CBSE Previous Year Solved Papers Class 12 Physics Delhi 2009
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 weightage. You have to attempt only one of the choices in such questions.
- You may use the following values of physical constants wherever necessary:
Note: Except for the following questions, all the remaining questions have been asked in Set-I and Set-II, 2009.
Question.1. What is sky wave propagation?
Answer : The type of propagation in which radio waves are 5. transmitted towards the sky and are reflected by the ionosphere towards the desired location on earth is called sky wave propagation.
Question.2.Write the following radiations in ascending order with respect to their frequencies :
X-rays, microwaves, UV rays and radio waves.
Answer : The given radiations can be arranged in ascending order with respect to their frequencies as :
Radio waves <Microwaves<UV rays<X-rays
Question.3.Magnetic field lines can be entirely confined within the core of
a toroid, but not within a straight solenoid. Why?
Answer : Magnetic field lines form closed loops around a current-carrying wire. The geometry of a straight solenoid is such that magnetic field lines cannot loop around circular wires without spilling over to the outside of the solenoid. The geometry of a toroid is such that magnetic 6. field lines can loop around electric wires without spilling over to the outside. Hence, magnetic field lines can be entirely confined within the core of a toroid, but not within a straight solenoid.
Question.4.You are given following three lenses. Which two lenses will you use as an eyepiece and as an objective to construct an astronomical telescope?
Answer : For constructing an astronomical telescope, the
objective should have the maximum diameter. Of the three lenses given, L1 has the maximum diameter. The eyepiece should have the highest power for better magnification. Therefore, we use lens L3.
Question.5.If the angle between the pass axis of polarizer and the analyser is 45°, write the ratio of the intensities of original light and the transmitted light after passing through the analyser.
Question.6.The figure shows a plot of three curves a, b, c, showing the variation of photocurrent Vs. collector plate potential for three differential intensities I1, I2 and I3 having frequencies v1, v2 and V3 respectively incident of a photosensitive surface.
Point out the two curve for which the incident radiations have same frequency but different intensities.
Answer : The stopping potential increases with increasing frequency of radiations. Thus curves a and b have the same frequency but different intensities.
Question.7. What type of wavefront will emerge from a (i) point source, and (ii) distance light source? .
Answer : (i) For a point source, wavefront will be spherical, (ii) For a distant light source, the wavefronts will be plane.
Question.8.Two nuclei have mass numbers in the ratio 1: 2. What is the
ratio of their nuclear densities?
Answer : Nuclear density is independent of mass number. Hence, both the atoms have the same nuclear density.
i.e.,p1 : p2 = 1:1
Question.9. A cell of emf ‘E’and internalresistance ‘V’is connected across a variable resistor ‘R’. Plot a graph showing the variation of terminal potential ‘V’ with resistance R. Predict from the graph the condition under which ‘V’ becomes equal to ‘E’.
Answer : V becomes equal to E when no current flows
Question.10. (i) Can two equipotential surfaces intersect with each other? Give reasons.
(ii) Two charges —q and +q are located at points A (0, 0,-a) and B (0, 0, +a) respectively. How much work is done in moving a test charge from point P (7, 0, 0) to Q(-3, 0, 0)?
Answer : (i) Two equipotential surfaces cannot intersect with each other because when they will intersect, the electric field will have two directions, which is impossible.
(ii) Potential at P(7, 0, 0)
Question.11. By what percentage will the transmission ranges of TV tower be affected when the height of the tower is increased by 21%?
Question.12.Derive an expression for drift velocity of free electrons in a conductor in terms of relaxation time.
For all the electrons in the conductor, average value of v1 is zero. The average of vt is vd or drift Velocity.
This is the average velocity experienced by an electron in an external electric field.
There is no fixed time after which each collision occurs. Therefore, we take the average time after which one collision takes place by an electron.
Let this time, also known as relaxation time, be t. Substituting this in equation (1)
Question.17. Why are high frequency carrier waves used for transmission?
What is meant by term modulation’? Draw a block diagram of a simple modulator for obtaining an AM signal.
Answer : For transmitting a signal, the antenna should have a size comparable to the wavelength of the signal (at least λ/4 in dimension), where X is the wavelength.
If the frequency of the signal is small, then its wavelength becomes very large and it is impractical to make those large antennas for the corresponding large wavelengths. For higher frequencies, wavelength is smaller, which is the reason why high frequency carrier waves are used for transmission.
The process of superimposing information containing in a low frequency signal on a high frequency signal is called modulation.
