XII PHYSICS RBSE FLASHBACK 2016

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1 XII PHYSICS RBSE FLASHBACK 16 Time: 3.15 Hs. Max. Maks: 56 Geneal Instuctions: 1. Candidate must wite fist his/he Roll No. on the question pape compulsoily.. All the questions ae compulsoy. 3. Wite the answe to each question in the given answe-book only. 4. Fo questions having moe than one pat, the answes to those pats ae to be witten togethe in continuity. 5. If thee is any eo/diffeence/contadiction in Hindi & English vesions of the question pape, the question of Hindi vesion should be teated valid. 6. Q. Nos. Maks Pe Question Thee ae intenal choices in Q. Nos. 1 and 7 to Use of calculato is not allowed in the examination. Q.1 Wite the definition of electic dipole moment. A.1 Electic Dipole moment is the poduct of magnitude of eithe of the chage and vecto displacement fom negative to positive chage. p= q(a) = qa Q. Daw a diagam of equipotential suface fo a single chage. A. 1 Q.3 Detemine the esistance of a esisto with the help of given V-I gaph.

2 A.3 Fom the gaph, slope of gaph at v = Volts, I =.1A R= = Ω.1 V = = R I Q.4 A chage q entes pependiculaly with the diection of a magnetic field B with a velocity V. What would be the foce acting on this chage? A.4 Since v is pependicula to B, So foce, F= q(v B) o = qυ Bsin9 F = qυ B Q.5 Wite the Faaday s law of electomagnetic induction. A.5 The magnitude of the induced emf in a cicuit is equal to the time ate of change of magnetic flux though the cicuit. Mathematically, the induced emf is given by dφb ε= dt The negative sign indicates the diection of ε and hence the diection of cuent in a closed loop. Q.6 A light bulb is ated at 1W fo a V supply. Find the peak voltage of the souce. A.6 As pe question, Vms = V V = V max ms = = 311V. Q.7 Which electomagnetic waves ae used in emote contolle (switches)? A.7 Infa ed waves. Q.8 The adius of cuvatue of concave mio is 4 cm. Detemine its focal length. A.8 Radius of cuvatue, R = 4cm (given) R This focal length f = = cm. Fo concave mio, f = cm. (By sign convention) Q.9 What is fasightedness o hypemetopia? A.9 Hypemetopia is that defect of eye due to which it can see fa off objects distinctly but cannot do so fa those objects which ae lying neae than a cetain distance. Q.1 How much enegy is equied to excite a hydogen atom fom its gound state to second excited state? A.1 Enegy of an electon in n th obit is given by : 13.6z En = ev n Fo Hydogen atom in gound state, z = 1, n = 1 E 1 = 13.6 ev Fo second excited state n = 3

3 13.6 E3 = = 1.51eV 3 Thus enegy equied is E= E3 E 1. = 1.51 ( 13.6) = 1.9 ev Q.11 Daw the logic symbol of AND gate. A.11 Q.1 Wite the two examples of accepto impuities. A.1 Boon, Aluminium Q.

3 E3 = = 1.51eV 3 Thus enegy equied is E= E3 E 1. = 1.51 ( 13.6) = 1.9 ev Q.11 Daw the logic symbol of AND gate. A.11 Q.1 Wite the two examples of accepto impuities. A.1 Boon, Aluminium Q.13 Wite the SI unit of cuent density. A.13 The SI unit of cuent density is A/m (Ampee mete ) Q.14 Wite two chaacteistics of a mateial to constuct a pemanent magnet. Give two examples of such mateials. A.14 Fo pemanent magnet mateial should have high etentivity and high coecivity. Examples : Alnico, cobalt Steel and Ticonal. Q.15 Daw a labelled diagam of an altenating cuent Geneato. Detemine the induced electomotive foce by the otation of coil in it. A.15 When the coil is otated with a constant angula speed ω, the angle θ between the magnetic field vecto B and the aea vecto A of the coil at any instant t is θ = ω t (assuming θ = º at t = ). As a esult, the effective aea of the coil exposed to the magnetic field lines changes with time, and the flux at any time t is φ B = BA cos θ= BA cos ω t Fig. : AC Geneato Fom Faaday s law, the induced emf fo the otating coil of N tuns is then, dφb d ε= N = NBA (cosω t) dt dt Thus, the instantaneous value of the emf is ε = NBA ω sin ωt

