CBSE Previous Year Solved Papers Class 12 Physics Outside Delhi 2016

CBSE Previous Year Solved  Papers  Class 12 Physics Outside Delhi 2016

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions:

1. 1.  All questions are compulsory. There are 26
      questions in all.
   2.  This question paper has five sections : Section A, Section B, Section C, Section D and Section E.
   3. Section A contains five questions of one mark each, Section B contains five questions of two marks each, Section C contains twelve questions of three marks each, Section D contains one value based question of four marks and Section E contains three questions of five marks each.
   4.  There is no overall choice. However, an internal choice has been provided in one question of two marks, one question of three marks and all the three questions of five marks weightage. You have to attempt only one of the choices in such questions.
   5. You may use the following values of physical constants wherever necessary:

SET I

SECTION – A

Question. 1. A charge ‘q’ is moved from a point A above a dipole of dipole moment ‘p’ to a point B below the dipole in equatorial plane without acceleration. Find the work done in the process.

Answer: Work done, W=q(V_B -V_A )=q x 0=0. W=0

Question. 2. In what way is the behaviour of a diamagnetic material different from that of a paramagnetic, when kept in an external magnetic field ?
Answer : The magnetic field lines pass through the paramagnetic material while the magnetic field lines move away from the diamagnetic material.

Question. 3. Name the essential components of a communication system.
Answer: The essential components are: Transmitter, communication channel and receiver

Question. 4. Why does sun appear red at sunrise and sunset?
Answer: Sun appears red at sunrise and sunset due to the scattering of light.

Question. 5. The plot of the variation of potential difference across a combination of three identical cells in series, versus current is shown below. What is the emf and internal resistance of each cell ?

SECTION-B

Question.6. Define modulation index. Why is it kept low ? What is the role of a band pass filter ?
Answer : Modulation Index : It is the ratio of the amplitude of the modulating wave to that of the carrier wave.
So, m_a =E_m /E_c.
The modulation index is kept low because when Em is greater than Ec, the carrier is over-modulated, i.e., ma> 1, which causes distortion during the reception.The main function of a bandpass* filter is to limit the bandwidth of the output signal to the band allocated for the transmission.

Question.7. A ray PQ incident normally on the refracting face BA is refracted in the prism BAC made of material of refractive index 1.5. Complete the path of ray through the prism. From which face will the ray emerge ? Justify your answer.

Answer: The ray PQ suffers the total internal reflection at the other face AC and will come out along that face only. This means that the ray will strike the other face AC just at the critical angle, C.

Question.8. Calculate the de-Broglie wavelength of the electron orbitting in the n = 2 stage of hydrogen atom.

Question. 9. Define ionization energy.
How would the ionization energy change when electron in hydrogen atom replaced by a particle of mass 200 times that of the electron but having the same charge?
OR
Calculate the shortest wavelength of the spectral lines emitted in Balmer series. [Given Rydberg constant, R=10⁷m^-1]
Answer: Ionization energy is defined as the amount of energy needed to remove the valence electron of an isolated gaseous atom.

Question. 10. A battery of emf 12V and internal resistance 2Ω connected to a 4Ω is resistor as shown in the figure.
(a) Show that a voltmeter when placed across the cell and across the resistor, in turn, gives the same reading.
(b) To record the voltage and the current in the circuit, why is voltmeter placed in parallel and ammeter in series in the circuit ?


So, a voltmeter when placed across the cell and across the resistor, gives the same reading.
(b) An ammeter is connected in series because it has very low resistance. So, when, an ammeter is connected in series, then there is not much increase in the resistance of the Circuit and hence the current through the circuit is unchanged.
A voltmeter is connected in parallel because it has resistance. It draws a very small current from the circuit.

SECTION – C

Question. 11. Define an equipotential surface. Draw equipotential surfaces:
(i) in the case of a single point charge and
(ii) in a constant electric field in Z-direction. Why the equipotential surfaces about a single charge are not equidistant ?
(iii) Can electric field exist tangential to an equipotential surface ? Give reason.

The electric field due to single charge is not constant, this is the reason why the equipotential surfaces about a single charge are not equidistant,
(iii) No, electric field cannot exist tangential to an equipotential surface. If happen so then a charged particle will experience a force along the tangential line and can move along it. As a charged particle can move only due to the potential difference means along the direction of charge of potential, this contradicts the concept of an equipotential surface.