Block diagram of a simple modulator :
Question.18. A radioactive nucleus A’ undergoes a series of decays according to the following scheme :
The mass number and atomic number of A are 180 and 72 respectively. What are these numbers for A4?
Question.19. A thin conducting spherical shell of radius R has charge Q spread uniformly over its surface. Using Gauss’s law, derive an expression for an electric field at a point outside the shell. Draw a graph of electric field E(r) with distance r from the centre of the shell for 0≤ r ≤ ∞.
Answer : According to Gauss law,
Question.20. Three identical capacitors C1 ,C2 and C3 of capacitance 6 μF each are connected to a 12 V battery as shown.
(i) charge on each capacitor
(ii) equivalent capacitance of the network
(iii) energy stored in the network of capacitors
Answer : Existing diagram can be redrawn as follows :
Question.21. (a) The energy levels of an atom are as shown below- Which of them will result in the transition of a photon of wavelength 275 nm?
(b) Which transition corresponds to emission of radiation of maximum wavelength?
Answer : (a) Energy transitions for A, B, C, and D are :
Thus, B will result in transition of a photon of wavelength of 275 nm.
E = hc/λ
For maximum wavelength, energy transition should be minimum.
A undergoes minimum energy transition.
A = 2 eV
Thus, photon in A will have the maximum wavelength.
Question.22. A proton and an alpha particle are accelerated through the same potential. Which one of the two has (i) greater value of deBroglie wavelength associated with it, and (ii) less kinetic energy? Justify your answers.
Question.23. In a single slit diffraction experiment, when tiny circular obstacle is placed in path of light from a distance source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why?
State two points of difference between the interference pattern obtained in Youngs double slit experiment and the diffraction pattern due to a single slit.
Answer : A bright spot is observed when a tiny circular object is placed in path of light from a distant source in a single slit diffraction experiment because light rays flare into the shadow region of the circular object as they pass the edge of the tiny circular object. The light from all the edges of the tiny circular object are in phase with each other. Thus they form a bright spot at the centre of the shadow of the tiny circular object.
The two differences between the interference patterns obtained in Youngs double slit experiment and the diffraction pattern due to a single slit are as follows:
- The fringes in the interference pattern obtained from diffraction are of varying width, while in case of interference, all are of the same width.
- The bright fringes in the interference pattern obtained from diffraction have a central maximum followed by fringes of decreasing intensity, whereas in case of interference, all the bright fringes are of equal width.
Question.24. (a) Define self inductance. Write its S.I. units.
(b) Derive an expression for self inductance of a long solenoid ‘ of length l, cross-sectional area A having N number of turns.
Answer : (a) The phenomenon in which emf is induced in a single isolated coil due to change of flux through the coil by means of varying the current through the same coil is called self inductance. S.I. unit of inductance is henry.
Question.25. The figure shows experimental set up of a meter bridge. When the two unknown resistances X and Y are inserted, the null point D is obtained 40 cm from the end A. When a resistance
of 10 Ω is connected in series with X, the null point shifts by 10 cm. Find the position of the null point when the 10 Ω resistance is instead connected in series with resistance ‘Y’. Determine the values of the resistances X and Y.
Answer:For meter bridge:
Question.26. Derive the expression for force per unit length between two long straight parallel current carrying conductors. Hence define one ampere.
Explain the principle and working erf a cyclotron with the help of a schematic diagram. Write the expression for cyclotron frequency.
Answer : Two long’parallel conductors a and b separated by a distance d and carrying currents Ia and If, respectively are shown below.
Fba = -Fab
1 ampere is the value of that steady current which when maintained in each of the two very long, straight, parallel conductors of negligible cross-section and placed one metre apart in vacuum, would produce on each of these conductors a force equal to 2 x 10-7 Newton per metre of length.
Cyclotron is a machine used to accelerate charged particles or ions to high energies. It uses both electrical and magnetic fields in combination to increase the speed of the charged particles. The particles move in two semi-circular containers D1 and D2, called Dees. Inside the metal box, the charged particle is shielded from external electric fields.
When the particle moves from one dee to another, electric field is acted on the particle.
The sign of the electric field is changed alternately, in tune with the circular motion of the particle. Hence, the particle is always accelerated by the electric field. As the energy of the particle increases, the radius of the circular path increases.
The above expression is the expression for cyclotron frequency. The oscillator applies an ac voltage across the dees and this voltage must have a frequency equal to that of cyclotron frequency.
Question.27. Three light rays red (R), green (G) and blue (B) are incident on a right angled prism ‘abc’ at face ‘ab’. The refractive indices of the material of the prism for red, green and blue
wavelengths are 1.39, 1.44 and 1.47 respectively. Out of the three which colour ray will emerge out of face ‘ac’ ? Justify your answer. Trace the path of these rays after passing through face ‘ab’.
