srjc pre alevel paper3

8/14/2019 SRJC Pre Alevel Paper3

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SERANGOON JUNIOR COLLEGEGeneral Certificate of Education Advanced LevelHigher 2NAMECG INDEX NO.

PHYSICSJC2 Pre 'A' Level ExaminationPaper 3 Longer structured euestionsCandidates answer on the euestion paper.No additional materials required.

9745t0316 October 20092 hours

READ THESE INSTRUCTIONS FIRSTWrite your Civics group number, index number and name on all the work you hand in.Write in dark blue or black pen on both sides of the paper.You may use a soft pencil for any diagrams, graphs or rough working.Do not use staples, paper clips, highlighters, glue or correction fluid.Section AAnswer all questions.Section BAnswer any two questions.You are advised to spend about one hour on each section.At the end of the examination, fasten all your work securelytogether.The number of marks is given in brackets [ ] at the end ofeach question or part question.

This document consists of 21 printed pages and 1 blank page.

For Examiner's Use1

234567

Total

2009 9745/03/Pre A Leveu2009 ffurn Over


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3Section AAnswer ALL questions in this section

Fig 1.1 shows a vertical current carrying wire I of 2.O A (flowing out of paper) insideuniform hodzontal magnetic field (shown along plane of the paper).N+=

Fig. l.1

The direction of the magnetic field is at bearing 340" (or 20' west of north). The magneticflux density B of the field is 0.040 T. The wire carrying the currenl 1 is 30.0 cm long.(a) Draw on Fig 1.1 the magnetic field at P due to the current and label it "a". tll(b) Give the bearing of the magnetic field at P due to the current L

bearing = ' tlt(c) lf the resultant magnetic field al P points exactly north, find the ratio

Bmagnetic flux density at P due to the current

ratio = l2l

(d) Draw onsRJC 2009

Fig 1.'l the magnetic force acting on the current and label it "d". 111[urn Over745/03iPre A Level/2009


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4(e) Determine the magnetic force acting on the current and give its bearing.

magnetic force =bearing =

(0 lf a second vertical wire carrying a current flowing into the paper is now placedgive the bearing of the resultant force acting on this second wire.

bearing =

N [1]' t1l

at P,

" t1l

9745/03/P.e'A' Levei/2009RJC 200S lTum Over


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5(a) Explain why the potential difference across the terminals of a battery is always lowerthan its e.m.f when it is connected to a circuit.

State the condition under which the potential difference acrcss a baftery,s termjnalsequal to its e.m.f.

,"r r r r r t"*" "

r;;-, ; ;;";

;";,, ;;;; "; ;; ;";",;; ;, " :;j:'in series wilh a cell Z of e.m.f. 3.0 V wilh internal resistance 0.5 O.

Fig.2.1Calculate the potential difference between X and y.

potential difference = v l2l

sRJC 200S 9745/03/Pre'A' Level/2009 ffurn Over


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6

(c) Another circuit consisting of a cell W in series with .1.0 O and 2.0 Oconnected to positions X and p which are 8O.O cm apart. This is illustrated resistors isin Fig. 2.2.

(0 lf the galvanometer registersshown on the ammeterFig.2.2

a null deflection at position P, determine the reading

(ii) Calculate the power loss in cellZ.

ammeter reading = A 121

w t2l

1.0 () 2.o a

9745/03/Pre'A' Level/2009power loss =

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7(iii)Wire XY is replaced with another wire of the same material and length but with asmaller cross-sectional area.

State and explain the changes (if any) in1. the balance length,

.................... ... t212. the final ammeter reading at balance.

9745/03/Pre'A' Level/2009RJC 2009 [furn Over


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B

3 (a) When Uranium-23s nuclei are fissioned by slow moving neutrons, the foilowingreaction takes place: \2{iu*,jn >'lj r+'!'.v+ot.Oldentify the particle c and state-1he number b of such particles produced in the reaction.= lqLna{ - Re*r"rn,rI\>t Lr\u,rJ. (l4/lyg*( - lfu.,r )c' o = ....3.............. trl"=....1t1........... trtJ----\ - t^r^n"r'a(b) The blngng enQrgy per nucleon of U-235, I-131 and Y-102 are 7.6 lvlev, 8.5 MeV and8.6 MeV respectively. Calculate the enerqy released bV '1.0 kq of Uranium.T. A^ nt v4k",,c = liEr + BEy * BFw

-- ( rrr x 9-g) 1(luz'1'u)- 1rr


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sln scientific research, spectroscopists often use the emission line spectrum of gases touniquely identify the elements that constitute the gases. The emission line speitrum ofhydrogen is shown in Fig 4.1.

