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NYJC 2020 9749/01/J2Prelim/20 [Turn over

NANYANG JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2

CANDIDATE NAME

CLASS

TUTOR’SNAME

CENTRE NUMBER S

INDEX NUMBER

PHYSICS 9749/01

Paper 1 Multiple Choice 17 September 2020

1 hour

Additional Materials: Multiple Choice Answer Sheet

READ THESE INSTRUCTIONS FIRST

Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. Write your name, class, Centre number and index number in the spaces at the top of this page. There are thirty questions on 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 Answer Sheet. Read the instructions on the Answer Sheet 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. The use of an approved scientific calculator is expected, where appropriate.

This document consists of 15 printed pages.

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NYJC 2020 9749/01/J2Prelim/20

Data

speed of light in free space c = 3.00 × 108 m s−1

permeability of free space = 4 × 10−7 H m−1

permittivity of free space = 8.85 × 10−12 F m−1

(1 / (36)) × 10−9 F m−1

elementary charge e = 1.60 × 10−19 C

the Planck constant h = 6.63 × 10−34 J s

unified atomic mass constant u = 1.66 × 10−27 kg

rest mass of electron me = 9.11 × 10−31 kg

rest mass of proton mp = 1.67 × 10−27 kg

molar gas constant R = 8.31 J K−1 mol−1

the Avogadro constant NA = 6.02 × 1023 mol−1

the Boltzmann constant k = 1.38 × 10−23 J K−1

gravitational constant G = 6.67 × 10−11 N m2 kg−2

acceleration of free fall g = 9.81 m s−2

Formulae uniformly accelerated motion 21

2s ut at

2 2 2v u as work done on / by a gas W p V hydrostatic pressure p gh gravitational potential /Gm r

temperature / K / C 273.15T T

pressure of an ideal gas 21

3

Nmp c

V

mean translational kinetic energy of an ideal molecule 3

2E kT

displacement of particle in s.h.m. 0 sinx x t

velocity of particle in s.h.m. 0 cosv v t

2 20x x

electric current I Anvq

resistors in series 1 2 . . .R R R

resistors in parallel 1 21/ 1/ 1/ . . .R R R

electric potential 04

QV

r

alternating current/voltage 0 sinx x t

magnetic flux density due to a long straight wire

0

2IBd

magnetic flux density due to a flat circular coil

0

2NIBr

magnetic flux density due to a long solenoid 0B nI

radioactive decay 0 exp( )x x t

decay constant 12

ln2

t

3


1 What is the number of SI base units required to express electric field strength and magnetic flux density?

electric field strength magnetic flux density

A 3 3

B 4 3

C 3 4

D 4 4

2 The viscosity μ of a fluid can be determined by measuring the terminal velocity vt of a sphere when it descends in the fluid. The fluid has a density ρf while the sphere has a density ρs and a diameter d. The viscosity can then be calculated using the equation

25

9s f

t

dv

The quantities measured are

vt = (1.60 ± 0.04) m s

ρs = (2700 ± 20) kg m

ρf = (900 ± 10) kg m

d = (20.0 ± 0.4) mm

What is the percentage uncertainty in the value of ?

A 6.2 % B 7.1 % C 8.2 % D 30 %

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NYJC 2020 9749/01/J2Prelim/20

3 A driver decelerates uniformly as he approaches a junction, makes a U-turn, and accelerates at the same rate that he decelerated earlier. Which of the following velocity-time and acceleration time graphs best represents the motion?

A

B

C

D

v / m s–1

t / s

a / m s–2

t / s

v / m s–1

t / s

a / m s–2

t / s

v / m s–1

t / s

a / m s–2

t / s

v / m s–1

t / s

a / m s–2

t / s

5


4 The diagram shows a laboratory experiment in which a feather falls from rest in a long evacuated vertical tube of length L.

The feather takes time T to fall from the top to the bottom of the tube.

How long will the feather take to fall 0.50 L from the top of the tube?

A 0.25 T B 0.29 T C 0.71 T D 0.75 T

5 The diagram shows two blocks of mass m and 2m connected by a light inextensible cord passing

over a light, free-running pulley. At what angle θ must the smooth slope be inclined such that the block with mass 2m accelerates down the slope at 1.7 m s-2?

A 14° B 41° C 49° D 76°

m

2m

θ

1.7 m s-2

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NYJC 2020 9749/01/J2Prelim/20

6 The diagram shows two spherical masses approaching each other head-on at the same speed u. One has mass 2m and the other has mass m. The masses then collides elastically.

Which diagram shows the speeds of the masses after the collision?

A B

C D

7 A beam, the weight of which may be neglected, is supported by three identical springs. When

a weight W is hung from the middle of the beam, the extension of each spring is x.

The middle spring and the weight is removed. What is the extension when a weight of 2W is hung from the middle of the beam?

A 4

3

x B

3

2

x C 2x D 3x

6

u 7

6

u

5

3

u

7


8 A gas at a pressure of 5.0 105 Pa is enclosed in a cylinder fitted with a piston.

The gas expands by 4.0 m3 against a constant exernal pressure of 1.0 105 Pa.

How much work does the gas do against the external pressure?

A 4.0 105 J B 12 105 J C 16 105 J D 20 105 J

9 A ball rolls down the circularly curved track shown below.

Which of the following correctly describes its speed, acceleration along the slope and kinetic energy?

speed acceleration kinetic energy

A increases increases increases

B decreases decreases decreases

C increases decreases increases

D increases constant increases

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NYJC 2020 9749/01/J2Prelim/20

10 A track cycling competiton consists of a 4.0 km race and a one hour record race. Cyclists move around a circular track with a circumference of 1.00 km.

In the 4.0 km race, the top cyclist finishes the race in 4 mins and 1.934 s. In a one hour record race, the top cyclist travelled a distance of 55.089 km in one hour.

What is the difference between the angular speed of these top cyclists of each race?

A 0.0012 rad s-1 B 0.0077 rad s-1 C 0.014 rad s-1 D 0.92 rad s-1

11 The section of a cycling track where sharper turns are made has a horizontal radius of 20.0 m, and is banked at 40o to the horizontal.

Given that the maximum speed of the cyclist of mass 80 kg is 60 km h-1, what is the difference in frictional force that the bicycle tyres must provide compared to an unbanked track?

A 140 N B 450 N C 760 N D 1200 N

12 A planet has an average density 4950 kg m-3. Given that the acceleration of free fall near the surface of the planet is 0.9 g, where g is the acceleration of free fall near Earth’s surface, the radius of the planet is approximately

A 260 km B 2500 km C 2800 km D 6400 km

13 Two satellites, A and B, orbiting around Earth have the same kinetic energy. Satellite A has a larger mass than satellite B. Which of the following statements is true?

A Satellite A has a larger period.

B Satellite A has a smaller total energy.

C Satellite A has a smaller orbital radius.

D Satellite A has a larger angular velocity.

angle of bank is 40o at sharp turning

20.0 m

9


14 Fig. 14.1 shows the variation of velocity v with displacement x of a simple harmonic oscillator. The state of motion of point P is shown in Fig. 14.1.

Fig. 14.1

Fig. 14.2 shows the variation with time t of the acceleration a of the oscillator.

Fig. 14.2

Which of the points on Fig. 14.2 correspond to the state of motion represented by point P?

15 Five particles have speeds of 2u, 5u, 10u, 11u and 15u. Which of the following statements is

correct?

A The root mean square speed is equal to the mean speed.

B The root mean square speed exceeds the mean speed by more than 1u.

C The root mean square speed exceeds the mean speed by less than 1u.

D The mean speed exceeds the root mean square speed by more than 1u.

16 Latent heat of vaporisation is the energy required to

A separate the molecules of the liquid

B separate the molecules and to push back the atmosphere

C increase the average molecular speed in the liquid state compared to that in gaseous state

D separate the molecules and to increase the average molecular speed in the liquid state compared to that in gaseous state.

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NYJC 2020 9749/01/J2Prelim/20

17 The diagram below shows a sound wave traveling from left to the right in air at a particular instant. Air particles P and Q are the centres of a rarefaction and a compression respectively. Which of the following describes the directions of motion of P and Q at this instant?

Particle P Particle Q

A to the right to the left

B to the right at rest

C to the left to the right

D to the left at rest

18 A vertical tube is completely filled with water. A small sound source of constant frequency is held

a little above the open upper end and water is run out from the lower end.

A number of resonance positions are detected. The second position was detected when the water surface is 12 cm below the top of the tube and another occurs at 36 cm.

At which of the following distances should resonance also be detected?

A 24 cm B 26 cm C 44 cm D 48 cm

19 A diffraction grating with 6.0 × 105 lines per metre is placed at right angles to a ray of white light.

The first and second order spectra produced are shown in the diagram.

The angle between the red and violet ends of the spectrum is for the first order spectrum and for the second order spectrum. How do and compare?

A < ½ B = ½ C = D >

Q

Direction of wave propagation

P

grating

white light

red

red

violet

violet

First order spectrum

Second order spectrum

11


20 In which of the following cases does an electric field do positive work on a charged particle?

A A positive charge moves to a point of higher electric potential.

B A negative charge moves opposite to the direction of the electric field.

C A positive charge completes one circular path around a stationary positive charge.

D A negative charge moves perpendicular to the electric field between two parallel and oppositely charged plates.

21 A beam of protons passes through a velocity selector, set up by an electric field and a magnetic field that are perpendicular to each other. The electric field is set up by a pair of charged plates with a plate separation of 2.0 cm as shown in the diagram below. The magnetic flux density, B is 1.5 T and directed into the plane of the paper.

If protons travelling at 2.0 × 107 m s-1 pass through undeflected, what would be the direction and magnitude of the electric field?

direction magnitude

A downwards 6.0 × 105 N C-1

B upwards 6.0 × 105 N C-1

C downwards 3.0 × 107 N C-1

D upwards 3.0 × 107 N C-1

2.0 cm

beam of protons

v = 2.0 × 107 m s-1

B = 1.5 T

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NYJC 2020 9749/01/J2Prelim/20

22 A battery supplies a current of 0.025 A for 80 s. During this time it produces 18.0 J of electrical energy while resistor M receives 11.0 J and resistor N receives 4.0 J.

What is the e.m.f. of the battery and its internal resistance L? e.m.f. of battery / V internal resistance L of battery /

A 1.5 60

B 9.0 60

C 9.0 360

D 16.5 360

23 Five identical resistors are connected as shown in the circuit. If the power dissipated by X is 2.0 W, what is the total power supplied by the battery?

A 2.5 W B 3.5 W C 7.5 W D 10 W

X

13


24 The diagram shows four long straight current-carrying conductors placed at the corners of a square at W, X, Y and Z. The current in each conductor has the same magnitude.

Point O is the intersection of the diagonals of the square. If the magnetic flux density at point O is zero, which of the following is true?

A The current in Y must be in the opposite direction as that in W and the current in X must be in the opposite direction as that in Z.

B The current in Y must be in the same direction as that in W and the current in X must be in the opposite direction as that in Z.

C The current in Y must be in the same direction as that in X and the current in W must be in the opposite direction as that in Z.

D The current in Y must be in the same direction as that in X and the current in W must be in the same direction as that in Z.

25 In an experiment to record electrical events of short duration, a student drops a bar magnet

through a very thin, horizontal coil, as shown.

Which graph best represents how the induced e.m.f. E in the coil varies with time t?

A B C D

thin

falling bar magnet

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NYJC 2020 9749/01/J2Prelim/20

26 A uniform magnetic field directed at 50 from the vertical passes through a circular metal ring of diameter 0.50 m and resistance 3.0 .

The magnetic flux density through the ring decreases by 4.0 × 10–5 T at a constant rate in 2.0 s. During this change, the current induced in the ring

A remains constant at 1.0 µA.

B remains constant at 1.3 µA.

C decreases from zero to 1.0 µA.

D decreases from zero to 1.3 µA.

27 When a light bulb is connected across an a.c. source of peak voltage 170 V, the power

dissipated is 40 W. Two such light bulbs are now connected in series to the electrical mains of 220 V r.m.s. What will be the total power dissipated?

A 33.5 W B 67.0 W C 80.0 W D 134.0 W

metal ring

magnetic field

50

15


28 Light of wavelength is incident on a metal surface in a vacuum. Photoelectrons are emitted from the surface of the metal.

Which of the following best shows the variation with of the maximum kinetic energy EK of the

emitted electrons?

A

B

C

D

29 An atom makes a transition from an energy level of energy E1 to a level of energy E2, emitting a photon of wavelength .

Which expression gives this wavelength in terms of Planck constant h and the speed of light c?

