Physics (Theory) Exam Paper PDF - 55/1/1
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H$moS> Z§. Code No. 55/1/1 amob Z§. Roll No. ZmoQ> NOTE (I) H¥$n`m Om±M H$a...
H$moS> Z§. Code No. 55/1/1 amob Z§. Roll No. ZmoQ> NOTE (I) H¥$n`m Om±M H$a b| {H$ Bg àíZ-nÌ _o§ _w{ÐV (I) Please check that this question n¥ð> 23 h¢ & paper contains 23 printed pages. (II) àíZ-nÌ _| Xm{hZo hmW H$s Amoa {XE JE H$moS (II) Code number given on the right >Zå~a H$mo N>mÌ CÎma-nwpñVH$m Ho$ _wI-n¥ð> na hand side of the question paper {bI| & should be written on the title page of the answer-book by the candidate. (III) H¥$n`m Om±M H$a b| {H$ Bg àíZ-nÌ _| (III) Please check that this question >37 àíZ h¢ & paper contains 37 questions. (IV) H¥$n`m àíZ H$m CÎma {bIZm ewê$ H$aZo go (IV) Please write down the Serial nhbo, CÎma-nwpñVH$m _| àíZ H$m H«$_m§H$ Number of the question in the Adí` {bI| & answer-book before attempting it. (V) Bg àíZ-nÌ H$mo n‹T>Zo Ho$ {bE 15 {_ZQ >H$m (V) 15 minute time has been allotted to g_` {X`m J`m h¡ & àíZ-nÌ H$m {dVaU read this question paper. The nydm© _| 10.15 ~Oo {H$`m OmEJm & question paper will be distributed 10.15 ~Oo go 10.30 ~Oo VH$ N>mÌ Ho$db at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the students will read the àíZ-nÌ H$mo n‹T>|Jo Am¡a Bg Ad{Y Ho$ Xm¡amZ question paper only and will not do CÎma-nwpñVH$m na H$moB© CÎma Zht {bI|Jo & write any answer on the answer-book during this period. ^m¡{VH$ {dkmZ (g¡ÕmpÝVH$) PHYSICS (Theory) {ZYm©[aV g_` : 3 KÊQ>o A{YH$V_ A§H$ : 70 Time allowed : 3 hours Maximum Marks : 70.55/1/1 1 P.T.O. gm_mÝ` {ZX}e : {ZåZ{b{IV {ZX}em| H$mo ~hþV gmdYmZr go n{‹T>E Am¡a CZH$m g™Vr go nmbZ H$s{OE : (i) `h àíZ-nÌ Mma IÊS>m| _§| {d^m{OV {H$`m J`m h¡ – H$, I, J Am¡a K & (ii) Bg àíZ-nÌ _| 37 àíZ h¢ & g^r àíZ A{Zdm`© h¢ & (iii) IÊS> H$ – àíZ g§»`m 1 go 20 VH$ A{V bKw-CÎmar` àíZ h¢, àË`oH$ àíZ 1 A§H$ H$m h¡ & (iv) IÊS> I – àíZ g§»`m 21 go 27 VH$ bKw-CÎmar` àíZ h¢, àË`oH$ àíZ 2 A§H$m| H$m h¡ & (v) IÊS> J – àíZ g§»`m 28 go 34 VH$ XrK©-CÎmar` àH$ma Ho$ àíZ h¢, àË`oH$ àíZ 3 A§H$m| H$m h¡ & (vi) IÊS> K – àíZ g§»`m 35 go 37 VH$ ^r XrK©-CÎmar` àH$ma Ho$ àíZ h¢, àË`oH$ àíZ 5 A§H$m| H$m h¡ & (vii) àíZ-nÌ _| H$moB© g_J« {dH$ën Zht h¡ & VWm{n, EH$-EH$ A§H$ Ho$ Xmo àíZm| _|, Xmo-Xmo A§H$m| dmbo Xmo àíZm| _§o, VrZ-VrZ A§H$m| dmbo EH$ àíZ _§o VWm nm±M-nm±M A§H$m| dmbo$ VrZm| àíZm| _§o Am§V[aH$ {dH$ën {X`m J`m h¡ & Eogo àíZm| _| Ho$db EH$ hr {dH$ën H$m CÎma Xr{OE & (viii) BgHo$ A{V[aº$, Amdí`H$VmZwgma, àË`oH$ IÊS> Am¡a àíZ Ho$ gmW `Wmo{MV {ZX}e {XE JE h¢ & (ix) Ho$ëHw$boQ>am| AWdm bm°J Q>o~bm| Ho$ à`moJ H$s AZw_{V Zht h¡ & (x) Ohm± Amdí`H$ hmo, Amn {ZåZ{b{IV ^m¡{VH$ {Z`Vm§H$m| Ho$ _mZm| H$m Cn`moJ H$a gH$Vo h¢ : c = 3 108 m/s h = 6.63 10–34 Js e = 1.6 10–19 C 0 = 4 10–7 T m A–1 0 = 8.854 10–12 C2 N–1 m–2 1 = 9 109 N m2 C–2 4 0 BboŠQ´>m°Z H$m Ðì`_mZ (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–1.55/1/1 2 General Instructions : Read the following instructions very carefully and strictly follow them : (i) This question paper comprises four Sections – A, B, C and D. (ii) There are 37 questions in the question paper. All questions are compulsory. (iii) Section A – Questions no. 1 to 20 are very short answer type questions, carrying one mark each. (iv) Section B – Questions no. 21 to 27 are short answer type questions, carrying two marks each. (v) Section C – Questions no. 28 to 34 are long answer type questions, carrying three marks each. (vi) Section D – Questions no. 35 to 37 are also long answer type questions, carrying five marks each. (vii) There is no overall choice in the question paper. However, an internal choice has been provided in 2 questions