X X Figure 1 (B) (D) The loss of power in a transformer can be reduced by: (A) Increasing the number of turns in primary.

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1 PHYSICS Paper-I (Thery) (Three hurs) (Candidates are allwed additinal 15 minutes fr nly reading the paper, They must NOT start writing during this time) Answer all questins in Part I and six questins frm Part II, Chsing tw questins frm each f the Sectins A, B and C. All wrking including rugh wrk shuld be dne n the same sheet as, and adjacent t, the rest f the answer. The intended marks fr questins r parts f questins are given in brackets [ ]. (Material t be supplied: Lg tables including Trignmetric functins) A list f useful physical cnstants is given at the end f this paper. PART I Answer all questins Questin 1 A Chse the crrect alternative A, B, C r D fr each f the questins given [5] belw: (i) Electric field intensity E at a pint P (Figure 1) at a perpendicular distance r frm an infinitely lng line charge X X having linear charge density λ is given by: P r X X Figure 1 (A). (ii) (B) (iii) (A) (C) 1 λ E = 4πε r 1 λ E = 4πε r (B) (D) 1 λ E = 4πε r 1 λ E = 4 πε r A mving cil galvanmeter can be cnverted int a vltmeter by cnnecting: (A) A lw resistance in series (B) A lw resistance in parallel with its cil with its cil (C) A high resistance in parallel with its cil (D) A high resistance in series with its cil The lss f pwer in a transfrmer can be reduced by: (A) Increasing the number f turns in primary. (B) Increasing ac vltage applied t primary. (C) Using slid cre made f steel. (D) Using a laminated cre f sft irn.

2 (D) (iv) Which equatin represents the emissin f a beta particle by a radiactive nucleus: (A) n 1H 1e ν (B) 0 0 γ 1e 1e (C) H He 1e (D) He 7 N 8O 1H (A) (v) An imprtant cmpnent f Michaelsn s methd t determine speed f light is: (A) a NICOL prism (B) a bi prism (C) a grating (D) an ctagnal mirrr (D) B Answer all questins briefly and t the pint: [15] (i) In an electric diple, what is the lcus f a pint f zer ptential? The perpendicular bisectr f the length f the diple. (ii) What is the resistance f a carbn resistr whse clured bands are shwn belw: Yellw Red Brwn White A carbn resistr Figure R = 4*10 1.±10% = 40±4 = 64 Ω (The frth clur is nrmally refered t as silver. In the questin it is given as white) (iii) A part f an electric circuit is shwn belw (Figure 3): B Ω I 3A 5A A 4Ω A 6Ω Figure 3 Using Kirchff s nd Law, find the current I flwing thrugh the 4Ω resistr. Fr the mesh ACBA: 14I6=0 4I=6 I = 3/=1.5A. (iv) Figure 4 belw shws a graph f emf e generated by an ac generatr verses time t. C

3 Y e 00 (Vlt) t (secnd) Y Figure 4 What is the frequency f the emf? f = 1/T =1/0.1 =10Hz. (v) Arrange the three types f magnetic materials, i.e. paramagnetic, diamagnetic and ferrmagnetic materials, in decreasing rder f their magnetic susceptibility. ferrmagnetic materials, paramagnetic materials, diamagnetic materials. (vi) Which electrmagnetic wave is lnger than X ray but shrter than light wave? Ultra vilet. (vii) Calculate the critical angle fr glass and water pair. (The refractive index fr glass is 1.50 and refractive index fr water is 1.33) μ g = = = 1.18 μ w SinC SinC = = 0.887, C = Sin = The critical angle = 6.5. (viii) Name an ptical device, which, when used with a spectrmeter, can determine the wavelength f the given mnchrmatic light. A diffractin grating. (ix) Yung s duble slit experiment was perfrmed with mnchrmatic light f blue clur. The experiment was then repeated, first with light f red clur and then, with light f yellw clur. Which clur prduces interference pattern with maximum fringe separatin (i.e. fringe width)? Dλ β = λ. Since red light has highest value f wavelength, it gives the maximum d fringe width. (x) Calculate the dispersive pwer f glass up t three decimal places, frm the fllwing data: Refractive index f glass fr red clur =1.60 Refractive index f glass fr yellw clur =1.61 Refractive index f glass fr vilet clur =1.6 μ v μ r ω = = = μ y (xi) De Brglie wavelength f electrns f kinetic energy E is λ. What will be its value if kinetic energy f electrns is made 4E?

