physics01- model papers of kerala board of higher secondary education, xiith class

7/29/2019 Physics01- Model papers of Kerala Board of Higher Secondary Education, XIIth Class

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1

CO:1

To get an idea about frictional electricity charges and their properties through simple experi-

ments problem solving discussion and I.T.

Question Text:

Q. No:1

When a rubber sheet is rubbed with woolen carpet, the carpet is found to acquire a positive charge of 810-7C

a) In the above process charging is by

(i) conduction (ii) friction (iii) induction (iv) polarization

b) Among Rubber shoe and woolen carpet which one acquire both mass and charge during rubbing?

Explain.

c) The charge acquired by the carpet is quantized. Justify this statement.

Hcp Infn ]c-h-Xm-\n-bn d jq Dc--tm ]c-h-Xm-\nv t]mkn-ohv Nmv e`n- -XmbnI p .

a) GXv {]Xn- mkw Imc-W-amWv ]c-h-Xm-\nv Nmv e`n- -Xv.

(i) I - (ii) {^nIvj (iii) CUIvj (iv) t]mfssdtk-j

b) d jqhn\mtWm ]chXm\nmtWm Nmpw amkpw e`nXv? Fs\bmWvCXvkw`hnXvFv hni-ZoIcn-p-I.

c) ]c-h-Xm-\nv e`n Nmv tImsskvUv BWv. Cu {]kvXm-h\ kan-p-I.

Score (1+2+2 = 5), Time : 10 mintsScoring key

a) friction 1 Score (MP-1, 2)

b) Rubber shoe, Rubber shoe gains electrons, so it gains mass and charge 2 Score (MP - 1,2,7)

c) charge is an integral multiple of e (1.6 10-19)

q ne= ,e

q=n , 19-

-7

101.6

108

= 12105 2 Score (MP 8, 7, 5)

CO: To understand Coulombs law and extract it to find the forces due to multiple electronic

charges and continues charge distribution through discussions and solving numerical prob-lems in groups

Question Text :

Q. No: 2

Two point charges q1and q

2are separated by a distance r in space

(a) what happens to the force between the charges;

(i) When the magnitude of the point charges increases

(ii) When the distance of separation decreased

Hence establish the law governing it.

(b) Three point charges +2C each are placed at the corners of an equilateral triangle of side 1m. Find the

magnitude of the force between

PHYSICS

QUESTION BANK FOR PLUS TWO STUDENTS

q2

q1

q3

A

B C


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charge q1and q

2[F

12]

and q1

and q3

[F13

]

(c) Draw12F

r

and13F

r

at A and find the total force acting on q1. Hence establish super position principle

(21/2)

(d) The law governing force between two point charges can be established using charged pith balls

Comment on the statement. (1/2)

[Total Score : 6]

Time : 8 mts

q1, q

2Fo t]mbn v Nmp-I r F AI-e- n shn-cn-p-p.

a) (i) Nmp-IfpsS tIhe-aqeyw IqSp-tm Ah Xn-ep _e- n\v F v kw -hnpw?

(ii) Nmp-I Xn-ep AIew Ipdbp-tm Ah Xn-ep _e-n\v F v kw -hnpw

b) 2C tIhe-aq-ey-ap aqv t]mbn v Nmp-I 1m hi-ap Hcp ka`p-P{Xn-tIm-Wns

aqv aqe-Ifn-embn hn-cn-p-p. q1Dw q

2Dw Xnepw (F

12) q

1Dw q

3Dw Xnepw (F

13) D _e-

ns tIh-e-aqeyw Ip-]n-Sn-p-I.c) A bn

12Fr

Dw13F

r

Dw hcv BsI _ew Ip-]n-Sn-p-I. XpSv kqs]m-kn-j {]nkn-

fn Fn-t-cp-I.

d) cv Nmv sNbvX ]nv t_mfp-I D]-tbm-Knv c v t]mbn v Nmp-I Xn-ep_e-s -p-dn- p \nbaw {]kvXm-hn-m Ignbpw A`n-{]m-b Ipdnsgp-Xp-I.

Scoring Key

Score

a) (i) increases (1/2)

(ii) increases (1/2)Coulombs law statement (1/

2+ 1/

2) (MP - 1,2)

b) F =2

219

r

qq109 (1/

2)

Result (1/2) (MP - 1,2,5)

c) Figure (1/2)

1312 FFFrrr

+= (1/2)

Cos60F2FFFF 13122

13

2

12 ++= (1

/2) (MP - 1,2,5,7)

Answer (1/2)

Super position (1/2)

Principle (1/2)

d) No (1/2) (MP = 1,2,5,7,10)

CO-3

To develop an idea about electric field and to study the electric field due to a point charge

and dipole through discussions and solving numerical problems in group.

Q. Text

Q. No. 3

A metal sphere of radius R carrying q (+ve) charges is shown in the figure.


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3

a) Draw the electric lines of force related to this metal sphere.

b) Derive the strength of electric field at a distance r from the centre of sphere.

c) A sphere of radius 1cm can hold a charge of 1 coulomb Comment on the statement.

[Hint:- The dielectric strength of air is 3 mv/m]

Total Score 4

R Bc-ap Hcp teml-tKmfw q (t]m-k-ohv) Nmv hln-p-p.

a) Cu teml-tKm-fhp-ambn _-s Ce-Iv{SnIv _e-tcJI hcbvpI

b) teml-tKm-f- ns ay- n\nvr Zqcn-ep Hcp _np-hn A\p- -hsSp Ce-Iv{SnIv_e-tcJbpsS Xo{hX ImWp--Xn-\p kahmIyw cq]oIcn-p-I.

c) tem-l- tKm-f- n\v 1 Ipfv Nmv hln-p-hm Ign-bpw. Cu {]kvXm-h-\-sb-p-dnv A`n-{]m-bIpdnsgpXpI.

(kqN\ : hmbphns ssU Ce-Iv{SnIv Xo{hX 3mv/mBWv)

Time : 7 min

Score : 4

Scoring key

a)

(MP - 1,2)

b)

oqE.ds=

o

2

qr4E =

2o r

q

41E

= (1/2) (MP - 1,2)

c) No, the charge will leak through the air

(MP - 1, 5, 7, 10)

CO 4

To understand the behavior of dipole in a uniform electric field through discussion and IT

R

r

R

r


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4

Q. Text

Q. No. 4

An electric dipole is placed in a uniform electric field of Intensity E as shown in the figure.

a) What are the forces on +q and -q?

b) What is the net force on the system?

c) Copy the diagram, mark the forces and derive an expression for torque.

d) If the dipole is placed in a nonuniform electric field what nature of motion does it show?

Hcp Ce-Iv{SnIv ssUtm E Xo{hXbp Hcp bqWnt^mw Ce-Iv{SnIv ^oUn shn-cn-p-p.

a) +q Nmnepw -q Nmnepw GsXmw _e--fmWv A\p- -h-s-Sp--Xv.

b) ssUtm-fn A\p- -hsSp BsI _e-sa{X

c) Nn{Xw hcv ssUt]m-fn A\p- -hsSp _e- AS-bm-fsSp- p-Ibpw tSmns kahmIyw cq]o-I-cn-p-Ibpw sNp-I.

d)ssUt]m shp-Xv t\m bqWnt^mw Ce-Iv{SnIv ^oUn-em-sWn GXp-Xcn-ep-Ne-\-am-bn-cnpw kw -hn-pI?

