CBSE Previous Year Solved Papers Class 12 Physics Outside Delhi 2012
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.Two wires of equal length, one of copper and the other of manganin have the same resistance. Which wire is thicker?
Answer : Since the resistivity of alloy is greater than the resistivity of its constituents. We have resistivity of manganin greater than resistivity of copper metal.
Question.2. What are the directions of electric and magnetic field vectors relative to each other and relative to the direction of propagation of electromagnetic waves?
Answer: Since electromagnetic waves are transverse in nature. We have electric and magnetic fields associated with an electromagnetic wave perpendicular to each other and perpendicular to the direction of propagation of electromagnetic waves.
Let the direction of electric field and magnetic field is along Y and Z-axis then the direction of propagation of EM waves will be along positive X-axis.
Question.3.How does the angular separation between fringes in single¬slit diffraction experiment change when the distance of separation between the slit and screen is doubled?
Answer : We know angular separation is given as
Since 9 is independent of D i.e., the distance of separation between the screen and the slit, so when D is doubled, angular separation would remain same.
Question.4.A bar magnet is moved in the direction indicated by the arrow between two coils PQand CD. Predict the directions of induced current in each coil
Answer : According to Lenz’s law the polarity of the induced emf is such that it opposes the change in magnetic flux responsible for its production.
Since North pole of bar magnet is receding away from the coil so the right end of the coil will develop South pole i.e., induced current as seen from the left end will be anticlockwise.
Question.5. For the same value of angle of incidence, the angles of refraction in three media A, B and C are 15°, 25° and 35° respectively. In which medium would the velocity of light be minimum?
Answer : As light travels from a rarer to denser medium it ‘ bends towards the normal as its speed decreases. So, if the
bending is more, the speed of the light would be less in that medium, compared to other media. As the angle of refraction is measured with respect to the normal, the ray making the
least angle of refraction would bend more and the speed oflight would be minimum in that case. So, the correct option is medium A where refracting angle is 15°.
Question.6. A proton and an electron have same kinetic energy. Which one has greater de-Broglie wavelength and why?
Answer : Since de-Broglie wavelength X in terms of kinetic energy is given as
Question.7. Mention the two characteristic properties of the material suitable for making core of a transformer.
Answer : Two characteristic properties of material:
(i) Low hysteresis loss.
(ii) Low coercivity.
Question.8.A charge ‘q’ is placed at the centre of a cube of side /. What is the electric flux passing through each face of the cube?
Answer : By using Gauss’s Law,
Question.9. A test charge ‘q’ is moved without acceleration from A to C along the path from A to B and then from B to C in electric field E as shown in the figure, (i) Calculate the potential difference between A and C. (ii) At which point (of the two) is the electric potential more and why?
Answer : (i) Since work done is independent of the path therefore, we may directly move from A to C.
Potential difference between A and C is given by,
Question.10. An electric dipole is held in a uniform electric field.
(i) Show that the net force acting on it is zero.
(ii) The dipole is aligned parallel to the field. Find the work
done in rotating it through the angle of 180°.
Answer: (i) Consider an electric dipole consisting of two equal and opposite point charges, -q at A and +q at B, separated by a small distance 2a.
Question.11. State the underlying principle of a transformer. How is the large scale transmission of electric energy over long distances done with the use of transformers?
Answer : Transformer Principle : It is a device which converts high voltage AC into low voltage AC and vice-versa. It is based upon the principle of mutual induction. When alternating current is passed through a coil, an induced emf is set up in the neighbouring coil.
Transformers are used for transmission of electrical energy over long distances.
It step up the output voltage of power plant using step up transformer which reduce the current through cables and hence reduce resistive power loss. Then a step down transformer is used at consumer end to step down the voltage.
Question.12. A capacitor of capacitance ‘C’ is being charged by connecting it across a dc source along with an ammeter. Will the ammeter show a momentary deflection during the process of charging? If so, how would you explain this momentary deflection and the resulting continuity of current in the circuit? Write the expression for the current inside the capacitor.
