F.Y.

515

March 2013

Part- III PHYSICS :vi:aximum: 60 Scores Time : 2 Hours Cool off time: 15 Minutes General Instructions to Candidates :

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There :':::: a 'Cool off time' of 15 minutes in addition to the writ1ng time of 2 hours.

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neither allowed to write your answers nor to discuss anything with d.ccring the 'cool off time':

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the 'cool off time' to get familiar with questions and to plan your answers.

0

Read the questions carefully before answering.

• ·All questions are compulsory and only internal choice is allowed. •

When you select a question, all the sub-questions must be answered from the same question itself.

•

Calculations, figures and graphs should be shown in the answer sheet itself.

•

Malayalam version of the questions is also provided.

•

G~ ·.· 2

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Electronic devices except noriprogrammable calculators are not allowed in the Examination Hall.

2:: ,:.2. ::ic:-:.s v;11ereyer necessary.

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K-20

1

(P.T.O.)

515 Pick out the fundamental unit from the following :

1

Second, m/s, Newton, Joule

1.

®06l':l' o.JOffi.lJ,mJOJ
(1)

6)(1\)cOOc§', OJI RiD I 6l
(1)

Match the following.

2

r---.---------~----------------------+---~----~--------~-------------~

SL

A

B

:

~K_o.+-----------------------+-----------------·--------~ . Newton's first law

1

Change in momentum

ffiJJ§6lr(lj 63cm::Jo me.Jm m1CWOJo

. @'@c00(5Y0)16leJ OJJ®J::Jffilo

Conservation of linear momentum

2

Action B

. Newton's third law

3

Law of inertia

I g-Q~mi0n91JO mlcwOJo .

ffiJJ§6lr(lj OlJmJ::lo me.Jm m1WOJo

Impulse

4

reaction

Momentum before collision = .:\1omentum after collision 6lcB:>O§In91Jm OlJ,cibCTL.J~ ®'g}cOOo ~(fl)n910lJ,§§

3

3.

· There are three distinct modes of heat transfer.

= 8c3::8~- c·.5< :'lu

®'g)cOOo

o.D"lB LS::Jrill
a) The main mode of transmission

of heat by which the sun heats the surface of the earth is :

ffi@cfuJ,mJJ,:

i) Conduction

i) .0J::Je.Jmo

ii) Convection

ii)

ffiloOJo.Dmo

iii) Radiation

iii)

n.flcB:>1Cll61Do

iv)

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iv) None of these b) Explain briefly, the occurrence

b) ®Oo.J 6l6l
of a sea breeze based on heat


transfer. 4

(1)

(1)

. illJ,CllJ,c001 nJl(!()BQJOcOOJ,
A small water drop placed on a lotus leaf is· spherical in. shape.

4.

(1)

b) Why does the· small water drop acquire a spherical shape?

(1)

K-20

(1)

anoOJro
a) Define surface tension.

(1)

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(1)

(1

2

515 5.

Fill in the blanks.

. (6x'!2=3)

a)

•••••••• ~••• ~-<";~ ••••••••••

F·S F·S

scalar quantity

b)

mass. m

momentum, p

KE= ........ '" .. ..

((l'f]l)cB{))o' P

w®1~cfuoro~o

......... collision

momentum ·conserved

energy conserved

......... 6)cfu0~n.5::i®

((l'f]l)cB{))o rruo
~ 'D ffi~o rruo
c)

d)

unit of power o..J OJ ol6l CVl

e)

f)

••••• ;

Body of mass, m

Power, P nJOJffi,

6.

746 watt

w 46!Tl1 8

OJCQ;J6o116lCVl 1110%)",

p

m

•••• 0. 0

•••

~.

PE= .................... .

~wcno,

m.illOm1<8cfuOfil~o

h

= ........... .

P= ........................ .

scalar product

P·= ........................ .

m)tgcfue.J& 6llnJOMce;'S

A student plucks at the center of a stretched string arid observes the wave pattern produced.

b) Plot the above wave pattern pictorially. Label the nodes and the antinodes on the pattern.

6.

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63(()2,

cfum.JlCW66lS

®(()o(f)6013 cfO

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a) O{J3®6 ®rn<0'®1e.J6~ ®rnoW6013cfO ((l'f]l)6Tfl

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( 1)


b) ®
o..J0<8RmO

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(2)

g)2'D

nJ0<8Rsm1 arn

((l'f]l) CVl1 <8mo M 6
rJ eJ 6lli a5Q 6) .aJ ~6&3:,.