Question. 12. (i) State law of Malus.
(ii) Draw a graph showing the variation of intensity (I) of polarised light transmitted by an analyser with angle (θ) between polariser and analyse.
(iii) What is the value of refractive index of a medium- of polarising angle 60°?
Answer : (i) Malus discovered that when a beam of completely plane polarized light is passed through the analyser, the intensity T of transmitted light changes directly as the square of the cosine of the angle 0 between the transmission directions of polarizer and analyzer. This is known as the law of Malus.

Then, the refractive index of the material becomes 1.73.

Question.13. Sketch the graphs showing variation of stopping potential with frequency of incident radiations for two photosensitive materials A and B having threshold frequencies v_A> v_B.
(i) In which case is the stopping potential more and why ?
(ii) Does the slope of the graph depend on the nature of the material used ? Explain.
Answer: (i) The stopping potential of metal A is more than that of metal B because the stopping potential increases linearly with the increase in frequency of incident radiation.

(ii) No, the slope of the graph tells us the value of h/e which is same for both the materials. So, it does not depend on the nature of the materials.

Question.14.

Question.15. (i) Derive an expression for drift velocity of free electrons.
(ii) How does drift velocity of electrons in a metallic conductor vary with increase in temperature ? Explain.
Answer: (i) Consider a conductor in which an electric field E is produced. Let a free electron experience a force (-eE) in this electric field. So, the acceleration of free electron is


(ii) The drift velocity of free electrons in a metallic conductor decreases with increase in temperature, because, if we increase the temperature of the metallic conductor the collision between the electrons and ions increases, which decrease in the relaxation time. Hence, drift velocity decreases.

Question.16. (i) When an AC source is connected to an ideal inductor show that the average power supplied by the source over a complete cycle is zero.
(ii) A lamp is connected in series with an inductor and an AC  source. What happens to the brightness of the lamp when the key is plugged in and an , iron rod is inserted inside the inductor ? Explain.


So, when an ac source is connected to an ideal inductor, the average power supplied by the source over a complete cycle is zero.
(ii) The brightness of the lamp will decrease. When the key is plugged in and the iron rod is inserted inside the inductor, it decreases the inductance. Hence, the reactance of the inductor (X_L = coL) decreases. So, the impedance of the circuit increases, which decreases the current in the circuit.

Question.17. (i) Explain with the help of a diagram the formation of depletion region and barrier potential in a p-n junction.
(ii) Draw the circuit diagram of a half wave rectifier and explain its working.
Answer: (i) During the formation of p-n junction, the holes diffuse from p-type semiconductor to the n- type, and electrons diffuse from n-type to p-type. This is because of the concentration gradient across p-side and n-side.
When a hole diffuse from p to n type, it leaves an immovable negative charge. Similarly, when an electron diffuses from n to p type, it leaves an unmovable positive charge. When the diffusion of holes and electrons takes place continuously across the junction, a layer of unmovable positive and negative charges are developed on either side of the junction. This layer is called the depletion layer or the depletion region.
(ii) Half wave rectifier:

Working : when an input ac voltage is applied across the primary coil, a potential difference is developed across the ends of the secondary coil. Consider that in half cycle of input ac signal, the end A acts as the +ve end and B acts as the-ve end of the battery.
So, the diode is in forward bias and we get output across the ends of the load resistance RL.
In the second half cycle, ends A and B reverse in polarity. Now, A acts as the-ve end and B acts as the +ve end. So, the diode D is in reverse bias and no output is obtained due to the high resistance offered by the diode.

So, in this process, we get output alternately, and hence the diode is called the half wave rectifier.

Question.18. (i) Which mode of propagation is used by short wave broadcast service having frequency range from a few MHz up to 30 MHz ? Explain diagrammatically how long distance communication can be achieved, by this mode.
(ii) Why is there an upper limit to frequency of waves used in this mode ?
Answer: (i) Sky wave propagation is used by short wave broadcast services having frequency range from a few MHz up to 30 MHz.

The sky waves reach the receiver after reflection from the ionosphere The oscillating electric field of electromagnetic wave changes the velocity of the electrons in the ionosphere which changes the effective dielectric constant and hence refractive index. In a single reflection from the ionosphere, radio-waves cover a distance of not less than 4000 km. In this way, a very long distance communication is possible with the help of sky waves.
(ii) There is an upper limit to frequency of waves used in this mode because waves above 40 MHz get refracted through ionosphere and escape.

