2010 H2 Physics Ans - RI

Question Answer

1 D

2 B

3 D

4 C

5 C

6 B

7 C

8 B

9 B

10 C

11 B

12 C

13 C

14 D

15 C

16 B

17 B

18 D

19 C

20 D

21 B

22 A

23 B

24 B

25 A

26 C

27 B

28 D

29 A

30 C

31 D

32 D

33 C

34 B

35 D

36 A

37 B

38 A

39 D

40 C

www.erwintuition.com


1

1 (a) vernier callipers

zero error (do not accept parallax)

(b)

dV h

V d h

V d h

V

V

2

3

3

3

=π4

= 964.665 cm

∆ 2∆ ∆= +

2×0.01 0.1∆ = + ×964.665

8.50 17.0

= 8 cm

= (965 ±8) cm

2 (a) U W T

ρVg

V

U

ρ

1 1

1

-3 3

2

2 -3

-3

= -

= 50 - 40

=10 N

10 =

10=

1000×9.81

=1.02×10 m

= 50 -34

=16 N

16=

1.02×10 ×9.81

=1600 kg m

(b) (i)

(ii)

s

s

Taking moments about O,

5.00×500+(70×9.81× )=1000sin53.0°×10.0

=7.99 m

(iii)

R T

R T

R

x

y

2 2

At equilibrium, the net vertical and horizontal forces must be zero.

= cos53.0°= 602 N

= 500+(70×9.81) - sin53.0°= 388 N

= 602 +388

= 716 N

2010 JC2 Prelim H2 Physics Paper 2 Suggested Solutions

weight of

beam force by worker

on beam

tension reaction

force



2

3 (a) Gravitational field strength acts in the direction of decreasing potential / As distance

from the centre of a spherical mass increases, gravitational field strength decreases

but the gravitational potential increases.

(b) Total energy on surface = Total energy at infinity where PE=0 & KE=0 (since

spacecraft launched with minimum kinetic energy)

−

E s

E

GM m

Rmin

KE + = 0

∴

-11 24

min

11

6.67×10 ×5.98×10 ×2250KE =

6370000

=1.41×10 J

(c) Gravitational potential energy heat (air resistance) + light (fire over its body) +

sound energy (due to rapid vibrations of the body of the spacecraft)

(d) True weightlessness occurs in a situation where the astronaut is remote from the

gravitational field of celestial bodies i.e. gravitational force is zero / net gravitational

field strength is zero / experiences no net acceleration due to zero gravitational

force on him.

The astronaut and the spacecraft he is in are both accelerating with the same

acceleration, hence the normal contact force on him is zero / the gravitational force

on the astronaut entirely provides for the centripetal force for him to be able to

move in circular orbit, hence the normal contact force on him is zero. Here,

weightlessness is just a sensation as the gravitational force on him still exists.

4 (a) It is the sum of the kinetic energies and potential energies of the particles in the

gas.

(b) (i)

=pV nRTUsing ,

−× × ×

= =×

=

pVT

nR

5 31.0 10 5.0 10

0.20 8.31

300 K



3

(ii)

5 (a) (i) 0.25

sin 14.48 14.5R

Rθ θ= ∴ = = °

Sine rule and cosine rule are also accepted.

(ii)

(iii)

( )( )

( ) ( ) ( )

29

3

2

0

tan

15 1010.5 10 9.81 tan 14.5

4 0.5

0.0799 m

8.0 cm

EF mg

R

R

θ

πε

−

−

=

×= ×

=

=

(b) θ will be smaller.

R R

0.50 R

hemispherical bowl

2θ

Normal reaction

force due to the

bowl, N

Electrostatic

Force due to

other ball, FE

Weight of ball, W



4

6 (a)

Magnetic flux

through small

coil

time

Induced

e.m.f in

small coil

.

time

(b) Magnetic flux through the small coil is proportional to the magnetic flux density

which is proportional to the alternating current in the large coil. Hence the magnetic

flux-time graph through the small coil has the same shape as that of the current-

time graph for the large coil.

Induced e.m.f. in the small coil is proportional to the rate of change of magnetic flux

linkage which is obtained from the negative of the slope of the magnetic flux- time

graph.

(c) When the planes of the two coils are at 900 to each other, the magnetic field due to

the current in the large coil is parallel to the plane of the small coil. Hence there is

no magnetic flux linkage with the small coil and no e.m.f. will be induced in the

small coil.

