PHYSICS (Theory) Past Paper PDF

Set-1 Series RPQS1/1 àíZ-nÌ H$moS> Q.P. Code 55/1/1 amob Z§....

Set-1 Series RPQS1/1 àíZ-nÌ H$moS> Q.P. Code 55/1/1 amob Z§. narjmWu àíZ-nÌ H$moS> >H$mo CÎma-nwpñVH$m Ho$ Roll No. _wI-n¥ð >na Adí` {bIo§ & Candidates must write the Q.P. Code on the title page of the answer-book. ^m¡{VH$ {dkmZ (g¡ÕmpÝVH$) PHYSICS (Theory) :3 : 70 Time allowed : 3 hours Maximum Marks : 70 ZmoQ> NOTE (I) H¥$n`m Om±M H$a b| {H$ Bg àíZ-nÌ _o§ _w{ÐV (I) Please check that this question paper contains 27 printed pages. n¥ð> 27 h¢ & (II) H¥$n`m Om±M H$a b| {H$ Bg àíZ-nÌ _| (II) Please check that this question paper contains 33 questions. >33 àíZ h¢ & (III) àíZ-nÌ _| Xm{hZo hmW H$s Amoa {XE JE (III) Q.P. Code given on the right hand side of the question paper should be written on àíZ-nÌ H$moS >H$mo narjmWu CÎma-nwpñVH$m Ho$ the title page of the answer-book by the _wI-n¥ð> na {bI| & candidate. (IV) H¥$n`m àíZ H$m CÎma {bIZm ewê$ H$aZo go (IV) Please write down the serial number of the question in the answer-book before nhbo, CÎma-nwpñVH$m _| àíZ H$m H«$_m§H$ attempting it. Adí` {bI| & (V) Bg àíZ-nÌ 15 {_ZQ (V) 15 minute time has been allotted to read this question paper. The question paper >H$m g_` {X`m J`m h¡ & àíZ-nÌ H$m will be distributed at 10.15 a.m. From 10.15 ~Oo {H$`m OmEJm & 10.15 a.m. to 10.30 a.m., the students will 10.15 ~Oo go 10.30 ~Oo VH$ N>mÌ Ho$db read the question paper only and will not àíZ- write any answer on the answer-book during this period. do CÎma-nwpñVH$m na H$moB© CÎma Zht {bI|Jo & 11-55/1/1 1 P.T.O.  : : (i) 33 (ii) (iii) 1 16 1 (iv) 17 21 2 (v) 22 28 3 (vi) 29 30 4 (vii) 31 33 5 (viii) (ix) (x) : c=3 108 m/s h = 6.63 10 34 Js e = 1.6 10 19 C 0 =4 10 7 T m A 1 0 = 8.854 10 12 C2 N 1 m 2 1 =9 109 N m2 C 2 4 0 (me) = 9.1 10 31 kg Ý`yQ´>m°Z H$m Ðì`_mZ = 1.675 10 27 kg àmoQ>m°Z H$m Ðì`_mZ = 1.673 10 27 kg AmdmoJmÐmo g§»`m = 6.023 1023 à{V J«m_ _mob ~moëQ²>µO_mZ {Z`Vm§H$ = 1.38 10 23 JK 1 11-55/1/1 2 General Instructions : Read the following instructions carefully and follow them : (i) This question paper contains 33 questions. All questions are compulsory. (ii) This question paper is divided into five sections Sections A, B, C, D and E. (iii) In Section A Questions no. 1 to 16 are Multiple Choice type questions. Each question carries 1 mark. (iv) In Section B Questions no. 17 to 21 are Very Short Answer type questions. Each question carries 2 marks. (v) In Section C Questions no. 22 to 28 are Short Answer type questions. Each question carries 3 marks. (vi) In Section D Questions no. 29 and 30 are case study-based questions. Each question carries 4 marks. (vii) In Section E Questions no. 31 to 33 are Long Answer type questions. Each question carries 5 marks. (viii) There is no overall choice given in the question paper. However, an internal choice has been provided in few questions in all the Sections except Section A. (ix) Kindly note that there is a separate question paper for Visually Impaired candidates. (x) Use of calculators is not allowed. You may use the following values of physical constants wherever necessary : c=3 108 m/s h = 6.63 10 34 Js e = 1.6 10 19 C 0 =4 10 7 T m A 1 0 = 8.854 10 12 C2 N 1 m 2 1 =9 109 N m2 C 2 4 0 Mass of electron (me) = 9.1 10 31 kg Mass of neutron = 1.675 10 27 kg Mass of proton = 1.673 10 27 kg 6.023 1023 per gram mole Boltzmann constant = 1.38 10 23 JK 1 11-55/1/1 3 P.T.O.  IÊS> H$ 1. R Ho$ d¥ÎmmH$ma db` Ho$ ê$n Bgo Amdoe KZËd -joÌ H$m (A) (B) eyÝ` (C) (D) 2 0R 4 R 4 0R 0 2. Xg g§Ym[aÌ, 1 F h¡, H$mo 100 V Ho$ òmoV {H$`m J`m h¡ (A) 10 2 J (B) 10 3 J (C) 0·5 10 3 J (D) 5·0 10 2 J 3. JE n[anW na {dMma H$s{OE & BgHo$ {~ÝXþ A Am¡a B Ho$ ~rM {d^dmÝVa h¡ : (A) 6V (B) 8V (C) 9 V (D) 12 V 4. H$moB© nme, {Oggo Ymam I X{jUmdV© àdm{hV hmo ahr h¡, x y , z-Aj Ho$ AZw{Xe Bg nme H$s àd¥{Îm hmoJr : (A) x-Aj Ho$ AZw{Xe J{V H$aZo H$s (B) y-Aj Ho$ AZw{Xe J{V H$aZo H$s (C) {gHw$ Zo H$s (D) 5. H$moB© 10 cm bå~m Vma y-Aj Ho$ AZw{Xe aIm h¡ & Bggo YZmË_H$ y-{Xem _| 1.0 A H$s ^ Ymam àdm{hV hmo ahr h¡ & Bg joÌ _| H$moB© Mwå~H$s` joÌ B = (5 mT) j (8 mT) k^ {dÚ_mZ h¡ & Bg Vma na ~b h¡ : ^ ^ (A) (0·8 mN) i (B) (0·8 mN) i ^ ^ (C) (80 mN) i (D) (80 mN) i 11-55/1/1 4  SECTION A 1. A thin plastic rod is bent into a circular ring of radius R. It is uniformly charged with charge density. The magnitude of the electric field at its centre is : (A) (B) Zero (C) (D) 2 0R 4 R 4 0R 0 2. Ten capacitors, each of capacitance 1 µF, are connected in parallel to a source of 100 V. The total energy stored in the system is equal to : (A) 10 2 J (B) 10 3 J (C) 0·5 10 3 J (D) 5·0 10 2 J 3. Consider the circuit shown in the figure. The potential difference between points A and B is : (A) 6V (B) 8V (C) 9 V (D) 12 V 4. A loop carrying a current I clockwise is placed in x y plane, in a uniform magnetic field directed along z-axis. The tendency of the loop will be to : (A) move along x-axis (B) move along y-axis (C) shrink (D) expand 5. A 10 cm long wire lies along y-axis. It carries a current of 1.0 A in ^ ^ positive y-direction. A magnetic field B = (5 mT) j (8 mT) k exists in the region. The force on the wire is : ^ ^ (A) (0·8 mN) i (B) (0·8 mN) i ^ ^ (C) (80 mN) i (D) (80 mN) i 11-55/1/1 5 P.T.O. 6. à{VamoY G Ho$ {H$gr J¡ëdoZmo_rQ>a H$mo 0 go I A n[aga Ho$ Eo_rQ>a _| n[ad{V©V {H$`m J`m h¡ & `{X Bg J¡ëdoZmo_rQ>a _| àdm{hV Ymam I A H$s 0·1% h¡, Vmo Eo_rQ>a H$m à{VamoY h¡ : G G G G (A) (B) (C) (D) 999 1000 1001 100·1 7. Amd¥{Îm Ho$ {H$gr ac òmoV go g§`mo{OV Ym[aVm C Ho$ {H$gr g§Ym[aÌ H$s à{VKmV X h¡ & `{X g§Ym[aÌ H$s Ym[aVm H$mo Xmo JwZm Am¡a òmoV H$s Amd¥{Îm H$mo VrZ JwZm H$a {X`m OmE, Vmo à{VKmV hmo OmEJr : X 2 3 (A) (B) 6X (C) X (D) X 6 3 2 8. Mma joÌ I, II, III VWm IV _| {dÚwV-joÌ {ZåZ{b{IV ê$n _| d{U©V h¢ : joÌ I : Ex = E0 sin (kz t) joÌ II : Ex = E0 joÌ III : Ex = E0 sin kz joÌ IV : Ex = E0 cos kz {H$g joÌ _| {dñWmnZ Ymam hmoJr ? (A) I (B) IV (C) II (D) III 9. _| ~m_a loUr H$s Xÿgar , g§JV hmoVm h¡ : (A) nf = 2 VWm ni = 3 (B) nf = 3 VWm ni = 4 (C) nf = 2 VWm ni = 4 (D) nf = 2 VWm ni = 10. Ge H$m _mXZ As Ho$ gmW {H$`m J`m h¡ & BgHo$ H$maU : (A) Ge Ho$ OmbH$ H$s g§aMZm {dH¥$V hmo OmVr h¡ & (B) MmbZ Bbo Vr h¡ & (C) hmobm| H$s g§»`m _| d¥{Õ hmoVr h¡ & (D) MmbZ Bbo m h¡ & 11. Xmo nw§O, A Am¡a B {OZHo$ µ\$moQ>m°Zm| H$s D$Om© H«$_e: 3·3 eV Am¡a 11·3 eV h¢, H«$_dma {H$gr YmpËdH$ n¥îR> (H$m`©-\$bZ 2·3 eV) H$mo àXrßV H$aVo h¢ & nw§O A Ho$ H$maU CËg{O©V H$m nw§O B H$s A{YH$V_ Mmb go AZwnmV h¡ : 1 1 (A) 3 (B) 9 (C) (D) 3 9 11-55/1/1 6 6. A