Question.19. (i) Identify the part of the electromagnetic spectrum which is:
(a) suitable for radar system used in aircraft navigation,
(b) produced by bombarding a metal target by high speed electrons,
(ii) Why does a galvanometer show a momentary deflection at the time of,charging or discharging a capacitor ? Write the necessary expression to explain this observation.
Answer: (i) (a) Microwave (b) X-rays
(ii) During the charging and discharging of a capacitor, a flow of charges take place from the battery to the plates of the capacitor. This produces a conduction current in the circuit and hence the galvanometer shows a momentary deflection.

This expression is called the generalized expression of Ampere’s law.

Question.20. For a CE-transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 2 V. Suppose the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1 kΩ.

= 2/2000 x 100 = 10 µA.

Question.21. Define the term wave front. State Huygen’s principle. Consider a plane wave front incident on a thin convex lens. Draw a proper diagram to show how the incident wave front traverses through the lens and after refraction focusses on the focal point of the lens, giving the shape of the emergent wave front.
OR
Explain the following, giving reasons:
(i) When monochromatic light is incident on a surface separating two media, the reflected and refracted light both have the same frequency as the incident frequency.
(ii) When light travels from a rarer to a denser medium, the speed decreases. Does this decrease in speed imply a reduction in the energy carried by the wave ?
(iii) In the wave picture of light, intensity of light is determined by the square of the amplitude of the wave. What determines the intensity in the photon picture of light ?
Answer: Wave Front: A wave front is the locus of all the points in space that reach a particular distance by a propagating wave in same phase at any instant. Huygen’s Principle: It is based on two assumptions:
(a) Each point of the wave front behaves like a source of secondary wavelets. These secondary wavelets spread out in all directions with the same speed as that of the original wave.
(b) When we draw an envelope in the forward direction of the secondary wavelets at any instant, then this envelope tells the new position of the wave front at that instant.

OR

1. Both the reflection and refraction takes place due to the interaction of light with the atoms at the surface of the separation. Light incident on these atoms, force them to vibrate with the frequency of light. But, the light emitted by these charged atoms is equal to their own frequency of oscillation. So, both the reflected and refracted lights have same frequency.
2. The energy carried by a wave depends on the amplitude of the wave. It does not depend on the speed of the wave propagation. Hence the energy of the wave remains same and does not decrease.
3. The intensity of light is determined by the number of photons incident per unit area around the point at which intensity is to be determined.

Question.22. Use Biot-Savart law to derive the expression for the magnetic field on the axis of a current carrying circular loop of radius R.
Draw the magnetic field lines due to a circular wire carrying current I.
Answer: Imagine a circular coil of radius R with centre O. let the current flowing through the circular loop be I. Let the circular coil be made of a large number of small elements of current, each having a length of dl.

According to Biot-Savart’s law, the magnetic field at the centre of the circular loop due I dl will be

SECTION -D

Question.23. Ram is a student of class X in a village school. His uncle gifted him a bicycle with a dynamo fitted in it. He was very excited to get it. While cycling during night, he could light the bulb and see the objects on the road. He, however, did not know how this device works. He asked this question to his teacher. The teacher considered it an opportunity to explain the working to the whole class.
Answer the following questions:
(a) State the principle and working of a dynamo.
(b) Write two values each displayed by Ram and his schoolteacher.
Answer: (a) A dynamo works on the principle of electro magnetic induction

A dynamo includes a coil attached to a small turbine fitted with a plastic cap.
The coil is placed in a magnetic field. When the plastic cap comes in contact with moving tyres of the bicycle, the coil placed between the poles of a ‘ magnet rotates, thus, the flux through the coil changes continuously. This induces a current in the coil which is connected to a bulb which lights up.
As long as the bicycle is moving, the coil keeps on rotating, and hence, the flux keeps on changing. At a steady rate, we get a steady current and hence a light of steady intensity.
(b) The values shown by Ram:
(i) He doesn’t hesitate in asking the questions.
(ii) Curious to know the scientific reasons behind the working of dynamo.
The values shown by the school teacher:
(i) Good command in his subject and
(ii) Helping other students also in understanding the concept.