Time : 7 min

Score : 4

Scoring Key

Score

a) +qE, -qE (1/2) (MP - 1,2)

b) Zero (1/2) (MP - 1,2)

c) fig, Proof )EP( = 2 (MP - 1,2,6)

d) Both translation and rotational motion (1/2) (MP = 1, 2, 6, 7)

CO 5

To develop an idea about electric potential, p.d and to study the electric potential due to a

point charge, dipole and system of charges through group discussion and solving numerical

problems in groups.

Text :

No. 5

A charge of +5C is placed in free space P

Q

5C

5mm

3mm

E

-q

+q


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5

a) The workdone to bring a + IC charge from infinity to a point P is called

(i) capacitance (ii) dielectric constant (iii) potential energy (iv) electric potential

(b) Calculate the workdone in above process

(c) Calculate the workdone to move a +IC charge from P to Q

Hcp 5C Nmv iq\y-X-bn shn-cn-p-p.

a) A\- Xbn\npw +1C Nmns\ P F _np hsc sImp-h-cp-hm-\m-h-iy-amb {]hr-nsb ]dbp t]cv.

(i) Imknkv (ii) ssU Ce-IvSnIv tIm v

(iii) s]mjy F\Pn (iv) Ce-Iv{SnIv s]mjy

b) takq-Nn- n {]{In-b-bvm-hiyamb {]hrn IW-m-p-I.

c) +1CNmvP F _nphn\npw Q F _np-hsc sImp-h-cp-hm-\m-h-iy-amb {]hrnIW-m-p-I.

Score (1+1 1/2+1 1/

2= 4)

Time 8 mts

Scoring Key

a) Electric potential (1) (MP -1,2)

b) Workdone = potential =o4

1

r

q

J1015103

1051096

3

69

=

(11/2) (MP - 1,2,5,6)

c) Workdone p.d =o4

1

21

11rr

q

=

33

69

105

1

103

1105109

=

15351059 6 =

1521059 6 = J106 6 (1

1/2) (MP - 1,2,5,6,7)

CO 6

To derive a mathematical relation for electric potential energy of a system of point charges

and electric dipole through general discussion and numerical problems

Question Text:

Q.No: 6

The above fig shows a charge q1placed at a distance r

1from the origin.

a. What is the work done to bring a charge q2at P from infinity

b) Derive equation for work done to bring another charge q3on the point Q (q

1and q

2are present) from

infinity.c) What is the total work done.

To arrange the three changes in the system and name the work done

q1

r2

r1

P

Q

O

12rr

3rr


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Nn{Xn q1F Nmv HPn-\n\npw r

1AIe- n shn-cn-p-p.

a) CXn q2

F NmvP F _np-hn sImp-h-cp--Xn\v th {]hn-bpsS AfhvIW-m-p-I.

b) q1Dw q

2Dw AXns m\mbncnptm asmcp Nmmb q

3,Q

F _np-hn sImp -

hcp-Xn\v Bhiy-amb {]hn-bpsS Afhv F{X

c) aqv Nmp-Itfbpw Hcp system Bbn \ne-\np--Xn\v Bhiy-amb {]hn-bpsS AfhvF{X. Cu {]hn- GXv t]cn-emWv Adn-b-s-Sp--Xv.

Score 3

Time 5 min

Scoring Key

a) Potential of P due to q1=

12

1

o r

q

41 (

1/2) (MP - 1,2)

Work done to bring q2from infinity to P =

12

21

o r

qq

41 (

1/2) (MP - 1,2,5)

b. Potential at Q due to q1and q

2=

23

2

o13

1

o r

q

41

r

q

41

+ (1/

2)

Workdone to bring q3from infinity to Q = 3

23

2

13

1

o r

q

r

q

4

1q

+

(1/

2) (MP - 1,2,5)

c. Total workdone to make the system =

++

23

32

13

31

12

21

o r

qq

r

qq

r

q

41 q

Total workdone is equal to potential energy (1/2) (MP - 1,2,5,6,7)

CO:7

To understand Gausss Theorem and apply it to find electric field due to an infinitely long

straight wire uniformly charged infinite plane sheet and uniformly charged thin spherical

shell through general discussion, group discussion and solving numerical problems in groups.

Question Text

No: 7

A spherical surface enclosing a charged spherical shell of radius R and charge density is shown

a) which law establishes the total flux related to this surface and total charge on the spherical shell? State

the law (11/2

)

b) Using this law find the field at any point on the spherical surface. (2)

c) If the entire charge on the shell is brought to its centre, does the field on the spherical surface change?

Justify. (11/2)

R

r


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7

Hcp kvs^dn- k^-kn-\p-n-en-cn-p Nmv sNbvX Hcp teml-tKm-f-amWv Nn{X- nImWn- n-cn-p--Xv. Cu teml-tKm-fns tdUn-bkvRDw Nmv sUknn bpw BWv.

a). kvs^dn- k^-kpambn _-s-n-cn-p ^vfIvkn-t\bpw, teml-tKm-f- nse NmPn-t\bpw _-sSp- p \nbaw GXm-Wv. AXv {]kvXm-hn-p-I.

b). Cu \nbaw D]tbm-K-sSpn kvs^dn- k^knse GsXnepw Hcp t]mbnnse Ce-Iv{SnIv ^oUv Ip-]n-Sn -p-I.

c). Cu teml-tKm-f-nse apgp-h Nmpw tKmf-ns tI{-n-tev sImp-h--Xmbn Icp-XpI. Cu Ahbn kvs^dn k^knse Hcp t]mbnnse oUv BZytXn\nvamdptam? hni-Zam-p-I.

Time : 9 min

Score : 5

Scoring Key

a) Gausss Theorem (1/2

) (MP - 1)

Statement (1) (MP - 1,2)

b.

= =z

r

r

E ds R. 42

0

(1) (MP - 1,2,5)

and result (1)

c. Does not change (1/2)

Proof (1 ) (MP- 1,2,5,7)

CO: 8

To get an idea about dielectrics and electric polarization through discussions

Question Text :

Q.No: 8

Classify the following molecules in to polar and nonpolar and justify your answer with proper explanation.

Nn{X- n ImWn- n-cn-p Xm-{X-Isf t]mfm, t\mt]m-fm Fv ho-I-cn-p-I. \n-fpsS Dcw hni-ZoIcnv km[q-Icn-p-I.

H H

H2

O

H

OCO2

H

Cl

HCl

O

H H

H2O


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8

Time 5

Score 2

Scoring Key Score

Polar HCl, H2O (1/

2)

Non - Polar - H2CO

2(1/

2) (MP - 1, 3)

Explanation based on dipole moment (1/2+1/2) (MP - 1, 3, 4, 7)

CO:9

To get an idea about dielectrics and electric polarization through discussions.

Question Text :

Q.No: 9

The figure below shows a non polar dielectric slab placed in between the plates of an uncharged parallel

plate capacitor.

Area of each plate = A

Distance of separation between the plates = d

a) If the dielectric slab is absent and the capacitor is charged to a surface charge density , the electric field

in between the plates is

(i) E =o (ii) E =

o

2

(iii) E =o

(iv) E =

o2

b) Redraw the given figure, which shows the alignment of the nonpolar molecules, when the capacitor is

charged

c) Derive an expression for capacitance of the above capacitor with the dielectric slab in between the

plates

d) Dielectric constant of a conductor is infinity - Justify

Nmv sNm Hcp Im-kn--dns tp-In-S-bn hn-cn-p Hcp t\m t]mfmssU Ce-Iv{Sn-ns Nn{Xw Xmsg sImSp-n-cn-p-p.

a) ssU Ce-Iv{SnIv m_v sUkn-n-bn-tev Nmv sNp-tm, tp-In-S-bn-ep Ce-Iv{Snv oUv

(i) E =o (ii) E =

o

2

(iii) E = o

(iv) E = o2

b) Im-kn- Nmv sNp-tm, t\mt]m-fm tamfn-Iyq-fp-I Fs\ {IaoIcn--sSp-p-shv Nn{Xw ]pXp-n-h-cv ImWn-p-I.


