1

Preliminary Examinations 2010

JC2 Physics H2 Solutions

Paper 1 Solutions 1 B 11 A 21 A 31 A 2 B 12 D 22 A 32 B 3 D 13 C 23 D 33 C 4 D 14 C 24 C 34 C 5 D 15 D 25 C 35 A 6 C 16 D 26 B 36 D 7 B 17 B 27 A 37 C 8 B 18 B 28 B 38 A 9 A 19 B 29 A 39 D 10 C 20 A 30 C 40 D 1 Solution: B

A LW

AA L W

A AL W

A L W

=∆ = +

∆ = +

∆ =

l w

l w

w + l

2 Solution: B

Frequency of radio wave ranges from 3 Hz to 300 GHz Question: 500 µHz = 500 x 10-6 Hz---- too small Mass of lightest atom; Hydrogen-1 is 1.007825u; ~ 1.66 x 10-27 kg Mass of heaviest stable atom; Lead-208 is 207.9766521u ~ 3.45 x 10-25 kg Question: 500 pg = 500 x 10-12 g = 5 x 10-10 kg (too heavy) Acceleration due to free fall 981 mm s-2 (too small--- should be 9810 mm s-2)

3 Solution: D

v2=u2+2as 49.12 = 14.72 + 2(9.81)s s= 111.86 m; 112m

2

4 Solution: D

The time taken for the flight up to the highest point should be shorter than the time taken for the flight down.

5 Solution: D

To find T1, take pivot at rightmost end of bridge

1

1

( ) ( )2

( )2

(1 )2

+ − =

+ − =

= + −

b c

b c

bc

lW W l x T l

lW W l x

Tl

W xW

l

To find T2, take pivot at leftmost end of bridge

2

2

( ) ( )2

( )2

( )2

+ =

+ =

= +

b c

b c

bc

lW W x T l

lW W x

Tl

W xW

l

6 Solution: C

As linear momentum is conserved, there is total transfer of momentum during collision as all 3 discs are identical.

7 Solution: B

Upthrust = Weight of balloons + Weight of house ρairgVhelium = ρheliumgVhelium + Weight of house Weight of house = 1.2 x 9.81 x 0.17 x 20000 – 0.18 x 9.81 x 0.17 x 20000 = 34000 N

8 Solution: B

By Hooke’s Law, the component of the sphere’s weight down the incline causes the spring to compress by a value e.

mm 4.29

m 0294.0500

30sin)81.9)(00.3(or

sin0

=

≈=

=

e

kemg θ

3

9 Solution: A

The kinetic energy is ½mv2 and remains constant as it reaches constant velocity, Rate of change of GPE = mgh / t = mg(h/t) = mgv

10 Solution: C

( )

2spore spore

2seoul seoul

o 2spore

sporeo

seoul

cos37.5

11.26

cos37.5

a r

a r

r

a

a

ω

ω

ω

=

=

=

= =

11 Solution: A

By the principle of conservation of energy,

2

2

-1

10 0

21

80.0(9.81)(30.0 30.0cos50 ) (80.0)214.5 m s

i i f f

o

PE KE PE KE

mgh mv

v

v

+ = +

+ = +

− =

=

By the principle of conservation of momentum,

-1

( )

80.0(14.5) 0 (80.0 45.0)

9.28 ms

T T J J T Jm u m u m m v

v

v

+ = ++ = +

=

12 Solution: D

( )

-1

A C C

2 2A C

2 2

2 22

c

At point C, particle's velocity is entirely horizontal.

i.e. 3.0 m s

By COE,

KE GPE KE

1(2 )

25 3

0.408 m4

322.1 m s

0.408

v

m v v mg R

Rg

va

R−

= −

= +

− =

−= =

= = =

4

13 Solution: C

( )

2 3

32 3mercury mercury2

neptuneneptune

32 2

neptune

neptune

5.79450

4500.241

5.79

165 years

T R

T R

RT

T

T

∝

= =

=

=

14 Solution: C

16

89.73 (1)2300

97.14 (2)600

(1) 89.73 600

(2) 97.14 2300

7.41 148095

19985.83 m

3.00 10 kg

GMR

GMR

GMR

RR

R

R

M

φ = −

− = −+

− = −+

+= =+

==

= ×

15 Solution: D

( )

o

1

o

o

o

11

At equilibrium position,

mg

0.10 1010 N m

0.10

When extended distance from equilibrium position,

mg

10 N m10 rad s

0.10 kg

cos( )

