CBSE Class12 - Physics Paper Solved - 2015...1) Interference is the superposition of waves originating from two coherent sources of light whereas diffraction is the superposition of

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CBSE Class12 - Physics Paper Solved -2015

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Question 1. What happens when a forward bias is applied to a p-n junction?

Solution. When p-n junction is forward biased, the holes are accelerated and they cross the barrier to reach in region of N type of semiconductor. This increases the current in the p-n junction circuit. The width of potential barrier deceases.

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Question 2. Write the important characteristic features by which the interference can be distinguished from the observed diffraction pattern ?

Solution. Difference Between Interference And Diffraction:1) Interference is the superposition of waves originating from two coherent sources of light whereas diffraction is

the superposition of the wavelets coming from different parts of wave fronts of a wave.

2) The interference pattern is of uniform brightness throughout , whereas in diffraction the brightness of fringes keeps decreasing with the rise in the diffraction pattern.

3) The fringe width of interference pattern may be equal or not, but in case of diffraction fringes are always of unequal width.

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Question 3. Explain the basic differences between the construction and working of a telescope and a microscope ?

Solution. Basic Difference In Between Microscope And Telescope:(1) The compound microscope has convex lens of longer focal length as objective and eyepiece of shorter focal.

(2) Tube Length Of Microscope, L= VO + Ue, since v is constant whereas ue can be varied. In case of telescope, length of the tube L= FO+ Ue,

(3) A microscope is used for viewing a tiny object, whereas telescopes are used for watching distant and heavenly objects.


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Question 4. Arrange the following electromagnetic wave in the order of their increasing wavelength:(a) γ- rays(b) Microwaves(c) X-rays(d) Radio wavesHow are infrared waves produced? What role dose infra-red radiation play in (i) maintains the Earth’s warmth and (ii) physical therapy?

Solution. Following is the arrangement of electromagnetic wave in the increasing order of their wavelength:a) Gamma raysb) X-raysc) Microwavesd) Radio waves(i) When sunlight falls on the atmosphere of earth, a major part of the ultraviolet radiation of sunlight is reflected in thespace. The infrared radiations of sunlight are absorbed by the atmosphere after multiple reflection because infrared radiations are not energetic enough to go out of earth’s atmosphere. Infrared radiations has hot radiations so after multiple reflection, they heat up the atmosphere. Infrared radiations falling on earth and warm atmosphere keeps the surface of earth warm.(ii) In physical therapy the warmth and heat energy of infrared radiations are used. These radiations provide a soothing effect in our muscles. They repair and heal up the weak cells and tissues.


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Question 5. The focal length of an equiconvex lens is equal to the radius of curvature of either face. What is the refractive index of the material of the lens?

Solution. The formula of lens maker formula is given as,

1

𝑓= 𝜇 − 1

1

𝑅1

−1

𝑅2

, but we are given R1 = R2 = f = R, therefore,

1

𝑅= 𝜇 − 1

1

𝑅+

1

𝑅,

Or 𝜇 =3

2= 1.5


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Question 6. Write the important characteristic features by which the interference can be distinguished from the observed diffraction pattern.Explain the basic differences between the construction and working of a telescope and a microscope.

Solution. Difference Between Interference And Diffraction:1) Interference is the superposition of waves originating from two coherent sources of light whereas diffraction is the superposition of the wavelets coming from different parts of wave fronts of a wave.2) The interference pattern is of uniform brightness throughout whereas in diffraction the brightness of fringes keeps decreasing with the rise in diffraction pattern.3) The fringe width of interference pattern may be equal or not but in case of diffraction fringes are always of unequal width.

ORBasic Difference In Between Microscope And Telescope:

1) The compound microscope has convex lens of longer focal length as objective and eye piece of shorter focal2) Tube Length Of Microscope, L= VO +Ue , since v0 is constant whereas ue can be varied. In case of telescope, length of the tube L= FO+ Ue,3) Microscope is used for viewing tiny object whereas telescopes are used for watchingdistant and heavenly objects


Physicsmodels.inQuestion 7. State Bohr’s postulate of hydrogen atom which successfully explains the emission lines in the spectrum of hydrogen atom. Use Rydberg formula to determine the wavelength of Hα line.[Given: Rydberg constant R = 1.03 × 107 m–1]

Solution. Bohr’s Postulates Of Hydrogen Atom:1) Positive charges reside in the nucleus of hydrogen atom and electrons revolve around the nucleus. There is electrostatic force of attraction between positive and negative charges. Electrons are orbiting in the stationary orbits due to centripetal force.2) Electrons revolve in the quantized or preferred orbits, the angular momentum of which is integral multiple of h/2π, where h is Plank’s constant i.e. L = m v r = h/2π3) When electrons are energized, they jump up the higher orbit and remain in the excited orbit for about 10–4 s then return to their ground state releasing difference of energy in the form of light.

hν = E2 – E1

For Determination of Wavelength of Hα

R = 1.03 × 107 m-1 (given)Now 1/ƛ = R ( 1/n1

2 – 1/n22 ),for Hα, n1 = 2 and n2 = 3.