4 4 Q.16 Cuent in a cicuit falls fom 5.A to zeo in.1s. If an aveage emf of 1 Volt is induced then calculate self-inductance of an inducto in the cicuit. A.16 Initial cuent I i = 5A Final Cuent I f = A Change of time t =.1sec. (given) Induced Emf E = 1V Let the self inductance be L. di E = L dt L( 5) 1 =.1 L = Heny. Q.17 What is displacement cuent? Obtain an expession of displacement cuent fo a chaged capacito. Wite Ampee-Maxwell s law. A.17 Displacement cuent is the cuent which comes into existence due to the change of electic flux with espect to time (when the applied field is vaying). Let us conside a situation when a paallel plate capacito is being chaged and Q is the instantaneous value of chage on the capacito. If the plates of capacito have an aea A and the magnitude of Q electic field E between the plates is. Aε 1Q Q Then electic flux φ E = E A= A= ε A ε. Now if the chage Q changes with time, then, dφ E 1 dq = dt ε dt dφe o id =ε dt whee i d is displacement cuent. Ampee Maxwell Law :- B.d =µ (i + i d) dφ dt E (i ) =µ +ε Q.18 A eflected light becomes completely polaised when coesponding incident angle is 6. When light is incident on the substance. Detemine the velocity of efacted ay in the medium. A.18 As pe question eflected light is completely polaized when angle of incidence is 6. Thus, Bewste s angle i B = 6. Thus efactive index of medium is :- µ= tan ib = tan 6 = Thus the velocity of light in medium v= = m/s. 3 Q.19 The theshold fequency fo caesium metal is Hz. Detemine its wok function in ev. 14 A.19 Theshold fequency ν T = Hz Thus wok function φ = hν T

5 ( )( ) = = J = ev =.18 ev Q. What is stopping voltage (o cut off voltage)? Plot a gaph of vaiation of photoelectic cuent with collecto plate potential fo two incident adiations of same fequency and diffeent intensities. A. The minimum negative (etading) potential V fo which the photocuent stops o becomes zeo is called the cut-off o stopping potential. Q.1 Wite the statement of Boh s Second postulate of quantisation. Detemine the wavelength of fist spectal line in the Lyman seies of the hydogen spectum. [Rydbeg constant R = m -1 ] Wite the law of adioactive decay. A adioactive nucleus is decaying in the following way. Detemine the mass numbe and atomic numbe of final poduct X 4 when initial nucleus has mass numbe A = 38 and atomic numbe Z = 9. A α β α γ ZX X1 X X3 X4 A.1 Boh s second postulate: Boh s second postulate defines the stable obits. This postulate states that the electon evolves aound the nucleus only in those obits fo which the angula momentum is some integal multiple of h/π whee h is the Planck s constant (= J s). Thus the angula momentum (L) of the obiting electon is quantised. That is L = nh/π Fo fist line in Lyman Seies, = R 1 n λ = ( ) 1 λ 4 7 λ= λ= m