of 1 mark, 2 questions of 2 marks, 1 question of three marks and all the 3 questions of five marks. You have to attempt only one of the choices in such questions. (viii) In addition to this, separate instructions are given with each section and question, wherever necessary. (ix) Use of calculators and log tables is not permitted. (x) 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 Avogadro’s number = 6.023 1023 per gram mole Boltzmann constant = 1.38 10–23 JK–1.55/1/1 3 P.T.O. IÊS> H$ ZmoQ> : ZrMo {XE JE àË`oH$ àíZ _| g~go A{YH$ Cn`wº$ {dH$ën Mw{ZE : 1. ¶{X {H$gr ~ÝX n¥îR> go JwµOaZo dmbm ZoQ> {dÚwV² âb³g eyݶ h¡, Vmo Bggo ¶h {ZîH$f© {ZH$mbm Om gH$Vm h¡ {H$ 1 (A) Bg n¥îR> go H$moB© ZoQ> Amdoe n[a~Õ Zht h¡ & (B) Bg n¥îR> Ho$ ^rVa EH$g‘mZ {dÚwV²-joÌ {dÚ‘mZ h¡ & (C) Bg n¥îR> Ho$ ^rVa EH$ {~ÝXþ go Xÿgao {~ÝXþ VH$ {dÚwV² {d^d {dM[aV H$aVm h¡ & (D) n¥îR> Ho$ ^rVa Amdoe CnpñWV h¡ & 2. Xÿar L Ho$ n¥WH$Z VWm + q Am¡a – q Amdoem| go ~Zm H$moB© {dÚwV² {ÛY«wd {H$gr EH$g‘mZ {dÚwV²-joÌ E ‘| ñWm¶r gmå¶mdñWm ‘| h¡ & Bg {ÛY«wd H$s pñWa-d¡ÚwV pñW{VO D$Om© h¡ 1 (A) qLE (B) eyݶ (C) – qLE (D) – 2 qLE 3. H$moB© nmoQ>p¡ Ýe¶mo‘rQ>a {H$gr gob H$m {d.dm. ~b (emf) ‘mn gH$Vm h¡ ³¶m|{H$ 1 (A) nmoQ>¡pÝe¶mo‘rQ>a H$s gwJ«m{hVm A{YH$ hmoVr h¡ & (B) g§VwbZ Ho$ g‘¶ gob go H$moB© Ymam Zht br OmVr h¡ & (C) g§VwbZ Ho$ g‘¶ nmoQ>p¡ Ýe¶mo‘rQ>a Vma go H$moB© Ymam àdm{hV Zht hmoVr h¡ & (D) gob Ho$ AmÝV[aH$ à{VamoY H$s Cnojm H$a Xr OmVr h¡ & 4. {H$gr ~¡Q>ar Ho$ {gam| go 4 Am¡a 6 Ho$ Xmo à{VamoYH$m| R1 Am¡a R2 H$mo nmíd© ‘| g§¶mo{OV {H$¶m J¶m h¡ & BZ XmoZm| à{VamoYH$m| ‘| e{³V j¶ H$m AZwnmV P1 : P2 hmoJm 1 (A) 4:9 (B) 3:2 (C) 9:4 (D) 2:3.55/1/1 4 SECTION A Note : Select the most appropriate option from those given below each question : 1. If the net electric flux through a closed surface is zero, then we can infer 1 (A) no net charge is enclosed by the surface. (B) uniform electric field exists within the surface. (C) electric potential varies from point to point inside the surface. (D) charge is present inside the surface. 2. An electric dipole consisting of charges + q and – q separated by a distance L is in stable equilibrium in a uniform electric field E. The electrostatic potential energy of the dipole is 1 (A) qLE (B) zero (C) – qLE (D) – 2 qEL 3. A potentiometer can measure emf of a cell because 1 (A) the sensitivity of potentiometer is large. (B) no current is drawn from the cell at balance. (C) no current flows in the wire of potentiometer at balance. (D) internal resistance of cell is neglected. 4. Two resistors R1 and R2 of 4 and 6 are connected in parallel across a battery. The ratio of power dissipated in them, P1 : P2 will be 1 (A) 4:9 (B) 3:2 (C) 9:4 (D) 2:3.55/1/1 5 P.T.O. 5. {H$gr Ymamdmhr Hw$ÊS>br H$m Mwå~H$s¶ {ÛY«wd AmKyU© {ZåZ{b{IV ‘| go {H$g na {Z^©a Zht H$aVm h¡ ? 1 (A) Hw$ÊS>br ‘| ’o$am| H$s g§»¶m (B) Hw$ÊS>br H$s AZwàñW-H$mQ> H$m joÌ’$b (C) Hw$ÊS>br ‘| àdm{hV Ymam (D) Hw$ÊS>br Ho$ ’o$am| H$m nXmW© 6. {H$gr IJmobr¶ XÿaXe©H$ Ho$ A{^Ñí¶H$ b|g H$m ~‹S>m ÛmaH$ 1 (A) XÿaXe©H$ H$s {d^oXZ j‘Vm ‘| d¥{Õ H$a XoVm h¡ & (B) à{V{~å~ H$s M‘H$ H$mo H$‘ H$a XoVm h¡ & (C) à{V{~å~ Ho$ gmBµO ‘| d¥{Õ H$a XoVm h¡ & (D) XÿaXe©H$ H$s bå~mB© H$‘ H$a XoVm h¡ & 7. AndV©Zm§H$ 1·47 Ho$ H$m±M go ~Zm H$moB© C^¶moÎmb b|g {H$gr Ðd ‘| Sy>~m h¡ & ¶h b|g AÑí¶ hmo OmVm h¡ Am¡a H$m±M H$s g‘Vb n{Å>H$m H$s ^m±{V ì¶dhma H$aVm h¡ & Bg Ðd H$m AndV©Zm§H$ h¡ 1 (A) 1·47 (B) 1·62 (C) 1·33 (D) 1·51 8. ZrMo {XE JE {H$g dU© Ho$ àH$me Ho$ {bE {H$gr H$m±M Ho$ {àµÁ‘ Ho$ AënV‘ {dMbZ H$moU H$m ‘mZ g~go H$‘ hmoJm ? 1 (A) bmb dU© (B) Zrbm dU© (C) nrbm dU© (D) ham dU© 9. aXaµ’$moS>© ‘m°S>b Ho$ AZwgma {ZåZ{b{IV _| go H$m¡Z-gm H$WZ ghr Zht h¡ ? 