narjmWu H$moS >H$mo CÎma
-nwpñVH$m Ho$ _wI
-n¥ð
>na Adí` {bIo§ &
Candidates must write the Code on the title page of the answer-book.
Series OSR
H$moS> Z§.
42/1 (SPL)
Code No.
amob Z§.
Roll No.
^m¡{VH$ {dkmZ (g¡ÕmpÝVH$)
PHYSICS (Theory)
{ZYm©[aV g_`
: 3KÊQ>o
A{YH$V_ A§H$
: 70Time allowed : 3 hours Maximum Marks : 70
H¥$n`m Om±M H$a b| {H$ Bg àíZ
-nÌ _o§ _w{ÐV n¥ð>
16h¢ &
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| &
H¥$n`m Om±M H$a b| {H$ Bg àíZ
-nÌ _| >
30àíZ h¢ &
H¥$n`m àíZ H$m CÎma {bIZm ewê$ H$aZo go nhbo, àíZ H$m H«$_m§H$ Adí` {bI| &
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 &
Please check that this question paper contains 16 printed pages.
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.
Please check that this question paper contains 30 questions.
Please write down the Serial Number of the question before
attempting it.
15 minutes 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.
gm_mÝ` {ZX}e
:(i)
g^r àíZ A{Zdm`© h¢ &
(ii)
Bg àíZ
-nÌ _| Hw$b
30àíZ h¢ & àíZ g§.
1go
8VH$ Ho$ àíZ A{V
-bKwCÎmar` àíZ h¢
Am¡a àË`oH$ EH$ A§H$ H$m h¡ &
(iii)
àíZ g§.
9go
18_| àË`oH$ àíZ Xmo A§H$ H$m h¡, àíZ g§.
19go
27_| àË`oH$ àíZ VrZ
A§H$ H$m h¡ Am¡a àíZ g§.
28go
30_| àË`oH$ àíZ nm±M A§H$ H$m h¡ &
(iv)
VrZ A§H$m| dmbo àíZm| _| go EH$ _yë` AmYm[aV àíZ h¡ &
(v)
àíZ
-nÌ _| g_J« na H$moB© {dH$ën Zht h¡ & VWm{n, Xmo A§H$m| dmbo EH$ àíZ _|, VrZ A§H$m|
dmbo EH$ àíZ _| Am¡a nm±M A§H$m| dmbo VrZm| àíZm| _| AmÝV[aH$ M`Z àXmZ {H$`m J`m h¡ &
Eogo àíZm| _| AmnH$mo {XE JE M`Zm| _| go Ho$db EH$ àíZ hr H$aZm h¡ &
(vi)
H¡$bHw$boQ>a Ho$ Cn`moJ H$s AZw_{V Zht h¡ & VWm{n, `{X Amdí`H$ hmo Vmo Amn bKwJUH$s`
gmaUr H$m à`moJ H$a gH$Vo h¢ &
(vii)
Ohm± Amdí`H$ hmo Amn {ZåZ{b{IV ^m¡{VH$ {Z`Vm§H$m| Ho$ _mZm| H$m Cn`moJ H$a gH$Vo h¢ :
c = 3 108 m/s h = 6.63 10–34 Js e = 1.6 10–19 C o = 4 10–7 T mA–1 o 4 1 = 9 10 9 N m2 C–2 me = 9.1 10–31 kg General Instructions :
(i) All questions are compulsory.
(ii) There are 30 questions in total. Questions No. 1 to 8 are very short answer type questions and carry one mark each.
(iii) Questions No. 9 to 18 carry two marks each, questions No. 19 to 27 carry
three marks each and questions No. 28 to 30 carry five marks each.
(iv) One of the questions carrying three marks weightage is value based question.
(v) There is no overall choice. However, an internal choice has been provided in one question of two marks, one question of three marks and all the three questions of five marks each weightage. You have to attempt only one of the choices in such questions.
(vi) Use of calculators is not permitted. However, you may use log tables if necessary.
(vii) You may use the following values of physical constants wherever necessary : c = 3 108 m/s h = 6.63 10–34 Js e = 1.6 10–19 C o = 4 10–7 T mA–1 o 4 1 = 9 10 9 N m2 C–2 me = 9.1 10–31 kg
1.
{H$gr namd¡ÚwV _mÜ`_ Ho$ ‘d¡ÚwV Y«wdU’ nX H$mo n[a^m{fV H$s{OE & BgH$m Eg.AmB©.
(S.I.)_mÌH$ {b{IE &
1Define the term ‘electric polarisation’ of a dielectric medium. Write its S.I. unit.
2.
{H$gr H$U na
‘q’Amdoe h¡, `h H$U
‘v’doJ go {H$gr ~mø Mwå~H$s` joÌ _| J{V H$aVm
h¡ & {H$Z eVm] (à{V~§Ym|) Ho$ AÝVJ©V Bg na bJZo dmbm ~b
(i)eyÝ` hmoJm,
(ii)A{YH$V_
hmoJm
? 1Under what conditions does a particle of charge ‘q’ moving with velocity ‘v’, experience (i) no force, (ii) maximum force, in an external magnetic field ?
3.
{H$gr MmbH$ _| Amdoe dmhH$m| H$m Andmh doJ {H$Z Xmo H$maH$m| na {Z^©a H$aVm h¡
?CëboI H$s{OE &
1Write any two factors on which drift velocity of charge carriers in a conductor depends.
4.
ì`mnH$sH¥$V Eopån`a Ho$ n[anW {Z`_ H$m J{UVr` ì`§OH$ {b{IE &
1Write the mathematical expression for the generalized Ampere’s circuital law.
5.
Ý`ypŠbAm°Zm| Ho$ EH$ `w½_ H$s pñW{VO D$Om© VWm CZHo$ ~rM H$s Xÿar Ho$ \$bZ Ho$ ê$n _|
AmaoI It{ME &
1Draw a plot showing variation of potential energy of a pair of nucleons as a function of their separation.
6.
g§Mma ì`dñWm _| ‘nwZamdV©H$’ ([anrQ>a) H$m àH$m`© Š`m hmoVm h¡
? 1State the function of a ‘Repeater’ in a communication system.