6 6 Law of Radioactive delay: In any adioactive sample, which undegoes α, β o γ -decay, it is found that the numbe of nuclei undegoing the decay pe unit time is popotional to the total numbe of nuclei in the sample. A α β α γ ZX X1 X X X4 Step I: α Decay A α A 4 ZX Z X1 Step II : β Decay A 4 β A 4 Z X1 Z 1X Step III : α Decay A 4 α A 8 Z 1X Z 3X3 Step IV : γ Decay A 8 A 8 Z 3X3 Z 3X4 So Atomic No = Z 3 = 9 3 = 89 Mass No. = A 8 = 38 8 = 3 Q. Wite two dawbacks of Ruthefod s atomic model. A. (i) Accoding to classical electomagnetic theoy, an acceleating chaged paticle emits adiation in the fom of electomagnetic waves. The enegy of an acceleating electon should theefoe, continuously decease. The electon would spial inwad and eventually fall into the nucleus. Thus, such an atom can not be stable. (ii) Accoding to the classical electomagnetic theoy, the fequency of the electomagnetic waves emitted by the evolving electons is equal to the fequency of evolution. As the electons spial inwads, thei angula velocities and hence thei fequencies would change continuously, and so will the fequency of the light emitted. Thus, they would emit a continuous spectum, in contadiction to the line spectum actually obseved. Q.3 Explain skywave popagation by making a diagm. A.3 In the fequency ange fom a few MHz up to 3 to 4 MHz, long distance communication can be achieved by ionospheic eflection of adio waves back towads the eath. This mode of popagation is called sky wave popagation and is used by shot wave boadcast sevices. The ionospheic laye acts as a eflecto fo a cetain ange of fequencies (3 to 3 MHz). Electomagnetic waves of fequencies highe than 3 MHz penetate the ionosphee and escape. These phenomena ae shown in the Fig. The phenomenon of bending of em waves so that they ae diveted towads the eath is simila to total intenal eflection in optics.

7 7 Fig. : Sky wave popagation. Q.4 Wite definitions (i) Tansduce (ii) Modulation A.4 (i) Tansduce A device that convents one fom of enegy to anothe. (ii) Modulation A pocess in which low fequency modulating signal is supeimposed on high fequency caie wave. Q.5 Wite Kichoff s fist ule. A battey of 1V and negligible intenal esistance is connected to the diagonally opposite cones of a cubical netwok consisting of 1 esistos each of esistance Ω. Detemine the equivalent esistance of the netwok. A.5 Kichoff s Fist Rule At any junction, the sum of the cuents enteing the junction is equal to the sum of cuents leaving the junction The paths AA, AD and AB ae obviously symmetically placed in the netwok. Thus, the cuent in each must be the same, say, I. Futhe, at the cones A, B and D, the incoming cuent I must split equally into the two outgoing banches. In this manne, the cuent in all the 1 edges of the cube ae easily witten down in tems of I, using Kichhoff s fist ule and the symmety in the poblem. Next take a closed loop, say, ABCC EA, and apply Kichhoff s second ule: IR (1/)IR IR + ε = whee R is the esistance of each edge and e the emf of battey. Thus, 5 ε= IR The equivalent esistance Req of the netwok is

8 8 ε 5 Req = = R 31 6 Fo R = Ω R eq = (5/3) Ω Q.6 Why eflecting telescope is supeio in compaison to efacting telescope? Wite two easons. Magnifying powe of a telescope is 8. When it is adjusted fo paallel ays the distance between eyepiece and objective lens is 18cm. Detemine the focal length of both the lenses. A.6 (a) Thee is no chomatic abeation in a mio. (b) If a paabolic eflecting suface is chosen, spheical abeation is also emoved. (ii) Magnifying Powe m = 8 given f m= = 8 fe.(1) Fo paallel ay adjustment, distance between eyepiece and objective lens = 18 cm fo+ fe = 18cm () Thus 8fe+ fe = 18 [(fom(1)] and fo fe = cm = 16cm. Q.7 Detemine the foce acting between two paallel cuent caying conducto wies. Wite theoetical definition of ampee on this basis. Establish diffeent elations between magnetisation M, magnetic intensity H, Magnetic susceptibility χ and elative magnetic pemeability µ. Define the magnetic susceptibility χ. A.7 Conside two long paallel conductos a and b sepaated by a distance d and caying (paallel) cuents I a and I b, espectively. The conducto a poduces, the same magnetic field B a at all points along the conducto b. The ight-hand ule tells us that the diection of this field is downwads (when the conductos ae placed hoizontally). Its magnitude is given by µ I Ba = π d The conducto b caying a cuent I b will expeience a sideways foce due to the field B a. The diection of this foce is towads the conducto a. We label this foce as F ba, the foce on a segment L of b due to a. The magnitude of this foce is given by Fba = IbLBa µ II a b = L πd The ampee is the value of that steady cuent which, when maintained in each of the two vey long, staight, paallel conductos of negligible coss-section, and placed one mete apat in vacuum, would poduce on each of these conductos a foce equal to 1 7 newtons pe mete of length.

