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. 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$m
g_{Xm J`m h¡ & àíZ-nÌ H$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
CBSE Class 12 Physics Question Paper 2020
Set 55 / 1 /
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 Jm 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¢, àËoH$ àíZ **_1_** A§H$ H$m h¡ & _(iv)_ **IÊS> I** _–_ àíZ g§»m 21 go 27 VH$ bKw-CÎmaràíZ h¢, àËoH$ 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¢, àËoH$ àíZ **_3_** A§H$m| H$m h¡ & _(vi)_ **IÊS> K** _–_ àíZ g§»m 35 go 37 VH$ ^r XrK©-CÎmaràH$ma Ho$ àíZ h¢, àËoH$ 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$ën {Xm Jm h¡ & Eogo àíZm| | Ho$db EH$ hr {dH$ën H$m CÎma Xr{OE & (viii) BgHo$ A{V[aº$, AmdíH$VmZwgma, àËoH$ 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$ àmoJ H$s AZw{V Zht h¡ & (x) Ohm± AmdíH$ hmo, Amn {ZåZ{b{IV ^m¡{VH$ {ZVm§H$m| Ho$ _mZm| H$m Cn`moJ H$a gH$Vo h¢ :
c = 3 108 m/s h = 6.63 10 –^34 Js e = 1.6 10 –^19 C 0 = 4 10 –^7 T m A–^1 0 = 8.854 10 –^12 C^2 N–^1 m–^2
= 9 109 N m^2 C–^2
BboŠQ´>m°Z H$m Ðì_mZ (me) = 9.1 10 –^31 kg ÝyQ´>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 {ZVm§H$ = 1.38 10 –^23 JK–^1
IÊS> H$
ZmoQ> : ZrMo {XE JE àËoH$ àíZ _| g~go A{YH$ Cnwº$ {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$moB© nmoQ>¡pÝe¶mo‘rQ>a {H$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>¡pÝe¶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$m| R 1 Am¡a R 2 H$mo nmíd© ‘| g§¶mo{OV
{H$¶m J¶m h¡ & BZ XmoZm| à{VamoYH$m| ‘| e{³V j¶ H$m AZwnmV P 1 : P 2 hmoJm 1 (A) 4 : 9 (B) 3 : 2 (C) 9 : 4 (D) 2 : 3
. 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 R 1 and R 2 of 4 and 6 are connected in parallel across a battery. The ratio of power dissipated in them, P 1 : P 2 will be 1 (A) 4 : 9 (B) 3 : 2 (C) 9 : 4 (D) 2 : 3
. 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· (B) 1· (C) 1· (D) 1· 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.
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$aVr h¡? 1 18.^2713 Al H$s Zm{^H$s¶ {ÌÁ¶m 3·6 ’$‘u h¡ & 6429 Cu H$s Zm{^H$s¶ {ÌÁ¶m kmV H$s{OE & 1
AWdm {H$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
19. Xmo {d{^Þ àH$me-gwJ«mhr n¥îR>m| M 1 Am¡a M 2 na Amn{VV àH$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«$‘e… K 1 Am¡a K 2 h¢, go ^am J¶m h¡ & XmoZm| hr àH$aUm| ‘| g§Ym[aÌ H$s Ym[aVm g‘mZ h¡ & K, K 1 Am¡a K 2 ‘| g§~§Y àmßV H$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$s AY©-Am¶w T 1 Am¡a T 2 VWm {H$gr jU na CZ‘| eof ~Mo hþE na‘mUwAm| H$s g§»¶m H«$‘e… N 1 Am¡a N 2 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 M 1 and M 2 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 K 1 and K 2 respectively as shown in the figure. The capacitance of the capacitor is same in the two cases. Obtain the relationship between K, K 1 and K 2. 2
(Case 1) (Case 2)
23. Define the term ‘Half-life’ of a radioactive substance. Two different radioactive substances have half-lives T 1 and T 2 and number of undecayed atoms at an instant, N 1 and N 2 , respectively. Find the ratio of their activities at that instant. 2
. 55/1/1 13 P.T.O.
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 Using lens maker’s formula, derive the thin lens formula u
–^1
v
f 1 for a 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
IÊS> J
28. (a) Xmo gobm| Ho$ {d.dm. ~b (emf) E 1 Am¡a E 2 VWm BZHo$ AmÝV[aH$ à{VamoY H«$‘e… r 1 Am¡a r 2 h¢ & O~ BZHo$ 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 I 0 (I 0 > 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$m à^mdr à{VamoY ³¶m h¡? 3 30. V = V 0 sin t Ho$ {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$Q> bm¶m J¶m h¡ &
32. (a) Amno{jH$ {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$ 23592 Y Omo Amaå^ ‘| {dam‘ ‘| h¡, EH$ -H$U H$mo CËg{O©V H$aHo$ 23190 X ‘|
Anj{¶V hmo OmVm h¡ &
Y X He 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 23592 Y, initially at rest, decays into 23190 X by emitting an
-particle
Y X He 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 19 P.T.O.
(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 q 1 and q 2 located at r 1 and r 2 , respectively in an 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.
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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
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