1 NAME: ___________________________________________ CLASS: ___________ INDEX: __________

CATHOLIC JUNIOR COLLEGE JC2 Preliminary EXAMINATIONS Higher 2

PHYSICS 9646/01 Paper 1 02 Sep 2014 1 hour 15 minutes Additional Materials: Multiple Choice Answer Sheet 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

D B B D C C B A A B C A A D D C B B B A

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

C C D A B C D B C C A A A B D D D C C B

1 The Helmholtz Energy of a system, H, is given by the equation,

H = U – TS where U is the internal energy of the system, T is the thermodynamic temperature and S is the entropy of the system. What are the SI base units of entropy, S?

A K-1 B J K-1 C kg m s-2 K-1 D kg m2 s-2 K-1

Answer: D

2 An electrician needs to know the effective resistance of a circuit consisting of two resistors connected in parallel. Using an ohmmeter, he finds that the resistances of the two resistors are (3.00 ± 0.05) Ω and (0.60 ± 0.05) Ω.

What is the effective resistance of the circuit, expressed with its appropriate uncertainty?

A (0.50 ± 0.02) Ω B (0.50 ± 0.04) Ω

C (0.50 ± 0.10) Ω D (3.60 ± 0.10) Ω

1-2-2

1-

K s m kg

K J ][

]K[ J

]][[ ][

][ ][ ][

=

=

=

=

==

S

S

STU

TSUH

2

Answer: B Effective resistance: Using the extreme value method: Uncertainty of Reff

3 The graph below describes the motion of an object rebounding from a horizontal surface after being released from a point above the surface.

The quantity represented on the y-axis is the ball’s

A acceleration B displacement C momentum D velocity

Answer: B The point of release is taken to be y = 0 and the displacement from the horizontal surface from which the ball rebounds from is represented by the dotted line on the graph. The displacement of the ball is observed to decrease with time for subsequent bounces due to loss of energy.

y

time 0

Ω=

+=+=

0.50

600

1

003

1111

21

eff

eff

R

RRR ..

Ω=

+=+=

0.54

650

1

053

1111

:maximum

21

eff

eff

R

RRR ..

Ω=

+=+=

0.46

550

1

952

1111

:minimum

21

eff

eff

R

RRR ..

Ω=

−=∆

0.04

2

460540 ..effR

3

4 A metal sphere is held just below the surface of a deep tank of liquid and released. Which of the following best illustrates how the acceleration a varies with time t after release?

A

B

C

D

Answer: D

Assuming that drag is proportional to v Net force acting on sphere is Fnet = mg – kv ma = mg –kv –kv = ma – mg Differentiating the above equation with respect to time t,

-kdt

dv= m

dt

da [

dt

dv= a]

-ka = mdt

da

dt

da= -

m

ka

At time t = 0, sphere is just released from rest and v = 0, a = g. a = 0, is when sphere reaches terminal velocity (where v is a maximum constant value).

Note that dt

darepresents the gradient of the a-t graph. The negative sign indicates that the graph is

downward sloping. The value of gradient dt

dadecreases with time since a decreases from g to 0.

Hence a-t graph is a curve whose gradient is negative and decreases with time.

4

5 Two blocks, one made of wood and the other of iron, are arranged at rest on the ground as

depicted in combination I and II below.

Which one of the following statements is correct?

A The force by the iron block on the wooden block in I is greater than that by the wooden block

on the iron block in II. B The force by the wooden block on the iron block in I is the same as that by the iron block on

the wooden block in II by virtue of Newton’s 3rd law. C The force by the wooden block on the iron block is equal to the weight of the wooden block

in I while the force by the iron block on the wooden block is equal to the weight of the iron

block in II.

D The force by the ground on the iron block in I is greater than the force by the ground on the

wooden block in II because the iron block, being denser than the wooden block, exerts more force on the ground.

Answer: C

Consider FBD of the wooden block in I. Since it is at equilibrium, its weight is equal to the normal

contact force on the wooden block by the iron block for I. (Note: Both are not an action-reaction pair)

Consider FBD of the iron block in II, since it is at equilibrium, its weight is equal to the normal

contact force on the iron block by the wooden block for II. (Note: Both are not an action-reaction pair)

Note: Weight of wooden block is LESSER than weight of iron block.

A is wrong as force by iron block on wooden block in I should be LESSER not greater.

B is wrong as force by wooden block on the iron block in I is LESSER than force by iron block on

the wooden block in II. For D, treat the iron and wooden block as one object. At equilibrium, their combined weight is

equal to the force by the ground on the iron block for I. At equilibrium, their combined weight is

equal to the force by the ground on the wooden block for II. The statement should be the force by

the ground on the iron block in I is the SAME as the force by the ground on the wooden block in II since their combined weights are the same for both scenarios.

iron

wood iron

wood

combination I combination II

5

6 A ladder ST, resting on a rough floor and learning against a rough wall, is on the point of slipping. The ladder has weight W. The contact forces exerted on the ladder by the wall and floor are P and Q respectively. Which one of the following diagrams correctly shows the directions of these forces?

A

B

C

D

Answer: C

All the lines of action of forces acting on the ladder intersect at a single point when it is in equilibrium.

For option B, though the line of actions of the forces act at a single point, the force P is made up of both the normal and frictional forces and the frictional force should act upwards.

6

7 A stationary polonium nucleus (A = 210, Z = 84) emits an α particle with kinetic energy Eα. Which of the following is the kinetic energy of the recoiling nucleus?

A αE

105

2 B αE

103

2 C αE

5

2 D αE

103

42

Answer: B

Let x be the recoiling nucleus. By conservation of momentum: total initial momentum = total final momentum

M.0 = mx v + mα Vα

v = xm

Vm αα−

KE of X = ½ mx v2 = ½ mx

2

−

xm

Vm αα

= ½ mα Vα2

2

x

x

m

mmα = Eα

4-210

4 = Eα

103

2

8 A light inextensible string is wound, as shown, over a frictionless, light pulley. What is the tension

in the string between the 2.0 kg and the 3.0 kg masses when the system is released?

A 6.5 N B 16 N C 20 N D 39 N

Answer: A Let T be the tension in between the 3.0 kg and 1.0 kg masses and a be the acceleration of the system (The direction of a is assumed to be in the downward direction of the 2 kg and 3 kg masses). Consider the 3.0 kg and 2.0 kg masses as one system, taking downwards as positive 5.0g – T = 5.0a (1) Consider the FBD of the 1.0 kg mass, taking upwards as positive T – 1.0g = 1.0a (2) Solving simultaneously, a = (2/3)g Consider the FBD of the 2.0 kg mass, Let T’ be the tension between the 3.0 kg and 2.0 kg masses, 2.0g – T’ = 2.0a T’ = 6.5 N

7

9 A body of mass 1.0 kg initially at rest slides down an inclined plane that is 1.0 m high and 10.0 m long as shown in the figure below.

If the body experiences a constant resistive force of 0.5 N while travelling on the slope, what is the kinetic energy of the body at the base of the plane?

A 4.8 J B 9.3 J C 10 J D 15 J

Answer: A Lost in G.P.E = Gain in K.E + Work against friction 1(1)(9.81) = K.E + 0.5(10) K.E = 4.8 J

10 The figure shows a wheel which is driven by an electric motor. A rope is fastened at one end to a

spring balance. The rope passes over the wheel and supports a freely hanging load. When the wheel is turning anticlockwise at a steady speed, the balance reading is constant. What is the output power of the motor?

A 0.3 kW B 1.2 kW C 1.5 kW D 1.8 kW

Answer: B Treat the load and rope attached to load as one system, (assume massless rope) To keep the load at a steady speed tension due to spring balance + additional force due to wheel on the rope connected to load = mg additional force due to wheel = 100 – 20 = 80 N Velocity of rim of wheel = distance / time = circumference x 50 rev s-1

= 0.30 x 50 = 15 m s-1 Output power = Force x velocity = 80 N x 15 = 1200 W = 1.2 kW

Load = 100 N

50 rev s-1

Spring balance

Reading = 20 N

Circumference of wheel = 0.30 m

8

11 The clock on Big Ben has a minute hand of length 4.3 m and an hour hand of length 2.7 m. What is the ratio of the centripetal acceleration of a point on the tip of the minute hand to a point on the tip of the hour hand?

A 1.6 B 150 C 230 D 5700

Answer: C

12 Which of the following statements is not true for a stone attached to a string and swung in a uniform vertical circular motion?

A The magnitude of resultant force acting on the stone changes depending on the position of the stone in the circle.