A E E

hc hc1 2 B

hc hc

E E

1 2

C hc hc

E E

2 1

D hc

E -E1 2

30 A proton and an alpha particle ( 42He Helium nucleus) have the same de Broglie wavelength.

The ratio speed of proton

speed of alpha particle is

A 1

4 B

1

2 C 2 D 4

End of Paper

EK

EK

EK

EK



CANDIDATE NAME

CLASS

TUTOR’SNAME

CENTRE NUMBER

S INDEX NUMBER

PHYSICS 9749/02Paper 2 Structured Questions 2 September 2020

2 hoursCandidates answer on the Question Paper.

No Additional Materials are required.


Write your name, class, Centre number and index number in the spaces at the top of this page. Write in dark blue or black pen on both sides of the paper. You may use a HB pencil for any diagrams, graphs. Do not use staples, paper clips, glue or correction fluid. The use of an approved scientific calculator is expected, where appropriate. Answer all questions. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.

For Examiner’s Use

1 / 8

2 / 10

3 / 7

4 / 9

5 / 11

6 / 8

7 / 8

8 / 19

Total / 80


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NYJC 2020 9749/02/J2Prelim/20

Data




(1 / (36)) × 10−9 F m−1












2s ut at

2 2 2v u as work done on / by a gas W p V

hydrostatic pressure p gh gravitational potential /Gm r temperature / K / C 273.15T T


3

Nmp c

V


2E kT



2 20x x





QV

r



0

2IBd


0

2NIBr



decay constant 12

ln2

t

3


1 During a wet weather, a stationary rain droplet falls from great height and eventually reaches a terminal velocity of 9.00 m s-1.

(a) Explain how the droplet achieved terminal velocity.

[3] (b) The droplet eventually hits the side window of a moving train. The relative velocity, v of the

droplet makes an angle of θ with the vertical, as shown in Fig. 1.1.

Fig. 1.1

Sketch a graph of the train speed against θ using the grid in Fig. 1.2. [3]

Fig. 1.2

motion of the train θ

v

θ / °

train speed / m s-1

0 20 40 60 80 100

40.0

30.0

20.0

10.0

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NYJC 2020 9749/02/J2Prelim/20

(c) The train is travelling at a speed of 125 km h-1. On reaching the station, the brake is applied such that the train decelerates uniformly. During the deceleration, the train travels another 850 m before coming to a rest.

Determine the time taken for the train to stop.

time taken = s [2]

[Total: 8]

2 A water jetpack consists of a backpack consisting of a motor, two thrusters and a pipe for constant supply of water. A diagram of one thruster is shown in Fig. 2.1.

Fig. 2.1

Water is ejected vertically downwards through the nozzle. The thruster moves vertically upwards. The density of water is 1.00 × 103 kg m-3. The nozzle has a circular cross-section of radius 10.0 cm. The thruster ejects 30.0 kg of water in 0.20 s. Assume that the water leaving the nozzle has the shape of a cylinder of radius 10.0 cm.

(a) (i) Show that the water has a constant speed of 4.8 m s-1 relative to the thruster. [2]

water density 1.00 × 103 kg m-3

5


(ii) Show that the force exerted on water ejected by the jetpack is 1.4 × 103 N. [1] (iii) Determine the maximum combined mass of a jetpack with its user that can be lifted

with an acceleration of 9.81 m s-2. Explain your answer.

mass = kg [3]

(b) When in use, the jetpack takes in a constant volume of water from the environment and has a constant mass. To move horizontally, the user tilts the thrusters slightly using the handles connected. Fig. 2.2 shows one such jetpack in use above a lake.

Fig. 2.2

(i) Determine, for a user with a jetpack of combine mass 80 kg, the maximum angle with respect to the vertical which the thrusters can be tilted without the user losing altitude. Assume that the force exerted by the water pipe on the user is negligible.

maximum angle = ° [2]

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NYJC 2020 9749/02/J2Prelim/20

(ii) Calculate the power delivered to the water for a stationary user just above the water.

power delivered = W [2]

[Total: 10]

3 (a) Two inclined planes RA and LA each have the same constant gradient. They meet at their

lower edges, as shown in Fig. 3.1.

Fig. 3.1

A small ball moves from rest down plane RA and then rises up plane LA. It then moves down plane LA and rises up plane RA to its original height. The motion repeats itself.

Explain why the motion of the ball is not simple harmonic.

[1]

7


(b) A small ball rests at point P on a curved track of radius r, as shown in Fig. 3.2.

Fig. 3.2

The ball is moved a small distance to one side and is then released. The horizontal displacement x of the ball is related to its acceleration a towards P by the expression

r

gxa

where g is the acceleration of free fall.

(i) With reference to the expression provided, show that the ball undergoes simple harmonic motion.

[2]

(ii) The radius r of curvature of the track is 28 cm. Determine the time interval between the ball passing point P and then returning to

point P.

= s [2]

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NYJC 2020 9749/02/J2Prelim/20

(c) The variation with time t of the displacement x of the ball in (b) is shown in Fig. 3.3.

Fig. 3.3

Some moisture now forms on the track, causing the ball to come to rest after approximately several oscillations.

On the axes of Fig. 3.3, sketch the variation with time t of the displacement x of the ball for

the first two periods after the moisture has formed. Assume the moisture forms at time t = 0. [2]

[Total: 7]

4 (a) Define electric field strength.

[1] (b) Two charged metal spheres A and B are situated in a vacuum, as illustrated in Fig. 4.1.

The shortest distance between the surfaces of the spheres is 12.0 cm.

A movable point P lies along the line joining the centres of the two spheres, a distance x from the surface of sphere A.

The variation with distance x of the electric field strength E at point P is shown in Fig. 4.2.

T 2T T/2 3T/2

sphere A sphere B

12.0 cm

x

P

Fig. 4.1

9


(i) Using Fig. 4.2, comment on the magnitude and polarity of the charge of sphere A and sphere B.

[2] (ii) A proton is at point P where its acceleration is at a minimum. Use data from Fig. 4.2 to

determine the acceleration of the proton.

acceleration = m s-2 [2]

0

0.1

0.2

0.3

0.4

0.5

0.6

0 2 4 6 8 10 12

Fig. 4.2

E / kV m-1

x / cm

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NYJC 2020 9749/02/J2Prelim/20

(c) The variation with distance x of the electric potential V at point P is shown in Fig. 4.3.

(i) State, in words, the relationship between electric potential at point P and electric field strength at point P.

[1] (ii) A proton is initially at rest on the surface of sphere A and moves along the line joining

the centres of the two spheres to reach maximum speed at point P.

Use data from Fig. 4.3 to determine the maximum speed of the proton.

maximum speed = m s-1 [3] [Total: 9]

-15

-10

-5

0

5

10

0 2 4 6 8 10 12

V / V

x / cm

Fig. 4.3

11


5 A cell P, a fixed resistor R and a uniform resistance wire AB are connected in a circuit as shown in Fig. 5.1.

Fig. 5.1 Cell P has e.m.f. 4.0 V and internal resistance 0.75 . Wire AB has length 1.50 m and resistance

5.5 . The voltmeter reads 1.3 V.

(a) Show that the potential difference across AB is 2.4 V. [2]

(b) A cell Q and a sensitive ammeter are connected to the circuit in Fig. 5.1, as shown in Fig. 5.2.

Fig. 5.2 Cell Q has e.m.f. E and internal resistance 0.25 . The ammeter reads zero when the length

of AC is 0.56 m.

4.0 V 0.75

R

cell P

A B

V

5.5

1.50 m

C

4.0 V 0.75

R

cell P

A B

V

E 0.25

cell Q

sensitive ammeter

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NYJC 2020 9749/02/J2Prelim/20

(i) Determine E.

E = V [2]

(ii) There is a reading on the ammeter when the connection C is shifted closer to A. State and explain the direction of the current across cell Q.

[2]

(d) The resistance wire AB is detached from the circuit and coiled in a circular manner as shown

in Fig. 5.3. Metallic fasteners with negligible resistance are used to secure the wire at X and Y, where XY is the diameter of the coil.

Fig. 5.3 (i) Determine the resistance of wire AB when coiled in this manner.

resistance = [3] (ii) An e.m.f. source is again connected across AB. State and explain whether the drift

velocity of the electrons is greater in AX or in XY.

[2]

[Total: 11]

A B

15 cm 15 cm

X Y

13


6 W is a long straight wire of length 1.2 m. A beam of electrons is projected with a speed of 2.1 ×107 m s1 parallel to W at a distance of 20 cm from it, as shown in Fig. 6.1.

Fig. 6.1

(a) When the switch is closed, the electron beam is deflected laterally by the current in W. Sketch the appearance of the deflected beam in Fig. 6.1. [1]

(b) The current in W is 1.8 A. Calculate the flux density due to the current at a distance of 20 cm

from W.

flux density = T [2]

(c) Assuming that the lateral deflection of the electron beam is much less than 20 cm, show that the deflected path has a radius of 66 m.

[2]

(d) Calculate the lateral deflection of the beam at Q in mm.

deflection = mm [3] [Total: 8]

electron beam

wire W

Q

20 cm

P

1.2 m

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NYJC 2020 9749/02/J2Prelim/20

7 A 240 V a.c. mains is connected to a step-down transformer, a full wave rectifier and a motor as shown in Fig. 7.1. The electric motor requires an operating peak power of 6.6 W.

Fig. 7.1

The variation of potential difference across the 50 Ω resistor with time is shown in Fig. 7.2.

Fig. 7.2

(a) Determine the turns ratio of primary coil to secondary coil in the step-down transformer.

turns ratio = [2] (b) (i) For half a period T, the potential at X is higher than that at Y. On Fig. 7.1, draw an arrow to indicate the direction of current, if any, across the four

diodes A, B, C and D during this half of a period. [1]

240 V mains

Step-down Transformer

IP

X

Y

Full Wave Rectifier

electric motor50

Ω

T 2T

-12.2

12.2

potential difference / V

time / s

A B

C D

15


(ii) Sketch, on Fig. 7.3, the variation of potential difference across the electric motor with time. Numerical calculation is not required. [1]

Fig. 7.3

(iii) Hence, sketch, on Fig. 7.4, the variation of output power of the step-down transformer with time when the electric motor is operating at its normal operating power. Include appropriate values in the output power axis.

Fig. 7.4

[2] (c) Given that the transformer is 80% efficient, determine the r.m.s. current flowing through the

primary coil.

current = A [2]

[Total: 8]


time / s

output power / W

time / s

T 2T

T 2T

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NYJC 2020 9749/02/J2Prelim/20

8 In a refrigeration system, a fluid known as a refrigerant undergoes alternating changes in its physical state and volume to produce cooling. In this question, you will be guided to design a simple refrigeration system using the data provided.

Fig. 8.1 is a pressure vs volume graph showing the refrigerant cycling through four states P, Q, R and S. The four phases of the cyclic process are evaporation, condensation, compression and throttling (pressure releasing).

Fig. 8.1

(a) Using the information provided in Fig. 8.1, match each phase in the cycle with its description in the chart below. PQ has been matched.

Phase Description

PQ Evaporation

QR Condensation

RS Compression

SP Throttling (Pressure release)

[2]

Pre

ssur

e

Volume

P

Q

(liquid)

(vapour)

(liquid + vapour)

S

R

17


Fig. 8.2 is a table showing the thermal properties of the refrigerant as a saturated vapour and as a saturated liquid at various pressures.

Pressure P

(kPa)

Temperature T

(K)

Enthalpy h

(kJ kg1)

Entropy s

(kJ kg1 K1) Saturated

Liquid Saturated

Vapour Saturated

Liquid Saturated

Vapour 100 246.8 17.3 234.5 0.072 0.952 200 263.1 38.5 244.5 0.155 0.938 300 273.9 52.8 250.9 0.208 0.931 400 282.1 64.0 255.6 0.248 0.927 500 288.9 73.4 259.3 0.280 0.924 600 294.8 81.5 262.4 0.308 0.922 700 299.9 88.8 265.1 0.332 0.920 800 304.5 95.5 267.3 0.354 0.918 900 308.7 101.6 269.3 0.374 0.917

1000 312.6 107.4 271.0 0.392 0.916 1200 319.5 117.8 273.9 0.425 0.913 1400 325.6 127.3 276.2 0.453 0.911 1600 331.1 136.0 277.9 0.479 0.908 1800 336.1 144.1 279.2 0.503 0.905 2000 340.7 151.8 280.1 0.525 0.902

Fig. 8.2

A saturated vapour is one which a small compression will cause it to condense.

(b) Deduce what a saturated liquid is.

[1]

(c) By referring to Fig. 8.1, identify the state (P, Q, R or S) at which the refrigerant is a saturated vapour, and another at which it is a saturated liquid.

saturated vapour: , saturated liquid: [2]

Fig. 8.3 is an incomplete table showing the thermal properties of the refrigerant at states P, Q, R and S in the cyclic process.