1 (A) {H$gr na‘mUw Ho$ ^rVa CgH$m A{YH$m§e ^mJ Imbr h¡ & (B) Zm{^H$ Ho$ Mmam| Amoa Bbo³Q´>m°Z CZ na H$m¶©aV Hy$bm°‘r ~b Ho$ à^md ‘| n[aH«$‘m H$aVo h¢ & (C) na‘mUw H$m A{YH$m§e Ð춑mZ VWm CgH$m Hw$b YZmdoe CgHo$ Ho$ÝÐ na gm§{ÐV hmoVm h¡ & (D) Bg ‘m°S>b Ûmam na‘mUw Ho$ ñWm{¶Ëd H$mo ñWm{nV {H$¶m J¶m &.55/1/1 6 5. The magnetic dipole moment of a current carrying coil does not depend upon 1 (A) number of turns of the coil. (B) cross-sectional area of the coil. (C) current flowing in the coil. (D) material of the turns of the coil. 6. Larger aperture of objective lens in an astronomical telescope 1 (A) increases the resolving power of telescope. (B) decreases the brightness of the image. (C) increases the size of the image. (D) decreases the length of the telescope. 7. A biconvex lens of glass having refractive index 1·47 is immersed in a liquid. It becomes invisible and behaves as a plane glass plate. The refractive index of the liquid is 1 (A) 1·47 (B) 1·62 (C) 1·33 (D) 1·51 8. For a glass prism, the angle of minimum deviation will be smallest for the light of 1 (A) red colour. (B) blue colour. (C) yellow colour. (D) green colour. 9. Which of the following statements is not correct according to Rutherford model ? 1 (A) Most of the space inside an atom is empty. (B) The electrons revolve around the nucleus under the influence of coulomb force acting on them. (C) Most part of the mass of the atom and its positive charge are concentrated at its centre. (D) The stability of atom was established by the model..55/1/1 7 P.T.O. 10. 0·5 eV H$m¶©’$bZ Ho$ {H$gr YmpËdH$ n¥îR> na 1 eV Am¡a 2 eV D$Om©Am| Ho$ µ\$moQ>m°Z H«$‘mJV AmnVZ H$aVo h¢ & BZ XmoZm| àH$aUm| ‘| A{YH$V_ D$Ou¶ àH$m{eH$-Bbo³Q´>m°Zm| H$s J{VO D$Om©Am| H$m AZwnmV hmoJm 1 (A) 1:2 (B) 1:1 (C) 1:3 (D) 1:4 ZmoQ> : Cn`wº$ CÎma go [aº$ ñWmZm| H$s ny{V© H$s{OE : 11. n¥Ïdr na {H$gr ñWmZ na Mwå~H$s¶ joÌ Am¡a Z{V H$moU H«$‘e… 0·3 G Am¡a 30 h¢ & Bg ñWmZ na n¥Ïdr Ho$ Mwå~H$s¶ joÌ Ho$ D$Üdm©Ya KQ>H$ H$m ‘mZ _________ hmoJm & 1 12. {H$gr Q´>mÝg’$m°‘©a Ho$ H«$moS> ‘| _________ YmamAm| H$mo {ZåZV‘ H$aZo Ho$ {bE nQ>{bV bmoho H$s erQ>m| H$m Cn¶moJ {H$¶m OmVm h¡ & 1 13. {H$gr n[aZm{bH$m H$s bå~mB© Am¡a CgH$s AZwàñW-H$mQ> Ho$ joÌ’$b ‘| {~Zm H$moB© n[adV©Z {H$E Cg‘| ’o$am| H$s g§»¶m XþJwZr H$a Xr JB© h¡ & Bg n[aZm{bH$m H$m ñd-àoaH$Ëd _________ JwZm hmo OmEJm & 1 14. ~moa Ho$ na‘mUw ‘m°S>b Ho$ AZwgma Bbo³Q´>m°Z H$s H$jm H$s n[a{Y gX¡d Xo ~«m°½br Va§JX¡¿¶© H$s _________ JwUO hmoVr h¡ & 1 AWdm -j¶ ‘| OZH$ Am¡a g§V{V Zm{^H$m| ‘| _________ H$s g§»¶m g‘mZ hmoVr h¡ & 1 15. {H$gr H$m±M Ho$ g‘~mhþ {àµÁ‘ go JwµOaVr hþB© {H$gr àH$me {H$aU ‘| Cg {àµÁ_ Ho$ H$moU Ho$ ~am~a AënV‘ {dMbZ hmoVm h¡ & Bg {àµÁ‘ Ho$ nXmW© Ho$ AndV©Zm§H$ H$m _mZ _________ h¡ & 1 ZmoQ> : {ZåZ{b{IV Ho$ CÎma Xr{OE : 16. Eopån¶a-‘¡³gdob n[anWr¶ {Z¶‘ Ho$ J{UVr¶ ê$n H$mo {b{IE & 1 17. ‘mXZ gm§ÐVm ‘| d¥{Õ {H$g àH$ma {H$gr p-n g§{Y S>m¶moS> Ho$ õmgr ñVa H$s Mm¡‹S>mB© H$mo à^m{dV H$aVr h¡ ? 1 27 64 18. Al H$s Zm{^H$s¶ {ÌÁ¶m 3·6 ’$‘u h¡ & Cu H$s Zm{^H$s¶ {ÌÁ¶m kmV H$s{OE & 1 13 29 AWdm {H$gr Bbo³Q´>m°Z Am¡a {H$gr àmoQ>m°Z H$s Mmb g‘mZ h¢ & BZgo g§~Õ Xo ~m°½br Va§JX¡¿¶m] H$m AZwnmV kmV H$s{OE & 1.55/1/1 8 10. Photons of energies 1 eV and 2 eV are successively incident on a metallic surface of work function 0·5 eV. The ratio of kinetic energy of most energetic photoelectrons in the two cases will be 1 (A) 1:2 (B) 1:1 (C) 1:3 (D) 1:4 Note : Fill in the blanks with appropriate answer : 11. The magnetic field and angle of dip at a place on the earth are 0·3 G and 30, respectively. The value of vertical component of the earth’s magnetic field at the place is ___________. 1 12. Laminated iron sheets are used to minimize ___________ currents in the core of a transformer. 