4 h p = me = ; Thus: λ h h m 4E = = ' λ λ 1 λ ' = λ The new wavelength λ = λ/. (xii) Hw much energy will be created if 1 g f matter is destryed cmpletely? E = m c. E = 10-3 x (3x10 8 ).=9.0 x J (xiii) Cmplete the fllwing nuclear reactin: n 6C 5 B n 6C 5 B 1H 0n. (xiv) State ne imprtant use f a Zener dide. Zener dide is used as a vltage regulatr. (xv) Draw a labelled graph f vltage verses time fr a signal vltage used in digital circuits. Vltage (V) Time(μs) PART II Answer six questins in this part, chsing tw questins frm each f the Sectins A, B and C. SECTION A Answer any tw questins Questin (a) An islated 16μF parallel plate air capacitr has a ptential difference f 1000 V (Figure 5 a). A dielectric slab having relative permittivity (i.e. dielectric cnstant) =5 is intrduced t fill the space between the tw plates cmpletely (Figure 5 b). Calculate: (i) the new capacitance f the capacitr. (ii) the new ptential difference between the tw plates f the capacitr. Dielectric slab 16μF A I R V=1000V Figure 5 a Figure 5 b [] (i) C m = κ C.= 5x16 = 80μF (ii) Since the charge is cnserved, the new p.d = 1/κ times the riginal p.d = 1000/5 = 00 V.

5 (b) An electrn revlves rund the nucleus f hydrgen atm in a circular rbit f radius 5 x m. Calculate: (i) intensity f electric field f the nucleus at the psitin f the electrn. (ii) electrstatic ptential energy f the hydrgen nucleus and electrn system. [4] (i)frce between the prtn and the electrn, F = F = 11 4πε (5 10 ) The intensity f electric field at the psitin f the electrn = E =F/e E = 9 10 = N / C 11 (5 10 ) 19. Questin 3 (a) e e (ii) U = = 9 10 = J πε (5 10 ) (5 10 ) The negative sign shws that the nucleus and the electrn frms a bund system. (c) (i) What is Peltier effect? State ne difference between Peltier effect and Seebeck effect. (ii) Explain the statement: Temperature cefficient f resistance f a metal is 4 x10 3 / C. (i)when a current is allwed t flw thrugh a thermcuple, it is fund that heat is absrbed at ne junctin and heat is liberated at the ther junctin. This phenmenn is called Peltier effect. In Seebeck effect current flws thrugh the thermcuple when ne junctin is kept at a cld junctin and the ther junctin at a ht regin. Where as in Peltier effect the thermcuple must be cnnected t a surce f electricity. In fact Peltier effect is the reverse phenmenn f Seebeck effect. (ii) It means that the rati f change in resistance t the riginal value f resistance per unit degree f temperature is 4x10-3. In the circuit shwn belw (Figure 6), PQ is a unifrm metallic wire f length 4 m and resistance 0Ω. Battery B has an emf f 10V and internal resistance f 1Ω. J is a jckey r slide cntact. Resistance f the ammeter A and cnnecting wires is negligible. [4] Figure 6 X=1Ω Z G 10Ω Y=4Ω P 4m,0Ω J A Q B