400 nm(a) Using the concept of electronspectrum is formed.

700 nmin atoms, explain how an emission line

Fig 4.1energy levels

t2l(b) Fig 4.2 below represents the lowest energy levels of the electron in a hydrogen atom,with the principal quantum number n and the corresponding valuls of energyassociated with each level shown.

0.54 eV-0.85 eV.1.5'1 eV-3.40 eV

-n=2

n=5n=4n=3

Fig 4.2Calculate the longest possible wavelength of the spectral line observed when anelectron undergoes a transition from the energy levels shown in Fig 4.2 to the groundstate (n =1).

wavelength =9745/03/Pre'A' Level/2oos

m [3]

-13.6 eV

2009 lfurn Over


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10(c) The electron transition in part (b)(i) is an example of spontaneous emission.

ln contrast, in lasers the dominant process that leads to that formation of a welFcollimated intense beam is stimulated emission. The word LASER is an acronym forLight Amplification by Stimulated Emission of Radiation.By drawing a suitable energy level diagram, explain the process of stimulaledemission.

(d) Explain why the energy of electrons in solids is represented by energy bands, whereasthose in isolated gas atoms are represented by discrete energy levels.4rr, or-on^r a* i{r,."eJ. ............J........'7 o\,t?-hnur+ e- & ,r.ot iu-^i tr ,-g. o_ "t&-oi'. eC,tLt r- lpr e"pryl..... . .... ...-...... . .I ...... c' AC,^iL l!,r'at,,fr!}.. :.. e:Fr:.. ?3..s".W f:*t!:{.t31v ' Pal,, tc 6e (u-'- f.\',o@:L- Low gt W;r z 2o-


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't1Section B

Answer two questions from this section.5 (a) (i) Explain what is meant by the intemalenetgy of agas.

t21

itst3I

(ii) The pressure of ideal gas, p, is related to its density, p, by the equationp = ir(c')wnere (c') istne mean scuare speed of the molecules.Show that the internal energy of an ideal gas is directly proportional tothermodynamic temperature.

(b) Exprain. using kinetic theory "ffi$l_-z-a l,u{r'4 I",, ,&^*r (0,^ ;(i) the specific latent heat of iApoi.--rsation is higher than the specitrc lalent heifusion for the same substance,Ge-,

sRJC 2009 9745/03/Pre'A' Level/2009 lfurn Over


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132. Determine the final pressure ofthe gas.

t1i

finar pressure of ga" = ... lr..?.1.1.1.y.'-.... p" t t(iii)During the power stroke of the engine, the gas expands by doing 62 J of work,

while no lhermal energy enters or leaves the gas.'L State the tirsf /aw of thermodynamics.

2. By applying the law to this process, calculate the change in the internal energyof the gas during the power stroke.

change in intemal energy = J I2I

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(ii)15

Use Fig. 6.2 to explain how a magnetic fieldsimultaneously in the region without changing the and an electric fieldmotion of the charged may existparticle.

Fig.6.2t)

o-t !0. rY

2. I

fuu=Fc'FE fr- fuu+" lt v-{.rttz.feJu1, r3l

15.0 cm

0.r


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16(ii) Calculate the rate of increase of ihe area of toop ABDCA.

rate of increase of area =(iii)As the rods AB and CD move,

Explain why an electrical

m' st 121an electrical signal is generated in the coil ABDCA.signal is generated using Faraday's law ofelectromaonetic induction'b -G

121

(iv)Calculate the e.m.f. induced jn the loop ABDCA., r= l-gJ)- $ru^- A&= 0.({ (

(d=)

v't go-- o,rgo J-l \ 1611),-1BAu)id (?vi

e.m.f. induced = v L2l

sRJC 2009 9745/03/Pre A' Level/2009 lturn Over


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17(v) Describe and explain the direction of the induced cunent in the loop ABDCA.

l2'l(vi)Rod AB is now being pulled at 3.0 m sl in the same direction as rod CD as shownin Fig. 6.4. Explain whether the e.m.f. induced will have a smaller or largermagnitude compared to that calculated in (cxiv).