Answer: Yes, the ammeter shows a momentary deflection during the process of charging because changing electric field produce displacement current between the plates of the capacitor.
The resulting continuity of current in the circuit is because inside the capacitor, displacement current exist and in the wire conduction current flow.
Question.13. An object AB is kept in front of a concave mirror as shown in the figure.
(i) Complete the ray diagram showing the image formation of the object.
(ii)How will the position and intensity of the image be affected if the lower half of the mirrors reflecting surface is
Image formed will be inverted, between focus and center of curvature and small in size.
(ii) If the lower half of the mirror’s reflecting surface is painted black, the position of image will be same but its intensity gets reduced.
Question.14. Draw a labelled ray diagram of a reflecting telescope. Mention its two advantages over the refracting telescope.
Answer : Reflecting Telescope :
Advantages over refracting telescope :
- It reduces the spherical aberration by using parabolic mirror and forms a clear focussed image.
- The resolving power of a large aperture mirror is high and hence minute details of distant stars can be obtained.
Question.15. Describe briefly with the help of a circuit diagram, how the flow of current carriers in a p-n-p transistor is regulated with emitter-base junction forward biased and base-collector junction reverse biased.
Answer : In p-n-p transistor, the emitter base junction is always forward biased with voltage VEE and the collector base junction is always reverse biased with voltage Vcc
The holes in the emitter are pushed into the base by the positive terminal of battery of voltage VEE- Since, base is thin and lighdy doped so only few holes combine with electrons in the base. Thus, base current IB is small. Since Vcc is quite large almost 99% of holes coming from emitter are collected by collector. For each hole reaching the collector, an electron is released from the negative terminal of the collector base battery to neutralize thf hole. For each hole consumed in collector, a bond breaks in emitter and electron is released that enters positive terminal of emitter-base battery. Thus, we can say the current is carried by holes inside the transistor and by electrons in external circuit.
Question.16. In the given block diagram of a receiver, identify the boxes labelled as X and Y and write their functions.
X → IF stage → Intermediate frequency stage to change the , carrier frequency to lower frequency.
Y → Amplifier → It amplifies the signal i.e., it increase the amplitude of the detected signal to compensate the attenuation of the signal. The detected signal may not be strong enough to use.
Question.17. A light bulb is rated 100 W for 220 V ac supply of 50 Hz. Calculate:
(i) The resistance of the bulb;
(ii) The rms current through the bulb.
An alternative voltage given by V = 140 sin 314t is connected across a pure resistor of 50Ω. Find
(i) the frequency of the source.
(ii) the rms current through the resistor.
Question.18. A circular coil of‘N’ turns and radius ‘R’ carries a current ‘I’. It is unwound and rewound to make another coil of radius ‘R/2’, current T remaining the same. Calculate the ratio of the magnetic moments of the new coil and original coil.
Answer:The magnetic moment m of a current loop,
Question.19. Deduce the expression for the electrostatic energy stored in ( a capacitor of capacitance ‘C’ and having charge ‘Q’. How will the (i) energy stored and (ii) the electric field inside the capacitor be affected when it is completely filled with a dielectric material of dielectric constant ‘K’?
Answer: Energy stored in a charged capacitor: The energy of a charged capacitor is measured by the total work done in charging the capacitor to a given potential. We know that,
When dielectric material of dielectric constant ‘K’ is introduced inside the capacitor then
Question.20. Calculate the value of the resistance R in the circuit shown in the figure so that the current in the circuit is 0.2 A. What would be the potential difference between points B and E?.
Question.21. You are given three lenses L1, L2 and L3 each of focal length 20 cm. An object is kept at 40 cm in front of L1, as shown. The final real image is formed at the focus T of L3. Find the separations between L1, L2 and L3
But we have seen above that image by L1 is formed at 40 cm on the right of L1 which is at 20 cm left of L2 (focus of L2). So, x1= distance between L1 and L2 = (40 + 20) cm = 60 cm Again distance between L2 and L3 does not matter as the image by L2 is formed at infinity.
