PHYSICS CLASS - XII CBSE BOARD PAPER SOLUTION (SET 2) λ = 4 10 m

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1 () PHYSICS CLSS - XII CBSE BD PPER SOLUTION (SET ). Sl. Diamagnetic materials have a tendency t repel magnetic field lines while paramagnetic materials attract magnetic field lines.. Sl. Let e = emf f each cell r = internal resistances f each all ccrding t graph, 3e = 6V and e r ( ) 3 3 = s, e = V and r = Ω 3. Sl. t the time f sunset and sunrise, sunlight travels mre distance thrugh atmsphere. S frm sunlight, blue light gets scattered, at the time f sunset and sunrise. Hence due t absence f blue clr in direct sunlight, sunlight appears red. 4. Sl. Since the charge is taken alng equiptential surface, the wrk dne by electric field (r external agent) will be zer. 5. Sl. basic cmmunicatin system cnsist f infrmatin surce, (ii) transmitter, (iii) receiver, (iv) link between transmitter and receive i.e. the cmmunicatin channel. 6. Sl. Frm Bhr's Pastulates, and S, π r = nλ fr th n rbit. r =.59n fr ( ) ( ) π.59 λ = th n rbit in Hydrgen. = Sl. (a) Let the current cnstituted be i, as shwn. Then using Kirchff's lp law, i V 4 Ω Ω V ( i) ( i) 4 + = i = Ptential difference acrss resistr = 4 i = 8V Ptential difference acrss cell terminals = = = 8V ( ) Hence bth readings are cming ut t be same. (b) T recrd vltage between tw given pints & B, ptential difference between and B shuld be same as that f vltmeter. T recrd current at pint P in wire, the current at P must be same as that f current in ammeter. Hence vltmeter is cnnected in parallel and ammeter is cnnected in series. 8. Sl. Inisatin Energy: The minimum energy required an electrn frm atm t infinity. We knw that Energy f electrn in n th shell is directly prprtinal t mass f electrn. S under given cnditin, inizatin energy will becme times. Here it is assumed that even by increasing mass f electrn, mtin f nucleus is negligible. Shrtest wavelength f spectral lines f Balmer series crrespnds t n = t n =. 7 S, = m λ 7 λ = 4 m 9. Sl. The rati f amplitude f mdulating signal t the amplitude f carrier wave is knwn as mdulatin index. m = c CTJEE Raman Niwas (Near kashvani), Mehmrganj, Varanasi. Ph.: (54) website: catjee.in

2 () The value f mdulatin index () is kept lw i.e. t avid distrtin. The band pass filter rejects lw and high frequencies, and allws a particular band f frequencies t pass.. Sl. On first surface B, n bending will ccur (as the ray is nrmal t B) On surface C, snell's law gives, (b) 6 z y C B + x + v 5 β + 6 C C + + v x =, y = 5, z = (ii) β C C Ne He a =, b =, c = 4 3. Sl. The required graph is as shwn : V S B C (Stpping Ptential) v B v.5sin 6 = sin ( r ) sin ( r ) =.3 which is nt pssible. S ray will underg TIR n surface C and will emerge ut frm surface BC.. Sl. If τ represents mean relaxatin time. ee acceleratin f element =. me where e = electrnic charge. E = External Electric field. me = Mass f electrn. Final velcity achieved This final velcity averaged ver time = ee me τ = ee me τ ee vd = me τ (ii) The drift velcity will decrease n increase in temperature as increase in temperature will decrease mean relaxatin time.. Sl. (a) β + emissin reactin. + + P n + e + v + Q where P + = prtn n = neutrn e + = psitive v = neutrin Q = Energy Z X Z Y + β + + v v (frequency) The stpping ptential depends n Kinetic Energy f mst energetic electrn. If incident light has mre frequency, such Kinetic energy will als be mre. T stp mst energetic electrn, stpping ptential required will als be mre. Hence stpping ptential will be mre fr light with frequency v. (ii) N, the slpe is independent f material used. This is evident frm fllwing relatin between V S and v. ev = hv φ V S S = h φ v e e [φ = Wrk functin] S slpe is a fundamental cnstant h e. 4. Sl. If a plane plarized light is incident n a plarid such that plane f plarizatin makes angle θ with transmissin axis f plarid, then intensity f transmitted light becmes cs θ times incident light's intensity. This is Malus Law. (ii) The required graph is: Intensity I O π π θ 3π (iii) Using Brewster's law = tanθ p = tan 6 S, = 3 CTJEE Raman Niwas (Near kashvani), Mehmrganj, Varanasi. Ph.: (54) website: catjee.in

