CBSE Previous Year Solved Papers Class 12 Physics Outside 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:
Question.1. What is the electrostatics potential due to an electric dipole at an equatorial point?
Answer : The electric potential due to an electric dipole at an equatorial point is zero.
Question.2. Name the EM waves used for studying crystal structure of solids. What is its frequency range?
Answer : X-rays are used to study crystal structure of solids. The frequency range is 1 nm to 10-3 nm.
Question.3. An electron does not suffer any deflection while passing through a region of uniform magnetic field. What is direction of the magnetic field?
Answer : The magnetic field is given to be in the direction in which the electron moves. The force is given by
Question.4. How would the angular separation of interference fringes in Young’s double slit experiment change when the distance between the slits and screen is doubled?
Question.5. Two thin lenses of power + 6D and -2D are in contact. What is the focal length of the combination.
Question.6. The stopping potential in an experiment on photoelectric effect is 1.5 V. What is the maximum kinetic energy of the photoelectrons emitted?
Question.7. Two nuclei have mass numbers in the ratio 1: 8. What is the ratio of their nuclear radii?
Question.8. Give the logic symbol of NOR gate.
Question.9. Draw 3 equipotential surfaces corresponding to a field that uniformly increase in magnitude but remains constant along z-direction. How are these surfaces different from that of a constant electric field along z-direction?
Answer : For constant electric field in z-axis equipotential surfaces will be plane parallel to xy- planes .
If field that increases in magnitude, the equipotential surfaces will be planes parallel to XY plane But, as the field increases such planes will get closer.
Question.10. Define electric flux. Write its S.I. unit A charge q is enclosed by a spherical surface of radius R. If the radius is reduced to half, how would the electric flux through the surface change?
Answer: Electric flux is the total number of electric field lines crossing an area. Its SI units in Nm /C. The electric flux through a spherical surface of radius R for a charge q enclosed by the surface is If radius is reduced to half, the electric flux remains the same.
Question.11. Define refractive index of a transparent medium. A ray of light passes through a triangular prism. Plot a graph showing the variation of the angle of deviation with the angle of incidence.
Answer : Refractive Index : The ratio of velocity of light in vacuum to the velocity of light in medium is called absolute refractive index of the medium.
Graph : The plot of angle of deviation versus angle of 0 incidence for a triangular prism is shown below :
Question.12. Calculate the current drawn from the battery in the given network.
Question.13. Answer the following questions:
(a) Optical and radio telescopes are built on the ground while X-ray astronomy is possible only from satellites orbiting the Earth. Why?
(b)The small ozone layer on top of the stratosphere is crucial for human survival. Why?
(a) X-rays from celestial objects cannot reach the earth’s surface. So for X-ray astronomy to be possible the satellite has to be present in space. However, visible light and lower wavelengths of light emitted by celestial objects reach the surface of the earth. So these objects can be observed with optical and radio telescopes.
(b) The ozone layer is crucial for human survival as they help to block UV radiations and other high frequency harmful radiations and prevent them from reaching the earth’s surface.
Question.15. Define the term ‘linearly polarised light’.
When does the intensity of transmitted light become maximum, when a polaroid sheet is rotated between two crossed polaroids?
Answer: ‘Linearly polarized light’ is an electromagnetic wave, in which the vibrations of electric field are restricted to a single plane. ‘
The intensity of transmitted light is given by I =Im cos2 θ
The intensity of transmitted light is maximum, when θ =0° or 180°, or the polarizing axis of the two polaroids is parallel to each other.
Question.16. A wire of 15 Ω resistance is gradually stretched to double
its original length. It is then cut into two equal parts. These parts are then connected in parallel across a 3.0 volt battery. Find the current drawn from the battery.
Answer : When any resistor is stretched to double its original length, the new resistance becomes four times of original resistance.
Question.17. (a)a nucleus in its ground state is always less than the total mass of its constituents — neutrons and protons. Explain.
(b) Plot of graph showing the variation of potential energy of a pair of nucleons as a function of their separation.