Hence, the distance between L2 and L3 can have any value.
Question.22. Define the terms (i) ‘cut-off voltage’ and (ii) ‘threshold frequency’ in relation to the phenomenon of photoelectric effect.
Using Einstein’s photoelectric equation show how the cut-off voltage and threshold frequency for a given photosensitive material can be determined with the help of a suitable plot/ graph.
Answer : When light of suitable frequency is incident on a metal surface, electrons are ejected from the metal. This phenomenon is called the photoelectric effect.
(i) The cathode is illuminated with light of some fixed frequency v and fixed intensity lj. A small photoelectric current is observed due to few electrons that reach anode just because they have sufficiently large velocity of emission. If we make the potential of the anode negative with respect to cathode then the electrons emitted by cathode are repelled. Some electrons even go back to the cathode so that the current decreases. At a certain value of this negative potential, the current is completely stopped. The least value of this anode potential which just stops the photocurrent is called cut off potential or stopping potential.
(ii) For a given material, there is a certain minimum frequency that if the incident radiation has a frequency below this threshold, no photoelectric emission will take place, howsoever intense the radiation may be falling.
According to Einstein’s photoelectric equation, maximum K.E. is given as
We can read the value of threshold frequency from graph. From equation (1), we can find the value of stopping potential (V0).
Question.23. A series LCR circuit is connected to an ac source. Using the phasor diagram, derive the expression for the impedance of the circuit. Plot a graph to show the variation of current with frequency of the source, explaining the nature of its variation.
Answer : Let an alternating emf E = E0 sin ωt is applied to a series combination of inductor L, capacitor C and resistance R. Since all three of them are connected in series the current through them is same. But the voltage across each element has a different phase relation with current.
With increase in frequency, current first increases and then decreases. At resonant frequency, the current amplitude is maximum.
Question.24. Mention three different modes of propagation used in communication system. Explain with the help of a diagram how long distance communication can be achieved by ionospheric reflection of radiowaves.
Answer: Three modes of propagation of electromagnetic waves,
(a) Ground waves,
(b) Sky waves
(c) Space waves.
Sky wave propagation is used for long distance communication by ionospheric reflection of radio waves.
In the ionosphere of the Earth’s atmosphere, there are a large number of charged particles (ions). The ionosphere is situated about 65 km — 400 km above the surface of the Earth. The ionization of molecules occurs due to the absorption of the ultraviolet rays and high energy radiation from the sun. The ionosphere acts as a reflecting layer for certain range of frequencies (3 MHz — 30 MHz)!’The transmitting antenna sends the EM signals of this frequency range towards the ionosphere. When the EM waves strikes the ionosphere, it is reflected back to the Earth. A receiving antenna at a remote location on the Earth receives these reflected signals.
Question.25. Draw a plot of potential energy of a pair of nucleons as a functions of their separations. Mark the regions where the nuclear force is
- attractive and
- repulsive. Write any two characteristic features of nuclear forces.
Answer : Potential energy of a pair of nucleons as a function of their separation :
Here, Part AB represents repulsive force and part BCD , represents attractive force.
ro is the distance at which potential energy is minimum.
For a separation greater than ro, the force is attractive and for separations less than ro the force is strongly repulsive. Characteristic features of nuclear forces are :
- Nuclear forces are much stronger then Coulomb forces acting between charges or the gravitational forces between masses.
- The nuclear force between neutron-neutron, proton- neutron and proton-proton is approximately the same. The nuclear force does not depend on the electric charge.
Question.26. In a Geiger-Marsden experiment, calculate the distance of closest approach to the nucleus of Z = 80, when a-particle of 8 MeV energy impinges on it before it comes momentarily to rest and reverses its direction.
How will the distance of closest approach be affected when the kinetic energy of the a-particle is doubled?