0.020 cos(10

ω

ω

−

−−

=×= = =

− + =

− − =

= −

⇒ = = =

∴ ==

o

kx

mgk

x

x

k x x ma

mg kx kx ma

ka x

m

km

x x t

t )

5

16 Solution: D

Ek = ½ mw2xo

2sin2wt 17 Solution: B

2λ = 0.056 v = fλ (3.0 x 108) = f(0.056/2) f = 10.7 GHz

18 Solution: B

I α A2

For maximum intensity, Io α (3A)2

For minimum intensity, I α (A)2

Io/I = (3A/A)2 I = Io/9

19 Solution: B

20 Solution: A

For option B and D, we require two progressive waves to create interference pattern and standing wave. Option C is not true because light wave only travels at the speed of light, not at any velocity. A wave can also behave as a particle. As such, it can carry a momentum.

21 Solution: A

For option B and C, it is only true if the system is an ideal gas. For option D, different states of the system will have different internal energy. Therefore, it is only true if the two system are at the same state. According to 1st law of thermodynamics, the internal energy of a system depends on the heat absorbed and the work done on the system. Thus, the internal energy of the system can also be changed without heating the system.

22 Solution: A

pV = NkT p = (N/V)kT p = NvkT 0.50p = N’k(2.0T) N’ = 0.25 Nv

6

23 Solution: D

To remain stationary, mg = eE = e(V/d). The equation can also be arranged as mg = 2e(0.5V/d) when the oil drop acquire an additional negative charge.

24 Solution: C

Electric potential is a scalar. It add up to zero at point X and Y. The vector sum of the electric field at point X due to + 5 µC and - 5 µC point upwards. The electric field at point Y due to + 5 µC points in Northeast direction while that due to -5 µC points Northwest direction. As a result, the resultant field at point Y points upwards.

25 Solution: C

Before voltage supply is reversed, diode is forward-bias.

Total current in circuit = V / RX = 12 / 2 = 6.0 A

Since diode is ideal, hence current across resistor Y or Z = 6.0 / 2 = 3.0 A

Potential difference across resistor Y or Z = I (R) = 3.0 x 2 = 6.0 V

Hence, potential difference across voltage supply = 12 + 6.0 = 18 V

After voltage supply is reversed, diode is reverse-bias, no current will flow through the diode. Effective resistance = 2.0 + 2.0 = 4.0 Ω Total current in circuit = V / R = 18 / 4.0 = 4.5 A Hence, voltmeter reading = Iacross X (R) = 4.5 x 2 = 9.0 V

2.0 Ω 2.0 Ω

D

2.0 Ω

V

X

Y Z

7

26 Solution: B

p.d across the 14 Ω resistor = 3 x 0.5 = 1.5 V I = V / R = 1.5 / 1.4 = 0.11 A (or 0.107 A) ε = 3.5 x 0.5 = 1.75 V Using ε = V +Ir 1.75 = 1.5 + 0.107r r = 2.3 Ω

27 Solution: A

Only A is the conclusion we can draw from the information given. If the filament of the fifth bulb has broken, the voltmeter will register the voltage across the transformer and not a zero reading. We cannot conclude that C is correct. B and D are possible but further testing need to be done.

28 Solution: B

The equivalent networks are: and

11 15 2

1 2 52.5 5(2 )

2.5

YZRR

RR

R

− = +

+=

= Ω

1

1

1 15.0

1 12.5 7.5

1.88

XYRR R

−

−

= + +

= +

= Ω

29 Solution: A

B field within coil P will be directed upwards. Using LHR, force on wire will be directed towards TS. Hence rod will move towards TS with acceleration i.e. increasing speed.

X Y

R

R 5.0 Ω R R

Y Z

5.0 Ω

8

30 Solution: C

oB

2o

2oB

B

2o

7 2

7 2

2

2

2

Since the wire is suspended,

2

4 10 100.0(0.020)(9.81)

2

2 10 100.00.020 9.81

0.0102 m

1.02 cm

IF I L

d

I L

d

IFL d

F mgL L

I mg

d L

d

d

µπ

µπµπ

µππ

π

−

−

= × ×

=

=

=

=

× × =

× ×=×

==

31 Solution: A

Magnet (N-pole) approaches coil, there would be changing magnetic flux linking coil. According to Faraday’s law an induced emf is generated. Since, circuit is closed, an induced current is formed. The direction of current would produce a ‘N’ pole. When magnet is inside the coil, there is no change in flux linkage and no emf and current is induced. When magnet (S-pole) leaves the coil, there would be changing flux linkage and an induced emf and current is produced. According to Lenz’s law, the current flow would produce a ‘N’ pole on the right end of the coil. This is opposite to when it approaches. Hence, diagram A.