7.
namdVu XÿaXe©H$ _| EH$ b|g Ho$ ñWmZ na AdVb Xn©U H$mo A{^Ñí`H$ Ho$ ê$n _| à`moJ
H$aZo H$m EH$ H$maU {b{IE &
1Give one reason for using a concave mirror, rather than a lens, as an objective in a reflecting type telescope.
8.
{H$gr dmhH$ Va§J
c(t)H$mo Bg àH$ma ì`º$ {H$`m J`m h¡
: c(t) = 3 sin (4t) volt_m°Sw>bZ {g½Zb
m(t)Xem©E AZwgma EH$ dJ© Va§J h¡ &
BgH$m _m°Sw>bZ gyMH$m§H$ kmV H$s{OE &
1A carrier wave c(t) is given by c(t) = 3 sin (4t) volt
The modulating signal m(t) is a square wave as shown.
Find its modulation index.
9. (a)
dmZ S>o J«m\$ O{ZÌ Ho$ ~‹S>o Jmobr` MmbH$ Imob na Cƒ {d^dmÝVa H$mo$ EH$ gr_m
VH$ hr ~‹T>Zo XoZo Ho$ {bE H$maH$ H$s nhMmZ H$s{OE &
(b)
EH$ g_mÝVa ßboQ> (n{Å>H$m) g§Ym[aÌ H$s à^mdr Ym[aVm
1 Fh¡ VWm BgH$s Xmo
ßboQ>m| Ho$ ~rM H$s Xÿar
1 cmh¡ & BgH$s Xmo ßboQ>m| Ho$ ~rM _|
9 106 V/mnamd¡ÚwV gm_Ï`© H$m EH$ namd¡ÚwV _mÜ`_ ^a {X`m OmVm h¡ & Vmo g§Ym[aÌ na Amdoe
H$m {ZYm©aU H$s{OE &
2(a) In a Van de Graaff generator, identify the factor which limits the build-up of a high potential difference on a large spherical conducting shell.
(b) A dielectric medium of dielectric strength 9 106 V/m is filled between the plates of a parallel plate capacitor having effective capacitance of 1 F and plate separation 1 cm. Determine the charge on the capacitor.
10.
{ZåZ{b{IV J«mµ\$ (AmboI), Xmo n¥WH²$ loUr
LCRn[anWm| Ho$ {bE,
tHo$ gmW dmoëQ>Vm
(V)VWm Ymam
(I)Ho$ n[adV©Z H$mo Xem©Vo h¢ &
BZ_| go {H$g n[anW _| g§Ym[aÌ à{V~mYm
(XC),àoa{UH$ à{V~mYm
(XL)go
(i)A{YH$ hmoJr
(ii)
H$_ hmoJr
?AnZo CÎma Ho$ {bE H$maU {b{IE &
2The following graphs depict the variation of voltage (V) and current (I) versus (t) for two different series LCR circuits.
State, giving reason, in which of the circuits capacitive impedance (XC) is (i) greater, (ii) smaller than the inductive impedance (XL).
11. (a) + z-
Aj Ho$ AZw{Xe J_Z H$aVr hþB© EH$ {dÚwV²
-Mwå~H$s` Va§J H$s Amd¥{Îm
‘v’VWm
Va§JX¡¿`©
‘’h¡ & Bg Va§J go g§~Õ {dÚwV² VWm Mwå~H$s` joÌ Ho$ KQ>H$m| Ho$ {bE
J{UVr` {Zê$nU ({MÌU) H$mo {b{IE &
(b) z-
Aj Ho$ AZw{Xe g§MaU H$aVr hþB©, {H$gr g_Vb {dÚwV²
-Mwå~H$s` Va§J H$m,
CgHo$ Xmobm`_mZ {dÚwV² VWm Mwå~H$s` joÌ Ho$ KQ>H$m| Ho$ gmW EH$ aoIm{MÌ (ñHo$M)
~ZmBE &
2(a) Write the mathematical expressions for electric and magnetic field components associated with an electromagnetic wave of frequency ‘v’ and wavelength ‘’ propagating along the + z-axis.
(b) Draw the sketch of the plane electromagnetic wave propagating in the z-direction along with the components of oscillating electric and magnetic fields.
12.
Xn©U gyÌ H$m Cn`moJ `h Xem©Zo Ho$ {bE H$s{OE {H$ CÎmb Xn©U Ûmam ~Zm Am^mgr à{V{~å~
gX¡d AmH$ma _| N>moQ>m Am¡a Xn©U Ho$ Y«wd d \$moH$g Ho$ ~rM pñWV hmoVm h¡ &
2AWdm
`hm± AmaoI _| `§J Ho$ {Û{Par à`moJ Ho$ EH$ ì`dñWmnZ H$mo Xem©`m J`m h¡, {Og_| Ho$ÝÐr`
q\«$O
‘O’na h¡ &
nX} na ~Zo ì`{VH$aU n¡Q>Z© na Š`m à^md n‹S>oJm, `{X,
(i)òmoV
SH$mo
S1S2g_Vb Ho$
g_mÝVa D$na H$s Amoa {dñWm{nV H$a {X`m OmE VWm
(ii)òmoV
SH$mo ídoV àH$me Ho$ {H$gr
òmoV go à{VñWm{nV H$a {X`m OmE
? 2Use the mirror formula to show that the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.
OR
The figure shows an experimental set-up of Young’s double slit experiment with the central fringe at ‘O’.
How will the interference pattern on the screen be affected when (i) the source S is displaced upwards parallel to the plane S1S2 and (ii) the source S is replaced by white light ?
13.
{H$gr àH$me
-g§doXr n¥ð> (gVh) Ho$ {bE, {ZamoYr {d^d
(V0)VWm Amn{VV {d{H$aUm| H$s
Amd¥{Îm
(v)Ho$ nXm| _|, AmBÝñQ>mBZ Ho$ àH$me
-{dÚwV² g_rH$aU H$mo {b{IE & `h Xem©Zo Ho$
{bE EH$ Cn`wº$ J«mµ\$ (AmboI) ~ZmBE, {H$ Bg J«mµ\$ go
(i){H$gr nXmW© Ho$ {bE H$m`©
\$bZ VWm
(ii)ßbm§H$ {Z`Vm§H$ Ho$ _mZ Ho$ ~mao _| gyMZm H¡$go àmá H$s Om gH$Vr h¡ &
2Write Einstein’s photoelectric equation in terms of the stopping potential (V0) and the frequency of the incident radiation (v) for a given photosensitive surface. Draw a suitable graph to show how one can get the information about (i) the work function of the material and (ii) value of Planck’s constant from this graph.