B The tension in the string is lowest when the stone is at the highest point of the circular motion.

C The kinetic energy of the stone is constant throughout the entire circular motion.

D The resultant acceleration is always directed towards the centre of the circle.

Answer: A In a uniform circular motion, the magnitude of the tangential velocity and the radius of the object

remain constant. Therefore, since Fc = r

mv 2

,the centripetal force which is the resultant force of

the stone must remain the same.

23012360072

360034

2

and T

2 :Since

2

2

2

2

2

2

=×

==

=

==

)/(.

/.

/

/

)(

hh

mm

h

m

c

c

Tr

Tr

a

a

Tra

ra

π

ωπ

ω

9

13 A piece of a decommissioned satellite is initially at rest 620 km above the surface of the Earth, and begins falling towards the Earth’s surface. Given that Earth’s radius and mass are 6400 km and 6.0 × 1024 kg respectively, what is the velocity of the object when it hits the surface of Earth? You may assume that air resistance is negligible.

A 3.3 × 103 m s-1 B 4.5 × 103 m s-1 C 3.6 × 104 m s-1 D 1.1 × 107 m s-1

Answer: A GPEi + 0 = GPEf + KEf KEf = GPEi - GPEf = - (GPEf – GPEi)

14 In 1979, the space probe Voyager 1 was sent to capture images of Jupiter. At the point of closest approach to Jupiter, it was 3.49 × 108 m from the centre of Jupiter and 7.80 × 1011 m from the Sun as shown in the figure below (not drawn to scale).

Given that the mass of Jupiter is 1.90 × 1027 kg and the mass of the Sun is 1.99 × 1030 kg, what was the escape velocity of the Voyager 1 at the point of closest approach to Jupiter? The gravitational forces due to other objects are assumed to be negligible.

A 18.4 km s-1 B 19.6 km s-1 C 26.9 km s-1 D 32.6 km s-1

1-3

33

2411

2

s m 1033

106206400

1

106400

11006106762

112

11

2

1

×=

×+−

×××=

−=

−=

−−

−−=

−

.

)().)(.(

if

if

if

rrGMv

rrGMm

r

GMm

r

GMmmv

10

Answer: D Resultant potential at the point: Escape velocity:

15 A mass-spring system is placed in water and allowed to oscillate with decreasing amplitude until it

comes to a complete stop. Which diagram shows the variation of the displacement y with time t of the system?

A

B

C

D

Answer: D With light damping, there should be a progressive decrease in the amplitude of oscillation. Option A is wrong because the first two peaks are of the same height. Option B is wrong because the amplitude is not continuously decreasing. Option C is wrong because the graph is not sinusoidal. Option D is true as the amplitude decreases progressively.

( )

( )

1-

8

S

S2

s km 632

103352

2

2

1

.

).(

=

×=

+=

+=

JE

JE

v

mmv

φφ

φφ

1-8

8

2711

11

3011

S

kg J 10335

10493

1090110676

10807

1099110676

×−=

×

×××−+

×

×××−=

−+−=

+=

−−

.

.

..

.

..

J

J

S

S

Jres

r

GM

r

GM

φφφ

11

16 A block P, of mass m, attached to a spring with spring constant k, oscillates in simple harmonic motion on a frictionless horizontal surface. A block Q, of the same mass as P, is attached to a similar spring and oscillates vertically in simple harmonic motion. The angular frequency ω of the

oscillation is determined bym

kω = .

Given that both blocks oscillate with the same amplitude, which of the following statements is true?

A The maximum kinetic energy of P is smaller than the maximum kinetic energy of Q.

B The maximum kinetic energy of P is larger than the maximum kinetic energy of Q.

C The maximum elastic potential energy of P is smaller than the maximum elastic potential energy of Q.

D The maximum elastic potential energy of P is larger than the maximum elastic potential energy of Q.

Answer: C Option A and B are wrong because based on the equation, KEmax = ½mω2x0

2 the maximum KE for both horizontal and vertical systems are equal since k is the same, mass of P and Q are equal and the amplitude is the same. Option C is true since for block P, the spring is not stretched at equilibrium before it is allowed to oscillate whereas for block Q, the spring is stretched at equilibrium before it oscillates with the same amplitude as block P. Hence EPEmax for block P < EPEmax for block Q. Based on the reasoning for option C, option D is wrong.

17 Four different solids A, B, C and D of equal masses at 20 °C are separately heated at the same

rate. Their melting points and specific heat capacities are as shown in the table below.

Which of these solids will start to melt first?

Liquid Melting point / °C Specific heat capacity / J kg−1 K−1

A 80 1200

B 100 800

C 150 600

D 300 250

Answer: B Q = mc∆Ɵ = m(800)(100-20) = 64 000m for B. The heat required for liquid B to reach melting point is the lowest.

18 Container X contains neon gas and container Y contains argon gas. Container X has twice the

volume of container Y. The temperatures of the gases in both containers are the same. What is the ratio of the mean kinetic energy of a neon atom to the mean kinetic energy of an argon atom? [The relative atomic masses of neon and argon are 20 and 40 respectively.]

A 0.5 B 1 C 2 D 4

Answer: B KE α T for ideal gas, since T is the same for both gases, mean KE is the same for both gases.

12

19 A stationary sound wave is set up in a 4.0 m column, with both ends open, using a sound generator producing waves of frequency 400 Hz. The speed of the sound is 320 m s-1. If the sound generator doubles the frequency of the sound wave produced, how many more antinodes will be present in the stationary wave?

A 5 B 10 C 11 D 21

Answer: B Using v = fλ, λ = 320 / 400 = 0.80 m No. of wavelengths in tube = 4.0 / 0.80 = 5 No. of antinodes in 5 wavelengths = 11 If frequency is doubled, λ = 320 / 800 = 0.40 m No. of wavelengths in tube = 4.0 / 0.40 = 10 No. of antinodes in 10 wavelengths = 21 Thus, there are 10 more antinodes in the stationary wave.

20 The figure below shows two ideal polarisers A and B where their transmission axes are initially parallel to each other.

Polarised light of amplitude E0 and intensity I0 is incident on A with its electric field vector parallel to the transmission axis. Polariser B is then rotated so that its transmission axis makes an angle θ, as shown in the figure above. Which of the following graphs shows how the intensity of the transmitted light It varies with the angle θ?

13

A

B

C

D

Answer: A Option A is the answer as when θ is 0°, the transmission axes are perpendicular to one another and hence no light passes through. When θ is 90°, the axes are parallel to one another and maximum intensity of light passes through. In between these two angles, the intensity transmitted increases.

21 A blue laser light is used in a Young’s double-slit experiment. Which of the following will be observed when a change is made to the experiment?

Change to experiment Observation

A Covering one of the slits completely. No fringe pattern is seen.

B Moving the source of light nearer to the double slits.

Fringe separation will increase.

C Covering one of the slits with a polaroid. No change in position of central bright fringe.

D Replacing the blue laser light with a red laser light.

Central bright fringe will shift upwards.

Answer: C Option C is true as the maximum intensity point will not shift although the maximum intensity will decrease. Option A is wrong because a single-slit diffraction pattern will be seen. Option B is wrong because the fringe separation is not dependent on the distance between light source and slit. Option D is wrong because replacing the laser light with one of higher wavelength will cause the fringe separation to increase but will not shift the central bright fringe (0th order maximum).

14

22 The mercury spectrum contains two intense yellow lines with wavelengths of 577.0 nm and 579.1 nm respectively. Light from a mercury lamp is incident normally on a diffraction grating ruled with 4300 lines per centimetre. To distinguish the two yellow lines clearly, it is desirable that there should be an angular separation of at least 0.30° between the two lines. Which order of diffraction can be used?

A 2nd order B 3rd order C 4th order D 5th order

Answer: C

m102.3264300

10d 6

2−

−

×==

Using d sinθ = nλ for both wavelengths 577.0 nm and 579.1 nm for the different orders will give

the following:

n θ / ° for 577.0 nm θ / ° for 579.1 nm ∆θ / ° 2 29.744 29.864 0.120 3 48.090 48.323 0.233

4 (max order) 82.867 84.793 1.926 5 NA NA NA

15

23 Which graph correctly relates the electric potential in the field of a negative point charge with distance r from the charge?

A

B

C

D

Answer: D Electric potential at a point is defined as the work done per unit charge by an external agent to bring a positive test charge from infinity to that point without any change in KE.

r

QV

04πε=

since the point charge is negative in this case, the potential must be negative as well, i.e.