Therefore 1/ƛ = 1.03×107 ( 1/22 – 1/32) or 1/ƛ = 1.03 × 107 (1/4 – 1/9 ) or1/ƛ = 1.03 × 107 × 5/36 or ƛ = 36 × 10-7/5.15 =7 × 10 –7 m


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Question 8. State the two Kirchhoff's rules used in electric networks. How are these rules justified?

Solution. Kirchhoff’s Rules Of Electric Network

1) Algebraic sum of current at a junction is always zero i.e.

I1 + I2 +i3+ i4+ ..................... = 0 i.e. 𝑛=1𝑛 𝑖𝑛 = 0

2) The potential difference of the whole network is equal to the sum of the products of current and resistance of the individual network.

𝑛𝑉 = 1𝑛 𝐼𝑅

Justification: At a junction incoming current is equal to outgoing current i.e. there is no deposition of current at the junction.

According to Ohm’s law potential difference of a battery is equal to the product of current and resistance offered the electric network. So total voltage of the electric networks is equal to the total current multiplied by the total resistance of the network.


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Question 9. Given the ground state energy E0 = –13.6 eV and Bohr radius a0 = 0.53 oA . Find out how the de Broglie wavelength associated with the electron orbiting in the ground state would change when it jumps into the first excited state.

Solution: de-Broglie’s Wavelength

Given E1 = –13.6 × 1.6×10-19 J,

E2 = –13.6/22e V = –3.4 ×1.6×10-19 J and c = 3 × 108 ms-1

Now hc/ƛ = E2 – E1

or ƛ = hc/ E2 – E1 = 6.63 × 10-34 × 3 × 108 / (13.6 – 3.4) × 1.6 × 10-18

ƛ = 6.63 × 3 × 10-7 / 10.2 × 1.6 = 1.22 × 10-7 m.


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Question 10.a) Distinguish between ‘Analog’ and ‘Digital’ forms of communication.b) Explain briefly two commonly used applications of the ‘Internet’.

Solution. a) Difference Between Analog And Digital Signal Analog Communication: A continuous signal of current or voltage which is in the form of a sinusoidal wave is called analog communication signal. Digital Signal: A form of communication in which level of signal ( voltage or current ) is in the form of two level either 0 or 1. For example voltage signal either 0 or 5 V.

b) Two important applications of Internet are as follows:

i.) Mails can be transferred across the globe within no time using the internet. Yahoomail and gmail helps us to send mails with convenience within no time.ii.) Social networking sites like facebook and twitter helps people to interact with eachother and share their ideas within each other across the globe.


Physicsmodels.inQuestion 11. Plot a graph showing the variation of current density (j) versus the electric field (E) for two conductors of different materials. What information from this plot regarding the properties of the conducting material, can be obtained which can be used to select suitable materials for use in making (i) standard resistance and (ii) connecting wires in electric circuits ? Electron drift speed is estimated to be of the order

of mm s −1. Yet large current of the order of few amperes can be set up in the wire. Explain briefly.

Solution.

Above graphs show the plot of current density verses electric field . Graph A represents that material A is suitable for making connecting wires because in the electric field current density is higher for material A.

Material B is suitable for making resistance wires since in the electric field current density is low for material B.

Corresponding to low drift velocity mm s-1 the electron current flowing through the conductor placed in the electric field is much higher because there are much higher number of electrons per unit volume per unit time flowing through the conductor.


Physicsmodels.inQuestion 12. State Biot - Savart law. Deduce the expression for the magnetic field at a point on the axis of a current carrying circular loop of radius ‘R’ distant ‘x’ from the centre. Hence, write the magnetic field at the centre of a loop.

Solution.