14.
Cn`wº$ {MÌ H$s ghm`Vm go
p-ng§{Y Ho$ {Z_m©U _| _w»` à{H«$`mAm| H$m {ddaU Xr{OE &
õmgr joÌ Ho$ ~ZZo H$m g§jon _| dU©Z H$s{OE &
2Explain, with the help of a diagram, the basic processes involved in the formation of p-n junction. Write briefly how the depletion region is developed.
15. (a)
A§H$s` ({S>{OQ>b) VWm AZwê$n (EZmbm°J) g§Ho$Vm| Ho$ ~rM AÝVa {b{IE &
(b)
Cg _yb JoQ> H$m VH©$ àVrH$ VWm CgHo$ {bE gË`_mZ gmaUr ~ZmBE Omo {Zdoe H$m
ì`wËH«${_V ê$nmÝVa CËnÞ H$aVm h¡ &
2(a) Distinguish between digital and analogue signals.
(b) Write the logic symbol and truth table of the basic gate which produces an inverted version of the input.
16.
Xmo bå~o, nVbo g_mÝVa Vmam| H$m aoIr` Amdoe KZËd H«$_e…
2 10–9 C/mVWm
– 3 10–9 C/mh¡ & `o dm`w _| EH$
-Xÿgao go
20 cmXÿa aIo JE h¢ & BZ àË`oH$ Vmam| go
10 cmXÿa pñWV {H$gr {~ÝXþ na {dÚwV² joÌ H$m n[a_mU (_mZ) VWm CgH$s {Xem kmV
H$s{OE & $
2Two long thin parallel wires having linear charge density 2 10–9 C/m and – 3 10–9 C/m are kept 20 cm apart in air. Find the magnitude and direction of the electric field at a point 10 cm from each wire.
17.
EH$ BboŠQ´>m°Z VWm EH$ µ\$moQ>m°Z _| àË`oH$ H$m Va§JX¡¿`©
3.315 nmh¡ & Vmo BboŠQ´>m°Z H$s
J{VO D$Om©
(K)H$m µ\$moQ>m°Z H$s D$Om© go AZwnmV kmV H$s{OE &
2An electron and a photon each have a wavelength 3.315 nm. Find the ratio of the kinetic energy (K) of the electron to the energy of the photon.
18.
EH$ Amdo{eV H$U na
2 nCH$m Amdoe h¡ & `h H$U doJ
v = 105 ^i (m/s)go Mwå~H$s`
joÌ
B_| J{V H$aVm hþAm EH$ Mwå~H$s` ~b,
F = 2 10–5 (– ^j ) NH$m AZw^d
H$aVm h¡ & Bg Mwå~H$s` joÌ H$s {Xem VWm n[a_mU (_mZ) kmV H$s{OE &
2A charged particle having a charge of 2 nC moving in a magnetic field B with a velocity v = 105 ^i (m/s) experiences a magnetic force
F = 2 10–5 (–^j) N. Find the direction and magnitude of the magnetic field.
19.
Xmo g§Ym[aÌm| H$s Ym[aVm H«$_e…
C1VWm
C2h¡, Am¡a
C1 = 2C2&
(i)nhbo BZHo$ loUrH«$_
_| g§`moOZ H$mo Am¡a
(ii){\$a BZHo$ g_mÝVa
-H«$_ _| g§`moOZ H$mo, EH$ hr ~¡Q>ar go Omo‹S>m OmVm
h¡ & BZ Xmo XemAm| go {H$g_|
(a)g§{MV D$Om© VWm
(b)Cnm{O©V Amdoe, A{YH$ hmoJm
?AnZo CÎma H$s nw{ï> H$s{OE &
3Two capacitors of capacitances C1 and C2 such that C1 = 2C2 are connected in turn (i) in series and (ii) in parallel across the same battery. In which of the two cases will the (a) energy stored and (b) charge acquired be more ? Justify your answer.
20.
{H$gr EH$dUu òmooV go àH$me
0.2 mmMm¡‹S>r EH$b {Par (pñbQ>) na n‹S>Vm h¡, Am¡a Bggo
{Par Ho$ g_Vb go
1 mXÿa pñWV nX} na {ddV©Z n¡Q>Z© ~ZVm h¡ & `{X _w»` C{ƒð> H$s Hw$b
aoIr` Xÿar
4.8 mmh¡, Vmo AZwà`wº$ àH$me H$s Va§JX¡¿`©
()H$m n[aH$bZ H$s{OE &
Ho$
Bg _mZ Ho$ Cn`moJ Ûmam, {ÛVr` (Xÿgar) AXrá q\«$O H$s aoIr` Mm¡‹S>mB© H$m n[aH$bZ
H$s{OE &
3{H$gr {ddV©Z n¡Q>Z© _| q\«$Om| H$s Vrd«Vm _| H¡$go n[adV©Z hmoVm h¡
?Light from a monochromatic source falls on a single slit of width 0.2 mm to produce a diffraction pattern on a screen kept at a distance of 1 m from the plane of the slit. If the total linear width of the principal maxima is 4.8 mm, calculate the wavelength () of the light used. Using this value of , calculate the linear width of the second dark fringe.
How does the intensity of fringes in a diffraction pattern vary ?
21.
{H$gr AmYma ~m`{gV Q´>m§{µOñQ>a Ho$ {bE
CE{dÝ`mg _| AÝVaU A{^bjU$ H$mo EH$
AmboI _| Xem©BE & Bg A{^bjU H$m Cn`moJ Cg AdñWm joÌ H$m A{^{ZYm©aU H$aZo _|
H$s{OE, Ohm± Q´>m§{µOñQ>a H$m Cn`moJ
(i)àdY©H$ Ho$ ê$n _| VWm
(ii)pñdM Ho$ ê$n _| {H$`m Om
gH$Vm h¡ &
Q´>m§{µOñQ>a Ho$ pñdM Ho$ ê$n _| Cn`moJ H$s H$m`©{d{Y H$mo g§jon _| ñnï> H$s{OE &
3Draw the transfer characteristic of base biased transistor in CE configuration. Use this characteristic to identify the regions where transistor can be used as (i) an amplifier and (ii) a switch.