7.

density ( p) and modulus of

ffi)Jlffi® ( P

)

( 2)

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(E) CW2,6lSCW2,o <8Qle.JOsm" uo mJB<0'®16lCVl lnJ <8 OJ wo ( V) ((l'f]l)@co.fl!3Jl rn1 . cB€) 6cm®". cE cw 66ls il6lM6lQl®n,9:1smam <8n..OOffiQl2,e.J ML-l T- 2
elasticity (E). (The dimensional formula of E is 1'vfL~ 1 T-2 ). principle

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Velocity of sound depends on

a) State the homogeneity.

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0 ••• •• 0

at a height, h

a) What type of wave is produced in the string?

7

= ........... .

of (1)

(1)

lo..Jm)®OoJlcB€)J,c£lj.

b) Using the above principle, arrive at an expression for the velocity of sound. (Take K = 1).

.K-20

b)

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(P.T.O.)

515 8.

8.

Hooke's law states that stress rx

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a:mfiHlmlffi,!,nJ ::.l®(OJO)"i ~::16m"

strain.

cru"6161LSCQhfQ

a) What is the necessary condition

LoJcru"®::.JOJlc&ii),!,('ffi,!,.

for the above law to be valid?

O{j)rm"

@'@JOJCftiJQl::.JCQJ m16DJmDm ~enD'?

a:mJocJil 63m,J, rucru"®,!,rulm,!,6nSO
stress and strain for a given

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solid material under increasing (2)

l(f) 0 n.Ol61 CVl

63 (l) 6

6l1J ffil) 0

cruoD::.JCQJI1(0{0)::.JSJ,
a) State and explain the work done

9.

a)

®061':l'cJ:jOWJ,rm

in the following situations :

61 !2J w" ®

i) A person carrying a heavy

oJl ru ml cOOJ,cfu.

load wEtlks on a level road.

( 1)

b) 61 s cfO 6161 Wd m'O cru"61 Ls cru" ru fll BD1 <;81 c002,

graph, the relation between

9.

m1wco

a) nD,J,cfu"cru" mlwmo CTDQW,!,OJ::.lcfu,!,rm®lm"

(1)

b) Explain with the help of a

tensile stress,

oD,!,cfu"cru"

i)

(1)

!2JJ,RJ,oJOSJ,cfu~lm'O

LoJ OJ,\ (010)\

LoJ cru"® ::.l rul ~"

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OJJcfu"®l m1mcJ:j::.JCQJ a:o::.JrJJI m'Ocfu~s1

ii) A man spending his energy

( 1)

mScOO,J,rm,J,.

by pushing on a concrete wall.

ii) 63ffiJ, 11cfuOG'ffilLc£b"lff !51(0{0)1CQ51m'O (1)

(1)

b) 200N 61cVJ 63(06 C"U.S':ffi6YlieJo 63ffiJ,

displaces a body through 6 m in

ru cru"® 6rul61 m

the direction of the force. Find

5m

6'l'L 21 (010)161

cffl

B1 Cftl cw1 m'O

B~roo m"lcOO.;rm,J,. OJml®,J,o.!l61cVJI1alm'O

61!2JCQJ"® LoJOJ,)(OIO)l cfu6nSJ,oJl~c&ii),!,cfu.

(1)

1 0.

10. Kinetic theory of gases is based on

(1)

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m0c001w::.J611) ru::.J®cfu60B~61s w®la:cfuoro~

the molecular picture of matter.

cru1BD::.Jcrmo LnJmJ®::.JoJl~ro1cOOJ,rm®.

a) Write any two postulates of· kinetic theory of gases.

®~1

63m::.J::JD ~'DfO~o ..2.lleJOJ::.lcOO,J,rm6.

b) A constant force of 200N

the work done on the body.

( 2)

a) OJO®cfu60B~J,6!S.

(1)

W®lcfu

cru1 BD ::.1 crm (010)/61 ~ 1 eJ J,o ro 6n5"

b) Write short notes on : a:oJ::.JmlRJ,a:eJffml' ntl)':l',J,®J,cfu.

i)

Equipartition of energY

ii) Mean free path

(1)

b) cfuJ,0161cJ:j':l'J,®J,cfu:

(1)

i)

~cfuj1oJofil§1m1® 630n.O" Cl8)mfO@il

ii) m"lcfll Lo.di oJo
K-20

(1)

4

( 1)

(1)

515 11. The motion equation

repres~nted

by the

y(t) = A9os(wt+¢)

11.

y(t)=ACos(wt+¢) 1s

ffi)QlOJOcEh,lo

LoJ
61.2JCJtjJ,lTO

called simple harmonic motion

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(SHM).