3 (3) 5. Sl. Equiptential Surface : It is a surface such that all pints n the surface has same electrstatic ptential. Fr pint charge : spherical equiptential surfaces. Idl Idl db = = 4π r 4π ( a + x ) The directin f db is perpendicular t r in the plane paper i.e., alng PQ. Magnetic field due t all the current elements will add up t give zer magnetic field alng y-axis. S the resultant magnetic field is: (ii) Z Plane equiptential surface y (iii) N. Because if E has a cmpnent alng equiptential surface, wrk will be dne by electric field fr mving a charge alng the surface. S electrstatic ptential will change alng the surface, which is nt pssible. 6. Sl. Magnetic field at a pint n the axis f a circular cil carrying current: Cnsider a circular cil f radius 'a' with centre 'O', carrying current I. Suppse P is any pint n the axis f the circular cil at a distance x frm the centre, such that OP = x Cnsider tw small elements f length dl at C and D at diametrically ppsite current element f the cil PC = PD = r = a x x Which is the expressin fr magnetic field at axis f current carrying lp. Magnetic field lines due t a current carrying circular lp is as shwn: 7. Sl. Wavefrnt : If a wave is prpagating thrugh a medium then a surface in the medium such that wave disturbance at every pint n the surface is in same phase, is called a wavefrnt. Huygen's principal : This is the principle t find wavefrnt at t=dt if we knw wavefrnt at t =. We draw small spheres f radius cdt centered at every pint f wavefrnt at t =. The cmmn tangent t these small spheres represent wavefrnt at t = dt. Wavefrnt while refractin frm lens : F ccrding t Bit Savart's law, the magnitude f magnetic field at P due t current element dl at C is Idl sinθ db = 4π r dl and r are perpendicular θ = 9 When wavefrnt f incident light reaches n a surface, the pints n the surface behave as secndary surces. These secndary surces prduce reflected and refracted waves. Since the CTJEE Raman Niwas (Near kashvani), Mehmrganj, Varanasi. Ph.: (54) website: catjee.in

4 (4) reflected and refracted rays riginate frm same secndary surce, their frequencies are same. (ii) s energy f the phtn depends n its frequency (nt n wavelength) which remains same hence energy f the phtn des nt change. (iii) Intensity in phtn picture f light is determined by prbability f finding a phtn at a pint. Higher the prbability f finding a phtn, higher is the intensity. 8. Sl. VC ic = = = 3 R C 3 ic ic (ii) Nw β =, ib = = = i β (iii) Vi = ib Rb b 5 3 = 3 5 = vlt 9. Sl. (a) Micrwaves are used in radar system fr air craft navigatin. (b) X-rays are prduced by bmbarding a metal target by high speed electrns. (ii) t the time f charging r discharging, charge flws n either plates f the capacitr with a finite rate. fter steady state is achieved the current becmes zer and deflectin in galvanmeter becmes zer. The current as a functin f time (after beginning f charging r discharging prcess) is: (ii) The electrmagnetic waves f frequencies greater than 4 MHz penetrate the insphere and escape. This is the reasn that there is an upper limit t frequency f waves used in this mde.. Sl. Frmatin f Ptential barrier and depletin regin in a p-n junctin : In a p-n junctin dide, due higher cncentratin f hles n P-side and electrns n n-side, the diffusin f hles twards h-side and electrns twards p-side takes place. Hence a regin f unneutralized negative in n p-side and psitive ine n n-side is frmed near the junctin which is depleted f mbile charges. This regin in called depletin regin. In the depletin regin, a ptential difference V B is created is called ptential barrier as it creates an electric field which ppses the further diffusin f electrns and hles. (ii) Wrking f Half Wave Rectifier: Where is maximum current and represents time cnstant f the circuit.. Sl. Sky wave prpagatin in used by shrtwave brad cast services having frequency range frm a few MHz upt 3 MHz. The sky waves reach the receiver after reflectin frm the insphere. In a single reflectin frm the insphere, the radi-waves cver a distance nt less than 4km. That is why, very lng distance rund the earth cmmunicatin is pssible with the help f this mde. The input given t the rectifier will have bth psitive and negative cycles. The half rectifier will allw nly the psitive half cycles and mit the negative half cycles. S first we will see hw half wave rectifier wrks in the psitive half cycles. Psitive Half Cycle: The dide will allw current flwing in clck wise directin frm ande t cathde in the frward bias (dide cnductin will take place in frward bias) which will generate nly the psitive half cycle f the C. Negative Half Cycle: In the negative half cycle the current will flw in the anti-clckwise directin and the dide will g in t the reverse bias. In the reverse bias the dide will nt cnduct s, n current in flwn CTJEE Raman Niwas (Near kashvani), Mehmrganj, Varanasi. Ph.: (54) website: catjee.in

