SURA's Guides for 3rd to 12th Std for all Subjects in TM & EM Available MARCH 2017

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1 1 th STD [Time : 3 hours] PHYSICS [XII] (With key) [Marks : 150] Part - I Note : (i) Answer a the questions. (ii) Choose the most suitabe answer from the given four aternatives and write the option code and the correspoding answer. [30 1 = 30] 1. In Newton's experiment the radii of the m th and (m+4) th dark rings are respectivey (a) 50 A rms current (c) 5 A rms current (b) 5 A peak current (d) none of these. 5 mm and 7 mm. What is the vaue of m? 8. The radio-isotope used in agricuture is : (a) (b) 4 (a) 15 P 31 (b) 15 P 3 (c) 8 (d) 10 (c) 11 Na 3 (d) 11 Na 4. The resting frequency of FM transmitter is 98.5 MHz. The aowed minimum and 9. In a given thermocoupe, the neutra temperature : maximum frequency on either side of the (a) is a constant centre frequency are respectivey: (b) depends on the temperature of cod (a) MHz and MHz juction (b) MHz and MHz (c) depends upon the temperature of (c) MHz and MHz inversion (d) 98 MHz and 99 MHz. (d) both (b) and (c). 3. Arrange eectron (e), proton(p) and 10. In Raman effect, the waveength of the deutron(d) in the increasing order of their incident radiation is 5890 Å. The specific charge: waveengths of Stokes and anti-stockes ines are respectivey: (a) e, p, d (b) d, p, e (a) 5880 Å and 5900 Å (c) p, e, d (d) d, e, p (b) 5900 Å and 5880 Å 4. Phosphor-bronze wire is used for suspension in a moving coi gavanometer, because it has: (a) high conductivity (b) high resistivity (c) arge coupe per unit twist (d) sma coupe per unit twist 5. An LCR series circuit is connected to 40 V A.C. suppy. At resonance, the vaues of V R V L and V C are respectivey : (a) 80 V, 80 V and 80 V (b) 10 V, 60 V and 60 V (c) 40 V, 10 V and 10 V (d) 180 V, 40 V and 40 V. MARCH Transformer works on : (a) AC ony (b) DC ony (c) Both AC and DC (d) AC more effectivey than DC. 7. A DC of 5 produces the same heating effect as an A.C. of : (c) 5900 Å and 5910 Å (d) 5870 Å and 5880 Å 11. The photoeectric effect can be expained on the basis of : (a) corpuscuar theory of ight (b) wave theory of ight (c) eectromagnetic theory of ight (d) quantum theory of ight 1. A hoow meta ba carrying an eectric charge produces no eectric fied at points : (a) outside the sphere (b) on its surface (c) inside the sphere (d) at a distance more than twice. [1]

2 Sura s XII Std Physics March Question Paper with Answers 13. The emitter base junction of a given transistor is forward biased and its coector base 1. The mass defect of a certain nuceus is found to be 0.03 amu. Its binding energy is : junction is reverse biased. If the base current (a) 7.93 ev (b) 7.93 kev is increased, then its : (c) 7.93 MeV (d) 7.93 GeV (a) V CE wi increase (b) I C wi decrease. The coour code of a carbon resistor is, (c) I C wi increase (d) V CC wi increase Brown, Back, Brown and Red. The vaue 14. In a Bainbridge mass spectrometer positive of the resistor is: rays of the same eement produce different (a) 10 Ω ± 5% (b) 1 kω ± % traces. The traces correspond to : (c) 100 Ω ± % (d) 10 Ω ± % (a) isotopes (b) isobars (c) isotones 3. Light from a source is anaysed by an anayser. When the anayser is rotated, the (d) none of the above intensity of the emergent ight : 15. The energy of eectron in the first orbit of hydrogen atom is 13.6 ev. Its potentia (a) Does not vary energy is : (b) Remains uniformy dark (a) 13.6 ev (b) 7. ev (c) Varies between maximum and zero (c) 7. ev (d) 6.8 ev (d) Varies between maximum and minimum 16. Point charges q 1 and q are paced in air at a 4. The eectric fied at a point cm from an distance 'r'. The ratio of the force on charge infinite ine charge of inear charge density q 1 by charge q and force on charge q by 10 7 cm 1 is : charge q 1 is : q q (a) NC 1 (b) NC 1 1 (c) 9 10 (a) q (b) q NC 1 (d) NC Which of the foowing quantities is scaar? q 1 (a) dipoe moment (b) eectric force (c) 1 (d) q 17. The RF channe in a radio transmitter produces : (a) audio signas (b) high frequency carrier waves (c) both audio signa and high frequency carrier waves (d) ow frequency carrier waves. 18. The nucei 13 AI 7 and 14 Si 8 are exampes of: (a) isotopes (b) isobars (c) isotones (d) isomers 19. In a Copitt's osciator circuit : (a) capacitive feedback is used (b) tapped coi is used (c) no tuned LC circuit is used (d) no capacitor is used 0. In hydrogen atom, which of the foowing transitions produce a spectra ine of maximum frequency? (a) 6 (b) 1 (c) 4 3 (d) 5 (c) eectric fied intensity (d) eectric potentia 6. Since the input impedance of an idea operationa ampifier is infinite : (a) its input current is zero (b) its output resistance is high (c) its output votage becomes independent of oad resistance (d) it becomes a current controed device 7. The energy of a photon of characteristic X-ray from a Cooidge tube comes from : (a) the kinetic energy of the free eectrons of the target (b) the kinetic energy of the ions of the target (c) the kinetic energy of the striking eectron (d) an atomic transition in the target 8. The refractive index of the medium, for the poarising ange 60º is : (a) 1.73 (b) (c) 1.5 (d) 1.468