(SHM)

a) Which one of the following. examples closely represents SH;;rf? Substantiate your ans\\".er. (1) i) The rotation of the earth about its axis. ii) Oscillations of a swing. b) A vibrating simple pendulum of period Tis placed in a lift which 1s accelerating downwards. What is
a) ciD061':l'
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~BOn.DaJ6m60B~1affi

SHM

61m loJ®1m1womo

(1)

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i)

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ii) 63(()6 ~'D6'f1mj)OeJl61<"@ <1l'@§o. · b) ®O
nJl ro1 cw cw161 m

(1)

g)261®60B61ffi 6YlJOW1c006lTOJ,. c)


61c€h06n'56~

CYUDOffiOm>laJo

y = 2Cos(0.5nt + n/3)

y = 2Cos( 0.5n-t + n/3)

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Find the amplitude and period of the particle.

(1)

nJl ro1 cw cw 6o

(1)

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12. Figure given below shows the motion of a school bus starting from the point 0 and travels along a straight line.

12. 0

ntj)

ffi) 6r.!li ro1 c00 6rm

63 ro 6

m)ce:,~Ub 6YlJCfid161<"@ .2Je.Jffi!il06ID' .n.Jll®CIT®1m'll ce:,o6l1'51.!1[l ro1 c006rm®. ·

OS

28

48

68

88

108

128 •148

0

A

8

c

D

E

F

Om

10m

20m

30m

40m

50m

a) Complete the following table.

60m

G

?Om

a) ®061':!'
(1)

oJJiOCIT®1 CWO cOO2,c€h.

Time taken

Displacement from 0

0@52, (QYO) ffi) Ql m.Jo

ffi)LDOmomnroo

0 CQJlm'llm1mi'

Velocity LnJ
28

10-0 =10m

...........

10 s

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

5 m/s

I K-20

5

(P.T.O.)

515 b) Is the motion of the bus uniform or non uniform? Justify your answer.

(1)

(1)

c) Draw the position---time graph of the above motion.

6lo.J::xrulcUJ®-6!6!8o LCDOnl)u OJ(l)CQJucOOJ,cfu.

(1)

d)

d) A student in the school bus

611J
oJl 8 J 0 ill ®O'l

( 1)

611J lpdl6l CiJl

notices the speedometer of the cruuolla:NJomiRiO cBoo6ffi'lcOOz,rmml' n@®u

bus. Which type of speed 1s shown by the speedometer?

(1)

®ffio G:OJCD®CQJo6'f\)?

(1)

Answer either question 13A or 13B.

13A. According to l\ ewton' s law of motion,

13A.ffiJ,I§6)CVl __QjeJffi (Y)iCQJQ)o ®OffiJCIDCD~-±u 6X~o

the force depends on the rate of

@@) cOO

change of momentum.

ffi@®Wl.gJMcOOJ,®J,. a) 611J eJ o ®O ~cOOJ,®®kn crDru;o
a) Name the law that helps to measure force.

.cL 2J m

(l)

mCQJ IL (010): 6! cfil

b) Using the above law, deduce an

~ o.J a: cw ocnlg,f

611J eJ (010)'l mu

(2)

(2)

a:~ollcmffi'lcOOJ,cBo.

c) A man jumping out of a moving

c)

611J
.!:lJ eJ'l c00 6(If)

bus falls with his head forward.

o.JJOG:(OTO)O~{ !.lJOSJ,® lllffiJ,dl:lJ® ®eJ

What should he do in order to

_lllJ,®
land safely?

(1)

® 0

CQJ16! eJ 010) 0 ®

OR

6

®0 CQ) 0 ciD

otD \I'OlJ, (1)

6! !.lJ q)J6ffio ?

OR

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(1)

CE o.J 6! C?l cro)?

b) a: mam <;8 o 6'fUIO) mi CQ) mo

expression for force.

mnl a: cOO'16! m

OJ J ® J 0 cru (610)l@ CVl

515

OR

OR

~mill~ lOJ(OJOlOcBo~rm1cWleJ~

13B. 200 ill"1Rill

13B.1'he outer side of a circular track of radius 200m is raised to make an

63(()2, o_iO®CQJ2,6JS o..J60o ISOWo

angle of 15° with the horizontal.

illocWl 15° a)

a) Wh1ch

force

pro"\-ides

the

a car taking the circular track?