As light is incident normally on face ab, so no refraction occurs at the face ab: Now light is incident on face ac at i = 43° the face ac will not transmit light for which
Question.28. (a) Derive an expression for the average power consumed in , a series LCR circuit connected to a.c. source in which the phase difference between the voltage and the current in the circuit is Ø.
(b) Define the quality factor in an a.c. circuit. Why should the quality factor have high value in receiving circuits? Name the factors on which it depends.
(a) Derive the relationship between the peak and the rms value of current in an a.c. circuit.
(b) Describe briefly, with the help of labelled diagram, working of a step-up transformer.
A step-up transformer converts a low voltage into high voltage. Does it not violate the principle of conservation of energy? Explain.
Answer : (a) The rate of dissipation of energy in an electrical circuit is called the ‘power’. It is equal to the product of the emf and the current. The power of an alternating-current depends upon the phase difference between the emf and the current.
The instantaneous values of the emf and the current in an a.c.
Question.29.(i) Draw a circuit diagram to study the input and output characteristics of an n-p-n transistor in its common emitter configuration. Draw the typical input and output characteristics.
(ii) Explain, with the help of a circuit diagram, the working of n-p-n transistor as a common emitter amplifier.
How is a zener diode fabricated so as to make it a special purpose diode? Draw I-V characteristics of zener. diode and explain the significance of breakdown voltage.
Explain briefly, with the help of a circuit diagram, how a p-n junction diode works as a half wave rectifier.
When no a.c. signal is applied, the potential difference Vcc between the collector and emitter is given by
Vcc = VCE + ICRL
When an a.c. signal is fed to the input circuit, the forward bias increase during the positive half cycle of the input. This result in increase in IC and decrease inVcc Thus during the positive half cycle of the input, the collector becomes less positive.
During the negative half cycle of the input, the forward bias is decreased resulting in decrease in IE and hence IC Thus Vcc would increase making the collector more positive.
Hence in a common-emitter amplifier, the output voltage is 180° out of phase with the input voltage.
(i) Input signal voltage Vi= IBRB
(ii)Output signal voltage V0 = ICRL
(iii)Voltage gain (Av) of the amplifier is
Zener diode is fabricated such that both the p-type and the w-type are highly doped. This makes the depletion region thin. When an electric field is applied, a high electric field appears across the thin depletion region. When the electric field becomes very high, it knocks off electrons from the host atoms to create a large number of electrons. This results in a large value of current inside the circuit.
An ac current has a positive half cycle and a negative half cycle.
A pn junction allows current to pass only in one direction and – that is when it is forward biased.
When a positive half-cycle occurs, the p-side has a lower potential. Therefore, the diode is now forward biased and therefore, conducts and this positive cycle is available for the load.
When a negative half cycle occurs, the n-side he has a higher potential than the p-side. Hence, the diode is now reverse biased and thus, does not conduct. As a result, this positive half cycle also does not conduct. Therefore, it does not appear at the load and is cut-off.
We obtain a waveform, which has only positive half cycles and therefore it is called half-wave rectifier.
Question.30. Trace the rays of light showing the formation of an image due to a point object placed on the axis of a spherical surface separating the two media of refractive indices n1 and n2 Establish the relation between the distances of the object, the image and the radius of curvature from the central point of the spherical surface.
Hence derive the expression of the lens markers formula.
Draw the labelled ray diagram for the formation of image by a compound microscope.
Derive the expression for the total magnification of a compound microscope. Explain why both the objective and the eyepiece of a compound microscope must have short focal lengths.
In the given figure, image is I and object is denoted as O. The centre of curvature is C.
The rays are incident from a medium of refractive index n1 to another of refractive index n2.
We consider NM to-be perpendicular to the principal axis.
Note: Except for the following questions, all the remaining questions have been asked in Set-I, 2009.
Question.1. Name the electromagnetic radiation to which waves of wavelength in the range of10-2 m belong. Give one use of this part of EM spectrum.
Answer: Microwave – Range 0.1 m to 1 mm.
Use : Microwaves are used in Radar system for aircraft navigation.
Question.2. What is ground wave propagation?
Answer : Ground wave propagation is a mode of wave propagation in which the ground has a strong influence on the propagation of signal wave from the transmitting antenna
to receiving antenna.
Question.5. Unpolarized light is incident on a plane surface of glass of
refractive index p at an angle i. If the reflected light gets totally polarized, write the relation between the angle i and refractive index p.