3 Sura s XII Std Physics March Question Paper with Answers 3 9. The part of the AC generator that passes the current from the coi to the externa circuit is : (a) fied magnet (b) spit rings (c) sip rings (d) brushes 30. The threshod frequency of a photosensitive surface is Hz. Then which of the foowing wi produce photoeectric effect from the same surface? (a) Sodium vapour amp (b) Ruby aser (c) He-Ne aser (d) Both (b) and (c). Part - II Note : Answer any fifteen questions. [15 3 = 45] 31. What is an eectric dipoe? Define eectric dipoe moment. 3. Why is it safer to be inside a car than standing under a tree during ightning? 33. State Ohm's aw. 34. How much time 10 0 eectrons wi take to fow through a point in a conductor so that the current is 00 ma [e = C]? 35. State Faraday's aws of eectroysis. Wheatstone's bridge. 36. What are the characteristics of heating eement used in eectric heating device? 37. State Feming's right hand rue. 38. An a.c. generator consists of a coi of 10,000 turns and of area 100 cm. The coi rotates at an anguar speed of 140 rpm in a uniform magnetic fied of T. Find the maximum vaue of the emf induced. 39. Write any three uses of infrared radiations. 40. A 300 mm ong tube containing 60 cc of sugar soution produces a rotation of 9º when paced in a poarimeter. If the specific rotation is 60º. cacuate the quantity of sugar contained in the soution. 41. Write any three medica appications of X-rays. 4. The Rydberg constant for Hydrogen is m 1. Cacuate the short waveength imit of Lyman series. 43. State the postuates of specia theory of reativity. 44. Define curie. 45. Write any three properties of neutrons. 46. Define bandwidth of an ampifier. 47. Draw the circuit diagram of a summing ampifier using an operationa ampifier. 48. What is an intrinsic semi conductor? Give two exampes. 49. A gavanometer of resistance 100 Ω which can measure a maximum current of 1 ma is converted into an ohmmeter by connecting a battery of emf 1 V and a fixed resistance of 900 Ω in series. When an externa resistance is measured the current reading is 0.1 ma. Cacuate the vaue of the resistance. 50. What are the different types of radiowave propagation? Part - III Note : (i) Answer question number 5 4 compusoriy. (ii) Answer any six of the remaining 11 questions. [7 5 = 35] (iii) Draw diagrams wherever necessary. 51. Deduce an expression for the capacitance of a parae pate capacitor. 5. Obtain the condition for bridge baance in 53. How can e.m.f of two ces be compared using potentiometer? 54. A stream of deutrons is projected with a veocity of 10 4 ms 1 in XY-pane. A uniform magnetic fied of induction 10 3 T acts aong the Z-axis. Find the radius of the circuar path of the partice. (Mass of deutron is kg and charge of deutron is C). (OR) A circuar coi of radius 0 cm has 100 turns wire and it carries a current of 5 A. Find the magnetic induction at a point aong, its axis at a distance of 0 cm from the centre of the coi. 55. Obtain an expression for the sef-inductance of a ong soenoid. 56. State and expain Brewster's aw. 57. Write any five properties of cathode rays. 58. Derive an expression for de-brogie's waveength of matter waves. 59. Write any five appications of photo eectric ces.