(1)

6W eJ o

little

above

63ro6

cBooo1m"

mro'OcBo6(f()

6W eJ o

(1)

0()3®?

.

m1 ill (UIO) 2, (f()

·

.

~CQJill(I)'(Q)1


the

innerside.

a:o_~ocmocm

b) @(() 6 OJ ~ 6'flliO) o_i 0® Cill 66! S ~ cfO IS 0 (f)

outer side of a curved track is a

OJ~(I)'(Q)OcBo~Ml cWl e.i6~ o..JO@cWl@JcB{))~S1

@])OJ cro,)ill 0 cWl ro1 cB{))6(f() 6) ffi) em lS16) oJ R@0

b) Name the process by which the raised

~CQJill(I)'(Q)1cWlro1cB{))6(f()6·

cBoS(f()6

necessan- cen:ripetal force. for

®1ro~mru6

lo_1 lcBo1CQJ w"cB{))"

a: o_1 ro" ( 1)

. mro'OcBoc?,cBo.

(1)

c) gJyo a: cBo cru1 ro'O ® (f()1ro1cB{)) 6(f() a: cw R o

c) Using the data provided in this case, determine the maximum

63 '91 OJ ocB{))6(f()®1 m" ®Om 6OJ B ml cw Ql oCQJ

permissible speed to avoid

0()3 R OJ 6o

skidding. (Given J1 s = 0.25).

cro"o.ll cw"

m1ill~cWlcB{)),J,cBo. ( f..L s = 0.25)

(2) 14.

14. Figure given below depicts the sche~1:2. tic ::._~epresenta tion

cBo ~ s1 w

®06!'9

@(f()1ro1cB{))c?,(f() .2.llL®o 63(!J2,

Lo_i®1m1womo

of an

(2)

cg)cW~6Jm

6Jill~2,(f()J,.

engine.

Hot , Reservoir T1

a)

\Yllic~

type of engine is this, a

heat e::-:gine or a refrigerator?

a) (1)

nDIR

b) Write the -± steps of operation in the CarlJ.Ot cycle.

b) (2)

oB3®6 ®ffi~1e.J~ omcmm1mo~? ntl)
®OLOOJO ocoD1ma:oRill? . (1)

cBooillQlmo~r

6) 6) ffi)

cB{))1~161 ~

lo..JOJ ill(UIO)m(I)'(Q)16Je.J ffiOe.J,J, i2J2,0JSJ,cfucf0

c) A Carnot engine is working

(2)

08)'9J,®,J,cBo.

between temperatures of 27°C,

c) 63(1)6 cmoiOa:mo§"

and 327°C. Find its efficiency ( TJ ).

gQ®

~'Do9JQlooJ1mJ,o

(1)

omcr~m1cm

327°C

27°C

~'Dn91"illooJ1mJ,o

gQscWlmo lo_j OJ ill (I)'(Q)1 cB{))6(f() 6.

®0®16)~

08)nD1n'l1JciOcru1 ( TJ ) cBo6rrSJ,oJ1 s1 cB{))J,cm.

K-20

7

(1)

(P.T.O.)

515 ((ffi)6)~9:b1am

Answer either question 15A or 15B.

15A

15A Three vessels of different shapes

15A. OJJ®,)CTUu®

height 'h' and their bottom parts are to

o:lJim,J,

'h' am

6)0J~o

. m10~
manometers

measuring the pressure. The water lev~ls

(ll'@em~®1 CQJl eJ,J,~

nJ:::>L®6llB~1am 63
:tre filled with water to the same connected

15B nfD'l',J,®J,em.

m:::>a:m:::>o:JIRO,J,ems o.eJS1~D.z!1
in all the vessels remain the

ntj)~:::>

same.

nJ:::>L®6llB~1
g:JeJffi1
63Ql(l)QlnJ:)6)eJ ((ffi)OJ
A

a) Identify the above phenomenon.

(1) (1)

b) Predict the pressure level shown by the manometers.

b) moa:momiROi,emcill

(1)

em:::>6lTJ1cOO,!,tm

mro13(0'(0)16)
(1)

c) Blood pressure in humans is greater at the feet than at the brain. Explain why.