6 (i) When the jckey J des nt tuch the wire PQ, what is the reading f the ammeter A? (ii) Where shuld the jckey J be pressed n the wire PQ s that the galvanmeter G shws n deflectin? (i) When J in pen circuit: ttal resistance R = 4Ω 1 Ω 1 Ω = 6 Ω The ammeter reading ; I = V/R = 10/6 = 1.67 A. (ii) X r PJ PJ 1 =.where r is the resistance/m lengthf wire.i.e. = Y r JQ JQ 4 i.e PJ : JQ ::1: 4 Since length f the wire is 4.0m, the psitin f J is 0.8m frm end P t get n deflectin in the galvanmeter. (b) What is current density? Write the vectr equatin cnnecting current density J with electric field intensity E, fr an hmic cnductr. [] Current density is defined as the current passing per unit area f crss sectin f the cnductr. It is a vectr quantity. I V R V J = = =, Multiplying bth sides by the length f A A RA the cnductr; l Vl l 1 V 1 lj = = V ( ) = V ; J = RA RA ρ l ρ (c) J = Eσ ; whereσ is the cnductivity. r J r = σ E A small magnetic needle NS having magnetic diple mment P r m is kept in tw unifrm and perpendicular magnetic fields B r F and B r H as shwn belw (Figure 7). B r H Nrth West B r F East (i) (ii) Figure 7 Suth (i) What is the effect f each f the magnetic fields B r F and B r H n the needle? (ii) When the needle is in equilibrium, btain an expressin fr angle θ made by the needle with B r H in terms f B r F and B r H nly. B r H and B r F interact each ther and the needle NS will align in the directin f the resultant field, with its N- ple pinting between the Nrth and the east.

7 B H S θ N B R B F B H =B R Cs θ; B F = B R Sin θ B F /B H = B R Sin θ/b R Cs θ B F /B H = tan θ θ = tan-1(b F /B H ) (In this case a sketch is necessary) Questin 4 (a) Figure 8 belw shws tw very lng cnductrs PQ and RS kept parallel t each ther in vacuum at a distance f 0cm. They carry currents f 5A and 15A, respectively, in the same directin, as shwn. Find the resultant magnetic flux density B r R at a pint M which lies exactly midway between PQ and RS. P R I 1 =5A M I =15A 0cm Q Figure 8 S μi1 μi Magnetic field due t PQ at M =B 1 = = = 10 = 10 T πa 4πa 10 directed away frm the reader. Magnetic field due t RS at M = μi μi B = = = 10 = 3 10 T πa 4πa 10 directed twards the reader. The resultant magnetic flux density = B B 1 = x10-7 T; twards the reader. (b) With the help f a neatly drawn labelled diagram, prve that the magnitude f mtinal emf e is given by e=blv, where l is the length f a metallic rd and v is the velcity with which it is pulled in a transverse magnetic field B. A cnductr is bent in the frm FCDE and a straight cnductr XY can slide ver t make a lp in FCDE with length l tuching it. A unifrm magnetic field B

8 acting in the regin f lp with its directin pinting away frm the reader. X X F C B l v D Y Y E The cnductr initially ccupies the psitin XY and mves t X Y with a unifrm velcity v. The change in magnetic flux prduced in time dt = ' ' dφ = B AreaXYX Y = Blvdt. dφ e = = Blv dt (c) Plt a labelled graph shwing variatin in impedance Z f a series LCR circuit with frequency f f alternating emf applied t it. What is the minimum value f this impedance? Impedance Z R, the hmic resistance Frequency (f) The minimum value f impedance is the hmic resistance f the circuit. (Yu may be tempted t write the equatin fr impedance and s n; nt asked in the questin!) Questin 5 SECTION B Answer any tw questins (a) On the basis f Huygen s wave thery, shw that when light is incident n a plane mirrr bliquely, angle f reflectin is equal t angle f incidence. Accrding t Huygen s thery f secndary wavelets, every pint f a wavefrnt is a surce f secndary wavelets.