4 4 Sura s XII Std Physics March Question Paper with Answers 60. A reactor is deveoping energy at the rate of 3 MW. Cacuate the required number of Answers fissions per second of 9 U 35. Assume that Part - I energy per fission is 00 MeV. 1. (d) State and prove De Morgan's theorems. 6. A 10 MHz sinusoida carrier wave of. 3. (c) MHz and MHz (b) d, p, e ampitude 10 mv is moduated by a 5 khz 4. (d) sma coupe per unit twist sinusoida audio signa wave of ampitude 5. (c) 40 V, 10 V and 10 V 6 mv. Find the frequency components of 6. (a) AC ony the resutant moduated wave and their 7. (c) 5 A rms current ampitudes. 8. (b) Part - IV 15 P 3 Note : (i) Answer any four questions in 9. (a) is a constant detai. 10. (b) 5900 Å and 5880 Å (ii) Draw diagrams wherever necessary. 11. (d) quantum theory of ight [4 10 = 40] 1. (c) inside the sphere 63. State the principe and expain the construction 13. (c) I C wi increase and working of Van de Graaff generator. 64. Derive an expression for the magnetic 14. (a) isotopes 15. (c) 7. ev induction at a point due to an infinitey ong 16. (c) 1 straight conductor carrying current. Write 17. (b) high frequency carrier waves the expression for the magnetic induction 18. (c) isotones when the conductor is paced in a medium 19. (a) capacitive feedback is used of permeabiity 'µ'. 0. (b) A source of aternating e.m.f. is connected to a series combination of a resistor R, an 1. (c) 7.93 MeV inductor L, and a capacitor C. Obtain with. (c) 100 Ω ± % the hep of a vector diagram and impedance 3. (a) Does not vary diagram, an expression for (i) the effective 4. (b) NC 1 votage (ii) the impedance (iii) the phase reationship between the current and the votage. 66. Derive an expression for bandwidth of interference fringes in Young's doube sit experiment. 67. Draw a neat sketch of Ruby Laser. Expain its working with the hep of energy eve diagram. 68. Expain the construction and working of a Geiger-Muer Counter. 69. What is meant by feedback? Derive an expression for votage gain of an ampifier with negative feedback. 70. Expain the principe and working of RADAR with neat bock diagram. 5. (d) eectric potentia 6. (a) its input current is zero 7. (d) an atomic transition in the target 8. (a) (d) brushes 30. (a) Sodium vapur amp Part - II 31. Two equa and opposite charges separated by a very sma distance constitute an eectric dipoe. Two point charges +q and q are kept at a distance d apart. The magnitude of the dipoe moment is given by the product of the magnitude of the one of the charges and the distance between them. Eectric dipoe moment, p = qd of qd.

5 Sura s XII Std Physics March Question Paper with Answers 5 p -q +q d It is a vector quantity and acts form q to +q. The unit of dipoe moment is C m. 3. i) The meta body of the car provides eectrostatic shieding. ii) So Eectric fied inside car is zero. iii) During ightning the eectric discharge passes through the body of the car. 33. At constant temperature, the steady current fowing through a conductor is directy proportiona to the potentia difference between the ends of the conductor. I V V = IR 34. Given data : n = 10 0 ; e = C ; Soution: I = 00 ma E 0 = NAB ω = NAB pv = A Soution: ne t I= q = t ne t = = 3 I = = t = 80 s Nichrome which is an aoy of nicke and chromium is used as the heating eement for the foowing reasons. 1) It has high specific resistance ) It has high meting point 3) It is not easiy oxidized. 37. The forefinger, the midde finger and the thumb of the right hand are hed in three mutuay perpendicuar directions. If the forefinger points aong the direction of the magnetic fied and the thumb is aong the direction of motion of the conductor, then the midde finger points in the direction of the induced current. 38. Given data : N = 10,000 A = 10 cm = 10 m. V = 140 rpm = rps, B = T E 0 =? = π 7 3 E 0 = 5.75 V 39. (i) Infrared amps are used in physiotherapy. (ii) Infrared photographs are used in weather forecasting. (iii) As infrared radiations are not absorbed by air, thick fog, mist etc., they are used to take photograph of ong distance objects. 40. Given data : 35. (i) First Law : The mass of a substance iberated at an eectrode is directy proportiona to the charge passing through the eectroyte. m α E. (ii) Second Law : The mass of a substance iberated at an eectrode by a given amount of charge is proportiona to the chemica equivaent of the substance. = 300 mm = 30 cm = 3 decimeter. θ= 9º ; S = 60º ; υ = 60 cc; m =? Soution: θ θ S= c = ( m/ u) m = θ. u s = m= 3g