State Pressure P

(kPa) Temperature T

(K)

Enthalpy h

(kJ kg1)

Entropy s

(kJ kg1 K1)

P 300

Q 1200 322.0

R

S 0.443

Fig. 8.3

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NYJC 2020 9749/02/J2Prelim/20

(d) Using Fig. 8.2 and your answer to (c), fill in the missing values in Fig. 8.3 for state P. [2]

(e) At state Q, the refrigerant is a superheated vapour, and its thermal properties at various pressures are shown in the graphs in Fig. 8.4 (a) & (b).

Fig. 8.4 (a)

Fig. 8.4 (b)

Ent

halp

y /

kJ k

g1

Temperature (K)

1600 kPa1400 kPa1200 kPa1000 kPa800 kPa

Ent

ropy

/ k

J kg

1 K

1

Temperature (K)


19


The process PQ is isentropic (entropy s is constant) and the pressure of the refrigerant is increased to 1200 kPa.

1. Explain how Fig. 8.4 (a) and/or (b) verifies that the temperature of the refrigerant at state Q is 322.0 K.

[1]

2. Fill in the missing values in Fig. 8.3 for state Q. [1]

(f) The process QR is isobaric (pressure p is constant). Using Fig. 8.2 and your answer to (c), fill in the missing values in Fig. 8.3 for state R. [2]

(g) The process RS is isenthalpic (enthalpy h is constant). The process SP is isothermal (temperature T is constant) and isobaric. Fill in the missing values in Fig. 8.3 for state S.[1]

(h) Energy is inputted during the compression phase of the refrigeration process. The rate of work done by the compressor is calculated using

W = m ( hf hi )

where m is the mass flow rate of the refrigerant, and hf and hi are the values of enthalpy after and before the compression respectively.

If the compressor compresses 3.20 kg of refrigerant in 1 second, calculate W the rate of work done by the compressor.

W = kW [2]

20

NYJC 2020 9749/02/J2Prelim/20

(i) Cooling is achieved in the evaporation phase of the process. The rate of heat removal during this phase is calculated using

Q = m ( hf hi )

where m is the mass flow rate of the refrigerant, and hf and hi are the values of enthalpy after and before the evaporation respectively.

For the same compressor in (h), calculate Q the rate of heat removal by the refrigeration process.

Q = kW [2]

(j) The efficiency of a refrigeration system is measured using its coefficient of performance which is calculated using

C = Q / W

Calculate C the coefficient of performance for the refrigeration system you designed.

C = [1]

(k) A refrigeration system is regarded as efficient only if it has a coefficient of performance of above 4.

Explain with reference to the principle of conservation of energy, why, unlike the efficiency of a conventional mechanical system, a coefficient of performance of greater than 1 is possible.

[2]

[Total: 19]

End of Paper



CANDIDATE NAME

CLASS

TUTOR’SNAME

CENTRE NUMBER

S INDEX NUMBER

PHYSICS 9749/03Paper 3 Longer Structured Questions 14 September 2020




Write your name, class, Centre number and index number in the spaces at the top of this page. Write in dark blue or black pen on both sides of the paper. You may use a HB pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, glue or correction fluid.

The use of an approved scientific calculator is expected, where appropriate.

Section A Answer all questions. Section B Answer one question only. You are advised to spend one and a half hours on Section A and half an hour on Section B. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.


Section A

1 / 10

2 / 7

3 / 6

4 / 9

5 / 8

6 / 8

7 / 12

Section B

8 / 20

9 / 20

Total / 80


2

NYJC 2020 9749/03/J2Prelim/20

Data




(1 / (36)) × 10−9 F m−1












2s ut at




3

Nmp c

V


2E kT



2 20x x





QV

r



0

2IBd


0

2NIBr



decay constant 12

ln2

t

3


Section A

Answer all the questions in the spaces provided. 1 Ganymede is a moon of mass 1.50 x 1023 kg and radius 2.64 x 106 m orbiting the planet Jupiter

of mass 1.90 x 1027 kg as shown in Fig. 1.1. The orbital period of Ganymede is 171.6 hours.

Fig. 1.1

(a) (i) Show that the distance x between the centres of Jupiter and Ganymede is 1.07109 m. Explain your working.

[2]

(a) (ii) S is a point between Jupiter and Ganymede where the resultant gravitational field strength is zero. Determine the distance from the centre of Jupiter to point S.

distance = m [2]

Jupiter

Ganymede

rocketx

4

NYJC 2020 9749/03/J2Prelim/20

(a) (iii) On the axes below, sketch a graph from the surface of Jupiter to the surface of Ganymede to show the variation of gravitational field strength g with distance r from the centre of Jupiter. [2]

(b) A rock is positioned on the surface of Ganymede at the point furthest from Jupiter, as shown in Fig. 1.2. The rock is launched with an escape velocity v for it to reach infinity, far away from Jupiter and Ganymede.

Fig. 1.2

(i) Show that the gravitational potential at the position of the rock in Fig. 1.2

is 8 11.22 10 J kg .

[2]

g

r 0 surface of Jupiter

surface of Ganymede

x

Jupiter

Ganymede

rock x v

5


15 cm h

mass hanger and masses

beaker of water

pulleys

Fig. 2.1

(b) (ii) Calculate the escape velocity v.

v = m s−1 [2]

[Total: 10]

2 A metal block, of 15 cm height, was partially submerged in water of density 1.0 × 103 kg m-3.

The block has a uniform density, ρ, and a uniform cross-sectional area, A. The block is held up by an inelastic cord tied to a mass hanger, through two frictionless pulleys, as shown in Fig. 2.1.

Different quantities of masses were added to the mass hanger and the corresponding submerged height, h, of the block was measured.

The variation of total mass m of mass hanger and masses added, with the submerged height, h, is shown in Fig. 2.2.

6

NYJC 2020 9749/03/J2Prelim/20

(a) Using Fig. 2.2, determine the total mass m required to remove the metal block from the water.

m = kg [1]

(b) By considering the forces acting on the metal block, show that the gradient of the graph

in Fig. 2.2 is represented by 3

1

1.0 10 A

.

[2]

0.00

0.02

0.04

0.06

0.08

0.10

0.38 0.40 0.42 0.44 0.46 0.48 0.50

Fig. 2.2

m / kg

h / m

7


(c) Hence, determine the cross-sectional area, A.

A = m2 [2]

(d) Determine the density of the metal block, ρ.

ρ = kg m-3 [2]

[Total: 7]

3 (a) Explain the difference between accuracy and precision and state the type of error associated with each.

(i) Accuracy:

Type of error:

(ii) Precision:

Type of error: [2]

8

NYJC 2020 9749/03/J2Prelim/20

1 (b) Three digital clocks A, B and C are being tested in a laboratory. Using signals from the Global Positioning System, the displays on the clocks at the exact time of 12:00:00 on four successive days are shown in Fig. 3.1. The clocks are reset each day at 00:00:00.

Clock Day 1 Day 2 Day 3 Day 4

A 12:06:40 12:06:38 12:06:39 12:06:43

B 12:03:59 12:02:49 12:01:54 12:03:15

C 11:59:59 12:00:02 11:59:57 12:00:05

Fig. 3.1

State and explain,

(i) which clock is the most accurate,

[1]

(ii) which clock is the most precise.

[1]

(c) For clock B in (b), estimate the time displayed at 12:00:00 on day 5 with its associated uncertainty.

time = s [2]

[Total: 6]

9


4 (a) Explain what is meant by an ideal gas.

[2]

(b) A cylinder contains 3.16 mol of an ideal gas at a pressure of 4.81 105 Pa and a volume of 1.20 104 cm3. Heat is removed for a duration of 12 minutes at constant pressure such that its temperature decreased by 110 K.

(i) Calculate the initial temperature of the gas.

initial temperature = K [1]

(ii) Determine the new volume of the gas after heat is removed.

volume of the gas = m3 [2]

(iii) Determine the change in internal energy of the gas.

change in internal energy = J [1]

10

NYJC 2020 9749/03/J2Prelim/20

(iv) Determine the rate of heat loss.

rate of heat loss = W [3]

[Total: 9]

5 Fig. 5.1 and Fig. 5.2 show an amusement park ride that rotates passengers strapped onto a

platform in a vertical circular motion about a pivot. The platform is 8.4 m away from the pivot and there is a counter-weight at the other end of the rotating arm, 3.0 m away from the pivot. The mass of the counter-weight and fully loaded platform are 200 kg and 2000 kg respectively.

The platform is initially brought to the highest position shown in Fig. 5.1 where it is turned upside down and then released from rest. The counter-weight and platform subsequently swings about the pivot until the counter-weight is at the highest position as shown in Fig. 5.2. The speed of the counter weight at that instant is uc and that of the platform is up.

(a) Show that the ratio of . .2 8p

c

u

u

[1]

pivot

Fig. 5.1 Fig. 5.2

Counter-weight

pivot

uc

up

8.4 m

3.0 m

11


(b) Hence, show that the speed up of the platform when it is at the lowest position is 17.7 m s-1.

[2]

(c) Consider a man with a mass of 70 kg on the ride when the platform is at the lowest position. Given that the force strapping the man to the platform at this position is 100 N, calculate the reaction force which the platform exerts on this man. Explain your working.

reaction force = N [3]

(d) When the man is just about to reach the top of the path as shown in Fig. 5.3, his

handphone slips from his hands. State and explain the subsequent path of his handphone until it hits the ground.

[2]

[Total: 8]

pivot

Fig. 5.3

12

NYJC 2020 9749/03/J2Prelim/20

6 A rectangular coil PQRS of dimensions 14.0 cm by 12.0 cm, moves with a constant speed of 2.0 cm s-1 through a region of uniform magnetic field WXYZ of width 10.0 cm, as shown in Fig. 6.1. There is a magnetic flux density B of 1.5 T in WXYZ directed into the plane of the paper.

(a) State and explain the direction of induced current in coil PQRS when it enters the field.

[3]

(b) Calculate the maximum possible magnetic flux linkage during the motion.

maximum flux linkage = Wb [1] (c) The coil PQRS is moved from the position shown in Fig. 6.1 until side PS is aligned with

WZ. Sketch and label with appropriate values on the axes, the following graphs: (i) the variation with displacement of magnetic flux linkage through the coil, [2]

R

Q

2.0 cm s-1

P

S

14.0 cm

12.0 cm

Fig. 6.1

W

X

Y Z

B = 1.5 T

10.0 cm

displacement / cm

magnetic flux linkage / Wb

0

13


(ii) the variation with time of induced e.m.f. in the coil. [2]

[Total: 8]

time / s

e.m.f. / V

0

14

NYJC 2020 9749/03/J2Prelim/20

7 A light elastic band of length 20 cm is stretched to a length of 28 cm, and then allowed to return back to its original length. Fig. 7.1 shows how the length of the band varies with the force applied.

Fig. 7.1

(a) Explain how the graph shows that the band does not obey Hooke's Law.

[2]

(b) Estimate the work done by the applied force to stretch the band.

work done = J [2]

leng

th /

cm

force / N

stretching

releasing

15


(c) The elastic band is used in a catapult to project a pellet into the air. The band is stretched to a length of 28 cm and released. Explain using Fig. 7.1, why the kinetic energy of the pellet at the point of release is significantly smaller than your answer in (b).

[2]

(d) The initial kinetic energy of the projected pellet is 3.8 J. Calculate the efficiency of the band in accelerating the pellet.

efficiency = [2]

(e) Suggest one reason for the loss in efficiency.

[1]

(f) The elastic band is replaced with another new elastic band of the same length which obeys Hooke's Law. It is also stretched to a length of 28 cm and transfers 3.8 J of kinetic energy to the pellet.

(i) Calculate the spring constant of this band.

spring constant = N m1 [2]

(ii) Sketch the graph for this new elastic band on Fig. 7.1. [1]

[Total: 12]

16

NYJC 2020 9749/03/J2Prelim/20

Section B

Answer one question from this Section in the spaces provided.

8 The variation with time t of the displacement y of a wave A, as it passes a point P, is shown in Fig. 8.1.

The intensity of wave A is 2I.

(a) Use Fig. 8.1 to determine the frequency of wave A.

frequency = Hz [2]

(b) A second wave B with the same frequency and speed as wave A also passes point P.

Wave B has intensity I and is lagging wave A by 2

rad.

On Fig. 8.1, sketch the variation with time t of the displacement y of wave B. Label the graph B.

Show your working.

[3]

-3.0

-2.0

-1.0

0.0

1.0

2.0

3.0

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5

t / 10-15 s

y / cm

Fig. 8.1

17


(c) Explain why wave A and wave B are coherent.

[1]

(d) Wave R is the resultant wave due to the superposition of wave A and wave B. The amplitude of wave R occurs at t = 1.2 × 10-15 s.

Using your answer in (b) and Fig. 8.1, determine, in terms of I, the intensity of wave R at point P.

intensity = [2]

(e) After passing through point P, wave R is plane-polarised by passing it through a polarising filter. Determine, in terms of I, the intensity of the plane-polarised wave R.

intensity = [1]

(f) Wave R is made to pass through a double slit in an interference experiment, as shown in Fig. 8.2.

