GAN ENG SENG SCHOOL PHYSICS Sec 4 Express 6091/01

This paper consists of 16 pages including the cover page

GAN ENG SENG SCHOOLPreliminary Examination 2020

CANDIDATE NAME

CLASSINDEX NUMBER

PHYSICSPaper 1 Multiple Choice

Sec 4 Express

6091/0114 September 2020

1 hour

Additional Materials: OTAS

Calculators are allowed in the examination

READ THESE INSTRUCTIONS FIRST

Write in soft pencil.Do not use staples, paper clips, highlighters, glue or correction fluid.Write your name, class and index number on the OTAS.

There are forty questions in this paper. Answer all questions. For each question there are four possible answers A, B, C, and D.Choose the one you consider correct and record your choice in soft pencil on the separate OTAS.

Read the instructions on the OTAS very carefully.

Each correct answer will score one mark. A mark will not be deducted for a wrong answer.

Any rough working should be done in this booklet.

Total Marks

40

GESS 4EXP PHY P1 PRELIM 20 CHL

2

1 Which of the following vector addition shows correctly the addition of two forces? A B C D 2 Which of the following representing the longest length? A 400 μm B 0.004 Gm C 4 000 dm D 40 Mm 3 Object X falls freely from rest for 3 seconds and object Y also falls freely from rest for

6 seconds. Which of the following statements is true?

A Y falls half as far as X. B Y falls twice as far as X. C Y falls three times as far as X. D Y falls four times as far as X.


3

4 Joash threw a ball vertically upwards at an initial speed of 30 m s-1. What is the distance travelled by the ball 5 s after the throw?

A 75 m B 65 m C 45 m D 25 m

5 The graph below shows the force acting on a 1 kg block of wood which is originally

moving at 6 m s-1 on a table.

If the table has a friction of 2 N, what is the velocity-time graph of the wood?

Fapp/ N

t/ s 0 2 3 5

4

8

A B v / m s-1

6

14

20

v / m s-1

0

6 8

10

20

2 3 5

v / m s-1

6

20

8 10

v / m s-1

6

10

14

C D

2 3 5 0 t / s t / s

t / s t / s 5 3 2 2 3 5 0 0


4

6 Which of the following will experience the largest inertia? A A bowling ball that is rolling at 20 m/s. B A basketball that is falling at a rate of 10 m/s2. C An elephant that is at rest. D A car that is travelling at 100 km/h. 7 Hydrometers are used to measure the density of four different liquids, P, Q, R and S

as shown.

Which arrangement shows correctly the decreasing density of the four liquids? A Q, R, P, S B Q, R, S, P C S, P, Q, R D S, P, R, Q

8 Which of the following statements about gravitational field is correct? A Gravitational field strength of the earth is independent of location. B Gravitational field causes forces on objects because they are charged. C The gravitational field attraction of the Earth acts towards the centre of the Earth. D The gravitational field strength on the Moon is less than that on Earth because there is no atmosphere on the Moon.

P Q

R S


5

9 A 200 g metre rule is balanced at the 45 cm mark as shown below.

A load of 50 g is now hung at the 15 cm mark. What is the new pivot position that will allow the ruler to be balanced again?

A 39 cm mark B 37 cm mark C 35 cm mark D 30 cm mark 10 An object was spinning about its pivot point, P, due to a force F as shown below. Which direction should a force, 2F, be applied to stop the spinning?

pivot

P

F

P

A

B

C

D

x

x

x x

x/2

x/2

x/2

x/2


6

11 In the manometer as shown below, 0.4 m of liquid A was added together with liquid B. What is the density of liquid B if the density of liquid A is p? A 4p /3 B 3p /4 C 4p /7 D 3p /7 12 The diagram below shows a mercury barometer used to measure the surrounding

atmospheric pressure. What is the pressure measured at X? A 16 cm Hg B 22 cm Hg C 52 cm Hg D 62 cm Hg

Atmospheric pressure

A

10

20

30

40

50

60

70

80

90 vacuum

x

cm

Liquid A

Liquid B


7

13 Xavier was pushing a mass 50 kg to the top of the ramp. The gravitational field strength is 10 N/kg.

What is the efficiency of Xavier?

A 83.3 % B 13.3 % C 25.0 % D 8.3 %

14 Donovan was throwing a 500 g basketball upwards with an initial velocity of 20 m/s. The gravitational field strength is 10 N/kg and assuming no air resistance.

What is the gain in potential energy of the basketball after leaving Donovan’s hand for 1 s? A 100 J B 75 J C 25 J D 10 J

200 N

15 m 5 m


8

15 A gas is heated in a rigid sealed container. Which quantity does not change? A The average speed of the gas particles B The average force exerted on the walls of the container by the gas particles C The average distance between the gas particles D The frequency of collisions on the walls of the container by the gas particles 16 According to the kinetic theory, matter is made up of very small particles in a constant

state of motion. Which row best describes the particle behaviour in the liquid state?

forces between particles motion of particles

A strong move randomly at high speed B strong vibrate but are free to move position C very strong vibrate to and fro about a fixed position D weak move randomly at high speed

17 All the objects below are made of the same material and having the same thickness. A B C D Which object will cool down the fastest if they were all being heated to the same

temperature? 18 Which statement about the transfer of thermal energy is correct? A All metals conduct thermal energy equally well. B Convection can only occur in solids or liquids. C Convection occurs in liquids because hot liquid is denser than cold liquid. D The radiation that transfers thermal energy is a type of electromagnetic radiation.


9

19 When ice changes into water at 0 oC, I work is done in breaking the molecular structure of ice in solid state. II internal energy is increased. III energy is absorbed to raise the temperature. A I only B II only C I and II only D I, II and III 20 The diagram shows a graph of how the e.m.f., ε / V, of a thermocouple varies with

temperature, θ / °C. Why is the thermocouple inappropriate for measurement of temperature in the range as shown?

A The thermocouple produces an e.m.f. at 0 °C. B The relationship between ε and θ is non-linear. C The thermocouple does not always indicate a unique value of e.m.f.. D The e.m.f. has not been measured using temperatures using the Kelvin scale.

ε / V

θ / °C


10

21 The diagrams show the scale of a voltmeter connected to a thermocouple thermometer.

What is the temperature of the liquid?

A 75 oC B 100 oC C 120 oC D 125 oC

22 When Satish mixed 30 g of liquids Q and 50 g of liquid R together, he obtained a final

temperature of 45 oC. What is the ratio of heat capacity of liquid Q to heat capacity of liquid R if the initial

temperature of liquid Q and R are 20 oC and 90 oC respectively? A 9/5 B 5/9 C 1/3 D 3/1 23 Four different materials of equal mass were heated by the same heater. The

temperature-time graph of the four different materials were shown below. Which material has the lowest specific heat capacity?

thermocouple probe in melting ice

thermocouple probe in steam

thermocouple probe in liquid

temperature

time

A

B C D


11

24 Which of the following shows correctly the wavefront as it enters into a deeper region? A B C D 25 The diagram below shows a pressure-distance graph of a longitudinal wave. Which point indicates that it is the centre of rarefaction?

pressure

distance

B A

D

C

shallow deeper shallow deeper

shallow deeper shallow deeper


12

26 Ren Jie was looking through the mirror at his musical instrument. Where is the position of the image of musical instrument? 27 A beam of light shone at an angle of 55 o in the glass is shown below.

