CBSE Board Class XII Physics Set 1 Board Paper 2008 (Solution)

Transcription

1 CBSE Bard Class XII Physics Set 1 Bard Paper 2008 (Slutin) 1. The frce is given by F qv B This frce is at right angles t &. 2. Micrwaves. It is used in radar & cmmunicatin purpses. 3. Or As m e e m S, K e K 4. The lens will nt be visible if n refractin ccurs at the liquid glass interface. This means that the incident ray shuld g thrugh the glass withut any deviatin. Fr this cnditin t be fulfilled, the refractive index f the liquid must be equal t The 500 C charge is at the same distance frm all the crners f the square. The ppsite crners, say A and C, will have the same ptential VA = VC. Wrk dne in mving a charge q between pints A and C is given as: W = q(vc VA) = q 0 = 0 Hence, n wrk is dne in mving the charge between tw diagnally ppsite pints n the square. 6. Because heavy water mlecules d nt absrb fast neutrns but simply them. 7. D d When light enters a denser medium, its wavelength decreases by a factr 1.3 and hence the fringe width als decreases by a factr 1.3. The fringe width will decrease. 1

2 8. When n current is drawn frm the cell, V becmes equal t Ettal. E = V + ir i = 0, E = V Frm graph, when i = 0, V = 6V S, emf f each cell = Ettal /3 = 6/3 = 2 V 9. Let P be an axial pint at distance r frm the centre f the diple. Electric ptential at pint P is given as V V V 1 2 V and V are the ptentials at pint P due t charges +q and -q respectively q q q 2a V= 4 r a r a 4 r a 1 p r a Magnetic susceptibility: It is the rati f the intensity f magnetisatin (I) induced in the material t the magnetizatin frce (H) applied n it. Magnetic susceptibility is represented as: I m H Elements having psitive susceptibility: aluminum, sdium. Elements having negative susceptibility: antimny, cpper. The minus sign (-ve) signifies that the magnetic field is weakened in the presence f the material. 11. (i) Standard equatin f magnetic field is B B sin t kx T y Cmparing this equatin with the given equatin, we get z (ii) E B c V / m Accrding t right hand system f E, B, K, the electric field scillates alng negative z-axis, s equatin is 9 11 E sin(2 10 t 300x) V / m 2

3 12. We knw average pwer in an capacitive circuit is given by OR Let an inductr L and resistr R be cnnected in series t a surce f alternating emf as shwn in the fllwing figure. The maximum vltage acrss resistr R is given as: V R IR In a resistr circuit, vltage alng OX. The maximum vltage acrss inductr L is given as: V I X L L V R is in phase with current I. It is represented by OA In an inductive circuit, vltage V L acrss the inductr leads the current I by 90. It is represented by OB alng OY. The vectr sum f ptentials V R and V L is the vltage phasr (E ). It is represented by OK. 2 2 E OK OA OB 2 OA OB cs OA OB V V I R I X 2 2 L E I R X R L L The impendence f the circuit is given by: 2 2 E I R XL Z= I I 2 2 L Z R X 3

4 13. The reactin can be written as Ne Na Q where, where, Q = Kinetic energy f the daughter nucleus Ignring the rest mass f the anti-neutrin and the mass f the electrn, the maximum kinetic energy f the emitted electrn is given by, Na 14. An intrinsic semicnductr is a pure semicnductr where as a p-type semicnductr is a semicnductr dped with trivalent impurity atms like brn r gallium. An intrinsic semicnductr has same number f hles and electrns while in a p-type semicnductr the number f hles is greater than the number f electrns. Each hle is assciated with a nearby negative-charged dpant in, and thus semicnductr remains electrically neutral as a whle. 15. Reflecting type telescpe: Advantages f a reflecting telescpe ver a refracting telescpe: 1. Image frmed by a reflecting telescpe is brighter than that frmed by a refracting telescpe. 2. As the bjective in a reflecting telescpe is a mirrr, the image frmed by this telescpe des nt underg any chrmatic aberratin Given: i A 4 Fr triangular prism, A=60 3 i Using the prism frmula, we can btain the relatin fr the refractive index f the material f the prism as: 4

5 17. Fr NAND gates, the Blean expressin is written as: Y A.B A B Fr the given wave frm, Thus, the wavefrm fr utput Y can be shwn as 5

6 18. The maximum distance can be given as, m 46.51km 19. Wavefrnt: It is defined as the lcus f pints having the same phase f scillatin. Let a plane wavefrnt AB be incident n a refracting plane surface PQ separating a rarer medium f refractive index frm a denser medium f refractive index ( > ). During the time the disturbance frm B reaches B, the disturbance frm A must have travelled a distance AA = c2t, where c2 is the velcity f light in the rarer medium. With A as center and AA as radius, draw a sphere. Draw a sphere. Draw a tangent, t the sphere, frm pint B. B A will be the refracted wavefrnt. Let us nw cnfirm the validity f the refracted wavefrnt. Fr A EB t be the true refracted wavefrnt, the fllwing shuld be satisfied. 6