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c) tpIv CSbn ssU CeIv{SnIv m_v Dtm, Imkndns Imknkv Xcp kahmIyw cq]oIcn-p-I.

d) Hcp IU-IvS-dns ssU Ce-Iv{SnIv tIm v A\--am-Wv. km[q-I-cn-p-I.

Time 7 mts

Score 5

Scoring Key

Score

(a) Eo

(1/

2) (MP - 1)

(b) Figure showing polarisation (1) (MP - 1,2,5)

(c) E =

o

p

-(1/

2)

V = Ed (1/2)

Q = A (1/2)

p- ) = K1

(1/2)

Result C =d

KAo (1/2) (MP - 1,2,5,7)

d) Explanation based on ability to allow electric field lines to pass through (1) (MP 1,2,5,7)

CO-10

To develop an idea about capacitor, capacitance and series and parallel combination of ca-pacitor through discussion and solving numerical problems.

Question text

Q.No 10

Three capacitors C1, C

2and C

3are connected to a cell of emf V as shown in the figure.

a) The arrangement of these three capacitors are called.................(i) parallel combination (ii) series combination

(iii) LCR combination (iv) c-c combination

b) Find the effective capacitance of the above combination.

c)

1V

2V

4V

a b c d e f g

V

C2

C3C1

a b c d e f g


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10

The above graph shows the variation of potential in going from a to g. From the graph the relation among

C1, C

2and C

3is

(i) C1

= C2= C

3ii) 2C

1= 2C

2= C3

(iii) C1=C

2= 2C

3iv) C

1 C

2 C

3

Total score 4

Time 9minC

1, C

2, C

3Fo Im-kn-dp-I Hcp V thmv D skp-ambn _n- n- n-cn-p-Xv Nn{X-

n ImWn- n-cn-p-p.

a) Im-kn--dp-I Cu coXn-bn LSn- n-p--Xns\ ..............................-Fv ]d-bp-p.

(i) ]mc-e tImn-t\-j (ii) kocokv tImn-t\-j

(iii) LCR tImnt\j (iv) CC tImn-t\-j

b) Cu tImn-t\-j-\n BsI A\p-`-h-s-Sp Im-kn-kv F{X-bmWv Fv Is-p-I.

c) {Km^n Xn-cn-p-XvaF _np-hn\nvg F _np hscbp s]m-jyjns hyXn-bm-\-am-Wv. C

1, C

2, C

3Fo Im-kn--dp-I-fpsS _w {Km^n\nv Inp-Xv F-s\-bm-bn-

cn-pw.(a) Series Combination (1) (MP - 1,2)

(b) V - V1

+ V2

+ V3

(1)

321 C

Q

C

Q

C

Q

C

Q++= (1/

2)

321 C1

C1

C1

C1 ++= (1/

2) (MP- 1, 2, 3, 5)

c) Q is same, V1

= 1V, V2

= 1V, V3

= 2V

C1 : C2 : C3

21:

11:

11

= 2C1

= 2C2

= C3

(Score 1) (MP - 1, 2, 3, 5, 6, 7, 10)

CO: 11

To derive an equation for energy of a capacitor and capacitance and parallel plate capacitor

with and without dielectric medium between the plates through general discussion

Q. Text

Q. No: 11

Two metal plates X and Y of the area A are separated by a distance d, charged + and - respectively.

a) This arrangement is called ...............................

b) The arrangement stored energy in the ..............................

(Magnetic field, Electric field, Electromagnetic field, Gravitational field)c) Derive an expression for the energy stored the arrangement

d) When we increase separation between two plates by keeping V constant, what happens to total

energy stored in the system

X Y

V

d


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11

A hn-kvXo-ap X, YFo c v teml tp-I d AI-e- n shn-cn-p-p. AhbvvbYm-{Iaw +hpw hpw Nmp-I \In-bn-cn-p-p.

a) Cu {Iao-I-c-Ws ............ F-dn-b-s-Sp-p.

b) Cu {Iao-I-c-W- n Duw kw -cn-p-Xv ....... BWv.

(amKv\nIv oUv, Ce-Iv{SnIv oUv, Ce-Ivt{Sm-amKv\nIv oUv, {Kmhn-tj ^oUv)

c) Cu {Iao-I-c-W-n kw`-cn--s-Sp Duw Is-m-\p- ka-hmIyw cq]o-I-cn-p-I.d) thmtPv nc-ambn \nn-smv tp-I Xn-ep AIew hn-n-p-I-bm-sW-n

samw F\Pnv F v kw -hnpw?

Time : 10 min

Score : 5

Scoring Key Score

a) Capacitor (1) (MP - 1)

b) Electric field (1) (MP - 1,2)

c) E = 2

1

CV2

(Proof) (2) (MP - 1,2,4,5)

d) Since V is constant while decreasing C

Energy also decreases (Capacitance decreases

with increase in distance) (1) (MP - 1, 2, 4, 5,7)

CO. No: 12

To get an idea about principle and working of a vande graff generator through discussion and

IT

Question Text:

Q No: 12

a) Identify the device

(i) Moving coil Galvanometer

(ii) Cyclotron

(iii) Photovoltaic cell

(iv) Van-de-Graff electrostatic generator

b) Explain the construction and working of the above device

c) What happens if the upper metal sphere is replaced by a cubical shaped metal? Explain.

a) Nn{X- n ImWn- n-cn-p D]-I-c-W-ns t]sc v?

(i) aqhnwKv tImbn Kmht\m-ao

(ii) ssktmt{Sm (iii) t^mtmthmmbnIvsk

Metal sphere

Motor


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12

(iv) hmUn--{Km^v Ce-Ivt{Sm-m-nIv P\-td

b) apIfn ImWn- n-cn-p D]IcWns \nm-Whpw {]h -\-coXnbpw hni-ZoIcn-p-I.

c) D]I-cWns apI`m-K- p tdmfm-Ir-Xn-bn-ep teml- n\v ]Icw Iyq_m-Ir-Xn-bn-eptemlw D]-tbm-Kn- m F v kw -hnpw? hni-Zo-I-cn-p-I.

Score - (1 + 2 + 2 = 5)

Time 9 mtsScoring Key

a) (iv) Van-de-graff electrostatic generator (1) (MP - 1)

b) Construction (1)

Working (1) (M.P -1,2)

c) The device will not work (1) (MP 1, 2, 5)

The charge will get discharged (1) (MP-1, 2, 5, 7)

CO: 13

To establish the relation between drift velocity, mobility and electric current through general

discussion

Question Text

Q: No: 13

A rectangular conductor of length l and area of cross section A and electron density n; is shown below

a. When the face Y is given positive potential and X negative potential what will happen to the electrons

inside the block

b. What is meant by drift velocity? How is it related to the field inside the metal?

c. Deduce an expression connecting intensity of electric field and drift velocity.

d. Under the application of an electric field do all the electrons move in a same direction? Explain

l\ofap-Xpw A t{Imkv skIvj\ Gcn-bm-tbmSv IqSn-bXpw n Ce-Ivt{Sm km{Xbp--Xp-amb Hcp teml-IjvWamWv Nn{X- n sImSp- n-cn-p--Xv.

a) Y AS-bm-f-s-Sp- nb `mK v t]mkn-ohv s]mjyepw X AS-bm-f-s-Sp- nb `mK v s\K-ohv s]mjyepw sImSp- m teml- ns Dn-ep Ce-Ivt{Sm-Wp-Iv F v kw -hnpw?

b) {Un^vv shtem-knn F-Xp-sImv FmWv A-am-p-Xv? {Un^vv shtem-knn teml-nse Ce-Iv{SnIv ^oUn-t\mSv Fn-s\-bmWv _-sp InS-p-Xv?

c) {Un^vv shtem-kn-n-tbbpw Ce-Iv{SnIv Id v Ckn-n-tbbpw _-s-Sp- p ka-hmIywcq]oIcnpI.

d teml-I-jvW-n-\p-n Ce-Iv{SnIv ^oUv Ds-n AXn-\p-nse apgp-h Ce-Ivt{Sm-Wp-Ifpw Htc Zni-bn-emtWm k-cn-p-I. hni-ZoIcn-p-I.