What is the change in the direction of the light path if the refractive index of glass is 1.45?

A 56.3 o B 31.7 o C 23.3 o D 21.3 o

Glass

55 o

Air

Ren Jie

Musical instrument o

B x

A x

C x

D x


13

28 An object is placed in front of a thin converging lens as shown below. Which direction shows correctly the path ray P will take after it passes through the lens?

29 Which list shows electromagnetic waves in order of increasing frequency? A visible light, X-rays, Gamma rays B visible light, Gamma rays, X-rays C X-rays, Gamma rays, visible light D Gamma rays, X-rays, visible light 30 What is the wavelength of an electromagnetic wave that has a frequency of 6.0 x 105 GHz? A 5.0 x 10-13 m B 5.0 x 10-10 m C 5.0 x 10-7 m D 5.0 x 10-4 m 31 How will the amplitude and wavelength change as sound increases in loudness and

pitch? Amplitude Wavelength A increase increase B increase decrease C decrease increase D decrease decrease

F

A B

C

D P object


14

32 Which object will not attract tiny pieces of paper? A a metal knife sharpened with a knife sharpener B a ceramic spoon wiped with a kitchen cloth C a plastic comb combed through dry hair D a wooden pencil rubbed with an eraser 33 A stationary negative charge in an electric field experiences an electric force in the

direction shown.

What is the direction of the electric field? A horizontally to the left B horizontally to the right C vertically downwards D vertically upwards 34 An electron carries 1.6 x 10–19 C. What is the potential difference across a 5 resistor

if there are 1 x 1010 electrons flowing through it in 4 μs? A 0.002 V B 2.00 V C 3.2 x 10-14 V D 1.28 x 10-15 V 35 A circuit was set up with a lamp connected as shown below. What happens to the voltmeter reading as switch P is closed? A The voltmeter reading remains unchanged. B The voltmeter reading decreases. C The voltmeter reading increases. D The voltmeter reading shows zero reading.

P

force

-


15

36 Yida was spending 6 hours each night revising for his upcoming Physics examination. He had two 100 W light bulbs in his room. The weekly bill for his room was $3.00

How much saving would he have in a week if a unit of electricity cost 24 cents? A 270 cents B 201 cents C 199 cents D 98 cents 37 A compass is placed inside a closed iron cylinder on the table. The north pole of a

permanent magnet is brought near the cylinder as shown in diagram. In which direction does the needle of the compass point? A away from the permanent magnet B opposite direction C original direction D towards the permanent magnet 38 The diagram shows a two-pole single-coil electric motor.

The split-ring commutator reverses the current in the coil as it rotates. How many times did the current reversed as the coil completes one full revolution?

A 1 B 2 C 3 D 4

compass

Iron cylinder magnet


16

39 The diagram shows the front view of a loudspeaker, which consists of magnets and a coil amongst other items.

If the current is flowing in the speaker’s coil in an anticlockwise direction as shown, what is the direction the coil that it will move?

A To the right. B Out of the page. C To the left. D Into the page.

40 Two long, parallel conductors carrying current lie in a horizontal plane.

The two conductors attract one another.

Which statement is true?

A The two currents are in the same direction. B The two currents are in opposite direction. C The two currents are parallel to the Earth’s magnetic field. D The two currents are at 90° to the Earth’s magnetic field.

END OF PAPER

South pole

North pole

coil

This paper consists of 20 pages including the cover page

GAN ENG SENG SCHOOLPreliminary Examination 2020

CANDIDATE NAME

CLASSINDEX NUMBER

PHYSICSPaper 2

Sec 4 Express

6091/0201 September 20201 hour 45 minutes

Candidates answer on the Question Paper.

Calculators are allowed in the examination

READ THESE INSTRUCTIONS FIRSTWrite your class, 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 or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid/tape. Section A Answer all questions. Section B Answer all questions. Question 11 has a choice of parts to answer. Candidates are reminded that all quantitative answers should include appropriate units. Candidates are advised to show all their working in a clear and orderly manner, as more marks are awarded for sound use of Physics than for correct answers. At the end of the examination, fasten all your work securely. The number of marks is given in brackets [ ] at the end of each question or part question.

For Examiner’s Use

Section A 50

Section B

30

Total

80


2

For Examiner’s

Use

Section A [50 marks]

Answer all the questions in this section.

1 (a) The size of a virus is approximately the size of ultraviolet wavelength. State the wavelength of the ultraviolet in metres.

…………………………………………………………………………………………………[1]

(b) (i) Given that the thickness of a hair is about 103 times more than size of a virus, state a suitable apparatus to measure the thickness of hair.

…………………………………………………………………………………………………[1] (ii) State two necessary steps needed to ensure that accurate readings are obtained

when reading the thickness of hair.

………………………………………………………………………………………………… …………………………………………………………………………………………………

…………………………………………………………………………………………………

…………………………………………………………………………………………………[2]

2 Alex was running for the inter-school 5 km cross country competition. He ran a total displacement of 5 km round a reservoir. He completed the whole competition with a time of 20 minutes. Alex’s friend, Clement shared that his velocity is 15 km/h.

(a) State what is meant by displacement of 5 km.

………………………………………………………………………………………………...…

…………………………………………………………………………………………………[1]

(b) In the paragraph above, state and explain the two errors found.

………………………………………………………………………………………………..…

………………………………………………………………………………………………...… ………………………………………………………………………………………………...…

…………………………………………………………………………………………………[2]


3

For Examiner’s

Use

2 (c) The velocity time graph of Alex during the run was shown below.

(i) State and explain the period at which Alex was running with the greatest acceleration.

………………………………………………………………………………………………...…

………………………………………………………………………………………………...… ………………………………………………………………………………………………...…

…………………………………………………………………………………………………[2]

(ii) Calculate the greatest resultant force that Alex is experiencing. Take Alex’s mass as 50 kg.

Greatest resultant force = ………………………………..[2]

velocity/ m/s

time/ min

9

8

5

0 13 16 20


4

For Examiner’s

Use

3 An object of mass 0.80 kg is suspended and equally supported by two identical strings as shown in Fig. 3. (not drawn to scale). Fig. 3

(a) State two differences between mass and weight.

………………………………………………………………………………………………...… ………………………………………………………………………………………………...… ………………………………………………………………………………………………...…

………………………………………………………………………………………………...[2]

(b) Calculate the weight of the object. (Gravitational field strength = 10 N/kg)

weight = ………………………… [2] (c) By drawing a scale diagram, determine the magnitude of the tension in one string.