The separation between the slits is a. The fringes are viewed on a screen at a distance 2.00 m from the double slit. The fringe separation x is measured for different slit separation a.

Fig. 8.2

2.00 m

a

screen double slit

wave R

18

NYJC 2020 9749/03/J2Prelim/20

A graph of a against 1

xis shown in Fig. 8.3.

(i) Determine the wavelength of wave R.

wavelength = m [3]

(ii) State the effect, if any, on the appearance of the fringes observed on the screen when the following changes are made separately, at a fixed value of a:

1. wave R is replaced by a blue laser light,

[1]

0.00

0.10

0.20

0.30

0.40

0.50

0.60

1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

a / mm

1

x / cm-1

Fig. 8.3

19


2. the width of each slit is increased but the separation remains constant.

[2]

(iii) At a particular value of slit separation a, the variation with distance along the screen of the intensity of the image on the screen is shown in Fig. 8.4.

1. Using Fig. 8.4 and Fig. 8.3, determine the value of a.

a = mm [2]

2. Using Fig. 8.4, determine the width of each slit.

slit width = m [3]

[Total: 20]

0.0

0.2

0.4

0.6

0.8

1.0

-0.015 -0.010 -0.005 0.000 0.005 0.010 0.015

Fig. 8.4 distance / m

intensity

20

NYJC 2020 9749/03/J2Prelim/20

9 Two metal plates X and Y are contained in an evacuated container and are connected as shown in Fig. 9.1. Monochromatic electromagnetic radiation of blue light is incident on metal plate X.

The potential of X with respect to Y is varied by adjusting the position of the sliding contact from E to G. The variation of current I recorded at the ammeter with potential V of X is given by Fig. 9.2 below.

(a) Explain an evidence provided by the photoelectric effect experiment for the failure of the wave theory of light.

[2]

V

A

X Y

O F

E G

incident monochromatic radiation

Fig. 9.1

I / mA

V / V

Fig. 9.2

21


(b) Circle the portion of the graph on Fig. 9.2 that represents the variation of current as the sliding contact moves from F to G. [1]

(c) Sketch, on Fig. 9.2, the graph when electromagnetic radiation of the same intensity in

the ultraviolet region is used instead. [2]

(d) It is observed that photoelectrons are emitted from plate X when photons of wavelength 520 nm is illuminated on it. The metal plate has a work function of 1.40 eV.

(i) Explain what is meant by work function.

[1]

(ii) Determine the maximum speed with which a photoelectron leaves the surface of the plate. Explain your working.

maximum speed = m s-1 [3]

(iii) Calculate the minimum de Broglie wavelength associated with the photoelectrons that leaves the surface of the plate.

minimum wavelength = m [2]

22

NYJC 2020 9749/03/J2Prelim/20

(e) Photoelectrons with a de Broglie wavelength of 17.5 pm is allowed to strike a carbon film as shown in Fig. 9.3. Concentric circles are formed on a screen which is placed 20 cm away. The screen is parallel to the plane of the carbon film.

Fig. 9.3

(i) Explain the part played by diffraction in the production of the concentric circles.

[3]

Fig. 9.4 shows a scaled diagram of the concentric rings formed on the screen.

Fig. 9.4

20 cm

23


(ii) Show that the spacing between the carbon atoms is 140 pm. Explain your working clearly.

[3]

(iii) Metal plate X is replaced with another of lower work function

1. Sketch, on Fig. 9.4, the ring representing the first order maxima. [1]

2. Explain your answer to (iii)1.

[2]

[Total: 20]

End of Paper



CANDIDATE NAME

SOLUTIONS

CLASS

TUTOR’SNAME

CENTRE NUMBER S

INDEX NUMBER

PHYSICS 9749/01

Paper 1 Multiple Choice 17 September 2020

1 hour

Additional Materials: Multiple Choice Answer Sheet


Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. Write your name, class, Centre number and index number in the spaces at the top of this page. There are thirty questions on 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 Answer Sheet. Read the instructions on the Answer Sheet 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. The use of an approved scientific calculator is expected, where appropriate.


2

NYJC 2020 9749/01/J2Prelim/20

Data




(1 / (36)) × 10−9 F m−1












2s ut at




3

Nmp c

V


2E kT



2 20x x





QV

r



0

2IBd


0

2NIBr



decay constant 12

ln2

t

3


1 What is the number of SI base units required to express electric field strength and magnetic flux density?

electric field strength magnetic flux density

A 3 3

B 4 3

C 3 4

D 4 4

Ans: B

3 1

qE ma

maE

It

E kg m s A

2 1

qvB ma

maB

Itv

B kg s A

2 The viscosity μ of a fluid can be determined by measuring the terminal velocity vt of a sphere when it descends in the fluid. The fluid has a density ρf while the sphere has a density ρs and a diameter d. The viscosity can then be calculated using the equation

25

9s f

t

dv

The quantities measured are

vt = (1.60 ± 0.04) m s

ρs = (2700 ± 20) kg m

ρf = (900 ± 10) kg m

d = (20.0 ± 0.4) mm

What is the percentage uncertainty in the value of ?

A 6.2 % B 7.1 % C 8.2 % D 30 %

Ans: C

30 0.04 0.8

1800 1.60 20.0

100% 8.2%

4

NYJC 2020 9749/01/J2Prelim/20

3 A driver decelerates uniformly as he approaches a junction, makes a U-turn, and accelerates at the same rate that he decelerated earlier. Which of the following velocity-time and acceleration time graphs best represents the motion?

A

B

C

D

Ans: C

The driver decelerated in one direction and accelerated in the opposite direction at the same rate. Therefore the acceleration vector should be in the same direction before and after he made the U-turn.

v / m s–1

t / s

a / m s–2

t / s

v / m s–1

t / s

a / m s–2

t / s

v / m s–1

t / s

a / m s–2

t / s

v / m s–1

t / s

a / m s–2

t / s

5


4 The diagram shows a laboratory experiment in which a feather falls from rest in a long evacuated vertical tube of length L.

The feather takes time T to fall from the top to the bottom of the tube.

How long will the feather take to fall 0.50 L from the top of the tube?

A 0.25 T B 0.29 T C 0.71 T D 0.75 T

Ans: C

2

21

2

1

0.5

For 1st half,

0.5 0.5 --- (1)

For entire journey,

0.5 --- (2)

Solve (1) and (2)

0.71

s ut at

L at

L aT

t T

5 The diagram shows two blocks of mass m and 2m connected by a light inextensible cord passing

over a light, free-running pulley. At what angle θ must the smooth slope be inclined such that the block with mass 2m accelerates down the slope at 1.7 m s-2?

A 14° B 41° C 49° D 76°

m

2m

θ

1.7 m s-2

6

NYJC 2020 9749/01/J2Prelim/20

Ans: C Applying N2L to the 2-masses system, 2mg sin θ – mg = 3m(1.7) 2 sin θ – 1 = 3(1.7)/9.81 θ = 49° 6 The diagram shows two spherical masses approaching each other head-on at the same

speed u. One has mass 2m and the other has mass m. The masses then collides elastically.

Which diagram shows the speeds of the masses after the collision? A B

C D

Ans: A

Since collision is elastic, OR By conservation of momentum, u – (-u) = v2 – v1 2mu – mu = 2mv1 + mv2 2u = v2 – v1 (eliminates B, C & D) u = 2v1 + v2 (eliminates B, C & D)

7 A beam, the weight of which may be neglected, is supported by three identical springs. When a weight W is hung from the middle of the beam, the extension of each spring is x.

The middle spring and the weight is removed.

What is the extension when a weight of 2W is hung from the middle of the beam?

6

u 7

6

u

5

3

u

7


A 4

3

x B

3

2

x C 2x D 3x

Ans: D When W is doubled, x for each spring doubles, hence 2x. But when one spring is removed, the new extension is thus (2x) × (3/2) = 3x.

8 A gas at a pressure of 5.0 105 Pa is enclosed in a cylinder fitted with a piston.

The gas expands by 4.0 m3 against a constant exernal pressure of 1.0 105 Pa.

How much work does the gas do against the external pressure?

A 4.0 105 J B 12 105 J C 16 105 J D 20 105 J Ans: A The gas has to do work against the external pressure. Therefore, the value of the pressure used is the value of the external pressure, not the gas’s own pressure.

W = pV = 1.0 105 (4.0) = 4.0 105 J

9 A ball rolls down the circularly curved track shown below.

Which of the following correctly describes its speed, acceleration along the slope and kinetic energy?

speed acceleration kinetic energy

A increases increases increases

B decreases decreases decreases

C increases decreases increases

D increases constant increases

Ans: C

By conservation of energy, as GPE is lost, KE and hence speed is gained by the ball.

8

NYJC 2020 9749/01/J2Prelim/20

Weight of the ball is constant, but the component of weight along the slope decreases to zero. Hence acceleration along the slope decreases.

10 A track cycling competiton consists of a 4.0 km race and a one hour record race. Cyclists move around a circular track with a circumference of 1.00 km.

In the 4.0 km race, the top cyclist finishes the race in 4 mins and 1.934 s. In a one hour record race, the top cyclist travelled a distance of 55.089 km in one hour.

What is the difference between the angular speed of these top cyclists of each race?

A 0.0012 rad s-1 B 0.0077 rad s-1 C 0.014 rad s-1 D 0.92 rad s-1

Ans: B

First cyclist angular speed = (4 x 2π)/(241.934) = 0.10388 rad s-1

Second cyclist angular speed = (55.089 x 2π)/(60x60) = 0.09615 rad s-1

Difference = 0.0077 rad s-1

11 The section of a cycling track where sharper turns are made has a horizontal radius of 20.0 m, and is banked at 40o to the horizontal.

Given that the maximum speed of the cyclist of mass 80 kg is 60 km h-1, what is the difference in frictional force that the bicycle tyres must provide compared to an unbanked track?

A 140 N B 450 N C 760 N D 1200 N

Ans: C For a non-banked track, centripetal force is provided by the frictional force f1 = mv2 / r = (80)(60 x 1000/3600)2 / 20.0 = 1110 N For a 40o banked track, Ncos40o = mg + f2 sin40o Nsin40o + f2 cos40o = mv2 /r f2 = 347 N Difference = 1110 – 383 = 763 N = 760 N.

12 A planet has an average density 4950 kg m-3. Given that the acceleration of free fall near the surface of the planet is 0.9 g, where g is the acceleration of free fall near Earth’s surface, the radius of the planet is approximately

A 260 km B 2500 km C 2800 km D 6400 km

Ans: D

angle of bank is 40o at sharp turning

20.0 m

9


32 2

11

4 3

3 4

3 0.9 9.816400 km

4 6.67 10 4950

GM G gg g R R

R R GR

R

13 Two satellites, A and B, orbiting around Earth have the same kinetic energy. Satellite A has a larger mass than satellite B. Which of the following statements is true?

A Satellite A has a larger period.

B Satellite A has a smaller total energy.

C Satellite A has a smaller orbital radius.

D Satellite A has a larger angular velocity.

Ans: A

Since 21

2KE mv and MA > MB, then vA < vB.

Gravitational force on satellite provides centripetal force for satellite to orbit

2

2

2

GMm mv

r rGM

vr

Since vA < vB, then rA > rB. Hence C is false.

Since v

r , vA < vB and rA > rB, then A < B . Hence D is false.

Since 2

T

and A < B , TA > TB. Hence A is true.

From 2 GMv

R , 21 1 1

2 2 2 2

GMm GMmKE mv GPE

R R

.

Since KEA = KEB, then GPEA = GPEB and TEA = TEB. Hence B is false. 14 Fig. 14.1 shows the variation of velocity v with displacement x of a simple harmonic oscillator.

The state of motion of point P is shown in Fig. 14.1.

Fig. 14.1

Fig. 14.2 shows the variation with time t of the acceleration a of the oscillator.

10

NYJC 2020 9749/01/J2Prelim/20

Fig. 14.2

Which of the points on Fig. 14.2 correspond to the state of motion represented by point P? Ans: D

At point P, the displacement is positive and velocity is negative.

From the definition of SHM, acceleration and displacement are always oppositely directed. Thus, a should be negative. A and B not possible.

For v, sketch x-t graph and determine whether point C or D has negative velocity (i.e. where gradient of x-t graph is negative). (Using a=-2x, since a-t graph is a positive sine graph, x-t graph is a negative sine graph.)

15 Five particles have speeds of 2u, 5u, 10u, 11u and 15u. Which of the following statements is correct?

A The root mean square speed is equal to the mean speed.

B The root mean square speed exceeds the mean speed by more than 1u.

C The root mean square speed exceeds the mean speed by less than 1u.

D The mean speed exceeds the root mean square speed by more than 1u.

