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PHYSICS Paper-I (Theory) (Three hours) (Candidates are allowed additional 15 minutes for only reading the paper, They must NOT start writing during this time) Answer all questions in Part I and six questions from Part II, Choosing two questions from each of the Sections A, B and C. All working including rough work should be done on the same sheet as, and adjacent to, the rest of the answer. The intended marks for questions or parts of questions are given in brackets [ ]. (Material to be supplied: Log tables including Trigonometric functions) A list of useful physical constants is given at the end of this paper. PART I (Compulsory) Question 1 [20] Answer all questions briefly and to the point: (i) State Gauss s theorem of electrostatics. Ans:- Gauss s law states that the total flux emanating from a closed surface enclosing electric charges in air is equal to 1/ o times the net charge enclosed. (ii) An alpha particle is accelerated through a potential difference of 2000volt. What will be the increase in its energy in electron volt? Ans: Increase in K.E = loss in P.E = e V = 1.6x10-19x2000 =3.2x10-16J. (iii) Assuming the earth as an isolated spherical conductor of radius 6400km, calculate its electrical capacitance in farad. 22 Ans:- C = 4 oR= 4 8.85 x 10 -12 6400 10 3 = 7.120 10 -4 .F 7 (iv) If the potential difference applied across a variable resistor is constant, draw a graph between the current in the resistor and the resistance. Ans:- Current I = V/R. Here I 1/R R I (v) Find the equivalent resistance between the points a and c of the given network of resistors. (See fig.1). 1 b 6 9 a 7 4 c 6 d Fig.1 Ans:- It is a network satisfying Wheatstone bridge balance condition. Hence 7 can be removed. 1 1 1 = + R ac = 6 Rac 15 10 Thus the resistance between a and c = 6 . (vi) Show with a labelled graph how thermo-emf varies with the temperature of the hot junction of a thermocouple. Ans:Thermo emf e n (vii) i Temperature ( ) What is the function of shunt in an ammeter? Ans:- Since the shunt resistance is of very low value, major portion of the current will pass through it allowing only the required current through the galvanometer part of the ammeter. (viii) What is meant by Wattless current? Ans: In an LCR circuit with ac supply, the voltage and the current in the circuit will have a phase difference. The component perpendicular to the voltage does not contribute any power in the circuit. This component is called the wattles current. (ix) Name the physical principle and one advantage in the use of optical fibres. Ans:- Optical fibres works on the principle of total internal reflection. Optical fibres are used for faster communication purposes. (x) Can two independent monochromatic sources of light be coherent? Explain very briefly. Ans:- No. The phase difference between the sources will always differ due to the difference in transition time electrons from the excited states of the two sources. What is the relation between the refractive indices n1 and n2, if the behaviour of (xi) light rays is as shown in the following Fig.2? 2 n1 n2 n1 Fig.2 Here n1 and n2 are the refractive indices of the surrounding medium and the lens respectively. Ans:- Refractive index n2 must have lower value than n1 since the refracted ray in the second medium bends away from the normal at the point of incidence. 1 1 n2 1 ( It can be explained on the basis the lens makers equation = 1 f n1 R R 2 1 since the rays diverge the lens must have a negative focal length. For a convex lens 1 n n 1 is positive. Hence 2 1 must be negative. Thus 2 <1. There for R1 R2 n1 n1 n2<n1.) (xii) What is the principle of global communication by satellites? Ans:- A certain number (at least three) of geostationary satellites are required for global communication. When they receive messages from ground stations, after modifying the frequency and strength, send them back to the earth and the stationary satellites at other positions receive them and send the messages to other part of the earth. Thus the earth as a whole can be covered by communication. (xiii) State Brewster s law for polarisation of light. Ans:- Brewster s law states when reflection from a glass plate takes place ,at the angle of incidence (polarising angle), the tangent of the angle is equal to the refractive index of the medium (glass) (xiv) Two thin lenses- one convex and the other concave- of focal lengths 15 cm and 20cm respectively are put in contact. Find the focal length of the doublet lens. 1 1 1 = + f f1 f1 Ans:1 1 1 1 = + = f = 60cm. f 15 20 60 (xv) Two stars three and four light years away from the earth have same luminous intensity. Find the ratio of the illuminance (intensity of illumination) on the surface of the earth, normal to the starlight, produced by each