galvanometer of resistance G is converted into an ammeter of range 0 to I A. If the current through the galvanometer is 0.1% of I A, the resistance of the ammeter is : G G G G (A) (B) (C) (D) 999 1000 1001 100·1 7. The reactance of a capacitor of capacitance C connected to an ac source of frequency nce of the capacitor is doubled and the frequency of the source is tripled, the reactance will become : X 2 3 (A) (B) 6X (C) X (D) X 6 3 2 8. In the four regions, I, II, III and IV, the electric fields are described as : Region I : Ex = E0 sin (kz t) Region II : Ex = E0 Region III : Ex = E0 sin kz Region IV : Ex = E0 cos kz The displacement current will exist in the region : (A) I (B) IV (C) II (D) III 9. The transition of electron that gives rise to the formation of the second spectral line of the Balmer series in the spectrum of hydrogen atom corresponds to : (A) nf = 2 and ni = 3 (B) nf = 3 and ni = 4 (C) nf = 2 and ni = 4 (D) nf = 2 and ni = 10. Ge is doped with As. Due to doping, (A) the structure of Ge lattice is distorted. (B) the number of conduction electrons increases. (C) the number of holes increases. (D) the number of conduction electrons decreases. 11. Two beams, A and B whose photon energies are 3·3 eV and 11·3 eV respectively, illuminate a metallic surface (work function 2·3 eV) successively. The ratio of maximum speed of electrons emitted due to beam A to that due to beam B is : 1 1 (A) 3 (B) 9 (C) (D) 3 9 11-55/1/1 7 P.T.O. 12. >m°Z Am¡a {H$gr J{V_mZ àmoQ>m°Z go gå~Õ Va§Jm| H$m Va§JX¡¿`© g_mZ h¡ & BgH$m `h A{^àm` h¡ {H$ BZH$m/BZH$s g_mZ h¡ : (A) g§doJ (B) H$moUr` g§doJ (C) Mmb (D) D$Om© 13 16 (A) (R) (A) (R) (A), (B), (C) (D) (A) A{^H$WZ (A) Am¡a H$maU (R) XmoZm| ghr h¢ Am¡a H$maU (R), A{^H$WZ (A) H$s ghr ì¶m»¶m H$aVm h¡ & (B) A{^H$WZ (A) Am¡a H$maU (R) XmoZm| ghr h¢, naÝVw H$maU (R), A{^H$WZ (A) H$s ghr ì¶m»¶m H$aVm h¡ & (C) A{^H$WZ (A) ghr h¡, naÝVw H$maU (R) µJbV h¡ & (D) A{^H$WZ (A) µJbV h¡ VWm H$maU (R) ^r µJbV h¡ & 13. (A) : àH$me-{dÚwV à^md _|, CËg{O©V \$moQ>mo- Amn{VV àH$me H$s Vrd«Vm _| d¥{Õ Ho$ gmW d¥{Õ hmoVr h¡ & (R) : àH$me-{dÚwV Ymam Amn{VV àH$me H$s Va§JX¡¿`© na {Z^©a H$aVr h¡ & 14. (A) : O~ Xmo Hw$ÊS>{b`m| H$mo EH$-Xÿgao Ho$ D$na bnoQ>m OmVm h¡, Vmo CZHo$ ~rM AÝ`moÝ` àoaH$Ëd A{YH$V_ hmoVm h¡ & (R) : O~ Xmo Hw$ÊS>{b`m± EH$- g§~ÕVm A{YH$V_ hmoVr h¡ & 15. (A) : {H$gr ~¡Q>ar go loUr _| g§`mo{OV Am¡a ñdV§ÌVmnyd©H$ {Zb§{~V Xmo b§~o g_mÝVa Vma EH$-Xÿgao go Xÿa hQ>Vo h¢ & (R) : {dnarV {XemAm| _| àdm{hV Ymamdmhr Xmo Vma EH$-Xÿgao H$mo à{VH${f©V H$aVo h¢ & 16. (A) : g_Vb Xn©U Am¡a CÎmb Xn©U {H$gr ^r n[apñW{V _| dmñV{dH$ à{V{~å~ Zht ~Zm gH$Vo h¢ & (R) : H$moB© Am^mgr à{V{~å~ dmñV{dH$ à{V{~å~ ~ZmZo Ho$ {bE {~å~ H$s ^m±{V H$m`© Zht H$a gH$Vm h¡ & 11-55/1/1 8 12. The waves associated with a moving electron and a moving proton have the same wavelength. It implies that they have the same : (A) momentum (B) angular momentum (C) speed (D) energy Questions number 13 to 16 are Assertion (A) and Reason (R) type questions. Two statements are given one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false and Reason (R) is also false. 13. Assertion (A) : In photoelectric effect, the kinetic energy of the emitted photoelectrons increases with increase in the intensity of the incident light. Reason (R) : Photoelectric current depends on the wavelength of the incident light. 14. Assertion (A) : The mutual inductance between two coils is maximum when the coils are wound on each other. Reason (R) : The flux linkage between two coils is maximum when they are wound on each other. 15. Assertion (A) : Two long parallel wires, freely suspended and connected in series to a battery, move apart. Reason (R) : Two wires carrying current in opposite directions repel each other. 16. Assertion (A) : Plane and convex mirrors cannot produce real images under any circumstance. Reason (R) : A virtual image cannot serve as an object to produce a real image. 11-55/1/1 9 P.T.O.  IÊS> I 17. dh Vmn kmV H$s{OE {Og na {H$gr {gëda go ~Zo Vma H$m à{VamoY CgHo$ 20 C na à{VamoY H$m Xmo JwZm hmoJm & 20 C H$mo g§X^© Vmn _mZ br{OE Am¡a 20 C na {gëda H$m à{VamoY H$m Vmn JwUm§H$ = 4·0 10 3 K 1 br{OE & 2 18. (H$) Amd¥{Îm 5·0 1014 Hz H$m EH$dUu àH$me dm`w go AndV©Zm§H$ 1·5 dmbo _mÜ`_ _| J_Z H$aVm h¡ & XmoZm| _mÜ`_m| Ho$ AÝVamn¥îR> na (i) namd{V©V àH$me, VWm (ii) And{V©V àH$me H$s Va§JX¡¿`© kmV H$s{OE & 2 AWdm (I) 16 cm \$moH$g Xÿar H$m H$moB© g_VbmoÎmb b|g AndV©Zm§H$ 1·4 Ho$ nXmW© H$m ~Zm h¡ & b|g Ho$ d{H«$V n¥îR> H$s {ÌÁ`m n[aH${bV H$s{OE & 2 19. H$moB© {~å~ 40 cm dH«$Vm {ÌÁ`m Ho$ {H$gr AdVb Xn©U Ho$ gm_Zo 30 cm Xÿar na pñWV h¡ & (i) ~Zo à{V{~å~ H$s pñW{V VWm (ii) à{V{~å~ H$m AmdY©Z kmV H$s{OE & 2 20. J{VO D$Om© E Ho$ {H$gr Ý`yQ´>m°Z (Ðì`_mZ m) VWm g_mZ D$Om© Ho$ {H$gr µ\$moQ>m°Z na {dMma H$s{OE & _mZ br{OE Ý`yQ´>m°Z H$s Xo ~«m°½br Va§JX¡¿`© n VWm µ\$moQ>m°Z H$s Va§JX¡¿`© p h¡ & n Ho$ {bE ì`§OH$ àmßV H$s{OE & 2 p 21. nXmW© GaAs Ho$ {bE dmoëQ>Vm Ho$ gmW Ymam Ho$ {dMaU H$mo Xem©Zo dmbm J«m\$ Amao{IV H$s{OE & Bg J«m\$ na dh joÌ A§{H$V H$s{OE Ohm± : 2 (H$) à{VamoY G$UmË_H$ h¡, VWm (I) Amo_ Ho$ {Z`_ H$m nmbZ hmoVm h¡ & 11-55/1/1 10  SECTION B 17. Find the temperature at which the resistance of a wire made of silver will be twice its resistance at 20 C. Take 20 C as the reference temperature and temperature coefficient of resistance of silver at 20 C = 4·0 10 3 K 1. 2 18. (a) Monochromatic light of frequency 5·0 1014 Hz passes from air into a medium of refractive index 1·5. Find the wavelength of the light (i) reflected, and (ii) refracted at the interface of the two media. 2 OR (b) A plano-convex lens of focal length 16 cm is made of a material of refractive index 1.4. Calculate the radius of the curved surface of the lens. 2 19. An object is placed 30 cm in front of a concave mirror of radius of curvature 40 cm. Find the (i) position of the image formed and (ii) magnification of the image. 2 20. Consider a neutron (mass m) of kinetic energy E and a photon of the same energy. Let n and p be the de Broglie wavelength of neutron and the wavelength of photon respectively. Obtain an expression for n. 2 p 21. Plot a graph showing the variation of current with voltage for the material GaAs. On the graph, mark the region where : 2 (a) resistance is negative, and (b) 11-55/1/1 11 P.T.O.  IÊS> J 22. 