Tension = ……………………………. [4]

40 40


5

For Examiner’s

Use

4 Andres was trying to open a door at the AVA room as shown in Fig. 4.

Fig. 4

(a) State and explain two ways how Andres could open the door more efficiently.

………………………………………………………………………………………………...… ………………………………………………………………………………………………...… ………………………………………………………………………………………………...… ………………………………………………………………………………………………...…

…………………………………………………………………………………………………[2]

(b) State what is moment and its formula.

………………………………………………………………………………………………...… ………………………………………………………………………………………………...… ………………………………………………………………………………………………...…

…………………………………………………………………………………………………[2]

(c) State two reasons why moment is not measured in Joule.

………………………………………………………………………………………………...…

………………………………………………………………………………………………...…

………………………………………………………………………………………………...…

………………………………………………………………………………………………...…

………………………………………………………………………………………………...… …………………………………………………………………………………………………[2]


6

For Examiner’s

Use

5 A manometer was used to measure the pressure difference between the atmospheric pressure and the gas pressure in a chamber as shown in Fig. 5.1. The atmospheric pressure is 76 cm Hg and the density of mercury is 13 600 kg/m3.

Fig. 5.1

(a) State the pressure found in the gas chamber in Pa.

pressure = ……………………. [2]

(b) The liquid used in the manometer is now changed to liquid X. Liquid X has a density of 6 800 kg/m3 and the manometer increases in its diameter. State the new height difference found in the manometer. Support your explanation with numerical calculation.

height difference = …………………..[2]

mercury

Connection to gas chamber

Manometer

15 cm


7

For Examiner’s

Use

5 (c) Fig. 5.2 shows a hydraulic system works by exerting a force on the piston A which in turn pushes the piston B upwards with a larger force. Explain how the principle of conservation of energy is applicable in the hydraulic system.

Fig. 5.2 ……………………………………………………………………………………………… ………………………………………………………………………………………………. ……………………………………………………………………………………………… ………………………………………………………………………………………………. ……………………………………………………………………………………………… ………………………………………………………………………………………………. ………………………………………………………………………………………………. ……………………………………………………………………………………………… ………………………………………………………………………………………………. ………………………………………………………………………………………………. ……………………………………………………………………………………………… ……………………………………………………………………………………………[4]


8

For Examiner’s

Use

6 A simple pendulum of length 1.00 m has a bob of mass 0.20 kg. The bob is pulled aside by a horizontal distance of 0.3 m and is then released as shown in Fig. 6.

Fig. 6

(a) Calculate the gravitational potential energy at position B.

gravitational potential energy = ………………….. [2]

(b) Calculate the velocity at which the bob will pass through A.

velocity at A = …………………. [2]

(c) State the assumption made in (b).

………………………………………………………………………………………………. ……………………………………………………………………………………………[1]

B

0.30 m

1.00 m

A

0.95 m


9

For Examiner’s

Use

7

A light uncharged metal ball, X, is freely suspended and is in contact with an uncharged metal sphere, Y, as shown in Fig 7.1. A positive charged metal rod, Z, is moved towards Y as shown in Fig. 7.2

Fig 20.1 Fig 20.2 Fig. 7.1 Fig. 7.2

(a) Explain what is meant by electric field. …………………………………………………………………………………… ……………………………………………………………………….……………

[1]

(b) State the charges induced on spheres X and Y in Fig. 7.2 X : ……………………. Y : …………………….

[1]

(c) Sketch in Fig 7.1 the electric field pattern between Y and Z.

[1]

(d) When Z is brought to touch Y and then removed, suggest what will happen. …………………………………………………………………………………… ………………………………………………………………………………….… …………………………………………………………………………………… ………………………………………………………………………………….… …………………………………………………………………………………… ………………………………………………………………………………….… …………………………………………………………………………………… …………………………………………………………………………………… ………………………………………………………………………………….… ………………………………………………………………………………….…

[3]


10

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Use

8 In the circuit shown in Fig. 8.1, E is a battery of e.m.f. 20.0 V and negligible internal resistance. R1 is a variable resistor of maximum resistance 400 , R2 is a fixed resistor of resistance 200 , V1 is a voltmeter and A is an ammeter.

Fig. 8.1

(a) Determine the maximum and minimum readings of V1.

maximum reading = …………………..[1]

minimum reading = …………………..[1]

(b) Calculate the current through the ammeter when R1 is set at 250 Ω.

current = ………………[2]

E 20.0 V

R1

R2

200 Ω

A V1


11

For Examiner’s

Use

8 (c) A thermistor connected in series with a 1 200 Ω resistor is added to the circuit as shown in Fig. 8.2. V2 is the second voltmeter. When temperature is at 0 °C, the

resistance of the thermistor is 3 600 . When the temperature is at 100 °C, its resistance is 400 .

Fig. 8.2

Calculate the readings of the ammeter A and voltmeter V2 when R1 is set to 250 Ω and the thermistor is placed in steam from pure water boiling at standard atmospheric pressure.

ammeter reading = …………………[1]

voltmeter reading = …………………[1]

1 200 Ω

E 20.0 V

R1 250 Ω

R2

200 Ω V2

A V1


12

For Examiner’s

Use

Section B [30 marks]

Answer all the questions from this section. Answer only one of the two alternative questions in Question 11.

9 Fig. 9.1 shows the circuit diagram of a hairdryer. A motor-driven fan and a heating

element are used to generate warm air. The hairdryer is connected to a 230 V a.c. supply. Switch S can be connected to either contact J or K.

(a) The hairdryer is used to dry wet hair.

Explain, using kinetic theory of matter, how the hairdryer can increase the rate of evaporation of water from the wet hair.

……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………[3]

230 V mains J

motor heating element

M

K

S

R

cool air warm air

main switch

Fig. 9.1


13

For Examiner’s

Use

9 (b) During quality control tests of the hairdryer in the factory, switch S is first connected to contact J. Some measurements are made to obtain the data shown in Fig. 9.2.

resistance of the heating element 25 Ω

resistance of resistor R 40 Ω

temperature of air entering the hairdryer 28 C

temperature of air flowing out of the hairdryer 33 C

rate of air flow through the hairdryer 46 g/s

Fig. 9.2 Calculate the specific heat capacity of air.

Specific heat capacity of air = ………………….[2]

(c) Switch S is then connected to contact K. State and explain the change in the rate

of air flowing through the hairdryer and the temperature of the air flowing out of the hairdryer, as compared to when switch S is connected to contact J.

……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ..…………………………………………………………………………………………… ………………………………………………………………………………………………

……………………………………………………………………………………………… ………………………………………………………………………………………………

……………………………………………………………………………………………… ………………………………………………………………………………………………

……………………………………………………………………………………………[3]


14

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Use

9 (d) The cost of using the hair dryer for 25 minutes when the hair dryer is set to the higher air flow rate is 28 cents. Calculate the unit cost of the electricity used.