The trace on the c.r.o. will show the amplitude of the induced e.m.f. reduced to

zero.

7 (a) (i)

4

6

4

5.6 24 14 1881.6

1881.6 3000 60 60 2.0321 10 MJ

1kWh 1000 60 60 3.60 10 J

2.0321 10 MJ 5644.7 kWh

P UA T

E Pt

∆= = × × =

= = × × × = ×

= × × = ×

∴ × =

(ii) 1 2

9

( ) (5.6 3.2) 24 14 806.4

806.4 3000 60 60 8.7091 10 J 2419.2 kWh

Savings 2419.2 0.25 $604.80

P U U A T

E Pt

∆ ∆

∆ ∆

= − = − × × =

= = × × × = × =

= × =



5

(b) (i)

1 2 3

1

1 1 1 1 1 1 1

1.4 1.9 1.4

1 1 10.51154

1.4 1.9 1.4

0.51154 60 14 430

C

C

C

U U U U

U

P U A T W∆

−

= + + = + +

= + + =

= = × × =

(ii) t / mm P / W

50 430

100 250

150 170

200 130

(iii)

Note: Heat transfer via conduction dominates and the relationship between rate

of heat transfer P and thickness t of thermal insulation is given to be ∆

=kA T

Pt

,

where k is the thermal conductivity of the medium, A is the surface area normal

to direction of heat transfer and ∆T is the difference tempearture Thus, graph

should be an inverse curve.

300

400

200

100

0

x

x

x

x

t / mm

150 200 250 50 100

P / W



6

(iv) The extrapolation of the graph to 250 mm may not be accurate because:

- 250mm is outside of the range of data collected.

- the behaviour of the graph/relationship may change beyond 200mm.

- there are too few data-points to determine the shape of the curve accurately.

- it is difficult to extrapolate a non-linear graph accurately.

(c) (i) From the graph,

=

= =

= × = =

= = = =

ideal

hheating e

h

heating e

e

COPe

QCOP

W

Q tWP

t COP

mod

mod

0.094

1( ) 10.638

( ) 0.60 10.638 6.3828

/ 5000783 W

( ) 6.3828

(ii) A heat pump requires less power than actually needed,

thus they are more cost effective (although not efficient).

(iii) Refrigerator, air-conditioner



7

8

Diagram

Arrangement of LDR and electric light bulb:

Electric light bulb circuit:

LDR circuit:

Problem Definition

To investigate how the resistance of the LDR depends on the intensity of the

illumination incident on the LDR.

Dependent variable (R): resistance of LDR

Independent variable (i): intensity of illumination incident on the LDR

Controlled variables: distance between light source and LDR or light intensity

sensor

e.m.f. of dry cells

alignment of light source with LDR or light intensity sensor

black cardboard tube

to LDR circuit to electric light

bulb circuit

LDR / light

intensity sensor

electric light bulb

mA

1.5 V dry cell V

A

V 12.0 V battery



8

Apparatus and Material

electric light bulb

rheostat

light-dependent resistor

light meter with light intensity sensor

digital ammeter

digital milli-ammeter

digital voltmeter

dry cells

connecting wires

black cardboard tube

sticky tape

metre rule

Procedures

1. Set up the apparatus as shown in the diagrams above. Tape the LDR in

position at one end of the black cardboard tube with adhesive tape. Tape

the electric light bulb at the other end of the black cardboard tube with

adhesive tape, using a meter rule to ensure that it is aligned along the

same horizontal axis as the LDR.

2. Adjust the rheostat in the electric light bulb circuit to maximum resistance.

Close the circuit to switch on the light bulb.

3. Measure and record the p.d reading V on the voltmeter and current

reading I on the milli-ammeter in the LDR circuit.

4. Replace the LDR with the light intensity sensor, connected to a light meter,

at the same position. Record the intensity reading i on the light meter.

5. Increase the resistance of the rheostat and repeat steps 3 and 4 to obtain

at least 6 sets of readings.

6. The resistance of the LDR can be calculated using the equation

VR

I=

7. The distance between the electric light bulb and the LDR or light intensity

sensor is kept at a constant distance throughout the experiment by using

adhesive tape to fix them in position and using a metre rule to measure the

distance between them before each reading to ensure that it is constant.