0·1 m ^wOm H$m H$moB© KZ AmaoI _| Xem©E AZwgma Cg àXoe _| pñWV h¡ Ohm± {dÚwV-joÌ ^ E = 500 x i {dÚ_mZ h¡ & `hm± x _rQ>a _| VWm E, NC 1 _| h¡ & n[aH${bV H$s{OE : 3 (H$) (I) KZ Ho$ ^rVa Amdoe & 23. (H$) Ymam KZËd H$mo n[a^m{fV H$s{OE & `h A{Xe h¡ `m g{Xe ? {H$gr YmVw Ho$ MmbH$ _| H$moB© {dÚwV-joÌ E ~ZmE aIm J`m h¡ & `{X Bg MmbH$ Ho$ à{V EH$m§H$ Am`VZ _| Bbo m, Amdoe e) H$s g§»`m n h¡ VWm BgH$m {dlmpÝV H$mb h¡, Vmo `h Xem©BE {H$ Ymam KZËd j = E h¡, Ohm± ne2 = h¡ & 3 m AWdm (I) ìhrQ>ñQ>moZ goVw {H$go H$hVo h¢ ? ìhrQ>ñQ>moZ goVw Ho$ g§VwbZ Ho$ {bE Amdí`H$ eV] àmßV H$s{OE & 3 24. 1·3384 10 14 J J{VO D$Om© H$m H$moB© àmoQ>m°Z, nyd© {Xem Ho$ AZw{Xe {X{eH$ 2·0 mT Ho$ {H$gr EH$g_mZ Mwå~H$s` joÌ B _|, CÎma go X{jÊm H$s Amoa j¡{VOV: J{V_mZ hmoVo hþE àdoe H$aVm h¡ & n[aH${bV H$s{OE : 3 (H$) àmoQ>m°Z H$s Mmb (I) àmoQ>m°Z _| CËnÞ ËdaU H$m n[a_mU (J) àmoQ>m°Z Ho$ J_Z nW H$s {ÌÁ`m [àmoQ>m°Z Ho$ {bE (q /m) = 1·0 108 C/kg br{OE] 11-55/1/1 12  SECTION C 22. A cube of side 0·1 m is placed, as shown in the figure, in a region where ^ electric field E = 500 x i exists. Here x is in meters and E in NC 1. Calculate : 3 (a) the flux passing through the cube, and (b) the charge within the cube. 23. (a) Define current density. Is it a scalar or a vector ? An electric field E is maintained in a metallic conductor. If n be the number of electrons (mass m, charge e) per unit volume in the conductor and its relaxation time, show that the current density j = E, ne 2 where =. 3 m OR (b) What is a Wheatstone bridge ? Obtain the necessary conditions under which the Wheatstone bridge is balanced. 3 24. A proton with kinetic energy 1·3384 10 14 J moving horizontally from north to south, enters a uniform magnetic field B of 2·0 mT directed eastward. Calculate : 3 (a) the speed of the proton (b) the magnitude of acceleration of the proton (c) the radius of the path traced by the proton [Take (q /m) for proton = 1·0 108 C/kg] 11-55/1/1 13 P.T.O. 25. H$moB© àoaH$, H$moB© g§Ym[aÌ Am¡a H$moB© à{VamoYH$ {H$gr ac òmoV v = vm sin t go loUr _| g§`mo{OV h¢ & n[anW _| j{`V Am¡gV e{º$ Ho$ {bE ì`§OH$ ì`wËnÞ H$s{OE & n[anW H$s AZwZmX Amd¥{Îm Ho$ {bE ^r ì`§OH$ àmßV H$s{OE & 3 26. (H$) {dÚwV-Mwå~H$s` Va§J H$s Va§JX¡¿`© H$mo àm`: {d{H$aU CËnÞ H$aZo dmbo {ZH$m` Ho$ A{^bmj{UH$ gmBµO go g§~§{YV {H$`m OmVm h¡ & Bg H$WZ H$s nwpîQ> Ho$ {bE Xmo CXmhaU Xr{OE & (I) (i) bå~r Xÿ[a`m| Ho$ ao{S>`mo àgmaU Ho$ {bE bKw-Va§J ~¡ÊS> H$m Cn`moJ {H$`m OmVm h¡ & ? (ii) àH$m{eH$ Am¡a ao{S>`mo Xÿa~rZ (XÿaXe©H$) YaVr na ñWm{nV H$s OmVr h¢, naÝVw X-{H$aU IJmo{bH$s Ho$db n¥Ïdr H$s H$jm _| n[aH«$_m H$aZo dmbo ? 3 27. aXa\$moS>© Ho$ na_mUw _m°S>b H$s H${_`m± {b{IE & ~moa Zo CÝh| {H$g àH$ma Xÿa {H$`m ? `h Xem©BE {H$ ~moa na_mUw H$s {d{^Þ H$jmAm| Ho$ ~rM H$s Xÿ[a`m± g_mZ Zht hmoVr h¢ & 3 28. (H$) Zm{^H$ Ho$ {H$Ýht Xmo JwUm| H$mo {b{IE & (I) ? (J) `h Xem©BE {H$ g^r Zm{^H$m| Ho$ {bE Zm{^H$s` Ðì` H$m KZËd g_mZ hmoVm h¡ & 3 IÊS> K 29 30 29. H$moB© b|g Xmo n¥îR>m| go {Kam H$moB© nmaXeu _mÜ`_ hmoVm h¡ {OgH$m EH$ AWdm XmoZm| n¥îR> Jmobr` hmoVo h¢ & {H$gr b|g H$s \$moH$g Xÿar CgHo$ XmoZm| n¥îR>m| H$s dH«$Vm {ÌÁ`mAm| Am¡a b|g Ho$ nXmW© Ho$, Cg nXmW© Ho$ gmnoj {Oggo b|g {Kam h¡, AndV©Zm§H$ Ûmam {ZYm©[aV H$s OmVr h¡ & {H$gr b|g H$s j_Vm CgH$s \$moH$g Xÿar H$s ì`wËH«$_ hmoVr h¡ & `{X H$B© b|gm| H$mo gånH©$ _| aIm OmVm h¡, Vmo g§`moOZ H$s j_Vm CZ b|gm| H$s AnZr-AnZr j_VmAm| H$m ~rOJ{UVr` `moJ hmoVm h¡ & 11-55/1/1 14 25. An inductor, a capacitor and a resistor are connected in series with an ac source v = vm sin t. Derive an expression for the average power dissipated in the circuit. Also obtain the expression for the resonant frequency of the circuit. 3 26. (a) examples to justify this statement. (b) (i) Long distance radio broadcasts use short-wave bands. Why ? (ii) Optical and radio telescopes are built on the ground, but X-ray astronomy is possible only from satellites orbiting the Earth. Why ? 3 27. Write the drawbacks of Rutherfor them 3 28. (a) State any two properties of a nucleus. (b) Why is the density of a nucleus much more than that of an atom ? (c) Show that the density of the nuclear matter is the same for all nuclei. 3 SECTION D Questions number 29 and 30 are case study-based questions. Read the following paragraphs and answer the questions that follow. 29. A lens is a transparent medium bounded by two surfaces, with one or both surfaces being spherical. The focal length of a lens is determined by the radii of curvature of its two surfaces and the refractive index of its medium with respect to that of the surrounding medium. The power of a lens is reciprocal of its focal length. If a number of lenses are kept in contact, the power of the combination is the algebraic sum of the powers of the individual lenses. 11-55/1/1 15 P.T.O.  (i) {H$gr H$m±M Ho$ ~Zo C^`moÎmb b|g Ho$ XmoZm| \$bH$m| H$s dH«$Vm {ÌÁ`m R g_mZ h¡ VWm H$m±M H$m AndV©Zm§H$ n h¡ & Bg b|g H$s j_Vm h¡ : 1 2 (n 1) (2n 1) (A) (B) R R (n 1) (2n 1) (C) (D) 2R 2R (ii) j_Vm P Ho$ {H$gr C^`moÎmb b|g, {OgHo$ XmoZm| \$bH$m| H$s dH«$Vm {ÌÁ`m g_mZ h¡, H$mo CgHo$ _w»` Aj Ho bå~dV² Xmo g_mZ ^mJm| _| H$mQ>m J`m h¡ & b|g Ho$ EH$ ^mJ H$s j_Vm hmoJr : 1 P (A) 2P (B) P (C) 4P (D) 2 (iii) Cn`w©º$ b|g Ho$ XmoZm| ^mJm| H$mo AmaoI _| Xem©E AZwgma EH$ -Xÿgao Ho$ gånH©$ _| aIm J`m h¡ & Bg g§`moOZ H$s j_Vm hmoJr : 1 P P (A) (B) P (C) 2P (D) 2 4 (iv) (H$) j_Vm P Ho$ {H$gr C^`moÎmb b|g, {OgHo$ XmoZm| \$bH$m| H$s dH«$Vm {ÌÁ`m g_mZ h¡, H$mo CgHo$ _w»` Aj Ho$ AZw{Xe Xmo ^mJm| _| H$mQ>H$a XmoZm| ^mJm| H$mo AmaoI _| Xem©E AZwgma ì`dpñWV {H$`m J`m h¡ & Bg g§`moOZ H$s j_Vm hmoJr : 1 (A) eyÝ` (B) P P (C) 2P (D) 2 AWdm (I) 60 cm Am¡a 20 cm \$moH$g Xÿar Ho$ Xmo CÎmb b|gm| H$mo EH$-Xÿgao Ho$ gånH©$ _| g_mjV: aIm J`m h¡ & Bg g§`moOZ H$s j_Vm h¡ : 1 (A) 6·6 D (B) 15 D 1 1 (C) D (D) D 15 80 11-55/1/1 16  (i) A double-convex lens, with each face having same radius of curvature R, is made of glass of refractive index n. Its power is : 1 2 (n 1) (2n 1) (A) (B) R R (n 1) (2n 1) (C) (D) 2R 2R (ii) A double-convex lens of power P, with each face having same radius of curvature, is cut into two equal parts perpendicular to its principal axis. The power of one part of the lens will be : 1 P (A) 2P (B) P (C) 4P (D) 2 (iii) The above two parts are kept in contact with each other as shown in the figure. The power of the combination will be : 1 P P (A) (B) P (C) 2P (D) 2 4 (iv) (a) A double-convex lens of power P, with each face having same radius of curvature, is cut along its principal axis. The two parts are arranged as shown in the figure. The power of the combination will be : 1 (A) Zero (B) P P (C) 2P (D) 2 OR (b) Two convex lenses of focal lengths 60 cm and 20 cm are held coaxially in contact with each other. The power