, deduce the condition for (a) constructive, and (b) destructive interference at a point on the screen. Draw a graph showing variation of intensity in the interference pattern against position V on the screen.
(ii) Compare the interference pattern observed in Young’s double slit experiment with single slit diffraction pattern, pointing out three distinguishing features.
(i) Plot a graph to show variation of the angle of deviation as a function of angle of incidence for light passing through a prism. Derive ah expression for refractive index of the prism in terms of angle of minimum deviation and angle of prism.
(ii) What is dispersion of light ? What is its cause ?
(iii) A ray of light incident normally on one face of a right isosceles prism is totally reflected as shown in fig. What must be the minimum value of refractive index of glass? Give relevant calculations.
(ii) Compose of interference pattern observed in Young’s double slits and the single slits diffraction :
(i) It is a transparent medium having two planes and non-parallel refracting surface inclined to each other and three surfaces are not participating in refraction.
(ii) Dispersion of light : These colours are often observed as light passes Through a triangular prism upon passing through the prism, the white light is separated into its component colours : red, orange, yellow, green, blue, and violet. The separation of visible light into its different colours is known as dispersion. Dispersion occurs because for different colour of light a transparent medium will have different refractive indices (µ).
Question.26. (i) Define the term drift velocity.
(ii) On the basis of electron drift, derive an expression for resistivity of a conductor in terms of number density of free electrons and relaxation time. On what factors does resistivity of a conductor depend ?
(iii) Why alloys like constantan and manganin are used for making standard resistors ?
(i) State the principle of working of a potentiometer.
(ii) In the following potentiometer circuit AB is a uniform wire of length 1 m and resistance 10 Ω. Calculate the potential gradient along the wire and balance length AO (=l).
Answer : (i) Drift velocity is defined as the average velocity which the electrons are drifted towards the positive terminal under the effect of applied electric field. Thermal velocities are randomly distributed and average thermal velocity is zero.
Where P is the specific resistance or resistivity of the material of the wire. It depends on number of free electron per unit volume and temperature.
(iii) They are used to make standard resistors because :
(a) They have high value of resistivity
(b) Temperature coefficient of resistance is less.
(c) They are least affected by temperature.
(i) Principle of potentiometer : The basic principle of potentio meter is that when a constant current flows through a wire of uniform cross-section area and the composition of the potential drop across any length of the wire is directly proportional to that length.
A potentiometer is a device used to measure an unknown emf or potential difference and internal resistance of a cell accurately.
- A potentiometer consists of a long uniform cross-section of wire generally made of manganin or constantan.
- Usually, 1 m long separate pieces of wire are fixed on a wooden board parallel to each other.
- The wire are joined in series by thick copper strips.
- The ends A and B are connected to a battery (called driving all), a plug key and rheostat.
- A jockey J is provided with the help of which contact can be made at any point on the wire.
- This circuit sends a constant current I through the wire AB.
Principle : When a constant current flows through a wire of uniform cross sectional area and composition the potential drop across any length of the wire is directly proportional to that length.
Let V be the potential difference across the portion of the wire of length l whose resistance is R
By Ohm’s law,