(a) The mass of a nucleus in its ground state is always less than the total mass of its constituents because some mass is converted into energy, in accordance with the equation E = mc2. This difference in mass is called the mass defect and the energy corresponding to the mass defect is the binding energy. This
is the energy that has to be supplied*to the nucleus to break it up into its constituents..
(b) The graph is
Question.18. Write the function of (i) Transducer and (ii) Repeater in the context of communication system.
Write two factors justifying the need of modulation for transmission of a signal.
- Transducer converts energy from one form of another either at the input or at the output. For example sound signals are converted to electrical signals so that they can be transmitted through the communication channel.
- Repeater stations receive the signal, amplify it and then transmit it. So they are a combination of a receiver and , transmitter. This helps to increase the range of transmission of signals.
Two factors justifying the need of modulation for transmission of signal are :
- Modulation helps to increase the frequency of the signal. This helps to transmit it over larger distances. This is because the power radiated by an antenna is proportional to 1/λ2 The power radiated by the antenna increase for high frequencies.
- The size of the antenna required is proportional to (wavelength/4). So if waves of large wavelength are transmitted then the size of the antenna required is impractical. By increasing the frequency of the signal by modulation, its wavelength is decreased. So now shorter and more practical size antennas can be built.
Question.19. A positive point charge (+ q) is kept in the vicinity of an uncharged conducting plate. Sketch electric field lines originating from the point on to the surface of the plate. Derive the expression for the electric field at the surface of a charged conductor.
A parallel plate capacitor is charged by a battery. After some time the battery is disconnected and a dielectric slab of dielectric constant K is inserted between the plates. How would (i) the capacitance, (ii) the electric field between the plates and (iii) the energy stored in the capacitor, be affected? Justify your answer.
Answer : Representation of electric field :
The electric field due to a positive charge (+q) is represented as :
Electric field due to a point charge : Consider a point charge +q placed at the origin O of the coordinate frame. Let qo be any test charge placed at P.
Question.20. (i) State the principle of working of a meter bridge.
(ii) In a meter bridge balance point is found at a distance l1 t with resistance R.and S as shown in the figure.
- State Faradays law of electromagnetic induction.
- A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earths magnetic field at the location has a magnitude of 5 x 10-4 T and the dip angle is 30°?
- Faraday’s law of electromagnetic induction states that the induced emf is proportional to the rate of change of magnetic flux.
- The induced emf will be given by E = Blv where the magnetic field B is perpendicular to the length l. In the question the earth’s magnetic field is given and the angle of dip is 30 degrees, so the magnetic field perpendicular to the direction of the plane is B sin 30°. ’
Question.22. In Young’s double slit experiment, monochromatic light of wavelength 630 nm illuminates the pair of slits and produces an interference pattern in which two consecutive bright fringes are separated by 8.1 mm. Another source of monochromatic light produces the interference pattern in which the two consecutive bright fringes are separated by 7.2 mm. Find the wavelength of light from the second source. What is the effect on. the interference fringes if the’ monochromatic source is replaced by a source of white light?
When the monochromatic source is replaced by a source of white light, the fringe width is changed.
Question.23. Draw a schematic arrangement of the Geiger-Marsden experiment. How did the scattering of a-particles by a-thin foil of gold provide an important way to determine an upper limit on the size of the nucleus? Explain briefly.
Answer: Diagram of the Geiger Marsden experiment.
The alpha particles that are incident head-on with the gold nucleus experience a very large force of repulsion and undergo maximum deflection. Equating the kinetic energy of the incident alpha particle with the potential energy of the alpha particle and gold nucleus, the sum of the approximate radius of the gold nucleus and alpha particles, can be found.
Question.24. Distinguish between sky wave and space wave propagation. Give a brief description with the help of suitable diagrams indicating how these waves are propagated.
Answer : The sky waves are reflected from the ionosphere and received by a receiver. Space waves penetrate the ionosphere and are intercepted by a satellite. They can also be used for line of sight communication.