of the combination is : 1 (A) 6·6 D (B) 15 D 1 1 (C) D (D) D 15 80 11-55/1/1 17 P.T.O. 30. g§{Y S>m`moS> {XîQ>H$mar Ho$ ê$n _| : ac dmoëQ>Vm H$mo {XîQ>H$mar (dc) dmoëQ>Vm _| n[ad{V©V H$aZo H$s à{H«$`m H$mo {XîQ>H$aU H$hVo h¢ Am¡a Bg n[adV©Z H$mo gånÞ H$aZo dmbr `w{º$ H$mo {XîQ>H$mar H$hVo h¢ & {H$gr p-n g§{Y S>m`moS> Ho$ A{^bmj{UH$ go `h kmV hmoVm h¡ {H$ O~ p-n g§{Y S>m`moS> AJ«{X{eH$ ~m`{gV hmoVm h¡, Vmo CgH$m à{VamoY {ZåZ hmoVm h¡ VWm O~ níM{X{eH$ ~m`{gV hmoVm h¡, Vmo CgH$m p-n g§{Y S>m`moS> Ho$db AJ«{X{eH$ ~m`g hmoZo na hr MmbZ H$aVm h¡ & p-n g§{Y S>m`moS> H$m `hr JwU {XîQ>H$mar Ho$ ê$n _| BgHo$ Cn`moJ Ho$ {bE Bgo Cn`wº$ ~ZmVm h¡ & Bg àH$ma, O~ {H$gr ac dmoëQ>Vm H$mo {H$gr p-n g§{Y Ho$ {gam| na AZwà`wº$ {H$`m OmVm h¡, Vmo `h Ho$db CÝht àË`mdVu AY©-MH«$m| _| MmbZ H$aVr h¡ O~ `h AJ«{X{eH$ ~m`g hmoVr h¡ & dh {XîQ>H$mar Omo {H$gr ac dmoëQ>Vm Ho$ AY©-MH«$ H$m {XîQ>H$aU H$aVm h¡, Cgo AY©-Va§J {XîQ>H$mar H$hVo h¢ Am¡a Omo XmoZm| AY©-MH«$m| H$m {XîQ>H$aU H$aVm h¡, Cgo nyU©-Va§J {XîQ>H$mar H$hVo h¢ & (i) {H$gr nyU©-Va§J {XîQ>H$mar na AZwà`wº$ {H$gr àË`mdVu dmoëQ>Vm H$m dJ© _mÜ` _yb V0 _mZ h¡ & Vmo {XîQ>rH¥$V {ZJ©V dmoëQ>Vm H$m dJ© _mÜ` _yb _mZ h¡ : 1 2 V0 V02 (A) (B) 2 2 2 V0 V0 (C) (D) 2 2 2 (ii) {H$gr nyU©-Va§J {XîQ>H$mar _| àË`oH$ S>m`moS> go Ymam àdm{hV hmoVr h¡ : 1 (A) {Zdoe {g½Zb Ho$ nyU© MH«$ Ho$ {bE (B) {Zdoe {g½Zb Ho$ AY©-MH«$ Ho$ {bE (C) {Zdoe {g½Zb Ho$ AY©-MH«$ go H$_ Ho$ {bE (D) {Zdoe {g½Zb Ho$ Ho$db YZmË_H$ AY©-MH«$ Ho$ {bE (iii) {H$gr nyU©-Va§J {XîQ>H$mar _| : 1 (A) {H$gr {ZpíMV g_` na XmoZm| S>m`moS> AJ«{X{eH$ ~m`{gV hmoVo h¢ & (B) {H$gr {ZpíMV g_` na XmoZm| S>m`moS> níM{X{eH$ ~m`{gV hmoVo h¢ & (C) {H$gr {ZpíMV g_` na EH$ AJ«{X{eH$ ~m`{gV VWm Xÿgam níM{X{eH$ ~m`{gV hmoVm h¡ & (D) àW_ AY©-MH«$ _| XmoZm| AJ«{X{eH$ ~m`{gV hmoVo h¢ VWm {ÛVr` AY©-MH«$ _| níM{X{eH$ ~m`{gV hmoVo h¢ & 11-55/1/1 18 30. Junction Diode as a Rectifier : The process of conversion of an ac voltage into a dc voltage is called rectification and the device which performs this conversion is called a rectifier. The characteristics of a p-n junction diode reveal that when a p-n junction diode is forward biased, it offers a low resistance and when it is reverse biased, it offers a high resistance. Hence, a p-n junction diode conducts only when it is forward biased. This property of a p-n junction diode makes it suitable for its use as a rectifier. Thus, when an ac voltage is applied across a p-n junction, it conducts only during those alternate half cycles for which it is forward biased. A rectifier which rectifies only half cycle of an ac voltage is called a half-wave rectifier and one that rectifies both the half cycles is known as a full-wave rectifier. (i) The root mean square value of an alternating voltage applied to a V full-wave rectifier is 0. Then the root mean square value of the 2 rectified output voltage is : 1 V0 V02 (A) (B) 2 2 2 V0 V0 (C) (D) 2 2 2 (ii) In a full-wave rectifier, the current in each of the diodes flows for : 1 (A) Complete cycle of the input signal (B) Half cycle of the input signal (C) Less than half cycle of the input signal (D) Only for the positive half cycle of the input signal (iii) In a full-wave rectifier : 1 (A) Both diodes are forward biased at the same time. (B) Both diodes are reverse biased at the same time. (C) One is forward biased and the other is reverse biased at the same time. (D) Both are forward biased in the first half of the cycle and reverse biased in the second half of the cycle. 11-55/1/1 19 P.T.O.  (iv) (H$) {H$gr AY©-Va§J {XîQ>H$mar na 50 Hz Amd¥{Îm H$s H$moB© àË`mdVu dmoëQ>Vm AZwà`wº$ H$s JB© h¡ & Vmo {ZJ©V H$s C{_©H$m Amd¥{Îm hmoJr : 1 (A) 100 Hz (B) 50 Hz (C) 25 Hz (D) 150 Hz AWdm (I) AmaoI _| Xem©E AZwgma H$moB© {g½Zb {H$gr p-n g§{Y S>m`moS> na AZwà`wº$ {H$`m J`m h¡ & à{VamoY RL Ho$ {gam| na {ZJ©V H$s nhMmZ H$s{OE : 1 (A) (B) (C) (D) 11-55/1/1 20  (iv) (a) An alternating voltage of frequency of 50 Hz is applied to a half-wave rectifier. Then the ripple frequency of the output will be : 1 (A) 100 Hz (B) 50 Hz (C) 25 Hz (D) 150 Hz OR (b) A signal, as shown in the figure, is applied to a p-n junction diode. Identify the output across resistance RL : 1 (A) (B) (C) (D) 11-55/1/1 21 P.T.O.  IÊS> L> 31. (H$) (i) {H$gr ~mø EH$g_mZ {dÚwV-joÌ E _| {H$gr {dÚwV {ÛY«wd p H$s pñW{VO D$Om© Ho$ {bE H$moB© ì`§OH$ ì`wËnÞ H$s{OE & Bg {ÛY«wd H$s pñW{VO D$Om© H$~ (1) A{YH$V_, Am¡a (2) {ZåZV_ hmoVr h¡ ? (ii) H$moB© {dÚwV {ÛY«wd {~ÝXþ Amdoe 1·0 pC Am¡a + 1·0 pC Omo x y Vb _| H«$_e: (0, 0) Am¡a (3 mm, 4 mm) na pñWV h¢, go {_bH$a ~Zm h¡ & Bg 1000 V ^ àXoe _| {H$gr {dÚwV-joÌ E = i H$mo bJm`m J`m h¡ & {ÛY«wd m na H$m`©aV ~b-AmKyU© kmV H$s{OE & 5 AWdm (I) (i) 2a Xÿar Ho$ n¥WH$Z Am¡a q Am¡a q Amdoem| go ~Zm, H$moB© {dÚwV {ÛYw«d ({ÛYw«d AmKyU© p = p ^i ) x-Aj Ho$ AZw{Xe AnZo Ho$ÝÐ H$mo _yb-{~ÝXþ na aIVo hþE pñWV h¡ & `h Xem©BE {H$ Bg {ÛY«wd Ho$ H$maU {H$gr {~ÝXþ ^ 1 p. i x, (x >> a) na {d^d V,. h¡ & 4 0 x2 (ii) H«$_e: {ÌÁ`m 1 cm Am¡a 3 cm Ho$ Xmo {d`wº$ YmpËdH$ Jmobm| S1 VWm S2 H$mo Bg àH$ma Amdo{eV {H$`m J`m h¡ {H$ CZHo$ Amdoe KZËd g_mZ 2 10 9 C / m2 h¢ & BZ XmoZm| Jmobm| H$mo EH$-Xÿgao go AË`{YH$ Xÿar na aIH$a {H$gr nVbo Vma go g§`mo{OV {H$`m J`m h¡ & Jmobo S1 na Z`m Amdoe n[aH${bV H$s{OE & 5 32. (H$) (i) {H$gr ac òmoV v = vm sin t go {H$gr à{VamoYH$ Am¡a {H$gr g§Ym[aÌ H$mo loUr _| g§`mo{OV {H$`m J`m h¡ & n[anW H$s à{V~mYm Ho$ {bE ì`§OH$ ì`wËnÞ H$s{OE & (ii) H$moB© àoaH$ {H$gr n[anW _| MmbH$ Ho$ ê$n _| H$~ H$m`© H$aVm h¡ ? BgHo$ {bE H$maU Xr{OE & 11-55/1/1 22  SECTION E 31. (a) (i) Derive an expression for potential energy of an electric dipole p in an external uniform electric field E. When is the potential energy of the dipole (1) maximum, and (2) minimum ? (ii) An electric dipole consists of point charges 1·0 pC and + 1·0 pC located at (0, 0) and (3 mm, 4 mm) respectively in 1000 V ^ x y plane. An electric field E = i is switched on m in the region. Find the torque acting on the dipole. 5 OR ^ (b) (i) An electric dipole (dipole moment p = p i ), consisting of charges q and q, separated by distance 2a, is placed along the x-axis, with its centre at the origin. Show that the potential V, due to this dipole, at a point x, (x >> a) is equal ^ 1 p. i to.. 4 0 x2 (ii) Two isolated metallic spheres S1 and S2 of radii 1 cm and 3 cm respectively are charged such that both have the same 2 charge density 10 9 C / m2. They are placed far away from each other and connected by a thin wire. Calculate the new charge on sphere S1. 