Theanswersprovided in the marking scheme are for reference only. They are not the only possible answers.Alternativeanswers are acceptable so long as they are well reasoned.

Theexaminationemphasises the testing of understanding, the practical application of knowledge and the use oflfOCessingskills. Candidates are advised to study this document in conjunction with the examiner's commentsoocandidates'performance in this booklet.

foressay-typequestions, candidates are expected to demonstrate an understanding of the question, an ability toi1eployrelevant knowledge of the subject in response to the questions, and to present their answers logically andcoherently.

(ii) Let P be the pushing force and m be the mass of the player.Taking moment about 0:(0.47 m + 1.18 m) P cose = (1.18 m) mg sine

tane = l.65P ..... (I)l.18mg

QI Taking moment about G,(0.47 m) P cose = (1.18 m) R sine-(l.18 m)fcose

or Taking moment about shoulder,(0.47 m)mgsine+(0.47 m+ l.18 m)fcose= (0.47 m+ l.18 m) Rsine

Resolving forces horizontally and vertically:Horizontal: P =f (2)Vertical: R = mg (3)

Maximum friction = 0.25 mg .... (4)

From (I):l.65P

tane = ---1.18mg

tane = 1.65 x 0.25mgl.18mg

tane = 0.350Hence e = tan-I [0.350] = 19.30

Work done = gain in k.e. of the football = ~ mu2

2

w = ~ (0.42 kg)(20 m S-I)22

= 84 J

WAverage power = -

t84 J---0.15 s

= 560 W

1. (b) (ii) Horizontal: ucos(} (= x ... (I)

Vertical: y = usin() (+ ~ (-g) (2 ... (2)2

Solving (I) and (2), u = 20 m s-t, () =35" and x = 38 m(= 2.32 sy = -0.29 mTherefore, Roy should place his hands at h = 1.7 m above the ground.

Path difference between the two rays (ray CD reflected from the upper slide and ray ~reflected from the slide below) = 2t

Phase difference on reflection (at the slide below) = 1t or equivalent path difference = 2::.2

'A..'. 2( + - = m'A. where m = 1, 2, 3, for a bright fringe to be formed at 0

2

Difference in air-film thickness between adjacent (bright or dark) fringes = 2::.2

Th fi'A./2 1.0 mm

ere ore -- = ---d 60 mm

560x 10-9m 60 mmd= ----x---

2 1.0 mmd = 1.68 x 10-5 m

The air-film thickness increases by 2::. for a smaller fringe separation,2

the angle between the slides and hence the paper thickness are larger by ratio.

I ! I !I I I I

I I I I

I I I I

I I I I

I I I I

I I I I

I I I Ii i

wider fringe separationfringes more or less evenly spaced

to rhd.c. :

F: Force on RS upwards.I: Current in solenoids anticlockwise (view from the left)

copperwire frame

to rheostat and smoothd.c. supply

~(ii) F

(0.072 x 10-3 kg)(IO m S-2)

B

leB(2.0 A)(0.03 m) B0.012 T (or 12 mT /1.2 x 10-2 T)

Theoretical value B' = ~onI

(4n x 10-7 H m-I)(~)(2.0 A)0.30 m

0.0126 T

- the finite thickness of the copper wires- the finite width of the gap between the solenoids- the flux leakage of the soleniod- the arm RS may not be exactly perpendicular to the magnetic field- the theoretical value only gives an estimation of the B-field since it is valid for infinitely

long solenoid- the theoretical value is the B-field at a point on the axis inside an infinitely long solenoid- the effects of the earth's magnetic field- the effects of the magnetic field due to apparatus nearby- difficult to attain balance and hence there is error in I(Accept other reasonable answer)

(iii) At one end of a solenoid,B 0.012T72

= 0.006 T (or 6 mT / 0.6xlO-2 T)

MarksIAIA

ANYTWO@I

Marks Paper 1 Sect

4. (a) (i) Molecular kinetic energy <K. E.> = 2- kT (or 2- ~ 7)2 2 N A

= 2- (1.38 X 10-23J K-1)(273 K) 5. (a) 02 Q

= 5.65 X 10-21J lA Q(J

(ii) Mass of an atom m = (0.0202 kg mol-I) = 3.355 X 10-26k(b) T6.02 x 1023mol-I g

I 3 th2- m vnns = - kT

2 2 (c) T(3)(1.38x 10-23)(273) (2)(5.65 x 10-21)

using the answer from (a)(i))iii

Vnns =3.355 X 10-26

(or3.355xI0-26

= 580 m S-I IA(d) (i

(iii) N = nN A = (2.5 mol)(6.02 x 1023morl) = 1.505 x 1024

The internal energy isU = N <K.E.> = (1.505 x 1024)(5.65 x 10-21J) I

= 8503 J IA(i

(b) (i) PoVo = NkT(100 x 103 Pa)( Vo) = (1.505 x 1024)(1.38 X 10-23J K-1)(273 K) I

Vo = 0.0567 m3 IA

(ii) W= pt-.V= (100 X 103 Pa)(0.0567 m3) I= 5670 J IA

(iii) Internal energy at state Y is 8503 J (same isotherm as state X)

Temperature at state Z doubles that of state X because the volume is double as the pressureremains unchanged.

Ux = Uy = 8503 JUx = TxUz Tz

TUz = -!:- Ux = 2(8503 J)Tx

= 17006 JV VWork done by the gas = f pdV = nRT In ( ----Y- ) = Po Vo In ( ----Y- )Vx Vx

= (100 X 103 Pa)(0.0567 m3) In ( ~ )I

= 3930 JConsider path a:t-.U = Q + W (W = work done on the gas)

0= Q -3930 JQ = 3930 J

Therefore, heat flowing into the gas = 3930 J

Q = t-.U + W (W = work done by the gas)=0+3930J

Q = 3930 J

Draw a cross / letter / figure on the translucent paper.OrBlack out the laboratoryOrMake sure that the axis of the convex lens is perpendicular to the cardboard.(Accept other reasonable answers.)