6 6 Sura s XII Std Physics March Question Paper with Answers 41. (i) X-rays are used as a diagonistic too in medicine. (ii) It is used to study the crysta structure in soids. 4. Given data : R = m 1 For short waveength imit of Lyman Series, n 1 = 1, n =, λs =? Soution: The wave number for Lyman series is, v = n R n For short waveength imit, V I 1 = R R ex = V 1 = ex I = ohm v s = = R s 1 = R R λ ( ) ex = ohm or 1 1 λ S = = = A o G+R+R 1 = 10,000 ohm 7 R R 1 = = 9000 ohm. λ s = 911.6Å 50. Radiowave is propagated from the 43. i) The aws of Physics are the same in a transmitting to the receiving antenna in three inertia frames of reference. different ways depending on the frequency ii) The veocity of ight in free space is a of the wave. They are: constant in a the frames of reference.. (i) Ground wave propagation 44. Curie is defined as the quantity of a radio active substance which gives (ii) Space wave propagation disintegrations per second or (iii) Sky wave propagation. becquere. This is equa to the activity of one Part - III gram of radium. 45. i) Neutrons are neutra in nature. ii) Neutrons are stabe inside the nuceus. But outside the nuceus they are nonstabe. iii) They are neutra so they pentrate the nuceus easiy. 46. It is defined as the frequency interva between ower cut-off and upper cut-off frequencies. 47. B.W = f H f L. 48. A semiconductor which is pure and contains no impurity is known as an intrinsic semicircuar. In an intrinsic semiconductor the number of free eectrons and hoes are equa. Exampe: Pure germanium and siicon. 49. I g = 1 ma V = 1v G = 100 ohm R = 900 ohm G+R = 1000 ohm V I = R = = 10 3 A ex 51. Capacitance of a parae pate capacitor : The parae pate capacitor consists of two parae meta pates X and Y each of area A, separated by a distance d, having a surface charge density σ (figure). The medium between the pates is air. A charge +q is given to the pate X. It induces a charge q on the upper surface of earthed pate Y. When the pates are very cose to each other, the fied is confined to the region between them. The eectric ines of force starting from pate X and ending at the pate Y are parae to each other and perpendicuar to the pates. Summing ampifier

7 Sura s XII Std Physics March Question Paper with Answers 7 +q -q Parae pate capacitor By the appication of Gauss s aw, eectric fied at a point between the two pates is, E = σ ε o Potentia difference between the pates X and Y is V = 0 0 σ Edr = dr = εo d d d X Y σd εo The capacitance (C) of the parae pate capacitor q C = V = σa σd/ ε = εo A o d equation (1), () and (3) [since, σ = q A ] ε C = d The capacitance is directy proportiona to the area (A) of the pates and inversey proportiona to their distance of separation (d). 5. Wheatstone s bridge : An important appication of Kirchoff s aw is the Wheatstone s bridge figure. o A P I 1 A I R I B I G I 3 G D E I 4 Q S Wheatstone s bridge C Wheatstone s network consists of resistances P, Q, R and S connected to form a cosed path. A ce of emf E is connected between points A and C. The current I from the ce is divided into I 1, I, I 3 and I 4 across the four branches. The current through the gavanometer is I g. The resistance of gavanometer is G. Appying Kirchoff s current aw to junction B, I 1 I g I 3 = 0...(1) Appying Kirchoff s current aw to junction D I + I g I 4 = 0...() Appying Kirchoff s votage aw to cosed path ABDA I 1 P + I g G I R = 0...(3) Appying Kirchoff s votage aw to cosed path ABCDA I 1 P + I 3 Q I 4 S I R = 0...(4) When the gavanometer shows zero defection, the points B and D are at same potentia and I g = 0. Substituting I g = 0 in I 1 = I 3...(5) I = I 4...(6) I 1 P = I R...(7) Substituting the vaues of (5) and (6) in equation (4) I 1 P + I 1 Q I S I R = 0 I 1 (P + Q) = I (R+S)...(8) Dividing (8) by (7) I1( P + Q) I ( R + S ) = IP IR 1 P + Q R + S = P R Q S 1 + = 1 + P R Q S P = R or P R = Q S This is the condition for bridge baance. If P, Q and R are known, the resistance S can be cacuated.