5 32. (a) (i) A resistor and a capacitor are connected in series to an ac source v = vm sin t. Derive an expression for the impedance of the circuit. (ii) When does an inductor act as a conductor in a circuit ? Give reason for it. 11-55/1/1 23 P.T.O.  (iii) {H$gr {dÚwV b¡ån H$s A{^H$ënZm 110 V dc Am¡a 11 A {dÚwV Ymam na àMmbZ Ho$ {bE H$s JB© h¡ & `{X Bg b¡ån H$mo 220 V, 50 Hz Ho$ ac òmoV na loUr _§o {H$gr Hw$ÊS>br Ho$ gmW àMm{bV {H$`m OmVm h¡, Vmo Hw$ÊS>br H$m àoaH$Ëd kmV H$s{OE & 5 AWdm (I) (i) {H$gr Q´>mÝg\$m°_©a H$m Zm_m§{H$V AmaoI ItMH$a CgHo$ H$m`©H$mar {gÕmÝV H$m dU©Z H$s{OE & {H$gr dmñV{dH$ Q´>mÝg\$m°_©a _| hmoZo dmbo D$Om© õmg Ho$ {H$Ýht VrZ H$maUm| H$s ì`m»`m H$s{OE & (ii) H$moB© Q´>mÝg\ Vm _| n[ad{V©V H$aVm h¡ §KZ H$aVm h¡ ? ì`m»`m H$s{OE & (iii) {H$gr Q´>mÝg\$m°_©a H$s àmW{_H$ Am¡a {ÛVr`H$ Hw$ÊS>{b`m| _| \o$am| H$s g§»`m H«$_e: 200 Am¡a 3000 h¡ & àmW{_H$ Hw$ÊS>br H$mo Xr JB© {Zdoe dmoëQ>Vm 90 V h¡ & n[aH${bV H$s{OE : (1) {ÛVr`H$ Hw$ÊS>br Ho$ {gam| na {ZJ©V dmoëQ>Vm (2) `{X {ÛVr`H$ Hw$ÊS>br _| Ymam 2·0 A h¡, Vmo àmW{_H$ Hw$ÊS>br _| Ymam 5 33. (H$) (i) H$moB© àH$me {H$aU {H$gr {Ì^wOmH$ma {àµÁ_ go JwµOaVr h¡ & AmnVZ H$moU Ho$ gmW {dMbZ H$moU {H$g àH$ma {dMaU H$aVm h¡ ? J«m\$ Ûmam Xem©BE & AV: Ý`yZV_ {dMbZ H$moU n[a^m{fV H$s{OE & (ii) H$moB© àH$me {H$aU {àµÁ_ H$moU A Ho$ {H$gr {àµÁ_ Ho$ EH$ AndV©H$ \$bH$ na A{^bå~dV² AmnVZ H$aHo$ H$moU na {dM{bV hmoVr h¡ & {gÕ H$s{OE sin ( A ) {H$ {àµÁ_ Ho$ nXmW© H$m AndV©Zm§H$ n h¡ & sin A 11-55/1/1 24  (iii) An electric lamp is designed to operate at 110 V dc and 11 A current. If the lamp is operated on 220 V, 50 Hz ac source with a coil in series, then find the inductance of the coil. 5 OR (b) (i) Draw a labelled diagram of a step-up transformer and describe its working principle. Explain any three causes for energy losses in a real transformer. (ii) A step-up transformer converts a low voltage into high voltage. Does it violate the principle of conservation of energy ? Explain. (iii) A step-up transformer has 200 and 3000 turns in its primary and secondary coils respectively. The input voltage given to the primary coil is 90 V. Calculate : (1) The output voltage across the secondary coil (2) The current in the primary coil if the current in the secondary coil is 2·0 A. 5 33. (a) (i) A ray of light passes through a triangular prism. Show graphically, how the angle of deviation varies with the angle of incidence ? Hence define the angle of minimum deviation. (ii) A ray of light is incident normally on a refracting face of a prism of prism angle A and suffers a deviation of angle. Prove that the refractive index n of the material of the prism sin ( A ) is given by n. sin A 11-55/1/1 25 P.T.O.  (iii) {H$gr {àµÁ_ Ho$ nXmW© H$m AndV©Zm§H$ 2 h¡ & `{X {àµÁ_ H$m AndV©H$ H$moU 60 h¡, Vmo kmV H$s{OE : (1) Ý`yZV_ {dMbZ H$moU, VWm (2) AmnVZ H$moU & 5 AWdm (I) (i) hmBJoÝg H$m {gÕmÝV {b{IE & H$moB© g_Vb Va§J {H$gr namdVu n¥îR> na H$moU i na AmnVZ H$aVr h¡ & VXZwê$nr namd{V©V Va§JmJ« H$s aMZm H$s{OE & Bg AmaoI H$m Cn`moJ H$aHo$, {gÕ H$s{OE {H$ namdV©Z H$moU AmnVZ H$moU Ho$ ~am~a hmoVm h¡ & (ii) àH$me Ho$ H$bm- ? H$bm-g§~Õ òmoV Ho$ ê$n _| H$m`© H$a gH$Vo h¢ ? ì`m»`m H$s{OE & (iii) `§J Ho$ {Û{Par à`moJ _| {H$gr àH$me nw§O Ûmam, {Og_| Xmo Va§JX¡¿`©, EH$ kmV Va§JX¡¿`© 520 nm H$s VWm Xÿgar AkmV Va§JX¡¿`© H$s h¡, Xmo Eogo ì`{VH$aU n¡Q>Z© CËnÞ H$aVo h¢, {OZ_| AkmV Va§JX¡¿`© H$s MVwW© XrßV q\«$O kmV Va§JX¡¿`© H$s n§M_ XrßV q\«$O Ho$ g§nmVr h¡ & H$m _mZ kmV H$s{OE & 5 11-55/1/1 26  (iii) The refractive index of the material of a prism is 2. If the refracting angle of the prism is 60 , find the (1) Angle of minimum deviation, and (2) Angle of incidence. 5 OR (b) (i) State Huygens principle. A plane wave is incident at an angle i on a reflecting surface. Construct the corresponding reflected wavefront. Using this diagram, prove that the angle of reflection is equal to the angle of incidence. (ii) What are the coherent sources of light ? Can two independent sodium lamps act like coherent sources ? Explain. (iii) A beam of light consisting of a known wavelength 520 nm and an unknown wavelength experiment produces two interference patterns such that the fourth bright fringe of unknown wavelength coincides with the fifth bright fringe of known wavelength. Find the value of. 5 11-55/1/1 27 P.T.O.  Marking Scheme Strictly Confidential (For Internal and Restricted use only) Senior School Certificate Examination, 2024 SUBJECT NAME PHYSICS (Theory) (CODE 55/1/1) General Instructions: - 1 You are aware that evaluation is the most important process in the actual and correct assessment of the candidates. A small mistake in evaluation may lead to serious problems which may affect the future of the candidates, education system and teaching profession. To avoid mistakes, it is requested that before starting evaluation, you must read and understand the spot evaluation guidelines carefully. 2 “Evaluation policy is a confidential policy as it is related to the confidentiality of the examinations conducted, Evaluation done and several other aspects. Its’ leakage to public in any manner could lead to derailment of the examination system and affect the life and future of millions of candidates. Sharing this policy/document to anyone, publishing in any magazine and printing in News Paper/Website etc may invite action under various rules of the Board and IPC.” 3 Evaluation is to be done as per instructions provided in the Marking Scheme. It should not be done according to one’s own interpretation or any other consideration. Marking Scheme should be strictly adhered to and religiously followed. However, while evaluating, answers which are based on latest information or knowledge and/or are innovative, they may be assessed for their correctness otherwise and due marks be awarded to them. In class-X, while evaluating two competency-based questions, please try to understand given answer and even if reply is not from marking scheme but correct competency is enumerated by the candidate, due marks should be awarded. 4 The Marking scheme carries only suggested value points for the answers These are in the nature of Guidelines only and do not constitute the complete answer. The students can have their own expression and if the expression is correct, the due marks should be awarded accordingly. 