The ground state energy of hydrogen atom is -13.6 eV. If an electron makes a transition from an energy level —0.85 eV to -3.4 eV, calculate the wavelength of the spectral line emitted. To which series of hydrogen spectrum does the wavelength belong?
Answer : Let r0 be the centre to centre distance between the alpha-particle and nucleus.
Question.27. Define relaxation time of the free electrons drifting in a conductor. How is it related to the drift velocity of free electrons? Use this relation to deduce the expression for the electrical resistivity of the material.
Answer : Relaxation time (τ), is the time for which a free electron accelerates before it undergoes a collision with the positive ion in the conductor. Or, we can say it is the average time elapsed between two successive collisions. It is of the order 10-14 seconds. It decreases with increase of temperature and is given as
Question.28. (a) In Youngs double slit experiment, derive the condition for
(i) constructive interference and
(ii) destructive interference at a point on the screen.
(b) A beam of light consisting of two wavelengths, 800 nm and 600 nm is used to obtain the interference fringes in a Youngs double slit experiment on a screen placed 1.4 m away. If the two slits are separated by 0.28 mm, calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide. OR
(a) How does an unpolarized light incident on a polaroid get polarized? Describe briefly, with the help of a necessary diagram, the polarization of light by reflection from a transparent medium.
(b) Two polaroids ‘A’ and ‘B’ are kept in crossed position.
How should a third polaroid ‘C’ be placed between them so that the intensity of polarized light transmitted by polaroid B reduces to l/8th of the intensity of unpolarized light incident on A?
Answer : (a) Condition of constructive and destructive interference :
In the given figure, S is a monochromatic source of light. S1 and S2 are two narrow pin holes equidistant from S and they act as coherent sources. Cofasider a point P on the screen XY placed parallel to S1 and S2.
Let a1 be the amplitude of the waves from S1 and S2 that from S2. Let ø be the phase difference between the two waves reaching the point P. Let y1 and y2 be the displacements of the two wavews, arriving at P.
Clearly the maximum intensity is obtained in the region of superposition at those points where waves meet in the same phase or the phase difference between the waves is even multiple of π or path difference between them is the integral multiple of λ and maximum intensity is (a1 + a2)2 which is greater than the sum intensities of individual waves by an amount 2a1 + a2.
(a) Polaroid is made up of a special material which blocks one of the two planes of vibration of an electromagnetic wave. Because of its chemical composition it allows only those vibrations of the electromagnetic wave which are parallel to its crystallographic axis.
An ordinary beam of light on reflection from a transparent medium becomes partially polarised. The degree of polarization increases as the angle of incidence is increased. At a particular value of angle of incidence, the reflected beam becomes completely polarised. This angle of incidence is called the polarizing angle (ip).
Question.29. (a) Describe briefly, with the help of a diagram, the role of the two important processes involved in the formation of a. p-n junction.
(b) Name the device which is used as a voltage regulator. Draw the necessary circuit diagram and explain its working.
(a) Explain briefly the principle on which a transistor- amplifier works as an oscillator. Draw the necessary circuit diagram and explain its working.
(b) Identify the equivalent gate for the following circuit
and write its truth table.
Answer: (a) Two important process involved in the formation of a p-n junction are :
Diffusion and Drift: In n-type semiconductor, electrons are the majority carriers and holes are minority carriers. In the same way, in p-type semiconductor holes are majority and electrons are minority carriers. During the formation of p-n junction, due to concentration gradient, the holes diffuse from p side to n side and electrons diffuse from n side to p side. This motion gives rise to diffusion current across the junction.
When an electron diffuses from n to p side, it leaves behind a positive charge. In such a manner a positively charged layer forms on «-side of the junction.
Similarly, when a hole diffuses from p to n side, it leaves behind a negative charge and a negatively charged layer forms on p side of the junction.
This space charge region is known as depletion region. An electric field directed from positive charge towards negative charge develops. Due to this field, electrons on p side of the junction move to w-side and holes on n side of the junction move to p side. This motion of charge carriers due to the electric field is called drift.