9 9 Intensity of magnetizing field H : Any magnetic field in which a magnetic mateial is placed fo its magnetization is called magnetizing field. In a magnetizing field the atio of magnetizing field B to the pemeability of fee space is called intensity of magnetizing field, i.e., B H= o B =µ H and has unit Am 1. µ Intensity of magnetization M : When a magnetic mateial is magnetized by placing it in a magnetizing field, the induced dipole-moment pe unit volume in the specimen is called intensity of m magnetization, i.e., M = i.e., intensity of magnetization is numeically equal to the induced polestength pe unit V volume. Magnetic susceptibility χ m : The atio of magnitude of magnetization to that of magnetizing field is called magnetic susceptibility, i.e., M = χmh It is a scala with no units and dimensions and physically epesents the ease with which a magnetic mateial can be magnetized, i.e., lage value of χ m implies that the mateial is moe susceptible to the field and hence can be easily magnetized. Relative Pemeability µ : The atio of pemeability of a medium to that of fee space, i.e., ( µ / µ ) is called elative pemeability µ and is equal to the atio of magnitudes of total field inside the B µ H µ mateial to that of magnetizing field, i.e., µ = = = B µ H µ. Q.8 Wite Gauss s law. Detemine the electic field at the points which ae situated outside and inside of a unifomly chaged thin spheical shell. Daw necessay diagams of Gaussian sufaces. Wite the definition of electic potential. Calculate the electic potential due to a point chage Q at distance fom it. Daw a gaph between electic potential V and distance fo a point chage Q. A.8 Gauss Law : The flux of electic field though any closed 1 suface is times the total chage enclosed by suface. ε q Edscos θ=. ε Field due to a unifomly chaged thin spheical shell Let σ be the unifom suface chage density of a thin spheical shell of adius R (Fig.). The situation has obvious spheical symmety. The field at any point P, outside o inside, can depend only on (the adial distance fom the cente of the shell to the point) and must be adial (i.e., along the adius vecto). (i) Field outside the shell: Conside a point P outside the shell with adius vecto. To calculate E at P, we take the Gaussian suface to be a sphee of adius and with cente O, passing though P. All points on this sphee ae equivalent elative to the given chaged configuation. (That is what we mean by spheical symmety.) The electic field at each point of the Gaussian suface, theefoe, has the same magnitude E and is along the adius vecto at each point. Thus, E and S at evey point ae paallel and the flux though each element is E S. Summing