V r

1−α

24 Four charges are arranged at the corners of a square as shown. Point P is located at the centre of

the square. The diagonal of the square is 8.0 cm.

How much work is done by the field in bringing a −3.0 nC charge from infinity to point P without any change in its kinetic energy?

A −270 µJ B −135 µJ C +135 µJ D +270 µJ

0

potential

0

potential

0

potential

0

potential

P

+200 nC +300 nC

−400 nC

−500 nC

16

Answer: A

the potential at point P, VP = 1

1

4 r

Q

oπε+

2

2

4 r

Q

oπε+

3

3

4 r

Q

oπε+

4

4

4 r

Q

oπε

since r1 = r2 = r3 = r4 = 2

08. = 4.0 cm,

VP = ( )4321

14

1QQQQ

ro

+++πε

The work done by external agent in bringing the −3.0 nC charge to point P without any change in KE = EPE = VP x q. The work done by the field = − EPE

= ( ) qQQQQro

+++− 4321

14

1

πε

= ( ) ( )99

2

9

1003105004003002001004

109 −−

−×−

×−−+

×

×− .

.

= 41072 −×− . C

= −270 µJ

25 The graph below shows the variation with current I of the potential difference V across an electronic component.

Which of the following statement is correct?

A The electric component is a diode.

B The resistance is increasing with increasing potential difference.

C The resistance of the component is 950 Ω when the potential difference is 0.60 V.

D When the potential difference is 0.60 V, the power dissipated at the component is 0.90 W.

0.0

1.0

0.5

1.5

2.0

0.2

0.4

0.6

0.8

1.0

0.0

1.2 V / V

I / mA

17

Answer: B The resistance is increasing with increasing p.d. Hence, it cannot be a diode. The resistance of the component is 400 Ω when the potential difference is 0.60 V. (R = ratio of

V to I.) When the p.d. is 0.60 V, the power dissipated is 0.90 mW.

26 A strain gauge consists of a length of wire with uniform cross-sectional area. Its resistance is 4.000 kΩ. It is attached to a gas container. When the container expands, the strain gauge

changes its dimensions. Its length increases by 2.0% and diameter reduces by 1.0 %. What is the new resistance of the strain gauge?

A 3.842 kΩ B 4.121 kΩ C 4.163 kΩ D 4.897 kΩ

Answer: C

R 2d

L

A

Lαα , L2 = 1.02 L1 , and d2 = 0.99 d1

=

×=

=

22

2

1

1

2

1

2

990

1021

..

d

d

L

L

R

R1.0407

R2 = 4.163 kΩ.

27 The diagram shows a network of five resistors.

What is the effective resistance between P and S?

A 1.9 Ω B 2.1 Ω C 2.5 Ω D 3.6 Ω

Answer: D When a p.d. is applied across PS, the potential at Q = potential at R. Therefore, there is no potential difference across the 5.0 Ω resistor and it should not be considered when determining the effective resistance. Hence,

0603

1

0402

11

.... ++

+=

effR

Reff = 3.6 Ω.

P

Q

S R

2.0 Ω

6.0 Ω

3.0 Ω 4.0 Ω

18

28 A thermistor R1 is connected to a battery of constant e.m.f. with negligible internal resistance as shown in the figure.

Which of the following actions will cause an increase in the potential difference V measured by the

voltmeter? Assume that the voltmeter has infinite resistance.

A Close switch S

B Increase the light intensity at R3 with S open

C Remove the earth connection at M with S open

D Increase the temperature of the thermistor with S open

Answer: B By potential divider principle, voltmeter reading increase when effective resistance across thermistor is increased or resistance R3 is reduced. Greater light intensity reduces the resistance of the LDR.

29 A simple model of a current balance is assembled as shown in the diagram below:

When current, I, flows through the setup, the wire frame is balanced. If the strength of the magnetic field is halved, which of the following changes to the setup will ensure that the wire frame remains balanced?

R1

P Q

S

R3

R2

M V

19

A doubling the mass of the weights

B reversing the direction of the current

C shifting the pivot in the direction of the weights

D reducing the magnitude of the current, I, by half

Answer: C When the B-field is halved, the force on the right side of the frame reduces by half, causing the clockwise moments to be also halved. Any changes that should be made should either increase the force (and thus the moments) to the original magnitude, increase the clockwise moment or decrease the anti-clockwise moment. The mass of the weights will increase the anti-clockwise moment, so this is rejected. Reversing the current direction will change the direction of the force and lead to a net anti-clockwise moment in the frame. Halving the current will decrease the force on the wire, and thus decrease the clockwise moment further. Moving the pivot to the left will increase the clockwise moment and decrease the anti-clockwise moments, and this will help restore the balance in the frame.

30 A flat circular coil of diameter 40.0 mm has 600 turns and is placed so that the plane of the coil is perpendicular to a uniform magnetic field of flux density 30.0 mT. The magnetic flux density first reduces to zero and then increases to 30.0 mT in the opposite direction at a constant rate. The time taken for the whole operation is 60.0 ms. What is the average value of the e.m.f induced in the coil?

A 0.00 V B 0.377 V C 0.754 V D 3.02 V

Answer: C

E=∆ϕ

∆t

E=∆NBA

∆t=NA

∆B

∆t

E=600×π(0.0200)2×30-(-30)×10

-3

60×10-3

= 0.754 V

20

31 A new particle was discovered at CERN. When one such particle of velocity 1.5 × 107 m s-1 was passed through a 3.0 T magnetic field directed out of the paper, the path of the particle was as shown in the figure below.

Given that the mass of the particle has been predicted to be 2.48 × 10-28 kg, what is the charge of the particle?

A +1.6 × 10-19 C B -1.6 × 10-19 C C +3.2 × 10-19 C D -3.2 × 10-19 C

Answer: A Using Fleming’s Left Hand Rule, it can be deduced that the current is in the direction of the path, which implies that the charge of the particle is positive.

C 101.6)10(3.0)(7.75

)10)(1.510(2.48

Br

mvq

Bqvr

mv

19

3

728

2

−

−

−

×=×

××==

=

21

32 A flexible conductor loop, C is placed in the centre of a larger metal coil that is connected to a variable power supply as shown in the figure below. The plane of the flexible conductor loop is parallel to the plane of the larger metal coil. Initially, the current in the larger metal coil is kept constant and flows clockwise.

When the current in the larger metal coil is suddenly switched off, what is the direction of the induced current and the change in the area of the flexible conductor loop at that instance?

Direction Area

A clockwise increase

B clockwise decrease

C anticlockwise increase

D anticlockwise decrease

Answer: A The current in the larger outer loop generates a magnetic field into the page, within the loop where the smaller loop is. When the current is switched off, the magnetic flux density pointing into the page decreases. An e.m.f is generated in the flexible tube due to the change in magnetic flux linkage through the loop as predicted by Faraday's law. The current is induced in the closed loop. According to Lenz’s law, the induced current would want to reinforce the reduced magnetic field into the page, hence the direction of the induced current is clockwise. In addition, since the magnetic flux decreases as the current is switched off, the flexible loop would increase the area as a way to oppose that change.

CCurrent flow direction

22

33 An alternating current with potential difference V / V varies with time t / ms across a resistive load

as shown below.

What is the mean power dissipated in the resistive load if its resistance is 20 Ω?

A 0.340 W B 0.375 W C 0.450 W D 0.800 W

Answer: A

⟨V2⟩= -3 2+-1 2+2

2+4

2+2

2

5

⟨V2⟩=6.80 Vrms=2.61 V

⟨P⟩= Vrms2

R=

2.612

20= 0.340 W

34 What is the potential difference through which the electron must be accelerated from rest for it to have a wavelength of 2.4 x 10-10 m?

A 9.48 V B 26.2 V C 52.4 V D 9.48 x 1026 V

Answer: B Using the conservation of energy,

Gain of KE of electron=Loss of EPE of electron

p2

2m=qeV

h λ 2

2m=qeV

V=h λ

2

2mqe

V=6.63×10

-342

22.4×10-1029.11×10

-311.6×10-19

=26.2 V

t / ms

- 3

- 1

2

4

1 2 3 4 5 6 7 8 9 10 11

V / V

23

35 Which of the following statements about the Heisenberg Uncertainty Principle is true?

A This principle is not experimentally verified.

B This principle can only be applied to quantum particles.

C This principle states that if we were to improve our instrument’s precision, both the positions and the momentum measurements can be improved simultaneously.

D This principle states that no matter how we were to improve our measurement techniques, there will always be uncertainties about the concepts of length, momentum, energy and time measurements.