7 20. The given situatin can be shwn as: Let be the angle traced by the free end f the rd in time t. The area swept-ut by the rd in time t is given as: 7

8 Since the angle between the area vectr and the magnetic field vectr is zer, the magnetic flux linked t this area is given as: Accrding t Faraday s laws f electrmagnetic inductin, induced emf (e) is given as Hence, the current induced in the rd is given as: 21. (i) Dide used is Zener dide. (ii) Circuit diagram fr Zener dide as a vltage regulatr: (iii) Zener dide as a vltage regulatr: The Zener dide is cnnected in parallel t external lad resistance RL. Let the unregulated d.c. input vltage Vi is applied t the Zener dide, whse breakdwn vltage is VZ. if the applied vltage Vi > VZ, the Zener dide is in breakdwn cnditin. As a result, it easily cnducts current thrugh it. Thus depending upn the input vltage, the current in the circuit r thrugh the Zener dide may change but vltage acrss it remains unaffected by change in lad resistance. Hence the utput vltage acrss the Zener dide is a regulated vltage. 8

9 22. Here L = 200 mh, C = 500 F, R = 10, E = 100 V Pwer factr f the circuit = 1 S Z = R i.e. XL = XC (ii) Current amplitude Irms = ( Z = R) At resnance frequency, Z=R Current amplitude, I (iii) Q-factr V V Z R A 23. Cnsider a cnductr f length l and area f crss-sectin A, having n electrns per unit vlume, as shwn in the fllwing figure. 9

10 Vlume f the cnductr = Al Ttal number f electrns in the cnductr = Vlume Electrn density = Aln Since e is the charge f an electrn, the ttal charge cntained in the cnductr: Q = Alen Let a ptential difference V be applied acrss the cnductr. The resulting electric field in the cnductr is given by: E=V/I Hence, free electrns begin t drift in a directin ppsite t that f the electric field E. The time taken by the free electrns t crss-ver the cnductr is given as: l t v d vd is the drift velcity f free electrns. Current flwing thrugh the cnductr is given by: OR (i) The fllwing figure shws a simple circuit diagram that cnsists f n identical cells cnnected in series, each having an emf E and internal resistance r. The cells are charged by a dc surce f emf E using a lad f resistance R. Equivalent emf f n cells cnnected in series is given as: E = E + E + up t n terms = ne Equivalent internal resistance f n cells cnnected in series is given as: r = r + r + up t n terms = nr Ttal resistance f the resistance R is given by: R = r + R = nr + R 10

11 (ii) (a) Current flwing thrugh resistr R is given by: (b) Ptential difference acrss the cmbinatin f cells is given as: 24. (i) Here E1 = 1.5 V, l1 = 60 cm, l2 = 80 cm & E2 =? By the principle f ptentimeter (ii) The circuit will nt wrk because there will be little drp f ptential acrss the ptentimeter wire in cmparisn t the emf f the unknwn cell. Hence, the balance pint will nt be btained n the ptentimeter wire. Thus, fr the functining f the ptentimeter, the emf f the driver cell shuld be greater than the emf f the unknwn cell. (iii) The high resistance des nt affect the balance pint. Because balancing pint depends nly n emf f ptentimeter. 25. Given wrk functin. Using Einstein s phtelectric equatin, we have 11

12 26. The transitin will crrespnd t energy levels frm -13.6eV t -1.5eV i.e transitin D. Energy E= E2-E1= (-13.6) = 12.1eV. 27. The plt f variatin f amplitude w fr an amplitude mdulated wave is given belw: Mdulatin index:- It is defined as the rati f amplitude f mdulating signal t the amplitude f carrier wave. Mathematically it can be written as The quality f the transmitted signal is determined by the index f mdulatin. The variatin in the carrier amplitude Ac is small fr a small mdulatin index and vice-versa. Hence, fr effective mdulatin, the value f shuld be checked and it shuld be ensured that the mdulatin index is never greater than unity. 28. Cycltrn: Cycltrn is a device by which the psitively charged particles like prtns, deutrns, etc. can be accelerated. Principle: Cycltrn wrks n the principle that a psitively charged particle can be accelerated by making it t crss the same electric field repeatedly with the help f a magnetic field. 12