Ans: B r.m.s. speed = √((2u)2 + (5u)2 +(10u)2 + (11u)2 + (15u)2 / 5 ) = 9.75 u mean speed = (2u + 5u +10u + 11u + 15u) / 5 = 8.6 u

16 Latent heat of vaporisation is the energy required to

A separate the molecules of the liquid

B separate the molecules and to push back the atmosphere

C increase the average molecular speed in the liquid state compared to that in gaseous state

D separate the molecules and to increase the average molecular speed in the liquid state compared to that in gaseous state.

Ans: B As the potential energy of the liquid is increased for substance to change from liquid to gases, the volume has to increase and this energy will also include work done to push back the atmosphere.

11


17 The diagram below shows a sound wave traveling from left to the right in air at a particular instant. Air particles P and Q are the centres of a rarefaction and a compression respectively. Which of the following describes the directions of motion of P and Q at this instant?

Particle P Particle Q

A to the right to the left

B to the right at rest

C to the left to the right

D to the left at rest

Ans: C Take direction of displacement to the right to be positive. Since P and Q are centres of rarefaction and compression respectively, they both have zero

displacement at this instant. 18 A vertical tube is completely filled with water. A small sound source of constant frequency is

held a little above the open upper end and water is run out from the lower end.

A number of resonance positions are detected. The second position was detected when the water surface is 12 cm below the top of the tube and another occurs at 36 cm.

At which of the following distances should resonance also be detected?

A 24 cm B 26 cm C 44 cm D 48 cm

Ans: C The wavelength can be deduced to be 16 cm as 12 cm below the water surface is the 2nd

occurrence of resonance. 36 cm below water surface is the 5th occurrence. Other possible resonance positions include 4 cm, 20 cm, 28 cm, 44 cm, 52 cm below the water surface.

Q

Direction of wave propagation

P

P Q

Current instant

Next instant

12

NYJC 2020 9749/01/J2Prelim/20

19 A diffraction grating with 6.0 × 105 lines per metre is placed at right angles to a ray of white light. The first and second order spectra produced are shown in the diagram.

The angle between the red and violet ends of the spectrum is for the first order spectrum and for the second order spectrum. How do and compare?

A < ½ B = ½ C = D > Ans: A Wavelengths of white light range from 700 nm (red) to 400 nm (violet). d sin θ = nλ α = sin-1 (λred × 6.0 × 105) – sin-1 (λviolet × 6.0 × 105) = 11° β = sin-1 (2λred × 6.0 × 105) – sin-1 (2λviolet × 6.0 × 105) = 28° 20 In which of the following cases does an electric field do positive work on a charged particle?

A A positive charge moves to a point of higher electric potential.

B A negative charge moves opposite to the direction of the electric field.

C A positive charge completes one circular path around a stationary positive charge.

D A negative charge moves perpendicular to the electric field between two parallel and oppositely charged plates.

Ans: B Option B is the only one where the E-field does positive work. In option A and D, it is the external

agent that does positive work. For option C, since force and displacement is perpendicular, no work is done.

grating

white light

red

red

violet

violet

First order spectrum

Second order spectrum

13


21 A beam of protons passes through a velocity selector, set up by an electric field and a magnetic field that are perpendicular to each other. The electric field is set up by a pair of charged plates with a plate separation of 2.0 cm as shown in the diagram below. The magnetic flux density, B is 1.5 T and directed into the plane of the paper.

If protons travelling at 2.0 × 107 m s-1 pass through undeflected, what would be the direction and magnitude of the electric field?

direction magnitude

A downwards 6.0 × 105 N C-1

B upwards 6.0 × 105 N C-1

C downwards 3.0 × 107 N C-1

D upwards 3.0 × 107 N C-1

Ans: C

As proton is a positive charge, by Fleming’s Left Hand rule, it will experience an upward force. To remain undeflected, the proton must also experience a downward electric force. Hence the electric field must be directed downward.

7 7 1

Since

1 5 2 0 10 3 0 10 N C

E BF F

qE Bqv

E Bv . ( . ) .

22 A battery supplies a current of 0.025 A for 80 s. During this time it produces 18.0 J of electrical

energy while resistor M receives 11.0 J and resistor N receives 4.0 J.

What is the e.m.f. of the battery and its internal resistance L?

2.0 cm

beam of protons

v = 2.0 × 107 m s-1

B = 1.5 T

14

NYJC 2020 9749/01/J2Prelim/20

e.m.f. of battery / V internal resistance L of battery /

A 1.5 60

B 9.0 60

C 9.0 360

D 16.5 360

2

2

Ans: B

18.0 (0.025)(80)

9.0 V

18.0 11.0 4.0 0.025 (80)

60

batt

r

E VIt

E I rt

r

r

23 Five identical resistors are connected as shown in the circuit. If the power dissipated by X is 2.0 W, what is the total power supplied by the battery?

A 2.5 W B 3.5 W C 7.5 W D 10 W

Ans: A The four resistors that are not labelled are connected in parallel, thus their effective resistance is one quarter that of resistor X, which can be considered to be in series with resistor X. Since power is proportional to resistance for the same current, the power dissipated by the four resistors is going to be ¼ that of X. Thus power is 2.5 W.

24 The diagram shows four long straight current-carrying conductors placed at the corners of a square at W, X, Y and Z. The current in each conductor has the same magnitude.

X

15


Point O is the intersection of the diagonals of the square. If the magnetic flux density at point O is zero, which of the following is true?

A The current in Y must be in the opposite direction as that in W and the current in X must be in the opposite direction as that in Z.

B The current in Y must be in the same direction as that in W and the current in X must be in the opposite direction as that in Z.

C The current in Y must be in the same direction as that in X and the current in W must be in the opposite direction as that in Z.

D The current in Y must be in the same direction as that in X and the current in W must be in the same direction as that in Z.

Ans: D

If IY and IX are in the same direction, they will create BY and BX to produce a net B either upwards or downwards. Therefore, IW and IZ must also in the same direction as IY and IX to produce no net B at point O. i.e. current in all four wires must be in the same direction.

25 In an experiment to record electrical events of short duration, a student drops a bar magnet through a very thin, horizontal coil, as shown.

Which graph best represents how the induced e.m.f. E in the coil varies with time t?

A B C D

Ans: D

thin

falling bar magnet

16

NYJC 2020 9749/01/J2Prelim/20

When the north pole of the magnet approaches the coil, there is an increase in flux linkage. Hence an emf and current is induced in a direction to produce a B field pointing upward to oppose the increase I flux linkage. When the magnet leaves the coil, an emf and current is now induced in a direction to produce a B field pointing upward oppose the decrease in flux linkage. Answer has to be D because the induced emf, which is the rate of change of flux linkage should start from zero.

26 A uniform magnetic field directed at 50 from the vertical passes through a circular metal ring of diameter 0.50 m and resistance 3.0 .

The magnetic flux density through the ring decreases by 4.0 × 10–5 T at a constant rate in 2.0 s. During this change, the current induced in the ring

A remains constant at 1.0 µA.

B remains constant at 1.3 µA.

C decreases from zero to 1.0 µA.

D decreases from zero to 1.3 µA.

Ans: A

5 2

6

0.50(4.0 10 )sin50( ( ) )1 12 1.0 10 A

2.0 3.0

BAI

R t R

The current remains constant as the rate of change of flux is constant.

27 When a light bulb is connected across an a.c. source of peak voltage 170 V, the power dissipated is 40 W. Two such light bulbs are now connected in series to the electrical mains of 220 V r.m.s. What will be the total power dissipated?

A 33.5 W B 67.0 W C 80.0 W D 134.0 W

Ans: B

.

2

2

2

170240

361 25

orms

rms

VV

VP

R

RR

metal ring

magnetic field

50

17


, ..

22 22067 0

361 25 2Total rms

Total

VP W

R

28 Light of wavelength is incident on a metal surface in a vacuum. Photoelectrons are emitted from the surface of the metal.

Which of the following best shows the variation with of the maximum kinetic energy EK of the

emitted electrons?

A

B

C

D

Ans: C

K

K

hcE

hcE

Plotting EK against gives a graph of y = 1/x (both axes as asymptotes) which is then shifted downwards to touch the horizontal axis.

29 An atom makes a transition from an energy level of energy E1 to a level of energy E2, emitting a photon of wavelength .

Which expression gives this wavelength in terms of Planck constant h and the speed of light c?

A E E

hc hc1 2 B

hc hc

E E

1 2

C hc hc

E E

2 1

D hc

E -E1 2

Ans: D

EK

EK

EK

EK

18

NYJC 2020 9749/01/J2Prelim/20

1 2

1 2

hcE E

hc

E E

30 A proton and an alpha particle ( 42He Helium nucleus) have the same de Broglie wavelength.

The ratio speed of proton

speed of alpha particle is

A 1

4 B

1

2 C 2 D 4

Ans: D

By De Broglie’s relation,

( )

( )

1

4 1

4

proton

proton

proton proton

proton

proton

proton

p

p

m v

m v

u v

u v

v

v

End of Paper



CANDIDATE NAME

SOLUTIONS

CLASS

TUTOR’SNAME

CENTRE NUMBER

S INDEX NUMBER

PHYSICS 9749/02Paper 2 Structured Questions 2 September 2020




Write your name, class, Centre number and index number in the spaces at the top of this page. Write in dark blue or black pen on both sides of the paper. You may use a HB pencil for any diagrams, graphs. Do not use staples, paper clips, glue or correction fluid. The use of an approved scientific calculator is expected, where appropriate. Answer all questions. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.


1 / 8

2 / 10

3 / 7

4 / 9

5 / 11

6 / 8

7 / 8

8 / 19

Total / 80


2

NYJC 2020 9749/02/J2Prelim/20

Data




(1 / (36)) × 10−9 F m−1












2s u t a t


temperature /K / C 273.15T T


3

Nmp c

V


2E kT



2 20x x

electric current I Anvq resistors in series

1 2 . . .R R R

resistors in parallel 1 21 / 1 / 1 / . . .R R R


QV

r



0

2IBd


0

2NIBr



decay constant 12

ln2

t

3


1 During a wet weather, a stationary rain droplet falls from great height and eventually reaches a terminal velocity of 9.00 m s-1.

(a) Explain how the droplet achieved terminal velocity.

[3] (b) The droplet eventually hits the side window of a moving train. The relative velocity, v of the

droplet makes an angle of θ with the vertical, as shown in Fig. 1.1.

Fig. 1.1

Sketch a graph of the train speed against θ using the grid in Fig. 1.2.

motion of the train θ

v

As the droplet initially falls from rest, its velocity increases/accelerates downwards due to net downward force of weight.

Drag force due to air resistance acts upwards and it increases with increasing velocity.

So the net downward force (and hence by N2L, its acceleration) decreases until zero.

The droplet will then reach a constant velocity.

4

NYJC 2020 9749/02/J2Prelim/20

Fig. 1.2

[3]

(c) The train is travelling at a speed of 125 km h-1. On reaching the station, the brake is applied such that the train decelerates uniformly. During the deceleration, the train travels another 850 m before coming to a rest.

Determine the time taken for the train to stop.

time taken = s [2]

[Total: 8]

0.5( )

2 / ( )

2(850) / (125 / 3.6 0)

49 s

s u v t

t s u v

θ / °

train speed / m s-1

0 20 40 60 80 100

40.0

30.0

20.0

10.0

vA v

- vB

Terminal velocity of droplet = vA

Train speed = vB

Relative velocity v = vA – vB

θ

tanθ = vB/vA = vB /9

train speed vB = 9 tanθ

5


2 A water jetpack consists of a backpack consisting of a motor, two thrusters and a pipe for constant supply of water. A diagram of one thruster is shown in Fig. 2.1.

Fig. 2.1

Water is ejected vertically downwards through the nozzle. The thruster moves vertically upwards. The density of water is 1.00 × 103 kg m-3. The nozzle has a circular cross-section of radius 10.0 cm. The thruster ejects 30.0 kg of water in 0.20 s. Assume that the water leaving the nozzle has the shape of a cylinder of radius 10.0 cm.

(a) (i) Show that the water has a constant speed of 4.8 m s-1 relative to the thruster. [2]

(ii) Show that the force exerted on water ejected by the jetpack is 1.4 × 103 N. [1]

water density 1.00 × 103 kg m-3

2

2

2

-1

30.01000 0.100 1

0.20

4.8 m s

m

V

m r L

mr v

t

v

v

30 4.81 700 N

0.20

pF

t

Force on water by jet pack = 2 x 700 = 1400 N

6

NYJC 2020 9749/02/J2Prelim/20

(iii) Determine the maximum combined mass of a jetpack with its user that can be lifted with an acceleration of 9.81 m s-2. Explain your answer.

mass = kg [3]

(b) When in use, the jetpack takes in a constant volume of water from the environment and has

a constant mass. To move horizontally, the user tilts the thrusters slightly using the handles connected. Fig. 2.2 shows one such jetpack in use above a lake.

