PHYSICS 2004 (Delhi Region)

Transcription

1 PYSP 004 / Class X 1 PHYSCS 004 (Delhi Regin) Time llwed: 3 Hurs M. Marks: 70 General nstructins: (a) ll questins are cmpulsry. (b) There are 30 questins in ttal. Questin 1 t 8 carry ne mark each, questins 9 t 18 carry tw marks each, questins 19 t 7 carry three marks each and questins 8 t 30 carry five marks each. (c) There is n verall chice. Hwever, an internal chice has been prvided in ne questin f tw marks, ne questin f three marks and all three questins f give marks. Yu have t attempt nly ne f the given chices in such questins. (d) Use f calculatrs are nt permitted. (e) Yu may us the fllwing physical cnstraints whenever necessary. c = ms -1 h = Js e = C = Tm -1 = C N -1 m = N m C - m e = kg 1 ev = J 1 atmic mass unit (u) = kg 1 atmic mass unit (u) = MeV Mass f neutrn m n = kg ltzmann's cnstant K = J K -1 vgadr's number N = /mle 1. n the diagram shwn belw is a circular lp carrying current. Shw the directin f the magnetic field inside the lp with the help f lines f frce. The magnetic liens f frce f a circular lp carrying current are shwn belw:. Name the type f mdulatin scheme preferred fr digital cmmunicatin. Pulse Cde Mdulatin 3. Write the nuclear decay prcess fr -decay f 3 15 P P 16S 1e 4. What is the name given t that part f electrmagnetic spectrum which is used fr taking phtgraphs f earth under fggy cnditins frm great heights?

2 PYSP 004 / Class X nfrared rays. 5. 'Heavy water is ften used as a mderatr in thermal nuclear reactins'. Give reasn. Heavy water is used as a mderatr because its mass is nearest t that f a neutrn and it has negligible chances fr neutrn absrptin. 6. The graph shws the variatin f vltage, V acrss the plates f tw capacitrs and versus increase f charge, Q stred n them. Which f the tw capacitrs has higher capacitance? Give reasn fr yur answer. V Q C Q V C Q V ut V < V C > C s capacitr has a higher capacitance. 7. rm temperature (7.0 C) the resistance f a heating element is 100. What is the temperature f the element if the resistance is fund t be 117, given that the temperature cefficient f the material f the resistr is Here R 7 = 100, R t = 117. = C 1, t =? R t = R 7 [1 + (t 7)] 117 = 101[ (t 7)] 117 = (t 7) 117 = 101 = (t 7) 16 = (t 7) 16 t 7 = t 7 = 1000 t = 107 C 10 4 C

3 PYSP 004 / Class X 3 8. bulb and a capacitr are cnnected in series t an a.c. surce f variable frequency. Hw will the brightness f the bulb change n increasing the frequency f the a.c. surce? Give reasn. X C X C 1 1 C C 1 Therefre, brightness f the bulb will increase. 9. Explain with the help f diagram the terms (i) magnetic declinatin and (ii) angle f dip at a given place. (i) Magnetic inclinatin ( ): The angle between the gegraphic meridian and magnetic meridian at a place is called magnetic declinatin at that place. declinatin vertical Gegraphical meridian Gegraphic N-S H dip E V Hrizntal line Magnetic meridian (ii) ngle f dip ( ): The angle made by the earth s ttal magnetic field called angle f dip at any place. with the hrizntal is 10. Tw circular cils, ne f radius r and the ther f radius R are placed c-axially with their centers cinciding. Fr R >> r, btain an expressin fr the mutual inductance f the arrangement. Suppse a current flws thrugh the uter circular cil. Magnetic field at the centre f the cil is R Field may be cnsidered cnstant ver the crss sectinal area f the inner smaller cil. Hence r 1 r M R 1 r M R 11. Find the wavelength f electrmagnetic wave f frequency Hz in free space. Give its tw applicatins. = c m = m = Å