Answer: D Option A: There are phenomena that could be explained by the HUP (i.e diffraction of photons through a slit). Option B: This principle applies to all particles, not just quantum particles. Option C: We can only improve the measurements of one of the quantities rather than both. Option D: True.

36 Light of frequency f falls on a metal surface of work function energy φ and ejects electrons of maximum kinetic energy K. If the wavelength of this light is doubled and its intensity halved, what will be the maximum kinetic energy of the emitted electrons?

A K

B 2K

C 2 hf - φ

D

2

1hf - φ

Answer: D For the original set up,

hf=hc

λ=ϕ+K

When the wavelength is doubled the energy of each photon is halved The change in intensity (keeping the frequency constant) will change the rate of photon incident on the plate. Since the question only requires the final maximum KE of the photoelectrons, the rate of photon incidence is not important here.

Therefore, hf= hc

λnew=ϕ+Knew

Knew=hc

λnew

-ϕ

=hc

2λ-ϕ

=1

2hf-ϕ

24

37 Which of the following statements concerning semiconductors is false?

A In semiconductors, the number of holes in the valence band depends on the temperature of the semiconductor.

B In p-type semiconductors, the majority mobile charge carriers are holes and the minority carriers are electrons.

C In intrinsic semiconductors, the number of holes in the valence band is always equal to the number of electrons in the conduction band.

D Intrinsic semiconductors are not as good electric conductors as compared to extrinsic semiconductors because in intrinsic semiconductors, holes and electrons recombine in the presence of an applied electric field.

Answer: D Option A: The higher the temperature of the semiconductor, the higher the probability that the electrons in the valence band would be promoted to the conduction band leaving behind holes in the valence band. Option B: In the p-type semiconductor, the acceptor atoms will be creating more holes in the extrinsic semiconductor. Therefore, there will be more holes compared to conducting electrons that are mobile to conduct electricity. Option C: The electrons and the holes in an intrinsic semiconductor exist in pairs. Option D: Doping adds additional energy levels in the energy gap, thus allowing electrical conductivity to increase.

38 Which of the following statements concerning the semiconductor diode is correct?

A The size of the depletion region in the diode is fixed at all times.

B It is possible to make a diode using intrinsic semiconductors without dopants.

C The depletion region is formed due to the difference in the mobile charge carrier concentration in the n-type and the p-type semiconductor.

D The p-type side of the diode is positively charged and the n-type side is negatively charged. However, overall, the diode is electrically neutral.

Answer: C Option A: The depletion region size is changed, depending on the orientation of the connection of the e.m.f. in the external circuit. Option B: The diode is made from extrinsic semiconductors (dopants are added to intrinsic semiconductors). Option C: True. Option D: The diode’s depletion region is negatively charged in the p-type side and positively charged for the n-type side of the diode.

25

39 At time t, a sample of a radioactive substance contains N atoms of a particular nuclide. At time (t + ∆t), where ∆t is a short period of time, the number of atoms of the nuclide is (N - ∆N). Which expression is equal to the decay constant of the nuclide?

A

N

N∆ B

tt

NN

∆+

∆− C

tN

N

∆

∆ D

t

NN

∆

∆

Answer: C Definition of decay constant: It is the fraction of the total number of undecayed nuclei present which decays per unit time. Suppose N is the size of a population of radioactive atoms at a given time t, and dN is the amount by which the population decreases in time dt; then the rate of change

is given by the equation dt

dN= −λN, where λ is the decay constant.

dN

dt=-λN

λ=-dNdtN

=

∆N∆tN

=∆N

N∆t

40 Hydrogen and oxygen nuclei may result from the bombardment of nitrogen nuclei with helium nuclei. The reaction can be represented by the following nuclear equation:

7

14N+2

4He →8

17O +1

1H

The speed of light is c, and the masses of the particles are: Nitrogen : mN Helium : mHe Hydrogen : mH Oxygen : mO

What is the net energy released during such a reaction?

A [mH + mO – (mN + mHe)]c2

B [(mN + mHe) – (mH + mO)]c2

C (mH + mO + mN + mHe)c2

D mN+mHe-mH+mO

c2

Answer: B The net energy released

= ∆mc2 = [(mN + mHe) – (mH + mO)]c2

1

CATHOLIC JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS Higher 2

PHYSICS 9646/02 Paper 2 29 Aug 2014 Suggested Solutions 1 hour 45 minutes

1 (a) State Newton’s second law of motion. [1] It states that the rate of change of momentum of a body is proportional to the

resultant force acting on it, and the direction of momentum change takes place in the direction of the resultant force.

1

(b) A squash ball of mass 24 g hits the wall when it reaches its maximum height of 3.2 m. It leaves a racket which is 1.7 m above the ground as seen in Fig. 1.1. The ball is incident on the wall with a horizontal velocity of 15 m s-1 and rebounds in a horizontal direction with a velocity of 12 m s-1. The ball is in contact with the wall for 0.15 s.

Fig. 1. 1

(i) Calculate the initial vertical component of the ball’s velocity.

vertical velocity = ……………………. m s-1

[2]

Taking upwards as positive,

Using 2asuv 2y

2y +=

1.7) - (3.2 2(-9.81)u0 2y +=

1y s m 5.4u −

=

1

1

(ii) Determine the average force exerted on the wall during the ball’s collision with the wall.

magnitude of the force = ……………………….. N direction of force on the wall = ……………………........

[4]

Applying Newton’s 2nd law of motion, and taking rightwards as positive

Favg = ∆t

∆p,

2

Favg = 0.15

15)0.024(-12

t

u)m(v −=

−

= - 4.32 N (Force on ball by wall which acts to the left) [1 for magnitude of F] By Newton’s 3rd law of motion, Force on wall by ball = - Force on ball by wall = 4.32 N Force on wall acts to the right.

1 1 1 1

(iii) State and explain whether the collision of the ball with the wall is elastic. [1]

It is not elastic as the speed of the ball has decreased from 15 m s-1 to 12 m s-1 which means that the KE of the ball-wall system, which is ½ mv2, has decreased. Must mention how KE of system changes (it decreases).

1

(iv) Explain why the ball does not rebound to the point from where it was hit by the racket. [2]

As the horizontal velocity is reduced after collision, with the same time of flight before collision with the wall, the horizontal displacement during rebound will be reduced. (The ball will land closer to the wall.) Students must mention the reduction in both horizontal velocity and displacement.

1 1

2 (a) Explain what is meant by a progressive wave. [1] A progressive wave is a wave in which energy is transferred from one point to

another by means of vibrations or oscillations within the wave.

1

(b) Sound is propagated in air as a longitudinal progressive wave, in which there is a repeated sequence of displacements of the air particles. Fig. 2.1a illustrates nine particles, equally spaced along the line AB, in still air.

Fig. 2.1b

1 2 3 4 5 6 7 8 9

A B

P

P

Q

Q

d +ve

Fig. 2.1a

3

(i) A sound wave of wavelength equal to the distance between A and B is sent through the air in the direction of PQ. On line PQ on Fig. 2.1a, draw the possible positions of the nine particles in the wave relative to their undisturbed positions which they occupy in still air, when the sound wave propagates through the particles. [1]

For part (i), positions of particles must conform to 1 wavelength, with clear differences in displacements of particles.

1

(ii) Using (b)(i), sketch a graph showing how the displacement d of the particles from their undisturbed positions vary along PQ on Fig. 2.1b. Take direction to the right as positive. [1]

For part (ii), wave drawn must correspond to diagram in (i) and displacement of particles to the right is taken to be positive on the graph.

1

(iii) A sound wave can also be described in terms of a repeated sequence of changes in

pressure. On Fig. 2.1b, identify, and label with H, a point where the pressure is the highest. Justify your answer.

[2] Point H, a point of high pressure, occurs at where particle 5 is as this is where

compressions occur due to neighbouring air particles 4 and 6 coming closer to one another. Identifying and labelling of H correctly at particle 5.

1

1

(iv) A loudspeaker, generating a sound wave with a wavelength magnitude equal to AB,

and a screen are placed at A and B respectively. Describe and explain the changes, if any, to Fig. 2.1b over time as compared to the current scenario. [2]

As a stationary wave is being formed, the wave will not progress and each particle will oscillate with different amplitudes. In the current scenario, the wave will progress towards Q with each particle moving left and right, reaching the same amplitude over time.