5 The Head-Examiner must go through the first five answer books evaluated by each evaluator on the first day, to ensure that evaluation has been carried out as per the instructions given in the Marking Scheme. If there is any variation, the same should be zero after delibration and discussion. The remaining answer books meant for evaluation shall be given only after ensuring that there is no significant variation in the marking of individual evaluators. 6 Evaluators will mark( √ ) wherever answer is correct. For wrong answer CROSS ‘X” be marked. Evaluators will not put right (✓)while evaluating which gives an impression that answer is correct and no marks are awarded. This is most common mistake which evaluators are committing. 7 If a question has parts, please award marks on the right-hand side for each part. Marks awarded for different parts of the question should then be totaled up and written in the left- hand margin and encircled. This may be followed strictly. 55/1/1 Page 1 of 24 8 If a question does not have any parts, marks must be awarded in the left-hand margin and encircled. This may also be followed strictly. 9 If a student has attempted an extra question, answer of the question deserving more marks should be retained and the other answer scored out with a note “Extra Question”. 10 No marks to be deducted for the cumulative effect of an error. It should be penalized only once. 11 A full scale of marks 0 to 70 has to be used. Please do not hesitate to award full marks if the answer deserves it. 12 Every examiner has to necessarily do evaluation work for full working hours i.e., 8 hours every day and evaluate 20 answer books per day in main subjects and 25 answer books per day in other subjects (Details are given in Spot Guidelines).This is in view of the reduced syllabus and number of questions in question paper. 13 Ensure that you do not make the following common types of errors committed by the Examiner in the past:- Leaving answer or part thereof unassessed in an answer book. Giving more marks for an answer than assigned to it. Wrong totaling of marks awarded on an answer. Wrong transfer of marks from the inside pages of the answer book to the title page. Wrong question wise totaling on the title page. Wrong totaling of marks of the two columns on the title page. Wrong grand total. Marks in words and figures not tallying/not same. Wrong transfer of marks from the answer book to online award list. Answers marked as correct, but marks not awarded. (Ensure that the right tick mark is correctly and clearly indicated. It should merely be a line. Same is with the X for incorrect answer.) Half or a part of answer marked correct and the rest as wrong, but no marks awarded. 14 While evaluating the answer books if the answer is found to be totally incorrect, it should be marked as cross (X) and awarded zero (0)Marks. 15 Any un assessed portion, non-carrying over of marks to the title page, or totaling error detected by the candidate shall damage the prestige of all the personnel engaged in the evaluation work as also of the Board. Hence, in order to uphold the prestige of all concerned, it is again reiterated that the instructions be followed meticulously and judiciously. 16 The Examiners should acquaint themselves with the guidelines given in the “Guidelines for spot Evaluation” before starting the actual evaluation. 17 Every Examiner shall also ensure that all the answers are evaluated, marks carried over to the title page, correctly totaled and written in figures and words. 18 The candidates are entitled to obtain photocopy of the Answer Book on request on payment of the prescribed processing fee. All Examiners/Additional Head Examiners/Head Examiners are once again reminded that they must ensure that evaluation is carried out strictly as per value points for each answer as given in the Marking Scheme. 55/1/1 Page 2 of 24  MARKING SCHEME : PHYSICS (042) CODE :55/1/1 Q.NO. VALUE POINT/EXPECTED ANSWERS MARKS TOTAL MARKS Section A 1. (B) Zero 1 1 2. (D) 5.0 ×10-2 J 1 1 3. (B) 8V 1 1 4. (C) Shrink 1 1 5. (B) ( - 0.8 mN) î 1 1 6. G 1 1 (B)  1000 7. X 1 1 (A) 6 8. (A) I 1 1 9. (C) n f  2 and ni  4 1 1 10. (B) the number of conduction electrons increases 1 1 11. 1 1 1 ( C) 3 12. (A) momentum 1 1 13. (D) Assertion (A) is false and reason (R) is also false. 1 1 14. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the 1 1 correct explanation of the Assertion (A) 15. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the 1 1 correct explanation of the Assertion (A) 16. (D) Assertion (A) is false and reason (R) is also false. 1 1 Section B 17. Finding the temperature 2 R  R  1   T  T   ½ R = 2 R  [Given] 2 R   R  1   T  T   ½ On solving T  T  250 T  270C or 543K 1 2 55/1/1 Page 3 of 24 18. (a) Finding the wavelength of (i) Reflected Light 1 (ii) Refracted Light 1 (i) v=υλ 3×108 = 5×1014 × λ 1 λ = 600 nm or 6 ×10-7m (ii) air medium   600 nm medium  1.5 1 = 400 nm or 4×10-7m OR (b) Calculating the radius of the curved surface 2 1 1 1   (  1)    f  R1 R2  1 1 1 1  (1.4  1)    16 R  1 1  0.4  16 R R = 16 × 0.4 R = 6.4 cm 1 2 19. Finding the (i) position of the image formed 1 (ii) magnification of the image 1 ½ (i) + = + = On solving v = - 60 cm ½ 55/1/1 Page 4 of 24  (ii) m = - ½ −60 =-( ) −30 ½ = -2 2 20. Obtaining an expression for λn / λp 2 ½ E= => λp = ½ λn = = √( ) ½ = × √( ) = √( ) ½ 2 21. Plotting the graph 1 Marking the region where: (a) resistance is negative ½ (b) Ohm’s law is obeyed ½ 1+ ½ + ½ 2 55/1/1 Page 5 of 24  SECTION C 22. Calculating (a) the flux passing through the cube 2 (b) the charge within the cube 1 a) φL = 𝐸⃗. 𝐴⃗ = - [500 x 0.1] x [(0.1)2] = - 0.5 N m2 C -1 ½ φR = 𝐸 ⃗. 𝐴⃗ = [500 x 0.2] x [(0.1)2] = 1 N m2 C -1 ½ Net flux = φL + φR = 0.5 N m2 C -1 1 ½ b) flux, φ = charge, q = φ x εo = 0.5 εo ½ = 4.4 x 10-12 C 3 23. a)  Defining current density ½  Whether scalar or vector ½  Showing 𝚥⃗ = α 𝐸⃗ 2 Current density is the amount of charge flowing per second per unit area ½ normal to the flow. Alternatively: 𝑗= It is a vector quantity. ½ The amount of charge crossing the area A in time ∆t is I ∆t, where I is the magnitude of the current. Hence, ½ I ∆ t = ne A |vd| ∆t 55/1/1 Page 6 of 24  ½ I∆t= τ n ∆t |E| ½ I = |j|A ½ |j| = τ |E| 𝚥⃗ = α 𝐸⃗ OR b) Defining Wheatstone bridge 1 Obtaining balancing conditions 2 1 Alternatively: If the figure is explained in words full credit to be given. For loop ADBA: –I1 R1 + I2 R2 + Ig G = 0 (1) ½ For loop CBDC: I4R4 - I3 R3 - Ig G = 0 (2) ½ For balanced wheatstone bridge, Ig = 0 ½ And by applying Kirchoff’s junction rule to junction D and B, I1 = I3 & I2 = I4 From eqn (1) and (2) 55/1/1 Page 7 of 