4 inding Energy per nuclen (MeV) PYSP 004 / Class X 4 This wavelength crrespnds t X rays which are used (i) as a diagnstic tl (ii) as a treatment fr certain frms f cancer. 1. Draw the graph shwing the variatin f binding energy per nuclen with mass number. Give the reasn fr the decrease f binding energy per nuclen fr nuclei with high mass numbers. The graph shwing the variatin f binding energy per nuclen with mass number O C 1 He N 14 Fe 56 U Li H Mass number () The binding energy per nuclen fr nuclei with high mass numbers is lw due t the large culmb repulsin between the prtns inside these nuclei. 13. With the help f diagram, shw the biasing f a light emitting dide (LED). Give its tw advantages ver cnventinal incandescent lamps. The diagram is shwn belw: n n p Metallised cntact The fllwing are the tw advantages ver cnventinal incandescent lamps: (1) Lw peratinal vltage and less pwer. () Fast actin and n warm up time required. 14. cmpund micrscpe with an bjective f 1.0 cm fcal length and an eye-piece f.0 cm fcal length has a tube length f 0 cm. Calculate the magnifying pwer f the micrscpe, if the final image is frmed at the near pint f the eye. L D 0 5 m mme 50 f f 1 e OR

5 PYSP 004 / Class X 5 The magnifying pwer f an astrnmical telescpe in the nrmal adjustment psitin is 100. The distance between the bjective and the eye-piece is 101 cm. Calculate the fcal lengths f the bjective and f the eye-piece. f Here, M = fe = 100 and f = 100f e ut f + f e = 101 r 100f e + f e = 101 f e = 1 cm and f = 100 cm. 15. State Gauss's law in electrstatics. Using this therem, derive the expressin fr the electric field intensity at any pint utside a unifrmly charged thin spherical shell. 1 ccrding t Gauss therem, the electric flux thrugh a clsed surface is times the charge q enclsed by the clsed surface. q E.ds E S Cnsider a Gaussian surface f radius r (>R) utside the spherical shell. y Gauss s therem, q E.ds Eds E.4 r E 1 q E 4 r 16. Find the ttal energy stred in the capacitrs in the given netwrk. F 6 V 1 F 1 F F F F and F capacitrs are in series, their equivalent capacitance = 1 F Nw 1 F and 1 F capacitrs are in parallel, s their equivalent capacitance = = F Then we have F and F capacitrs in series, their equivalent capacitance = 1 F

6 PYSP 004 / Class X 6 Finally, we have 1 F capacitance in parallel with 1 F capacitance, their equivalent capacitance. C = = F = 10 6 F Ttal energy stred = 1 1 CV ( 10-6 )(6) = = J 17. What are eddy currents? Discuss briefly any ne applicatin f eddy current. Eddy current: The current induced in the bdy f a cnductr when the amunt f magnetic flux linked with the cnductr changes, if called eddy current. The magnitude f eddy current is nduced emf Re sis tance e R d / dt R The directin f eddy current is given by Lenz s law. The eddy current is als called Fucault current. pplicatin: Electrmagnetic brakes in an example f eddy current. strng magnetic field is applied t a metallic drum rtating with the axle cnnecting the wheels. The large eddy current is set up in the drum which ppse the mtin f the drum. Thus the train stps. 18. The diagram shws a piece f pure semicnductr, S in series with a variable resistr R, and a surce f cnstant vltage V. Wuld yu increase r decrease the value f R t keep the reading f ammeter () cnstant, when semicnductr S is heated? Given reasn. V S When the pure semicnductr S is heated, its resistance decreases. S the current in the circuit increases. T keep the reading f ammeter () cnstant, the value f R has t be increased. 19. State tw cnditins t btain sustained interference f light. n Yung's duble slit experiment, using light f wavelength 400 nm, interference fringes f width X are btained. The wavelength f light is increased t 600 nm and the separatin between the slits is halved. f ne wants the bserved fringe width n the screen t the same in the tw cases, find the rati f the distance between the screen and the plane f the interfering surces in the tw arrangements. The cnditins fr btaining sustained interference f light are (i) The tw light surces shuld be cherent. (ii) The tw light surces shuld be narrw and placed clse t each ther. Fringe width in first case, D D 400 d d Fringe width in secnd case is same. D' 600 D' 100 d/ d D 400 D' 100 d d