Let us consider a circular coil of radius a with centre O carrying current I. We take a point P at a distance X from the centre of

the circular coil such that AP = BP = r = 𝑎2 + 𝑥2 . Now in order to find magnetic field intensity at an axial point P, we apply Biot savart law due to current element Idl at point A as,

dB =𝝁𝟎𝟒𝝅

𝒙𝑰𝒅𝒍 sin 𝟗𝟎

𝒓𝟐, which is acting perpendicular to AP along PC. AP makes angle Φ with the axial line OP.

Similarly dB’ = 𝝁𝟎𝟒𝝅


𝒓𝟐, which is acting perpendicular to BP along PD.

dB = dB’ = 𝝁𝟎𝟒𝝅

𝒙𝑰𝒅𝒍

𝒓𝟐


Physicsmodels.inResolving dB and dB’ along and perpendicular to axial line., we get, dBsinφ along axial line OP and dB cosφperpendicular to OP. Magnetic field dB sinφ along axial line OP survives, other component is cancelled. Therefore magnetic field at P is given by,

dB sinφ =𝝁𝟎𝟒𝝅


𝒓𝟐= 𝝁𝟎𝟒𝝅

𝒙𝑰𝒅𝒍

𝒓𝟐x 𝒂

𝒓=

𝑰𝒅𝒍 𝒂

𝟒𝝅𝟑 𝒂𝟐+𝒙𝟐

Therefore magnetic field at P is given by,

B =𝝁𝟎𝑰 𝒂

𝟒𝝅𝟑 𝒂𝟐+𝒙𝟐. 𝑑𝑙 =

𝝁𝟎𝑰 𝒂

𝟒𝝅𝟑 𝒂𝟐+𝒙𝟐x 2𝜋𝑎

Case 1. The magnetic field at an axial point P is given by,

B at point P = 𝝁𝟎𝑰 𝒂

𝟒𝝅𝟑 𝒂𝟐+𝒙𝟐x 2𝜋𝑎2

Case 2. The magnetic field at the centre O, of the circular coil i.e. at x=0 is given by,

𝐵0 = 𝝁𝟎𝟒𝝅

𝒙2𝝅 𝑰 𝒂𝟐

𝒂𝟐 𝟑/𝟐 = =

𝝁𝟎𝟒𝝅

𝒙𝟐𝝅𝑰

𝒂= 𝝁𝟎𝟐𝒙

𝑰

𝒂


Physicsmodels.inQuestion 13. What dose a polaroid consist of? Show, using a simple polaroid, that light waves are transverse in nature. Intensity of light coming out of a polaroid does not change irrespective of the orientation of the pass axis of the polaroid. Explain why.

Solution. Polaroid is made of natural tourmaline crystal but we do not have large size tourmaline crystal naturally occurring, tourmaline is not suitable for large size light polarization. Therefore in practice artificial Polaroid is used. Artificial Polaroid is made from needle type crystal of iodoquinine sulphate known as herapathite crystals which are packed side by side, in layers with their axis parallel. Herapathite crystals are very fragile, when they are slightly disturbed. So they protected by keeping them in layers in between the parallel glass plates. In order to polarise and show that light is composed of transverse component, we take two Polaroid , one Polaroid T1 serves as polariser and other Polaroid T2 as analyser. When electromagnetic light is incident on the polariser, it gets polarised i.e. E. M. Wave motion is confined in one particular direction only. Now we place second Polaroid parallel to first polaroid’s axis and analyze the light emerging out from analyser on changing the its orientation about its own plane. We observe that intensity of light is maximum when axes of polariser and analyser are parallel and then goes on deceasing and finally becomes zero or minimum when axes of T1 and T2 are perpendicular to each other as shown in the diagrams. This experimental demonstration shows that light has transverse nature which can be polarised. Had light been longitudinal in nature then intensity of light would have not changed on changing orientation of analyser with respect to polariser.


Physicsmodels.inQuestion 14. How is a Zener diode fabricated? What causes the setting up of high electric field even for small reverse bias voltage across the diode? Describe, with the help of a circuit diagram, the working of Zener diode as a voltage regulator.

OR(a) Explain with the help of a diagram, how depletion region and potential barrier are formed in a junction diode.(b) If a small voltage is applied to a p-n junction diode, how will the barrier potential be affected when it is (i) forward biased, and (ii) reveres biased?

Solution.

Semiconductor diode is fabricated by proper varying level of doping of p type and n type extrinsic semiconductors by amalgamating them at the junction. Zener diode has the characteristic to operate in the break-down region. Zener diode has very sharp breakdown voltage.