1 13. The number of turns of a solenoid are doubled without changing its length and area of cross-section. The self-inductance of the solenoid will become ___________ times. 1 14. According to Bohr’s atomic model, the circumference of the electron orbit is always an ___________ multiple of de Broglie wavelength. 1 OR In -decay, the parent and daughter nuclei have the same number of _____________. 1 15. A ray of light on passing through an equilateral glass prism, suffers a minimum deviation equal to the angle of the prism. The value of refractive index of the material of the prism is ___________. 1 Note : Answer the following : 16. Write the mathematical form of Ampere-Maxwell circuital law. 1 17. How does an increase in doping concentration affect the width of depletion layer of a p-n junction diode ? 1 27 64 18. The nuclear radius of Al is 3·6 fermi. Find the nuclear radius of Cu. 1 13 29 OR A proton and an electron have equal speeds. Find the ratio of de Broglie wavelengths associated with them. 1.55/1/1 9 P.T.O. 19. Xmo {d{^Þ àH$me-gwJ«mhr n¥îR>m| M1 Am¡a M2 na Amn{VV àH$me H$s Amd¥{Îm (v) Ho$ gmW {ZamoYr {d^d (Vo) H$m {dMaU AmaoI ‘| Xem©E AZwgma h¡ & BZ‘| go A{YH$ H$m¶©’$bZ dmbo n¥îR> H$s nhMmZ H$s{OE & 1 20. Ñí¶ LED Ho$ {daMZ ‘| h‘ Si Am¡a Ge H$m Cn¶moJ ³¶m| Zht H$a gH$Vo h¢ ? 1 IÊS> I 21. {H$gr ‘rQ>a goVw H$s H$m¶©{d{Y Ho$ {gÕmÝV H$s ì¶m»¶m H$s{OE & BgHo$ Cn¶moJ Ûmam {H$gr AkmV à{VamoY Ho$ ‘mZ H$mo {ZYm©[aV H$aZo Ho$ {bE n[anW AmaoI It{ME & 2 22. {H$gr g‘mÝVa n{Å>H$m g§Ym[aÌ H$s n{Å>H$mAm| Ho$ ~rM Ho$ [a³V ñWmZ H$mo Xmo T>§Jm| go nyU©V… ^am J¶m h¡ & nhbo àH$aU ‘|, Bgo namd¡ÚwVm§H$ K Ho$ JwQ>Ho$ go ^am J¶m h¡ & Xÿgao àH$aU ‘|, Bgo AmaoI ‘| Xem©E AZwgma g‘mZ ‘moQ>mB© Ho$ Xmo JwQ>H$m|, {OZHo$ namd¡ÚwVm§H$ H«$‘e… K1 Am¡a K2 h¢, go ^am J¶m h¡ & XmoZm| hr àH$aUm| ‘| g§ Ym[aÌ H$s Ym[aVm g‘mZ h¡ & K, K1 Am¡a K2 ‘| g§~§Y àmßV H$s{OE & 2 (nhbm àH$aU) (Xÿgam àH$aU) 23. nX ao{S>¶moEop³Q>d nXmW© H$s ‘AY©-Am¶w’ H$s n[a^mfm {b{IE & Xmo {d{^Þ ao{S>¶moEop³Q>d nXmWm] H$s AY©-Am¶w T1 Am¡a T2 VWm {H$gr jU na CZ‘| eof ~Mo hþE na‘mUwAm| H$s g§»¶m H«$‘e… N1 Am¡a N2 h¡ & Cg jU BZH$s g{H«$¶VmAm| H$m AZwnmV kmV H$s{OE & 2.55/1/1 10 19. The variation of the stopping potential (Vo) with the frequency (v) of the light incident on two different photosensitive surfaces M1 and M2 is shown in the figure. Identify the surface which has greater value of the work function. 1 20. Why cannot we use Si and Ge in fabrication of visible LEDs ? 1 SECTION B 21. Explain the principle of working of a meter bridge. Draw the circuit diagram for determination of an unknown resistance using it. 2 22. The space between the plates of a parallel plate capacitor is completely filled in two ways. In the first case, it is filled with a slab of dielectric constant K. In the second case, it is filled with two slabs of equal thickness and dielectric constants K1 and K2 respectively as shown in the figure. The capacitance of the capacitor is same in the two cases. Obtain the relationship between K, K1 and K2. 2 (Case 1) (Case 2) 23. Define the term ‘Half-life’ of a radioactive substance. Two different radioactive substances have half-lives T1 and T2 and number of undecayed atoms at an instant, N1 and N2, respectively. Find the ratio of their activities at that instant. 