Time: 10 mts

Score : 5

X Y


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13

Scoring Indicator

a. Drifts towards +ve and -ve plate 1/2Score (MP -1)

b. Defenition 1 Score (MP- 1)

Relation 1/2

Score

c. Derivation 2 Score (MP - 1,2,5)

d. No 1/2 Score (MP -1,2,5)Explanation 1/

2Score (MP - 1,2,5,7)

CO: 14

To understand ohms law and concept of electrical resistance resistivity and conductivity

through experiments, simple projects and graphical analysis.

Question Text:

Q: No: 14

To study the relation between potential difference and current in an electrical circuit, a student is

provided with a resistance wire, a cell and a key.

a) Draw a circuit which allows current flow through the resistance wire

b) Modify the circuit by introducing an ammeter, Voltmeter and a rheostat for varying the potential differ-

ence across the resistance and to measure that potential difference and the corresponding current.

c) Let in the above experiment the student obtained the following data

Draw a graph connecting V and I using above data. Then establish the relation between V and I as a

law.

d) Instead of the resistance wire if the student uses a p-n junction diode in the forward biased condition

how the relation between V and I changes? Justify.

Hcp Ce-Iv{SnIv kIyq-nse Idpw s]mjy Un^-dkpw Xn-ep _w ]Tn-p-hm, Hcp Ipn-v, sdkn-kv hb, sk, Io Fnh \In-bn-cn-p-p.

a) dkn-kv hb-dn-eqsS Idv Hgp-p-\v thn-bp Hcp kIyqv Ub{Kw hc-p-I.

b) d-kn-kv hb-dns c-- Xn-ep s]mjy Un^-dkv hyXym-k-s -Sp- n, B

s]mjy Un^-dkp-Ifpw AXn\v A\p-kyq-X-amb Idpw Af-p--Xn\v thn, HcpAo, thmv ao, dntam-mv Fnh DsSpn apIfn ]d kIyq-ns\ amn hcp-I.

c) apIfn ]d ]co-Wn Hcp Ipnv e`n UmIfmWv Xmsg \In-bn-cn-p--Xv.

Cu hnh-c- D]-tbm-KnvV, I Fnh Xn-ep _w ImWn-p Hcp {Km^v hcp-I.

V-IFnh Xn-ep _s Hcp \nb-aambn {]kvXm-hn-p-I.d) dknkv hbdn\v ]Icw t^mthUv _bkv Bbn Hcp UtbmUv D]tbmKnm V-I Fnh

Xn-ep _w Fs\ hyXym-k-s-Spp? \ymbo-I-cn-p-I.

Voltage (V)

Current (I) A

0

0

2

0.1

4

0.2

6

0.3

8

0.4

10

0.5

Voltage (V)

Current (I)

0

0

2

0.1

4

0.2

6

0.3

8

0.4

10

0.5


14/30

14

Score and time (1 + 1 + 2 + 2 = 6)

Time 10 mts

Scoring Key

(a)

(1) (MP - 1,2)

(b)

(1) (MP - 1,2)

(c) A graph connecting V and I , which is a straight line statement of ohms law. (Score 2) Mp (8, 9)

(d) Forward characteristics of a diode the graph is not a straight line V and I are not directly proportional,

up to knee voltage (Score 2) (MP - 1,2,6,8,9)

CO: 16

To get an idea about classification of materials based on conductivity through discussions

CO: 61To familiarize intrinsic and extrinsic semi conductors through discussions and II

Question Text:

Q.No.15

Table given below shows the resistivity and temperature coefficient of resistivity of certain materials.

a) Classify the materials into conductors, semiconductors and insulators and justify your answer

b) Temperature coefficient of resistance of silicon and carbon are negative. What does it mean? Justifyc) Carbon can not be used to make a pn junction diode. Why?

Nne ]ZmfpsS sdknnhnnbpw sStd tImb^njpw ]nIbn sImSpncnpp.

E Key

R

V

A

R

Rh

E

K

Material

Copper

Carbon

Platinum

Glass

............

Silicon

Resistivity

m at OoC

1.7 10-8

3.5 10-5

11 10-8

1010 10-14

1014

2300

Temperature coefficient

of resistance / (OoC)

0.0068

-0.0005

0.0039

-

-

0.07


15/30

15

a) ]Zmsf IU-IvtSgvkv, skan IU-IvtSgvkv, Ckp-te-tgvkv Fv hoIcn-p-I.\n-fpsS Dcw km[q-I-cn-p-I.

b) knen--n-sbpw Im_-Wn-tbpw sS-td- tImb-^n-j v Hm^v sdkn-kv s\K-ohvBWv. FmWv CXv A-am-p--Xv. Dcw km[q-I-cn-p-I.

c) Hcp pn PwKvj UtbmUv Dm-m Im_ D]-tbm-Kn-m-dn -. Fp-sImv?

Time 6 mtsScore: 4

Scoring key Score

(a) Classification (1/2+ 1/

2+ 1/

2) (MP - 1)

(b) Resistivity decreases with increasing (1/2)

temperature (MP-1,2)

(c) Four valance electrons of carbon is in the 2nd orbit,

where as in Silicon they are in the 3rd orbit.

Hence difficult to get free electrons in the case of Carbon. (1) (MP - 1,2,5,7)

CO: 16

To get an idea about colour code for resistors through demonstration and discussion.

Question Text :

Q. No: 16

A, B, C and D are four rings on a carbon resistor.

A = yellow, B = violet, C = Yellow, D, Silver

(a) What is the value of resistance of above resistor?

(b)

The combined resistance of the above two resistor is(i) 120 (ii) 45 (iii) 165 (iv) 35

A, B, C, D Fnh Im_ sdkn--dn D 4 hfb- BWv. a, hb-e-v, a, knh

a) apIfn sImSp- n-cn-p sdkn--dns sdkn-kv F{Xbm-Wv.

b) apI-fn sImSp-n-cn-p cv sdkn--dp-Ifpw IqSn tNm-ep sdkn-kv F{X-bmWv?

(i) 120 (ii) 45 (iii) 165 (iv) 35

Time : 4

Scoring key Score : 2

(a) 470 x 103 10% (Score 1) (MP- 1, 2)

(b) 165 (Score 1) (MP -1, 2, 5)

A B C D

Brown Brown Yellow Black

Red Green


16/30

16

CO: 17

To establish the law of combination of resistor in series and parallel through experiment and

discussion.

Question Text:

Q: No: 17A cell and two resistors R

1and R

2are provided to you.

a) Draw different combinations of resistors using R1, R

2and the cell.

b) Derive an expression for the effective resistance of the circuit in which current is the same in both

resistors

c) If R1= 4 and R2 = 6 ., in which combination effective resistance is minimum? find its value?

