ANNUAL EXAMINATION - ANSWER KEY II PUC - PHYSICS

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ANNUAL EXAMINATION - ANSWE KEY -08 II PUC - PHYSICS I.. PAT A What is an an equipotential suface? Suface with same potential at evey point on it.. Define dift velocity of fee electons.

1 ANNUAL EXAMINATION - ANSWE KEY -08 II PUC - PHYSICS I.. PAT A What is an an equipotential suface? Suface with same potential at evey point on it.. Define dift velocity of fee electons. Effective velocity of an electon due to applied extenal electic field. 3. Give an application of cycloton. Cycloton is used to acceleate chaged paticle. 4. State Faaday s law of electo magnetic induction. Induced emf in a cicuit is diecly popotional to ate of exchange of magnetic flux linked with it. 5. If the peak value of a.c. cuent is 4.4 A, what is its oot mean squae value? I ms 6. I A Mention one powe loss in tansfome. Flux leakage/ hysteesis loss. 7. Two lenses of powe +.5D and -0.5D ae kept in contact on thei pincipal axis. What is the effective powe of the combination? P = P + P = = D 8. The decay of poton to neuton is possible only inside the nucleus. Why? Mass of poton is less than mass of neuton and hence spontaneous disintegation is not possible outside the nucleus / Exchange of Mesons is possible fo small distance between nucleons / Nuclea foce is exchange foce / Nuclea foce is shot ange foce. 9. What is depletion egion in a semi conducto diode? egion between p type and n-type mateial with immobile chage caies/egion between p and g-type mateials, depleted of mobile chage caies. 0. What is the output of this combination? Ans :

2 II.. PAT B Mention any two factos on which the capacitance of a paallel plate capacito depends. (a) Aea of plates (b) gap between plates (c) dielectic constant of medium between plates. State Kichhoff s laws of electical netwok. (a) Algebaic sum of cuents at a node is equal to zeo. (b) Algebaic sum of emf s and potential dops in closed loop of cicuit is zeo. 3. Define magnetic declination and dip at a place. (a) Angle between geogaphic meidian and magnetic medium. (b) Angle between between eath s total magnetic field at a given place and hoizontal along magnetic meidian. 4. Wite the expession fo magnetic potential enegy of a magnetic dipole kept in a unifom magnetic field and explain the tem.s Potential enegy, U = -M B cos, M = Magnetic moment B = Magnetic fied = angle between magnetic axis and extenal magnetic field 5. Give any two applications of X-ays. (a) X ay photogaphy to identify dislocation/factue etc (b) Used in the teatment of cance (c) Used in X ay cystallogaphy 6. What is myopia? How to ectify it? Inability to see objects at fa distance, is called myopia o shot sightedness. Concave lens is used to ectify myopia. 7. Daw the diagam epesenting the schematic aangement of Geige-Masden expeimental set up fo the alpha paticle scatteing. Geige- Masden expeimental set up fo the alpha paticle scatteing. Lead slit Lead cavity ZnS sceen Micoscope Thin gold foil

3 8. Wite any two chaacteistics of nuclea foces. (a) shot ange foce (b) Exchange foces (c) Chage independent foce (d) Mass independent foce PAT C 9. Give thee popeties of electic chage. (a) Chage is additive (b) Chage is quantized (c) Chage is conseved 0. States Ampee s cicuital law and aive at the expession fo the magnetic field nea a staight infinite cuent caying wie. Law state that line integal of magnetic field aound a I closed path in vacuum is equal to poduct of absolute pemittivity and cuent theading the closed path. Conside a long staight conducto caying cuent I. Let P be a point at a pependicula distance fom the p dl B conducto. Conside a cicle of adius aound the conducto passing though P as an Apeian loop. Let d be a small line element of the loop at P. The magnetic field B at P due to cuent caying conducto is along the tangent to this cicule (Fom ight hand thumb ule) and paallel to d. Applying Ampee Cicuital law to the loop. B dl o I Bd cos o I Since = 0 and B is same at all the points on cicula loop, B dl 0 I B ( ) = 0I B. dl, cicumfeence of the loop. 0 I What is hysteesis? Define the tems coecivity and etentivity of a feomagnetic mateial. Lagging of magnetic field in a feomagnetic substance behind magnetic intensity applied is called hysteesis. (a) coecivity is the evese magnetic field applied to a feomagnetic substance to destoy the esidual magnetism. (b) etontivity: is the magnetic field emaining in a feomagnetic substance, in the absence of applied extenal magnetic field. 3