Briefly explain the working of a transistor as a switch.
22.
g§jon _| dU©Z H$s{OE {H$ AmH$me Va§J g§MaU {d{Y Ûmam ao{S>`mo Va§Jm| H$m g§MaU H¡$go hmoVm
h¡ &
EH$ àofr E|Q>rZm H$s D±$MmB©
20 mVWm A{^J«mhr E|Q>rZm H$s D±$MmB©
80 mh¡ &
LOS{dYm
_| g§VmofOZH$ g§Mma Ho$ {bE XmoZm| Ho$ ~rM H$s A{YH$V_ Xÿar H$m n[aH$bZ H$s{OE &
[{X`m
J`m h¡, n¥Ïdr H$s {ÌÁ`m
= 6.4 106 m] 3Describe briefly how radio waves are propagated in space wave mode of propagation.
A transmitting antenna has a height of 20 m and the height of the receiving antenna is 80 m. Calculate the maximum distance between them for satisfactory communication in LOS mode.
[Given, radius of Earth = 6.4 106 m]
23.
VrZ gobm| Ho$ {dÚwV² dmhH$ ~b (B©.E_.E\$.)
1 = 1.5 V, 2 = 2.0 VVWm
3 = 3 Vh¢, BZHo$ AmÝV[aH$ à{VamoY H«$_e…,
r1 = 0.3 , r2 = 0.4 VWm
r3 = 0.6 h¢ & BZ
VrZm| gobm| H$mo g_mÝVa
-H«$_ _| g§`mo{OV {H$`m J`m h¡ & `{X Bg g§`moOZ Ho$ ñWmZ na EH$
gob {b`m OmE, Vmo Cg gob H$m Vwë` {dÚwV² dmhH$ ~b (B©.E_.E\$.) VWm Vwë` à{VamoY
kmV H$s{OE &
3AWdm
Xem©E JE d¡ÚwV OmbH«$_ _| {H$aImoµ\$ Ho$ {Z`_m| Ho$ Cn`moJ Ûmam {dÚwV² YmamAm|
I1VWm
I2Ho$ _mZ kmV H$s{OE &
3Three cells of e.m.f., 1 = 1.5 V, 2 = 2.0 V and 3 = 3 V, having internal resistances r1 = 0.3 , r2 = 0.4 and r3 = 0.6 respectively are connected in parallel. Find out the equivalent e.m.f. and the equivalent resistance of a cell which can replace this combination.
OR
Use Kirchhoff’s rules to find out the values of the currents I1 and I2 in the electrical network as shown.
24.
boÝO Ho$ {Z`_ H$mo {b{IE & EH$ gab CXmhaU H$m dU©Z H$s{OE, Omo Xem©E {H$ `h {Z`_
D$Om©
-g§ajU {gÕmÝV Ho$ AZwê$n h¡ &
Bg {Z`_ Ho$ Cn`moJ Ûmam, AmaoI _| Xem©E JE Xmo g_Vbr` bynm| (nmem|) _| ào[aV {dÚwV²
Ymam H$s {XemE± H$maU XoH$a kmV H$s{OE•& `o byn (nme) {H$gr Eogo EH$g_mZ Mwå~H$s`
joÌ _| {dÚ_mZ h¢ {OgH$s {Xem nmem| Ho$ g_Vb Ho$ A{^bå~dV² VWm nmR>H$ (Amn) H$s
Amoa h¡ & nme
‘abcd’Bg Mwå~H$s` joÌ go ~mha {ZH$b ahm h¡ VWm nme
‘pqrs’Mwå~H$s`
State Lenz’s law. Describe a simple example to show that this law is in conformity with the principle of conservation of energy.
Using this law, predict, giving reasons, the directions of the induced currents in each of the two planar loops shown in the figure. The loop ‘abcd’ is moving out whereas the loop ‘pqrs’ is moving into the region of a uniform magnetic field directed normal to the planes of the loop and towards the reader.
25.
{ZåZ{b{IV H$mo H$maU g{hV ñnï> H$s{OE
:(a)
`{X {H$gr gm_mÝ` òmoV (O¡go gmo{S>`_ b¡ån) go àH$me {H$gr nmoboam°BS> erQ> go
hmoH$a JwµOaVm h¡, Vmo BgH$s Vrd«Vm AmYr hmo OmVr h¡ Am¡a nmoboam°BS> H$mo Kw_mZo go
nmaJV Vrd«Vm na H$moB© à^md Zht hmoVm &
(b)
{H$gr CÎmb b|g H$mo `{X Eogo _mÜ`_ _| Sw>~m {X`m OmE {OgH$m AndV©Zm§H$, b|g
Ho$ nXmW© Ho$ AndV©Zm§H$ go A{YH$ h¡, Vmo `h CÎmb b|g EH$ Angmar b|g H$s
^m±{V H$m`© H$aVm h¡ &
(c)
g§`wº$ gyú_Xeu _| A{^Ñí`H$ VWm Zo{ÌH$m XmoZm| hr Aën µ\$moH$g Xÿar Ho$ hmoVo h¢ &
3Explain the following giving reason :
(a) If light from an ordinary source (like a sodium lamp) passes through a polaroid sheet, its intensity is reduced to half and rotating the polaroid has no effect on the transmitted intensity. (b) A convex lens when immersed in a medium whose refractive index
is more than that of the material of the lens, behaves like a diverging lens.
(c) Both the objective and the eyepiece of a compound microscope have short focal lengths.
26.
EH$ loUr
LCR(Eb gr Ama) n[anW Ho$ {bE _mÜ` e{º$ j` H$m ì`§OH$ e{º$ JwUm§H$ Ho$
nX _| ì`wËnÞ H$s{OE & AV… Xem©BE {H$ _mÜ` e{º$ j` H$m _mZ AZwZmX H$s pñW{V _|
A{YH$V_ hmoVm h¡ &
3Deduce the expression for the average power dissipated in a series LCR circuit in terms of power factor. Hence show that average power dissipated at resonance is maximum.