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When Zener diode is forward biased , it acts as ordinary diode but when Zener diode is reverse biased as shown in the diagram ,it becomes special diode. This diode works at reverse low voltage. When vz is negative and low, corresponding to –vz, there is sharp increase in the current at breakdown voltage. At vz, the resistance shown by Zener diode, becomes zero. So the negative voltage at which current increases sharply, is called Zener’s breakdown voltage.

The cause of setting high electric field is sharp increase of charge carriers across the junction at breaking voltage.

The characteristics of Zener diode to conduct large current while maintaining low voltage across it, finds application in voltage regulator.

OR

(a) When we diffuse p and n type of semiconductor to form junction diode, at the junction the diffusion of holes from p region and electrons from n region takes place. In this case some of the electrons cross the barrier to reach p region and some of the holes come in the n region. In the p region a layer of electrons and in the p region a layer of holes is formed which repel further movements of charge carriers across the junction as a result of which a potential barrier is formed across the junction. The thickness of this potential barrier is called depletion layer and this region is calleddepletion region as shown in the diagrams.(b) When p n junction diode is forward biased, holes get repelled by positive polarity at the p region and holes reach in the n region to get attracted by negative polarity acting on n region. So junction diode conducts, which reduces potential barrier’s thickness and the thickness of depletion layer also deceases. On the other hand when junction diode is reverse biased, the diode conducts poorly as a result of which thickness of depletion layer and potential barrier increases.


Physicsmodels.inQuestion 15. Light of intensity ‘I’ and frequency ‘v ’ is incident on a photosensitive surface and causes photoelectric emission. What will be the effect on anode current when(i) the intensity of light is gradually increased,

(ii) the frequency of incident radiation is increased, and(iii) the anode potential is increased? In each case, all other factors remain the same. Explain, giving justification in each case.

Solution.(a) Effect Of Intensity (I) Of Light On Anode Current: When we keep increasing the intensity of light falling on photosensitive

surface, the photoelectric current also keep increasing but the energy of each photoelectron remains the same. In doing so, we are increasing number of photons falling on photo metal and photoelectric current is directly proportional to number of photoelectrons emitted by incident photons.

(b) Effect Of Frequency (ν) Of Light On Anode Current: On increasing the frequency of falling light on photo metal, the kinetic energy of photoelectrons emitted from photosensitive surface increases but the photoelectric current remains the same. The kinetic energy of photoelectrons depends on frequency of incident light i.e. higher is the frequency of photons,higher will be the kinetic energy of photoelectrons.

(c) Effect Of Anode Potential On Anode Current: On increasing anode potential, initially anode current increases as more and more number of photoelectrons are attracted by the anode. When we keep on increasing anode potential, after some time anode current does not increase. Soon saturation of anode current approaches because there are no morephotoelectrons left to be collected by increased potential of anode.


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Question 16. When is a transistor said to be in active state? Draw a circuit diagram of a p-n-p transistor and explain how it works as a transistor amplifier. Write clearly, why in the case of a transistor (i) the base is thin and lightly doped and (ii) the emitter is heavily doped.

Solution. When input voltage Vi is greater than 0.6V for silicon semiconducting chip then the transistor conducts. When emitter current increases corresponding to it collector current also increases and if Vi is equal to 1.0V, then transistor is said in the on state. This is called active state of transistor.

For the given amplifier circuit, Ie = Ib + Ic and VCB = VC + ICRC OR VC = VCB – ICRC.

When the positive half cycle of a. c. comes in the input circuit, the emitter base circuit is forward biased and the p n p transistor conducts. The emitter current Ie increases corresponding to which collector current increases and collector voltage decreases means VC becomes less negative, as a result of which output voltage VO gets magnified i.e. outputvoltage increases.


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When negative cycle of input voltage comes then less input current Ie flows. In turn less collector current Ic flows i.e. Ic

decreases. Thus VC becomes more negative and output voltage varies with the input a. c. signal.

i) Base of p-n-p transistor is lightly doped so that hole-electron combination is low and less base current in the emitter-base circuit.

ii) Collector region is heavily doped so that majority charge carrier holes are collected by the collector in the collector-base circuit.


Physicsmodels.inQuestion 17.(a) State three important factors showing the need for translating a low frequency signal into a high frequency wave before transmission.(b) Draw a sketch of a sinusoidal carrier wave along with a modulating signal and show how these are superimposed to obtain the resultant amplitude modulated wave.