2.55/1/1 11 P.T.O. 24. J{Verb Va§J Ho$ Va§JmJ« H$s n[a^mfm {b{IE & hmBJoÝg {gÕmÝV H$m Cn¶moJ H$aHo$, Cg pñW{V ‘| O~ àH$me {H$gr gKZ ‘mܶ‘ go {dab ‘mܶ‘ ‘| J‘Z H$aVm h¡, {H$gr g‘Vb AÝVamn¥îR> na AndV©Z H$m {Z¶‘ àmßV H$s{OE & 2 AWdm 1 1 1 b|g _oH$a gyÌ H$m Cn`moJ H$aHo$ {H$gr nVbo C^`moÎmb b|g Ho$ gyÌ – H$mo f v u ì`wËnÞ H$s{OE & 2 25. AmaoI ‘| Xem©E AZwgma Xmo bå~o grYo g‘mÝVa Vma A Am¡a B _|, Omo EH$-Xÿgao go d Xÿar na pñWV h¢, go g‘mZ {Xem ‘| g‘mZ Ymam I àdm{hV hmo ahr h¡ & (a) BZ Vmam| Ho$ ~rM {H$gr EH$ Vma go Xÿar x na pñWV {H$gr {~ÝXþ P na Mwå~H$s¶ joÌ kmV H$s{OE & (b) Xÿar x Ho$ gmW, 0 < x < d Ho$ {bE, Mwå~H$s¶ joÌ ‘| {dMaU H$mo Xem©Zo Ho$ {bE J«m’$ It{ME & 2 26. ~moa Ho$ na‘mUw ‘m°S>b H$m Cn¶moJ H$aHo$, hmBS´>moOZ na‘mUw H$s ndt H$jm ‘| n[aH«$‘m H$aVo hþE Bbo³Q´>m°Z H$s {ÌÁ¶m Ho$ {bE 춧OH$ ì¶wËnÞ H$s{OE & 2 AWdm (a) àH$me-{dÚwV² à^md à¶moJ Ho$ CZ Xmo ‘w»¶ àojUm| H$mo {b{IE {OZH$s ì¶m»¶m Ho$db AmB§ñQ>mBZ H$s àH$me-{dÚwV² g‘rH$aU Ûmam hr H$s Om gHo$ & (b) {H$gr àH$me gob Ho$ EoZmoS> {d^d Ho$ gmW àH$me {dÚwV² Ymam Ho$ {dMaU H$mo Xem©Zo Ho$ {bE J«m’$ It{ME & 2 27. {H$gr p-n g§{Y S>m¶moS> Ho$ {bE nX ‘õmgr ñVa’ Am¡a ‘{d^d àmMra’ H$s ì¶m»¶m H$s{OE & O~ {H$gr p-n g§{Y H$mo AJ«{X{eH$ ~m¶g H$aVo h¢, Vmo (a) õmgr ñVa H$s Mm¡‹S>mB©, VWm (b) {d^d àmMra H$m ‘mZ {H$g àH$ma à^m{dV hmoVm h¡ ? 2.55/1/1 12 24. Define wavefront of a travelling wave. Using Huygens principle, obtain the law of refraction at a plane interface when light passes from a denser to rarer medium. 2 OR 1 1 1 Using lens maker’s formula, derive the thin lens formula – for a f v u biconvex lens. 2 25. Two long straight parallel wires A and B separated by a distance d, carry equal current I flowing in same direction as shown in the figure. (a) Find the magnetic field at a point P situated between them at a distance x from one wire. (b) Show graphically the variation of the magnetic field with distance x for 0 < x < d. 2 26. Using Bohr’s atomic model, derive the expression for the radius of nth orbit of the revolving electron in a hydrogen atom. 2 OR (a) Write two main observations of photoelectric effect experiment which could only be explained by Einstein’s photoelectric equation. (b) Draw a graph showing variation of photocurrent with the anode potential of a photocell. 2 27. Explain the terms ‘depletion layer’ and ‘potential barrier’ in a p-n junction diode. How are the (a) width of depletion layer, and (b) value of potential barrier affected when the p-n junction is forward biased ? 2.55/1/1 13 P.T.O. IÊS> J 28. (a) Xmo gobm| Ho$ {d.dm. ~b (emf) E1 Am¡a E2 VWm BZHo$ AmÝV[aH$ à{VamoY H«$‘e… r1 Am¡a r2 h¢ & O~ BZHo$ nmíd© g§¶moOZ H$mo {H$gr ~mø à{VamoY R go g§¶mo{OV {H$¶m OmVm h¡, BZHo$ Vwë¶ {d.dm. ~b (emf) VWm Am§V[aH$ à{VamoY Ho$ {bE 춧OH$ ì¶wËnÞ H$s{OE & ¶h ‘m{ZE {H$ XmoZm| gob EH$-Xÿgao H$s ghm¶Vm H$a aho h¢ & (b) Cg àH$aU ‘| O~ XmoZm| gob gd©g‘ h¢ Am¡a à˶oH$ H$m {d.dm. ~b (emf) E = 5 V VWm Am§V[aH$ à{VamoY r = 2 h¡, R = 10 Ho$ ~mø à{VamoY Ho$ {gam| na dmoëQ>Vm n[aH${bV H$s{OE & 3 29. (a) {H$gr Ymamdmhr d¥ÎmmH$ma Hw$ÊS>br, {OgH$s {ÌÁ¶m r VWm ’o$am| H$s g§»¶m N h¡, go Ymam (I) àdm{hV hmo ahr h¡ & Bg Hw$ÊS>br go g§~Õ Mwå~H$s¶ joÌ Ho$ {bE 춧OH$ {b{IE & (b) ¶h ‘m{ZE {H$ Cn`w©º$ Hw$ÊS>br H$mo YZ Vb ‘| BgHo$ Ho$ÝÐ H$mo ‘yb-{~ÝXþ na aIVo hþE pñWV {H$¶m J¶m h¡ & {~ÝXþ (x, 0, 0) na Bg Hw$ÊS>br Ho$ H$maU CËnÞ Mwå~H$s¶ joÌ Ho$ _mZ Ho$ {bE 춧OH$ ì¶wËnÞ H$s{OE & 3 AWdm (a) {H$gr J¡ëdoZmo‘rQ>a H$s Ymam gwJ«m{hVm H$s n[a^mfm Xr{OE Am¡a BgHo$ {bE 춧OH$ {b{IE & (b) {H$gr J¡ëdoZmo‘rQ>a H$m à{VamoY G Am¡a BgH$s nyU© n¡‘mZm {djonU Ymam Ig h¡ & (i) Bg J¡ëdoZmo‘rQ>a H$mo I0 (I0 > Ig) VH$ H$s Ymam ‘mn gH$Zo dmbo Eo‘rQ>a ‘| {H$g àH$ma n[ad{V©V {H$¶m Om gH$Vm h¡ ? (ii) Bg Eo‘rQ>a H$m à^mdr à{VamoY ³¶m h¡ ? 3 30. V = V0 sin t Ho$ {H$gr òmoV go {H$gr à{VamoY R Am¡a g§Ym[aÌ C H$mo loUr ‘| g§¶mo{OV {H$¶m J¶m h¡ & (a) (i) à{VamoY Ho$ {gam| Am¡a (ii) g§Ym[aÌ Ho$ {gam| na {eIa dmoëQ>Vm H$m ‘mZ kmV H$s{OE & (b) AZwà¶w³V dmoëQ>Vm Am¡a Ymam Ho$ ~rM H$bmÝVa kmV H$s{OE & BZ‘| go H$m¡Z AJ« h¡ ? 