9 P R U T M Q i r S N MN is the reflecting surface. QT is the incident plane wavefrnt; making an angle f incidence i. By Huygen s principle, every pint n QT is a surce f secndary wavelet and all are in the same phase. When the disturbance frm T reaches S, the secndary wavelets frm Q might have reached a distance equal t TS. With Q as centre and TS radius draw an arc. Similar arcs can be drawn between Q and S. Jin the envelpes f all these arcs by a tangent SU; the reflected wavefrnt. Frm the right angled ΔQUS and ΔQTS : QU = TS USQ = TQS and QS is cmmn; hence ΔQUS ΔQTS are cngruent : i.e. i = r. Thus: angle f reflectin is equal t angle f incidence. (b) (i) What is a cntinuus emissin spectrum? Name ne surce f light which prduces such a spectrum. (ii) Explain in brief why dark lines are bserved in the slar spectrum. (i) When matter in bulk is excited, they emit all wavelengths frm red t vilet with ut any bundary and the spectrum is called a cntinuus spectrum. The spectrum frm carbn arc is cntinuus in nature. (ii) As slar energy passes thrugh gaseus media in its path, gases will absrb certain parts f the energy frm the spectrum which are characteristic emissin frequencies f thse gases. Thus certain clurs will be absent in the cntinuus nature f the slar spectrum. The absrbed regins in the slar spectrum are viewed as dark lines. (c) An equicnvex lens f glass, having fcal length f 10cm is split int [] tw identical plan-cnvex lenses each having fcal length f 1 as shwn belw (Figure 9) and Radius R Radius R F=10cm Find the value f f 1. Figure 9 Plane Surface f 1 f 1

10 Fr the full lens: = ( μ 1) ( = f R R ( μ 1) ( μ 1) ; R R = 1...(1) 0 Fr the half lens: ( μ 1) = ( μ 1) = f1 R R f = 0cm. 1 = 1... frm(1) 0 Questin 6 (a) An illuminated pint O kept 0cm frm a thin cnvex lens L 1 f fcal length 15 cm as shwn belw. A thin diverging lens L f fcal length 5cm is kept c-axial with the first lens and 35 cm frm it, as shwn in Figure 10. [4] L 1 L O Object 0cm 35cm Figure 10 Cmmn Principal axis Find the psitin f the final image frmed by this cmbinatin f lenses. The real image frmed by L 1 at v: u= 0cm. f = 15cm fu 0 15 = v = = = 8.6cm f u v u f 0 15 Thus the bject distance fr L =35cm 8.6cm= 6.4 cm. Let the distance f the final image be v. u = 6.4cm. f = 5cm fu = v = = = 1.8cm f u v u f Thus the final image is frmed at a distance f 1.8cm frm L twards left. (b) (i) What are cherent surces? [] (ii) In Yung s duble slit experiment, what is path difference between the tw light waves frming 5 th bright band (fringe)) n the screen? (i) Cherent light surces have the same frequency and a cnstant phase Difference (r same phase) between them always. (ii) Path difference = 5λ

11 (Path difference: xd /D = nλ, where n is the rder f the bright band.) (c) State ne similarity and ne difference between interference f light and diffractin f light. Bth interference and diffractin phenmena prduces alternate dark and bright bands. Interference is frmed by tw cherent surces f light, where as nly ne surce f light frms diffractin under suitable cnditins. [] Questin 7 (a) A ray EF f mnchrmatic light is incident n the refracting surface AB f a regular glass prism (refractive index=1.5) at an angle f incidence f i =55 (Figure 11). If it emerges thrugh the adjacent face AC, calculate the angle f emergence e. A F i E D B Figure 11 G e C Sini Sini Sin55 1 Sin55 μ = Sin( DFG) = = DFG = Sin ( Sin( DFG) μ In the triangle EDG: Angle D = (Angle A =60 fr a regular prism). ) = 33 DGF = {180 Sin e = 1.5 Sin( DGF) e = Sin 1 (10 0 (0.68) = )} = 7 Sin e = 1.5 Sin7 = 0.68 (b) (i) In case f plarized light, what is meant by the plane f [ plarisatin? (ii) Find refractive index f glass if its plarizing angle is 60. (i) The plane n which the electric vectr lies is called plane f plarisatin. (ii) μ = tan i p = tan 60 = (c) (i) Explain the statement: Angular magnificatin f a cmpund [3 micrscpe in nrmal use is 30. (ii) State hw the reslving pwer f an astrnmical telescpe can be increased. (i) It means that if the angle subtended at the unaided eye is 1 degree, then the angle subtended by the image thrugh the cmpund micrscpe is 30 degrees.( It means that the micrscpe can magnify an bject 30 times.)