Solution. (a) Following are three important factors in choosing to convert low frequency signal into high frequency signal before transmission.1. Low frequency signals are weak to travel long-long distances.2. High frequency signals are energetic enough to travel long distances.3. There is distortion in nature of low frequency signal due to resistances present in atmosphere as a result of which fidelity of the signals is lost.(b) The modulating signals which are weak to travel long distances are made to ride on the energetic carrier waves called superimposition of modulating signal on carrier wave signal. In this process amplitude modulation is one of the methods of superimposition. In amplitude modulation, carrier wave is clipped according to the amplitude of modulating signal andmodulating signal is superimposed on carrier wave as shown in diagrams. This arrangement is called amplitude modulated signal. This modulated signal is energized and amplified before transmission.


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Question 18. You are given three circuit elements X, Y and Z. When the element X is connected across an a.c. source of a given voltage, the current and the voltage are in the same phase. When the element Y is connected in series with X across

the source, voltage is ahead of the current in phase by 𝝅

𝟒. But the current is ahead of the voltage in phase by

𝝅

𝟒when Z

is connected in series with X across the source. Identify the circuit elements X, Y and Z.

When all the three elements are connected in series across the same source, determine the impedance of the circuit.

Draw a plot of the current versus the frequency of applied source and mention the significance of this plot.

Solution. When elements X, Y, Z are connected separately with source of a. c. voltage, according to given conditions, element X is resistor, element Y is inductor and element Z is a capacitor.

On connecting elements X, Y and Z with source of a. c. mains, a. c. circuit becomes resonant circuit and net impedance of the circuit is Z=R, because XL = XC i.e. ω L= 1/ω C.

Alternating current versus frequency graph is given below:


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Question 19. Write symbolically the nuclear β+ decay process of 𝟔𝟏𝟏𝑪 Is the decayed product X an isotope or isobar of 𝟔

𝟏𝟏𝑪 ?

Given the mass values m ( 𝟔𝟏𝟏𝑪 ) = 11.011434 u and m (X) = 11.009305 u. Estimate the Q value in this process.

Solution. ß-decay nuclear equation is given below:

611𝐶 -> 6

11𝐶 + −10𝑒 + 𝐸𝛽

The daughter element X is an isotope of carbon 611𝐶

611𝐶 = 11.011434 𝑢 and 6

11𝐶 = 11.009305 𝑢

Mass difference = (11.011434 -11.009305) u = .002029 u = .002029 × 931.5 M e V =

1.946835 M e V = 1.946835 × 1.6 × 10-13 = 3.114936× 10-13 J.


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Question 20. An object is placed 15 cm in front of a convex lens of focal length 10 cm. Find the nature and position of the image formed. Where should a concave mirror of radius of curvature 20 cm be placed so that the final image is formed at the position of the object itself?

Solution. The focal length of convex lens f = 10 cm.The position of object from the lens u = –15 cm.The lens formula: 1/f = 1/v – 1/u.Therefore 1/10= 1/v – 1/-15 or 1/v = 1/10 – 1/15 or 1/v = 1/30or v =30 cm

Therefore image is formed at a distance of 30 cm from the convex lens on the other side of the lens, it is inverted, real and magnified.Let the concave mirror is placed at a distance u to form image at the same point where object is placed i. e. V = - (30+15) = -45 cm.Focal length of concave mirror f = -10 cm.Mirror formula 1/f = 1/u + 1/v.Therefore 1/u = 1/f – 1/v = 1/-10 – 1/-45. 1/u = (-9 + 2)/90 = –7 / 90or u = –90 / 7 = –13 cm approximately.

Therefore the mirror must be placed at a distance of approximately 13 cm from image formed by the convex lens.


Physicsmodels.inQuestion 21. Arrange the following electromagnetic wave in the order of their increasing wavelength:(a) γ- rays(b) Microwaves(c) X-rays(d) Radio wavesHow are infra-red waves produced? What role dose infra-red radiation play in (i) maintain the Earth’s warmth and (ii) physical therapy?

Solution. Following is the arrangement of electromagnetic wave in the increasing order oftheir wavelength:a) Gamma raysb) X-raysc) Micro wavesd) Radio waves

i) When sun light falls on the atmosphere of earth, major part of the ultraviolet radiation of sun light is reflected in the space. The infrared radiations of sunlight are absorbed by the atmosphere after multiple reflection because infrared radiations arenot energetic enough to go out of earth’s atmosphere. Infrared radiations has hot radiations so after multiple reflection, they heat up the atmosphere. Infrared radiations falling on earth and warm atmosphere keeps the surface of earth warm.

ii) In physical therapy the warmth and heat energy of infrared radiations are used. These radiations provide soothing effect in our muscles. They repair and heal up the weak cells and tissues.