7 PYSP 004 / Class X 7 r D 3 D' 1 3 : 1 0. Give the mass number and atmic number f elements n the right-hand side f the decay prcess 0 86 Ru P He The graph shws hw the activity f a sample f Radn - 0 changes with time. Use the graph t determine its half-life. Calculate the value f decay cnstant f Radn Ru 84 P He Mass number f P = 16 tmic number f P = 84 Mass number f He = 4 tmic number f He = Data are nt prvided with the given graph. 1. circular cil f N turns and radius R, is kept nrmal t a magnetic field, given by = cs t. Deduce an expressin fr emf induced in this cil. State the rule which helps t detect the directin f induced current. Flux linked with the N turns f the cil = N = N cs t R = N R cs t nduced emf is given by d d E N R cs t dt dt N R sin t The directin f induced current can be determined by Lenz s law which states that the directin f induced current is such that it ppses the cause which prduces it.. n a series R - C circuit, R = 30, C = 0.5 F, V = 100 V and = 10,000 radian per secnd. Find the current in the circuit and calculate the vltage acrss the resistr and the capacitr. s the algebraic sum f these vltages mre than the surce vltage? f yes, reslve the paradx. Here R = 30, C = F, V rms = 100 V, = 10,000 rad/s 1 1 Z R 30 C rms Vrms 100 Z VR R V rms rms

8 PYSP 004 / Class X 8 VC X rms rms C 0.5 rms c 6 100V Yes, the algebraic sum f vltages acrss R and C is mre than the surce vltage. This is because the vltage acrss C lags behind the vltage acrss R by Define the term 'wrk functin' f a metal. The threshld frequency f a metal is f. When the light f frequency f is incident n the metal plate, the maximum velcity f electrns emitted is v 1. When the frequency f the incident radiatin is increased t 5f, the maximum velcity f electrns emitted as v. Find the rati f v 1 t v. The minimum amunt f energy required t eject an electrn frm a metal surface (withut imparting it any kinetic energy) is called wrk functin f the metal. s f is the threshld frequency, s W = hf Frm Einstein s phtelectric equatin, the maximum kinetic energy f emitted electrn is given by 1 mv 1 h f W h f hf hf and 1 mv h 5f hf 4hf 1 mv 1 hf 1 mv 4hf v1 1 r 1: v 4. Draw the circuit diagram f a cmmn-emitter amplifier using n-p-n transistr. Draw the input and utput wave frms f the signal. Write the expressin fr its vltage gain. The circuit diagram f a cmmn-emitter amplifier using an n-p-n transistr is shwn belw: V i C 1 nput ~ V R C E R C V CC Output V The vltage gain f a CE amplifier is given by V V R R V R R C C C i where is the current amplificatin factr f the transistr. 5. Write the symbl and truth table f an ND gate. Explain hw this gate is realized in practice by using tw dides. The symbl f an ND gate is Y

9 PYSP 004 / Class X 9 when is anded with, we write Y =.. The truth table is Y Realizatin f ND gate: Fr tw input ND gate, we use t tw dides D 1, D as shwn in figure. The resistr R is cnnected t psitive terminal f a 5 V battery. Operatin : (i) When bth and are grunded, bth D 1 an D get frward biased and cnduct. Output Y will be 0 since it neutralized by 5 V battery sending the current in ppsite directin. (ii) When is earthed (0) and is cnnected t 5 V battery (1), D 1 cnducts and D des nt cnduct. The utput Y = 0. (iii) When is cnnected t 5 (1) and grunded (0), D cnducts and D 1, des nt, utput Y = 0. (iv) when and are bth at high (1) level, nne f the dides cnduct. The utput is equal t the battery vltage f 5 V. i.e. Y = With the help f a blck diagram, explain the principle f an ptical fiber cmmunicatin system. Give its tw advantages ver cable cmmunicatin system. lck diagram f an ptical cmmunicatin system is given belw: Transmitter nput signal Optical Mdulatin Surce Receiver Demdulatin Optical Detectr Output Signal Principle: The input analg signal is first sampled and mdulated using PCM which prduces discrete pulses and gives cded steam, f but (0 s and 1 s) representing the signal, crrespnding t this stream f bits, the ptical beam frm a surce is mdulated. This mdulated signal is prpagated in medium called ptical fibre. t the receiver, an ptical detectr detects the signal which in turn, is cnverted back t electrical signal. This is decded and riginal analg signal is recnstructed. dvantages f ptical cmmunicatin system ver cable cmmunicatin system are (i) Wide channel bandwidth and large channel carrying capacity. (ii) Law transmissin lss. OR With the help f relevant diagrams, explain the fllwing terms:

