Cbse physics Sample paper 2020-21, Assignments of Physics

Download Cbse physics Sample paper 2020-21 and more Assignments Physics in PDF only on Docsity! .55/1/1 1 P.T.O. H$moS> Z§. Code No. amob Z§. Roll No. ZmoQ> NOTE (I) H¥$n`m Om±M H$a b| {H$ Bg àíZ-nÌ _o§ _w{ÐV n¥ð> 23 h¢ & (I) Please check that this question paper contains 23 printed pages. (II) àíZ-nÌ _| Xm{hZo hmW H$s Amoa {XE JE H$moS >Zå~a H$mo N>mÌ CÎma-nwpñVH$m Ho$ _wI-n¥ð> na {bI| & (II) Code number given on the right hand side of the question paper 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 _| >37 àíZ h¢ & (III) Please check that this question paper contains 37 questions. (IV) H¥$n`m àíZ H$m CÎma {bIZm ewê$ H$aZo go nhbo, CÎma-nwpñVH$m _| àíZ H$m H«$_m§H$ Adí` {bI| & (IV) Please write down the Serial Number of the question in the answer-book before attempting it. (V) Bg àíZ-nÌ H$mo n‹T>Zo Ho$ {bE 15 {_ZQ >H$m g_` {X`m J`m h¡ & àíZ-nÌ H$m {dVaU nydm©• _| 10.15 ~Oo {H$`m OmEJm & 10.15 ~Oo go 10.30 ~Oo VH$ N>mÌ Ho$db àíZ-nÌ H$mo n‹T>|Jo Am¡a Bg Ad{Y Ho$ Xm¡amZ do CÎma-nwpñVH$m na H$moB© CÎma Zht {bI|Jo & (V) 15 minute time has been allotted to read this question paper. The question paper will be distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the students will read the question paper only and will not 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 CBSE Class 12 Physics Question Paper 2020 Set 5 /1/1 .55/1/1 2 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$Vo h¢ : c = 3  108 m/s h = 6.63  10–34 Js e = 1.6  10–19 C 0 = 4  10 –7 T m A–1 0 = 8 .854  10–12 C2 N–1 m–2 0 4 1  = 9  109 N m2 C–2 BboŠQ´>m°Z H$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 5 P.T.O. 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 6 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 H o$ ’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$s {d ôXZ j‘Vm ‘| d¥{Õ H$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 7 P.T.O. 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 10 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 11 P.T.O. 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 12 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 b|g _oH$a gyÌ H$m Cn`moJ H$aHo$ {H$gr nVbo C^`moÎmb b|g Ho$ gyÌ u 1 – v 1 f 1  H$mo ì`wËnÞ H$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$ma à^m{dV hmoVm h¡ ? 2 .55/1/1 15 P.T.O. SECTION C 28. (a) Two cells of emf E1 and E2 have their internal resistances r1 and r2, respectively. Deduce an expression for the equivalent emf and internal resistance of their parallel combination when connected across an external resistance R. Assume that the two cells are supporting each other. (b) In case the two cells are identical, each of emf E = 5 V and internal resistance r = 2 , calculate the voltage across the external resistance R = 10 . 3 29. (a) Write an expression of magnetic moment associated with a current (I) carrying circular coil of radius r having N turns. (b) Consider the above mentioned coil placed in YZ plane with its centre at the origin. Derive expression for the value of magnetic field due to it at point (x, 0, 0). 3 OR (a) Define current sensitivity of a galvanometer. Write its expression. (b) A galvanometer has resistance G and shows full scale deflection for current Ig. (i) How can it be converted into an ammeter to measure current up to I0 (I0 > Ig) ? (ii) What is the effective resistance of this ammeter ? 3 30. A resistance R and a capacitor C are connected in series to a source V = V0 sin t. Find : (a) The peak value of the voltage across the (i) resistance and (ii) capacitor. (b) The phase difference between the applied voltage and current. Which of them is ahead ? 3 31. What is the effect on the interference fringes in Young’s double slit experiment due to each of the following operations ? Justify your answers. 3 (a) The screen is moved away from the plane of the slits. (b) The separation between slits is increased. (c) The source slit is moved closer to the plane of double slit. .55/1/1 16 32. (a) Amno{jH$ {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 {dÚwV ²-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$ Y235 92 Omo Amaå^ ‘| {dam‘ ‘| h¡, EH$ -H$U H$mo CËg{O©V H$aHo$ X231 90 ‘| Anj{¶V hmo OmVm h¡ & HeXY 4 2 231 90 235 92  + D$Om© OZH$ 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 17 P.T.O. 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 33. The nucleus Y 235 92 , initially at rest, decays into X 231 90 by emitting an -particle HeXY 4 2 231 90 235 92  + energy. 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 20 (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 Ama oI 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 21 P.T.O. (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 22 (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$aU AmaoI It{ME VWm Bggo Xn©U g‘rH$aU, v 1 u 1 f 1  ì¶wËnÞ H$s{OE & (b) H$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