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Question 22. Explain briefly the process of charging a parallel plate capacitor when it is connected across a d.c. battery.A capacitor of capacitance ‘C’ is charged to ‘V’ volts by a battery. After some time the battery is disconnected and the distance between the plates is doubled. Now a slab of dielectric constant, 1 < k < 2, is introduced to fill the space between the plates. How will the following be affected?(a) The electric field between the plates of the capacitor(b) The energy stored in the capacitor Justify your answer by writing the necessary expressions.

Solution. A parallel plate capacitor is charged by connecting it with battery in series with a resistor and a key. The charging of capacitor takes place through resistor R and it is exponential growth of charges through the capacitor. When capacitor of capacitance C is fully charged to its maximum then expression for charging is given below:

q = q0 ( 1- 𝑒𝑡

𝑅𝐶 )

In the above equation of charging of a capacitor, q0 is maximum charging and RC is time constant. Therefore time constant T = RC.When a parallel plate capacitor is connected to a battery of voltage V and after charging battery is removed then potential difference across the plates is maintained to V but on separation of plates to double the previous separation and filling the space by dielectric constant 1 < 𝑘 < 2, the capacitance C decreases.


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a ) Electric field E = 𝑞

∈𝐴=

𝐶𝑉

∈𝐴

Therefore electric field decreases as capacitance of the capacitor decreases.

b ) The energy stored in the capacitor

U = 1

2∈ 𝐸2 =

1∈𝑉2

2 𝑑2

Therefore the energy stored decreases as d increases.


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Question 23. Ajit had a high tension tower erected on his farm land. He kept complaining to the authorities to remove it as it was occupying a large portion of his land. His uncle, who was a teacher, explained to him the need for erecting these towers for efficient transmission of power. As Ajit realized its significance, he stopped complaining. Answer the followingquestions:(a) Why is it necessary to transport power at high voltage?(b) A low power factor implies large power loss. Explain.(c) Write two values each displayed by Ajit and his uncle.

Solution.a ) It is necessary to transmit power at high voltage because at high voltage transmission, power loss and expenditure is less and the conducting metallic required will not cost much. Power company will have to construct less number of towers, so cost of construction is less.

b ) cos∅ is called power factor. In case ∅ is less than 900 then power consumed is more, so there is loss of power.

c) Ajit wanted to get power transmission tower removed because he wanted to utilize his plough able land for crop procurement. The value he exhibited is better land utilization and nurturing of nature. His uncle wanted electricity for future development and progress.


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Question 24.(a) Define self-inductance of a coil. Obtain an expression for the energy stored in a solenoid of self-inductance ‘L’ when the current though it grows from zero to ‘I’.(b) A square loop MNOP of side 20 cm is placed horizontally in a uniform magnetic field acting vertically downwards as shown in the figure. The loop is pulled with a constant velocity of 20 cm s−1 till it goes out of the field.

(i) Depict the direction of the induced current in the loop as it goes out of the field. For how long would the current in the loop persist?(ii) Plot a graph showing the variation of magnetic flux and induced emf as a function of time.

OR


(a) Draw the magnetic field lines due to a circular loop area A carrying current I. Show that it acts as a bar magnet of magnetic moment m = IA.

(b) Derive the expression for the magnetic field due to a solenoid of length ‘2l’, radius ‘a’ having ’n’ number of turns per unit length and carrying a steady current ‘I’ at a point on the axial line, distance ‘r’ from the centre of the solenoid. How does this expression compare with the axial magnetic field due to a bar magnet of magnetic moment ‘m’?

Solution.(a) Self inductance is defined as induced emf produced per unit rate change of current through the inductance coil i.e.

L = -𝑒𝑑𝑖

𝑑𝑡

or L = - e ; 𝑑𝑖

𝑑𝑡= 1 As-1

(b)(i) So far square loop MNOP is inside the magnetic field ,there is no induced current produced but when coil moves out

of magnetic field , induced current is produced in the clockwise direction of square loop MNOP. This induced current lasts as soon as change in magnetic flux lasts.