4 . Aive at Snell s law of efaction, using Huygen s pinciple fo efaction of a plane wave. Snell s law using Huygens Pinciple. B V-t i X A i C Y Vt D Let XY epesent the suface sepaating medium and medium. Let v and v be the Speed of light in medium and medium espectively. Conside plane wave font AB incident in Medium at an angle i. Let t be the time taken by the seconday waves to tavel the distance BC. Then BC = vt. The seconday waves fom A will tavel a distance vt. Daw an ac at D in medium with AD as adius. The tangent fom C touches the ac at D. Then CD epesent the efacted wavefont. Let be the angle of efaction. In ABC, sini = BC () AC In ACD, sin AD AC Eqn()/()..() sin i BC vt v.(3) sin AD vt v If n and n ae the absolute efactive indices of medium and medium espectively and if c epesents the speed of light in vacuum, then, n c c and n v v n v n v..(4) Fom (3) and (4) n sin i sin n n sini = n sin This is the Snell s law of efaction. 3. Wite Boh s postulates of hydogen atom model. 4

5 Boh s postulates:. Electons evolve ound the nucleus only in cetain stable stationay obits without the emission of adiant enegy.. Electons evolve aound the nucleus only in those obits in which the angula momentum of the electon is integal multiple of i.e. Angula momentum, mv n h whee h is Planck s constant. h whee n =,, 3, whee n is called pincipal quantum numbe. 3. When an electon jumps fom obit of highe enegy to anothe of lowe enegy it emits a photon having enegy equal to the diffeence in enegy of the electon in oute and the inne obits. (Boh s fequency condition) i.e. hv = E E Whee E and E ae the enegies of the electon in oute and inne obits. 4. Deive the expession fo the half-life of a adio active nuclide. Half life of a adioactive element is the time taken fo 50% of the initial numbe of atoms to disintegate. Numbe of nuclei in the sample at any instant of time t is N = N0e- t () When t = T ( half life), numbe of nuclea pesent, N In equation (), N0 N 0 e T N0 e T T = loge =.303 log0 = = Half life 5. T Wite any thee distinctions between p-type and n-type semi conducto. n - type p- type It is obtained when a pue It is obtained when a pue semiconducto is doped with semiconducto is doped with tivalent pentavalent impuities. impuities Numbe of fee electons is geate Numbe of holes is geate than than numbe of holes numbe of fee electons Conductivity is mainly due to Conductivity is mainly due to holes electons Majoity chage caies ae in the Majoity chage caies ae in the conduction band valence band 5

6 6. Daw the block diagam of genealized communication system. Infomation message Tansmitte Signal Channel Signal signal Souce Message eceive Use Noise 7. Define and electical potential due to a point chage and aive at the expession fo electic potential at a point due to a point chage. Electic potential at a given point is defined as wok done in moving a unit positive chage fom infinity to that point, against electic field. +q O + C E B dx A P x Conside a point P at a distance fom a point chage q at O. A unit positive chage is placed at A, distant x fom O. The electic field at A is E q along OA. 4 0 x The wok done in moving a unit positive chage fom A to B though a small distance dx is dw = - E dx dw = (F = E and -ve sign indicates that F and E ae in opposite diections) q dx 4 0 x The total wok done in moving the unit positive chage fom infinity to P against the electic x W field is dw x q W 4 0 x W q 4 0 x W Potential at P is V 8. dx ( W q 4 0 x dx dx ) x x q q Obtain the expession fo the effective emf and the effective intenal esistance of two cells connected in paallel such that the cuents ae flowing in the same diection. E, Eeq, eq E, I I = I V 6 V I