SECTION – E

Question.24. (i) Draw a labelled diagram of a step-down transformer. State the principle of its working.
(ii) Express the turn ratio in terms of voltages.
(iii) Find the ratio of primary and secondary currents in terms of turn ratio in an ideal transformer.
(iv) How much current is drawn by the primary of a transformer connected to 220 V supply when it delivers power to a 110 V — 550 W refrigerator ?
OR
(a) Explain the meaning of the term mutual inductance. Consider two concentric circular coils, one of radius r₁ and the other of radius r₂ (r₁ <r₂)Placed co axially with centres coinciding with each other. Obtain the expression for the mutual inductance of the arrangement.
(b) A rectangular coil of area A, having number of turns N is rotated at f revolutions per second in a uniform magnetic field B, the field being perpendicular to the coil. Prove that the maximum emf induced in the coil is 2 πf NBA

Principle: A transformer works on the principle of mutual induction. Whenever the amount of magnetic flux, linked with a coil changes, an emf is induced in the neighbouring coil.
Working : When alternating current source is connected to the ends of primary coil, the current changes continuously in the primary coil, due to which magnetic flux linked with the secondary coil changes continuously. Therefore, the alternating emf of same frequency is developed across the secondary terminals. According to Faraday’s laws the e.m.f. induced in the primary coil.


OR
(a) Mutual Inductance: It is a phenomenon, when the current in an inductor changes, flux varies and it cuts any other inductor nearby, producing induced voltage in both inductors.
Consider two coils A and B and a time varying current I is flowing through the coil A then,

(b) Let N be the number of turns of the rectangular coil and A be its cross-sectional area placed in a magnetic field B, then, the magnetic flux linked with the coil,

Question.25. (i) Derive the mathematical relation between refractive indices n₁ and n₂ of two media and radius of curvature R for refraction at a convex spherical surface. Consider the object to be a point since lying on the principle axis in rarer medium of refractive index n₂ and a real image formed in the denser medium of refractive index n₂. Hence, derive lens maker’s formula.
(ii) Light from a point source in air falls on a convex spherical glass surface of refractive index 1.5 and radius of curvature 20 cm. The distance of light source from the glass surface is 100 cm. At what position is the image formed ?
OR
(a) Draw a labelled ray diagram to obtain the real image formed by an astronomical telescope in normal adjustment position. Define its magnifying power.
(b) You are given three lenses of power 0.5 D, 4 D and 10 D to design a telescope.
(i) Which lenses should he used as objective and eyepiece ? Justify your answer.
(ii) Why is the aperture of the objective preferred to be large ?
Answer:


In deriving Len’s maker formula, we adopt the coordinate geometry sign convention and make the assumptions:
(i) The lens is thin so that the distances measured from the poles of its two surfaces can be taken as equal to the distances from its optical centre.
(ii) The aperture of the lens is small.
(iii) The object is a point-object placed on the principal axis of the lens.
(iv) The incident and the refracted rays make small angles with the principal axis.


This is the lens maker’s formula. It has been derived for a convex lens forming a real image, but it is equally applicable to a convex lens forming a virtual image and to a concave lens which forms only virtual images, comparing eq. (iii) and (iv), we have


Magnifying power : The magnifying power of a refracting type astronomical telescope is defined as the ratio of angle subtended by the final image at eye to the angle subtended by the object at eye.
(b) (i) For higher magnification of the telescope, the value of f₀ > f_e
For this, the power should be the least. So, the objective lens should be 0.5D.
(ii) The aperture of the objective lens is made larger so, that it receives as much as light coming from the distant object and the . resolving power of the telescope increase.

Question.26. (i) Use Gauss’s law to find the electric field due to a uniformly charged infinite plane sheet. What is the direction of field for positive and negative charge densities ?
(ii) Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors C₁ and C₂ with their capacitances in the ratio 1: 2 so that the energy stored in the two cases becomes the same.
OR
(i) If two similar large plates, each of area A having surface charge densities + σ and — σ are separated by a distance in air, find the expressions for
(a) field at points between the two plates and on outer side of the plates. Specify the direction of the field in each case.
(b) the potential difference between the plates.
(c) the capacitance of the capacitor so formed.
(ii) Two metallic spheres of radii R and 2R are charged so that both of these have same surface charge density σ. If they are connected to each other with a conducting wire, in which direction will the charge flow and why ?

Consider a thin infinite uniformly charged plane sheet having the surface charge density of σ. The electric field is normally outward to the plane sheet and is same in magnitude but opposite in direction. Now, draw a Gaussian surface in the form of cylinder around an axis. Let its cross-sectional area be A. The cylinder is made from three surfaces A, S₂, and A’ and the electric flux linked with S₂ is 0. So, the total electric flux linked through the Gaussian surface is


OR
(i) (a) Consider a parallel plate capacitor with two identical plates X and Y, each having an area of A, and separated by a distanced.Let the space between the plates be filled by a dielectric medium with its dielectric constant as K and a be the surface charge density on each of the plates.