Answer : μ = tan i
Question.6. Draw a diagram to show refraction of a plane wave front incident in a convex lens and hence draw the refracted wave front.
Question.8.Two nuclei have mass numbers in the ratio 1 : 3. What is die ratio of their nuclear densities?
Answer: The nuclei density is same for all nuclei. Therefore, the ratio of their nuclear densities is :
f1: f2 = 1:1
Question.11. The output of a 2-input AND gate is fed to a NOT gate. Give the name of the combination and its logic symbol. Write down its truth table.
Answer: Name of the combination : NAND gate
Question.19. The equivalent capacitance of the combination between A and B in the given figure is 4μF.
(i) Calculate capacitance of the capacitor C.
(ii)Calculate charge on each capacitor if a 12V battery is connected across terminals A and B.
(iii) What will be the potential drop across each capacitor?
... Potential drop across 20μF capacitor = 2.4V
and potential drop across 5μF capacitor = 9.6 V.
Question.20. State Gauss’s law in electrostatics. Using this law derive an expression for the electric field die to a uniformly charged infinite plane sheet.
Answer : Gauss’s law : The net electric flux through any closed surface is equal to 1/εo times the net electric charge enclosed within that closed surface.
Gauss’s law may be expressed as :
Where, is the electric flux through a closed surface S enclosing any volume V, Q is the total charge enclosed within S, and Go is the electric constant.
Electric Field due to A Uniformly Charged Infinite Plane Sheet : Suppose a thin non-conducting infinite sheet of uniform surface, charge density σ.
Electric field intensity E on either side of the sheet must be perpendicular to the plane of sheet having same magnitude at all points equidistant from sheet.
Question.22. A electron and a proton are accelerated through the same potential. Which one the two has (i) greater value of deBroglie wavelength associated with it and (ii) less momentum? Justify your answer.
Answer : (i) The deBrogile wavelength associated with potential V is
Note: Except for the following questions, all the remaining questions have been asked in Set-1 and Set-II, 2009.
Question.2. At what angle of incident should a light beam strike a glass
slab of refractive index √3 such that the reflected and the refracted rays are perpendicular to each other?
Question.3. What is space wave propagation?
Answer : Spice wave propagation is the mode of propagation in which the radiowaves from the transmitting antenna reach the receiving antenna either direcdy or by reflection from the troposphere.
Question.5. Name the part of electromagnetic spectrum which is suitable for
(i) radar system used in aircraft navigation
(ii) treatment of cancer tumours
Answer: (i) Microwaves
(ii) Gamma rays
Question.6. Two nuclei have mass numbers in the ratio 2 : 5. What is the
ratio of their nuclear densities?
Answer : Since the nuclear density is same for all nuclei.
Ratio of their nuclei densities are 1 : 1.
Question.7. Differentiate between a ray and a wave front.
Answer : The locus of all the particles of the medium, which at any instance are vibrating in the same phase,is called the Wave front.
An arrow drawn normal to the wave front and pointing in the direction of propagation of the disturbance is called a Ray.
Question.12. (i) Sketch the output wave form from an AND gate for the inputs A and B shown in the figure.
(ii) Combination of AND and NOT is called NAND gate.
Question.15. (i)’ Can two equipotential surface intersect each other? Give reason.
(ii) Two charges +q and —q are located at points A (0, 0, -2) and B (0, 0, 2) respectively. How much work will be done in moving a test charge from point P (4, 0, 0) and Q. (-5, 0, 0)?
Answer : (i) Two equipotential surfaces cannot intersect each other because when they will intersect, the electric field will have two directions, which is impossible.
(ii) Potential at P (4, 0, 0)
Question.19. State Gauss Law in electrostatics. Use this law to derive an expression for the electric field due to an infinitely long straight wire of linear charge density A cm-1.
Answer: Gauss’ law: The net electric flux through any closed surface is equal to 1/εo times the net electric charge enclosed within that closed surface.
Gauss’s law may be expressed as :
Where, ΦE is the electric flux through a closed surface S . enclosing any volume V, Q is the total charge enclosed within S, and εo is the electric constant.
Electric field due to an identify long straight wire: Consider an infinitely long line charge having linear charge density λ. To determine the electric field at distance r, consider a cylindrical Gaussian surface of radius r and length l coaxial with the charge. By symmetry, the electric field E has same magnitude at each point of the curved surface S1 and is directed radially outward.
Question.23.Two parallel place 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 4μF
(ii)Calculate the potential difference between the plates of X andY.
(iii)What is the ratio of electrostatic energy stored in X and Y?
Answer: (i) Since capacitance C of the parallel plate capacitor is given by