10 1 ove all S, the flux though the Gaussian suface is Gauss s law E 4π. The chage enclosed is σ 4π R. By σ E 4π = 4πR ε σr q O, E = = ε 4πε whee q = 4 p R s is the total chage on the spheical shell. Vectoially, q E= ˆ 4 πε The electic field is diected outwad if q > and inwad if q <. This, howeve, is exactly the field poduced by a chage q placed at the cente O. Thus fo points outside the shell, the field due to a unifomly chaged shell is as if the entie chage of the shell is concentated at its cente. (ii) Field inside the shell: In Fig.(b), the point P is inside the shell. The Gaussian suface is again a sphee though P cented at O. The flux though the Gaussian suface, calculated as befoe, is E 4π. Howeve, in this case, the Gaussian suface encloses no chage. Gauss s law then gives E 4π = i.e., E = ( < R ) that is, the field due to a unifomly chaged thin shell is zeo at all points inside the shell. Electic Potential of any point is the wok done by an extenal foce in binging a unit positive chage fom infinity to that point. Potential due to a point chage Conside a point chage Q at the oigin (as show in figue). Fo definiteness, take Q to be positive. We wish to detemine the potential at any point P with position vecto fom the oigin. Fo that we must calculate the wok done in binging a unit positive test chage fom infinity to the point P. Fo Q >, the wok done against the epulsive foce on the test chage is positive. Since wok done is independent of the path, we choose a convenient path-along the adial diection fom infinity to the point P. At some intemediate point P ' on the path, the electostatic foce on a unit positive chage is Q 1 ' ˆ' (1) 4πε whee ˆ' is the unit vecto along OP '. Wok done against this foce fom ' to ' + ' is Q W = ' () ' 4πε The negative sign appeas because fo ' <, W is positive. Total wok done (W) by the extenal foce is obtained by integating Eq. () fom ' = to ' =, Q Q Q W = d' = = (3) ' 4πε 4πε' 4πε This, by definition is the potential at P due to the chage Q Q V() = (4) 4 πε

11 11 Q.9 Descibe the constuction of a compound micoscope. Deive an expession fo its total magnification. Daw a ay diagam fo the fomation of image by a compound micoscope. What is intefeence? Deive the conditions to obtain constuctive and destuctive intefeences. If a white light souce is used in place of monochomatic souce of light in young s double slit expeiment then what will be the effect on intefeence finges. A.9 The lens neaest the object, called the objective, foms a eal, inveted, magnified image of the object. This seves as the object fo the second lens, the eyepiece, which functions essentially like a simple micoscope o magnifie, poduces the final image, which is enlaged and vitual. The fist inveted image is thus nea (at o within) the focal plane of the eyepiece, at a distance appopiate fo final image fomation at infinity, o a little close fo image fomation at the nea point. Clealy, the final image is inveted with espect to the oiginal object. That the (linea) magnification due to the objective, namely h /h, equals. h L mo = = h fo whee we have used the esult s h h tanβ= = fo L Hee h is the size of the fist image, the object size being h and f o being the focal length of the objective. The fist image is fomed nea the focal point of the eyepiece. The distance L, i.e., the distance between the second focal point of the objective and the fist focal point of the eyepiece (focal length f e ) is called the tube length of the compound micoscope. when the final image is fomed at the nea point, is D me = 1+ f e when the final image is fomed at infinity, the angula magnification due to the eyepiece is m e = (D/f e ) Thus, the total magnification when the image is fomed at infinity, is

12 1 m m m L D = o e = f fe Redistibution of light enegy due to supeposition of two waves is called intefeence. Conside two Coheent souces S 1 & S as shown in figue: Spheical waves emanating fom S 1 and S will poduce intefeence finges on the sceen GG ' fo an abitay point P on the line GG ' [Fig.] to coespond to a maximum, we must have Fig. : Young s aangement to poduce intefeence patten. SP S1P = n λ ; n =,1,... Now, d d (SP) (S1P) = D + x + D + x = xd whee S1S = d and OP = x. Thus xd SP SP 1 = SP SP + 1 If x, d<<d then negligible eo will be intoduced if S P + S 1 P (in the denominato) is eplaced by D. Thus, xd SP SP 1 D Hence we will have constuctive intefeence esulting in a bight egion when nλd x = x n = ;n =, ± 1, ±,... d On the othe hand, we will have a dak egion nea 1 λd x = xn = n + ;n =, ± 1, ±. d The intefeence pattens due to diffeent component colous of white light ovelap (incoheently). The cental bight finges fo diffeent colous ae at the same position. Theefoe, the cental finge is white. Fo a point P fo which S P S 1 P = λ /, whee o b λb ( 4 A) epesents the wavelength fo the blue colou, the blue component will be absent and the finge will appea ed in colou. Slightly fathe away whee SQ SQ 1 =λ b =λ / whee λ ( 8A) is the wavelength fo the ed colou, the finge will be pedominantly blue. Thus, the finge closest on eithe side of the cental white finge is ed and the fathest will appea blue. Afte a few finges, no clea finge patten is seen. o