1 1

(c) A stereo system in a large hall has two identical speakers, S1 and S2, 1.2 m apart. The

amplitude of the output of each speaker is proportional to the voltage across its terminals. The voltage input to each speaker is adjusted by means of a balance control. The arrangement is shown in Fig. 2.2.

Fig. 2.2 (not to scale)

balance control

S1

S2

1.2 m

15 m

A

B

y

O

C

4

Initially, the speakers are emitting signals of frequency 1000 Hz which are in phase. The balance control is set such that there is a voltage of 6 V r.m.s. across each speaker. An

observer hears a loud sound of intensity Imax at A. As he moves along the line AC, 15 m away from the speakers, he observes that the intensity first falls to zero at point B, a distance y from A. The speed of sound in air is 330 m s-1.

(i) Determine the distance y. y = ……………. m [2]

Using a

Dx

λ= where

f

v=λ

).(

))((

221

151000

330

×=y , x is twice that of y (x = 2y) since B is at minimum intensity

y = 2.1 m

1 1

(ii) Determine the next higher frequency of the speaker such that point B would also be

a point of zero intensity. frequency = ……………. Hz [3] Since B is a point of minimum intensity, the path difference of S1 and S2 at point B,

mBB - SS 16502

330

2 = 21 .

.==

λ

For other wavelengths λ, point B will be at zero intensity if the path difference is odd

multiple of 2

λ.

Therefore,

m 0.11λ

1n wherem0.1652

λ1)(2n

=

==+

Using λ

vf = ,

λ

330=f

f = 3000 Hz

1 1 1

(iii) With the speakers emitting the original signal frequency of 1000 Hz, the balance

control is now adjusted such that the voltages across S1 and S2 are 3.0 V r.m.s. and

9.0 V r.m.s. respectively. In terms of Imax, determine the new intensity at point B. intensity = ……………. [3] Since amplitude ∝ voltage and amplitude2 ∝ intensity,

initially at A, constructive interference occurs as the two waves meet in phase to

give Imax. Thus amplitude proportional to 6 V + 6 V = 12 V,

hence Imax ∝12 V)2 = 144 V2 At B, destructive interference occurs as two waves meet in antiphase,

Thus amplitude proportional to 6 V - 6 V = 0 V giving Imin = 0. When the speakers are adjusted to 9 V and 3 V, the resultant amplitude at B is proportional to 9 V – 3 V = 6 V.

Hence Imin ∝6 V)2 = 36 V2 Imin= ¼ Imax

1 1 1

5

3 (a) Define potential difference between two points on an electric circuit. [1]

Potential difference between two points on a circuit is the amount of energy converted per unit charge from electrical to non electrical between the two points.

1

(b) A cell of e.m.f. 4.5 V and internal resistance of 0.70 Ω is connected in series with a

resistor R, as shown in Fig. 3.1. Resistor R is made of metal wire and the ammeter reads 200 mA.

Fig. 3.1

Determine the resistance of R. resistance of R = …………………. Ω [2]

ε= IR +Ir IR = ε - Ir (0.20)R = 4.5 – (0.70 x 0.20)

R = 21.8 Ω

1 1

(c) A second similar cell is now connected in series with the cell in (b) and the resistor R. The current in the circuit is 350 mA and the resistance of R changes.

Calculate the new resistance of R. new resistance of R = ……………………. Ω

[2]

Note that the total internal resistance is sum of the individual internal resistance (1.40 Ω).

ε= IR +Ir IR = ε - Ir (0.35)R = (4.5+4.5) – (1.40 x 0.35)

R = 24.3 Ω

1 1

0.70 Ω

A

4.5 V

6

(d) The cells in (c) are now connected in series with a fixed resistor of resistance 2500 Ω

and a thermistor, as shown in Fig. 3.2.

Fig. 3.2

The thermistor has resistance 5000 Ω at 0 °C and 2250 Ω at 20 °C.

(i) Determine the maximum and the minimum values of the readings of the voltmeter as the temperature of the thermistor is varied from 0 °C to 20 °C. The internal resistance of the cells can be assumed to be negligible and the voltmeter has a very high resistance.

[2]

By using potential divider, maximum reading is obtained when the resistance of the thermistor is 5000 Ω.

maximum reading = 0925005000

5000.×

+= 6.0 V (2 s.f.)

minimum reading is obtained when the resistance of the thermistor is 2250 Ω.

minimum reading = 0925002250

2250.×

+= 4.3 V

1 1

(ii) In one particular application of the circuit shown in Fig. 3.2, it is desired that the

potential difference across the fixed resistor should range from 3.6 V at 0 °C to 7.2 V at 20 °C. Determine whether, by substituting a different fixed resistor in the circuit of Fig. 3.2, it is possible to achieve this range of potential.

[3]

At 0 °C, resistance of thermistor is 5000 Ω. let R be the resistance of the fixed

resistor

3.6 = 095000

.×+R

R

Solving, R = 3330 Ω.

At 20 °C,

7.2 = 092250

.×+R

R

Solving, R = 9000 Ω.

The values of resistance of the fixed resistor for the two conditions are not the same. Hence, it is not possible to substitute a single fixed resistor to meet the requirements.

1 1 1

2500 Ω

V

7

4 (a) Define electric field strength at a point. [2] Electric field strength at a point is defined as the electric force per unit charge acting

on a unit positive test charge placed at that point.

1 1

(b) A and B are two identical conducting spheres. Sphere A carries a charge of +2Q while sphere B carries a charge of +Q. They are placed in a vacuum with their centres a distance d apart as shown in Fig. 4.1.

Fig. 4.1

(i) On Fig. 4.1, sketch electric field lines to represent the electric field around the two spheres.

[3]

Marking point: The field lines have to meet the surface of the spheres perpendicularly. The field lines must also be distributed such that the parts of surface of each sphere that is further from the other sphere have more field lines. (notice the purple dashed line that shows the perpendicularity) The field lines have to reasonably indicate the strength of the field at points in the region of electric field.(notice the purple dashed arrows that indicates strength). The field lines have to show symmetry about the axis connecting the centres of the sphere AND the direction of the field lines has to be radiating out of the spheres. null point is shown clearly, whereby the distance between the null point from the

centre of A is 2 times the distance between the null point and centre of B.

Any point mentioned is worth 1 mark. Maximum 3 marks to be awarded.

1 1 1 1

5 Ultraviolet-Visible (UV-Vis) Absorption technology uses the unique properties of discrete

energy levels of atoms and molecules to identify substances and determine the concentration of substances. The components of a UV-Vis absorption spectrometer are shown in Fig. 5.1.

Fig. 5.1

Sample Solution

Incident monochromatic

photon beam of intensity I0 Transmitted photon beam of

intensity I

Photodetector

8

At any one time, an incident monochromatic photon beam shines through a sample solution.

A photodetector measures the intensity of the light after passing through the sample. The

intensity of the incident photon beam is I0 and the intensity of the transmitted photon beam

is I. The extent of how much light of a given wavelength is absorbed is known as the absorbance, A. The higher the amount of light absorbed, the higher the absorbance. Various wavelengths of the incident photon beam, typically from 300 nm to 1000 nm shines through the sample to obtain a spectrum of absorbance against wavelengths for a certain material. Fig. 5.2 shows an example of a UV-Vis spectrum.

Fig. 5.2 (a) (i) Explain what is meant by 1. a photon, [1] A photon is a packet (or quantum) of electromagnetic radiation energy.

1

2. discrete energy levels. [2] Energy levels are the allowed energy values that the electrons are allowed to have

in an atom. Discrete means that the electrons energy values are not continuous, and there is no allowed energy values between these levels.

1 1

(ii) State two features of the UV-Vis spectrum in Fig. 5.2 that show that the energy of atoms and molecules exists as discrete levels. [2]

There are four peaks of high absorbance at certain wavelengths of incident photons. There is almost no absorbance between 460 nm to 560 nm.

1 1

(b) For an incident photon beam at a fixed wavelength, the absorbance, A, is given by

A = ln

An experiment is conducted to analyse nitrite concentration in water. The nitrite concentration, c, is varied and the corresponding absorbance, A, is recorded as shown in Fig. 5.3.

Concentration, c / g m-3 Absorbance, A

1.000 0.161

0.500 0.082

0.200 0.038

0.100 0.019

9

0.050 0.009

0.000 0.002

Fig. 5.3

(i) State a relationship between the absorbance of nitrite, A, and nitrite

concentration, c, in water with an incident beam of fixed wavelength. [1] Any one of the following:

• Absorbance of nitrite is directly proportional to its concentration in water

• As nitrite concentration increases, absorbance increases 1 (ii) Plot the point corresponding to nitrite concentration, c = 0.500 g m-3 on Fig. 5.4. [1] (iii) Draw a best fit calibration graph of the absorbance, A, against nitrite

concentration, c, for all the data points on Fig. 5.4.

