1 hour 15 minutes - Weeblyscore-in-chemistry.weebly.com/uploads/4/8/7/1/48719755/pjc_2015_prelim.pdf1.66u 10 27 kg rest mass of electron, 9.11u10 31 m e kg rest mass of proton, 1.67

2015/PJC/PHYSICS/9646 [Turn over

PIONEER JUNIOR COLLEGE JC2 Preliminary Examination

PHYSICS 9646/01 Higher 2 Paper 1 Multiple Choice 25 September 2015 1 hour 15 minutes Additional Material: Multiple Choice Answer Sheet

READ THESE INSTRUCTIONS FIRST

Write in soft pencil. Do not use staples, paper clips, highlighters, glue or correction fluid. Write your name, class and index number on the Answer Sheet in the spaces provided. There are forty 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.

This document consists of 21 printed pages.

Name Class Index Number

2

2013/PJC/PHYSICS/9646

Data

speed of light in free space, 81000.3 c m s–1

permeability of free space, 70 104 H m–1

permittivity of free space, 120 1085.8 F m–1

910361 F m–1

elementary charge, 191060.1 e C

the Planck constant, 341063.6 h J s

unified atomic mass constant, 271066.1 u kg

rest mass of electron, 311011.9 em kg

rest mass of proton, 271067.1 pm kg

molar gas constant, 31.8R J K–1 mol–1

the Avogadro constant, 231002.6 AN mol–1

the Boltzmann constant, 231038.1 k J K–1

gravitational constant, 111067.6 G N m2 kg–2

acceleration of free fall, 81.9g m s–2

3


Formulae

uniformly accelerated motion, 2

2

1atuts

asuv 222

work done on/by a gas, VpW

hydrostatic pressure, ghp

gravitational potential, r

Gm

displacement of particle in s.h.m., txx sin0

velocity of particle in s.h.m., tvv cos0

22

0 xx

mean kinetic energy of a molecule kTE2

3

of an ideal gas,

resistors in series, ...21 RRR

resistors in parallel, .../1/1/1 21 RRR

electric potential, r

QV

04

alternating current/voltage, txx sin0

transmission coefficient, T α exp 2kd where 2

28

h

EUmk

radioactive decay, )exp(0 txx

decay constant,

2

1

693.0

t

4


1 To determine the spring constant k, a student makes the following measurements of a

mass-spring system.

period, 2.8 0.1T s

mass, 50 1m g

The spring constant is determined using the equation 2m

Tk

.

What is the value of k and its associated uncertainty?

A 0.25 0.01 Nm−1

B 0.25 0.02 Nm−1

C 250 20 Nm−1

D 252 20 Nm−1

2 An object changes its velocity from 4 ms−1 due East to 7 ms−1 due South. What is its change in velocity? A 3 ms−1 at a direction of 30° East of South B 3 ms−1 at a direction of 30° West of South C 8 ms−1 at a direction of 60° South of West D 8 ms−1 at a direction of 60° South of East

5


3 The acceleration-time graph of an object moving in a straight line is as shown.

If the object starts its motion from rest, at which points on the graph does the object have the greatest velocity and the greatest displacement?

greatest velocity greatest displacement

A P Q

B P S

C Q R

D Q S

4 A cyclist takes off horizontally from a point 2.0 m above the surface of a puddle of water,

which is 2.5 m wide. What is the cyclist’s minimum take-off speed in order to clear the puddle of water? A 2.8 ms−1

B 3.9 ms−1

C 5.5 ms−1 D 6.1 ms−1

P

Q

R

S time

acceleration

0

2.0 m 2.5 m

puddle of water

6


5 A man in an elevator measures his weight with a weighing balance and finds that his

weight is reduced by 10%. Which of the following is a correct description of the motion of the elevator?

A Moving upwards and accelerating at 10

g.

B Moving upwards and decelerating at 10

g.

C Moving downwards and decelerating at 10

g.

D Moving downwards with uniform velocity.

6 A spacecraft of mass 44.0 10 kg is travelling at a constant velocity 1500 m s−1 with its

engine switched off. The engine is then switched on for 5.0 seconds to provide a thrust

of 51.0 10 N at right angles to the direction of travel of the spacecraft.

What is the angle of deviation of the spacecraft from its original direction of travel? A 0.092°

B 0.33° C 0.48°

D 0.92°

7 A uniform beam of length l is supported by two pivots A and B as shown below. The forces exerted on the beam by A and B are FA and FB respectively.

What is the ratio of B

A

F

F?

A 0.33 B 0.50 C 1.3 D 2.0

B A

FA FB

7


8 A metal ball is tied to two strings X and Y. String X is inclined at an angle to the vertical and has tension TX. String Y is horizontal and has tension TY.