3 31. {ZåZ{b{IV à˶oH$ àMmbZ Ho$ H$maU ¶§J Ho$ {Û{Par à¶moJ ‘| ì¶{VH$aU {’«$ÝOm| na ³¶m à^md hmoJm ? AnZo CÎmam| H$s nw{îQ> H$s{OE & 3 (a) nX} H$mo {P[a¶m| Ho$ Vb go Xÿa bo Om¶m J¶m h¡ & (b) {P[a¶m| Ho$ ~rM Ho$ n¥WH$Z ‘| d¥{Õ H$a Xr JB© h¡ & (c) òmoV {Par H$mo {Û-{Par Ho$ Vb Ho$ {ZH$Q> bm¶m J¶m h¡ &.55/1/1 14 SECTION C 28. (a) Two cells of emf E1 and E2 have their internal resistances r1 and r2, respectively. Deduce an expression for the equivalent emf and internal resistance of their parallel combination when connected across an external resistance R. Assume that the two cells are supporting each other. (b) In case the two cells are identical, each of emf E = 5 V and internal resistance r = 2 , calculate the voltage across the external resistance R = 10 . 3 29. (a) Write an expression of magnetic moment associated with a current (I) carrying circular coil of radius r having N turns. (b) Consider the above mentioned coil placed in YZ plane with its centre at the origin. Derive expression for the value of magnetic field due to it at point (x, 0, 0). 3 OR (a) Define current sensitivity of a galvanometer. Write its expression. (b) A galvanometer has resistance G and shows full scale deflection for current Ig. (i) How can it be converted into an ammeter to measure current up to I0 (I0 > Ig) ? (ii) What is the effective resistance of this ammeter ? 3 30. A resistance R and a capacitor C are connected in series to a source V = V0 sin t. Find : (a) The peak value of the voltage across the (i) resistance and (ii) capacitor. (b) The phase difference between the applied voltage and current. Which of them is ahead ? 3 31. What is the effect on the interference fringes in Young’s double slit experiment due to each of the following operations ? Justify your answers. 3 (a) The screen is moved away from the plane of the slits. (b) The separation between slits is increased. (c) The source slit is moved closer to the plane of double slit..55/1/1 15 P.T.O. 32. (a) Amno{jH$ {dÚwV²erbVm r VWm Amno{jH$ Mwå~H$erbVm r Ho$ {H$gr Ðì¶mË‘H$ ‘mܶ‘ ‘| àH$me H$s Mmb Ho$ {bE 춧OH$ {b{IE & (b) {ZåZ{b{IV ‘| Cn¶moJ hmoZo dmbr {dÚwV²-Mwå~H$s¶ Va§Jm| Ho$ Zm‘ Am¡a Va§JX¡¿¶© n[aga {b{IE : (i) aoS>ma àUm{b¶m| ‘| {d‘mZ MmbZ (nW-àXe©Z) ‘| (ii) µ\$gbm| H$s d¥{Õ Ho$ àojU Ho$ {bE n¥Ïdr Ho$ CnJ«hm| _| 3 33. Zm{^H$ 235 92 Y Omo Amaå^ ‘| {dam‘ ‘| h¡, EH$ -H$U H$mo CËg{O©V H$aHo$ 231 90 X ‘| Anj{¶V hmo OmVm h¡ & 235 231 4 Y X He + D$Om© 92 90 2 OZH$ Zm{^H$, g§V{V Zm{^H$ Am¡a -H$U H$s ~§YZ D$Om© à{V ݶyp³bAm°Z H«$‘e… 7·8 MeV, 7·835 MeV Am¡a 7·07 MeV h¢ & ¶h nyd©YmaUm aIVo hþE {H$ ~ZZo dmbm g§V{V Zm{^H$ CÎmo{OV AdñWm ‘| Zht h¡ VWm A{^{H«$¶m H$s D$Om© ‘| CgH$s ^mJrXmar H$s Cnojm H$aVo hþE CËg{O©V -H$U H$s Mmb kmV H$s{OE & 3 (-H$U H$m Ð춑mZ = 6·68 10–27 kg) 34. (a) {H$gr µOoZa S>m¶moS> Ho$ I-V A{^bmj{UH$ H$s ghm¶Vm go, n[anW AmaoI ItMH$a, BgH$s dc dmoëQ>Vm {Z¶§ÌH$ H$s ^m±{V H$m¶©{d{Y H$s ì¶m»¶m H$s{OE & (b) {H$gr µOoZa S>m¶moS> Ho$ p- Am¡a n-’$bH$m| H$m A˶{YH$ ‘mXZ H$aZo H$m ³¶m CÔoí¶ h¡ ? 3 IÊS> K 35. (a) JmCg {Z¶‘ H$m Cn¶moJ H$aVo hþE, R {ÌÁ¶m Ho$ EH$g‘mZ Amdoe {dVaU Ho$ Jmobr¶ Imob Ho$ H$maU BgHo$ Ho$ÝÐ go Xÿar x Ho$ {H$gr {~ÝXþ na {dÚwV²-joÌ Ho$ {bE 춧OH$ ì¶wËnÞ H$s{OE, O~{H$ (i) 0 < x < R, Am¡a (ii) x > R..55/1/1 16 32. (a) Write the expression for the speed of light in a material medium of relative permittivity r and relative magnetic permeability r. (b) Write the wavelength range and name of the electromagnetic waves which are used in (i) radar systems for aircraft navigation, and (ii) Earth satellites to observe the growth of the crops. 3 235 231 33. The nucleus Y , initially at rest, decays into X by emitting an 92 90 -particle 235 231 4 Y X He + energy. 