The e.m.f. of the dry cells is checked by connecting a voltmeter across them

before each reading to ensure that they remain at a constant value. The alignment

and orientation of the light bulb and LDR or light intensity sensor are kept constant

throughout the experiment by using adhesive tape to fix them in position.

Analysis

Assume that

nR k i=

Where i is the intensity of the illumination incident on the LDR,

R is resistance of the LDR , and

k and n are constants

Taking lg on both sides, +lglg lg kR n i=

.

Plot a graph of lg R against lg k.

If the above relationship is true, a straight line graph will be obtained where the

gradient is equal to n and the y-intercept is equal to lg k.

Hence 10ck = where c is the y-intercept.



9

Safety Precautions

1. Do not look directly at the bright light source. Wear polaroid protective

glasses.

2. Do not touch the hot light source with bare hands. Wear gloves when

handling the light bulb after use.

3. Do not handle electrical circuits with wet hands.

Producing Reliable Results / Additional Details

1. As the resistance of the LDR is quite large, the current reading will be

small. Hence a milli-ammeter should be used to measure the current in the

LDR circuit.

2. The voltmeter should be placed to measure the potential difference across

both the ammeter and LDR when the resistance of the LDR is high. This

will ensure that the ammeter will measure the small current through LDR.

3. The electric light bulb and LDR or light intensity sensor are placed in a

black cardboard tube to minimise light from the surroundings from

reaching the LDR or light intensity sensor.

4. Wait for intensity and multimeter meter readings to stabilise before

recording.

Other accepted variations of answer

1. The multimeter can be used as an ohmmeter to measure the resistance of

the LDR (note that in such a case, the ohmmeter is connected directly to

the LDR, no e.m.f. source is required).

2. Intensity of the light incident on LDR can be varied by varying the distance

between light source and LDR. Power of the light source must be kept

constant in such a case.



1

Physics Prelim Paper 3 Marking Scheme

SECTION A

1 (a) It is the motion of a body where its acceleration is directly proportional to its

displacement from a fixed point and is always directed towards that fixed point.

(b) (i) B is being forced to oscillate because P is heavier.

(ii) 1. Increases the damping of the rod B.

2. Decreases the natural frequency of rod B.

3. Increases the coupling between pendulum P and rod B.

Increases the amplitude of the driving force on rod B.

(iii)

2 (a) - The electrical resistance R of a conductor is defined as the ratio of the p.d. V

across it to the current I through it.

[No definition of electrical resistance – minus 1M]

- The electrical resistivity ρ of a material is the constant of proportionality relating

the electrical resistance R to the dimensions of the material (length and area).

OR word definition of L

RA

ρ= or A

RL

ρ =

OR R depends on dimensions while ρ is a material characteristic/property

(b)

( )

( )

22

2 2

2 2

2

2

4

4

OR

0.350.00.15

30.0

BC BCAB AB AB

BC BC BCAB AB

L L LR

A dd

d dR L L

R L Ld d

d

d

ρρ ρ

ππ= = =

= × ×

= × =



2


(c) (i) Terminal p.d. for the 2.00 V battery:

2.00 1.00 V2

r BJ

rV V

r= × = = (null deflection)

Since balance length1

2BJ BCL L= , 2 2 2.00 VBC BJ rV V V= = = .

(ii) Method 1

From part (c)(i),

2.00 VBCV =

Method 2

From part (c)(i),

2.005.00

0.400

BCBC

VR = = = Ω

I

From part (b), 0.15

AB AB

BC BC

R V

R V= =

2.00 0.15 0.30 V

0.30 2.00 2.30 V

AB

AC

V

V

= × =

= + =

OR

2.00 1.15 2.30 VACV = × =

( )2.50 2.30 0.400

0.500

ACE V IR

R

R

= +

= +

= Ω

From part (b), 0.15AB

BC

R

R=

5.00 0.15 0.75

0.75 5.00 5.75

AB

AC

R

R

= × = Ω

= + = Ω

OR

1.15 5.00 5.75 ACR = × = Ω

2.50

0.400

0.500

T ACR R R

R

= = +

= Ω

(d) Over-estimate. 0.20 V is actually the p.d. across R as well as that of the ammeter.