8 8 Sura s XII Std Physics March Question Paper with Answers 53. Comparison of emfs of two given ces using potentiometer : The potentiometer wire AB is connected in series with a battery (Bt), Key (K), rheostat (Rh) as shown in the figure. This forms the primary circuit. The end A of potentiometer is connected to the termina C of a DPDT switch (six way key doube poe doube throw). The termina D is connected to the jockey (J) through a gavanometer (G) and high resistance (HR). The ce of emf E 1 is connected between terminas C 1 and D 1 and the ce of emf E is connected between C and D of the DPDT switch. A I Bt E 1 ( ) K J Rh B C 1 D 1 C D G HR (OR) C D E comparison of emf of two ces x = 0 cm = 0 10 m ; B =? Let I be the current fowing through the primary circuit and r be the resistance of the potentiometer wire per metre ength. The DPDT switch is pressed towards C 1, D 1 so that ce E 1 is incuded in the secondary circuit. The jockey is moved on the wire and adjusted for zero defection in gavanometer. The baancing ength is 1. The potentia difference across the baancing ength 1 = Ir. Then, by the principe of potentiometer, E 1 = Ir...(1) The DPDT switch is pressed towards E. The baancing ength for zero defection in gavanometer is determined. The potentia difference across the baancing ength is = Ir, then E = Ir...() Dividing (1) and () we get, E E = 1 1 If emf of one ce (E 1 ) is known, the emf of the other ce (E ) can be cacuated using the reation, E = E Given data : v = 10 4 ms 1, B = 10 3 T, m = kg e = C, r =? Soution: Bev= mv r 7 10 mv r = = = Be r = 0.08 m 54. Given data : a = 0 cm = 0 10 m ; n = 100 ; I = 5A Soution: Magnetic induction at a point, aong its axis at a distance from the centre of the coi is B = µ 0na I µ 0na I = 3 / 3/ ( a + x ) ( a ) { a = x = 0 10 m µ na I µ ni 0 0 = = 3 ( a ) ( a) = 7 5 4π π = B = T 55. Consider a soenoid of N turns with ength and area of cross section A. It carries a current I. If B is the magnetic fied at any point inside the soenoid, then Magnetic fux per turn = B area of each turn

9 Sura s XII Std Physics March Question Paper with Answers 9 µ o But, B NI From Figure, = i p +90º + r = 180º µ o Magnetic fux per turn = NIA r = 90º i p From Sne s aw, Hence, the tota magnetic fux (f) inked with sin i p the soenoid is given by the product of fux sin r = µ through each turn and the tota number of where μ is the refractive index of the medium turns. µ o φ= NIA (gass) N Substituting for r, we get sin i p sin ip µ o i.e. φ= N IA = µ ; = µ sin(90 ip) cos ip... (1) tan i p = µ. If L is the coefficient of sef induction of the The tangent of the poarising ange is soenoid, then numericay equa to the refractive index of f = LI... () the medium. From equations (1) and (), µ o LI = N IA 57. Cathode rays have the foowing properties: (i) They trave in straight ines. (ii) Cathode rays possess momentum and µ L = o NA kinetic energy. (iii) Cathode rays produce heat, when aowed If the core is fied with a magnetic materia to fa on matter. of permeabiity µ, (iv) Cathode rays produce fuorescence when µ NA they strike a number of crystas, mineras then, L =. and sats. 56. Brewster s aw : (v) When cathode rays strike a soid substance Sir David Brewster conducted a series of of arge atomic weight, X-rays are experiments with different refectors and produced. found a simpe reation between the ange of poarisation and the refractive index of the 58. de Brogie s waveength of matter waves : medium. It has been observed experimentay de Brogie equated the energy equations of that the refected and refracted rays are at right Panck (wave) and Einstein (partice). anges to each other, when the ight is incident For a wave of frequency ν, the energy associated at poarising ange. with each photon is given by Panck s reation, E = hν (1) where h is Panck s constant. According to Einstein s mass energy reation, a mass m is equivaent to energy, E = mc...() where c is the veocity of ight. Poarisation by refection If, hν = mc

10 10 Sura s XII Std Physics March Question Paper with Answers hc λ = mc h First theorem: (or) λ = mc...(3) The compement of a sum is equa to the c product of the compements. If A and B are (since v = λ ) the inputs, then. A + B = A. B For a partice moving with a veocity v, if c = v Truth tabe of NOR gate from equation (3) h h Inputs Outputs λ = = mv p (4) A B Y = A + B where p = mv, the momentum of the partice These hypothetica matter waves wi have appreciabe waveength ony for very ight partices (i) Photoeectric ces are used for reproducing sound in cinematography. (ii) They are used for controing the temperature of furnaces. (iii) Photoeectric ces are used for automatic switching on and off the street ights. Second theorem: The theorems can be proved, first by considering the two variabe cases and then extending this resut as shown in the tabe Truth tabe to prove De-Morgan s theorems (iv) Photoeectric ces are used in the study of temperature and spectra of stars. A B A B A. B A + B A + B A B (v) Photoeectric ces are aso used in obtaining eectrica energy from sunight during space trave Given data : Required energy = J per second Energy per fission (00 MeV) = ev = J 6. Given data : Frequency of the carrier = f c = 10 MHz Soution: Frequency of the signa = f s = 5 khz Let N be the number of fissions per second to = MHz produced the required power Ampitude of the carrier signa = E c = 10 mv Energy per fission N = Required energy. Ampitude of the audio signa = E N = s = 6 mv 6 Frequency components of moduated wave =? N = = Ampitude of the components in the moduated wave =? N = Soution: 61. The two De Morgan s theorems are very important in deaing with NOR and NAND gates. They state that a NOR gate that performs the A + B function is equivaent to the function A. B and NAND gate, that performs the A. B function is equivaent to the function A. B. The moduated carrier wave contains the foowing frequencies: i) Origina carrier wave of frequency = f c = 10 MHz