32 Solution: B

Reasoning: As the ring moves closer to the electromagnet, it has more flux linkage. Faraday’s law states that an emf will be induced and Lenz’ law tells us that there will be an opposing force acting on the ring to slow it down. This external force will damp the system. The best answer is B.

9

33 Solution: C

Fig A, Mean Power, W =

2

02 2

022

rms

VV V

R R R

= =

Fig B, Vrms = 2area under the curve over aperiod−V t

T

= ( )2

0

0

12 0

2 2V T

VT

+=

Mean power for Figure B, PB = ( )2

2 20 0

2 24rms

VV VW

R R R= = =

34 Solution: C

Irms = P/V = 8.333 A to 10.9 A Corresponding I0= 11.79 A to 15.4 A Expression for I= I0 sinwt

35 Solution: A

T ∝ , where k = Thus if T’ = T ⇒ k’d’ = kd

W’ = W

W’ = W

W’ = W W’ = 0.577 W

10

36 Solution: D

∆E = 1% × 150 keV = 1.5 keV = 2.4 × 10-16 J ∆E ∆ t ≥ h/4π ∆t ≥ h/∆E4π = 2.2 × 10-19 s

37 Solution: C

Option A describes Stimulated Absorption – An atom can be ‘excited’ from a lower energy state (or ground state) E1 to a higher energy state E2 when one of its electron absorbs a photon of energy corresponding to hf = __ E2 – E1 . Option B should be: Spontaneous emission occurs more rapidly when the lifetime of the excited state is short. Option D should be: spontaneous emission competes more strongly with stimulated emission in transitions that emit blue or ultraviolet light than for transitions that emit red or infrared radiation (revisit the visible spectrum and compare the frequencies of the two radiations).

38 Solution: A

When the battery is reversed, its current drops to zero: This indicates that the semiconductor device is acting like a PN junction in reverse bias.

39 Solution: D

2413

2ln

0

30

−

−

=

=

eA

eAA tλ

CiA 61034.8 −×= (this is the activity of radio isotope in the body after 24 hours)

Percentage found in the thyroid = 4

100% 48%8.34

× =

40 Solution: D

A significant majority of the alpha particles will pass through the gold foil undeflected (0o). Deflection by large angles (up to 180o) is possible but likelihood is low and decreases with magnitude of angle.

11

12

Paper 2 Solutions 1 (a) The Principle of Superposition states that when two waves of the same kind

meet at a point in space, the resultant displacement at that point is the vector sum of the displacements that the two waves would separately produce at that point. [B2]

(b) (i) Must be a sine wave i.e. correct phase at start [B1] Amplitude = 1 div [B1] Period = 8 div [B1]

(ii)

Period = 8 div [B1] Two big peaks, two small peaks Approximate correct heights/times [B2]

(iii) The frequency of wave W1 is 4.2 x 1015 Hz.

The frequency of wave W2 is 2.1 x 1015 Hz. [M1] Frequency of the resultant wave is the same as the wave with a lower frequency i.e. the frequency of resultant wave is 2.1 x 1015 Hz. [A1]

(iv) Must have two waves of the same wavelength for coherence. [B1] OR Do not have a constant phase difference. [B1]

(c) θ = 45.70/2 = 22.85o [C1] d sin θ = nλ d sin 22.85o = 2(485.6 x 10-9) [M1] d = 2.5 x 10-6 m [M1] number of lines per millimeter = 1/d = 400 [A1] 2 marks out of 4 for using θ = 45.70

13

2 (a) (i) The average height of an adult is about 1.75 m.

The potential difference between the head and feet, V= Ed = 1.0 x 102 x 1.75 = 1.75 x 102 V acceptable range of average height of an adult: 1.50 m to 2.00m

[A1]

(ii) 175 V is a relatively high potential difference. It would have given a

nasty electric shock if it were applied by the power mains. This is because the power mains has an infinite supply of electric charge in the forms of electrons. A large current would thus have conducted through the body. However, the amount of free electrons in the atmosphere is very small. Thus, although there is a large potential difference between the head and feet, this potential difference is not able to deliver sufficient current through the body. Therefore, an electric shock is not experience.