Drift current is opposite in direction to the diffusion current. Electron drift E Electron diffusion
Initially, diffusion current is large and‘drift current is small. Space-charge region on either side increases as the diffusion process continues. This increases the electric field and hence the drift current. This process continues until the diffusion current equals the drift current. Thus, a p-n junction is formed.
Voltage regulator converts an unregulated dc voltage into a constant regulated dc voltage using zener diode. The un regulated voltage is connected to the zener diode through a series resistance Rs such that the zener diode is reverse biased. If the input voltage increases, the current through Rs and zener diode also increases. This increases the voltage drop across Rs without any change in voltage drop across zener diode. This is because in the breakdown region, zener voltage remain constant even though the current through zener diode changes.
Similarly, if the input voltage decreases, the current through Rs and zener diode decrease. The voltage drop across Rs decrease without any change in the voltage across the zener diode. Thus any change in input voltage results the change in voltage drop across Rs without any change in voltage across the zener diode.
Thus, zener diode acts as a voltage regulator.
(a) Transistor amplifier as an oscillator: In an oscillator, the output at a desired frequency is obtained without applying any external input voltage.
The common emitter n-p-n transistor as an oscillator is shown in the following figure.
A variable capacitor C of suitable range is connected in parallel to coil T2 to give the variation in frequency.
Oscillator action: As in an amplifier, the base-emitter junction is forward biased while the base collector junction is reverse biased. When the switch S is put on, charge of collector flows in the coil T2. The inductive coupling between coil T2 and T1 cause a current to flow in the emitter circuit i.e. feedback from input to output. As a result of positive feedback, the collector current reaches at maximum. When there is no further feedback from T2 to T1, the emitter current begins to fall and collector current decreases. Therefore, the transistor has reverted back to its original state. The whole process now repeats itself. The resonance frequency (f) of the oscillator is given by:
(a) Explain giving reasons, the basic*difference in converting a galvanometer into (i) a voltmeter and (ii) an ammeter.
(b) Two long straight parallel conductors carrying steady currents Ii and I2 are separated by a distance ‘tP. Explain briefly, with the help of a suitable diagram, how the magnetic field due to one conductor acts on the other. Hence deduce the expression for the force acting between the two conductors. Mention the nature of this force.
The direction of electric and magnetic forces are in opposite direction. Their magnitudes are in such a way that they cancel out each other to give net force zero and so the charge particle does not deflect.
(b) When an electric current flows in closed loop of wire, placed in a uniform magnetic field, the magnetic forces produce a torque which tends to rotate the loop so that area of the loop is perpendicular to the direction of the magnetic field.
(a) (i) In converting a galvanometer into a voltmeter, a very high suitable resistance is connected is series to its coil. So, r that galvanometer gives foil scale deflection.
(ii) In converting a galvanometer itjto an ammeter, a very small suitable resistance is connected in parallel to its coil. The remaining pair of the current i.e. (I – Ig) flows through the resistance.
(b) Assumption : Current flows in the same direction.
Using Right hand thumb rule, the direction of the magnetic field at point P due to current I2 is perpendicular to the plane of paper and inwards.
Similarly, at point Q on x2Y2, the direction of magnetic field due to current I1 is perpendicularly outward. ‘
Using Flemings left hand rule we can find the direction of forces F12 and F21 which are in opposite directions thus,
By Amperes circuital law, we have,
From above we get the magnitude of forces F12 and F21 are equal but in opposite direction. So,F12 = -F21 Therefore, two parallel straight conducting carrying current in the same direction attract each other. Similarly, we can prove if two parallel straight conductors carry currents in opposite direction, they repel each other with the same magnitude as equation (1).
Note : Except for the following questions, all the remaining questions have been asked in Set-I 2012.
Question.8. In a single-slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?