®eJcill

(1)

em:::>amnJ : :> B6lll3~1 eJ :::>GID. ntj)OID,J,6)em:::> 61"@~

d) Pick the odd one out.

(1)

oJ10J
Dentist chair, hydraulic brake, hydraulic press, venturi meter.

emJS,J,®am

d) emJ3(0'(0)1 am 6)nJ s:::>(Q'(O)@ nfDS,J,6)(Q'(O)'l'6®,J,em. (1)

B rrml ~u

OR

6)D.J (Q) (0'

6) 6)o.D~i emu


(1)

OR

K-20

8

OR 15B.The flow of an ideal fluid in a pipe of varying cross sect~on is shown.

15 B. OJ,\®J
OR ~ lce:, ocrou

ce:, 6 cpeJ1eJJ6lSCl2!6~

6l cru ce:,un9:l m 6ce:, 63
cm

63 Ill 6

6lnQCUl1Cl2!am

o..Ou~Cl2JlCUl16lc(q 63':P?,ttJ£) cfu06ID1.~,dl
B.

~

a) Differentiate between streamline flow and turbulent flow. b) State and prove Bernoulli's principle.

~~~As P ..

a) (1)

o..Ou~§.OCl2!J,O Cl sill 6YlJ!,eJ

frDulSlo 6l6leJcill o..Ou~§O.Cl2!J,O

®11ID1eJ6~

c(ll

OJj®,\OfrDo

(1)

nQ':PJ,®J,c&.

b) 6)6YlJ (/) ~ 6ITJ 0 gll ffi)u

(3)

J

(Q) (Q) 0

loJ cruu(Q) 0 OJ1 d1£) 6

. cfuCl2!J,o 6l®§.1Cl2!ld1£lJ,cfuCl2!J,o 6l.2Jgt)J,cEb.

16. ln an experiment with a bicycle rim, keeping the rim in the vertical position with both the strings in one hand, put the wheel in fast rotation (see figure). When string B is released, the rim keeps rotating in a vertical plane and the plane of rotation turns around the string A.

a) :Mention the law that explains the above result.

K-20

16.

.2J
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(1)

(P.T.O.)

515 b) Explain the practical example (shown in the figure) based on the law mentioned in a).

b)

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(1)

(1)

c) How \Vill you distinguish a hard boiled egg and a raw egg by spi:cming each on a table top?

C)

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(1)

d) A cylinder of mass 20 kg rotates about its axis with an angular speed of 100 rad s- 1 • The ~·adius of the cylinder is 0.25 m. What th'e magnit11de of angular momentum of the . cylinder about its axis?

d) 20kg OJJcr.J ·eo' (nJ 62C.. 100 Tad

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17. An athlete jurnps at an angle of 30° with a maximum speed of 9.4 mls.

17,

a) What is the shape of the path followed by the athlete in the jump?

(1) mmmre.Jp,~. (())06!rs'1w

b) Obtain an expressiOn to calculate. the horizontal range covered by the athlete.

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(2)

c)

c) Find the range covered by him in the above jump. Suggest the angle by which the athlete can · attain the maximum range.

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10

515 Ans\ver either question 18A or 18B.

18A ®06l~ffi:>1Cll'O

18A. The escape speed for an object from

18A.
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the earth is 11.2 km/s. a) What is meant by escape speed?

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a)

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18B %)Y'6®6cBo.

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.

6)

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®Oill®.OQl0c:006(ffi6l®mDV? b) Arrive at an expression for the escape speed from the earth. c)

b)
d! r1e

earth

contains

c) (1)

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an d)

atmosphere while the moon does not. Give the reason.

(2)

%) (0)0)1
speed depends on the mass of not.

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Explain whether the escape

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. (1)

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OR OR 18B.
18B.'I'he acceleration due to gravity (g)

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a) Cf)J,COJ,®JOcBoillnS:I6m

a) Define the acceleration due to gravity (g).

(g)

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(1)

(1)

b) @@Y'o 'd' 6lcfu061"'@ 'g' mllCll'O OJCOOOJt,cm b) Derive an expression for the Yc:,~·iation

the surface of earth. c1

_-\:

or

6

(3)

c) 'g'

the earth?

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(3) 'g'

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'h'

be half of that on the

sun ace

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,_,-:-cat height 'h' will the value 0

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of g at a depth 'd'

O{j) ()'U)?

(2)

(2)

11

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