R1, R

2Fo cv sdkn--dp-Ifpw Hcp skpw \nv Xn-cn-p-p.

a) dkn--dp-Ifpw skpw D]tbm-Knv km[y-amb tImn-t\-j-\p-I cq]oIcn-p-I.

b) c v sdkn--dp-I-fn-eq-sSbpw Xpey Id v Hgp-Ip--Xmb tImn-t\-js k^-e-{]-Xn-tcm[w

Is X -p -I .c) R

1= 4 , R2 = 6 BsWn GXv tImn-t\-j-\n-em-bn-cnpw k^e-{]Xn-tcm[w Ipd-hm-bn-

cn -p-I, AXns aqeyw Is-p-I.

Time : 7 min

Score: 4

Scoring key Score

a) parallel and series combination of resistors (1/2+ 1/

2) (MP - 1)

b) Series combination, R = R1+ R

2(Proof) 2 (MP - 1,2)

c) Parallel Combination R = 6464

+

1 (MP - 1,2,3,5)

CO: No: 19

To study temperature dependence of resistance through simple projects, discussion and graphi-

cal analysis.

Qn: Text -

Q. No: 18

a) State whether the following statement is true or false

The value of resistance of a metal increase with the rise of temperature

b) Explain the reason

c)

With the help of the above graph, match the followingI II

AB

C

Manganin

Iron

Wood

Carbon

Resistance

Temperature

A

B

C


17/30

17

d) Alloys like manganin, eureka, constantan etc are used in making standard resistance coils. Why?

Xmsg sImSp- n-cn-p {]kvXm-h\ icntbm sXtm Fv {]kvXm-hn-p-I.

Hcp teml- ns Dujvamhv IqSp--Xn-\-\p-k-cnv AXns {]Xn-tcm-[- ns Afhpw IqSpw.hniZampI.

apI-fn sImSp- n-cn-p Nn{X- ns klm-b-m Xmsg sImSp- n-cn-p-Xv tNcpw-]Sn

tNp-I.

d) mtUUv {]Xn-tcm-[- Dm-p-hm amKm-\n, bptd-, tImm Fo k-c-tem-l- D]-tbm-Kn-p-p. Imc-W-sa v?

Score :

TimeScoring key:-

a) True (1) (MP - 2)

b) As temperature increase the frequency of collision of free electrons increases. This reduces relaxation

time. (11/2) (MP - 2,5)

c)

(11/2) (MP - 2, 5, 6)

d) Due to high resistivity and low temperature coefficient of resistance. (1) (MP - 2, 5, 6, 9)

CO No: 20

To get an idea about internal resistance, Pd and emf of cell through experiment using poten-

tiometer and discussion.

Question Text:

Q No: 19

a) Name the pd between terminals of the cell when (i) key K is open (ii) K is closed

b) What is the reason for the difference in potential in the above two cases?

c) What happen to terminal potential if current is increased (i) For an ideal cell (ii) for ordinary cell

a) i) Key Xpdn-cn-p-tm (ii) Io AS-n-cn-p-tmgpw s]mjy hyXymkw GXv t]cn-emWvAdn-b-sSp--Xv.

b) c v case epw s]mjy hyXymkw hyXy-kvX-am-Im ImcWw F v

c) i)Hcp ideal cell ii) Hcp ordinary cell Fn-h-bn Id v IqSp-tm Dm -Ip hyXymkwFs\ Bbn-cn-pw.

I II

A

B

C

Carbon

Manganin

Iron

E

r

R

K

I II

A

B

C

amKm\nCcpvXSnIm_


18/30

18

Scoring key

a. (i) emf (ii) Terminal potential difference Score 1/2+ 1/

2( MP - 1,2)

b. Internal resistance Score 1/2

(MP (detects similarities) (MP - 1,2,3)

c. For ideal cell r = O, V = E, for ordinary cell I increases , V decreases

Score 2 MP (Cause - effect relations) (MP -1,2,3,5,6)

Question text :

Q. No: 20

a. Potentiometer is better than voltmeter for measuring emf because

i) It is cheap ii) Easy to handle

iii) Its measurement uses null method

b. Give the basic principle of potentiometer,

c.

If K2 is open balancing length is 600cm, if K2 is closed 350cm is balancing length. Calculate theinternal resistance

a) emf A-f-p--Xn\v s]mtjym-ao- thmvao--dn-t\-m \-Xm-sWv ]d-bm ImcWw

i) AXv hne-Ip-d--Xm-b-Xp-sImv (ii) D]-tbm-Kn-m Ffp--am-b-Xp-sImv (iii) Afm nullmethod D]-tbm-Kn-p--Xp-sIm- v .

c) Nn{Xn ImWn ncnp ]coWn K2open Bbncptm balancing length 600 cm

Dw closed Bbn-cp--tm 350 cm e`n-pp. skns internal resistance IW-m-p-I.

Score : 4

Time : 7a) iii Score (1) MP -1

b) Principle Score (1) (MP- 1, 2 )

c) r =( )

2

21

l

llR Score 1/

2MP (1, 2, 5)

Substitution and answer 11/2

CO 21

To understand Kirchoffs law, wheatstones bridge and metre bridge through experiments

and discussions.

Question Text:

Question No:21

K2

E

J

G

Rh

K1

3


19/30

19

A resistor network connected to a source of emf is shown

Applying Rule

a. What is the current through the resistance R2shown in the figure

b. If the current through the resistance R5is zero show that

4

3

2

1

R

R

R

R=

c) In the above circuit if R1is doubled and R

2is halved what change should be made in R

4. So that the

current through the resistance R5remains zero.

Hcp sshZyp-X-t{km-X-n-t\mSv LSn- n sdkn-kv s\v hmWv Nn{X- n ImWn-n-cn-p-Xv.

a) R2

F dkn-kn-eqsS Hgp-In-sm-n-cn-p Id v F{X-bmWv?

b) R5eqsS-bp Id v ]qPy-am-hp-tm

4

3

2

1

R

R

R

R= Fv ImWn-p-I.

c) apIfn sImSp kIyqnse R1

Cc-n-bm-p-Ibpw R2]Ip-Xn-bm-p-Ibpw sNp-p. Cu

kml-N-cy-n R4 F v amw hcp-n-bm R5eq-sS-bp Id v ]qPy- nXs \ne-\nm Ign-bpw.

Time : 7

Score : 4

Scoring Key

a. I- I1, 1 score (MP - 1,2)

b. Voltage across R1

1/2Score

Voltage across R2

1/2Score

Kirchoffs II Law 1/2Score

4

3

2

1

R

R

R

R= 1/

2Score (MP - 1,2,5,6)

c. Substitution 1/2Score

Becomes one fourth 1/2Score (MP - 1,2,5,6,7,10)

CO: 22

To understand heating effect of electric current and Joules law through experiments and

discussion.

Question text:Q.No:22

When electric current is passed through a resistance wire, it get heated up

V

R4

R3

R1

R2

I1 R

5

I


20/30

20

a) Name the law associated with this phenomenon.

b) What happens to the heat energy developed, if the current through the wire is doubled.

c) What happens to the heat energy developed in the wire, if the wire is stretched to double its length?

Explain

d) Length of a fuse wire is immaterial, comment on this statement.

Hcp dkn-kv hb-dn-eqsS Idv IS-n-hn-Sp-tm AXv NqSp-]n-Sn-p-p.

a) Cu {]Xn- m-k-hp-ambn _-s \nbaw GXv.

b) hb-dn-eq-sS-bp Idv Ccn Bn-bm, Dm-Ip Xm]-n\v Fv kw`-hnpw?

c) hb-dns \ofw hen-p-\on Cc-n-bm-n-bm Dm-Ip Xm]-n\v Fv kw`-hnpw?

d) Hcp ^yqkv hb-dns \ofn\v henb {]m[m-\y-an Cu {]kvXm-h\-tbmSv {]Xn-Icn-p-I.