Answer : In single-slit diffraction experiment fringe width is given as
If d is doubled the width of the central maximum is halved. Thus there is a reduction in the size of the central diffraction band. Intensity of central band of the diffraction pattern varies square of the slit width so as the slit gets double, intensity will get four times.
Question.19. You are given three lens L1, L2, L3 each of focal length 15 cm. An object is kept at 20 cm in front of L1, as shown. The final real image is formed at the focus ‘I’ of L3. Find the separations between L1, L2 and L3
Here, image by L3 is formed at focus. So the object should lie at infinity of L3. Hence, L2 will produce image at infinity. So, we can conclude that object for L2 should be at its focus.
But, we have seen above the image by L1 is formed at 60 cm right of L1 which is at 15 cm left of L2 (focus of L2).
So, x1 = distance between L1 and L2 = (60 +15) cm = 75 cm
Again distance between L2 and L3 does not matter as the image by L2 is formed at infini ty so X2 can take any value.
Question.27. In a Geiger Marsden experiment, calculate the distance of closest approach to the nucleus of Z = 75, when an oc-particle of 5 MeV energy impinges on it before it comes momentarily to rest and re verse its direction.
How will the distance of closest approach be affected when the kinetic energy of the oc-particle is doubled ?
The ground state energy of hydrogen atom is -13.6 eV. If an electron make a transition from an energy level -0.85 eV to -1.51 eV, calculate the wavelength of the spectral line emitted. To which series of hydrogen spectrum does this wavelength belong ?
Answer : Let ro be the centre to center distance between the alpha-particle and nucleus when the a-particle is at its stopping point.
Note : Except for the following questions, all the remaining questions have been asked in Set-I and Set-II 2012.
Question.1. How does the fringe width, in Young’s double-slit experiment change when the distance of separation between the slits and screen is doubled?
Answer: The fringe width becomes double when the distance of separation between the slits and screen is doubled.
Answer : The frequencies of electromagnetic waves have its inherent characteristics. When an electromagnetic wave travels from one medium to another, its wavelength changes but frequency remains unchanged.
Question.7. A proton and an electron have same kinetic energy. Which one has smaller de-Broglie wavelength and why ?
Answer : In terms of kinetic energy, wavelength is given by
an electron, thus a proton has smaller de-Broglie wavelength than a electron for the same kinetic energy.
Question.9.A circular coil of closely wound N turns and radius r carries a current I. Write the expression for the following:
(i) the magnetic field at its centre
(ii) the magnetic moment of this coil
Question.12.A light bulb is rated 150 W for 220 V AC supply of 60 Hz. Calculate
(i) the resistance of the bulb
(ii) the rms current through the bulb
An alternating voltage given by V = 70 sin 100π is connected across a pure resistor of 25Ω. Find
(i) the frequency of the source. (ii) the rms current through the resistor.
Question.20. Explain briefly the following terms used in communication system:
- Transducer : Any device that converts one from of energy to another can be termed as transducer. An electrical transducer may be defined as a device that coverts some physical variable in the electrical signals.
- Repeater : A repeater is a combination of a receiver and a transmitter. A repeater picks up the signal from the transmitter, amplifies and retransmits it to the receiver sometimes with a,change in carrier frequency.
- Amplification : It is a process of increasing the amplitude of a signal using an electronic circuit called the amplifier. Amplification is done at a place between the source and the destination.
Question.22. You are given three lens L1, L2 and L3 each of focal length 10 cm. An object is kept at 15 cm in front of L1, as shown. The final real image is formed at the focus ‘I’ of L3. Find the separations between L1, L2 and L3 .
Here, image by L3 is formed at focus. So the object should lie at infinity for L3. Hence, L2 will produce image at infinity. So, we can conclude that object for L2 should be at its focus. But, we have seen above that image by L1is formed at 30 cm right of L1which is at 10 cm left of L2 (focus of L2 ).
So, x1= distance between L1 and L2 = (30 +10) cm = 40 cm Again distance between L2 and L3 does not matter as the image by L2 is formed at infinity so X2 can take any value.