11 Sura s XII Std Physics March Question Paper with Answers 11 ii) Upper side band frequency, f c + f s = = MHz iii) Lower side band frequency f c f s = = MHz The moduation factor is, m = E 6 s = = 0.6 Ec 10 Ampitude of USB = Ampitude of LSB m Ec = = = 3mV Thus the machine, continuousy transfers the positive charge to the sphere. The eakage of charges from the sphere can be reduced by encosing it in a gas fied stee chamber at a very high pressure. The high votage can be used to acceerate positive ions for the purpose of nucear disintegration. Part - IV 63. It is a device which produces arge eectrostatic potentia difference of the order of 10 7 V. Principe: Eectrostatic induction and action of points. Construction: It consists of a hoow metaic sphere A mounted on insuating piars. A puey B is mounted at the centre of the 64. Magnetic induction due to infinitey ong sphere and another puey C is mounted straight conductor carrying current: XY is near the bottom. an infinitey ong straight conductor carrying A bet made of sik moves over the a current I (Figure). P is a point at a distance a pueys. from the conductor. AB is a sma eement of Two comb shaped conductors D and E ength d. q is the ange between the current are mounted near the pueys. eement I d and the ine joining the eement The comb D is maintained a positive d and the point P. According to Biot-Savart potentia of the order of 10 4 vot. aw, the magnetic induction at the point P due The upper comb E is connected to the to the current eement Id is inner side of the hoow metaic sphere. Working: Because of the high eectric fied near the comb D, the air gets ionized. The negative charges in air move towards the needes and positive charges are repeed towards the bet due to action of points. The +ve charges stick to the bet moves up end reaches near the comb E. E acquired negative charge and the sphere acquires positive charge due to eectrostatic induction. µ o Id sinθ db =... (1) The acquired +ve charge disturbed on the 4π r outer surface of the sphere.

12 1 Sura s XII Std Physics March Question Paper with Answers AC is drawn perpendicuar to BP from A. OPA = φ, APB = dφ In ABC, sin q = AC = AC AB d AC = d sin q... () From APC, AC = rdφ... (3) From equations () and (3), rdφ = d sin q... (4) Substituting equation (4) in equation (1) µ o Ird φ µ o Id φ db = =... (5) 4π r 4π r In OPA, cos φ = a r a r = cos φ Substituting equation (6) in equation (5)... (6) db = µ o 1 cos f df 4π a The tota magnetic induction at P due to the conductor XY is B = φ φ µ I o db = cos φ dφ 4πa φ1 φ1 B = µ I o 4π a [sin f 1 + sin f ] For infinitey ong conductor, f 1 = f = 90 o µ o B = I πa If the conductor is paced in a medium of permeabiity m, µ I B = π a. 65. Resistor, inductor and capacitor in series: An aternating source of emf e is connected to a series combination of a resistor of resistance R, inductor of inductance L and a capacitor of capacitance C Figure (a). (a) RLC sereis circuit (b) Votage phasor diagram Let the current fowing through the circuit be I. The votage drop across the resistor is,v R = I R (This is in phase with I) The votage across the inductor coi is, V L = I X L (V L eads I by π ) The votage across the capacitor is, V C = IX C (V C ags behind I by π ) The votages across the different components are represented in the votage phasor diagram Figure (b). V L and V C are 180 o out of phase with each other and the resutant of V L and V C is (V L V C ), assuming the circuit to be predominanty inductive. The appied votage V equas the vector sum of V R, V L and V C. OB = OA + AB ; V = V R + (VL V C) V = V + (V V ) R L C