Score ( 1/2+ 1 + 1 1/

2+ 2 = 5)

Time 10 mts

Scoring Key

a) Joules law (MP - 1,2)

b) H = I2

Rt, when I = 2I, H = 4H , Heat becomes four times (Score -1) (MP - 1,2,5)c) H = I2Rt, when L = 2L, R = 4R and H = 4H Heat becomes four times (Score -1/

2) (MP - 1, 2, 5)

d) The statement is true

Heat produced per unit surface area =I

r

2

2 32

-density, r -radius

Thus heat produced per unit area is independent of length of conductor

CO 23To get an idea about seebeck effect, thermocouple, thermo emf and application of seebeck

effect through project, experiment and discussions.

Question text :

Question No: 23

Data given below shows the measured thermo emf with temperature of the hot junction in the case of

Ni-Cu thermocouple.

Temperature

of hot junction OC

0

100

200

300

400

500

600

700

800

9001000

Thermo e.m.f

V

0

1525

2840

3945

4940

5525

6000

6265

6320

61355800


21/30

21

a) Draw a graph connecting temperature of hot junction and thermo emf of cold junction temperature is

found to be 0OC. (graph paper is suppled)

b) Extrapolate the graph to get temperature of inversion and hence note the values of neutral temperature

and temperature of inversion. What is the relation between temperature of cold junction, neutral tem-

perature and temperature of inversion?

c) Observing the shape of the graph, give a relation between thermo emf and temperature difference

between hot and cold junction.

d) If the temperature of cold junction is increased, what happens to neutral temperature.

Ni-Cu sXtam-In-fn, tlmv PwKvjs Dujvam-hn\v A\p-]m-Xambn Afv Innb sXtamC.F.F^v ]n-Ibn sImSp- n-cn-p-p.

a) sXtam C.-Fw.-F- pw, tlmv PwKvjs Dujvamhpw _n- n-p {Km^v hc-p-I. tImUvPwKvjs Dujvamhv0O Bbn ]cn-K-Wn-p-I. ({Km^v t] Xn-p- v)

b) sStd Hm^v Cthj In- --hn[w {Km^v \on, \yq{S sStd-dpw, sStdHm^v Cthj\pw Isn Fgp-Xp-I. sS-td- Hm^v tImUv PwKvj\pw \yq{S sS-tddpw sStd Hm^v Cthj\pw Xn-ep _w Fgp-Xp-I.

c) {Km^ns cq]w {inv sXtam C.-Fw.-F^,v tlmv PwKvj-\pw tImUv PwKvj\pw Xn-ep-Dujvamhv hyXym-khpw Xn-ep _w kqNn-n-p kahmIyw Fgp-Xp-I.

d) tImUv PwKvjs Dujvamhv Iqn-bm, \yq{S sS-td--dn\v F v kw -hnpw?

Time 9 mts

Score : 5

Scoring Key Score

a) Graph (11/2) (MP - 1,2,3)

b) Values of each relation (1/2+1/

2+ 1/

2)) (MP - 1,2,3,5)

c) Parabola, E =

2

2

1

+ (1

/2 +1

/2) (MP - 1,2,5,6)

d) Neutral temperature remaining constant (1) (MP - 1,2,5,6,10)

CO 25

To understand Biat-Savarts law and to apply it to find magnetic field due to an infinitely long

current carrying straight wire and circular loop through general discussion, problem solving

and graphical analysis.

Q. Text

Q. No. 24

Consider a conductor carrying current i, P is a point at a distance r away from the conductor.

a) What is the direction of magnetic field at P?

b) What are the factors affecting magnetic field at P due the element dl carrying current i?

c) Derive an expression for magnetic field at P, if the current carrying conductor has infinite length?d) Draw a graph connecting Intensity of magnetic field and distance.

. Pdl

I


22/30

22

Hcp Nme-I- n-eqsS i Id v Hgp-Ip-p. p F _np Nme-In\npw r AI-se-bmWv.

a) pF _np-hnse ImnI a-e- ns Zni GXv

b) i Id v ISpt]mIp Fensa vdl aqew p bnep-m-Ip Imn-I-a--es kzm[o-\n-pLS-I- Fsmw?

c) sshZypXNmeIns \ofw A\amIpIbmsWn p bn-ep - m -Ip Imn-I-a -ew Is -m-\p ka-hmIyw cq]o-I-cn-p-I.

d) ImnIaens Xo{hXbpw NmeIn\np Zqchpw _nnp {Km v hcpI.

Time : 11

Score : 5

Scoring key

MP Score

a) Inward to the plane of paper ( ) 11/

2

b) idl, r and sin 2 1

c) Proof - r2

Io

6 21/2

7 1

d.

CO: 25

To understand Biat-Savart law and to apply it to find magnetic field due to an infinity long

current carrying straight wire and circular loop through general discussion, problem solving

and graphical analysis

Quetion text

Question No: 25

A current flows through a circular loop of radius r is shown in the figure

a) What is the direction of magnetic field at o ?

b) Derive an equation for magnetic field at o due to the circular loop carrying current i?

c) If the loop splits into two equal halves as shown in figure.

What will be the magnetic field at the centre o.

B

r

or


23/30

23

tdUn-bkvr Bbn-p Hcp kp-e eq^n-eqsS i Id v IS-p-t]m-Ip-p.

a) eq^ns tI{-n-ep Imn-I-a--e-ns Zni Ip-]n-Sn-p-I.

b) O F _np-hnse Imn-I-a--e-ns Xo{hX Is-m-\p ka-hmIyw cq]o-I-cn-p-I.

c) kp-em eq^ns\ Htc Zni-bn sshZypXn {]h-ln-p cv Xpey-`m-K--fm-p-I-bm-sW-n (Nn{Xw {in-p-I) O F _np-hnse Imn-I-a-ew Is-p-I.

Time 11

Score : 41/2

Scoring key MP Score

a) Outward to the plane of the paper 1 1/2

b) Proof - B = 3/222

2

o

)x(r4

Ir

+

4 3

c) Zero 5 1

CO: NO: 26

To understand Amperes circuital law and its application to find magnetic field due to straight

and toroidal solenoids through general discussion, problem solving, It etc.

Qn. Text :-

Question No: 26

The figure shows a long straight conductor carrying a current I. A magnetic field is produced around the

conductor.

a) What is the magnitude of the magnetic induction at a point P distant x from the conductor?

b) What is the shape of the magnetic line of force?

c) If a small element of length dl of this line of force is considered, what is the angle between B and dl?

d) Deduce the law relating current I and magnetic induction B, by integrating the line integral of mag-

netic induction of the element

e) The above law is nothing but Biot-Savart law Justify.\nh Hcp \ofap Nme-InqSn I B-b Idv IS-n-hn-Sp--XmWv Nn{X-nImWn- n-cn-p--Xv. Nme-I- n-\p-Nppw Hcp Imn-I-a-ew cq]-s-Spw.

i

i

i

O i

r

PI

dl

x


24/30

24

a) Nme-In\npw r AIse-bp Hcp _np-hnse Imn-It{]cWns Xo{hX FmWv?

b) Imn-I-_-e-tc-J-I-fpsS cq]-sa- mWv? Nn{Xw hcvB bpsS Zni tcJ-s-Sp- p-I.

c) Imn-I-_-e-tc-J-bn-ep Hcp sNdnb `mK-ns \ofw dl BWv. B bpw dlbpw Xn-eptIm F{X?

d) Cu sNdnb `mK-ns Imn-I-t{]-cWw Ip-]n-Snv C-t{Kv sNbvXv IdvIDw Imn-I-t{]-

cWw BDw Xn-ep _w Isn AXns\ kw_-n-p \nb-a-n-se-p-I.e) apIfn ]dn-cn-p \nb-aw, _tbmvk-hmv \nbaw Xs-bm-Wv. kan-p-I.