( )

0.500 0.500

AC A

A

E V I R R

R R R

− = +

+ = Ω ⇒ < Ω

To show that the calculated R is an overestimate, there must be some statement

relating the resistance of the ammeter to either

- pd calculated for R actually includes the pd across the ammeter

- resistance calculated of R actually includes the resistance of the ammeter

3 (a) (i) E1 Ground state

E2 Metastable state

E3 Excited state

B2 for any 2 correct answers



3


(ii) Electrons are pumped up / excited from E1 to E3.

Electrons at E3 will quickly spontaneously decay from E3 to E2. Since E2 is

the metastable state, a population inversion is created between E2 and E1.

When one electron falls spontaneously from E2 to E1, a photon will be

emitted. This photon will go on to stimulate emission of electrons at E2.

These photons form the laser light.

(iii)

3

2

4

4 (3.0 10 )

26.53

minimum 26.5

E t

hE

t

hh f

f Hz

E f fE hf E hf h f

E f f

f Hz

π

π −

∆ ∆ ≥

∆ ≥∆

∆ ≥×

∆ ≥

∆ ∆ ∆ = ⇒ = ⇒ ∆ = = ∆

∆ =

(b)

(i) Occurs when the most energetic electrons are stopped completely and all

their kinetic energy is converted to photon energy.

(ii) Since

min

hc

e Vλ =

∆, minimum wavelength will increase when the accelerating

potential is decreased.

The intensity at all wavelengths will decrease because the speeds / kinetic

energy of the electrons are decreased.

minλ

0

Intensity

Kβ

K

α

Wavelength



4


4 (a) In the magnetic field,

2

mv mv

Bqv rr Bq

= ⇒ =

Since Tv

=2πr

,

2mvT T

v Bq Bq

mπ ∴ = ⇒ =

2π

(b) (i) 2T

π∴ = =

-27

-7

-19

)1.54x10 s

0.85 x 2 x 1.6 x 10

(6.68 x 10

(ii) In order for the nucleus to accelerate when it crosses the gap, freq. of the

alternating voltage = orbital freq. of the nucleus

1f∴ = =

6

-76.49 x 10 Hz

1.54 x 10

(iii) KE after one revolution = Work done by e-field on helium nucleus = 2qV = 2(2e)V = 4eV

(iv) The gain in KE after each rev. = 4eV

The gain in KE after five rev. = 20eV

2120

2mv eV∴ =

5406 57 10 -1

m s.eV

vm

= = ×

Comments:

(b)(ii) Common mistake is the failure to realize that in order for the nucleus to accelerate

when it crosses the gap, freq. of the alternating voltage = orbital freq. of the nucleus

because the ion crosses the gap twice in one revolution. A handful of students halved

the period of the ion and took the reciprocal to calculate the frequency of the voltage

supply which is incorrect.

(b)(iii) Common mistake 1: KE = 2eV.

Common mistake 2: KE = 2e (450 V) = 900 eV (eV is not electron-volt !)

[B1 – for sub]

[A1 for final answer]



5


SECTION B

5 (a) N = W cos θ

= (3.00 x 9.81) cos 30.0o

= 25.5 N

(b) a// = g sin θ

= 9.81 sin 30.0o

= 4.91 m s-2

(c) h/s// = sin θ

s// = h / sin θ

=0.500 / sin 30.0o

= 1.00 m

v//

2 = u//

2 + 2 a// s//

= 02 + 2 x 4.91 x 1.00

= 9.82 or 9.81

v// = √9.82 = 3.13 (shown)

(d) sy = uy t + ½ ay t2

2.00 = (3.13 sin 30.0o) t + ½ (9.81) t

2

4.905 t2 + 1.565 t – 2.00 = 0

t = 0.499 s

sx = ux t + ½ ax t2

= (3.13 cos 30.0o) (0.499) + 0

= 1.35 m

(e) (i) Impulse = area under F-t graph

= ½ (0.200) (360)

= 36.0 Ns

(ii) vy = uy + ay t

= (3.13 sin 30.0o) + (9.81) (0.499)

= 6.46 m s-1



6


∆py = m [vy – uy]

36.0 = 3.00 [vy – (–6.46)]

vy = 5.54 m s-1

(iii) Since the KE or speed after collision is smaller, the collision is inelastic.