Unit cost = …………….. [2]

10 (a) Fig. 10.1 shows an object, O placed at 10 cm away from the lens.

Fig. 10.1

(i) A virtual image of 1.5 times the size of object is formed. Draw 2 rays in

Fig. 10.1 to locate the position of the image and label it as I. [3]

(ii) State the distance of the image from the lens.

distance of image from lens = …………………….. [1]

O


15

For Examiner’s

Use

10 (b) Fig. 10.2a represent the positions of equally spaced ‘dots’ of air molecule

before a sound wave passes through air. The dots in Fig. 10.2b represents the positions of the same ‘rows’ at one particular instant as the sound wave passes.

Fig. 10.3

(i) In Fig. 10.2b, mark out accurately the amplitude and the wavelength of the sound wave and labelled it as a and respectively. [2]

(ii) In Fig. 10.3, sketch accurately the displacement distance graph of the sound wave. Positive displacement is to the right.

[2]

(iii) State how will the displacement distance graph in Fig. 10.3 change when the sound created is now louder and of a higher pitch. …………………………………………………………………………………………. …………………………………………………………………………………………. ………………………………………………………………………………………..[2]

Dis

plac

emen

t

Distance

Fig. 10.2a

Fig. 10.2b


16

For Examiner’s

Use

11 EITHER

(a) Fig. 11.1 shows a drop tower found in an amusement park. A 5 kW motor was used to lift up a gondola carrying riders to the top of the vertical structure. It then released to free fall down the tower. Magnetic brakes were activated to slow down the gondola as it approached the bottom of the ride. Take acceleration due to free fall as 10 m/s2.

Fig. 11.1

(i) Calculate the height of the tower in Fig. 11.1 if the gondola has a mass of 200 kg and it takes 30 s to reach to the top of the tower.

height of tower = ………………… [2]

(ii) Calculate the time taken for the gondola to reach the bottom of the tower when released.

time taken = ………………… [2]


17

For Examiner’s

Use

11 (b) Daryl did a mockup of the drop tower using two electromagnets, A and B as shown in

Fig. 11.2. The electromagnet A is used to hold the mass, a magnetic material, in place while the electromagnet B is used to create a braking force.

Fig. 11.2

(i) If the falling mass has a mass of 20 g, suggest with reason the magnetic force that the electromagnet B has in order to allow the falling mass to come to rest instantly.

…………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. ……………………………………………………………………………………….[2]

(ii) Suggest two ways the electromagnet can increase its strength if a larger falling mass is used.

…………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. ……………………………………………………………………………………….[2]

(iii) Daryl intended to increase the height of the fall of the falling mass. State, with reason, if the strength of electromagnet B need to change to break the fall.

…………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. ……………………………………………………………………………………….[2]

Falling mass

Electromagnet A

Electromagnet B


18

For Examiner’s

Use

OR 11

(a)

Fig. 11.3 shows a type of motor. PQ and RS are solenoids. The solenoids and the coil ABCD are connected in parallel to a battery.

Fig. 11.3

(i) State the polarity at end Q.

……………………………………………………………………………………[1]

(ii) State the direction of rotation of the coil as seen by the observer.

……………………………………………………………………………………[1]

(iii) Explain why the coil will rotate continuously. ……………………………………………………………………………………

…………………………………………………………………………………… ……………………………………………………………………………………

……………………………………………………………………………………

…………………………………………………………………………………… ……………………………………………………………………………………

…………………………………………………………………………………… ……………………………………………………………………………………[2]


19

For Examiner’s

Use

11 (b) Fig 11.4 shows how the current varies with the potential difference for light bulbs

J and K.

Fig 11.4

(i) State, with reason, if the filament of light bulb J and K are ohmic conductors.

………………………………………………………………………………………… ……………………………………………………………………………………….[2]

0 1 2 3 4 5

0.1

0.2

0.3

0.4

0.0

Potential Difference / V

Current / A

J

K


20

For Examiner’s

Use

11 (b) (ii) Estimate the resistance of bulb J and the resistance of bulb K when they

are operating at V = 0.20 V.

resistance of bulb J = ………….

resistance of bulb K = ………… [2]

(iii) The filaments of both bulbs are coiled tungsten wire. Describe two differences between the filaments of the bulbs that would cause their difference in resistance.

……………………………………………………………………………………

…………………………………………………………………………………… ……………………………………………………………………………………

…………………………………………………………………………………… ……………………………………………………………………………………

…………………………………………………………………………………… …………………………………………………………………………………… ..………………………………………………………………………………..[2]

END OF PAPER

GESS 3EX Pure Physics 6091/01 MYE19 chl Page 1

Answer for 2020 Sec 4Ex Pure Phy Prelim P1 Ans

1 C 2 D 3 D 4 B 5 D 6 C 7 A 8 C 9 A 10 B11 A 12 C 13 A 14 B 15 C 16 B 17 D 18 D 19 B 20 C21 D 22 A 23 B 24 C 25 B 26 C 27 D 28 B 29 A 30 C31 B 32 A 33 C 34 A 35 C 36 D 37 C 38 B 39 D 40 A

1 Addition of vectors must ensure resultant point from starting point of first force and end off at the ending point of last force

C

2 0.004 Gm = 4 x 106 m while 40 Mm = 4 x 107 m D

3 Distance fall by X = ½ (3)(30) = 45 m Distance fall by Y = ½ (6)(60) = 180 m

D

4 Distance covered in the first 3 s = 45 m Distance covered in the next 2 s = 25 m Total distance covered = 65 m

B

5 In the first 2 s, there’s a net force of 2 N bringing about an acceleration of 2m/s2

In the next 1 s, there’s a net force of -2 N bringing about an acceleration of -2 m/s2

In the last 2 s, there’s net force of 6 N bringing about an acceleration of 6 m/s2

D

6 Elephant has the largest mass thus highest inertia C

7 Liquid with hydrometer floating the most is the highest density A

8 Gravitational field attraction is acting towards the centre of the Earth C

9 Let new pivot point be x cm 50 ( x – 15) = 200 ( 45 – x ) x = 39 cm

A

10 Force at B will create an anticlockwise moment equal to the clockwise moment created by F due to a shorter perpendicular distance from pivot point

B

11 Pressure at A = Pressure at B Density of A x g x (0.4) = Density of B x g x (0.3) Density of B = (4/3) Density of A

A

12 Marking at X -= 22 cm Marking over mercury level in barometer = 74 cm Height of mercury above X = 74 – 22 = 52 cm

C

13 Total work done = 200 x 15 = 3000 J Useful work = 500 (5) = 2500 J Efficiency = (2500/3000) 100% = 83.3 %

A

14 The velocity after 1 s = 10 m/s Gain in PE = Loss in KE = ½ (0.5)(20)2 – ½ (0.5)(10)2 = 75 J

B

15 As gas is in a rigid sealed container, there’s no change in its volume thus the average distance remain the same.

C

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16 Liquid has strong molecular bonding and they are free to move their position B

17 Letter ‘M’ cool down the most as it has the largest surface area D

18 Infra red radiation is used to transfer thermal energy D

19 During melting, there is no change in its temperature and work is done to overcome forces of attraction and not break.