When the angle is increased with string Y remaining horizontal, A both TX and TY increase.

B both TX and TY decrease. C TX increases but TY decreases.

D both TX and TY remain constant.

9 Using a spring balance, an object weighs 25 N in air and 10 N when completely

immersed in a liquid of density 900 kg m3. What is the density of the object?

A 3900 kg m

B 31500 kg m

C 32300 kg m

D 315000 kg m

10 A man is pushing a cabinet of mass 30.0 kg up a slope inclined at 30.0 to the horizontal with a constant speed of 3.00 m s-1. When the cabinet moves up the slope by 5.00 m, 600 J of heat is produced due to friction.

What is the work done by the man in moving the cabinet up the slope by 5.00 m?

A 1340 J

B 1470 J

C 1870 J

D 2070 J

X Y

8


11 A cart is being pulled by two motorised vehicles as shown in the diagram below. The cart

is moving with a constant speed of 16.0 m s in the horizontal direction and the power

required to maintain this motion is 1500 W. What is the tension in the cables connecting the vehicles and the cart?

A 0 N

B 125 N

C 144 N

D 250 N

12 A theme park ride has a number of chairs each suspended from a cable from the edge of

a circular canopy. The canopy revolves so that the chairs swing outwards as they move round in circles.

The canopy has radius 4.0 m. The cables are 5.0 m long. For safety, the angle θ of the

cables with the vertical must not exceed 60º. The diagram above shows a chair swung outwards as the canopy revolves at the maximum safe speed.

What is the maximum angular speed of rotation? A 1.42 rad s−1

B 1.62 rad s−1 C 1.98 rad s−1

D 2.61 rad s−1

cart

vehicle

vehicle

60

4.0 m

5.0 m

60º

9


13 Two planets X and Y have masses 2M and M respectively. The centres of the two planets are separated by a distance 3d. Point P is on the line joining the centres of the two planets and is at a distance d from planet Y as shown. The mass of each planet is

assumed to be concentrated at its centre. What is the magnitude of the gravitational field strength at point P due to the two

planets? A zero

B 22

GM

d

C 2

GM

d

D 2

3

2

GM

d

14 Mercury is 75.79 10 km away from the Sun and it takes 0.241 years for Mercury to

make one revolution around the Sun.

If Saturn is 91.43 10 km away from the Sun, what is the period of its orbit around the

Sun? A 1.43 years B 2.04 years

C 12.0 years

D 29.6 years

planet X planet Y

P

d

3d

10


15 Planets P and Q have equal mass. The radius of P is twice that of Q. The minimum kinetic energy needed by a body to escape from the surface of planet P is KP.

What is the minimum kinetic energy needed by the same body to escape from the

surface of planet Q?

A 0.25KP

B 0.5KP

C 2KP

D 4KP

16 A block of ice is heated at a constant rate by a 0.25 kW heater. The graph below shows how the temperature of the ice and subsequently water changes with time. Assume that all the energy supplied is used to heat the ice.

Given that the specific latent heat of fusion of water is 3.3 × 105 J kg−1, what is the

original mass of the block of ice? A 0.50 g B 8.3 g

C 0.50 kg

D 8.3 kg

time / minutes

temperature / ºC

0

−5

−10

−15

15

10

5

2 4 6 8 10 14 12

11


17 The temperatures of two beakers of water of different amounts are measured using the

same thermometer. The temperatures registered are both 35 ºC and are steady. The water in beaker B is then poured into the beaker A. Which one of the following statements is correct? A The final temperature of the water will be 35 ºC.

B The final temperature of the water will be less than 35 ºC since the water from beaker

B has less internal energy. C The final temperature of the water will be more than 35 ºC since the total volume of

water in beaker A has less internal energy. D Since the mass and the specific heat capacity of water are unknown, the final

temperature of the water cannot be determined. 18 An ideal gas is contained in a cylinder fitted with a piston, as shown.

In Process 1, the piston is moved inwards slowly so that the volume of the gas is

reduced slowly to half its initial volume. The work done on the gas is W1 and the final temperature of the gas is θ1.

In Process 2, the piston is moved inwards suddenly so that the volume of the gas is

reduced to half its initial volume. The work done on the gas is W2 and the final temperature of the gas is θ2.

Which of the following correctly describes work done on the gas and the final

temperature of the gas in Processes 1 and 2?

work done on gas final temperature of gas

A W1 is the same as W2 θ1 is the same as θ2

B W1 is the same as W2 θ1 is less than θ2

C W1 is less than W2 θ1 is less than θ2

D W1 is less than W2 θ1 is greater than θ2

beaker A beaker B

cylinder

piston gas

12


19 The frequency of oscillation of an object undergoing simple harmonic motion is f.

If the frequency of oscillation is reduced to3

f, what is the frequency of the variation of its

kinetic energy?