24  = and = ½  = 3 24. Calculating a) the speed of the proton 1 b) the magnitude of the acceleration of the proton 1 c) the radius of the path traced by the proton 1.. ½ a) v = √( ) = 4 x 106 m/s ½ b) acceleration = qvB / m ½ = 8 x 1011 m/s2 ½ c) r = mv / Bq ½ = 20 m ½ 3 25. Deriving an expression for the average power dissipated in series LCR circuit 2 Obtaining expression for the resonant frequency 1 v = vm sinωt i = im sin(ωt+φ) Power, P = v i = ( vm sinωt ) x [ im sin(ωt+φ)] ½ = [ cos φ – cos(2ωt+φ)] (1) ½ The average power over a cycle is given by the average of the two terms in RHS of eqn (1). It is only the 2nd term which is time dependent. It’s average is zero. Therefore, ½ P= cos φ 55/1/1 Page 8 of 24  P = V I cos φ OR ½ P = I2 Z cos φ At resonance, XC = XL 1 = 𝜔𝐿 ½ 𝜔𝐶 ω= √( ) => υ= ½ 3 √( ) 26. a) Two examples 1 b) (i) Reason for use of short waves bands 1 (ii) Reason for x-ray astronomy from satellites 1 a) (Any Two)  Gamma radiation having wavelength of 10–14 m to 10–15 m, typically originate from an atomic nucleus.  X-rays are emitted from heavy atoms.  Radio waves are produced by accelerating electrons in a circuit. A transmitting antenna can most efficiently radiate waves having a wavelength of about the same size as the antenna. ½+½ b) (i) Ionosphere reflects waves in these bands 1 (ii) Atmosphere absorbs x-rays, while visible and radio waves can penetrate it 1 Note: Full credit to be given for part (b) for mere attempt. 3 27.  Drawbacks of Rutherford’s atomic model 1  Bohr’s explanation 1  Showing different orbits are not equally spaced 1 Drawbacks: i) According to classical electromagnetic theory, an accelerating charged particle emits radiation in the form of electromagnetic waves. The energy of an accelerating electron should therefore, continuously decrease. The electron would spiral inward and eventually fall into the nucleus. Thus, such 55/1/1 Page 9 of 24  an atom cannot be stable. ii) As the electrons spiral inwards, their angular velocities and hence their frequencies would change continuously. Thus, they would emit a continuous spectrum, in contradiction to the line spectrum actually observed. 1 Bohr postulated stable orbits in which electrons do not radiate energy 1 Alternatively: Bohr’s postulates (Any ONE of the three) (i) An electron in an atom could revolve in certain stable orbits without the emission of radiant energy. (ii) The electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of h/2π (iii) An electron might make a transition from one of its specified non- radiating orbits to another of lower energy. When it does so, a photon is emitted having energy equal to the energy difference between the initial and final states. The radius of the nth orbit is found as 1 rn α n2 Alternatively: Difference in radius of consecutive orbits is rn+1 – rn = k [(n+1)2 – n2)] = k (2n + 1) which depends on n, and is not a constant 3 28. a) Stating two properties of a nucleus 1 b) Why density of a nucleus is much more than that of an atom 1 c) Showing that density of nuclear matter is same for all nuclei 1 a) (Any TWO) (i) The nucleus is positively charged (ii) The nucleus consists of protons and neutrons (iii) The nuclear density is independent of mass number (iv) The radius of the nucleus, R = Ro A1/3 ½+½ b) Atoms have large amount of empty spaces. Mass is concentrated in 1 nucleus. 55/1/1 Page 10 of 24  c) Density = Mass / Volume = = = 1 So, density is independent of mass number 3 SECTION D 29. ( ) 1 (i) (A) (ii) (D) P/2 1 (iii) (B) P 1 (iv) a) (C) 2P 1 OR b) (A) 6.6 D 4 30. 1 (i) (A) √ (ii) (B) half cycle of the input signal 1 (iii) (C) One is forward biased and the other is reverse biased at the 1 same time (iv) a) (B) 50 Hz 1 OR b) (D) 4 55/1/1 Page 11 of 24  Section E 31. (a) (i)  Deriving the expression for potential energy 2  Maximum & Minimum value of potential energy ( ½ + ½ ) (ii) Finding the torque. 2 (i) The amount of work done in rotating the dipole from θ =  0 to θ = 1 by the external torque 1 ½ W=   dext o 1 ½ =  pE sin  d o W = pE (cos  0  cos 1 ) ½  For 0  and 1   2  = pE (cos  cos  ) 2 U( )   pE cos  ½ = - 𝑝⃗.𝐸⃗ (1) Potential energy is maximum when:   p is antiparallel to E ½ Alternatively:  = 180° or π radians 55/1/1 Page 12 of 24  (2) Potential energy is minimum when:   p is along to E ½ Alternatively:  = 0° (ii)   pE sin  ½ ½  (2aq ) E sin  4  (5  10 3  1  10 12 )103  5 ½ 12  4  10 Nm Direction is along –ve Z direction. ½ OR (b) (i) Deriving expression for potential 2½ (ii) New charge on Sphere S1 2½ (i) 2a -q O +q P 𝚤̂ ½ x 1 q V 4 0 r ½ V  V  q  V q 55/1/1 Page 13 of 24  1  q q  ½ V  4 0  (x  a ) (x  a )   q x  a  x  a   4 0  (x 2  a 2 )  q 2a p V   4 0 (x  a ) 4 0 (x 2  a 2 ) 22 As p is along x-axis, so  1 p. iˆ ½ V  4 0 (x 2  a 2 ) If x>>a  ½ 1 p. iˆ V  4 0 x 2 Alternatively: 1 q q  V    ----- (i) ½ 4 0  r1 r2  55/1/1 Page 14 of 24  By geometry r12  r 2  a 2  2ar cos r22  r 2  a 2  2ar cos  2acos a 2  r12  r 2 1   2  r r   2a cos   ½  r 2 1    r  2a cos   ½  r2 2  r 1  2  Similarly,  r  a Using binomial theorem & retaining terms upto the first order in ; we r obtain 1  1 1 2a cos   2 1 a   1    1  cos   ----- (ii) r1 r  r  r r  1  1 1 2a cos   2 1 a   1    1  cos   ----- (iii) r2 r  r  r r  Using equations (i) ,(ii) & (iii) & p = 2qa q 2a cos  p cos  V  4 0 r2 4 0r 2 ½  p cos   p. rˆ  As r is along the x – axis.   ½  p. rˆ  p. iˆ  ˆ  V  1 p.i 4 0 x 2 55/1/1 Page 15 of 24  (ii) Charge on sphere S1 : Q1 = surface charge density  surface Area 2  =  109   4 (1  10 2 )2   = 8  1013 C ½ Charge on sphere S2 : Q2 = surface charge density  surface Area 2  =  109   4 ( 3  102 )2    = 72  1013 C ½ When connected by a thin wire they acquire a common potential V and the charge remains conserved. Q1  Q 2  Q1  Q 2 ½  C1V  C 2V Q1  Q 2  (C1  C2 )V Q1  Q 2 Common potential(V)  C1  C 2 1 1 C1  4 0r1   10 2   10 11 F 9  10 9 9 1 1 C 2  4 0r2   3  10 2   10 11 F 9  10 9 3 13 80  10 V   1.8V 1 1 11 ½     10 9 3 1 𝑄  C1V   10 11  1.8 9 ½ 𝑄  2  10 12 C 55/1/1 Page 16 of 24  Alternatively: Charge on sphere S1 : Q1 = surface charge density  surface Area 2  =  109   4 (1  10 2 )2   = 8  1013 C ½ Charge on sphere S2 : Q2 = surface charge density  surface Area 2  =  109   4 ( 3  102 )2   = 72  1013 C ½ When connected by a thin wire they acquire a common potential V and the charge remains conserved. Q1  Q 2  Q1  Q 2 ½ = ½ On solving, 𝑄  2  10 12 C ½ 5 32. (a) (i) Deriving expression for impedance 2 (ii) Reason 1 (iii) Inductance of coil 