27.
emo_m EH$ {Z_‚mZ N>‹S> (am°S>) Ûmam H$m±M Ho$ EH$ nmÌ _| nmZr J_© H$aZo H$m à`ËZ H$a ahr
Wr (`h am°S> grYo hr _oZ nm°da gßbmB© go Ow‹S>r Wr) & O~ `h OmZZo Ho$ {bE {H$ nmZr
n`m©á J_© hþAm `m Zht, dh AnZr C±Jbr nmZr _| S>mbZo Om hr ahr Wr Vmo CgHo$ {ejH$ Zo
Cgo Eogm Z H$aZo Ho$ {bE gMoV {H$`m &
Bg KQ>Zm na AmYm[aV {ZåZ{b{IV àíZm| H$m CÎma Xr{OE
:(i)
`h gbmh XoH$a emo_m Ho$ {ejH$ Zo {H$g _yë` H$m àXe©Z {H$`m
?(ii)
O~ nmZr _| {dÚwV² àdm{hV H$s Om ahr h¡ Vmo Cgo Ny>Zm ˜VaZmH$ Š`m| hmoVm h¡
?`{X
emo_m Zo AmgwV Ob {b`m hmoVm, Vmo Š`m `h A{YH$ gwa{jV hmoVm
?(iii)
EH$ CXmhaU Xr{OE {Og_| AmnZo ñd`§ H$mo `m {H$gr AÝ` ì`{º$ H$mo dmñV{dH$
OrdZ _| Bg àH$ma Ho$ _yë` H$m àXe©Z H$aVo hþE nm`m hmo &
$
3Shoma was trying to heat water by means of immersion rod (connected directly to the main power supply) in a glass vessel containing water. As she tried to put her finger into the water to check if the water was warm enough, her teacher cautioned her against doing it.
Based on the above information, answer the following questions : (i) What value did Shoma’s teacher display by giving this advice ? (ii) Why is it considered dangerous to touch water when current is
being passed into it ? Will it be safer if Shoma had used distilled water ?
(iii) Give an example where you have displayed or observed someone else displaying similar value in real life situation.
28. (a)
Cn`wº$ AmaoI H$s ghm`Vm go {H$gr EH$g_mZ Mwå~H$s` joÌ
B_| pñWV,
A joÌ\$b H$s EH$ Am`VmH$ma
IYmamdmhr Hw§$S>br na bJZo dmbo ~bAmKyU© (Q>m°H©$)
Ho$ {bE EH$ ì`§OH$ (g{Xe ê$n _|) àmá H$s{OE &
(b)
CZ eVm] H$m C„oI H$s{OE {OZ_|
(i)D$na {XE JE ~b
-AmKyU© H$m ì`§OH$
Mb Hw§$S>br J¡ëd¡Zmo_rQ>a _| Cn`wº$ hmoVm h¡
; (ii)Mb Hw§$S>br J¡ëd¡Zmo_rQ>a H$s
gwJ«m{hVm ~‹T> gH$Vr h¡ &
5AWdm
(a)
n¥Ïdr Ho$ n¥ð> Ho$ {H$gr {~ÝXþ na Mwå~H$s` joÌ H$m {ddaU XoZo Ho$ {bE Amdí`H$
VrZ am{e`m| (Ad`dm|) H$m g§{já dU©Z, Amdí`H$ AmaoIm| Ho$ Cn`moJ Ûmam
H$s{OE & AZwà`wº$ Amdí`H$ gyÌ ^r {b{IE &
(b)
à{VMwå~H$s` VWm AZwMwå~H$s` nXmWm] H$mo EH$g_mZ Mwå~H$s` joÌ _| aIm J`m h¡ &
BZ nXmWm] Ho$ g_rn Mwå~H$s` joÌ aoImAm| H$m ì`dhma {MÌ Ûmam àX{e©V H$s{OE &
AnZo CÎma H$s nwpîQ> H$s{OE &
5(a) Derive, with the help of a suitable diagram, the expression (in vector form) for the torque acting on a rectangular loop of area A carrying current I, placed in a uniform magnetic field .B
(b) Mention the conditions under which (i) the above expression for the torque is applied in moving coil galvanometer; (ii) the sensitivity of the moving coil galvanometer is increased.
OR
(a) Give a brief description using necessary diagrams, of the three elements (quantities) required to specify the magnetic field of the Earth at a point on its surface. Also write the necessary formulae used.
(b) Specimens of a paramagnetic material and a diamagnetic material are placed in a uniform magnetic field. Draw the behaviour of magnetic field lines near these specimens. Justify your answer.
29.
{H$gr {ÌH$moUr` {àµÁ_ go hmoH$a JwµOaVr hþB© àH$me H$s {H$aU H$m _mJ© Xem©Zo Ho$ {bE EH$
{H$aU AmaoI ~ZmBE & Bg AmaoI Ho$ Cn`moJ go, {àµÁ_ Ho$ H$moU VWm Ý`yZV_ {dMbZ H$moU
Ho$ nXm| _|, {àµÁ_ Ho$ nXmW© Ho$ AndV©Zm§H$ Ho$ {bE gå~ÝY àmá H$s{OE & {H$gr {àµÁ_ Ho$
{bE, AmnVZ H$moU VWm {dMbZ H$moU Ho$ ~rM J«mµ\$ H$s àH¥${V H$mo Xem©BE &
5AWdm
(a)
Va§JmJ« H$s n[a^mfm {b{IE & {H$gr g_`
t = 0na {H$gr Va§JmJ« H$m AmH$ma
g_Vb Va§J Ho$ ê$n _| h¡ & hmBJoÝg H$s aMZm Ûmam Xem©BE {H$
(i)Hw$N> g_`
tHo$
níMmV², {ÛVr`H$ Va§{JH$mAm| Ho$ EZdobn (C^`{Zð> ñne©H$), g_Vb Va§J H¡$go
CËnÝZ H$aVo h¢ VWm
(ii)CÎmb b|g go JwµOaZo Ho$ níMmV² {ZJ©V Va§JmJ« {H$g àH$ma
JmobmH$ma hmo OmVm h¡ Am¡a µ\$moH$g na A{^g[aV hmo OmVm h¡ &
(b)
hmBJoÝg Ho$ {gÕmÝV Ho$ Cn`moJ Ûmam {H$gr gKZ _mÜ`_ go {dab _mÜ`_ H$mo
g§M[aV hmoVr hþB© g_Vb Va§J Ho$ AndV©Z Ho$ ñZ¡b Ho$ {Z`_ H$m gË`mnZ H$s{OE &
5Draw a ray diagram to show the passage of a ray of light through a triangular prism. Use this diagram to obtain the relation for the refractive index of the material of the prism in terms of the angle of minimum deviation and the angle of the prism. Plot the nature of the graph for the angle of deviation versus the angle of incidence in a prism.