13 Cnstructin: The cnstructin f a simple cycltrn is shwn in figure abve. It cnsists f tw-semi-cylindrical bxes D1 and D2, which are called dees. They are enclsed in an evacuated chamber. The chamber is kept between the ples f a pwerful magnet s that unifrm magnetic field acts perpendicular t the plane f the dees. An alternating vltage is applied in the gap between the tw dees by the help f a high frequency scillatr. The electric field is zer inside the dees Wrking and thery: At a certain instant, let D1 be psitive and D2 be negative. A prtn frm an in surce will be accelerated twards D2, it describes a semicircular path with a cnstant speed and is acted upn nly by the magnetic field. The radius f the circular path is given by,. Frm the abve equatin we get, (I) The perid f revlutin is given by, (Frm I) The frequency f revlutin is given by, Frm the abve equatin it fllws that frequency f is independent f bth v and r and is called cycltrn frequency. Als if we make the frequency f applied a.c. equal t f, then every time the prtn reaches the gap between the dees, the directin f electric field is reversed and prtn receives a push and finally it gains very high kinetic energy. The prtn fllws a spiral path and finally gets directed twards the target and cmes ut frm it. OR (a) Cnsider a current element AB f a thin curved cnductr XY thrugh which a cnstant current I is maintained. Let db be the magnitude f the magnetic field at P due t this current element f length dl. Accrding t Bit-Savart s law which was frmulated empirically in 1820, (i) db I (ii) db dl (iii) db sin, 13

14 where is the angle between and. is the psitin vectr f the bservatin pint P with respect t the center O f the current element. The directin f 1 (iv) db. r 2 is the directin f flw f current. (The field falls ff inversely with the square f the diatance between the surce f the field and the pint at which the field is t be measured. It is als called inverse square law.) where r is the distance f the bservatin pint P frm the mid-pint O f the current element Cmbining all the fur factrs, we get Where k is a cnstant f prprtinality whse value depends n the medium between the bservatin pint and the current element. (b) Magnetic field at the center f a circular current-carrying cil- Cnsider a circular cil f radius r thrugh which current I is flwing. Let AB be an infinitesimally small element f length dl. Accrding t Bit-Savart s law, the magnetic field db at the center P f the lp is given by Where is the angle between and. In the case f a circular lp, = 90. The net magnetic field at P is given by 14

15 29. Plane plarized light: In plane plarized light vibratins f electric vectrs are taking place in a particular plane nly. As the new Plarid is placed between tw crssed Plarid bisecting the angle between them, it means that it has been placed at an angle 45 0 as cmpared t first Plarid. If intensity f light cming ut f 1 st Plarid be I, then intensity f light transmitted thrugh secnd Plarid is I = I Cs 2 = I Cs 2 45 I and the light transmitted thrugh the third Plarid I = = I Cs 2 = Cs I 1 I (b) N, cnsider light which is made up f electric vectrs Ex, Ey with a 90 0 phase difference but equal amplitudes. The tip f electric vectr executes unifrm circular mtin at a frequency f light itself. This kind f light said t be circularly plarized. When such light is passes thrugh a Plarid, which is rtated, the transmitted average intensity remains cnstant. 15

16 (a) By Snell s law OR Fr i = ic, r = 90 0 n sin ic = 2 n1 (b) Optical fibres are used in telephne and ther transmitting cables. Light incident n ne end f ptical fiber at a small angle passes inside and underges repeated ttal internal reflectins inside the fibre. It finally cmes ut f the ther end, even if the fibre is bent r twisted in any frm. There is n lss f light thrugh the sides f the fibre. The nly cnditin is that angle f incidence f light must be greater than the critical angle fr the fibre material. 30. (a) Electric field intensity at any pint utside a unifrmly charged spherical shell: Cnsider a thin spherical shell f radius R and with centre O. Let charge + q be unifrmly distributed ver the surface f the shell. Let P be any pint n the Gaussian sphere S1 with centre O and radius r, as shwn in the fllwing figure. Accrding t Gauss s law, we can write the flux thrugh ds as: 16

17 At any pint n the surface f the shell, r = R Field lines when charge density f the sphere is psitive Field lines when charge density f the sphere is negative 17

18 (b) Diameter f the sphere = 2.5 m 2.5 Radius f the sphere, R 1.25m Charge density, 100 C /m C /m (i) Ttal charge, q 4R C (ii) Ttal electric flux, 3 q Nm C E OR (a) The figure given belw shws an electric diple f charges +q and q which are separated by distance 2a. Expressin fr the trque: The abve arrangement frms a cuple. The cuple exerts a trque which is given by, =Frce x Perpendicular distance between the tw frces =qe x 2a.sin =pe.sin, where p=qx2a=diple mment. Since the directin f trque is perpendicular t equatin as, we can rewrite the abve = (b) The wrk dne will be equal t the ptential energy f the system 18