2 55/1/1 Page 17 of 24  (i) VC + VR =V ½ v m2  v rm 2  v cm 2 vr m  im R vcm  im X c ½ v m2  (im R )2  (im Xc )2 i m2 R 2  X c2  = ½ vm  im  R 2  X c2  Impedance Z  R 2  X c2 ½ (ii) For direct current (dc), an inductor behaves as a conductor. As XL = ωL = 2π ν L 1 For dc ν = 0  XL= 0 Alternatively: - LdI Induced emf (ε) = - dt For dc; dI = 0  ε=0 55/1/1 Page 18 of 24  110 ½ (iii) R= = 10 Ω 11 v r ms 220 ir ms   R 2  X L2 100  X L2 220 ½ 11  100  X L2 220 100  X L2   20 11 Squaring both sides:  100  X L2  400 ½  X L2  300  X L  10 3  X L  2 fL  10 3  2  50  L ½ 3 L = H 10 OR (b) (i) Labelled diagram of step – up transformer 1 Describing working principle ½ Three causes 1½ (ii) Explanation 1 (iii) (1) Output voltage across secondary coil ½ (2) Current in primary coil ½ 55/1/1 Page 19 of 24  (i) OR 1 The working principle of transformer is mutual induction. When an alternating voltage is applied to the primary, the resulting current ½ produces an alternating magnetic flux which links the secondary and induces an emf in it. Causes of energy losses (Any three) (a) Flux leakage (b) Resistance of the windings (c) Eddy currents 55/1/1 Page 20 of 24  ½+½+½ (d) Hysteresis ½ (ii) No ½ Current changes correspondingly. So, the input power is equal to the output power. (iii) (1) Vs N s  VP N P Ns 3000 Vs   VP   90 NP 200 ½ Vs  1350 V (2) IP Ns  Is NP ½ 5 3000 IP   2  30 A 200 33. (a) (i) Graph showing variation of angle of deviation with angle of incidence 1 Defining angle of minimum deviation 1 sin( A   ) n (ii) Proof of refractive index sin A 1 (iii) (1) Finding angle of minimum deviation 1 (2) Angle of Incidence 1 55/1/1 Page 21 of 24  (i) 1 Minimum deviation angle is defined as the angle at which angle of 1 incidence is equal to the angle of emergence. Alternatively At minimum deviation refracted ray inside the prism becomes parallel to the base of the prism. (ii) At the face XZ :- ½  sin i  1  sin r ----- (1) r=i+δ [ from diagram] ----- (2) In ΔXMN ; A+( 90 –i) + 90 =180 55/1/1 Page 22 of 24   A=i ----- (3) Putting eq. (3) & (2) in eq. (1)  sin A  sin ( A   ) ½ sin ( A   )  sin A (iii)  A  m  sin    2  (1)  A sin 2  60  m  sin    2  2 ½ sin 30  60   m  1  sin    sin 45  2  2 60   m ½  45   m  30 2 A  m ½ (2) i 2 60  30 i  2 i  45 ½ OR (b) (i) Statement of Huygens’ Principle ½ Construction of reflected wave front ½ Proof of angle of reflection is equal to angle of incidence 1 (ii) Definition of coherent sources ½ Explanation 1 (iii) Finding the unknown wavelength 1½ (i) Each point of the wavefront is the source of a secondary disturbance and the wavelets emanating from these points spread out in all directions with the spread of the wave. Each point of the wavefront is the source of a 55/1/1 Page 23 of 24  secondary disturbance and the wavelets emanating from these points spread out in all directions with the speed of the wave. These wavelets emanating from the wavefront are usually referred to as secondary ½ wavelets and if we draw a common tangent to all these spheres, we obtain the new position of the wavefront at a later time. ½ ΔEAC is congruent to ΔBAC; so i  r 1 (ii) Two sources are said to be coherent if the phase difference between ½ them does not change with time. No, two independent sodium lamps cannot be coherent. ½ Two independent sodium lamps cannot be coherent as the phase between them does not remain constant with time. ½ (iii) 4 2  5 1 D  D 4  5  known ½ d d 5     known 4 5   520 4 = 650 nm 1 5 55/1/1 Page 24 of 24  Marking Scheme Strictly Confidential (For Internal and Restricted use only) Senior School Certificate Examination, 2024 SUBJECT NAME PHYSICS (Theory) (CODE 55/1/2) General Instructions: - 1 You are aware that evaluation is the most important process in the actual and correct assessment of the candidates. A small mistake in evaluation may lead to serious problems which may affect the future of the candidates, education system and teaching profession. To avoid mistakes, it is requested that before starting evaluation, you must read and understand the spot evaluation guidelines carefully. 2 “Evaluation policy is a confidential policy as it is related to the confidentiality of the examinations conducted, Evaluation done and several other aspects. Its’ leakage to public in any manner could lead to derailment of the examination system and affect the life and future of millions of candidates. Sharing this policy/document to anyone, publishing in any magazine and printing in News Paper/Website etc may invite action under various rules of the Board and IPC.” 3 Evaluation is to be done as per instructions provided in the Marking Scheme. It should not be done according to one’s own interpretation or any other consideration. Marking Scheme should be strictly adhered to and religiously followed. However, while evaluating, answers which are based on latest information or knowledge and/or are innovative, they may be assessed for their correctness otherwise and due marks be awarded to them. In class-X, while evaluating two competency-based questions, please try to understand given answer and even if reply is not from marking scheme but correct competency is enumerated by the candidate, due marks should be awarded. 4 The Marking scheme carries only suggested value points for the answers These are in the nature of Guidelines only and do not constitute the complete answer. The students can have their own expression and if the expression is correct, the due marks should be awarded accordingly. 5 The Head-Examiner must go through the first five answer books evaluated by each evaluator on the first day, to ensure that evaluation has been carried out as per the instructions given in the Marking Scheme. If there is any variation, the same should be zero after delibration and discussion. The remaining answer books meant for evaluation shall be given only after ensuring that there is no significant variation in the marking of individual evaluators. 6 Evaluators will mark( √ ) wherever answer is correct. For wrong answer CROSS ‘X” be marked. Evaluators will not put right (✓)while evaluating which gives an impression that answer is correct and no marks are awarded. This is most common mistake which evaluators are committing. 7 If a question has parts, please award marks on the right-hand side for each part. Marks awarded for different parts of the question should then be totaled up and written in the left- hand margin and encircled. This may be followed stric