5 (5) frm ande t cathde, and we cannt get any pwer at the lad resistance. (ii) Withut hesitatin he meet his teacher t slve his dubts. Values displayed by teacher. He has knwledge f physics principles. (ii) He is interested in sharing his knwledge.. Sl. Let the ac surce has emf given by = sin ( ωt + φ ) We knw that fr ideal inductr, current in it lags by π. S, sin π i = i ωt + φ where i = = X ωl L nd Pwer = P = i sin( ωt + φ ) cs( ωt + φ ) i = + sin ( ωt φ ) T T verage Pwer = sin ( ωt + φ ) i ω i π dt π = sin + ω Since the value f sin ( ωt φ ) ( ( ωt φ )) dt + cmpletes tw cycles ω frm t, the abve integratin must be zer. π Hence, verage supplied by inductr = (ii) Pwer dissipated by bulb = csφ = R Z Z where Z = Impedance f circuit csφ = Pwer factr. If irn rd is inserted in slenid it's inductance will increase. Hence Z will increase. S, pwer dissipated by resistr will decrease. 3. Sl. (a) It is based n principle f electrmagnetic inductin. When a cil is rtated in magnetic field, its magnetic flux changes. ccrding t Faradays law f electrmagnetic inductin, an is induced in it. This can be used t drive an electrical device. (b) Values displayed by ram. Curius t learn 4. Sl. Lens maker's frmula : The image f pint bject O is frmed in tw steps. The first reflecting surface 'C ' frms the image I f the bject O. The image I act as a virtual bject fr frmatin f image I by the secnd surface 'C '. Fr the first surface BC 'C ', n n π n + = OB BI BC n n n n... + = u v R Fr the secnd surface DC 'C '. n n n n + = DI DI DC Fr a thin lens, BI = DI, n n n n + = (ii) BI DI DC On adding equatins and (ii), we get n n + = ( n n ) + OB DI BC DC If the bject is at infinity, then OB = and I is at the fcus f the lens DI = f (fcal length f cnvex lens) n n + = ( n n ) + f BC DC By the sign cnventin, BC = + R and DC = R n = ( n n ) f R R n n r = f n n R R r = ( n ) f R R = ( n ) f R R CTJEE Raman Niwas (Near kashvani), Mehmrganj, Varanasi. Ph.: (54) website: catjee.in

6 (6) n where [ n = = n, is the refractive index f n material f the lens]. (ii) =.5, R = cm, u = cm v u R.5.5 v'.5.5 v' O C I =, = v' 4 v'.5 3, ' v ' = v = cm t secnd surface v u R.5.5 v , = + v 6 4 v = =, v = cm v I' Magnifying pwer f such telescpe is defined as, the rati f the angle subtended at the eye by the image t the angle subtended at the eye by the bject. (b) We knw that magnifying pwer depends n f. fe S bjective must be f high fcal length and eyepiece shuld be f lw fcal length. Hence, lens t be used fr eyepiece shuld be f pwer D nd fr bjective pwer used shuld be.5d (ii)perture f bjective is preferred t be large fr better reslutin f final image t be btained. If the aperture will be small the final image will be blurred due t diffractin phenmena. 5. Sl. Electric Field f a Unifrmly Charged Plane Cnsider an infinite plane which carries the unifrm charge per unit area. Suppse that the plane cincides with the plane (i.e., the plane which satisfies ). By symmetry, we expect the electric field n either side f the plane t be a functin f nly, t be directed nrmal t the plane, and t pint away frm/twards the plane depending n whether is psitive/negative. Diagram shwing real image frmed by an astrnmical telescpe in nrmal adjustment psitin. (ii)perture f bjective is preferred t be large fr a better reslutin f the final image t be btained. If the aperture will be small the final image will be blurred due t diffractin phenmena. Here, Figure: The electric field generated by a unifrmly charged plane. We draw a cylindrical Gaussian surface, whse axis is nrmal t the plane, and which is cut in half by the plane. Let the cylinder run frm frm t, and let its crss-sectinal area be. ccrding t Gauss' law, E( a) = σ CTJEE Raman Niwas (Near kashvani), Mehmrganj, Varanasi. Ph.: (54) website: catjee.in

7 (7) where is the electric field strength at. Here, the left-hand side represents the electric flux ut f the surface. It fllws that E σ Which is the expressin fr electric field due t large plane sheet f unifrm charge. (ii) Let the capacitance be C & C. C Equivalence f series cmbinatin = 3 Equivalance f parallel cmbinatin = 3C. If the ptential differences applied are V & V. C V 3 C V 3 9 V V = V 3 ( )( ) = V = (b) Ptential Difference σ = V = d. (c) If Q is magnitude f charge n each plate, Q d = V Capacitance (ii) Q = = V d R Ptential f the shell π σ σ 4 r V = k = r r V r Therefre sphere B is at higher ptential cmpared t, hence charge will flw frm B t. R 6. Sl. Diagram f Step-Dwn Transfrmer: B +σ σ P Q (a) Fr a pint between the plates (P) E P d σ σ σ = + = Because electric field due t bth plates are in same directin. Fr a pint utside plates (Q), the individual electric fields will be ppsite t each ther. E Q σ σ = t P electric field will be frm psitive t negative plate. Principle f wrking: it is based n the principle f Faraday s law f Electrmagnetic Inductin. Changing flux f the primary winding causes t get induced in secndary winding. (ii) (iii) Frm cnservatin f energy, S, Rati f primary t secndary current (iv) Here utput pwer = 55 W It must be same as input pwer. Hence t find input current, CTJEE Raman Niwas (Near kashvani), Mehmrganj, Varanasi. Ph.: (54) website: catjee.in