This mode of propagation is used by short-wave broadcast service.
The space waves are the radiowaves of very high frequency (i.e. between 54 MHz to 4.2 MHz). The space waves can travel through atmosphere from transmitter antenna to- receiver antenna either directly or after reflection from ground in the earth’s troposphere region. That is why space wave propagation in also known as tropbspherical propagation. The space waves travel in straight line from transmitting
antenna to receiving antenna. Therefore, the space waves are used for line of sight communication such as television
broadcast, microwave link and satellite communication.
Question.26. Give a circuit diagram of a common emitter amplifier using an n-p-n transistor. Draw a input and out waveforms of the single. Write the expression for its voltage gain.
Answer. Circuit diagrame of a common emitter amplofire using an n-p-n- transisstor.
Question.27. Draw a plot showing the variation of binding energy per nucleon versus the mass number A. Explain with the help of this plot the release of energy in the processes of nuclear fission and fusion.
Answer: Binding energy curve : The variation of average B.E per nucleon with mass number A, shown in figure below.
A graph between the average binding energy per nucleon and the mass number A of different nuclei is a curve, called ‘binding energy curve’.
- For nuclei having mass number A=50 to A=80 are almost stable.
- For nucler having mass number above 80, the average BE/nucleon decreases slowly and drops to about 7.6 MeV for uranium (A = 238). This lower value of binding energy per nucleon fails to overcome the Coulombian repulsion.
- For nuclei having mass number below 50 also, the average BE/nucleon decreases and below 20, it decreases sharply, e.g. : For heavy hydrogen (H2), it in only about 1.1 MeV. This shows that the nuclei having mass number below 20 are comparatively less stable.
- Below A = 50, the curve does not fall continuously, but has subsidiary peaks at 8o16, 6C12 and 2He4. This shows that these nuclei are more stable than their immediate neighbours.
- Binding energy per nucleon is small for both light and heavy nuclei. When light nuclei fuse to form a heavy nucleus, high value of B.E is released in Nuclear Fusion. When a heavy nucleus splits into light nuclei, high value of B.E is released in Nuclear fission.
Question.28. Draw a schematic sketch of a cyclotron. Explain briefly how it works and how it is used to accelerate the charged particles.
- Show that time period of ions in a cyclotron is
independent of both the speed and radius of circular path. .
- What is resonance condition? How is it used to accelerate the charged particles?
(a) Two straight long parallel conductors carry currents Ii and I2 in the same direction. Deduce the expression for the force per unit length between them.
Depict the pattern of magnetic field lines around them,
(b) A rectangular current carrying loop EFGH is kept in a uniform magnetic field as shown in the figure.
- What is the direction ,of the magnetic moment of the current loop?
- When is the torque acting on the loop (A) maximum, (B) zero?
Answer : Diagram of Cyclotron
Charged particles are introduced between the dees. An alternating voltage applied between the dees accelerates, these particles by an electric field. A magnetic field that is perpendicular to the plane of the dees exerts a force of the particles that is given by F = q (V x B). This cause the particles to follow a circular trajectory. As’ they reach the dees the polarity is reversed and the particles are once more accelerated. This continues and highly energetic beams of charged particles are obtained. Magnetic field is perpendicular to dees.
Hence, the time period is independent of speed and radius of circular path.
(ii) Resonance condition : When the angular frequency of the rotating charged particle and the angular frequency of the
alternating voltage applied across the dees of the cyclotron match. The charged particles are only accelerated by the electric field. The magnetic field only keeps it moving along a circular tra.ck. The acceleration happens when the charged particle crosses the gap between the two dees. At this instant the field between the dees has to be reversed so that the electric field can accelerate the charged particle.
Question.29. (a) What are eddy currents ? Write their two applications.
(b) Figure shows a rectangular conducting loop PQSR in which arm RS of length ‘l’ is movable. The loop is kept in a uniform magnetic field ‘B’ directed downward perpendicular to the plane of the loop. The arm RS is moved with a uniform speed V.