B

20 The relationship between and does not give a unique value C

21 8 division rep 100oC 10 division rep (100/8) x 10 = 125oC

D

22 Heat gained by Q = Heat lost by RCQ (45-20) = CR (90- 45) CQ / CR = 9/5

A

23 The rise in temperature is the fastest for a material with low specific heat capacity B

24 As wavefront enters into the deeper region, its speed increase and its wavelength increases. The direction of wave is bending away from normal.

C

25 Centre of rarefaction is the point with lowest pressure B

26 Image of object is perpendicular distance away from the mirror (double reflection) C

27 1.45 = sin(i) / sin (35)i = 56.3change in direction = 56.3 – 35 = 21.3o

D

28 After locating the image of object, draw the ray P towards the tip of the image arrow B

29 Gamma has the highest frequency with visible light having the longest wavelength. A

30 v = f x = (3 x 108) / (6 x 105 x 109) = 5 x 10-7 m

C

31 Loudness is proportional to amplitudePitch is proportional to frequencyFrequency is inversely proportional to wavelength

B

32 Metal is a good conductor of electricity. Once there’s an excess charges, it will be discharged away easily. The other options will result in charging object by rubbing as they are non metal.

A

33 As the negative charge attracted to positive charge, there must be presence of positive charge at the top and thus the electric field is downwards. Do not use Fleming’s Left Hand rule.

C

34 Q = It I = (1 x 1010 x 1.6 x 10-19 )/4 x 10-6 = 0.0004 A V = IR = 0.0004 x 5 = 0.002 A

A

35 With switch P closed, the effective resistance of the 3 resistors will become lesser and thus the total resistance of the whole circuit will decrease leading to an overall increase in the current in the circuit. Thus the pd across lamp will increase.

C

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36 Electrical energy = 2 x 0.1 x 6 x 7 = 8.4 kWh in a weekTotal cost = 8.4 x 24 = 201.6 centsTotal saving = 300 – 201.6 = 98.4 cents

D

37 The compass needed is not being influenced by the field strength of the permanent magnet.

C

38 Every half rotation, the current direction will be reversed in the coil B

39 Using Fleming’s Left Hand rule, the coil will move into the paper D

40 Same directional current conductor will attract each other. A

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e to

max

imis

e th

e di

stan

ce a

way

from

the

hing

e so

as

to b

ring

abou

t a lo

wer

forc

e ne

eded

to p

rodu

ce th

e sa

me

mom

ent.

He

coul

d al

so p

ush

the

door

at p

erpe

ndic

ular

lyso

that

the

dist

ance

mea

sure

d is

the

furth

est f

rom

the

hing

e

[1]

[1]

Mos

t can

sta

te a

t a fu

rther

di

stan

ce fr

om h

inge

but

man

y di

d no

t men

tion

abou

t pu

shin

g at

90

deg

to th

e do

or

to h

ave

a hi

gher

pe

rpen

dicu

lar d

ista

nce.

So

me

over

look

that

effi

cien

cy

mea

ns u

se a

less

er fo

rce

and

men

tion

abou

t usi

ng a

hig

her

forc

e to

pus

h

4bM

omen

t is

the

turn

ing

effe

ctof

forc

e.

It ca

n be

foun

d by

usi

ng p

rodu

ct o

f for

ce a

pplie

d an

d th

e pe

rpen

dicu

lar d

ista

nce

from

th

elin

e of

act

ion

of fo

rce

to th

e pi

vot.

[1]

[1]

Mos

t can

do

it

prin

g ba

lanc

ee

[1[1]]

Mo

[11]]

[1[1]]

Mos

t can

d

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effe

cttoff

ffororcece

.

oduc

t off

oo

the

Que

Ans

Mar

ksM

arke

r’s C

omm

ents

4cM

omen

t is

mea

sure

d by

the

prod

uct o

f the

per

pend

icul

ar d

ista

nce

from

the

pivo

t poi

nt

and

the

forc

e ap

plie

d. F

or m

easu

rem

ent t

o be

in J

oule

, it h

as to

be

the

prod

uct o

f dis

tanc

e tra

velle

din

the

sam

e di

rect

ion

as th

e fo

rce

appl

ied.

Mom

ent i

s a

vect

or q

uant

ity(in

Nm

) whi

le w

ork

done

is a

sca

lar q

uant

ity( i

n J)

[1]

[1]

Mos

t cou

ld n

ot s

tate

that

m

omen

tis

a ve

ctor

whi

le

wor

k do

ne is

a s

cala

r

5aPr

essu

re o

f atm

osph

ere

is lo

wer

than

gas

pre

ssur

e by

15

cm H

gPr

essu

re g

as =

76

+ 15

=91

cm H

g

Pres

sure

= p

gh =

136

00 x

10

x 0.

91=

123

760

Pa=

124

000

Pa

[1]

[1]

Mos

t can

do

it ex

cept

for

som

e w

ho u

ses

15 c

m

inst

ead

of 0

.15

m in

the

calc

ulat

ion

5bPr

essu

re d

iffer

ence

= 0

.15

x 10

x 1

3 60

0

With

the

new

liqu

id X

, the

pre

ssur

e di

ffere

nce

still

stay

s0.

15 x

10

x 13

600

= h

x 1

0 x

6800

Ther

efor

e h

= 0.

30 m

[1]

[1]

Mos

t can

do

it5c

Sinc

e liq

uid

is in

com

pres

sibl

e, th

e vo

lum

e of

liqu

id d

ispl

aced

at p

isto

n A

has

to b

e th

e sa

me

volu

me

disp

lace

d at

pis

ton

B.

Ala

rge

cros

s se

ctio

nala

t pis

ton

B w

ill cr

eate

a s

mal

l dis

plac

emen

tat p

isto

n B

due

toth

e di

spla

cem

ent a

t pis

ton

A in

ord

er to

obt

ain

the

sam

e vo

lum

e di

spla

ced.

As th

e pr

essu

re c

reat

ed a

t pis

ton

A is

the

sam

e as

the

pres

sure

at p

isto

n B,

pis

ton

B is

of

a la

rger

cro

ss s

ectio

nal a

rea

than

pis

ton

A, it

will

crea

te a

larg

er fo

rce

at p

isto

n B

Giv

en th

at w

ork

done

= F

orce

x d

ista

nce

trave

lled

in th

e di

rect

ion

of fo

rce.