Scoring Time

Score

Scoring key

a) B =r2

Io(1) (MP -1)

b) Concentric Circle (MP - 1)

c) Zero (1/2) (MP - 1, 2, 7)

d) Proof (2) (MP - 1, 2, 5, 9)

e) Biot-Savart law is the differential form of magnetic induction B or magnetising form H and amperes

circuital law is the integral form ofBr

.

CO No: 27

To got an idea about force an moving change in uniform electric field, and working of cyclo-

tron through general discussion and IT.

Question Text .

Question No: 27

When a changed particle entering normal to a uniform magnetic field, it take a circular path.

a. Name the particle accelerator using this principle.

b) Explain the working of that particle accelerator with relevant theory.

c) The neutrons cant be accelerated using this principle accelerator. Why?

Hcp NmPp IWw Hcp uniform Bb Imn-Ia-e- n\v ew_-ambn hm AXns ]mXhrm-Im-cam-Ip-p.

a) Cu XXzw D]-tbm-Knv {]hn-p particle acclearator s t]sc v

b) Cu particle accelerator s {]h\w A\p-tbm-Py-amb Theory tbm-Sp-IqSn hnhcn-p-I.

c) \yqt{Sm-Wp-IfpsS acceleration Cu D]-I-cWw D]-tbm-Knv hn- n-m Ign-bn-. Fp-sIm-v .

Score 3

Time 5 min

I

B


25/30

25

Scoring key

a) Cyclotron Score 1 (MP - 1)

b) Explanation Score 1 Cyclotron frequency Score 1

maximum velocity of particle Score 1 (MP - 1,3)

c) No charge on neutron Score 1 (MP - 1,3,5)

Question text :Question No: 28

Path of a chargeded particle in a particle accelerator is shown in the above figure

a) Name the particle accelerator

b) Write the principle behind the working of this particle accelerator

c) A particle of mass 6.65 10-27 kg having positive charge equal to two times of electron, moves with

a speed of 6 105 m/s in a direction perpendicular to that of a given magnetic field of flux density 0.4

weber/m2. Find the acceleration of the particle

Hcp particle accelerator Hcp charged particle k-cn-p ]mX-bmWv Nn{X- n ImWn- n-cn-p-Xv

a) Cu particle accelerator s t]sc v

b) Cu partice accelerator GXv XXz- ns ASn-m-\- n-emWv {]hn-p-Xv

c) Ce-Ivt{Sm-Wns NmPn-t\-m cn-cn Nmp Hcp t]mk-ohv IW-ns ]nw

6.65 10-27kgBWv. Cu IWw 6 105 m/s thKXbn 0.4 weber/m2 inbp Hcp amKv\nIv^oUn-eqsS ew_-ambn k-cn-p-tm Dm-Ip FIvk-e-td-j Ip-]n-Sn-p-I.

Score 6

Time 8 min

Scoring key

a. Cyclotron Score 1 MP 1, 2

b. Principle of cylcotron Score 2 MP 1, 2

c. Formula F = BqV

ma = BqV Score 1

substitution with SI unit Score 1 MP 1, 2, 5, 6

Answer with SI unit Score 1

CO: 28

To get an idea about force on a current carrying conductor and torque on current loop in a

magnetic field and electric field through demonstrations, discussion and IT

Question Text .

Question No: 29

A rectangular loop ABCD made of copper suspended in a magnetic field B is shown in this figure.

Here the conductors AD and BC are perpendicular to the field. A current I is passed through the loop.


26/30

26

a. Write down the expression for the force acting on a current carrying conductor placed in a magnetic

field.

b. Using the above expression find the force loop and torque acting on this loop.

c. What is the difference in torque acting on a rectangular loop when it is rotated in a radial and parallel

magnetic field.

sNp - sIm - p - m - nb ABCD F NXpc eqvB F ImnIt{Xn Xqnbnncnp

-Xm-Wv. Nn{X- n ImWn-n-cn-p--Xv. ChnsS AD, BC Fo mK- Imn-I-t-{X- n\vew_-am-Wv. Cu eqn-eqsS I F Id v IS- n-hn-Sp-p.

a) Id v {]h-ln-p-sIm-n-cn-p Hcp NmeIw Hcp Imn-I-t-{X-n shm AXns apI-fn A\p-`-h-s-Sp _ew Ip-]n-Sn-m klm-bn-p ka-hmIyw Fgp-Xp-I.

b) Cu ka-hmIyw D]-tbm-Knv apI-fn ImWn eqns apI-fn-ep tSmIv F{X-bm-sWvIp -]n -S n -p -I.

c) Id v {]h-ln-p-sIm-n-cn-p Hcp eqv Hcp tdUn-b amKv\nIv ^oUnepw Hcp ]mc-eamKv\nIv ^oUnepw Id-n-sm-n-cn-p-tm AXn\v apI-fn-ep tSmv GXv hn[-n-emWv hyXym-k-s-n-cn-p-Xv?

Time : 8 min, Score : 4

a. Expression 1 Score (MP 1)

b. Forces on four sides 1 Score (MP 1, 2, 5)

Torque 1 Score

c. Constant torque 1/2Score (MP - 1, 2, 5, 7 )

varying torque 1/2

Score

CO: 29

To understand the construction and working of moving coil galvanometer and its conversionto ammeter and volt meter through project and discussion

Question text :

Question No: 30

Moving coil Galvanometer is a device used for detecting very feeble current

a) What is the working principle of a moving coil galvanometer?

b) Describe the construction and working of a moving coil galvanometer

c) When a high current is passed through a moving coil galvanometer , it will get destroyed. How?

in Ipd Idv Xncn--dn-bm-\mWv aqhnwKv tImbn Kmh-t\m-ao- D]-tbm-Kn-p--Xv.a) Hcp aqhnwKv tImbn Kmht\m-aodns {]h\ XXzw F v?

b) aqhnwKv tImbn Kmht\m-aodns LS-\bpw {]h -\hpw hni-ZoIcn-pI

A

B

C

D

B( magnetic field)


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c) in IqSnb Idv aqhnwKv tImbn Kmh-t\m-ao-dneqsS IS-n-hn-m AXn\v tISv hcp-p. Fs\?

Score and Time

Scoring key

a) Working principle (Score 1 ) (MP - 1)

b) Construction and working (Score 4) (MP 1, 4, 5, 7)

c) High torque may distroy the suspension fibreHigh current may burn out the coil (Score 1) (MP - 1, 4, 5, 7,9)

CO: 30

To get an idea about how current loop acts as a magnetic dipole, its moment and torque

acting on a magnetic dipole in a uniform electric field through demonstration, discussion and

IT.

Question Text

Question No: 31

The magnetic field along the axis of a circular coil is found to be B = 3/222

2

o

)a2(r

Ia

+

a) What is the magnetic field along the axis if r>>a

b) Compare the above magnetic field with the electric field along the axis of an electric dipole and explain

what is magnetic dipole moment?

c) If the above circular coil is placed in a radial magnetic field, a torque will act. If the radial field is

replaced by a nonuniform magnetic field which is not radial, what will happen?