star. Ans:- (This portion is no more in your syllabus) (xvi) The historic experiment on the diffraction of electrons confirmed a new nature of electrons. What is this new nature of electrons? Who had proposed it? Ans:- Electrons can behave like waves and the wave associated with it is called matter 3 waves. Louis De Broglie Proposed that particles too can exhibit wave nature. (xvii) Explain briefly why there is a maximum frequency for the x- rays produced by an x-ray tube operating at a certain voltage. Ans:- The accelerating potential produces potential energy on the electron equal to eV. Under favourable condition the whole energy will be given to the target and thus (X-rays) photons of energy h max will be produced. The maximum eV frequency of the X-rays produced max = h (xviii) Draw circuit diagrams to illustrate forward biasing and reverse biasing of a diode. Ans:- Forward biasing P N P N + Reverse biasing + ( The diagrams must be drawn separately.There is no need to write any explanation.) (xix) Name two alternative sources of energy and mention where they can be used. Ans:- ( Not in the syllabus any more) (xx) The half-life period of a radioactive substance is 16 hours. After how much time will 6.25% of the material remain undecayed? Ans:N Time(hrs) It takes 64 hours to come to 6.25% of undecayed material 100 0 50 16 25 32 12.5 48 6.25 64 ( You may do it by using the formula ; but a little harder: 4 n n n N 1 6.25 1 6.25 100 1 N = N0 = i.e = , log = n log(1 / 2) log = n log 2 No 2 100 2 100 6.25 2 100 log 6.25 = 4; n = log 2 Where n is the number of half lives elapsed. Thus the total time elapsed = 4x16 =64hrs.) PART II Answer six questions in this part, choosing two questions from each of the Sections A, B and C. SECTION A (Answer any two questions) Question 2 (a) Derive an expression for the electric potential at a point due to a point [4] charge. Ans:- Let P be a point at a distance of r from the point charge q. The force at P on a unit positive test charge = F =k. q/r2. Let dr be an infinitesimal distance move towards the charge q. The work done = F dr. q r P dr The potential at P is defined as the work done in moving aunt positive charge from infinity to r. r r q 1 kq q = = V = k 2 dr = qk r r 4 o r r V = q 4 o r [3] (b) An electric flash lamp has 20 capacitors each of capacitance 5 F connected in parallel. The lamp is operated at 100 volt. Calculate how much energy will be radiated in a flash. Ans:- The equivalent capacitance of the combination is given by: C = C '+C '+...20terms = nC ' C = 20 5 = 100 F . Energy radiated in a flash = CV2.= x 100x10-6x(100)2.=0.5 J. (The power generated will be enormous since the flash takes place in the rage of milliseconds and microseconds) (c) Name one material whose resistivity decreases with rise in temperature. [2] Explain briefly on the basis of free electron theory why the resistivity decreases. Ans:- Silicon. As temperature increases more hole-electron pairs will be produced 5 and consequently intrinsic current of semiconductors increases. Thus their resistivity decreases as temperature increases. Question 3 (a) State and explain Kirchoff s laws for electric circuits. What are the conservation laws implied in each law? State the sign convention for current and emf. Use the Fig.3 given below for your explanation. + R1 [4] E1 Fig.3 R2 + R3 E2 Ans:+ A R1 G I1 B E1 I2 C R2 I3 F R3 + E2 D i) Junction law: The algebraic sum of currents meeting at a junction is equal to zero. For junction G: I1 I2 + I3 = 0. The convention used is: Current coming to a junction is taken as positive and that leaving the junction is taken as negative. The first law implies the law of conservation of electrical charges. ii) Mesh law (Loop law):- It state that the algebraic sum of the product of currents and resistances and the algebraic sum of emf s in a closed loop is equal to zero. For the loop GCBAG: Move in the anticlockwise direction: I2R2 I1R1+E1 = 0. The convention used is: If current and motion are in the same direction the IR product is negative and vice versa. For emf s: if the entry is from the negative side take its sign as positive and vice versa. The second law implies the law of conservation of energy. (The named circuit diagram is required to be drawn in this case since reference has to be given in the explanation. If a diagram is required in the explanation for any answer, do not hesitate to do it. This must be a common dictum.) (b) An electron is moving vertically upwards in the uniform magnetic field [3] of the earth. The velocity of the electron is 2.0 x106 m/s and the horizontal component of earth s magnetic field is 0.32 x 10-4 tesla. Obtain the magnitude and direction of the force on the electron. 