13 Sura s XII Std Physics March Question Paper with Answers 13 V = (IR) (IXL IX C) = I R + (XL X C) V =Z= R + (X L X C ) I The expression R + (XL X C) is the net effective opposition offered by the combination of resistor, inductor and capacitor known as the impedance of the circuit and is represented by Z. Its unit is ohm. The vaues are represented in the impedance diagram (Figure (c)). (c) Impedance diagram Phase ange f between the votage and current is given by: VL VC IXL IXC tan φ= = V R X tan L XC net reactance φ= = R resistance f = tan 1 XL XC R I 0 sin (ωt ± f) is the instantaneous current fowing in the circuit. 66. Young s Doube Sit Experiment: Light from a narrow sit S, iuminated by a monochromatic source, is aowed to fa on two narrow sits A and B paced very cose to each other. The width of each sit is about 0.03 mm and they are about 0.3 mm apart. Since A and B are equidistant from S, ight waves from S reach A and B in phase. So A and B act as coherent sources. According to Huygen s principe, waveets from A and B spread out and overapping takes pace to the right side of AB. When a screen XY is paced at a distance of about 1 metre from the sits, equay spaced IR aternate bright and dark fringes appear on the screen. These are caed interference fringes or bands. Using an eyepiece the fringes can be seen directy. At P on the screen, waves from A and B trave equa distances and arrive in phase. These two waves constructivey interfere and bright fringe is observed at P. This is caed centra bright fringe. Young's doube sit experiment When one of the sits is covered, the fringes disappear and there is uniform iumination on the screen. This shows ceary that the bands are due to interference. Expression for Bandwidth: Let d be the distance between two coherent sources A and B of waveength. A screen XY is paced parae to AB at a distance D from the coherent sources. C is the mid point of AB. O is a point on the screen equidistant from A and B. P is a point at a distance x from O, as shown in Figure. Waves from A and B meet at P its phase or out of phase depending upon the path difference between two waves. Interference band width Draw AM perpendicuar to BP The path difference d = BP AP AP = MP d = BP AP = BP MP = BM.

14 14 Sura s XII Std Physics March Question Paper with Answers In right anged DABM, BM = d sin θ. 67. The Ruby Laser consists of a singe crysta If q is sma, sin q = q of ruby rod of ength 10 cm and 0.8 cm in The path difference d = q. d diameter. A ruby is a crysta of auminium In right anged triange COP, tan q = OP x = oxide A O 3, in which some of auminium CO D ions (A1 3+ ) are repaced by the chromium xd ions (Cr For sma vaues of q tan q = δ= 3+ ). The opposite ends of ruby rod xd D are fat and parae; one end is fuy sivered The path difference δ=. and the other is partiay sivered (i.e.) semi D transparent. The ruby rod is surrounded by a Bright Fringes: By the principe of heica xenon fash tube which provides the interference, condition for constructive pumping ight to raise the chromium ions to interference is the path difference = n upper energy eve (figure). In the xenon fash xd = nλ tube, each fash asts severa miiseconds and D in each fash a few thousand joues of energy where n = 0, 1,... indicate the order of is consumed. bright fringes. D x= nλ d This equation gives the distance of the n th bright fringe from the point O. Dark Fringes: By the principe of interference, condition for destructive interference is the Ruby aser λ path difference = (n 1) where n = 1,, 3... indicate the order of the dark fringes. D λ x= ( n 1) d This equation gives the distance of the n th Energy eve diagram for ruby aser dark fringe from the point O. Thus, on the The simpified energy eve diagram of screen aternate dark and bright bands are chromium ions in a ruby aser, indicating seen on either side of the centra bright band. appropriate excitation and decay is shown in Bandwidth (b) figure. In norma state, most of the chromium The distance between any two consecutive ions are in the ground state E 1.When the ruby bright or dark bands is caed bandwidth. rod is irradiated by a fash of ight, the 5500 Å The distance between (n+1) th and n th order radiation (green coour) photons are absorbed consecutive bright fringes from O is given by: by the chromium ions which are pumped to D D D x( n+ 1) xn= ( n+ 1) λ nλ = λ the excited state E 3. The excited ion gives d d d up part of its energy to the crysta attice and D Bandwidth, β= λ decay without giving any radiation to the d meta stabe state E. Since, the state E has Simiary, it can be proved that the distance a much onger ifetime (10 3 s), the number between two consecutive dark bands is aso of ions in this state goes on increasing. Thus equa to D. Since bright and dark fringes popuation inversion is achieved between dλ the states E and E 1. When the excited ion are of same width, they are equi-spaced on from the metastabe state E drops down either side of centra maximum.