W

ork

done

at A

= F

orce

at A

x d

ista

nce

trave

lled

at A

Wor

k do

ne a

t B =

For

ce a

t B x

dis

tanc

e tra

velle

d at

B

Asm

all f

orce

at A

with

larg

e di

spla

cem

enta

t Aw

ill cr

eate

a la

rge

forc

eat

B w

ith a

sm

all d

ispl

acem

ent.

Thus

the

prin

cipl

eof

con

serv

atio

n of

ene

rgy

is a

pplic

able

in

hydr

aulic

sys

tem

.

[1]

[1]

[1]

[1]

Mos

t are

not

abl

e to

exp

lain

fu

lly w

hen

the

prin

cipl

e of

co

nsev

atio

n of

ene

rgy

is v

alid

in

hyd

raul

ic s

yste

m. A

lot o

f po

ints

are

mis

sing

in th

eir

expl

anat

ion

[1]

[11]]

[11]]

Mos

som

ein

stea

d ca

lcul

atio

ffere

nce

stititiitiiititiiititiiitittitititititittttttitttttttttttttttttttlllsssssssssssssssssssssssssss

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00

[1[1]]

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em

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prprinin

cicipl

eof

c

[1]

Que

Ans

Mar

ksM

arke

r’s C

omm

ents

6aG

PE =

mgh

GPE

= 0

.2 (1

0) (0

.05)

= 0

.1 J

[1]

[1]

Mos

t can

do

it

6bLo

ss in

GPE

= G

ain

in K

E 0.

1=

½ (0

.2) v

2

v =

1 m

/s[1

][1

]

Mos

t can

do

it

6cEn

ergy

is n

ot b

eing

lost

as

the

pend

ulum

sw

ings

from

B to

A.

All t

he g

ravi

tatio

nal p

oten

tial e

nerg

y lo

st is

equ

al to

the

kine

tic e

nerg

y ga

ined

Al

l gra

vita

tiona

l pot

entia

l ene

rgy

is c

onve

rted

to k

inet

ic e

nerg

y

[1]

Mos

t can

do

it

7aAn

ele

ctric

fiel

d is

a re

gion

in w

hich

a (u

nit p

ositi

ve)c

harg

e ex

perie

nces

an

elec

tric

forc

e.[1

]

Mos

t can

do

it7b

X:p

ositi

ve

Y :

nega

tive

[1]

Mos

t can

do

it

7c[1

]

Mos

t can

do

it

7dAs

Z to

uche

s Y,

neg

ativ

e ch

arge

s in

Y w

ill m

ove

to Z

.

As Z

is re

mov

ed, b

oth

X an

d Y

will

bot

h be

pos

itive

lych

arge

dan

d

X is

repe

lled

from

Yas

like

cha

rges

repe

l.

[1]

[1]

[1]

Som

e ov

erlo

ok a

nd m

entio

n th

at th

e po

stiv

e ch

arge

s m

ove

and

not e

lect

rons

w

hich

is th

e w

rong

con

cept

.

8aM

axim

um re

adin

g =

20 V

Min

imum

read

ing

= 2

00/6

00 x

20V

=6.

67V

(to 3

sf)

[1]

[1]

Mos

t can

do

it

V

0 x

20V

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gai

ned

gy

[1[1]]

Mos

ve)c

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]]

Mo

Mos

tstca

n do

[1[1

]]M

oM

ost c

and

[1[1]]

Que

Ans

Mar

ksM

arke

r’s C

omm

ents

8bI =

V /

R=

20 /

(250

+ 2

00)

=0.

0444

A (to

3sf

)[1

][1

]M

ost c

an d

oit

8cI t

hrou

gh th

erm

isto

r = V

/ R

= 2

0 / (

1 20

0 +

400)

= 0

.012

5 A

(to 3

sf)

Amm

eter

read

ing

= 0.

0444

A +

0.0

125

A

= 0

.056

9A

(to 2

sf)

Rea

ding

on

V 2

= 4

00 /

(400

+ 1

200

) x 2

0=

5 V

[1]

[1]

Mos

t can

do

it

Sect

ion

B9a

Hai

r dry

er p

rodu

ces

fast

mov

ing

hot a

ir m

olec

ules

whi

ch c

ollid

e w

ith th

e w

ater

m

olec

ules

on

the

hair

and

trans

fer t

herm

al e

nerg

y to

it.

Mor

e w

ater

mol

ecul

es in

crea

se it

s K

E an

d th

ey m

ove

mor

e vi

goro

usly

.

Ther

e’s

an in

crea

sed

in th

enu

mbe

r of f

ast m

ovin

g su

rfac

e w

ater

mol

ecul

esth

at w

illsu

cces

sful

ly b

reak

the

forc

es o

f attr

actio

n be

twee

n th

e re

mai

ning

mol

ecul

esan

d ov

erco

me

atm

osph

eric

pre

ssur

e to

bec

ome

gase

ous

stat

e th

us in

crea

ses

the

rate

of

evap

orat

ion.

[1]

[1]

[1]

Mos

t can

exp

lain

with

som

e ov

erlo

oked

that

the

hair

drye

r w

ill pr

oduc

e fa

st m

ovin

g ai

r m

olec

ules

use

d to

hea

t up

the

wat

er m

olec

ules

9bi

P =

V2 / R

= 2

302

/25

=21

16W

In 1

s, i

t will

prod

uce

2116

J of

ther

mal

ene

rgy.

Ener

gy p

rodu

ced

by h

eate

r in

1 s

= En

ergy

gai

ned

by a

irin

1 s

2116

= m

c

2116

= 4

6(c

) (33

-28)

c =

9.20

J/g

o C

[1]

[1]

Mos

t are

not

abl

e to

use

the

resi

stan

ce o

f hea

ting

elem

ent

in th

e ca

lcul

atio

n an

d in

stea

d us

es re

sist

or R

too!

The

y ov

erlo

oked

that

the

who

le

circ

uit i

s a

para

llel c

ircui

t with

th

e he

atin

g el

emen

t get

ting

the

full

emf i

nste

ad o

f hav

ing

to s

hare

with

the

resi

stor

R

[1]

[1[]]

[1[1]]

[1[1]

sw

hichh

cococooooocoocooooooooococococoooocococoooccocococcccocoooccoollid

e w

ithhhhhhhhhhhhhhh th

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it.