6 Ans:- The force on the electron F = Bev =0.32 x 10-4 x 1.6x10-19 x 2.0x106 =1.024x10-17N. The direction is vertically down. (The electron moves in a vertical circular path.) (c) Permanent magnets are made of special alloys while the core of [2] temporary magnets is made of soft iron. Why? Ans:- Coercivity of special alloys are very high so that larger demagnetizing field is required to demagnetize a permanent magnet made of them. Soft iron has very low coercivity so that they can loose their magnetism as soon as the magnetizing field is removed. ( Compare the properties of steel and soft iron: Steel: High residual magnetism Large coercivity Can withstand large temperature changes. Soft-iron: Large intensity of magnetization. High susceptibility Low hysteresis loss.) Question 4 (a) Define self inductance Obtain the expression for the self inductance of a solenoid, explaining steps with the help of a diagram. [3] Ans:-Self inductance of a coil is defined as the magnetic flux associated with it when unit current passes through it. r I I l Let l be the length of the solenoid, n, and the number of turns per unit length. A steady current I flows through the coil. N is the total number of turns. The magnetic flux density associated with the solenoid ; B= onI 7 B = o nI The magnetix flux through each turn : = B. A = o nI r 2 N I 2 2 2 N r N o N r N L = o n r 2 N = o = Q n = l l l Total flux through N turns = N = o nI r 2 N Q L = L = o N 2 r 2 L == l o r N 2 (b) What is meant by back emf in a dc motor? The back emf in a dc motor, delivering 3 kilowatts of mechanical power, is 180 volts when operating on 220 volt line. Determine the armature current and the motor resistance. Ans:- d.c motor is taken off the syllabus. (c) In an LCR circuit with all components connected in series, the emf and current flowing in the circuit are given by the following equations:E = 200 sin(314t + /6) volt I = 5 sin 314 t ampere. Obtain: the peak values of current and emf. (i) the frequency of the ac source. (ii) (iii) the phase difference between current and emf. [3] [3] Ans:- (i) E = 200 sin(314t + /6) volt (1) E = Eo sin( t + ) volt .(2) Comparing (1) and (2): peak emf =200V; frequency = 314/2 = 50Hz. I = 5 sin 314 t ampere (3) I = Io sin t ampere ..(4) Comparing (1),(3) and (4): peak current = 5.0 A. Comparing (2) and (4); the phase difference; = /6.rad. SECTION B (Answer any two questions) Question 5 (a) State Huygen s principle. Explain with the help of a diagram the phenomenon of [3] refraction of waves on the basis of this principle. Ans:- According to Huygen s theory of secondary wavelets a source of light is a source of disturbance and they travel out in space and the locus of the points of disturbance at an instant is the wavefront at that instant and every point of a wavefront is a source of secondary wavelets. Let MN be the surface of separation between two mediums(media). AC represents the incident wave front and i, the angle of incidence. Vi and V2 are the speed of light in the two media 8 respectively. According to Huygen s wave theory the point A becomes a point of disturbance at the surface before the disturbance at C reaches the surface MN. Let t be the time taken by the disturbance at C reaches B. By this time the disturbance at A will reach a distance of AD in the second medium. With A as centre and AD =V2 t draw a semi-circle. Draw a straight line from B to touch the envelope of the semicircle. BD will be the refracted wavefront. C V1 i M r A D B N V2 From the diagram: CB V1t = ...........(1) In ABC : Sini = AB AB AD V2 = ...........(2) In ABD : Sinr = AB AB Sini V1 = = Sinr V2 ie.Sini/Sinr = speed of light in the first medium/the speed of light in the second medium which is the Snell s law. (It seems to be a very long answer for 3 marks. Explanation can be reduced in the actual answer. The answer is long here for your understanding.) (b) What is chromatic aberration in lenses? State the necessary conditions for an [3] achromatic doublet. How these conditions are practically achieved? Ans:- As a white image is observed through a lens, due to dispersion all the colours of the visible spectrum are focused at different points on the principal axis. Thus the image formed will be coloured. This defect is called chromatic aberration. The necessary condition for achromatic doublet is : fY = ' f ' Y ; where and ' , are the dispersive powers of two lenses in the doublet And f Y and f Y' are the focal lengths of the two lenses for the colour yellow.Due to the ngative sign lenses must be one convex and the other concave. (c) Fraunhofer diffraction from a single slit of width 1.0 m is observed with light of wavelength 500nm. Calculate the half angular width of the central maximum. Ans:-For small angles: =n /d. For central maximum n=1 = 500x10-9/10-6=0.5radian. 