7 Cells ae said to be in paallel if negative teminals of all the cells ae connected togethe to one teminal and positive teminals ae connected to the othe teminal. Conside two cells having emf s E, E and intenal esistances, espectively connected in paallel. Let I and I ae the cuents though the cells espectively, then the main cuent is I = I + I.() If V is the teminal p.d acoss each cell then fo the fist cell, V = E = I E V o I And fo the second cell, V = E I o I E V Substituting I and I in eqn. () I I E V E V E E V () Let the paallel gouping of cells be eplaced with a single cell of emf E eq and intenal esistance eq then, V = Eeq I eq o I Eeq eq V...(3) eq Compaing () and (3), Eeq eq E E and o eq eq E E E E E E. Eeq eq 9. Deive an expession fo magnetic field on the axis of a cicula cuent loop, using Blot-Savat s law. Conside a cicle coil having N tuns with main adius. Let P be a point on the axis, at distance x fom the cente O. Id sin db 0 4 a along PM.() B A The magnetic field at P due to AB is, I x O whee a is the distance between the point and the element AB. A db cos B P db cos db db sin db Id db 0 4 a 7

8 ( 90 ). Similaly magnetic field at P due to the element A B is Id db 0 along PN..() 4 a The magnetic field db is esolved into hoizontal and vetical components, dbsin and dbscos espectively, whee is the angle between line joining the point to the element and axis of the coil. The vetical components of magnetic fields being equal and opposite cancel each othe and hoizontal components add up. Hence magnetic field at P due to two cuent elements = dbsin The esultant magnetic field at P due to one tun of the coil = db sin o Idl 4 a I o 4 a sin fom () d sin I o sin Since the field is due to elements, 4 a I o 3 4 a Fom OP, sin I o 3/ 4 ( x ) Fom AOP, dl Cicumfeence a a = ( + x)/ o a3 = ( + x)3/ o N I 3/. 4 ( x ) Fo N tuns of the coil B The diection of the magnetic field is along the axis passing though the cente and away fom the obseve facing the coil if the cuent in the coil is clockwise and towads the obseve if the cuent is anticlockwise. 30. Aive at the expession fo the impendance of a seies LC cicuit using phaso diagam method and hence wite the expession fo the cuent though the cicuit. Conside a wie of esistance, coil of inductance L and capacito of capacitance v L C connected in seies with a sinusoidal AC souce. V = V0sin t () VC VL C ~ v Let, I = I0sin( t - ) -----() be the cuent, which is same though the cicuit elements at any time. Whee is the phase diffeence between the voltage and cuent in the cicuit. If V, VL and VC ae the p.d acoss, L and C espectively, then we have, 8

9 V = I, VL = IXL and VC = IXC. V is in phase with cuent I, VL leads I by / and VC lags behind cuent I by /. If V is the p.d acoss the combination, (applied voltage) then V is the phaso sum of V, VL and VC as shown in the figue. V, VL and VC ae epesented by the vectos OA, OB and OC espectively. If VL > VC, the esultant of these two values is V L VC epesented by the vecto OD. Fom the phaso diagam, V = V + (VL VC) V = (I) + (IXL IXC) V ( I) ( IX L IX C ) V I ( X L X C ) V=I Z Z ( X L X C ) Impedance which is the effective opposition offeed by the cicuit fo the flow of AC though it. Also tan VL VC IX L IX L X L X C V I X XC tan L. Cuent I = 3. V V0 sin t I 0 sin t Z Z Deduce the elation between n, u, v, and fo efaction at a spheical suface, whee the symbols have thei usual meaning. In the diagam, XPY epesents a spheical suface of adius of cuvatue, sepaating two media of efactive induces n and n such that n > n. O is the luminous point object placed on the pincipal axis in ae medium. A ay OP incident along the pincipal axis poceeds without deviation. N Anothe ay OA incident at an angle I is efacted i along AI. Let be the angle of efaction. The efacted ays eally convege at I, which is the eal image of the object O. Let AOP =, AIP = and ACP = In AOC, + = i In ACI, () + = o = - () A X O P ae (n) C I Dense(n) Y Fo small apetue AP can be consideed to be staight and pependicula to pincipal axis. Now, tan = AP AP AP, tan, tan PO PI PC 9