13 Q.3 What is ectification? Daw the cicuit diagam of half wave ectifie and explain its woking. Show the input ac voltage and output voltage wavefoms fom the ectifie cicuit.

13 13 Q.3 What is ectification? Daw the cicuit diagam of half wave ectifie and explain its woking. Show the input ac voltage and output voltage wavefoms fom the ectifie cicuit. What is extinsic semiconducto? How many types of these ae? Wite thei names. Explain the pocesses which ae occued duing the fomation of a P-N junction. Detemine the electic field poduced at a P-N junction when width of depletion laye is 1 micomete and baie potential is.7 volt. A.3 Pocess of conveting ac into dc is called ectification. If an altenating voltage is applied acoss a diode in seies with a load, a pulsating voltage will appea acoss the load only duing the half cycles of the ac input duing which the diode is fowad biased. The seconday of a tansfome supplies the desied ac voltage acoss teminals A and B. When the voltage at A is positive, the diode is fowad biased and it conducts. When A is negative, the diode is evese-biased and it does not conduct. Theefoe, in the positive half-cycle of ac thee is a cuent though the load esisto R L and we get an output voltage, as shown in Fig.(b), wheeas thee is no cuent in the negative halfcycle. In the next positive half-cycle, again we get the output voltage. Thus, the output voltage, though still vaying, is esticted to only one diection and is said to be ectified. A semiconducto whose conductivity is mainly due to impuity (doping) is called Extinsic semiconducto. These ae of types (a) n-type (b) p-type Pocesses taking place duing fomation of p-n junction : (i) Diffusion - Duing the fomation of p-n junction, and due to the concentation gadient acoss p-, and n- sides, holes diffuse fom p-side to n-side (p n) and electons diffuse fom n-side to p-side (n p). This motion of chage caies gives ise to diffusion cuent acoss the junction. When an electon diffuses fom n p, it leaves behind an ionized dono on n-side. This ionised dono (positive chage) is immobile as it is bonded to the suounding atoms. As the electons continue to diffuse fom n p, a laye of positive chage (o positive space-chage egion) on n-side of the junction is developed. Similaly, when a hole diffuses fom p n due to the

14 14 concentation gadient, it leaves behind an ionised accepto (negative chage) which is immobile. As the holes continue to diffuse, a laye of negative chage (o negative space-chage egion) on the p- side of the junction is developed. (ii) Dift - Due to the positive space-chage egion on n-side of the junction and negative space chage egion on p-side of the junction, an electic field diected fom positive chage towads negative chage develops. Due to this field, an electon on p-side of the junction moves to n-side and a hole on n-side of the junction moves to p-side. The motion of chage caies due to the electic field is called dift. Thus a dift cuent, which is opposite in diection to the diffusion cuent (see figue) stats. As pe question, depletion laye width, d = 1 µ m Baie potential, V =.7V V.7 5 So, E = 7 1 v/m. 6 d = 1 1 =

SAMPLE PAPER I. Time Allowed : 3 hours Maximum Marks : 70

SAMPLE PAPER I. Time Allowed : 3 hours Maximum Marks : 70 SAMPL PAPR I Time Allowed : 3 hous Maximum Maks : 70 Note : Attempt All questions. Maks allotted to each question ae indicated against it. 1. The magnetic field lines fom closed cuves. Why? 1 2. What is