During collision, the KE of the ball is converted into sound energy, thermal

energy and/or elastic PE as the ball deforms.

6 (a) Any two:

1. The waves must be coherent.

2. The waves must have approximately the same amplitude.

3. The waves must be unpolarised or polarised in the same plane (for

transverse waves).

4. The waves must interfere to give regions of maxima (constructive

interference) and minima (destructive interference).

(b) Intensity required at 12 km away,

( )

( )

11

7 -2

4

2.5 105.0 10 W m

0.50 10

eye

eye

PI

A

−

−

−

×= = = ×

×

Consider light guide,

.

light

light

PI

A

P I A

=

=

( ) ( )27

5.0 10 4 12000π− = ×

905 W=

(c) (i) Shorter wavelengths means the anti-nodal lines will be closer to one another.

Hence, aircrafts may “lock on” to the wrong line of maxima / difficult to identify

the central line of maxima / difficult to differentiate the lines of maxima.

(ii) Since the two radio waves are in phase, along centre-line, path difference is

always zero / phase difference is always zero / P & Q are equidistant from

any point on the centre-line. Hence constructive interference occurs.

(d) (i) P is nearer to the aircraft.

Hence intensity (or amplitude) of signal should be higher.



7


(ii) From Fig. 6.3, phase difference

2

π= rad

(iii) Since phase difference of signals

2

π= ,

path difference 4

λ=

Distance from P to plane 2 2 2

4800 180 480 4827.30 m= + + =

Distance from Q to plane 2 2 2

4800 230 480 4829.42 m= + + =

Hence ( )4829.42 4827.304

λ− =

8.49 mλ =

635.3 10 Hz

cf

λ= = ×

Note:

Do not accept if student uses formula D

xa

λ= as it is not a 2-D problem.

Accept if student choose path difference as 5

4

λ,

9

4

λ, etc

(e) (i) If aircraft is on the central anti-nodal line,

it should detect maximum signals from both frequencies / the maximum

signal will be stronger

OR

If aircraft is on wrong anti-nodal line,

only one of the frequencies will show a strong signal.

(ii) If the ratio is an integer ratio,

It means that higher orders of maxima will still coincide/overlap

Hence the aircraft could still detect maximum signals from both frequencies

even though it is not on the central anti-nodal line.



8


(f) Advantage:

Can still work under low visibility conditions / Use of detector to align aircraft is

more accurate than using visual inspection.

Disadvantage:

Possible interference of signals from other sources (e.g. radio stations,

telecommunication base stations, etc) / It is more costly to install the emitters and

receivers on every airplane.

7 (a) Binding energy is defined as the amount of energy needed to split a nucleus into

its individual nucleons.

(b) (i) 4 3 1

2 2 0He He n→ +

(ii)

( ) ( )

Difference in total BE

4 6.8465 3 2.2666

20.5862 MeV

Q =

= −

=

(iii) ( ) ( ) ( )

( ) ( )

He-3 He-4931.494

He-4 He-3

20.58621.0087u

931.494

0.9866u

Qm n m m

m m

+ − =

−

= −

=

(iv) Energy is supplied in order to conserve mass-energy.

(c) (i) Helium-2 is unstable and cannot exist in a bound state.

(ii) Large coulomb repulsion between the protons.

(d) (i) Half life is the time taken for a sample of radioactive atoms to decay to half

its initial number.



9


(ii) ( )

( )

( ) ( )

.......... 1

........... 2

1 / 2 :

ln2

9ln2

6.149

0.682

P P

S S S

P P

S S S

P

S

S

P

P S

S

A N

A N

A N

A N

NT

NT

TN

T N

N N

λ

λ

λ

λ

=

=

=

=

= ×

=

(iii) 0

ln2

6.14

0

ln2

0

ln2

6.14

0

ln2

0

ln2 ln2

6.14

Let the total initial activity be

0.1

0.9

0.19

0.9

81

1 1ln2 ln81

6.14

ln81108 days

1 1ln2

6.14

P

P

P

P

P P

tT

S

tT

P

tT

S

tP T

t tT T

P P

P P

A

A A e

A A e

A A e

AA e

e

tT T

t

T T

−

−

−

−

−

=

=

= =

=

− =

= =

−

(iv) It can be used to trace a plant’s fertilizer uptake.



Documents

2010 H2 Physics Ans - RI