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(ii)

ORa )


Physicsmodels.inWhen current I passes through the circular coil of area A, the coil behaves as a bar magnet and magnetic dipole moment M of the coil is perpendicular to the plane of the coil. It has observed that :

M ∝ I, M ∝ A. Therefore M ∝ I A OR M = K I A, where K is proportionality constant. If I = 1, M = 1 and a = 1 then K = 1. Therefore M = I A.

b ) We know, the magnetic field intensity due current (I) carrying circular coil of radius a at its axial point P at a distance r from its centre is given by:

B= 𝝁𝟎𝟒𝝅

𝒙𝟐𝝅𝑰𝒂𝟐

𝒂𝟐+ 𝒓𝟐𝟑

𝟐

if there are n number of turns per unit length of a long solenoid then there are n dx number of turns in dx length of solenoid.Therefore magnetic field due to these turns is given by,

dB= 𝝁𝟎𝟒𝝅

𝒙𝟐𝝅𝑰𝒂𝟐𝒏𝒅𝒙

𝒂𝟐+ 𝒓𝟐𝟑

𝟐

Now, sin 𝜃 =𝑎

𝑧= 𝑧𝑑𝜃

𝑑𝑥or dx =

𝑎𝑑𝜃

sin 𝜃

Putting these values in the formula of Db, we get,

dB= 𝝁𝟎𝟒𝝅

𝒙𝟐𝝅𝑰𝒂𝟐𝒏 𝒂𝒅𝜽

𝒛𝟑 sin 𝜽, but sin 𝜃 =

𝑎

𝑧, therefore


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dB =𝝁𝟎 𝒏 𝑰𝒂

𝟐 𝒅𝜽

𝟐 𝒛𝟑 sin 𝜽= 𝝁𝟎 𝒏 𝑰 sin 𝜽 𝒅𝜽

𝟐or

dB =𝝁𝟎 𝒏 𝑰

𝟐 𝜽𝟏𝜽𝟐sin 𝜽𝑑𝜃 =

𝝁𝟎 𝒏 𝑰

𝟐−cos 𝜃

𝜃1

𝜃2

B = 𝝁𝟎 𝒏 𝑰

𝟐cos 𝜃1 − cos 𝜃2 , if θ1 = 0 and θ2 = π 𝑡ℎ𝑒𝑛

B = 𝝁𝟎 𝒏 𝑰

𝟐1 − (−1) = 𝜇0 𝑛 𝐼

This is the magnetic field due to long solenoid of length 2l at axial point at a distance r.


Physicsmodels.inQuestion 25.(a) In young’s double slit experiment, deduce the conditions for obtaining constructive and destructive interference fringes. Hence, deduce the expression for the fringe width.(b) Show that the fringe pattern on the screen is actually a superposition of slit diffraction from each slit.(c) What should be the width of each slit to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern, for green light of wavelength 500 nm, if the separation between two slits is 1 mm?

OR(a) Two thin convex lenses L1 and L2 of focal lengths f1 and f2, respectively, are placed coaxially in contact. An object is placed at a point beyond the focus of lens L1. Draw a ray diagram to show the image formation by the combination and hence derive the expression for the focal length of the combined system. (b) A ray PQ incident on the face AB of a prismABC, as shown in the figure, emerges from the face AC such that AQ = AR.

Draw the ray diagram showing the passage of the ray through the prism. If the angle and refractive index of the material of prism is 60° of the prism is 3 , determine the values of angle of incidence and angle of deviation.


Physicsmodels.in

a) Condition For Constructive And Destructive Interference:Let us consider waves originating from two coherent sources as,𝑦1 = 𝑎 sin 𝜔𝑡 ---(1)𝑦2 = 𝑏 sin(𝜔𝑡 + 𝜑) ---(2)

where a, b and 𝜑 are amplitudes and phase difference respectively.

According to superposition principle, the resultant wave ‘s displacement is given by,

y = 𝑦1 + 𝑦2= 𝑎 sin 𝜔𝑡 + 𝑏 sin(𝜔𝑡 + 𝜑)y = [𝑎 sin 𝜔𝑡 + bsin 𝜔𝑡 𝑐𝑜𝑠𝜑 + 𝑏𝑐𝑜𝑠𝜔𝑡𝑠𝑖𝑛𝜑] = sin 𝜔𝑡 [a + 𝑏𝑐𝑜𝑠𝜑] + b𝑐𝑜𝑠𝜔𝑡𝑠𝑖𝑛𝜑.