10 Fo small angles, tan, tan, tan AP AP AP,, PO PI PC Accoding to Snell s law n sini = n sin Fo small angles, ni = n n ( + ) = n ( - ) Fom () and ) AP AP AP AP n n PI PO PC PC n n n n PO PC PC PI...(3) Accoding to the sign convention, PO = -u, object distance PI = +v, image distance and PC = +, adius of cuvatue. In eqn(3) 3. n n n n u v n n n n. When n=na = and n = n, we have u v n n u v What is a ectifie? With suitable cicuit descibe the action of a full wave ectifie by dawing input and output wavefoms. ectification is the pocess of convesion of AC into DC. A ectifie is said to be full wave ectifie, if the pulsating dc passes though load esistance duing both the halves of input AC cycles. The cicuit consists of two diodes D and D with thei P-egions connected to the two ends of seconday of a tansfome. The N-egions of the diodes ae connected to one end of load esistance L whose othe end is connected to the cente tap. time time Duing the positive half cycles D conducts as it is fowad biased and D does not conduct since it is evese biased in effect. Thus pulsating de passes though load esistance fom A to B. Duing the negative half cycles D is evese biased and D is fowad biased in effect and hence D conducts. Again pulsating dc passes though load esistance fom 0

11 A to B. Thus cuent though load esistance is unidiectional, but fluctuating in each cycle as diodes D and D conduct altenatively esulting in full wave ectification. 33. Thee chages each equal to +4nC ae placed at the thee cones of a squae of side cm. Find the electic field at the fouth cone. E at D due to 4nC at A + 4nC A q E = = 9 0 = 9 04 NC- 4πε 0 (0.0) E at D due to 4nC at B 0.0 m 0.0 m + 4nC B 0.0 m -9 E = = 9 04 NC- (0.0) E E3 E E E E cos 90 E E (9 0 ) ( 9 0 ) E4 D 0.0 m C + 4nC E 8 = 779. NC-, bisects the angle be E and E have along BD E at D due to electicity at B E NC (0.08) Net electic field + E3 = = NC- =.73 05N/S bisectos the angle between E and E mg mass of nichome metal is dawn into a wie of aea of coss-section 0.05 mm. 3 0 kgm -3 Calculate the esistance of this wie. Given density of nichome 8.4 and esistivity of the mateial as. 0-6 m Given: Mass = g -3 A = 0.05 (0 ) m Density = kgm-3 Density = mass volume Volume mass Density V = m3 =. 0-6 m l A Volume = ℓ A = = 5.74 V A ℓ = m

12 35. A cicula coil of adius 0 cm and 5 tuns is otated about its vetical diamete with an angula speed of 40 ad S-, in a unifom hoizontal magnetic field of magnitude 5 0-T. Calculate the emf induced in the coil. Also find the cuent in the coil if the esistance of the coil is 5. Given: adius = 0 cm W = 40 ad s E=nAB - E = n( ) B B = 5 0- T = (0 0-) E=? Peak value of emf, E =.57 V = Peak value of Cuent I E A 5 In Young s double slit expeiment the slits ae sepaated by 0.8 mm and the sceen is placed at a distance of.4 m away fom the slits. The distance between the cental bight finge and the fifth dak finge is measued to be.35 cm. Calculate the wavelength of the light used. Also find the finge width if the sceen is moved 0.4 m towads the slits, fo the same expeimental set up. Given: d = m D =.4 m n5 = m =? xn = (n-) = ( 5 -) D d = m If sceen is moved 0.4 m towads the slits. D = = m D ' d =.4 0-4m 37. =.4 mm Light of fequency Hz is incident on a metal suface. Electons with thei maximum speed of ms- ae ejected fom the suface. Calculate the theshold fequency fo photo emission of electons. Also find the wok function of the metal in electon volt (ev). Given Planck s constant h = Js and mass of the electon Kg.

13 Given: Fequency of incident adiation V = Hz Vmax = ms- Theshold fequency, 0 =? Wok function, 0 =? h= Js me = kg h = 0 + mvmax 0 hv mvmax = [ ( ) ] = = 0-9 [ ] 0 = J ev = h 0 Theshold fequency V = hz 34 h ***************** 3

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