A 6

f B

3

f C

2

3

f D

4

3

f

20 A particle oscillates with simple harmonic motion along a straight line with amplitude A.

When the displacement of the particle from its equilibrium position is 2

A, its speed is u.

What is the speed of the particle when passing the equilibrium position?

A 2

3u

B 2 u

C 3 u

D 4 u

21 A sound wave of frequency 400 Hz is travelling in a gas at a speed of 320 m s1.

What is the phase difference between two points 0.20 m apart in the direction of travel of

the wave?

A 4

rad

B 5

2 rad

C 2

rad

D 5

4 rad

13


22 Two wave generators S1 and S2 produce water waves of wavelength 2.0 m. They are

placed 6.0 m apart as shown and are operated in phase. A sensor D which measures the amplitude of water waves is 7.0 m away from S1 as shown in the diagram below.

The shortest distance D could be moved along the straight line S1D in order to detect the largest amplitude of the resultant wave is

A 1.0 m towards S1.

B 3.0 m towards S1. C 1.0 m away from S1.

D 3.0 m away from S1. 23 When a two-slit arrangement is set up to produce interference fringes on a screen using

a monochromatic source of green light, the fringes are found to be too close to be clearly

observed.

Which of the following is a possible way to increase the separation of the fringes?

A Increase the width of each slit.

B Increase the distance between the two slits.

C Decrease the distance between the screen and the slits.

D Replace the light source with a monochromatic source of red light.

24 Light of wavelength 550 nm is incident normally on a diffraction grating having 400 lines per millimetre. What is the maximum number of bright fringes that can be observed?

A 4

B 5

C 8

D 9

S1 S2

D

6.0 m

7.0 m

14


25 Which of the following statements about electric potential is correct?

A Alternative units for electric potential are the joule and the volt.

B Unit potential gradient exists between any two points, if one joule of work is done in

transporting an electron between the points. C Two points in an electric field are at the same potential when a unit positive charge

placed anywhere on the line joining them remains stationary. D The electric potential due to a system comprising a point positive charge and a point

negative charge is given by the sum of the potentials due to the individual charges. 26 The diagram shows the electric field near a point charge and two electrons X and Y.

Which row of the table below describes the forces acting on X and Y?

direction of force magnitude of force on X

A radially inwards less than force on Y

B radially inwards greater than force on Y

C radially outwards less than force on Y

D radially outwards greater than force on Y

27 A uniform metal rod of length L has a square cross-section with sides of length W. It has

resistance R measured across its ends. The rod is subsequently stretched uniformly in a manner such that its volume V is kept constant throughout the process until its new resistance is 5R.

What is the new length of the rod?

A 5

L B 5L C 5L D 25L

X

Y

15


28 A cell of e.m.f 1.5 V and internal resistance 0.30 is connected to a variable resistor R as shown below. The current in the circuit is 0.50 A and passes though R for a time period of 4.0 minutes.

What is the total energy transferred by the battery? A 0.0031 J B 3.0 J

C 180 J

D 360 J

29 In the arrangement below, all the resistors have identical resistance R. The connecting

wires are assumed to have negligible resistance. What is the effective resistance across XY?

A 1

5R B

3

5R C

5

8R D

3

2R

0.30 1.5 V

R

X Y . .

. .

.

16


30 The circuit below shows three identical resistors each of resistance 1.0 connected to a

battery of e.m.f. 15 V with negligible internal resistance.

When an external circuit containing a similar resistor and cell of e.m.f. E is connected

across XY, the galvanometer shows a null reading.

What is the e.m.f. E of the cell?

A 3.8 V B 5.0 V C 10 V D 15 V

X Y

1.0

1.0

1.0 15 V

. .

. .

E

. .

1.0

17


31 A long solenoid of length l is connected to a cell of e.m.f. E and no internal resistance. The magnetic flux density at the centre of the solenoid is Bs.

The solenoid is subsequently cut to a length of 3

l and is reconnected to the same cell

shown below.

The magnetic flux density at the centre of a solenoid is equal to onI, where n is the

number of turns per unit length and I is the current through the coil.

What is the magnetic flux density at the centre of the shortened solenoid?

A 3

sB

B Bs

C 3Bs D 6Bs

l E

E

18


32 The graph shows the variation of the magnitude of a magnetic field B with time t for a

magnetic field that passes perpendicularly through the plane of a conducting rectangular frame.