Fig. 2.2

(i) Determine, for a user with a jetpack of combine mass 80 kg, the maximum angle with respect to the vertical which the thrusters can be tilted without the user losing altitude. Assume that the force exerted by the water pipe on the user is negligible.

maximum angle = ° [2]

By Newton’s 3rd law of motion, the force on jet pack with user by water is equal in magnitude and opposite in direction to the force on water by the jet pack with user. [1]

Applying Newton’s 2nd law of motion on the jet pack with user,

9.81 9.81 1400 1 71 kg 1

jet user jet userjet user

on jet user jet user jet user jet user

jet user jet user on jet user

jet user

F m a

F m g m a

m a g F

m m

0

cos 0

80 9.81cos 1

140056 1

yF

F mg

7


(ii) Calculate the power delivered to the water for a stationary user just above the water.

power delivered = W [2] [Total: 10]

3 (a) Two inclined planes RA and LA each have the same constant gradient. They meet at their

lower edges, as shown in Fig. 3.1.

Fig. 3.1

A small ball moves from rest down plane RA and then rises up plane LA. It then moves down plane LA and rises up plane RA to its original height. The motion repeats itself.

Explain why the motion of the ball is not simple harmonic.

[1]

(b) A small ball rests at point P on a curved track of radius r, as shown in Fig. 3.2.

Fig. 3.2

The ball is moved a small distance to one side and is then released. The horizontal displacement x of the ball is related to its acceleration a towards P by the expression

r

gxa

2130.0 4.8

2 2 [1]0.20

3400

gain in KE of water

W [1]

Pt

On either side of the plane, the (magnitude of) acceleration is constant. [M1] Hence

it is not simple harmonic since acceleration is directly proportional to displacement.

[A1]

8

NYJC 2020 9749/02/J2Prelim/20

where g is the acceleration of free fall.

(i) With reference to the expression provided, show that the ball undergoes simple harmonic motion.

[2]

(ii) The radius r of curvature of the track is 28 cm. Determine the time interval between the ball passing point P and then returning to

point P.

= s [2]

(c) The variation with time t of the displacement x of the ball in (b) is shown in Fig. 3.3.

Fig. 3.3

Some moisture now forms on the track, causing the ball to come to rest after approximately several oscillations.

On the axes of Fig. 3.3, sketch the variation with time t of the displacement x of the ball for

the first two periods after the moisture has formed. Assume the moisture forms at time t = 0. [2]

[Total: 7]

T 2T T/2 3T/2

g and r are constant so a is (directly) proportional to x [1]

negative sign shows a and x are in opposite directions [1]

2

2

2

2 9.81

0.28

1.06

gand

r T

T

T s

[1]

= T/2 = 0.53 s [1]

9


4 (a) Define electric field strength.

[1] (b) Two charged metal spheres A and B are situated in a vacuum, as illustrated in Fig. 4.1.

The shortest distance between the surfaces of the spheres is 12.0 cm.

A movable point P lies along the line joining the centres of the two spheres, a distance x from the surface of sphere A.

The variation with distance x of the electric field strength E at point P is shown in Fig. 4.2.

sphere A sphere B

12.0 cm

x

P

Fig. 4.1

Electric field strength at a point in the electric field is the force per unit positive

charge acting on a test charge at that point.

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NYJC 2020 9749/02/J2Prelim/20

(i) Using Fig. 4.2, comment on the magnitude and polarity of the charge of sphere A and sphere B.

[2] (ii) A proton is at point P where its acceleration is at a minimum. Use data from Fig. 4.2 to

determine the acceleration of the proton.

Fig. 4.2

Since the E-x graph is always positive, it shows that the field strength is acting in the

same direction (to the right), therefore sphere A (positive) and sphere B (negative) must

have different polarity. Sphere B has a larger magnitude of charge than sphere A.

From graph, minimum E is 0.088 kV.

By Newton’s 2nd Law of motion,

19 39 2

27

1.60 10 (0.088 10 )8.4 10 m s

1.67 10

F ma eE ma

eEa

m

11


acceleration = m s-2 [2]

(c) The variation with distance x of the electric potential V at point P is shown in Fig. 4.3.

(i) State, in words, the relationship between electric potential at point P and electric field strength at point P.

[1]

Fig. 4.3

Electric field strength at point P is the negative potential gradient at point P.

12

NYJC 2020 9749/02/J2Prelim/20

(ii) A proton is initially at rest on the surface of sphere A and moves along the line joining the centres of the two spheres to reach maximum speed at point P.

Use data from Fig. 4.3 to determine the maximum speed of the proton.


[Total: 9]

5 A cell P, a fixed resistor R and a uniform resistance wire AB are connected in a circuit as shown in Fig. 5.1.

Fig. 5.1 Cell P has e.m.f. 4.0 V and internal resistance 0.75 . Wire AB has length 1.50 m and resistance

5.5 . The voltmeter reads 1.3 V.

(a) Show that the potential difference across AB is 2.4 V. [2]

4.0 V 0.75

R

cell P

A B

V

5.5

1.50 m

From graph, largest ΔV is – 13.2 – 6.0 = – 19.2 V.

By Conservation of Energy, Gain in KE = Loss in Elec PE

2

194 1

27

10

2

2 2(1.60 10 )(19.2)6.07 10 m s

1.67 10

f

f

mv e V

e Vv

m

p.d. across R = 1.30 V

Using potential divider rule,

4.0 1.30.75 5.5

3.0

R

RR

Hence,

5.5p.d. across AB 4.0

0.75 3.0 5.52.4 V

13


(b) A cell Q and a sensitive ammeter are connected to the circuit in Fig. 5.1, as shown in Fig. 5.2.

Fig. 5.2 Cell Q has e.m.f. E and internal resistance 0.25 . The ammeter reads zero when the length

of AC is 0.56 m. (i) Determine E.

E = V [2]

(ii) There is a reading on the ammeter when the connection C is shifted closer to A. State and explain the direction of the current across cell Q.

[2]

(d) The resistance wire AB is detached from the circuit and coiled in a circular manner as shown

in Fig. 5.3. Metallic fasteners with negligible resistance are used to secure the wire at X and Y, where XY is the diameter of the coil.

C

4.0 V 0.75

R

cell P

A B

V

E 0.25

cell Q

sensitive ammeter

A B

15 cm 15 cm

X Y

1.50 0.562.4

1.501.5 V

E

The potential at C is higher than that at B. As C is shifted closer to A, the potential

at C increases, thus increasing the potential difference between BC. [1]

Since the potential between BC will become larger than the terminal p.d. of cell

Q, current will now flow through cell Q from C to B. [1]

14

NYJC 2020 9749/02/J2Prelim/20

Fig. 5.3 (i) Determine the resistance of wire AB when coiled in this manner.

resistance = [3] (ii) An e.m.f. source is again connected across AB. State and explain whether the drift

velocity of the electrons is greater in AX or in XY.

[2]

[Total: 11]

6 W is a long straight wire of length 1.2 m. A beam of electrons is projected with a speed of 2.1 ×107 m s1 parallel to W at a distance of 20 cm from it, as shown in Fig. 6.1.

Fig. 6.1

(a) When the switch is closed, the electron beam is deflected laterally by the current in W. Sketch the appearance of the deflected beam in Fig. 6.1. [1]

(b) The current in W is 1.8 A. Calculate the flux density due to the current at a distance of 20 cm

from W.

electron beam

wire W

Q

20 cm

P

1.2 m

15.5 / 150 0.0367 cm [1]

1 150 2 152 15 0.0367 0.0367 [1]

5 5

1.28

1.3 [1]

R

L

R

Since I = n A v q, drift velocity v is greater in AX than in XY

Method 1: as the current I is greater in AX than in either parallel section of XY.

Method 2: as the combined cross sectional area of XY is greater than AX.

B = I / 2 r

= 2107 1.8 / 0.20 [1]

= 1.8106 T [1]

15


flux density = T [2]

(c) Assuming that the lateral deflection of the electron beam is much less than 20 cm, show that the deflected path has a radius of 66 m.

[2]

(d) Calculate the lateral deflection of the beam at Q in mm.

deflection = mm [3] [Total: 8]

7 A 240 V a.c. mains is connected to a step-down transformer, a full wave rectifier and a motor as shown in Fig. 7.1. The electric motor requires an operating peak power of 6.6 W.

Fig. 7.1

The variation of potential difference across the 50 Ω resistor with time is shown in Fig. 7.2.

240 V mains

Step-down Transformer

IP

X

Y

Full Wave Rectifier

electric motor50

Ω

T 2T

-12.2

12.2


time / s

A B

C D

F = m a

B q v = m v2 / r [1]

r = m v / B q

= 9.11031 2.1107 / 1.8106 1.61019 [1]

Angle traversed = = L / r

= 1.2 / 66 [1] = 1.82102 rad

Deflection x = L tan(/2) = 1.2 0.91102 [1]

= 1.1102 m = 11 mm [1]

16

NYJC 2020 9749/02/J2Prelim/20

Fig. 7.2

(a) Determine the turns ratio of primary coil to secondary coil in the step-down transformer.

turns ratio = [2] (b) (i) For half a period T, the potential at X is higher than that at Y. On Fig. 7.1, draw an arrow to indicate the direction of current, if any, across the four

diodes A, B, C and D during this half of a period. [1] (ii) Sketch, on Fig. 7.3, the variation of potential difference across the electric motor with

time. Numerical calculation is not required. [1]

Fig. 7.3

(iii) Hence, sketch, on Fig. 7.4, the variation of output power of the step-down transformer with time when the electric motor is operating at its normal operating power. Include appropriate values in the output power axis.

Fig. 7.4

Total peak power output required from transformer

= Po across motor + Po across 50 Ω resistor

6.612.2

509.58 W


time / s T 2T

For sinusoidal input, peak 𝑉 240√2 339 V

∴.

27.8

output power / W

time / s T 2T 0

9.58

17


[2] (c) Given that the transformer is 80% efficient, determine the r.m.s. current flowing through the

primary coil.

current = A [2]

[Total: 8] 8 In a refrigeration system, a fluid known as a refrigerant undergoes alternating changes in its

physical state and volume to produce cooling. In this question, you will be guided to design a simple refrigeration system using the data provided.

Fig. 8.1 is a pressure vs volume graph showing the refrigerant cycling through four states P, Q, R and S. The four phases of the cyclic process are evaporation, condensation, compression and throttling (pressure releasing).

Fig. 8.1

(a) Using the information provided in Fig. 8.1, match each phase in the cycle with its description in the chart below. PQ has been matched.

Phase Description

PQ Evaporation

QR Condensation

RS Compression

SP Throttling (Pressure release)

[2]

Pre

ssur

e

Volume

P

Q

(liquid)

(vapour)

(liquid + vapour)

S

R

<P> = ½ Po = 4.7884 W

Since transformer is 80% efficient, 0.8 𝑃 𝑃

0.8 240 𝐼 , 4.7884

𝐼 0.0249 A

18

NYJC 2020 9749/02/J2Prelim/20

Fig. 8.2 is a table showing the thermal properties of the refrigerant as a saturated vapour and as a saturated liquid at various pressures.

Pressure P

(kPa)

Temperature T

(K)

Enthalpy h

(kJ kg1)

Entropy s

(kJ kg1 K1) Saturated

Liquid Saturated

Vapour Saturated

Liquid Saturated

Vapour 100 246.8 17.3 234.5 0.072 0.952 200 263.1 38.5 244.5 0.155 0.938 300 273.9 52.8 250.9 0.208 0.931 400 282.1 64.0 255.6 0.248 0.927 500 288.9 73.4 259.3 0.280 0.924 600 294.8 81.5 262.4 0.308 0.922 700 299.9 88.8 265.1 0.332 0.920 800 304.5 95.5 267.3 0.354 0.918 900 308.7 101.6 269.3 0.374 0.917

1000 312.6 107.4 271.0 0.392 0.916 1200 319.5 117.8 273.9 0.425 0.913 1400 325.6 127.3 276.2 0.453 0.911 1600 331.1 136.0 277.9 0.479 0.908 1800 336.1 144.1 279.2 0.503 0.905 2000 340.7 151.8 280.1 0.525 0.902

Fig. 8.2

A saturated vapour is one which a small compression will cause it to condense.

(b) Deduce what a saturated liquid is.

[1]

(c) By referring to Fig. 8.1, identify the state (P, Q, R or S) at which the refrigerant is a saturated vapour, and another at which it is a saturated liquid.

saturated vapour: , saturated liquid: [2]

Fig. 8.3 is an incomplete table showing the thermal properties of the refrigerant at states P, Q, R and S in the cyclic process.

State Pressure P

(kPa) Temperature T

(K)

Enthalpy h

(kJ kg1)

Entropy s

(kJ kg1 K1)

P 300 273.9 250.9 0.931

Q 1200 322.0 278 0.931

R 1200 319.5 117.8 0.425

S 300 273.9 117.8 0.443

Fig. 8.3

A saturated liquid is one which a small expansion will cause it to evaporate.