92 90 2 The binding energies per nucleon of the parent nucleus, the daughter nucleus and -particle are 7·8 MeV, 7·835 MeV and 7·07 MeV, respectively. Assuming the daughter nucleus to be formed in the unexcited state and neglecting its share in the energy of the reaction, find the speed of the emitted -particle. (Mass of -particle = 6·68 10–27 kg) 3 34. (a) Draw circuit diagram and explain the working of a zener diode as a dc voltage regulator with the help of its I-V characteristic. (b) What is the purpose of heavy doping of p- and n-sides of a zener diode ? 3 SECTION D 35. (a) Using Gauss law, derive expression for electric field due to a spherical shell of uniform charge distribution and radius R at a point lying at a distance x from the centre of shell, such that (i) 0 < x < R, and (ii) x > R..55/1/1 17 P.T.O. (b) H$moB© {dÚwV²-joÌ EH$g‘mZ h¡ VWm YZmË‘H$ x Ho$ joÌ ‘| ¶h + x {Xem Ho$ AZw{Xe H$m¶©aV h¡ & naÝVw BgHo$ gmW-gmW G UmË‘H$ x Ho$ joÌ ‘| – x {Xem Ho$ AZw{Xe ¶h g‘mZ n[a‘mU Ho$ gmW H$m¶©aV h¡ & Bg joÌ H$m ‘mZ E = 200 N/C x > 0 Ho$ {bE VWm E = – 200 N/C x < 0 Ho$ {bE h¡ & bå~mB© 20 cm Am¡a {ÌÁ¶m 5 cm H$m H$moB© bå~-d¥Îmr` {g{bÊS>a x-Aj Ho$ AZw{Xe Bg àH$ma pñWV h¡ {H$ BgH$m Ho$ÝÐ ‘yb-{~ÝXþ na VWm EH$ g_Vb ’$bH$ x = + 10 cm Am¡a Xÿgam ’$bH$ x = – 10 cm na h¡ & kmV H$s{OE : (i) {g{bÊS>a go hmoH$a JwµOaZo dmbm ZoQ> ~{h‘w©Ir âb³g & (ii) {g{bÊS>a Ho$ ^rVa CnpñWV Hw$b Amdoe & 5 AWdm (a) {H$gr ~mø {dÚwV²-joÌ E ‘| r Am¡a r na H«$‘e… pñWV Xmo {~ÝXþ Amdoem| 1 2 q1 Am¡a q2 Ho$ {ZH$m¶ H$s pñW{VO D$Om© Ho$ {bE 춧OH$ àmá H$s{OE & (b) {H$gr {d¶w³V {~ÝXþ Amdoe (– q) Ho$ H$maU g‘{d^d n¥îR> It{ME VWm BgH$s {dÚwV²-joÌ aoImE± {M{ÌV H$s{OE & (c) + 1 C, – 1 C Am¡a + 2 C Ho$ VrZ {~ÝXþ Amdoe Amaå^ ‘| EH$-Xÿgao go AZÝV Xÿar na h¢ & BZ g^r Amdoem| H$mo 10 cm ^wOm Ho$ {H$gr g‘~mhþ {Ì^wO Ho$ erfm] na ñWm{nV H$aZo ‘| {H$¶m J¶m Hw$b H$m¶© n[aH${bV H$s{OE & 5 36. (a) {H$gr J¡ëdoZmo‘rQ>a H$s Am¶VmH$ma Ymamdmhr Hw$ÊS>br na H$m¶©aV ~b-AmKyU© Ho$ {bE 춧OH$ ì¶wËnÞ H$s{OE & Mwå~H$s¶ joÌ H$mo Aar¶ ³¶m| ~Zm¶m OmVm h¡ ? (b) {H$gr -H$U H$mo 10 kV Ho$ {d^dmÝVa VH$ Ëd[aV {H$¶m J¶m h¡ VWm ¶h x-Aj Ho$ AZw{Xe J{V‘mZ h¡ & ¶h y-Aj Ho$ AZw{Xe H$m¶©aV B = 2 10–3 T Ho$ EH$g‘mZ Mwå~H$s¶ joÌ ‘| àdoe H$aVm h¡ & BgHo$ nW H$s {ÌÁ¶m kmV H$s{OE & (-H$U H$m Ð춑mZ = 6·4 10–27 kg br{OE) 5 AWdm (a) Zm‘m§{H$V AmaoI H$s ghm¶Vm go {H$gr CÀMm¶r Q´>mÝg’$m°‘©a H$s H$m¶©{d{Y H$s ì¶m»¶m H$s{OE & {ZåZ{b{IV H$s ì¶m»¶m Ho$ {bE H$maU Xr{OE : (i) Q´>mÝg’$m°‘©a H$m H«$moS> nQ>{bV hmoVm h¡ & (ii) bnoQ>m| ‘| Vm±~o (H$m°na) Ho$ ‘moQ>o Vma H$m Cn¶moJ {H$¶m OmVm h¡ &.55/1/1 18 (b) An electric field is uniform and acts along + x direction in the region of positive x. It is also uniform with the same magnitude but acts in – x direction in the region of negative x. The value of the field is E = 200 N/C for x > 0 and E = – 200 N/C for x < 0. A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the x-axis so that one flat face is at x = + 10 cm and the other is at x = – 10 cm. Find : (i) The net outward flux through the cylinder. (ii) The net charge present inside the cylinder. 5 OR (a) Find the expression for the potential energy of a system of two point charges q1 and q2 located at r and r , respectively in an 1 2 external electric field E. (b) Draw equipotential surfaces due to an isolated point charge (– q) and depict the electric field lines. (c) Three point charges + 1 C, – 1 C and + 2 C are initially infinite distance apart. Calculate the work done in assembling these charges at the vertices of an equilateral triangle of side 10 cm. 5 36. (a) Derive the expression for the torque acting on the rectangular current carrying