12 (ii) The reslving pwer can be increased by 1. Increasing the diameter f the bjective lens.. By decreasing the wavelength used fr bservatin. SECTION C Answer any tw questins Questin 8 (a) In Millikan s il drp experiment, the tw plates are cm apart. When a ptential difference f 355 V is applied between them, an il drp f radius 1μm is fund t remain suspended. Calculate the number f excess electrns n the drp. [Density f il = 900kg/m 3. Density f air may be ignred] When the drp is suspended: The weight f the drp = upward frce due t electric field n the charges V 4πr ρgl π r ρg = ne n = = = l 3Ve The number f excess electrns n the drp = (b) Figure1 belw shws a simple X ray tube. P1 and P are pwer supplies which generate 6V and 40,000 V respectively. Shw hw yu will cnnect these pwer supplies t the X ray tube s that it stars prducing X rays. F X ray tube Vacuum A T 6V P 1 Pwer supplies P [] Figure 1 40,000V 6V P 1 F X ray tube Vacuum 40,000V (N explanatin is required. Simply cpy the given diagram in the questin and make the cnnectins as required.) (c) (i) Write a balanced equatin shwing nuclear fissin f Uranium ( 35 9 U ) nucleus. A T P

13 (ii) In a nuclear reactr, what is the functin f: (1) Cadmium rds? () Graphite rds? U n 9U 56 Ba 36Kr n Q( energy 35 (i) ( ( ) ) (ii)(1)cadmium is called a cntrl rd which cntrls the fissin rate in the reactr. () Graphite rd is a mderatr which slws dwn fast neutrns t thermal nuetrns in the reactr. Questin 9 (a) λt Starting with N = N 0 e, btain a relatin between disintegratin cnstant λ f a radiactive element and its half life (T). Varius terms have their usual meaning. λt N N = N e...(1) When t = T then, N = N λt λt = N e e = ln λt ln e = ln λ = = T T (b) On an energy level diagram f hydrgen, shw by a dwnward r an upward arrw, a transitin which results in: (i) emissin line f Balmer series. (ii) emissin line f Lyman series. (iii) absrptin line f Lyman series. [] n=4 n=3 n= Emissin line f Lyman Series Absrptin line f Lyman Series Emissin line f Balmer Series n=1 (c) Calculate: (i) 4 Mass defect f Helium ( He) (ii) nucleus and Its binding energy in MeV. Mass f a prtn = u Mass f a neutrn = u

14 Questin 10 4 Mass f He nucleus = u (i) Δm = {x x } = u (ii) B.E = Δm * 931 =8.8MeV (a) Draw a labelled diagram f a cmmn emitter amplifier. What is the phase angle between the input and utput vltages? 3] Ic Ib b e c R Output ac amplified Ac input Ie Reverse bias Frward bias The phase angle between the input and utput is 180. (b) Threshld wavelength f a certain metal is 79 nm. What is the maximum kinetic energy f pht-electrns emitted by this metal if it is expsed t ultravilet light f wavelength 396 nm? hc hc hc hc 1 1 = K. Emax K. Emax = = hc{ } λ λ λ λ λ λ K. Emax = ( ) 10 = J (c) The fllwing cmbinatin f gates acts as a lgic gate. With the help f a truth table, find ut which lgic gate the cmbinatin represents. [] A B The truth table is given belw: Y 1 Y

15 A B Y 1 Y It is an OR gate. Useful Cnstants and Relatins: 1 Sped f light in vacuum (c) = 3.0 x 10 8 ms -1 Charge f a prtn (e) = 1.6 x C 3 Plank s cnstant (h) = 6.6 x Js 4 Acceleratin due t gravity (g) = 10ms - 5 Cnstant f prprtinality fr Culmb s Law 1 = 9 x10 9 mf -1 4πε 6 Cnstant f prprtinality fr Bit-Savart Law μ 4π = 10-7 Hm -1 7 Electrn vlt 1 ev = 1.6 x10-19 J 8 Unified Atmic Mass Unit (1 u) = 931 MeV