E an

d th

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viviviiiiiiiiiiiigogogggggggggggg

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ly.

num

bererrrrrrrrrrrrrrrrrrrrrrrrrrrerrrrrr

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movovovovovovovvovovovovvovovovovvvvvvvvvvvvvvvvv

inininininininininininininniinnninnnnnnnnnnnngggggggggggggggggggggggggggggg

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culeeeeeeeeeeeeeeeee

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at w

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ma

ma

mmmmmmmmmmmmmmmmmmmmmmmmmmmin

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mooooooooooooooooooooool

eleleleleleleeleleleeleeleeeleeeleleeleleleeeellellellcucucucucucucucucuuccccccccccccccccccccccccccccc

leleleleeleeleleleeeeeeeeeeeleeelessss

and

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pre

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tatatatatatatatatataaaaaaaaaaaaaaaaaaaaaaaaaaaaaae e e e e e ee ee

ooooof

[1[1]]

[1[1]]

[1[1]]

Mo

R =

233333333333333300000000000000000000

222222222222222222/2

55=

22222221222222222222222222216

WWW

llpr

ododododododododododododododododucuccccccccccc

eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee21111111

161616116161616161161161161616161616111611611611666666666666JJJJJJJJJ

of th

eeeeeeeeerrrmrmrrrrrrrrrr

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enennnnnnnnnnnrg

y.

ced

bybybbbbbbbbbbbbbbhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhe

ater

inininininininininninininiininnnnnnnnnnnnn11111111111

sssssssssssss =

Enererr

gygygygyygggggggggggggggggggggggggggggggggggggggggggga

iaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaneneneneneneneeeneneneneneenee

dd ddddddddddddddddbybybybybybybybybybyybyybyybyyyyyyyyyyyyyy

aaaaaaaaiririiiiiiiiiiiriiiiii

in 1

s

28)

Que

Ans

Mar

ksM

arke

r’s C

omm

ents

9cTh

ere

will

be re

duct

ion

to th

e te

mpe

ratu

reof

the

air t

hat i

s flo

win

g ou

t of t

he h

aird

ryer

.AN

D

The

flow

rate

of a

ir th

roug

h th

e ha

irdry

er in

crea

ses

The

flow

rate

incr

ease

s du

e to

a lo

wer

resi

stan

ce p

ath

alon

g th

e m

otor

lead

ing

to a

hi

gher

cur

rent

flow

ing

thro

ugh

the

mot

or.

As th

e flo

w ra

te o

f air

incr

ease

s an

d po

wer

of h

eate

r sta

ys u

ncha

nged

as it

is s

till i

n pa

ralle

l to

the

emf,

the

tem

pera

ture

of a

ir flo

win

g ou

t of t

he b

ody

drop

s.

}[1]

}

[1]

[1]

Mos

t can

exp

lain

that

the

tem

pera

ture

of a

ir flo

win

g ou

t de

crea

ses

but c

ould

not

re

late

to th

e flo

w ra

te to

ex

plai

n

9dA

high

erai

r flo

w ra

tew

ill be

whe

n th

e sw

itch

is a

t K.

E =

2.11

6x

(25/

60) =

0.8

817

kWh

Cos

t of e

lect

ricity

= 0

.881

7x

unit

cost

=28

cent

s

Uni

t cos

t = 3

1.8

cent

s

[1]

[1]

Mos

t cou

ld n

ot c

alcu

late

as

they

wer

e us

ing

resi

stor

R in

th

e ca

lcul

atio

n w

hich

is n

ot

need

ed if

you

nee

d a

high

er

flow

rate

of a

ir

10ai

[3]

Mos

t end

ed u

p dr

awin

g a

real

imag

e w

ihic

h is

1.5

tim

es

inst

ead

of v

irtua

l ray

and

the

who

le ra

y di

agra

m is

wro

ng

isst

ill in

[1[1]]

[1[1]]

re exp

s [1[1

]]

[1[1]]

Mos

tth

eyw

eth

e ca

lcu

need

ed if

yflo

ww ra

te o

f

[33]]

Mos

tr

Que

Ans

Mar

ksM

arke

r’s C

omm

ents

10ai

iD

ista

nce

of im

age

from

lens

= 1

5 cm

[1]

Poor

ly d

one

10bi

[1]

for a

mpl

itude

[1]

for

wav

elen

gth

Mos

t can

iden

tfy th

e am

pitu

de a

nd w

avel

engt

h co

rrect

ly.

10bi

i[1

]A

nega

tive

disp

lace

men

t mus

t be

dra

wn

first

[1] c

ompr

essi

on

and

rare

fact

ion

posi

tion

mus

t m

atch

es F

ig. 1

0.2b

Equa

l am

plitu

de

thro

ugho

ut

Mos

t can

iden

tify

the

wav

elen

gth

and

the

poin

t w

ith m

axim

um d

ispl

acem

ent

but c

ould

not

reco

gnis

e th

at

the

wav

efro

m m

ust s

tart

with

ne

gativ

e di

spla

cem

ent f

irst

befo

re d

raw

ing

the

posi

tive

disp

lace

men

t.

[1]

foforr a

mamplpl

ititudud

e

[11]]

foforr

wav

evelele

ngngthth

c

[1[1]]

A nee

gagativ

e didi

spla

cem

em

entnt

mmusus

t tbe

ddrara

wn

wn

ffirirsts

Mo

Mos

tstca

n id

ew

aw

avel

engt

hw

ithm

b

Que

Ans

Mar

ksM

arke

r’s C

omm

ents

10bi

iiTh

ere

will

be a

n in

crea

se in

am

plitu

dedu

e to

the

loud

ness

.

The

wav

elen

gth

will

be

shor

tera

s th

e hi

gher

pitc

h br

ings

abo

ut h

ighe

r fre

quen

cy a

nd

thus

sho

rter w

avel

engt

h.

[1]

[1]

Mos

t sta

ted

the

incr

ease

in

the

wav

elen

gth

BUT

over

look

th

at th

e ax

is u

sed

in F

ig. 1

0.3

is d

ista

nce

and

NO

T tim

e.

E11a

iTo

tal e

nerg

y in

put b

y m

otor

= P

x t

= 50

00 x

30

= 15

0 00

0 J

Ener

gy in

put b

y m

otor

= G

ain

in P

E 15

0 00

0 =

mgh

= (2

00) (

10) (

h)

h =

75 m

[1]

[1]

Mos

t can

do

it

E11a

iiAr

ea u

nder

vel

tim

e gr

aph

= 75

m

½ (v

)(t) =

75

½ (1

0 t)

(t) =

75

t = 3

.87

s

[1]

[1]

Mos

t can

do

itE1

1bi

The

wei

ght o

f fal

ling

wei

ght =

0.0

2 x

10 =

0.2

N

The

mag

netic

forc

e by

the

elec

trom

agne

t B m

ust b

e m

ore

than

0.2

Nso

that

it c

an c

reat

e

a de

cele

ratio

n ie

a n

egat

ive

net f

orce

so a

s to

brin

g to

an

inst

ant h

alt f

or th

e fa

lling

wei

ght.