[B1] [B1]

(b) (i) Although there is no current inside the tube, there is current flowing

through the external resistors. The current through the 6 resistors, I = V / R =1050 / 6R. The potential difference between the dynode 4 and the photocathode, V = IR = (1050 / 6R) x 4R = 700 V OR Using potential divider rule,

4p.d. across photocathode and dynode 4 1050 [M1]

6 700 V [A1]

RR

= ×

=

[M1] [A1]

(ii) With dynode 3 and 5 short-circuited, the resistance between them

is zero. Current will flow directly from dynode 5 to dynode 3, bypassing the two resistances between dynode 3 and 5 completely. Since there is no current flowing through the two resistors between dynode 3 and dynode 5, there is no potential difference across either of them. Hence, the potential at dynode 3 is equal to potential at dynode 4, which is also equal to the potential at dynode 5. The total resistance of the entire series dropped from 6R to 4R. The current I’ = V / R = 1050 / 4R. The potential difference between dynode 3 and the photocathode is V = IR = (1050 / 4R) x 3R = 788 V

[M1] [M1] [A1]

14

3 (a) 8

3pP

S S

VNN V

= = [M1]

Therefore, 83

n

p sV V =

8

230000 2403

n =

[M1]

n = 6.999 ~7 [A1] Assumption (one mark for any one of the following):

1) The transformers are ideal with 100% efficient. 2) The resistance of the primary and secondary coils pr and

sr

are zero, so that no energy is lost in the coils. 3) There are no magnetic flux losses and hence both coils

have the same flux through them.

(b) The purpose of the iron core is to confine the magnetic field lines to

ensure maximum magnetic flux linkage between the primary and secondary coils. [B1]

(c) 2 2240

Average Power = 993 W58

rmsV

R= = [M1, A1]

(d) Hair dryer

Vacuum cleaner Microwave oven Or any other appliances that has about 1000 W power rating. [B1]

4 A p-n junction is formed between slices of p-type and n-type semiconductor material

as shown in Fig. 4.1

Fig. 4.1

(a) See diagram above. (b) Boron is a Group III element. [C1]

When a boron atom replaces a silicon atom in the lattice, its three valence

p-type n-type

[B1]

15

electrons form covalent bonds with neighbouring silicon atoms, leaving an electron deficiency, a hole. [B1] This hole can carry current in the presence of an electric field by accepting electrons. Hence Boron creates an acceptor semiconductor/ p-type semiconductor. [B1]

(c) When the battery is connected in this manner, the applied voltage will cause the holes in the p-type region and the electrons in the n-type region to move into the depletion region. [M1]

This reduces the width of the depletion region. [A1] 5 (a) Binding energy of a nucleus is the work done on the nucleus to separate

it into its constituent neutrons and protons. [A1]

Other acceptable answers: Binding energy is the energy released when a nucleus is formed from its separate constituent neucleons (neutrons and protons).

(b) ∆m = (Zmp + Nmn) – mA Binding energy ∆E = ∆mc2 = [(90 x 1.0073) + (136 x 1.0087) -226.0249] u c2 [M1] = 1.8153 (934) = 1695 MeV [A1]

16

(c)(i) NOTE: (c) (i) and (ii) are marked together as the s ame part. 226

90Th 22288Ra* + 4

2 He Given that rest mass of 226

90Th = 226.0249 u

rest mass of 22288Ra = 222.0154 u

rest mass of 42 He = 4.0026 u

By COE: mTh c

2 = mRac2 + KERa + mα c2 + KEα

(226.0249 u) c2 = (222.0154 u) c2 + KEradium + (4.0026u) c2 + 2.38 x 1.6x10-19 KE radium Ra* = 4.06 MeV (ii) To find KE of γ :

In order to find the KE of γ need to find KE of radium Ra as well and take the two

equations as one and take Ra* would have decayed to Ra, and make assumption that momentum of γ ray is negligible. 22690Th 222

88Ra + 42 He +γ

0 = mRavra + mα vα (mRava)

2 = (mα vα)2

(½ mRavra2) mRa = ( ½ mα

2vα)mα 2

Ra Ra2

(½ m v )

½ m v α α

= m

m Ra

α

KE of Ra = 0.0429 MeV 22288Ra* 222

88Ra +γ By COE : mRa* c

2 + KERa* = mRac2 + KERa + KEγ

Since there is no difference in the masses of Ra* and Ra. KE of γ = 4.06 – 0.0429 = 4.02 MeV ……………………………………………………………………………………………. Alternatively: A number for students use COM for the first part would have found the KE of Ra instead of Ra*.