[1]

Fig. 5.4

1 mark – Plotted correctly to an accuracy of half smallest division 1 mark – Best fit straight line of positive gradient

1 1

(iv) Show that the calibration graph plotted indicates the relation

I=I0e-kc where c is the nitrite concentration in the sample and k is a positive constant. [3]

I=I0e-kc I0

I=ekc

Taking ln on both sides,

lnI0

I=kc

A=kc The calibration graph of absorbance, A, against concentration, c, plotted fits the equation of the form A=kc where k is the positive gradient of the graph and the graph has approximately 0 vertical intercept. 1 mark - for identify gradient and possible y-intercept 1 mark – relating plotted graph, linking the features (gradient and y-int) as stated

1

1 1

10

above (v) Determine the gradient of the best fit line drawn in b(iii). gradient =……………....... [2] Gradient of the calibration graph

y2-y

1

x2-x1

=0.159

Gradient of calibration graph = 0.159

1 1

(vi) Hence, state the value of k, together with a suitable unit for k. value of k =…………….......

unit of k =……………....... [2]

Value is to be the same as the gradient for the graph. Since kc has to have no units as A has no units, and the unit of c is g m-3, units of k is g-1 m3 (Accepting other prefixes for the units, e.g. kg-1 cm-3, provided that the value of k is correct)

1 1

(vii) Factory operators will receive warning letters from the authority if the river water

near to them has a nitrite concentration of higher than 0.550 g m-3. In a recent water analysis for nitrite concentration, the authority finds an absorbance, A, of 0.128. Using the graph drawn in Fig. 5.4, state whether you would advise the authority to issue warning letters to the operators. Justify your answer. [3]

The equation for the calibration graph is A=kc+D A=0.159c+0.003 When A = 0.128, 0.128=0.159c+0.003 c=0.788 The nitrite concentration in the water is 0.788 g m-3 which is higher than 0.550 g m-3. Therefore, I would advise the authority to issue warning letters to the operators. (c can be read off from the graph. Award 2 marks for the determination of 0.788 g m-3)

1 1 1

(c) From the table in Fig. 5.3, the experimental reading for absorbance is not zero even when the nitrite concentration is zero. Suggest two reasons why this is so. [2]

Any answers that are physically logical. Suggested solutions:

• Background noise of the machine

• Reflection of photon beam off the container reduces the light intensity

• Random fluctuations. When taking many readings, the average is a zero These will not be accepted:

• There are other substance absorbing light (when doing calibration, there should not be other substance that affects absorption)

• Faulty equipment (zero error does not mean equipment is faulty)

1 1

6 While a bar magnet is dropped vertically through a coil, there will be an induced e.m.f. in

the coil. The maximum e.m.f. E is induced as the magnet leaves the coil with speed v. Fig. 6.1 shows the coil wrapped around a vertical plastic tube with a magnet above it. It is suggested that the relation between E and v is

E = kvn

where k and n are constants.

11

Fig. 6.1 Design an experiment to determine the value of n. You may use any other equipment usually found in a Physics laboratory. You should draw a labelled diagram to show the arrangement of your equipment. In your account you should pay special attention to

(a) the equipment you would use, (b) the procedure to be followed, (c) the control of variables, (d) how the speed and induced e.m.f. would be measured, (e) any precautions that would be taken to improve the accuracy and safety of the

experiment.

Defining the problem (all need to be present for 1 mark) Speed, v is the independent variable. Induced e.m.f., E is the dependent variable Keep the number of turns on the coil, N constant.

V1

Diagram (Every component to be labelled – Total: 1 mark)

D1

12

Methods of data collection [4 marks] 1. Set up the apparatus as shown in the diagram. 2. Connect a voltage sensor to the coil which is connected to a data logger to

measure the induced e.m.f. as the magnet falls through the coil. Alternative: digital multimeter or CRO which can measure the peak value of a varying voltage.

3. Using a metre rule, measure the distance, h, travelled by the magnet from the point of its release till the point it leaves the tube.

4. Using ½ mv2= mgh (gain in KE = loss in GPE since magnet is falling), determine

the value of v from v = 2gh .

Alternative: Motion sensor + data logger. Have to explain how v can be obtained from the data logger.

5. In order to vary the speed v, change the distance h by raising the magnet higher above the tube.

Method of Analysis [2 marks]

1. Plot a graph of lg E against lg v. 2. Determine the value of n from the gradient of the graph.

Safety: [1 mark]

1. Use a g-clamp to clamp the retort stand to the table so that the set up does not topple over and fall on the experimenter. or

2. Collect the magnet with a box filled with cotton wool so that the magnet does not fall on the experimenter after passing through the tube.

Additional details: [3 marks – any 3 of these]

1. Repeat experiment for each value of v and then find the average. 2. Use same magnet or magnet of same strength to ensure that B-field of magnet is

constant. Alternatively, collecting magnet with cotton to prevent magnet being demagnetised and changing value of B-field.

3. To ensure magnet is vertical when dropping into the coil, use a non-metallic vertical guide to direct orientation of magnet.

4. To ensure coil is vertical, check that the tube is vertical with a spirit level. 5. Use of short magnet so that v is (nearly) constant.

P1 P1 P1

P1 P1 P1 S1 AD1 AD1 AD1

1

CATHOLIC JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS Higher 2

PHYSICS 9646/03 Paper 3 26 Aug 2014 Suggested Solutions 2 hours

Section A

1 (a) State what is meant by gravitational potential at a point. [1] The gravitational potential at a point is the work done per unit mass by an external

agent in bringing the unit mass from infinity to that point.

1

(b) The variation of the gravitational potential between the Earth and the Moon with displacement from the centre of the Earth along the line joining their two centres is shown in Fig. 1.1 (not to scale) below.

On the displacement axis of Fig 1.1, label a point P at which an object will experience no resultant gravitational force. Explain your reasoning.

[3]

The point P must be marked at the displacement corresponding to the turning point of

the graph. At the turning point, the gradient of the graph, which is the magnitude of the gravitational field strength, is zero. Hence, the net force per unit mass at that point is zero. Therefore, an object at that point will experience no resultant force.

1

1 1

Fig. 1.1

2

(c) The point P in part (b), where an object experiences zero resultant force is known as the neutral point.

(i) Let the distance from the centre of the Earth to point P be x, and the distance from the centre of the Moon to point P be y. Show that the expression that relates x and y, to the masses of the Earth, ME and the Moon, MM is

y

x=

M

E

M

M.

[2]

At the neutral point, the gravitational field strength due to Earth is equal to the gravitational field strength due to the Moon.

1 1

(ii) Given that the mass of the Moon is 7.40 × 1022 kg, the mass of the Earth is 5.97 × 1024 kg, and the distance between the centres of the Earth and the Moon is 3.84 × 105 km, calculate the distance from the centre of the Earth to point P.

distance = ………………….. km [2]

1 1

(d) Explain why it is not possible to place an object in a stable orbit around the Earth at the distance calculated in (c)(ii) from the centre of the Earth.

[2]

At the neutral point, an object experiences no resultant force.

However, an orbiting object requires a non-zero resultant gravitational force to provide for the centripetal force.

1 1

2 (a) State the First Law of Thermodynamics.

[2] Increase in internal energy of a system is the

sum of the work done on the system and heat supplied to the system.

1 1

(b) A sealed container contains air at a pressure of 1.0 x 105 Pa and a temperature of 27 °C. The volume of air is kept constant at 0.050 m3. The mass of one mole of air is 0.030 kg and air may be considered as an ideal gas.

(i) Calculate the mass of air in the container. mass = ………………. kg [3]

M

E

2M

2E

ME

M

M

y

x

y

GM

x

GM

gg

=

=

=

km 103.45 x

107.40

105.97

x103.84

x

M

M

y

x

5

22

24

5

M

E

×=

×

×=

−×

=

3

For an ideal gas, pV = nRT

RT

pVn =

273.15)(278.314

0.050101.0n

5

+×

××=

n = 2 mols mass = 2 x 0.030 = 0.060 kg (2 s.f.)

1 1

1

(ii) The container is heated to 200 °C. 1. Calculate the pressure of air at 200 °C.

pressure = ……………… Pa

Volume and amount of gas are kept constant.