P R

19


(d) Using Fig. 8.2 and your answer to (c), fill in the missing values in Fig. 8.3 for state P. [2]

(e) At state Q, the refrigerant is a superheated vapour, and its thermal properties at various pressures are shown in the graphs in Fig. 8.4 (a) & (b).

Fig. 8.4 (a)

Fig. 8.4 (b)

Ent

halp

y /

kJ k

g1

Temperature (K)


Ent

ropy

/ k

J kg

1 K

1

Temperature (K)


20

NYJC 2020 9749/02/J2Prelim/20

The process PQ is isentropic (entropy s is constant) and the pressure of the refrigerant is increased to 1200 kPa.

1. Explain how Fig. 8.4 (a) and/or (b) verifies that the temperature of the refrigerant at state Q is 322.0 K.

[1]

2. Fill in the missing values in Fig. 8.3 for state Q. [1]

(f) The process QR is isobaric (pressure p is constant). Using Fig. 8.2 and your answer to (c), fill in the missing values in Fig. 8.3 for state R. [2]

(g) The process RS is isenthalpic (enthalpy h is constant). The process SP is isothermal (temperature T is constant) and isobaric. Fill in the missing values in Fig. 8.3 for state S.[1]

(h) Energy is inputted during the compression phase of the refrigeration process. The rate of work done by the compressor is calculated using

W = m ( hf hi )

where m is the mass flow rate of the refrigerant, and hf and hi are the values of enthalpy after and before the compression respectively.

If the compressor compresses 3.20 kg of refrigerant in 1 second, calculate W the rate of work done by the compressor.

W = kW [2]

Entropy at P = Entropy at Q = 0.931.

From Fig 8.4 (b), when entropy is 0.931, temperature is 322.0 K.

W = 3.20 (278 – 250.9) = 86.4 kW

21


(i) Cooling is achieved in the evaporation phase of the process. The rate of heat removal during this phase is calculated using

Q = m ( hf hi )

where m is the mass flow rate of the refrigerant, and hf and hi are the values of enthalpy after and before the evaporation respectively.

For the same compressor in (h), calculate Q the rate of heat removal by the refrigeration process.

Q = kW [2]

(j) The efficiency of a refrigeration system is measured using its coefficient of performance which is calculated using

C = Q / W

Calculate C the coefficient of performance for the refrigeration system you designed.

C = [1]

(k) A refrigeration system is regarded as efficient only if it has a coefficient of performance of above 4.

Explain with reference to the principle of conservation of energy, why, unlike the efficiency of a conventional mechanical system, a coefficient of performance of greater than 1 is possible.

[2]

[Total: 19]

End of Paper

W = 3.20 (250.9 – 117.8) = 426 kW

C = 426 / 86.4 = 4.93

In a conventional mechanical system, part or all of the input energy is converted to

output energy. In a refrigeration system, internal energy of the refrigerant increases

in both compression and evaporation phases, thus the input energy (in the

compression phase) is not 'converted' to the heat energy that is removed (in the

evaporation phase). In a complete cycle, the internal energy returns to its original

value because internal energy of the refrigerant decreases in the other two phases.

Section A

Answer all the questions in the spaces provided.

1 Ganymede is a moon of mass 1.50 x 1023 kg and radius 2.64 x 106 m orbiting the planet Jupiter of mass 1.90 x 1027 kg as shown in Fig. 4.1. The orbital period of Ganymede is 171.6 hours.

Fig. 4.1

(a) (i) Show that the distance x between the centres of Jupiter and Ganymede is 1.07 109 m. Explain your workings. Gravitational force by Jupiter on Ganymede provides the centripetal force on Ganymede to orbit Jupiter

2

2

23

2

211 27

32

9

2

4

6.67 10 1.90 10 171.6 3600

4

1.07 10 m

GMmmx

x T

GMTx

x

x

[2]

(a) (ii) S is a point between Jupiter and Ganymede where the resultant gravitational field strength is zero. Determine the distance from the centre of Jupiter to point S.

2 2

27

9 23

9

0

1.90 10

1.07 10 1.50 10

1.06 10 m

G Jg g

Gm GM

sx s

s

s

s

s = m [2]

Jupiter

Ganymede

rocketx

(a) (iii) On the given axes below, sketch a graph from the surface of Jupiter to the surface of Ganymede to show the variation of gravitational field strength g with distance r from the centre of Jupiter. [2]

(b) A rock is positioned on the surface of Ganymede at the point furthest from Jupiter, as

shown in Fig. 4.2. The rock is launched with an escape velocity v for it to reach infinity, far away from Jupiter and Ganymede.

Fig. 4.2

(i) Show that the gravitational potential at the position of the rock in Fig. 4.2

is 8 11.22 10 J kg .

27 23

119 6 6

8 1

1.90 10 1.50 106.67 10

1.07 10 2.64 10 2.64 10

1.22 10 J kg

G G

GM Gm

x r r

[2] (b) (ii) Calculate the escape velocity v.

2 8

4 -1

loss in KE = gain in GPE

10 0 1.22 10

2

1.56 10 m s

r rm v m

v

v = m s−1 [2]

g

r 0 surface of Jupiter

surface of Ganymede

x

Minus [1] for each missing point - gon Jupiter surface>gon Ganymede surface

- g = 0 lies closer to Ganymede - graph lies between surfaces of Jupiter and Ganymede

- smooth curve

Jupiter

Ganymede

rock x v

15h

mass hanger

beaker of

pulle

2 A metal block, of 15 cm height, was partially submerged in water of density 1.0 × 103 kg m-3. The block has a uniform density, ρ, and a uniform cross-sectional area, A. The block is held up by an inelastic cord tied to a mass hanger, through two frictionless pulleys, as shown in Fig. 2.1.

Different quantities of masses were added to the mass hanger and the corresponding submerged height, h, of the block was measured.

The variation of total mass m of mass hanger and masses added, with the submerged height, h, is shown in Fig. 2.2.

0.00

0.02

0.04

0.06

0.08

0.10

0.38 0.40 0.42 0.44 0.46 0.48 0.50

Fig. 2.2

m / kg

h / m

(a) Using Fig. 2.2, determine the total mass m required to remove the metal block from the water.

m = kg [1]

(b) By considering the forces acting on the metal block, show that the gradient of the

graph in Fig. 2.2 is represented by 3

1

1.0 10 A

.

[2]

(c) Hence, determine the cross-sectional area, A.

A = m2 [2]

From the graph, when h is 0 cm, m is 0.486 kg.

Considering the forces acting on the metal block,

Upthrust + Tension = Weight of block

(where is the mass of block)

(since by considering the forces acting on the mass hanger)

(since weight of water

U T Mg M

U mg Mg T mg

U mg Mg

Ahg mg Mg U

fluid displaces by Archimedes' Principle)

1 1

water

water water

Ah m M

h m MA A

Therefore, gradient of graph =1

water A

= 3

1

1.0 10 A

[shown]

From graph,

0.080 0.010Gradient 0.833

0.390 0.474

3

3

3 2

10.833

1.0 10

1

1.0 10 ( 0.833)

1.2 10 m

A

A

(d) Determine the density of the metal block, ρ.

ρ = kg m-3 [2]

3 (a) Explain the difference between accuracy and precision and state the type of error associated with each.

(i) Accuracy:

(i)

(i) Type of error:

(ii) Precision:

(i) Type of error: [2]

1 (b) Three digital clocks A, B and C are being tested in a laboratory. Using signals from the Global Positioning System, the displays on the clocks at the exact time of 12:00:00 on four successive days are shown in Fig. 1.1. The clocks are reset each day at 00:00:00.

Clock Day 1 Day 2 Day 3 Day 4

A 12:06:40 12:06:38 12:06:39 12:06:43

B 12:03:59 12:02:49 12:01:54 12:03:15

C 11:59:59 12:00:02 11:59:57 12:00:05

Fig. 1.1

State and explain,

(i) which clock is the most accurate, and

[1]

. (ii) which clock is the most precise.

[1]

Accuracy refers to how closely a measured value agrees with the true value. systematic Precision refers to how closely individual measurements agree with each other (or average value) without reference to any true value. random

Clock C is the most accurate. The average is closest to the true value.

Clock A is the most precise. The variance is the least.

From (a),

3 33

mass of block, 0.486

0.486

0.486 0.4862.7 10 kg m

(1.2 10 )(0.15)

M

Ah

Ah

(c) For clock B in (b), estimate the time displayed at noon on day 5 with its associated uncertainty.

Time = s [2]

4 (a) Explain what is meant by an ideal gas.

[2]

(b) A cylinder contains 3.16 mol of an ideal gas at a pressure of 4.81 105 Pa and a volume of 1.20 104 cm3. Heat is removed for a duration of 12 minutes at constant pressure such that its temperature decreased by 110 K.

(i) Calculate the initial temperature of the gas.

initial temperature = K [1]

(ii) Determine the new volume of the gas after heat is removed.

volume of the gas = m3 [2]

(iii) Determine the change in internal energy of the gas.

Mean: 12h 180 59 120 49 60 54 180 15 / 4 seconds 12h 179s

Time: 12:02:59 Range/2 = 63 70 s OR Mean – Min = 65 70 s Time: 12:02:59 70 s

An ideal gas is one that obeys the ideal equation PV = nRT for all values of pressure, P, volume, V and thermodynamic temperature, T for a fixed mass of gas where R is the molar gas constant, n is number of moles of gas which is a constant for a fixed mass of gas.

PV=nRT, T1 = P1V1/nR = (4.81 105)( 1.20 10-2)/(3.16)(8.31) = 220K (219.8 K)

For a fixed mass of gas at constant pressure,

𝑉 𝑉 1.20 10 = 6.00 10 m-3

Loss in internal energy, ∆U = 3/2 nR∆T ( or 3/2 P∆V, ∆V is –ve due to a decrease in volume. )

∆𝑈 3.16 8.31 110 4330 J

change in internal energy = J [1]

(iv) Determine the rate of heat loss.

rate of heat loss = W [3]

[Total: 9]

5 Fig. 5.1 and Fig. 5.2 show an amusement park ride that rotates passengers strapped onto a platform in a vertical circular motion about a pivot. The platform is 8.4 m away from the pivot and there is a counter-weight at the other end of the rotating arm, 3.0 m away from the pivot. The mass of the counter-weight and fully loaded platform are 200 kg and 2000 kg respectively.

The platform is initially brought to the highest position shown in Fig. 5.1 where it is turned upside down and then released from rest. The counter-weight and platform subsequently swings about the pivot until the counter-weight is at the highest position as shown in Fig. 5.2. The speed of the counter weight at that instant is uc and that of the platform is up.

(a) Show that the ratio of . .2 8p

c

u

u

Since A = B

Hence uA/rA = uB/rB

uA/uB = rA/rB = 8.4/3.0 = 2.8 (shown)

pivot

Fig. 5.1 Fig. 5.2

Counter-weight

pivot

uc

up

8.4 m

3.0 m

From first law of thermodynamics, ∆U = Q + WD

𝑊𝐷 𝑃∆𝑉 4.81 10 1.20 10 6.00 10 =2890 J (2886J)

Q = -4330-2890 = - 7220 J (- 7222 J)

Since the Heat supplied is -7220J, Heat lost = + 7220 J

Rate of heat loss = 7220 / (12 60) = 10 W (10.0 W)

[1] (b) Hence, show that the speed up of the platform when it is at the lowest position is

17.7 m s-1.

[2]

(c) Consider a man with a mass of 70 kg on the ride when the platform is at the lowest position. Given that the force strapping the man to the platform at this position is 100 N, calculate the reaction force which the platform exerts on this man. Explain your working.

reaction force = N [3]

(d) When the man is just about to reach the top of the path as shown in Fig. 5.3, his handphone slips from his hands. State and explain the subsequent path of his handphone until it hits the ground.

[Total: 8]

pivot

Fig. 5.3

Loss in Ep(platform) = gain in Ek(counter wt) + gain in Ek(platform) + gain in Ep(counter wt)

mA g (16.8) = ½ mB uB2 + ½ mA uA

2 + mB g (6.0)

½ mB (uA / 2.8)2 + ½ mA uA2 = mA g (16.8 ) – mB g (6.0)

100 (uA / 2.8)2 + 1000 uA2 = 2000 (9.81) (16.8) – 200 (9.81) (6.0)

uA = 17.7 ms-1

For the man to undergo circular motion, the net force acting on the man towards the

centre of the circular motion is upwards.

R - force by strap - mg = mv2/r

R = mv2/r + mg + force by strap

= (70)(17.72/8.4 + 9.81) + 100 = 3400 N

At the point it leaves the man, the handphone will move off with the same speed as the platform in the tangential direction from that point. However due to gravity, it will subsequently move in a projectile motion and reach the ground on the right side of the ride.