9 [2] Question 6 (a) Draw a neat ray diagram of a simple microscope. Deduce the [3] formula for its angular magnification when the image is formed at the least distance of distinct vision. Ans:-A virtual image IM is M formed at the least distanceMPI= of distinct vision (D). L Angular magnification M= / = tan /tan I NPI= B N f O P u D PI (Q OB = IN )......(1) IN PO PI 1 1 1 1 1 = + = + (here sign convention is used) f u v PO D PI PI PI 1 1 1 = + = + (no need to use sign convention for PI : PO f D PO f D it is only a multiplcation factor) M = M = OB PO = D D + 1 = 1 + f f (b) A beam of monochromatic light of wavelength 500nm falls on two parallel slits. The distance between the slits is 0.15mm. Determine the width of the interference fringes on a screen placed at a distance of 1.5m from the slits. D 1.5 500 10 9 = 10 3 m = 1.0mm = Ans:- = 3 d 0.15 10 [3] (c) [2] Draw a sketch of electromagnetic spectrum, showing relative positions of UV, IR, X-ray and microwaves with respect to visible light. State approximate wavelengths of any two. Ans:- 10 Microwave Infra red Visible spectrum Ultra violet -7 X-rays -8 4*10 to 10 m 3*10-8 to 10-11 m Frequency increasing (Other ranges are: Microwave:3*10-1 to 1*10-3m. IR: 5*10-3 to 8*10-7m . Visible;4*10-7 to 8*10-7m.) Question 7 (a) With the help of a neat diagram describe with theory, Michelson s [4] method for the determination of the speed of light. What is the presently accepted nine digit value of the speed of light? Ans:- The main part of the apparatus is an octagonal mirror M which can be rotated at a constant frequency about its centre. A narrow beam of light produced by the slit P from an arc lamp S falls at the first face of the mirror M and after reflection from mirrors M1, M2, and the concave mirror C2 becomes parallel and travels about 35 km and returns its journey after reflection from another concave mirror C1 to the first concave mirror C2. Then the beam is reflected from the mirror M3 and falls on the 5th face of the octagonal mirror if the mirror is stationary. This beam is received at the low power telescopic eye piece V by the help of a total reflecting prism T. T V M4 6 C1 M5 M 5 7 8 P 1 M3 4 C2 3 2 M2 M1 S When the mirror M is rotated the beam of light reflected from the first face received in the eyepiece flickers if another side of the mirror does not come parallel to the first face by the time the beam comes back to the mirror after its journey. The smallest time in which the return beam can be seen through the eyepiece without flickering is when the 6th face of the mirror can occupy the place occupied by the 5th face since the mirror is rotated in a clockwise direction. The mirror is slowly rotated and the frequency f is noted for the first flickering of the 11 beam received in the eyepiece to stop. Time taken by the beam to travel a distance of 2d = 1/8f 2d Speed of light c = distance /time = = 16df , where d is the distance between the 1 8f observatories. The accepted value of speed of light in vacuum, c =299792458ms-1. (Note that the answer normally becomes lengthy for questions of this nature. This also you must accommodate.) (b) What is meat by pure and impure spectra? Explain briefly how you will [4] set up a spectrometer to obtain pure spectrum. Draw a diagram of a spectrometer and describe its parts. Ans:-If all the colours in spectra have distinct boundaries then the spectra is called pure spectra. On the other hand if the colours are overlapping then the spectra is called impure spectra. The setup in a spectrometer are as follows: 1. Focusing the cross-wire in the eyepiece of the telescope by drawing inward or outward.. 2. Adjusting the telescope to receive parallel beam of light by focusing a distant object at the cross-wire. 3. Adjust the slit in the collimator to produce a narrow beam of light. 4. Adjust the collimator to produce a parallel beam of light while looking through the telescope. 5. Adjust the prism table to be horizontal. Slit Collimator Prism table Vernier1 Vernier2 Telescope The spectrometer consists of the following parts: 1. A telescope with eyepiece. 2. A collimator with slit attached. 3. A prism table which can be freely rotated. 4. Two vernier scales are provided at the two diametrically opposite sides of the 12 spectrometer base and as the prism table is rotated corresponding angular measurements can be taken. (Note that this question requires points out of proportions to the marks allocated. You should be prepared to attempt such type of questions too!) SECTION C (Answer any two questions) Question 8 (a) The ionization potential of hydrogen atom is 13.6 volt. Draw the energy level diagram showing four levels. Calculate: (i) (ii) [4] the energy of the photon emitted when an electron falls from the third orbit to the second orbit. the wavelength of this photon. Ans:- Ionization potential 0.85 V 1.5 V n=4 n=3 3.4 V n=2 13.6 V n=1 (i) E = E2 E1 = (3.4 1.5)x 