coil of a galvanometer. Why is the magnetic field made radial ? (b) An -particle is accelerated through a potential difference of 10 kV and moves along x-axis. It enters in a region of uniform magnetic field B = 2 10–3 T acting along y-axis. Find the radius of its path. (Take mass of -particle = 6·4 10–27 kg ) 5 OR (a) With the help of a labelled diagram, explain the working of a step-up transformer. Give reasons to explain the following : (i) The core of the transformer is laminated. (ii) Thick copper wire is used in windings..55/1/1 19 P.T.O. (b) à{VamoY 0·1 H$s 20 cm bå~r H$moB© MmbH$ N>‹S> PQ CnojUr¶ à{VamoY H$s Xmo {MH$Zr g‘mÝVa nQ>[a¶m| AA Am¡a CC na pñWV h¡ & ¶h N>‹S> BZ nQ>[a¶m| na gaH$ gH$Vr h¡ VWm ¶h ì¶dñWm EH$g_mZ Mwå~H$s¶ joÌ B = 0·4 T CËnÞ H$aZo dmbo ñWm¶r Mwå~H$ Ho$ Y«wdm| Ho$ ~rM aIr JB© h¡ & AmaoI ‘| Xem©E AZwgma nQ>[a¶m±, N>‹S> VWm Mwå~H$s¶ joÌ VrZ nañna bå~dV² {XemAm| ‘| h¢ & ¶{X nQ>[a¶m| Ho$ {gam| A Am¡a C H$m bKwnWZ H$a {X¶m OmE, Vmo kmV H$s{OE (i) Bg N>‹S> H$mo v = 10 cm/s Ho$ EH$g‘mZ doJ go J{V H$amZo Ho$ {bE Amdí¶H$ ~mø ~b, Am¡a (ii) Eogm H$aZo Ho$ {bE Amdí¶H$ e{³V & 5 37. (a) Cg pñW{V Ho$ {bE {H$gr IJmobr¶ XÿaXe©H$ H$m {H$aU AmaoI It{ME {Og‘| A§{V‘ à{V{~å~ AZÝV na ~ZVm h¡ & Bg XÿaXe©H$ H$s {d^oXZ j‘Vm Ho$ {bE 춧OH$ {b{IE & (b) {H$gr IJmobr¶ XÿaXe©H$ Ho$ A{^Ñí¶H$ b|g H$s ’$moH$g Xÿar 20 m Am¡a BgH$s Zo{ÌH$m H$s ’$moH$g Xÿar 1 cm h¡ & (i) Bg XÿaXe©H$ H$m H$moUr¶ AmdY©Z kmV H$s{OE & (ii) ¶{X Bg XÿaXe©H$ H$m Cn¶moJ MÝБm H$mo XoIZo ‘| {H$¶m OmVm h¡, Vmo A{^Ñí¶H$ b|g Ûmam ~Zo à{V{~å~ H$m ì¶mg kmV H$s{OE & {X¶m J`m h¡ {H$ MÝБm H$m ì¶mg 3·5 106 m VWm MÝБm H$s H$jm H$s {ÌÁ¶m 3·8 108 m h¡ & 5 AWdm.55/1/1 20 (b) A conducting rod PQ of length 20 cm and resistance 0·1 rests on two smooth parallel rails of negligible resistance AA and CC. It can slide on the rails and the arrangement is positioned between the poles of a permanent magnet producing uniform magnetic field B = 0·4 T. The rails, the rod and the magnetic field are in three mutually perpendicular directions as shown in the figure. If the ends A and C of the rails are short circuited, find the (i) external force required to move the rod with uniform velocity v = 10 cm/s, and (ii) power required to do so. 5 37. (a) Draw the ray diagram of an astronomical telescope when the final image is formed at infinity. Write the expression for the resolving power of the telescope. (b) An astronomical telescope has an objective lens of focal length 20 m and eyepiece of focal length 1 cm. (i) Find the angular magnification of the telescope. (ii) If this telescope is used to view the Moon, find the diameter of the image formed by the objective lens. Given the diameter of the Moon is 3·5 106 m and radius of lunar orbit is 3·8 108 m. 5 OR.55/1/1 21 P.T.O. (a) H$moB© {~å~ {H$gr AdVb Xn©U Ho$ gm‘Zo pñWV h¡ & àojU H$aZo na ¶h nm¶m OmVm h¡ {H$ {~å~ H$m Am^mgr à{V{~å~ ~Zm h¡ & à{V{~å~ ~ZZm Xem©Zo Ho$ {bE {H$aU 1 1 1 AmaoI It{ME VWm Bggo Xn©U g‘rH$aU, ì¶wËnÞ H$s{OE & f u v (b) H$moB© {~å~ {H$gr g‘Vb-CÎmb b|g Ho$ gm‘Zo, {OgHo$ Jmobr¶ n¥îR> H$s dH«$Vm {ÌÁ¶m 20 cm h¡, Ho$ gm‘Zo 30 cm Xÿar na pñWV h¡ & ¶{X b|g Ho$ nXmW© H$m AndV©Zm§H$ 1·5 h¡, Vmo ~Zo à{V{~å~ H$s pñW{V Am¡a àH¥${V kmV H$s{OE & 5.55/1/1 22 (a) An object is placed in front of a concave mirror. It is observed that a virtual image is formed. Draw the ray diagram to show the 1 1 1 image formation and hence derive the mirror equation . f u v (b) An object is placed 30 cm in front of a plano-convex lens with its spherical surface of radius of curvature 20 cm. If the refractive index of the material of the lens is 1·5, find the position and nature of the image formed. 5.55/1/1 23 P.T.O.