6 A rectangular coil PQRS of dimensions 14.0 cm by 12.0 cm, moves with a constant speed of 2.0 cm s-1 through a region of uniform magnetic field WXYZ of width 10.0 cm, as shown in Fig. 6.1. There is a magnetic flux density B of 1.5 T in WXYZ directed into the plane of the paper.

(a) State and explain the direction of induced current in the coil PQRS when the coil enters the field. [3]

Method 1: As the coil enters the field, there is an increase in magnetic flux linkage into the plane of the coil. By Faraday’s Law, an e.m.f. [1] and a current are induced in the coil. By Lenz’s Law, the induced current flows in an anti-clockwise direction [1] so as to create a magnetic field pointing out of the plane of the coil to oppose the increase in flux linkage. [1]

Method 2: As the coil moves to the right, electrons along QR moves to the right. This

means that a current moves to the left [1], and using Fleming’s Left Hand Rule, the electrons in QR experience a downward magnetic force [1], causing the electrons to travel in a clockwise direction in the coil. This means that an induced current is travelling in an anti-clockwise direction in the coil. [1]

(b) Calculate the maximum possible magnetic flux linkage during the motion.

max 1.5 0.120 0.100 0.018 WbBA

maximum flux linkage =…………………………… Wb [1]

R

Q

2.0 cm s-1

P

S

14.0 cm

12.0 cm

Fig. 6.1

W

X

Y Z

B = 1.5 T

10.0 cm

(c) The coil PQRS is moved from the position shown in Fig. 6.1 until side PS is aligned with WZ. Sketch and label with appropriate values on the axes, the following graphs:

(i) the variation with displacement of magnetic flux linkage through the coil. [2]

Straight line starts from origin to (10, 1.8) [1] Horizontal line from (10, 1.8) to (14, 1.8) [1] Minus one mark if no/wrong label

(ii) the variation with time of induced e.m.f. in the coil.

[2]

Horizontal line with e.m.f. = -0.0036 V, starting from 0 to 5 s [1] Horizontal line with e.m.f. = 0 V, starting from 5 to 7 s [1] Minus one mark if no/wrong label

displacement / cm

magnetic flux linkage / Wb

0 10 14

0.018

time / s

e.m.f. / V

0

5 7

-0.0036

7 A light elastic band of length 20 cm is stretched to a length of 28 cm, and then allowed to return back to its original length. The graph in Fig. 7.1 below shows how the length of the band varies with the force applied.

Fig. 7.1

(a) Explain how the graph shows that the band does not obey Hooke's Law.

[2]

(b) Estimate the work done by the applied force to stretch the band.

work done = J [2]

(c) The elastic band is used in a catapult to project a pellet into the air. The band is stretched to a length of 28 cm and released. Explain using Fig. 7.1, why the kinetic energy of the pellet at the point of release is significantly smaller than your answer in (b)

leng

th /

cm

force / N

stretching

releasing

The graph is not a straight line/does not have a constant gradient [1]

This shows that force applied is not proportional to extension [1]. However, this proportionality relation is required of the band if it were to obey Hooke’s law.

Work done = Area between stretching graph and Length-axis [1]

= 15 squares x (0.02 m x 20 N)

= 6.0 J (0.3 J) [1]

Area between length axis and release curve is smaller that between the length axis and stretch curve [1]

This implies that only some of the stored elastic potential energy in the stretched band has been converted to kinetic energy of the pellet [1]

[2]

(d) The initial kinetic energy of the projected pellet is 3.8 J. Calculate the efficiency of the band in accelerating the pellet.

efficiency = W [2]

(e) Suggest one reason for the loss in efficiency.

[1]

(f) The elastic band is replaced with another new elastic band of the same length which obeys Hooke's Law. It is also stretched to a length of 28 cm and transfers 3.8 J of kinetic energy to the pellet.

(i) Calculate the spring constant of this band.

spring constant = N m1 [2]

(ii) Sketch the graph for this new elastic band on Fig 7.1. [1]

(Side working to determine final coordinate point for graph plotting:

At length = 28 cm, i.e. extension = 8 cm,

E = ½ F x

3.8 = ½ F x 0.08 F = 95 N)

Graph plotted should be a straight line starting from (0, 20) and ends at (95, 28)

[Total: 12]

efficiency = kinetic energy of pellet / initial EPE stored in band

= kinetic energy of pellet / WD in stretching band

= 3.8 / 6.0 = 0.63

Heat is produced when band is released.

E = ½ k x2

3.8 = ½ k (0.08)2 [1]

k = 1.2103 N m1 [1]

8 The variation with time t of the displacement y of a wave A, as it passes a point P, is shown in Fig. 8.1.

The intensity of wave A is 2I.

(a) Use Fig. 8.1 to determine the frequency of wave A.

frequency = Hz [2]

(b) A second wave B with the same frequency and speed as wave A also passes point

P. Wave B has intensity I and is lagging wave A by 2

rad.

On Fig. 8.1, sketch the variation with time t of the displacement y of wave B. Label the graph B.

Show your working.

[3]

Examiner’s comment:

-3.0

-2.0

-1.0

0.0

1.0

2.0

3.0

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5

t / 10-15 s

y / cm

Fig. 8.1

From graph, T = 2.00 × 10-15

f = 1/T = 1 / 2.00 × 10-15 = 5.00 × 1014 Hz

If intensity of 2I has amplitude of A, then intensity of I will have amplitude of (1/√2) A.

Amplitude of wave B = (1/√2) (3.0) = 2.1 cm.

sketch a graph with shape of – sin ωt with amplitude of 2.1 cm.

(c) Explain why wave A and wave B are coherent.

[1]

(d) Wave R is the resultant wave due to the superposition of wave A and wave B. An amplitude of wave R occurs at t = 1.2 × 10-15 s.

Using your answer in (b) and Fig. 8.1, determine, in terms of I, the intensity of wave R at point P.

intensity = [2]

(e) After passing through point P, wave R is plane-polarised by passing it through a polarising filter. Determine, in terms of I, the intensity of the plane-polarised wave R.

intensity = [1]

(f) Wave R is made to pass through a double slit in an interference experiment, as shown in Fig. 8.2.

The separation between the slits is a. The fringes are viewed on a screen at a distance 2.00 m from the double slit. The fringe separation x is measured for different

slit separation a. A graph of a against 1

xis shown in Fig. 8.3.

Fig. 8.2

2.00 m

a

screen double slit

wave R

They have a constant phase difference of π/2 rad between them.

From graph, at t = 1.2 × 10-15 s,

Amplitude of wave R = displacement of wave A + displacement of wave B = 2.4 + 1.3 = 3.7 cm

Intensity of wave R = (3.7/3.0)2 times intensity of wave A = (3.7/3.0)2 2I = 3.0 I

Intensity of plane-polarised wave R = ½ (Intensity of unpolarised wave R) = ½ (3.0 I) = 1.5 I

(i) Determine the wavelength of wave R.

wavelength = m [3]

(ii) State the effect, if any, on the appearance of the fringes observed on the screen when the following changes are made separately, at a fixed value of a:

1. wave R is replaced by a blue laser light,

[1]

2. the width of each slit is increased but the separation remains constant.

0.00

0.10

0.20

0.30

0.40

0.50

0.60

1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

a / mm

1

x / cm-1 Fig. 8.3

From graph, 3

62

(0.50 0.14) 10Gradient 1.16 10

(4.30 1.20) 10

x = λD / a → a = λD (1/x)

Therefore, λD = Gradient

λ = Gradient / D = 1.16 × 10-6 / 2.00 = 5.8 × 10-7 m (accept 580 – 600 nm)

Wavelength of blue light is shorter, so fringe separation will appear closer together.

More light goes through, so intensity of bright fringe will increase, increasing contrast between bright and dark fringes.

Also, diffraction will be less significant / less diffraction effect, (resulting in less overlapping for interference), hence fewer fringes will be observed on the screen.

(iii) At a particular value of slit separation a, the variation with distance along the screen of the intensity of the image on the screen is shown in Fig. 8.4.

1. Using Fig. 8.4 and Fig. 8.3, determine the value of a.

a = mm [2]

2. Using Fig. 8.4, determine the width of each slit.

0.0

0.2

0.4

0.6

0.8

1.0

-0.015 -0.010 -0.005 0.000 0.005 0.010 0.015

Fig. 8.4distance / m

intensity

From Fig. 8.4, x = [0.0073 – (– 0.0073) / 6 = 2.43 × 10-3 m

From Fig. 8.3, for 1/x = 1 / 2.43 × 10-3 = 412 m-1 = 4.12 cm-1

Reading off Fig. 8.3, a = 0.48 mm (accept 0.47 – 0.49 mm)

From Fig. 8.4,

w, width of central maximum = [0.0095 – (– 0.0095) = 0.019 m (accept 0.0085 to 0.0100)

tan θ = ½ w / D

Also, due to single slit diffraction, sin θ = λ / b

By small angle approximation, sin θ ≈ tan θ

½ w / D = λ / b

b = λD / ½ w = (5.80 × 10-7)(2.00) / ½ (0.019) = 1.2 × 10-4 m (accept 1.2 -1.4 × 10-4 m)

9 Two metal plates X and Y are contained in an evacuated container and are connected as shown in Fig. 9.1. Monochromatic electromagnetic radiation of blue light is incident on metal plate X.

The potential of X with respect to Y is varied by adjusting the position of the sliding contact from E to G. The variation of current I recorded at the ammeter with potential V of X is given by Fig. 9.2 below.

V

A

X Y

O F

E G

incident monochromatic radiation

Fig. 9.1

I / mA

V / V

Ans

Fig. 9.2

(a) Explain an evidence provided by the photoelectric effect experiment for the failure of the wave theory of light.

[2]

(b) Circle the portion of the graph on Fig. 9.2 that represents the variation of current as the sliding contact moves from F to G. [1]

(c) Sketch, on Fig. 9.2, the graph when electromagnetic radiation of the same

intensity in the ultraviolet region is used instead. [2]

(d) It is observed that photoelectrons are emitted from plate X when photons of wavelength 520 nm is illuminated on it. The metal plate has a work function of 1.40 eV.

(i) Explain what is meant by work function.[1]

(ii) Determine the maximum speed with which a photoelectron leaves the surface of the plate. Explain your working.


(ii) Calculate the minimum de Broglie wavelength associated with the photoelectrons that leaves the surface of the plate.

minimum wavelength = m [2]

1. Wave theory predicts that the photoelectric effect should occur for any frequency of the monochromatic incident light. If frequency of light is too low, its intensity can be increased to cause photoelectron emission. However, no electrons were emitted at all unless the frequency of the monochromatic incident light was greater than a minimum value (i.e. threshold frequency).

2. Wave theory predicts that at very low intensities, there will be a time delay in Emission of Photoelectrons as the electrons would need time to accumulate sufficient energy in order to escape from the metal surface.

3. Wave theory predicts that if the incident light intensity is increased, a greater amount of energy will be incident on the surface and hence electrons will escape with more energy from the surface of the metal. However, maximum kinetic energy of the photoelectrons and hence stopping potential is independent of the incident light intensity.

The work function Φ of the metal is the minimum amount of energy required to remove an electron from the surface of the metal.

By Einstein’s Equation for Photoelectric Effect or Conservation of Energy,

Maximum KE = 1.585 × 10-19 J

Maximum speed = 5.90 × 105 m s-1

Using de Broglie relation,

Minimum wavelength = 1.23 × 10-9 m

(e) Photoelectrons with a de Broglie wavelength of 17.5 pm is allowed to strike a carbon film as shown in Fig 9.3. Concentric circles are formed on a screen which is placed 20 cm away. The screen is parallel to the plane of the carbon film.

Fig. 9.3

(i) Explain the part played by diffraction in the production of concentric circles. [3]

Fig. 9.4 shows a scaled diagram of the concentric rings formed on the screen.

Fig. 9.4

Electrons exhibit wave properties (Wave-particle duality) and spread/bend/ not diffract when they pass through the slits formed by the carbon film. The (electron) waves interfere constructively to form bright concentric circles on the screen. [3]

20 cm

(ii) Show that the spacing between the carbon atoms is 140 pm. Explain your working clearly.

[3]

(iii) Metal plate X is replaced with another of lower work function

1. Sketch, on Fig. 9.4, the ring representing the first order maxima. [1]

2. Explain your answer to (iii)1.

[2]

Using tan = r / D,

When n = 1, diameter = 5.0 cm, r = 2.5 cm

tan = 2.5 / 20, = 7.13

dsin = n, dsin7.13 = (1) 17.5 × 10-12

d = 140 × 10-12 m or 140 pm

OR

When n = 2, diameter = 10.3 cm, r = 5.15 cm

tan = 5.15 / 20, = 14.4

dsin = n, dsin14.4 = (2) 17.5 × 10-12

d = 140 × 10-12 m or 140 pm

Work function decreases, Maximum KE of e increases

Momentum of e increases [1], de Broglie wavelength of e decreases

Using dsin = n, decreases [1]

Radius of ring is smaller

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