1.6x10-19.J =3.04x10-19.J (ii) E = h = hc/ ; i.e = hc/E= {(6.6x10-34)x(3x108)}/3.04x10-19 m =6.513x10-7 m. = 6513 . (b) Describe briefly with a diagram and theory an experiment to determine [4] e/m of electron. Ans:- The e/m of an electron is determined by J. J. Thomson s method. Principle:It is based on the principle that an electron experiences a force in electric and magnetic fields. The path of the electron in a uniform electric field is parabolic and its path in a uniform magnetic field is circular. Construction:- The apparatus consists of an electrically heated filament (F) to produce thermionic electrons and a cathode connected to the negative terminal of the accelerating potential. Anode (A) connected to the positive of the accelerating potential and a set of collimating cylinders are arranged to produce a narrow beam of electrons. Uniform electric and magnetic field are produced in the region (XX ). A zinc sulphide screen (S) is provide at the rear end of the tube, which is highly evacuated, to view the electron spot (S) 13 + F + C O A xBx S - XX T . Theory:The magnetic field is switched on first and the deflection of the electron beam is measured. The path is a part of a circle with radius r. The magnetic field is acting perpendicular to the plane of paper and away from the reader. mv 2 e v , = .........(1) , where v is the velocity of electrons. The value of r Bev = r m Br can be determined from the geometry of the apparatus and the deflection produced. Now the electric field is switched on and by adjusting the field (E) the electron beam can be brought back to O. When this happens the field are said to be crossed. Thus e E=Bev. Thus the velocity of the beam v = E/B (2) Substituting the value of v in equation (1) the value of e/m is determined. E e E V V = B= 2 = , where E = ; V is the potential difference between the 2 m Br d B r dB r electric plates and d the distance between them. Question 9 (a) What is Compton scattering? State briefly its importance. [2] Ans:- When X-rays fall on crystals, the scattered rays contain higher wavelength than the incident X-rays. This type of incoherent scattering is called Compton Scattering. Compton effect is one of the conclusive evidences of particle nature of radiation. (b) When ultraviolet light of wavelength 300nm is incident on a metal plate, [4] a negative potential of 0.54 volt is required to stop the emission of photo electrons. Calculate the energy of the incident photon and the work function for the metal in eV. Ans:h = h o + eVo , where Vo is the stopping potential & h o , the work function. Energy of the photon = h = hc = 6.6 10 34 3 10 8 6.6 10 34 3 10 8 J = eV 300 10 9 300 10 9 1.6 10 19 Energy of the photon = 4.125eV Work function = h eVo = 4.125eV 0.54eV = 3.585eV (c) Draw the circuit diagram of a half wave rectifier using a semiconductor diode. Explain briefly the function of each component. 14 [2] Ans:- P A.C input Voltage N D.C output R Step down transformer Voltage time time The circuit diagram is shown above. The step-down transformer sends the allowed ac supply in to the rectifier. The P-N junction diode conducts only when the P is positive (forward biased) and do not conduct when it is negative (reverse biased). The fluctuating d.c voltage is drawn across the load resistor R. Question 10 (a) Heavy water is a suitable moderator in a nuclear reactor. Explain [2] briefly why? Ans:-The mass of heavy hydrogen present in heavy water has the same order of mass with the neutrons. Thus the neutrons will be slowed down to thermal neutron class. If the moderator nucleus has larger or lighter mass than the neutron, it will bounce back without reducing its energy or pass on without reducing its energy much. Thus heavy water is one of the best materials as a moderator. [4] (b) Draw labelled diagrams to illustrate: energy bands of a conductor, semiconductor and insulator. (i) npn and pnp transistors. (ii) (iii) Transistor as an amplifier (common emitter). Ans:- (i) 15 Conduction Band Conduction Band Energy Gap Energy Gap Valance Band Insulators Valance Band Conductors Semiconductors (ii) NPN Collector PNP Base Collector Base Emitter Emitter (iii) Ic NPN b Ib Ac input c e Ie Output ac amplified R + Reverse bias + Forward bias (c) What do you mean by AND gate? How will you realize AND gate with junction diodes? [2] Ans:- AND gate is an electronic circuit which responds to logical statements by showing the presence of a voltage or its absence. (Realization of logic circuits are taken off the syllabus) [PHYSICAL CONSTANTS] 1/4 o = 9 x109 m/F 16 Plank s constant (h) = 6.6 x 10-34 Js Velocity of light (c) = 3 x108 ms-1 = 1.6 x 10-19 J 1 eV Mass of electron (me) = 9.0 x 10-31 Kg Permittivity of free space ( o) = 8.85 x 10-12 C2N-1m-2. 17
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