Deduce an expression for
(i) The emf iriduced across the arm ‘RS’,
(ii) The external force required to move the arm, and
(iii) The power dissipated as heat.
(a) State Lenz’s law. Give one example to illustrate this law. “The Lenz’s law is a consequence of the principle of conservation of energy”. Justify this statement.
(b) Deduce an expression for the mutual inductance of two long coaxial solenoids but having different radii and different number of turns.
Answer : (a) When a bulk piece of conductor is subjected to changing magnetic flux, the induced current developed in it is called eddy current.
Applications of eddy currents :
- Magnetic brakes in trains.
- Electromagnetic damping.
- Induction furnaces.
- Electric power meter.
(b) (i) The emf induced across the arm B lv
On account of the presence of the magnetic field, there will be a force on the arm RS. This force I (lx B), is directed outwards in the direction opposite to the velocity of the arm RS. The magnitude of this force is,
(a) Lenz’s Law : An induced electromotive force (emf) always gives rise to a current whose magnetic field opposes the original change in magnetic flux.
Lenz’s law is shown with the negative sign in Faraday’s law of induction :
Lenz’s law states that the current induced in a circuit due to a change or a motion in a magnetic field is so directed as to oppose the change in flux or to exert a mechanical force opposing the motion.
Explanation of conservation of energy : When a magnet is moved near a current carrying coil, the direction of the induced current opposes the motion of the magnet. When the north pole of the magnet is moved towards the coil, the induced current flows in a direction so that near face of the coil acts as a magnetic north pole. The repulsion between two poles opposes the motion of the magnet towards the coil. Similarly, when the north pole of the magnet is from the coil, the direction of the induced current is such as to make the near face of the coil a south pole. The attraction between the two poles opposes the motion of the magnet away from the coil. In either case, therefore, work has to be done in moving the magnet. This mechanical work appears as electrical energy in the coil. Thus Lenz’s law is in accordance with the principle of conservation of energy.
(b) Mutual Inductance of two long coaxial solenoids : Consider the following fig. which shows two long co-axial solenoids each of length l. Let the radius of the inner solenoid S1be r1 and the number of turns per unit length be n1.
Question.30. (a) (i) Draw a labelled ray diagram to show the formation of image in an astronomical telescope for a distant object, (ii) Write three distinct advantages of a reflecting type telescope
(b) A convex lens of focal length 10 cm is placed coaxially 5 cm away from a concave lens of focal length 10 cm. If an object is placed 30 cm in front of the convex lens, find the position of the final image formed by the combined system.
(a) With the help of a suitable ray diagram, derive the mirror formula for a concave mirror.
(b) The near point of a hypermetropic person is 50 cm
from the eye. What is the power of the lens required to enable the person to read clearly a book held at 25 cm from the eye?
(ii)Advantage of reflecting type telescope over a refracting
type telescope are :
- Lenses suffer from chromatic aberrations that are not there in mirrors.
- Lenses also have spherical aberration; a parabolic mirror will be free of spherical aberration.
- It is easier to support large mirrors as the back surface is nonreflecting, but a lens needs support around its rim.
(a) Mirror forumula for concave Mirror :
(b) The near point of the eye of this person is at 50 cm. The required lens must have a focal length such that the virtual image of the book placed at 25 cm is formed at 50 cm. Then image can be focussed by the eye.
Note: Except for the following questions, all the remaining questions have been asked in Set-I, 2009.
Question.1. What is work done in moving a test charge q through a distance of 1 cm along the equatorial axis of electric dipole?
Answer : Since potential difference for equipotential surface
ΔV = 0
Work done in moving a positive test charge q through a distance 1 cm is
w= qΔV = q x 0= 0
Question.5. Two thin lenses of power +4 D and — 2D are in contact. What is focal length of combination?
Question.6. Give the logic symbol of NAND gate?
Question.7.Two nuclei have mass number in the ratio 8 the ratio of their nuclear radii?
Question.8.The maximum kinetic energy of a photoelectron is 3 eV. What is its stopping potential?