15 Sura s XII Std Physics March Question Paper with Answers spontaneousy to the ground state E 1, it emits a photon of waveength 6943 Å. This photon traves through the ruby rod and is refected back and forth by the sivered ends unti it stimuates other excited ion and causes it to emit a fresh photon in phase with stimuating photon. Thus the refections wi amount to the additiona stimuated emission the so caed ampification by stimuated emission. This stimuated emission is the aser transition. Finay, a puse of red ight of waveength 6943 Å emerges through the partiay sivered end of the crysta. G.M. Counter The gain A is often caed as open-oop gain. Geiger Muer counter is used to measure the intensity of the radioactive radiation. When nucear radiations pass through gas, ionisation is produced. This is the principe of this device. Construction: The G.M tube consists of a meta tube with gass enveope (C) acting as the cathode and a fine tungsten wire (W) aong the axis of the tube, which acts as anode (figure). The tube is we insuated from the anode wire. The tube is fied with an inert gas ike argon at a ow pressure. One end is fitted with a thin mica sheet and this end acts as a window through which radiations enter the tube. A high potentia difference of about 1000 V is appied between the eectrodes through a high resistance R of about 100 mega ohm. Operation: When an ionising radiation enters the counter, primary ionisation takes pace and a few ions are produced. These ions are acceerated with greater energy due to the high potentia difference and they cause further ionisation and these ions are mutipied by further coisions. Thus an avaanche of eectrons is produced in a short interva of time. This avaanche of eectrons on reaching the anode generates a current puse, which when passing through R deveops a potentia difference. This is ampified by eectronic circuits and is used to operate an eectronic counter. The counts in the counter is directy proportiona to the intensity of the ionising radiation. The ionisation of the gas is independent of the type of the incident radiation. Hence, G.M. counter does not distinguish the type of radiation that enters the chamber. Wison s coud chamber is another type of partice detector. This was the first instrument to record the visua observation of the tracks of the charged partices, when they pass through matter. 69. Feedback is the process of adding a fraction of the output signa with the input signa. For an ordinary ampifier i.e. without feedback, et V 0 and V i be the output votage and input votage respectivey. If A be the votage gain of the ampifier, then V 0 A= Vi Feedback Ampifier The genera theory of feedback can be expained with the hep of bock diagram shown in figure. The feedback ampifier has two parts (i.e) ampifier and feedback circuit. The feedback circuit usuay consists of passive components (resistor, capacitor, inductor). A fraction (say β) of the output votage is feed back to the input through the feedback circuit. Let V'? be the output votge with feedback. Therefore, after feedback the input votag V' i becomes, V' i = V' i ± βv' 0... (1) For positive feedback, β is taken as positive. For negative feedback, β is taken as negative. For positive feedback, the input votage wi be V' i + βv' 0. When this is ampified A times by the ampifier, the output votage after feedback (V' 0 ) wi be A(V i ± βv' 0 ). V' 0 = A(V i + βv' 0 )... () V' 0 = (1 βa)=av i... (3)

16 16 Sura s XII Std Physics March Question Paper with Answers Then the votage gain of the ampifier with singe antenna is used for both transmission and feedback is reception and this is achieved with the use of V' o A Af = =... (4) TR switch (Transmitter Receiver Switch). This V 1 A β switching arrangement is caed as dupexer. This connects the antenna to the transmitter during Since 1 βa <1, A f > A. The positive feedback transmission and to the receiver during reception. increases the ampifier gain. Moreover, this switch isoates the sensitive For negative feedback, the feedback fraction is β receiver from the damaging effects of the high A A Af = = power transmitter. 1 ( Aβ ) 1+ Aβ Since 1+βA >1, A f < A.Therefore negative feedback reduces the ampifier gain. The term AB in caed oop gain and β in caed feedback ratio. 70. RADAR The term RADAR is an acronym for RAdio Detection And Ranging. It is a system which uses radio waves to detect and to fix the position of targets at a distance. Bock diagram of a radar system Principe of radar The transmitter is essentiay a high power Radar works on the principe of radio magnetron osciator which generates high echoes. The transmitter in a radar, radiates the power puses. This transmitter is turned on and high power eectrica puses into space. When these off with a periodic puse from the puser. Thus puses are incident on any distant target such as a the transmitter generates periodic puses of very mountain, ship or aircraft, they get scattered in a short duration. These short puses are fed to the directions. The transmitter antenna receives a part antenna which radiates them into the space. The of the scattered energy. This transmitter antenna antenna is highy directiona. aso acts as receiving antenna for the receiving If the transmitted puse hits any target, a weak echo puse. The puse traves with the speed of ight signa returns to the same antenna. But, now the TR ms 1. In other words, these puses cover switch puts the antenna in contact with the receiver. a distance of 300 metres for every micro second. This echo signa is ampified and demoduated Hence by measuring the time taken by the puse by the superhet receiver. The sensitivity of the to reach the target and back to the transmitter, receiver is very high. The detected output is sent the range or distance of the target can be easiy to the indicator. The indicator is a cathode ray tube. determined. To ocate the direction of the target, The CRT dispays the origina transmitted puse as directiona antennas are used. we as the detected echo puse aong a horizonta Transmission and Reception of Radar: The base ine. The synchronising puse generated by the bock diagram of a simpe radar system is shown timer is suppid to both transmitting and receiving in Figure. This bock diagram indicates that the systems. So, the indicator records the transmitted radar system consists of both the transmitting and puse as we as the returning puse simutaneousy. the receiving system. The returning echo puse appears sighty dispaced The transmitting system consists of a transmitter from the transmitted puse and this dispacement is and a puser. The receiving system consists of a a measure of the range of the target. receiver and an indicator. In most of the cases, a