Now let us assume R cosθ = a + 𝑏𝑐𝑜𝑠𝜑] ---(3)and R sinθ = b 𝑠𝑖𝑛𝜑 ---(4)


Physicsmodels.inPutting these substitutions in above equation, we get,

y = R sin [𝜔𝑡 + θ] ----(5)

Equation (5) represents the resultant wave, the amplitude of which is given by,

𝑅2 [𝑠𝑖𝑛2θ + 𝑐𝑜𝑠2θ] = [a + 𝑏𝑐𝑜𝑠𝜑]2 + [b 𝑠𝑖𝑛𝜑]2

R2 = a2 + b2 + 2ab cos𝜑 or R = a2 + b2 + 2ab cos𝜑

Condition For Constructive Interference : In order to have intensity of interfering waves maximum,

I ∝ 𝑅2 i. e. I ∝ a2 + b2 + 2ab cos𝜑 .

Cos𝜑 = MAX =+ 1. Therefore 𝜑 = 0, 2Π, 4Π, 6Π ---i.e. 𝜑 = 2𝑛Π, where n = 1,2, 3, 4.................Path difference = x = ƛ /2Π => 𝜑 = n ƛCondition For Destructive Interference:For this I should be minimum i. e. Cos𝜑 =minima = - 1. . Therefore 𝜑 = Π, 3Π ---i.e. 𝜑 = (2n − 1) Π, where n = 1,2, 3, 4.................


Physicsmodels.inPath difference = x = ƛ/2Π => 𝜑 =(2n − 1) ƛ /2.

Fringe Width: The interfering waves reach to point P from two coherent sources S1 and S2 , of light as shown in the diagram. The path difference between the two waves is given by,

Path difference = S2P – S1P = BP – AP

Now from triangles ΔAPE and ΔAPF

BP2 = BF2 + PF2 = D2 + (X +d/2 )2 = D2 +x2 + xd

OR BP = (D2 +x2 + xd )1/2 = D[ 1+ (x/D)2 + xd/ D2 ]1/2 = D[1+ (x/D)2/2 + xd/ D2 /2]

or BP = D+ x2/ 2D + xd / 2D.

Similarly AP = D+ x2/ 2D - xd / 2D.

Now Path difference = S2P – S1P =BP –AP = xd /D.

But for the bright fringe path difference = xd/D = n ƛ , where n = 1, 2, 3, 4 ----or x = nD ƛ /d.For dark fringe path difference = xd/D = (2n-1) ƛ /2 , where n = 1, 2, 3, 4 ----or x =( 2n- 1)D ƛ /2d.


Physicsmodels.inFringe Width: Let us consider fringe width of bright fringe.

Fringe width = nth fringe – (n-1)th fringe = nD ƛ /d – (n-1)D ƛ /d = D ƛ /d

Therefore fringe width = ß = D ƛ /d .

Same is the fringe width of dark fringe.

b) On looking for single slit diffraction pattern, we observe that diffraction pattern is the same as interference of waves by double slits experiment. The only difference is that in diffraction, we have central maxima and where maxima occurs in case of interference there occurs minima in diffraction pattern.

c) Width of single slit = 2x = D ƛ /a, where a is slit width of single slit .

And fringe width of double slit = ß = D ƛ /d .

Now ß = D × 10-9 /10-3, ( given d= 10-3m and ƛ = 10-9m )

But10 ß = 10× D × 10-9 /10-3 = D ƛ /a

10× D × 10-9 /10-3 = D× 10-9 /a or a = 10-4m.


Physicsmodels.inOR

a) Above is the required diagram for image formation by two lenses separated a distance a = d. The distance of object in front of lens L1 = u

The distance of image in front of lens L1 at position I1 = v’. The lens formula for L1 is given by,

1

f=

1

v′−1

u−−−−−− −(1)

The final image is formed at I by the lens L2 at distance = (v – d ) Therefore lens formula for L2 .i given by,


Physicsmodels.in

1

𝑓2=

1

𝑣−

1

𝑣′ − 𝑑−−−−−− − (2)

Now adding equations (1) and (2)

1

𝑓1+

1

𝑓2=

1

𝑣′−

1

𝑢+

1

𝑣−

1

(𝑣′−𝑑)but

1

f=

1

v′−

1

u, therefore

1

𝑓1+

1

𝑓2=

1

𝑓+

1

𝑣′−

1

(𝑣′−𝑑)or

1

𝑓1+

1

𝑓2=

1

𝑓+

𝑣′−𝑑 −𝑣′

𝑣′(𝑣′−𝑑)

1

𝑓1+

1

𝑓2=

1

𝑓-

𝑑

𝑓1𝑓2

This is the required formula


Thank You

Physicsmodels.in

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CBSE Class12 - Physics Paper Solved - 2015...1) Interference is the superposition of waves originating from two coherent sources of light whereas diffraction is the superposition of