17

Total momentum before decay = Total momentum after decay 0 = mRavra + mα vα (mRava)

2 = (mα vα)2 (½ mRavra

2) mRa = ( ½ mα2vα)mα

2Ra Ra

2

(½ m v )

½ m v α α

= m

m Ra

α [M1]

2

Ra Ra(½ m v )

2.38 = 4.0026

222.0154[M1]

KE of Ra = 0.0429 MeV [A1] This will give the answer of the KE of Ra. Acceptable as question is not clear. Then will need to find the KE of Ra* (see earlier method to find using COE).

18

6 (a) d/h = 394/450 = 0.876 ~ 0.90 [B1]

Since Chart No.49 has the correct fcu, fy and d/h, Chart No.49 is appropriate to be used for this design. [B1]

(b) (i)

(ii)

M/bh2 = 2.25 [M1] N/bh = 27.3 [M1] From Chart No.49, 100Asc/bh = 2.6 [M2 - allow errors up to 0.05 i.e. 2.55 < x < 2.65] (If 100Asc/bh = 2.7 or 2.5, allocate only one mark) Therefore, Asc = 2340 mm2 [A1] Provided four 32 mm reinforcement bars ~ 3220 mm2 > Asc, therefore the provided reinforcement steel is sufficient. [B1]

(c) N/bh > fcu [B1] Column would be crushed/break/crack. [B1]

(d) At 12 metres, the column is too slender with respect to its dimension. Might twist and buckle under load. [B1, any of the underlined words]

(e) From Chart No.39, corresponding N/bh = 21.0; allow 21.0 ± 0.5 [B1] From Char No.49, corresponding N/bh = 25.5; allow 25.5 ± 0.5 [B1] Decrease in N = 3.48bh kN Percentage decrease of N = 3.48/25.42 x100 = 13.7% [M1 and A1; ecf allowed]

19

Q7 Solution- Suggested marking schemes Basic Procedure [2]

This mark is scored as long as described procedures involve:

• Applying a force, Measuring an electrical signal

• Varying the force and repeating.

Ignore specifics – e.g. method of applying force, varying force,

what electrical signal, what it is measured with.

E.g. - Place slotted masses on the piezoelectric material,

measure p.d. across the face.

- Repeat for different total weight on it

Diagram shows [1] Diagram is married to the procedure and meant to be instructive.

E.g.

• How to force is to be applied/masses are to be placed.

• Appropriate electrical connections to the piezoelectric sample

(across opposing surfaces. If the ONLY issue is connections not

clearly shown, then award ½ )

• Appropriate circuit setup

- Reject circuits that have 1 lead to the DMM, or DMM

connected to CRO connected to computer.

- Reject circuits that have a EMF source, as the I, V now are

not due to electrical signals generated by the piezoelectric

sample

Workability of procedure is also assessed/ gauged here. If

suggested setup/connection is not feasible, withhold credit.

Mark can be allocated for different procedure/diagram, provided it

is clear how the setup is to

• Diagram shows piezoelectric sample with slotted masses

stacked on top of it, with

• DMM connected across the sample face

Method of measuring

/varying/ determining the

I.V. & D.V. [2]

• Measurement of the masses being placed on the sample

(I.V. to be measured with appropriate apparatus and not

assumed to be known values)

• Measurement of the p.d. developed across the sample

face (accept measurement of current as well, even though

unlikely material can sustain a current flow.

Accept power, P = IV as well)

Reject suggestions where force cannot be determined accurately

(e.g. dropping/throwing the weight onto the sample, poking it with

a finger, clamping it TAUT then resting the weight on it), as well

suggestions that

Control of Variables [max 2] • ACTIVELY Maintaining/monitoring the temperature, as

expansion or contraction due to temp change affecting

result.

(e.g. monitor with a surroundings with a thermometer,

perform in an climate controlled environment, e.g. aircon

20

room)

• ACTIVELY Monitor surrounding atmospheric pressure

during course of experiment, as changes in pressure might

affect the forces it is subjected to.

(e.g. with a barometer)

• If material is cut to size, then maintaining the size is valid.

(e.g. cut out a circular portion ~3-4 cm in diameter, about

the same size as the slotted masses)

• The arrangement/placement of masses on the

piezoelectric sheet.

Control of Other Variables refers specifically to less trivial items. If

student says use same material, A4 size, same wires, same DMM,

do not award credit.

Other details [max 4]

(enhanced

reliability/accuracy)

• Ensure good electrical contact between the DMM probes

and the surfaces of the sample (ensure clean from

oxide/oil, or solder)

• Wires/foil attached to the surface such that DMM probes

can be in electrical contact with the surface while allowing

even force applied on the surface.