10 PYSP 004 / Class X 10 (i) Pulse-psitin mdulatin (PPM) (ii) Pulse-duratin mdulatin (PDM) (i) n pulse-psitin mdulatin, the mdulating analg signal is sampled and the timing r psitin f the pulse is varied arund a fixed value cnfrming t the signal amplitude at the time f sampling. The psitive ging signal sets the pulse ahead and the negative ging signal sets the pulse behind the reference time. Mdulating wave PPM (ii) n pulse-duratin mdulatin, the mdulating analg signal is sampled and the pulse length r duratin arund a fixed value is varied. Mdulating wave 7. Name the device used fr data transmissin frm ne cmputer t anther. Justify the name. Using this device, draw the blck diagram fr data cmmunicatin and explain it briefly. PDM The device used t interface tw digital surce is called mdem. th mdulatin and Demdulatin functins are included in a mdem. The lck diagram fr a data cmmunicatin circuit using MODEMs is given belw: Mdem 1 Mdem Cmputer Mdulatr Demdulatr Cmmunicatin Channel Demdulatr Mdulatr Cmputer Transmitter side Receiver side The mdem at the transmitting statin changes the digital utput frm a cmputer t a frm which can be easily set thrugh a cmmunicatin channel. t the receiving mdem, the prcess is reversed. n transmitting mde, the mdem cnverts the digital t analg fr use in mdulating a carrier signal. t the receiver cmputer the carrier is demdulated t recver the data. 8. Derive an expressin fr the trque acting n a lp f N turns, area, carrying current i, when held in a unifrm magnetic field. With the help f circuit, shw hw a mving cil galvanmeter can be cnverted int an ammeter f given range. Write the necessary mathematical frmula.

11 PYSP 004 / Class X 11 (i) rectangular lp efgh is placed in a magnetic field at a certain rientatin with it. Using right hand rule we see that there are frces F l n the sides eh and fg, parallel t the axis f rtatin and cause n trque. ut there are frces n the sides ef and gh which cause trque. h F F e (h, g) m N g F (e, f) f (i) F Nte: When field is exactly perpendicular t the plane f the lp there is n turning effect. Let b be the length f the vertical side and be the current. These frces are F = b each. Let a be the length f fg and eh. Then the trque as shwn in figure (ii) is where Fasin is the smaller angle between and perpendicular t the area f the lp. = basin = absin = sin where is the area f the lp equal t ab. The magnetic mment f the lps is m, where is the vectr alng the utside nrmal accrding t the right hand rule and, therefre, = m sin. (ii) g G - g S galvanmeter can measure very small currents f the rder f r s. ut when larger current are desired t be measured, the galvanmeter is shunted with a resistance S. The resistance S allws the current t be by-passed. Suppse, the ammeter range is 0, then the current g which is the safe current fr full scale deflectin passes thrugh the galvanmeter resistance G and ( - g ) passes thrugh S. ccrding t Ohm's law S G g g g S G g OR Write an expressin fr the frce experienced by a charged particle mving in a unifrm magnetic field. With the help f a diagram, explain the principle and wrking f a cycltrn. Shw that cycltrn frequency des nt depend n the speed f the particle. cycltrn is a device develped by Lawrence and Livingstne by which the psitively charged particles likes prtn, deutrn etc. can be accelerated. Principle

12 PYSP 004 / Class X 1 The wrking f the cycltrn is based n the fact that a psitive charged particle can be accelerated t a sufficiently high energy with the help f smaller values f scillating electric field. Cnstructin t cnsists f tw D-shaped hllw evacuated metal chambers D 1 and D called the dees. These dees are placed hrizntally with their diametric edges parallel and slightly separated frm each ther. The dees are cnnected t high frequency scillatr which can prduce a ptential difference f the rder f 10 4 vlts at frequency 10 7 Hz. The tw dees are enclsed in an evacuated steel bx and are well insulated frm it. The bx is placed in a strng magnetic field prduced by tw ple pieces f strng electrmagnets N, S. The magnetic field is perpendicular t the plane f the dees. P is a place f inic surce r psitively charged particle. N H.F. Oscillatr ~ D P D1 E W Target S Wrking and Thery The psitive in t be accelerated is prduced at P. Suppse, at that instant, D 1 is at negative ptential and D is at psitive ptential. Therefre, the in will be accelerated twards D 1. On reaching inside D 1, the in will be in a field free space. Hence it mves with a cnstant speed in D 1 say v. ut due t perpendicular magnetic field f strength, the in will describe a circular path f radius r(say) in D 1, given by qv mv r where m and q are the mass and charge f the in. r mv q Time taken by in t describe a semicircular path is given by t r m v q q/ m = cnstant Thus this time is independent f bth the speed f the in and radius f the circular path. n case the time during which the psitive in describes a semicircular path is equal t the time during which half cycle f electric scillatr is cmpleted, then as the in arrives in the gap between the tw dees, the plarity f the dees is reversed i.e. D 1 becmes psitive and D negative. Therefre, the psitive in is accelerated twards D and it enters D with greater speed which remains cnstant in D. The in will describe a semicircular path f greater radius due t perpendicular magnetic field and again will arrive in a gap between the tw dees exactly at the instant, the plarity f the tw dees is reversed. Thus, the psitive in will g n accelerating