OR
(a) Define a wavefront. Given the shape of a wavefront as a plane wave at time t = 0, show using Huygens’ construction, (i) how the envelopes of secondary wavelets produce the plane wave at a later time t and (ii) how the emergent wavefront becomes spherical and converges to the focus after passing through a convex lens ?
(b) Verify using Huygens’ principle, Snell’s law of refraction of a plane wave propagating from a denser to a rarer medium.
30. (a)
hmBS´>moOZ na_mUw Ho$ {bE ~moa Ho$ {gÕmÝV Ho$ àW_ Xmo A{^J¥hrVm| H$mo {b{IE &
na_mUw H$s g§aMZm Ho$ dU©Z Ho$ {bE BZ A{^J¥hrVm| H$s Amdí`H$Vm H$mo g§jon _|
ñnï> H$s{OE &
(b)
~moa H$s V¥Vr` A{^J¥hrV Ho$ Cn`moJ Ûmam, hmBS´>moOZ na_mUw Ho$ ñnoŠQ´>_ Ho$ {bE
[aS>~J© H$m gyÌ {b{IE & Bg gyÌ H$s ghm`Vm go, hmBS´>moOZ ñnoŠQ´>_ H$s bmB_¡Z
loUr _| ñnoŠQ´>_r aoIm Ho$ àW_ gXñ` Ho$ Va§JX¡¿`© H$m n[aH$bZ H$s{OE &
([aS>~J© {Z`Vm§H$ H$m _mZ,
R = 1.03 107 m–1br{OE)
5AWdm
(a) (i)
Eoë\$m
()-j` VWm
(ii)~rQ>m
– (–)-j` _| àË`oH$ H$m EH$ CXmhaU Xr{OE & BZ
j`
-àH«$_m| H$mo gm§Ho${VH$ ê$n _| {b{IE &
(b)
Xr JB© Zm{^H$s` A{^{H«$`m,
MeV 27 . 3 n He H H 21 32 10 2 1 _| `Ú{n Ý`ypŠbAm°Zm| H$s g§»`m XmoZm| Amoa g§a{jV ahVr h¡, {\$a ^r D$Om© {Z_©wº$
hmoVr h¡ & ñnï> H$s{OE &
(c)
Xem©BE {H$ EH$ Zm{^H$ Ho$ {bE {dñV¥V ûm¥§Ibm _| Zm{^H$ KZËd pñWa ahVm h¡ VWm
Ðì`_mZ g§»`m
Ago ñdV§Ì hmoVm h¡ &
5(a) State the first two postulates of Bohr’s theory of hydrogen atom. Also explain briefly the necessity for invoking these postulates to describe the structure of the atom.
(b) Using Bohr’s third postulate, write the Rydberg formula for the spectrum of the hydrogen atom. With the help of this formula, calculate the wavelength of the first member of the spectral line in the Lyman series of the hydrogen spectrum.
(Take the value of Rydberg constant R = 1.03 107 m–1)
OR
(a) Give one example each for (i) -decay and (ii) –-decay by writing the decay processes in symbolic form.
(b) In a given nuclear reaction,
MeV 27 . 3 n He H H 21 32 10 2 1
although number of nucleons is conserved on both sides, yet energy is released. Explain.
(c) Show that nuclear density in a nucleus, on a wide range of nuclei is constant and independent of mass number A.
Page 1 of 13 Final draft 13/4/14 12:00noon
MARKING SCHEME SET 42/1 (SPL)
Q. No. Expected Answer / Value Points Marks Total Marks
1. Net dipole moment per unit volume developed in the dielectric medium in presence of an external electric field is called electric polarization.
SI unit : coulomb/m2
½
½ 1
2. i. When a charged particle is moving parallel/anti parallel to the magnetic field.
ii. When a charged particle is moving perpendicular to the magnetic field.
½
½ 1
3. Two factors on which drift velocity of charge carriers depends; Applied electric field and temperature of the conductor. (or any other correct 2 factors.)
½ ½ 1 4. = + Alternatively : = ( 1 1 5. 1 1 6. A repeater picks up the signal from the transmitter, amplifies and retransmits
it to the receiver.
1 1
7. i. There is no chromatic aberration. ii. Spherical aberration is reduced. iii. Easy mechanical support.
(Any one /any other one correct reason.) 1 1
8. =
= = 0.5
[Note: Award this one mark, if a student writes answer directly without calculation.]
½
½
1 9.
a) Limiting factor – break down field of air. b) c =
as dielectric strength E = =
½
(a) Identification of factor ½
(b) Determination of charge on the capacitor 1 ½ 1
Page 2 of 13 Final draft 13/4/14 12:00noon c = q= 9 x 10-2 C = 90mC ½ ½ ½ 2 10. i. For circuit I, Xc > XL
Reason: Current is leading the voltage in phase. ii. For circuit II, Xc < XL
Reason: Current is lagging the voltage in phase.
½ ½ ½ ½ 2 11. a) = = b) ½ ½ 1 2 12.
For a convex mirror, and for an object on left side i.e. , Using the mirror formula
=
(as
and
)
<
Image is between Pole and focus of the mirror, Linear magnification m= for
Hence, image is diminished
OR ½ ½ ½ ½ 2 Identification of circuits ½ + ½ Reasons ½ + ½
(a) Mathematical expressions for electric and magnetic field components ½ + ½ (b) Drawing of the sketch of the plane electromagnetic wave 1
Showing, that the virtual image produced by convex mirror is always
diminished and between focus & pole 2
Page 3 of 13 Final draft 13/4/14 12:00noon i. Fringe pattern on the screen will shift downwards with central fringe
below point 0, on the screen.
ii. Central fringe will be white, fringe closest on either side of central white fringe will be red and farthest will appear blue, after a few fringes, no clear pattern will be seen.