• Measuring p.d. for no weights (checking for possible “zero

error” p.d. due to sample’s own weight)

• Ensure expt performed on insulating surface

• Ensure absence of any other electrical devices or signals in

the vicinity

• Practicing loading/unloading with reasoning (e.g. test of

path dependence)

• Repeat entire experiment with different sample size/area

of force distribution/pressure (e.g. testing if indeed [force

∝ p.d.] or (pressure ∝ p.d.)

• Repeating and taking average + Plotting of an appropriate

graph to illustrate the relationship

Safety [max 1] • Safety precautions pertaining to handling heavy slotted

masses with due care

• Comment on relative safety of experiment [e.g. no high

voltages, toxic substances, high temperatures]

(suggestions on electric shocks or electrocutions or high

temperatures of wires are not very realistic. But if clear effort put in e.g. elaboration, > 1 non-ridiculous

suggestions made can award ½ mark to encourage active

consideration of safety)

21

Prelim H2 P2 Q7 Planning Training Task Suggested So lutions (Disclaimer: These solutions do not represent the only acceptable answer. Other methods, apparatus, procedures,

precautions may also be acceptable. It is best you check with your tutor regarding the acceptability of your solutions if they differ from what is presented below.)

Task – The Piezoelectric Effect A) Problem Interpretation:

• Independent Variable: The force applied to the piezoelectric sample. This will be practically done by placing variable

number of the slotted masses on it. • Dependent Variable: The potential difference developed across the piezoelectric sample, measured by a DMM. • Aim of the Experiment: To investigate how the potential difference developed across the piezoelectric sample varies with

the force it is subjected to.

• Control Variables: − Temperature of the surroundings/piezoelectric sample. − Size of the piezoelectric sample

B) Experimental Setup and Procedure

22

• Description of Procedure 1. Cut a small disc sized sample from the A4 flexible sheet of piezoelectric material, about 5

cm in diameter, using a ruler to measure out the diameter.

2. The apparatus is to be set up as shown in Fig 1 above.

3. Connect the probes of the DMM set to measure potential difference (prob ~mV or µV) across the faces of the piezoelectric sample. To ensure good electrical contact between the entire face of the piezoelectric sample and the DMM probes, place a sheet of metallic foil over the faces of it. The foil can extend beyond the face of the sample so that the probe can still be in contact with the sample bottom while still allowing even distribution of force on it. (see Fig 2) (if either the piezoelectric disc sample or the slotted mass is resting on the DMM probe [see Fig 3], the force will not be evenly distributed & the readings may not be accurate)

4. Measure the mass of a slotted masses with an electronic balance. Record the value, M. The

force on the piezoelectric disc sample can be calculated by W = Mg. 5. Place the slotted mass on the piezoelectric disc sample, and record the corresponding

potential difference given by the DMM. Record this as V.

6. Increase the number of slotted masses by one, measuring the total mass with the electronic balance, M. Repeat step 4 until all the slotted masses have been used for a total of 10 sets of data. Stack the slotted masses are properly on top of each other to ensure uniform distribution of force.

7. Repeat step 4 and 5 again, this time decreasing the number of slotted masses and thus the

force. The readings should not differ appreciably for the same M or W value. Record the corresponding DMM readings as V’.

8. Process and tabulate all data. Then plot a graph of force, W against potential difference, V to determine the relationship

between them. e.g.

W=Mg W ↑ W ↓ # M W/N V/V V’/V 1 2

23

Reliability considerations in the procedure are highlighted with an

• Additional reliability considerations 1. Monitor the temperature of the surroundings during the experiment, as it is identified

as a possible factor. Perform the experiment in a climate controlled environment if possible, such as an air-conditioned room. If not, monitor the temperature and only accept all the data provided the temperature does not change significantly over the course of the experiment.

2. Clean the surface of the piezoelectric disc sample of dirt, oils or other contaminants

before applying the foil. (e.g. use alcohol wipes). 3. Stack the slotted masses neatly and vertically over the sample to ensure even

distribution of weight. Do not let it tilt or skew to one side. 4. The experiment is to be conducted on an insulated surface to avoid electrical

interference to avoid inaccurate results. 5. Measure the p.d. across the sample even with no weights on it. This is to account for

the sample’s own weight which might cause possible “zero error”.

• Safety considerations 1. The experiment is relatively safe as it involves no heavy weights, high voltages, high

temperatures or harmful substances.

End of paper 2 solutions