- State the principle on which the working of an optical fiber is based.
- What are the necessary conditions for this phenomenon
- The phenomenon of reflection of light when the light travelling in a denser medium strike* the interface separating the denser medium and the rarer medium at an angle greater than the critical angle is called total internal reflection.
- Conditions :
(a) Light must travel from a denser medium to rarer medium.
(b) The angle of incidence in the denser medium must be greater than the critical angle for the two media in contact.
Question.10. (i) State the law that gives the polarity of the induced emf. (ii) A 15.0μF capacitor is connected to 220 V, 50 Hz source.
Find the capacitive reactance and the rms current.
(i) Lenz’s law : According to this law direction of induced emf Or current in a circuit is such that it opposes the cause or change that produces it.
Question.23. Use Gauss’s law to derive the expression for the electric field between two uniformly charged large parallel sheets with surface charge densities a and -a respectively.
(a) A charge +Q is placed on a huge spherical conducting shell
of radius R. Another small conducting sphere of radius r carrying charge is introduced inside the laige shell and is placed at its centre. Find the potential difference between two points, one lying on the sphere and the other on the shell.
(b) How would the charge between the two flow if they are
connected by a conducting wire? Name the device which works on this fact.
Answer : Electric field due to a uniformly charged infinite plane sheet: Suppose a thin non-conducting uniform surface
Electric field intensity E on the either side of the sheet must be perpendicular to the plane of sheet having same magnitude at all points equidistant from sheet.
Let P be any point at a distance r from the sheet. Let the small area elements are perpendicular on the surface of the imagined cylinder, so electric flux is zero. are parallel on the two cylindrical edges P and Q, which contributes electric flux.
... Electric flux over the edges P and Q of the cylinder is
(b) When both spheres are connected by a conducting wire then charge q on smaller sphere wifi flow onto the large sphere having charge Q. ,
Van de Graff generator works on this fact that charge to a hollow conductor is transferred to outer surface and is distributed uniformly over it.
Note : Except for the following questions, all the remaining questions have been asked in Set-I and Set-II, 2009.
1.Define the term ‘potential energy’ of charge q’ at a distance ‘r’ in an external electric field.
Answer: Potential energy of a single charge ‘q’ ara distance Y in an external field = q.V (r)
Question.4.The stopping potential in an experiment on photoelectric effect is 2V What is the maximum kinetic energy of the photoelectrons emitted?
Question.5.Two thin lenses of power + 5D and —2.5 D are in contact: What is the focal length of the combination?
Question.6.How would the angular separation of interference fringes in Young’s double slit experiment change when the distance between the slits and screen in halved?
Question.7. Give the logic symbol of AND gate.
Question.8.Two nuclei have mass numbers in the ratio 27 : 125. What is the ratio of the their nuclear radii?
Question.11. (i) What is the relation between critical angle and refractive index of a material?
(ii) Does critical angle depend on the colour of light? Explain.
Question.16. A wire of 20 Ω resistance is gradually stretched” to double its original length. It is then cut into two equal parts. These parts are then connected in. parallel across a 4.0 volt battery. Find the current drawn from the battery.
Answer: When any resistor is stretched to double its original length. The new resistance becomes four times of its original resistance.
Question.25. Explain with the help of a circuit diagram how a zener diode works as a DC voltage regulator. Draw its I-V characteristics.
Answer : Zener diode is fabricated such that both the p-type and the n-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 a large value of current inside the circuit.
Zener diode has a sharp breakdown voltage and this property of zener is used for voltage regulation.
Question.27. Define the activity of a radionuclide. Write its S.I. unit. Give a plot of the activity of a radioactive species versus time.
How long will a radioactive isotope, whose half life is T years, take for its activity to reduce to 1/8th of its initial value?
Answer : The total decay rate R of one or more radionuclide is called the activity of that sample. The S.I. unit for activity is becquerel.
1 becquerel = 1 Bq = 1 decay per second
Graph: The graph between the activity of a radioactive species and time is :