[1]

[1]

Mos

t did

not

real

ise

that

th

ere

is a

nee

d fo

r the

forc

e to

be

high

er th

an th

e w

eigh

t in

ord

er to

brin

g ab

out a

de

cele

ratio

n. IF

it is

the

sam

e as

the

wei

ght t

he o

utco

me

is

that

the

mas

ss w

ill fa

ll w

ith

cons

tant

vel

ocity

inst

ead

of

stop

ping

E11b

iiIn

crea

se th

e nu

mbe

r of t

urns

coilin

g th

e el

ectro

mag

net

Incr

ease

the

curr

ent f

low

ing

in th

e so

leno

id

[1]

[1]

Mos

t can

do

it

E11b

iiiTh

ere

is n

o ne

ed to

incr

ease

/ cha

nge

the

stre

ngth

of th

e el

ectro

mag

net B

as

the

amou

nt o

f for

ce n

eede

d is

stil

l the

sam

e ie

to o

verc

ome

the

wei

ght

[1]

[1]

Mos

t did

not

real

ise

that

the

forc

e is

the

sam

e as

the

wei

ght i

s st

ill th

e sa

me

desp

ite fa

lling

from

a h

ighe

r he

ight

O11

aiSo

uth

pola

rity

at Q

[1]

Mos

t can

do

it

nummmmmmmmmmmmmmmm

bebebeebebeebebeeebeebebebebeeeebrr rrrrrrrrrrrro

f turrrrrrrrrrrrr

nsnsnsnsnsnsnsnsnsnsnsnsnsnnsnsnsnsnnnsnsnsnnnnsnnnnssssssssssscocococococococococococco

iling

ththhhhhhhhhhhhhhhhhhhhhhhteeeeeeeeeeeeeee

eleeleleleleleleleleleeeeececeeeeeeeeeeeeeeeeeeeeeeeee

trtrrrrrrrrrrrrrromomomomomomommmomomommomommmomoooooooo

agagagagagagagagagagagagagggggggggggggggggnenenennnnnnnnnnnnnnnnn

t

rren

tntntttttttttttttt

[1[]]

[1[1]]

[[1]]

[1[1]]

[1[1]]

Mo

Mos

tstca

n do

10

= 0

.2 N

om

agne

t B m

uststsssssssssssssssssssssssssssssssssss

bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeem

om

om

om

om

om

om

om

oooom

oom

om

om

om

om

ooom

om

om

om

om

om

om

om

ooom

om

omm

ommmmmmmmmmmmm

rererereererererererereeerreerereeeererererrerereerereerrrrrertttttttttttttttttttttttth

ahahahhhahhahhhahahhahahahahahhhahhahahahahahahahhahahahhaahaaaaaaaaannn

0.00000000000000000000022222222 222222222222222222222222222

NNNNNNNNNNNNNNNNNNNNNNNNNNsososososososoosoososososososososooosososososososssssss

ttttttttttttttttttttttthahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

t it c

an c

reat

e

gativ

e ne

t for

cecececcececcecececececcceccceccccecccceecsososososososossosossosososososososososososososossososssoososoosoosoooosoosso

aaaaassssssssssssssssssssssssssstot

brin

ggttttttttttttto

oan

insnsssssnssnssnssssssssssssssnssnsssnssntaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

nt h

alt f

or r t

hththththththththhhhhththhthhththththhththhhthhthhththhttttttthe e ee ee e eeee ee eeeeee eeeeeeeee

fafafafafafafafafaafffffffffffffffffaffffffffflllllllllllllllllllinininiiniiinininiiniiiiin

g

[1[1]]

[1[1]

Mo

Mos

t did

nth

ere

t

flflflflflflflflfflflflflowowowowowowoowoowooooooooooowooooowooooowwwww

inininininnininininininninininnnnnggggggggggggggggggggggggggggggggggggggg

ininnnnnnnnnnnnnnnnnnnnnnnnnnntttttttttttttttttttttttttth

eheeheheheeeeeeheheeeheheeheeeeeeheeheh s

olollolllololololololollllllllllllllllloeneeneeneeneeneneeeneneneneeneeeeneeeeee

oioioiooioioioooioiooioooiooiooioiooioooiooooooddddddddddddddddddddddddddddddddd

o in

ccccccccccccrerererereerereererrerereeeereeeerereere

asaaasasasasasasasasasasasaaaaaaaaaae/e

cccccchahahhahahahahahahahahahahahahahahahahahahhaahahahhaahahaaahahaaangng

e th

e stststst

rerererengngngng

thhhhof

the

e

ededddddddddddddddddddddddddddd

iiiiiiiis s

till t

he e e e sasasasa

me

me

me

me

ie to

ov

Que

Ans

Mar

ksM

arke

r’s C

omm

ents

011a

iiTh

e co

il w

ill ro

tate

in a

clo

ckw

ise

man

ner

[1]

Mos

t can

do

it

011a

iiiAs

the

coil

rota

tes

past

the

verti

cal p

ositi

on, t

here

is a

reve

rse

in th

e di

rect

ion

of th

e cu

rren

t flo

win

g in

the

coil

due

to th

e pr

esen

ce o

f the

spl

it rin

g.

This

will

allo

w th

e co

il to

rota

te c

ontin

uous

ly a

s th

e fo

rce

actin

g on

the

left

side

of t

he

coil

is a

lway

s up

whi

le th

e rig

ht s

ide

of c

oil i

s al

way

s do

wn.

[1]

[1]

Mos

t did

not

rela

te w

ell t

he

resa

son

for c

ontin

uous

ly

rota

tion

but m

erel

y st

ate

ther

e is

pre

senc

e of

spl

it rin

g

011b

iBo

th b

ulb

J an

d K

are

not o

hmic

con

duct

ors

as th

e vo

ltage

is n

ot d

irect

ly p

ropo

rtio

nal t

o cu

rren

t

[1]

[1]

Som

e ov

erlo

oked

the

defin

ition

of o

hms

law

and

th

ink

that

bot

h ar

e oh

mic

as

it pa

sses

thro

ugh

the

orig

in01

1bii

At V

= 0

.2 V

Bu

lb J

cur

rent

= 0

.05

A th

us re

sist

ance

= V

/I =

0.2/

0.05

= 4

Bulb

K c

urre

nt =

0.0

2 A

thus

resi

stan

ce =

V/I

= 0.

2/ 0

.02

= 10

[1]

[1]

Mos

t can

do

it

011b

iiiTh

e lo

nger

the

leng

th o

f the

wire

use

d, th

e hi

gher

the

resi

stan

cean

d th

eth

icke

rthe

wire

, the

low

er th

e re

sist

ance

[1]

[1]

Mos

t can

do

it

fthe

[1]

[1[]]

th

[1[1]]

[1[1]

Som

defin

ith

ink

thpa

sses

t

0.2/

0.05

= 4

V/I=

0..2

/2/2/2/2//2/2/2/2/2/2/2/2/2//2/2/2/2/2/22/22/22222222222222200000000

0.0.0.0.0.0.000.000.00000.00.000000..000022222

====================1010000000000000

[1[1]]

[1[1]]

Mo

Mos

tstca

n do

eus

ed, t

he h

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Date post:	31-Jan-2023
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GAN ENG SENG SCHOOL PHYSICS Sec 4 Express 6091/01

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