15.0 cm

BFig. 6.4*Q,'^"aLg-',

- S X sb^;rte,v'

t-ums

121

SRJC 200S 9745/03/Pre'A' Level/2009 lfurn Over


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18Fig 7.1 below shows how x-rays are produced inside an x-ray tube. The electrons emittedal the calhode are accelerated from rest using an accelerating voltage to hit a targetmade of a metal at the anode and produce x-rays as a resull.

A graph of relative inlensity against wavelength of the emitted radiation shows anemission line spectrum superimposed on a continuous spectrum as shown in Fig. 7.2.RclsliveIntsxrty

Wavelaqtlr / l0'll ml'L Fig.,.z(a) (i) Determine the maximum energy of the x-ray photons emitted, and state how sucha photon arises. r* . J_cn v-

L ^.. - '.. - l. b6.r r9-'f 5maximum energy = lrk! {le-.11 J l2l

E'ectrons VacuumHeated f ilamenl calhode

Acceleratingvoltage %Fig.7.1

l.h.+ rnjYsRJC 2009 9745/O3/Pre'A' Level/2009 ffurn Over


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^,--\ 19(ii[ Shotv that the acceterating vottage Vd used is 104 kV.P7X*wtl e' - 4\/ = t'ro1,r,P''*V : . ,. .-- lrtgbV

t11(iii)The accelerating voltage is reduced to one lhird of its original value. Determine theminlmum wavelength of the x-rays emitted.(Tw4t y,:,r V nt*t V * vb *:,'^.1 .et--* lb f L1 hra-r.-a:e fa *06 tl."?L."e .

minimum wavelength =. ................... m til(iv)Sketch on Fig. 7.3 a graph to show how the new retative intensity of the x_raysemitted varies with wavelength.

Rclniyclntqrsity

t2l

(v) Calculate the speed at which the electrons hit the target when the acceleratingvoltage is 104 kV.

speed or erectrons = l1l..i.!9.1.. r",o,

Fis.7.3

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(vi)20

By making other suitable calculations, suggestelectrons are not observable in this experiment.tPIt

why the wave properties of

\ -- C-At xtD-'t"^

wJM{t-

4tl, v9J(

(b) (i).. ... f\ . .... . ......... ...................!.y. ...... . .. ... 121/\A e' I t--* svn'k" +t \r^ ; aloF{-1 9*Y * r*Explain how the following observation of the photoelectric "ff""t prouiO""B?if;*that electromagnetic radiation has a particulate nature:

There is no no-ticeable time lag between iltumination and the emission of electrons,irrespective of the light intensity.

e- &K"*,'o-r

liele-399::: SJe^,ttr^^g: #- pna;.+44 l3l0(4+\^}t t& at^. w^,L? .

(ii) Ultra-violet radialion of frequency 4.83 x 1015 Hz is incident on a metal surface andthe resulting photoelectrons have a maximum kinetic energy of 1.2g x 1O-18 J.1. Calculate the energy of a photon of this radiation.

ev [2]sRJC 2009 9745/03/Pre'A' Leveu2009

energy =

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212. Calculate the maximum kinetic_ energy of the photoetectrons produced byradiation of frequency 6.76 x 10'5 Hz incident on the same melal surface.

maximum kinetic energy = ................. J t313. By making suitable calculations, explain briefly what will happen when a veryhigh intensity radiation of frequency 2.50 x 101' Hz is incident on the samemetal surface.

t1lEND OF PAPER

Acknowledgement:With thanks toACJCJJCPJCSAJCTPJCYJC

9745/03/Pre'A' Level/2009RJC 2OO9 Over

srjc pre alevel paper3

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