Hcp kp-e tImbn-ens A- n-ep amKvs\nIv ôUns kahmIyw 3/2222o

)a2(rIa

+

BWv.

a) BsWn, A- n-ep amKv\nIv ôUns kahmIyw Fm-bn-cnpw?

b) Cu amKvs\nIv oUn-s\, Hcp Ce-Iv{SnIv ssUt]m-fns A- n-ep Ce-Iv{SnIv oUp-ambn Xmc-X-ay-s-Sp-n, amKvs\nIv ssUt]m apsa v Fm-sWv hni-Z-am-p-I.

c) tad kp-e tImbn Hcp tdUn-b amKvs\nIv ôUn hm Hcp tSmvDm -Ip -p. tdUn-b ôUns\ tdUn-b-e-m t\m bqWnt^mw amKvs\nIvôUvsImv amn-bm, tImbn-en\v Fv kw`-hnpw?

Time 6 mts

Score 4

Scoring key Score

a) B = 3

2

o

2r

Ia1 (MP - 1)

b) Comparison 1

Definition 1 (MP - 1,2,5)

c) Translation motion 1 (MP - 1,2,5,10)

CO 32

To get an idea about earths magnetic field and its source through general discussion and IT.


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28

Q Text.

Q. No 32

A freely suspended bar magnet aligned in north-south direction, i.e, north pole of magnet indicates north-

ern side of earth.

a) What is the direction of earths magnetic field?

b) Explain three magnetic elements of earth.c) Write any one of the causes of earths magnetism?

d) What in the alignment of a freely suspended magnet at geographic north pole of earth?

kzX{ambn Xqnbnncnp Hcp _m amKvs\v Fmbvtmgpw sXphSv Znibn \npp. AXm-bXv amKvs\ns t\m v t]m `qan-bpsS hS-p- m-K- ns\ kqNn- n-p-p.

a) `qan-bpsS Imn-Ia-e- ns Zni GXmWv?

b) `q-an-bpsS aqv amKv\nIv Fen-sa vkv hni-ZoIcn-p-I.

c) `qan-bpsS Imn-Ikz-`m-hns ImcWw Fgp-Xp-I.

d) `qan-bpsS D -c{[p-hn (Bn-Iv) kzX{ambn Xqn-bn-n-cn-p _mamKvs\ns Zni

F -c-n-em-bn-cnpw?Time : 7 Min

Score : 4

Scoring Key

Score

a) Magnetic North to Magnetic South or 1/2

(MP - 1)

almost in Geographic South to geographic north

b) dip, declination, Bh-Explanation 11/2

(MP - 1,3)

c) Any logical statement 1 (MP - 1,2,5)

d) Vertical with north pole of magnet downward 1 (MP - 1,2,3,5)

CO No: 33

To distinguish between dia, para and ferromagnetic substances through demonstration and

simple experiment.

Qn Text:

Question No: 33

A magnetic material contained in a curved glass plate, when placed in a nonuniform magnetic field, exhibits

a property as shown.

a) Which type of magnetic material is this? Explain the property

b) Write two examples for such a magnetic material. Explain how the property relates with temperature?

c) The susceptibility of a material is small what do you mean by this statement.

h{In-Xamb Hcp mkv {]Xe-n-en-cn-p Hcp Imn-IhkvXp, Hcp t\m--amKv\nIv ^oUnshn-cn-p-tm, Nn{X- n ImWn- n-cn-p kz`mhw ImWn-p-p.

N S


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29

a) CXv GXv Xcw Imn-IhkvXp-hmWv, khn-ti-jX hni-ZoIcn-p-I.

b) Ccw Imn-I-h-kvXp-v cv DZm-l-cWw Fgp-Xp-I. Cu khn-ti-j-Xbpw DujvamhpwFs\ _-s-n-cn-p-p.

c) Hcp hkvXp-hns kk-]vn-_n-enn hfsc Ipd-hm-Wv. Cu {]kvXm-h\ sImv Fm-Wv Dt-in-p--Xv.

ScoreTime

Scoring key

a) Paramagnetic - 1/2

Explanation - 1/2

(MP 2, 3, 4)

b. Aluminium, platinum - 1

As temperature increases paramagnetism decreases -1

(MP 2, 5, 6, 7)

c) The material cant be easily magnetized -1 (MP 2, 9, 10)

CO No: 34

To understand electromagnetic induction, Faradays laws induced emf, Lenzs law, Eddy cur-

rent, self and mutual inductance through experiments, general discussion group discussion,

by solving numerical problems

Question Text :

Q No: 34

When a magnet is moved inside the coil, the galvanometer shows a deflection which is shown in figure.

a. Name the phenomena involved in this process

b) Write the laws governing this phenomena

c) Explain how conservation of energy is satisfied in the phenomena

Hcp Imw solinoid\pn Nenptm galvanometer deflect sNpXmWvNn{Xn ImWnncnpXv.

a) GXv {]Xn- m-k-amWv CXn kw -hn-p--Xv.

b) Cu {]Xn- m-k- ns\ hni-Zam-p \nb-a {]m-hn-p-I.

c) Du-kwc-W\n-baw CXn Fs\ ]men-pp Fv hni-Zam-p-I.

Scoring Key

a. Electro magnetic induction Score 1 (MP - 1)

b. Laws of electromangetic induction Score 2 (MP - 1,2)

c. Explanation of conservation of energy Score 1 (MP - 1, 2, 5)

Question Text

N

S

A

B

G


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30

Question No: 35

When AC is switched on the thin metallic disc is found to thrown up in air.

a. Which makes the disc to thrown

b. How will you explain the mechanism behind the movement of disc

c. Write the working principle of induction heater.

Nn{Xn ImWnncnp sshZypXImntep AC on sNbvXm AXn\v apIfn ImWnn-p- I\wIp-d teml- -InSv sXdnv t]mIp-p.

a) teml- -InSv sXdn- p-t]m-Ip--Xn-\p ImcWw F v

b) Cu {]Xn- m-k-ns\ hni-ZoIcn-p--sXs\

c) Induction heaters {]h\ XXzw Fgp-Xp-I.

Scoring Key

a. Eddy current Score 1 (MP - 1, 2)

b. Explanation on the basis of eddy current Score 2 (MP - 1, 2, 5, 6)

c. Principle Score 1 (MP - 1, 2, 5, 6, 9)

Question text:

Q. No: 36

A solenoid is a insulated copper wire wound in the form of cylinder.

a) When current increases flux linked with the solenoid is ..............................

b) Derive an expression for inductance of a solenoid

c) Calculate the inductance and energy stored in the magnetic field of a air cored solenoid 50cm in

diameter and 50m in length wound with 1000 turns, if it is carrying a current of 9A.

Hcp cylinder BIr-Xn-bn Npn FSp Ih-NnX sNp-I-n-bmWvsolenoid

a) Solenoideq-sS-bp sshZypI {]hm-l-Xo{hX hn- n- m Npcp-fnse Imn-Iaew ....b) Hcp solenoid s inductanceIm-Wp--Xn-\p ka-hmIyw \nm-cWw sNp-I.

c) hmbp tIm Bbn-p Hcp t]mfn-t\m-bvSns \ofw 50m, hymkw 50am NppIfpsS Fw 1000Dw BWv. Cu tkmfn-t\m-bvUn-eqsS IS-p-t]m-Ip sshZypXn 9A BsWn CXns CU-Ivkpw CXn kw`-cn--s-Sp DuPhpw Is-p-I.

Scoring Key

a. Increases Score 1 (MP - 1,2)

b. Derivation Score 2 (MP - 1, 2, 3, 5, 6)

c. Formula Score 1/2

Substitution Score1

/2Answer with unit Score 1 (MP - 1, 2, 3, 5, 6, 9)

Metalic disc

Electro magnet

physics01- model papers of kerala board of higher secondary education, xiith class

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