The object distance u is shorter (or equal to) the focal length andtherefore no real image can be formed (or formed at infinity).

c) The (whole) image of the hole can still be seen (though it is dimmer) sincelight from all points of the hole can be focused on the screen by the uncovered part of the lens.

~) (i) For (u + v) at a minimum at 400 mm or 40 cm,v = u = 200 mm or 20 em

.fi. v image distancemagnl IcatlOn m = - (= b· d. )

u 0 ~ect Istance= I

~50mm

C= _1-jV

f= 400 Hz and V= 25 VFrom the graph,slope = 6.3 x 10-9 (A m) or 6.3 x 10-3 (IlA m)y-intercept = 0.5 x I0-6 (A) or 0.5 (IlA)

1 = 6.3 X 10-9 ~ + 0.5 x 10-{;d

.". C= ~ = _1_(6.3 X 10-9 ~+ 0.5 x 10-6)jV jV d

I (6.3 X 10-9 ~ + 0.5 x 10-6)(400Hz)(25V) d

= 6.3 X 10-13 ~ + 5.0 X 10-11 (d in m, C in F)d

(ii) Stray capacitance present due to capacitors formed by the metal plates with other conductorsnearby.

When a 2 mm metal plate is inserted, the set-up is equivalent to two capacitorsconnected in series. The capacitance of each is given by

GoA = 3 Co where Co is the capacitance before the metal plate is inserted.d /3

h . I .. C 3 CoT e eqUiva ent capacitance IS ' = --2

= 138.3 X 10-12F

The increase in capacitance t"!.C = 138.3 X 10-12F - 92.2 X 10-12 F= 46.1 X 10-12F

(11) The increase in charges t"!.Q = t"!.C x V = (46.1 x 10-12F)( I00 V)= 4.61 X 10-9 C

(III) The increase in energy stored M

= ..!.. t"!.C X V2 = (..!.. )(46.1 X 10-12F)(lOO V)22 2

= 2.31 X 10-7 J

(ii) Work done by the d.c. sourceW= t"!.Q x V= (4.61 X 10-9 C)(IOO V)

= 4.61 X 10-7 J

Part of the work done by the source becomes the energy stored in the capacitor andthe rest is for doing work on the 2 mm metal plate / dissipated in the connecting wires / circuitwhen current flows / heat generated in the wires.

VL or /(arbitraryunits)

(ii) The phase difference between Vs and VL decreases (VL shift to the right).Or The peak value of VL increases.

(i) The inductive reactance of the coil increases linearly with the source frequency f(ii) The inductive reactance is negligible compared to 100 kO.

/ = VR "" 6 Vo R 100 kO

6 X 10-5 A (i.e. 60 J..lA)

(iii) VL = /oXL = /0 (2rcfL)

VL ImV.. .. ... ... .... ... .. .. .. .. ....... ..... .. .. I···:·····'C .. .. .. .. .. .... .. -.- .; .- .'. .. ,. .. .. ..•. .. ... j., .., .. , ,

.. '.'..•.. , .. O' .. ... O' " ..0 O' .. ... ..• .. O' .. .... . .. .. 00

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"" "

j;.. , .~..

r... ..' ,r.

-.,. .. IM for COITectaxes labe Is , ,.. .. ; .. .. ..

... o· O'

.....

1M for graph Iotted /'., ... .. ..= COITect p :' ... ; .. , ...

.. ". ".. .. H, :. .. .. .. .... ;... .. ... : .. .. . .. ...... ,.; .. ..

~: . ;; , / , ..'.' ,

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.... .. .. ; -,.. ....•.. .. ; ;.. .. ...... .,.. ;.. .., .. ;. .,. .. i-'~' ... ,.. /' ",_. ,. -,.. ,.. '... ; ;.. .: ,.. ., .,.-, .

..... .~.... .. ..

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, '/ .. .~...~.".. ~

Slope of the graph VL Vsf is /0 (2rcL)(Since the peak voltage of the source is kept at 6 V andfis below 50 kHz, /0 remainsconstant at 6 x 10-5 A)

200 X 10-7 = (6 X 10-5 A)(2rcL)L= 5.3 x 10-4 H

force weight of columnpressure = -- =

area base area of column= pAhg

A=pgh

liquid columnof density p

(b) (i) Pressure at the upper surface PI = Po + P ghl

Pressure at the lower surface P2 = Po + P gh2pressure difference = P2 - PI

= pg (h2 - h,.)

(ii) Unchanged as it depends only on the thickness or difference in heights (h2 - hI) for acertain liquid.

(iii) By Bernoulli's principle and in this case VI= V2= 0PI + pg(H - hI) = P2 + pg(H - h2) where H is the depth of the liquid

P2 - PI =pg (h2 -hJ)

water of density 1000 kg m-3

net pressure force/upthrust

cavity at theblock's centre

weight of the block

(No need to indicate the pressure forces acting on other faces of the block)

pressure force = pressure difference x area= (1000 kg m-3)(10 m s-2)(0.06 m) x (0.10 m x 0.08 m)= 4.8 N (equals the weight of the block)

Let Vc be the volume of the cavity.mass of the block density x volume

ms = (2700 kg m-3)(0.1 0 x 0.08 x 0.06 m3- VC)

weight of the block ms g4.8 N (2700 kg m-3)(0.10 x 0.08 x 0.06 m3 - Vc)(10 m S-2)

VC 3.02 x 10-4 m3 or 302 cm3

(Qr using average density of the block = density of water)