T

p

V

nR=

i

i

f

f

T

p

T

p=

273.15)(200300.15

101.0T

T

pp

5

f

i

if +×

×=×=

pf = 1.58 x 105 Pa

2. Sketch on Fig. 2.1, a graph to represent the process of heating the air. Label this graph B.

Fig. 2.1

[2] 1

1

[1] 1 1

volume / m3

0 0.050

1.0

pressure / 105

Pa

B

C

4

(c) An oven of the same volume as the sealed container contains an identical amount of air as in part (b). The oven is heated from 27 °C to 200 °C at a constant pressure of 1.0 x 105 Pa. Air molecules can escape from the oven.

(i) Sketch on Fig. 2.1 a graph to represent the process of heating the air in the oven. Label this graph C. [1]

(ii) A greater amount of heat is required to increase the temperature of the air in the oven from 27 °C to 200 °C, compared to the air in the sealed container. Explain why this is so. [2]

For B, there is no work done by air since the volume is kept constant, thus all the heat supplied to air results in an increase in internal energy of the system. For C, the heat supplied to air is used to increase the internal energy and do work against the atmospheric pressure. Thus, to achieve the same increase in internal energy, more heat supply is needed.

1 1

3 An electron is projected into the magnetic field of a very long solenoid with a speed of

5.0 x 107 m s-1 at an angle of 30° to the axis of the solenoid. The magnetic flux density of the magnetic field is 7.50 mT and the direction of the magnetic field is along the axis of the solenoid, as shown in Fig. 3.1.

Fig. 3.1

(a) Given that the radius of the cross-section of the solenoid is 4.0 cm, determine quantitatively whether the electron will travel down the length of the solenoid or collide with the wall of the solenoid.

component of the electron’s velocity perpendicular to the axis of the solenoid = v sin 30° = 5.0 x 107 x sin 30° = 2.5 x 107 m s-1

let the radius of the helical path be r, resultant force perpendicular to the axis of solenoid = centripetal force

Bqv = r

mv 2

r = Bq

mv

r = 193

731

101.6107.5

102.5109.11−−

−

×××

×××

r = 1.9 cm diameter = 2r = 3.8 cm Since the diameter of the helical path is less than the radius of the solenoid, the electron will travel down the length of the solenoid.

1 1 1 1

[4]

5

(b) A uniform electric field is now produced in the same region and in the same direction as

the magnetic field. Deduce the changes, if any, to the motion of the electron when it is projected into the solenoid as in (a). You may draw a sketch to illustrate the path, if you wish.

[2]

There is an electric force opposing the motion of the electron along the axis of the solenoid. Hence, the component of the velocity of the electron which is directed along the axis of the solenoid will be reduced to zero before the electron accelerates in the opposite direction. This will result in the electron leaving the solenoid from the entry point, provided the solenoid is long enough. (any two of the underlined).

1 1

4 (a) Explain what is meant by the metastable state. [1] It is defined as a higher energy level where electrons can stay for a much longer

period of time than the usual 10-8s.

1

(b) Ruby is a crystal of aluminium oxide Al2O3 in which some aluminium ions are replaced by chromium ions. It contains 0.05% to 0.5% of chromium and its colour is pink. It is the energy levels of chromium, which takes part in lasing action to produce the distinct red colour laser light in the ruby laser. The energy level diagram of chromium is shown in Fig 4.1 below. (Note that the energy values in this diagram are relative to the ground state, E1, which is taken to be 0 eV.)

Fig. 4.1

(i) By showing suitable calculations, determine which energy level is the metastable state. Explain your reasoning clearly.

[4]

As the laser light is red, a suitable estimate for the wavelength of red light is

700 nm. Therefore, the energy transition involved should be about

∆E=hc

λred

=(6.63×10

-34)(3.0×108)

(700×10-9)(1.6×10

-19) = 1.78 eV

1

E4

E3

E2

E1

3.11 eV

2.26 eV

1.79 eV

0 eV

6

The most suitable state transition that corresponds with the above energy change is from E2 to E1. Since laser light is produced via stimulated emissions from the metastable state, and that population inversion is needed for lasing action to occur, the energy transition has to occur starting from the metastable state to a lower energy state. Therefore, the metastable state is E2 (1.79 eV).

1 1 1

(ii) The ruby laser delivers a 10.0 ns pulse of 1.00 MW average power. Estimate how

many photons are present within each pulse. number of photons = ……………………. [2]

Epulse=NphotonsEphoton

Nphotons=Epulse

Ephoton

=P∆t

∆E

= 1.0×10610.0×10

-91.79×1.6×10

-19

=3.49×1016

1

1

5 (a) By reference to band theory, explain how intrinsic semiconductors are able to conduct

electricity. [3] An intrinsic semiconductor has its valence band separated from the conduction band

by a small band gap. As a consequence of greater thermal energy supplied to the intrinsic semiconductor, some electrons are able to cross the band gap to enter the conduction band, leaving behind holes in the valence band. In the presence of a potential difference across the semiconductor, the electrons in conduction band and holes in valence band are mobile charge carriers. Therefore, electrical conduction will be possible.

1 1 1

(b) Explain how a p-n junction is able to carry a current under forward bias connection.

[3]

In a forward bias connection, the external emf makes the n-side of the component

(diode) a negative potential and p-type a positive potential (OR the external e.m.f. opposes the internal e-field). Under the influence of the external E-field, the electrons from n-side would be driven towards the p-n junction to the p-side; likewise, the holes from p-side to the n-side. The depletion region narrows and collapses, thus allowing mobile charge carriers, holes and electrons to move across the junction; holes to move from the p-side to the n-side and vice versa for the electrons, causing the resistivity to decrease.

1 1 1

7

Section B

6 (a) (i) Explain what is meant by the term resonance. [1] Resonance is a phenomenon when energy is added to a system without leakage.

The condition for this is that the driving frequency applied on an object matches its natural frequency, such that the object responds with maximum amplitude.

1

(ii) State what is meant by forced oscillation. [1] A forced oscillation is an oscillation when a periodic driving force is applied on a

system to cause it to oscillate. 1 (b) A composite block made of woods of different densities ρ1 and ρ2 floats in still water of

density ρw, as shown in Fig. 6.1. Each portion of wood is of height x and of a base with cross-sectional area A.

Fig. 6.1

(i) The block is floating as shown in Fig. 6.1. Determine the upthrust on the block in

terms of ρ1, ρ2, g, x and A. upthrust = …………………………………….. [2] When an object is floating, the object is in equilibrium and thus Fnet = 0.

Taking upwards as positive, U - W = 0 U = W U = ρ1(xA)g + ρ2(xA)g

1 1

(ii) The block is pushed down by a distance l, into the water without totally submerging

it, and is then released. 1. Show that the magnitude of acceleration a of the block is given by

x

ga w

)(

)(

21 ρρ

ρ

+= l

[2] When displaced slightly, Fnet = ma

Taking upwards as positive, U – W = ma ρwVwg – (ρ1(xA)g + ρ2(xA)g) = ma ρw(x+x/2+l)(A)g – (ρ1(xA)g + ρ2(xA)g) = (ρ1+ρ2)(xA)a Since ρw(x+x/2)(A)g = (ρ1(xA)g + ρ2(xA)g) from (i), ρw(l)(A)g = (ρ1+ρ2)(xA)a

lx)ρρ(

g)ρ(a w

21 +=

1

1

8

2. Explain why the block bobs up and down in simple harmonic motion. [2] The acceleration (upwards) is in the opposite direction to the displacement of

the block (downwards) and it can be seen from the equation that acceleration is directly proportional to the displacement. Therefore, the block moves in simple harmonic motion.

1

1

(iii) The depth h of the block is measured from below the water surface to the base of

the block as shown in Fig. 6.2.

Fig. 6.2

Fig. 6.3 shows the variation with time t of the depth h of the base of the block below the surface.

Fig. 6.3

1. Determine the amplitude of the oscillation of the block. amplitude = ……………. m [2]

From the graph, maximum displacement = 23.0 cm minimum displacement = 11.0 cm

amplitude = 2

011023 .. −

= 6.0 cm = 0.060 m

1

1

9

2. Hence, determine the height x of the wood as shown in Fig. 6.1. x = ……………. m [2] Since the amplitude of oscillation = 0.06 m,

equilibrium position is at 17 cm. x + x/2 = 0.17 x = 0.113 m or 11.3 cm x = 0.11 m OR taking the extreme depths x + x/2 + l = 0.23 x + x/2 + 0.06 = 0.23 x = 0.11 m

1 1

1 1

3. Determine the frequency of the oscillation. frequency = ……………. Hz [1] From graph,

T = 2.0 s

02

11

.Tf ==

f = 0.50 Hz

1

4. Hence, determine the speed of the block when h is 21.0 cm. speed = ……………. m s-1 [2] When h is 21.0 cm, displacement of the block is 21.0 – 17.0 = 4.0 cm

Using 220 xxωv −= ,

11

22

220

1401400

0400605002

2

−−==

−=

−=

sm.sm.v

..).)(π(v

xx)fπ(v

1 1

(iv) Surface water waves are then incident on the block and cause resonance in the

oscillation of the block. Damping is considered to be negligible.