13 PYSP 004 / Class X 13 every time it cmes int the gap between the dees and will g n describing circular path f greater and greater radius and finally acquires a sufficiently high energy. The accelerated in can be remved ut f the dees frm windw W, by applying the electric field acrss the deflecting plates E and F. Let v and r be the maximum velcity and the maximum radius f the circular path fllwed by the psitin in in cycltrn Then mv r qv r v qr m Max. KE = qr q r m 1 1 mv m m f T is the time perid f scillating electric field then, T t m q The cycltrn frequency is given by v 1 q t m t is als knwn as magnetic resnance frequency. 30. With the help f a ray diagram, shw the transfrmatin f image f a pint bject by refractin f light at a spherical surface separating tw media f refractive indices n 1 and n (n > n 1 ) respectively. Using this diagram, derive the relatin n n1 n n1 v u R Write the sign cnventins used. What happens t the fcal length f cnvex lens when it is immersed in water? M is a cnvex surface separating tw media f refractive indices n 1 and n (n > n 1 ). Cnsider a pint bject O placed n the principal axis. ray ON is incident at N and refracts alng N. The ray alng ON ges straight and meets the previus ray at. Thus is the real image f O. Frm Snell s law, n 1 sin i = n sin r r n 1 i = n r [ sin = as is very small] Frm NOC, i = + Frm NC, = r r r = n 1 ( + ) = n ( ) r n 1 + n = (n n 1 ) ut tan = tan = NP NP P M NP NP OP OM [P is clse t M]

14 PYSP 004 / Class X 14 tan = NP PC NP MC NP NP NP n1 n n n1 OM M MC n1 n n n1 r OM M MC Using Cartesian sign cnventin, OM u, M v,mc R n1 n n n1 u v R n n1 n n1 v u R This is the required relatin. OR 30. Deduce the cnditin fr balance in a Wheatstne ridge. Using the principle f Wheatstne ridge describe the methd t determine the specific resistance f a wire in the labratry. Draw the circuit diagram and write the frmula used. Write any tw imprtant precautins yu wuld bserve while perfrming the experiment. Wheatstne ridge is an arrangement f fur resistances P, Q, R and S jined tgether t frm a lp like the fur arms f a quadrilateral as shwn. t is said t be balanced when P/Q = R/S. n balanced cnditin, a cell jined between and sends a current i 1 in the branch P and i in the branch R, the same currents i 1 flws thrugh Q and i thrugh S. There is n current thrugh CD and hence V C = V D. P C i 1 Q i i 1 i i S R D pplying Kirchhff's law t the lp CD, i1p ir 0 i1p ir (1) and t the lp CD, i1q is 0 i1q is () Dividing with equatins, we get P/Q = R/S. The figure shws a meter bridge which cnsists f fur resistances (a) P which is unknwn (b) Q which is a resistance knwn frm a resistance bx (c) R which is the resistance f the length l f the meter bridge wire (d) S which is the resistance f the length (100 - l) f the meter bridge wire. The wire is a unifrm resistance wire and 100 cm lng. When the ptentials V C = V D there is n current in the galvanmeter.

15 PYSP 004 / Class X 15 P C Q R G D S This is a null cnditin. Pint D is called the null pint. ccrding t the principle f meter bridge xl PQ = R/S =, where x is the resistance per unit length f the wire. x 100 l The unknwn resistance, P = l P Q. 100 l xl Q. x 100 l Fr determinatin f specific resistance, the first step is t find the resistance. Then use the l frmula resistance =. The length l, be measured by a meter scale and area be measured by the micrmeter screw gauge. Precautins: (i) The resistance Q frm the resistance bx shuld be chsen s that the balance pint D cmes near t the center f the wire. (ii) Since the galvanmeter G is a sensitive meter, a high prtective resistr shuld be jined in series with it until a near balance is reached n the wire. When the near balance is reached this high resistr is shunted r remved and final balance pint is fund.