1
1 2
13
work function = -(e intercept on the y-axis) Planek’s constant = e slope of the curve.
½
½
½
½ 2
14.
Two processes involved during the formation of p-n junction are diffusion and drift. Due to the concentration gradient, across p and n sides of the
1 Effect on interference pattern when
i. Source S is displaced upwards. 1
ii. Source S is replaced by white light. 1
i. Einstein’s photoelectric equation ½
ii. Drawing of suitable graph 1 ½
Processes involved in the formation of p-n junction 2
Page 4 of 13 Final draft 13/4/14 12:00noon junction , holes diffuse from p n, and electrons from n p. This movement
of charge carriers leaves behind ionised acceptors on the p-side and donors on the n- side of the junction. This space charge region on either side of the junction, together, is known as depletion region.
1
2 15.
i. Analogue signals are in the form of continuos, time varying voltage
or current.
Digital signals are those which can take only discrete stepwise values
i.e. only two levels of voltage / current.
ii. ½ ½ ½ ½ 2 16.
Electric field at a point distant 10cm from each wire, = +
E = [ ]
= 9 x 109 x ( 2 )
= 9 x 5 x 2 x 10 N/C
= 900 N/C (towards the wire having negative linear charge density.) [Note : Give full credit, if a student doesn’t draw diagram.]
½ ½ ½ ½ 2 17. for electron ½
a) Distinction between digital & analogue signals 1 b) Logic symbol and truth table ½ + ½
Finding the magnitude and direction of the electric field. 2
Calculation of ratio of kinetic energy of electron and energy of photon. 2
Page 5 of 13 Final draft 13/4/14 12:00noon
K.E.
=
For photon, energy(E) =
=
66 x 10-4 ½ ½ ½ 2 18. = q( )Since, charged particle is moving along +x axis and experiences the force along –y axis, therefore magnetic field should be along +z axis . (using right hand thumb rule/ fleming’s left hand rule/ any relevant rule.)
i.e. should be along . F = qvBSin 2 x 10-5 = 2 x 10-9 x 105 B(as angle = ) = 0.1 T ½ ½ ½ ½ 2 19.
i. For series combination of the given capacitors
ii. For parallel combination of the given capacitors
a) As energy stored U , for same voltage source.
Energy stored in parallel combination of capacitors will be more, b) Charge acquired, Q for same Voltage source,
Hence charge acquired will be more in parallel combination. [Note : Also give full credit, for alternative methods.]
½ ½ ½ ½ ½ ½ 3 20.
Linear width of principal maxima(w)
4.8
=
4.8 x 10-7m = 480nm½
½ Calculation of wavelength ( ) of light used 1 ½
Calculation of linear width of second dark fringe ½
Variation of Intensity of fringes. 1
Finding the direction and magnitude of magnetic field. 2
Calculation of energy stored and charge acquired in given cases. 1½+1½
Page 6 of 13 Final draft 13/4/14 12:00noon Linear width of second dark fringe
=
=
half of the linear width of principal maxima=
2.4mmIntensity of bright fringes, falls rapidly as one moves from centre to the ends of the screen, and for the dark fringes intensity remains zero.
[Note : Also accept the intensity distribution curve of diffraction for this part.]
½ ½
1 3
21.
i. Region for transistor to be used as an amplifier : Active region. ii. Region to be used as a switch : cut off region and saturation region.
Working: when input voltage ( ) is low and unable to forward bias the transistor, is high. If is high enough to drive the transistor into saturation then is low. Therefore when transistor is not conducting, it is switched off and when it is driven into saturation it is switched on.
1 ½ ½ 1 3 22.
Space wave travels in a straight line from transmitting antenna to receiving antenna. Maximum distance = + = + = 2 x 8 x 103 + 4 x 8 x 103 = 48 x 103m = 48Km 1 ½ ½ ½ ½ 3 23.
=
=
½ ½ Draw of the transfer characteristic 1Identification of regions for amplifier and switch ½ + ½ Brief working of a transistor as a switch 1
Brief description of propagation of radio waves in space wave mode. 1
Calculation of maximum distance. 2
Calculation of equivalent e.m.f. 1 ½
Calculation of equivalent resistance 1 ½
Page 7 of 13 Final draft 13/4/14 12:00noon
=
=
=
= 5+5+5 = = 2.0VOR
In loop abcda -20 I1 + 40 – 40 I3 = 0 I1 +2 I3 = 2 ---(1) As I3= I1 + I2 I1 +2 I1 +2 I2 = 2 3 I1+ 2I2 = 2 ---(2) In loop adefa 40 I3 – 40 – 80 +20 I2 = 0 2 I3 + I2 = 6 ---(3) Substituting I3 = I1 + I2 2 I1 + 2I2 +I2= 6 2 I1 + 3I2 =6 ---(4)solving equation 2 and 4 I2= A And I1 = A ½ ½ ½ ½ ½ ½ ½ ½ ½ ½ 3 3 24.
The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it.
This law is in conformity with the principle of conservation of energy, consider a situation where induced current is in the direction opposite to that given by lenz’s law. Then in this situation kinetic energy will continuously increase without expanding any energy and hence one can construct a perpetual motion machine by a suitable arrangement, which violates the conservation of energy.
In irregular loop induced current will be in the sense. pqrsp, as induced current will oppose the increase in magnetic flux.
In other loop, induced current will be in the sense abcda, as the induced current will oppose the decrease in magnetic flux.
1
1
½ ½
3
Statement of Lenz’s law 1
Description of a simple example 1
Prediction of direction of induced current ½ + ½
Finding the value of I1 and I2 3
Page 8 of 13 Final draft 13/4/14 12:00noon 25.
a) It is because, light which is produced from an ordinary source (like a sodium lamp) is unpolarised, when an unpolarised light wave is passed through a Polaroid sheet, it gets linearly polarised with the electric vector oscillating along the direction perpendicular to the aligned molecules.
b) Ray of light parallel to the pricipal axis, gets deviated away from the normal at the first surface and towards the normal at the second surface since deviation at two surfaces is in same sense,
net deviation from both the surfaces make the ray diverge away from the principal axis.