1. Sketch a graph on Fig. 6.3, showing how h will vary due to the surface water waves. Label this graph R. [2]

10

Same shape with same period Higher amplitude due to resonance

1 1

2. As the block oscillates, oil is poured into the water. With reference to graph R,

describe and explain how the amplitude of the vertical oscillation changes. [3] Solution:

The amplitude of oscillation will decrease as the oil on the water will change the natural frequency of the block, hence resonance no longer occurs.

1 1 1

7 Fig. 7.1 shows a rectangular coil. The coil has 25 turns with dimensions of 15.0 cm

by 6.0 cm.

Fig. 7.1 (a) (i) Define magnetic flux. [1] The product of the magnetic flux density normal to the surface and the area of

the surface. 1 (ii) Initially, a uniform magnetic field of flux density 25 mT is at right angles to the plane

of the coil. Calculate the magnetic flux of the coil at this instance.

magnetic flux =…………….......Wb [1] =BA

=0.025×0.150×0.060 = 2.25 x 10-4 Wb (also accept -2.25 x 10-4) 1

15.0 cm coil of 25 turns

6.0 cm

11

(b) (i) State Faraday’s law and Lenz’s law of electromagnetic induction. [2]

Faraday's law of electromagnetic induction states that the magnitude of the induced electromotive force is directly proportional to the rate of change of magnetic flux linkage through the coil/solenoid. Lenz's law of electromagnetic induction states that the direction/polarity of the induced electromotive force is such that the induced current produces an effect that will oppose the change that causes it.

1

1 (ii) Fig. 7.2 shows the sinusoidal variation with time t of the magnetic flux density B that

is passing through the rectangular coil in part (a).

Fig. 7.2

Explain why the variation in magnetic flux density passing through the coil as shown in Fig. 7.2 leads to a generation of alternating e.m.f. [3]

As the sinusoidal magnetic flux density passing through the coil is changing, the

magnetic flux linkage through the coil is constantly changing. Hence, by Faraday's law, an e.m.f. is induced or generated in the coil. At times, the magnetic flux density is increasing while at other times it is decreasing. The polarity of the induced e.m.f. would be different when magnetic flux density increases as compared to the polarity of the e.m.f. as magnetic flux density decreases, thus generating an alternating current.

1

1

1 (iii) Show that the equation for the induced e.m.f. of the coil is

E=NAB0 2πTsin(2π

Tt)

where N is the number of turns in the coil, A is the area of coil, B0 is the maximum

magnetic flux density and T is the period of the change in magnetic flux density. [2] From Fig. 7.2, the magnetic flux density is given by,

cos 2

By Faraday's law and Lenz’ Law, the e.m.f. is given by

Ф , where is the magnetic flux linkage.

(1 mark awarded to the correct equation Ф with the working statement.)

cos 2

(1 mark awarded to for stating cos with substitution steps shown

explicitly)

1

1

B / mT

t / ms 6 14

25

- 25

2 10 4 8 12 16 18

12

Therefore,

2 sin 2

(iv) Hence, determine the maximum magnitude of the induced e.m.f. of the coil. maximum e.m.f. = ………………. V [2]

2 sin 2

Since the maximum magnitude of induced e.m.f. is when sin 1,

$%& 2 25 ( 0.150 ( 0.060 ( 0.025 (

., 4.42 V

1

1 (v) On Fig. 7.3, sketch a graph to show the variation with time of the e.m.f. induced in

the coil. [2] Solution:

Fig. 7.3

1 mark - Correct shape of graph (sin function, at least 2 waves) 1 mark - Labeling maximum and minimum E and period of waves.

(c) (i) The AC generated by the rectangular coil passes through a transformer. The primary coil of the ideal transformer has 15 turns. Assuming that the transformer is ideal, calculate the number of turns in the secondary coil if the value of the output root-mean-square potential difference at the secondary coil is 240 V.

number of turns =……………....... [3] Solution:

Root-mean-square voltage of input,

./$0 .√2

./$0 4.42√2 .2

32

.3.2

No. of turns in sec coil,

3 .3.2 2 240

4.42√2

( 15 240√24.42 ( 15

3 1150

1

1

1 (ii) In practice, transformers are not ideal, as energy is lost in the soft-iron core of the

transformer. Use Faraday’s law to explain why there are there energy losses in the soft-iron core

of a practical transformer. [2] According to Faraday's law, as the magnetic flux linkage in different parts of the

soft-iron core is constantly changing in the soft-iron core, e.m.f. are induced and hence electric currents are flowing in the soft iron core. (have to mention at least one current flowing in the core.)

1

E / V

t / ms

4.42

- 4.42

6 14 2 10 4 8 12 16 18

13

The currents induced in the soft-iron core will give rise to a joule heating effect as given by P = I2R, thus energy is lost as thermal energy in the soft-iron core.

1

(iii) A practical transformer has an input e.m.f. of 12 V while the secondary coil draws a

current of 0.25 A with a potential difference of 240 V. If the efficiency of the transformer is 81 %, determine the current in the primary coil.

current =…………….......A [2] For efficiency,

PoweroutputPowerinput ( 100%

PowerinsecondarycoilPowerinprimarycoil ( 100%

0.25 ( 24012A ( 100% 81%

Therefore, A 6.17A

1

1

8 (a) State what is meant by 1. isotope,

[2]

Solution: Two or more types of nuclei of the same element, having the same number of protons but different number of neutrons in their nuclei, hence possessing different mass numbers.

1 1

2. decay constant of a radioactive nucleus, providing a necessary condition for this

definition to hold.

[2]

Solution: It is the fraction of the total number of undecayed nuclei present which decays per unit time, provided the time for measurement is small (unit time is small).

1 1

(b) The isotopes Radium-224 , 22488 Ra, and Radium-226 ( 226

88 Ra) both undergo spontaneous

α-particle decay to form isotopes of Radon (Rn). The energies of the α-particles emitted from Radium-224 and Radium-226 are 5.68 MeV and 4.78 MeV respectively

(i) Write out the nuclear equations for the decays of Radium-226 and Radium-224 respectively.

decay of Radium-226

Ra88226

→ Rn+ α2486

222 (or He24 )

decay of Radium-224

Ra88224 → Rn+ α(or He)2

42

4

86

220

[2]

[2]

For each decay, 1 mark for correct nuclide, 1 mark for correct Z and A numbers. (ii) Suggest with a reason, 1. which of the two Radium isotopes has the larger decay constant.

[3]

An α-particle with greater energy implies that the parent nucleus from which it decayed from is a less stable nucleus. Therefore, the parent nucleus is

1

14

more likely to decay. Hence Radium-224 has a larger decay constant.

1 1

2. for Radium-224 decay, which nucleus has a larger binding energy per nucleon:

Radium-224 or its daughter nucleus, the Radon isotope.

[2]

The products of spontaneous radioactive decay are more stable than the decaying nucleus. Hence, the daughter nucleus, Radon-220 should have higher binding energy per nucleon.

1 1

(c) Radium-224 has a half-life of 3.6 days. (i) Calculate the decay constant of Radium-224, stating the unit in which it is

measured. decay constant = ………………………….

[2]

λ= ln2

t12E

= ln2

3.6

=0.193day-1

=2.23×10-6

s-1

1 1

(ii) Determine the activity of a sample of Radium-224 of mass 2.24 mg.

activity = ………………… Bq

[3]

N =mass of sample

molar mass×NA=

2.24×10-3

224×6.02×10

23

=6.02×1018

A=λN=2.23×10

-6×6.02×10

18

=1.3×1013

Bq

1

1 1

(d) Suggest two reasons why the results for the activity obtained from a sample may not

be accurate but can be precise.

[2]

Inaccuracy: Background radiation that was not eliminated from the calculations. OR Systematic errors (zero error, calibration errors) in the measuring instrument. Precision: Large population of the sample, with a short time to mrasure the activity,

1 1

-- End of Paper --