Alternatively: It is because the relative refractive index of lens
becomes less than 1 and
becomes –ve. Hence, lens behave like a diverging lens. [Note : Also accept the explanation through diagram.]
c) Due to short focal length of eye piece, angular magnification of eye piece increases and magnification of objective is large when is slighty greater than , since the microscope is used for viewing very close object therefore is small and hence , should be small.
1 1 1 3 26. V = I =
Instantaneous power ‘p’ supplied by the source P = VI = .
= [
Average power over a complete cycle
P = (because the average of time dependent term is zero.)
=
=
At resonance V and I are in same phase i.e. P = , which is maximum ½ ½ ½ ½ 1 3 27.
i. Concerned/ Caring(or any other one relevant value)
ii. Because, one can get electric shock, as water conducts electricity, Yes, relatively safer, because distilled water is a bad conductor of electricity.
iii. Any example displaying similar value in real life situation.
1 1
1 3
28.
Explanation of parts a, b and c 1 +1+1
Deduction of the expression for the average power 2 Showing the average power dissipation is maximum at resonance 1
Answer of the parts i, ii and iii 1+1+1
a) Derivation of the expression for torque 3 b) Mentioning the conditions for (i) & (ii) 1+1
Page 9 of 13 Final draft 13/4/14 12:00noon
a)
Forces on the arms BC and DA are equal, opposite and act along the axis of coil, being collinear along the axis, they cancel each other,
Forces on arms AB and CD are and which are equal and opposite, but not collinear.
= =IbB
therefore magnitude of torque acting on the loop
=
+(Area of the loop A = ab) = I X
b)
i. In moving coil galvanometer, this expression is applied at i.e.
=
NIABii. Senstivity of a galvanometer is increased by increasing number of turns/ increasing area of coil/ increasing magnetic field/ by decreasing torsional constant.
OR
a)
½ ½ ½ ½ ½ ½ 1 1 ½ 5a) Brief description of three elements ½ + ½ + ½
Necessary diagram ½
Necessary formula ½ + ½
b) Drawing the behavior of magnetic field lines ½ + ½
Justification ½ + ½
Page 10 of 13 Final draft 13/4/14 12:00noon Three physical quantities are Magnetic declination, angle of dip
(inclination) and horizontal component of Earth’s field.
The angle between the true geographic north and north shown by compass needle is called magnetic declination.
Dip is the angle that the total magnetic field of earth makes with the surface of Earth.
Component of total magnetic field of earth along the surface of earth is called horizontal component.
= , =
And tanI =
where and are horizintal and vertical component of the earth’s field and I is the inclination at a place.
b)
The field lines gets concentrated inside the material and field inside is enhanced.
The field lines are repelled or expelled and the field inside the material is reduced. ½ ½ ½ ½ ½ ½ ½ ½ ½ 5 29. a)
In the quadrilateral AQNR
---(1)
From the triangle QNR
+ ---(2)
Comparing these two equations
+ =A ---(3)
The total devation is the sum of derivations at the two faces. =
1
½
½
Ray diagram showing the passage of ray 1
Obtaining the relation for refractive index 3
Nature of graph. 1
Page 11 of 13 Final draft 13/4/14 12:00noon That is = i+ ---(4)
At the minimum derivation Dm, the refracted ray inside the prism becomes
parallel to its base.
= Dm, i = e which implies =
Equation (3) becomes 2r = A or r =
In the same way equation (4) becomes Dm = 2i-A, => i=
The refractive index of the prism is = =
OR
a) Wave front is the surface of constant phase
Alternatively: locus of points, which orscillate in phase.
i.
To produce the plane wave at a later time t, draw spheres of radius
vt
from each point on the plane wave front . By drawing the common tangent to all these spheres a plane wave at a later time t is obtained. (where
v
is the speed of waves in the medium)½ ½ ½ ½ 1 1 ½ ½ 5
a) Definition of wave front 1
Showing (i) and (ii) using Huygen’s construction 1+1
b) Verification of Snell’s law. 2
Page 12 of 13 Final draft 13/4/14 12:00noon ii.
The central part of the incident plane wave traverses the thickest portion of the lens and is delayed the most. Therefore the emerging wavefront has a depression at the centre and hence it becomes spherical and converges at the focus.
b)
= = = (Refractive index of rarer medium wrt denser medium.) ½ ½ 1 ½ ½ 5 30. a)
i. An electron in an atom revloves in certain stable orbit without the emission of radiant energy.
ii. Electron revolves around the nucleus only in those orbit for which the angular momentum is an integral multiple of .
Alternatively: where n is an integer
According to classical electromagnetic theory, an accelerated electron emits radiation in the form of electromagnetic waves. Therefore energy of electron would decrease continuously and electron would spiral inward and eventually fall into the nucleus, such an atom can not be stable. As the electrons spiral inwards, their frequency would change continuously, thus they would emit a continuous spectrum in contradiction to the line spectrum observed. To overcome above shortcomings, Bohr’s theory came into existence.
1 1
1 a) Statement of the first two postulates of Bohr’s theory. 1+1
Brief explanation of the necessity 1
b) Rydberg formula 1
c) Calculations of wavelength 1
Page 13 of 13 Final draft 13/4/14 12:00noon b) Bohr’s third postulate , gives the Rydberg Formula
= R
For first member of spectral line in lyman series,
= 129nm OR a) i. Alternatively
ii.
Alternatively:
b) Binding energy per nucleon of the fussed muclei is more than the binding energy per nucleon of the lighter nuclei i.e. final system is more tightly bound than the initial system therefor energy is released.
c) Let A be the mass number and m be the average mass of a nucleon, Nuclear density
=
=
Substituting R =
Which is constant and independent of Mass Number(A).
1 ½ ½ 1 1 1 ½ ½ ½ ½ 5 5
a) One example each for (i) and (ii) 1+1
b) Explanation of release of energy 1
c) Showing that nuclear density is independent of mass number 1