0 2 4 6 Which of the following ranks the e.m.f. generated in the frame from the smallest to the largest at the sections P, Q and R of the graph? A RQP B QPR C PQR D RPQ

33 A sinusoidal potential difference of peak voltage Vo is applied across a resistor of

resistance R and produces heat at a mean rate P.

What is the mean rate of heat production when the rectangular wave potential difference, as shown above is applied across the same resistor?

A P B 5 P C 5 P D 10 P

R

Q P

V / V

3Vo

-Vo t / s

0.6 0.4 0.2

0

B / T

t / s

19


34 A beam of monochromatic light is incident on a metal surface.

If the wavelength is halved but the intensity remains unchanged, which of the following

statements is/are correct? (1) The photon energy is doubled. (2) The photon momentum is doubled. (3) The rate of incidence of photons on the surface is halved. A (1) only

B (1) and (2) only C (2) and (3) only

D (1), (2) and (3) 35 The work function of sodium, zinc and platinum are 2.46 eV, 4.31 eV and 6.35 eV

respectively. How many of the above metals will emit photoelectrons when indigo light is incident on

them?

A 0 B 1 C 2 D 3

36 The figure below shows an X-ray spectrum.

If the accelerating potential difference is decreased and the target element is replaced by

an element of higher atomic number, what will be the changes to the X-ray spectrum?

minimum wavelength wavelength of characteristic X-rays

A decrease decrease

B decrease increase

C increase decrease

D increase increase

intensity

wavelength

20


37 Why is laser light coherent?

A The excited electrons are in a metastable state.

B The system is in a state of population inversion.

C Stimulated emission causes the emitted photon and the incident photon to be of the

same phase. D Photons of the same energy as that of the incident photons are emitted when the

electrons transit from a higher energy level to a lower energy level. 38 Which statement about conduction of electricity in solids is correct?

A Free electrons are found in both the conduction band and the valence band.

B Some metals can conduct due to overlapping of their energy bands and there are no

energy gaps at all. C In an insulator, the valence band is completely filled and the band gap is small

compared to a semiconductor. D In a p-type semiconductor, the majority of the charge carriers are holes and the

minority of the charge carriers are electrons. 39 A nucleus has a nucleon number A, a proton number Z, and a binding energy B. The

masses of the neutron and proton are nm and pm , respectively, and c is the speed of

light. Which of the following is the correct expression for the mass of the nucleus?

A n pAm Zm B

B 2n p

BAm Zm

c

C n pA Z m Zm B

D 2n p

BA Z m Zm

c

21


40 A mixture of Iodine-131 (half-life 8 days and initial activity 131A ) and Iodine-132 (half-life 2

hours and initial activity 132A ) has a combined initial activity of 5000 Bq. After 12 days the

combined activity of the mixture is 1000 Bq.

What is the ratio 132

131

A

A?

A 0.23

B 0.43

C 0.57

D 0.77

End of paper

1

Answers to 2015 JC2 Preliminary Examination Paper 1 (H2 Physics)

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

1 Given that 2m

Tk

,

350 10

2.8 2k

0.2518k N m−1

Since 2

2

4 mk

T

,

2k m T

k m T

1 0.1

20.2518 50 2.8

k

0.02k N m−1

Therefore, 0.25 0.02k N m−1.

Answer: B

2 f iv v v

f iv v v

2 2 2

f iv v v

2 2

7 4v

8v m s−1

f

i

v

vtan

4

tan7

30

Answer: C

2

3 The change in velocity can be determined from the area under the acceleration-time

graph. Point Q is the turning point of the velocity-time graph and so is the maximum value. Beyond point Q, the velocity decreases but remains positive. Hence, S is a point of maximum displacement.

Answer: D

4 Taking downwards as positive,

21

2y y ys u t a t

212.0 9.81

2t

0.639t s

x xs u t

2.5 0.639xu

3.92xu m s−1

Therefore, minimum speed is 3.9 m s−1. Answer: B 5 Taking upwards as positive, N mg ma

0.9mg mg ma

10

ga

Therefore acceleration is downwards.

Elevator is moving upwards and decelerating at 10

g.

Answer: B

6 5

4

1.0 102.5

4.0 10a

m s−2

change in velocity = (2.5)(5.0) = 12.5 m s−1

angle of deviation 1 12.5tan 0.48

1500

Answer: C 7 Taking moments about the centre of the beam,

0.2

24

A

B

AB

F

F

lF

lF

Answer: D

3

8 Resolving the tension in string X vertically and horizontally,

cos

cos

sin

X

X

Y X

T W

WT

T T

From the above equations for TX and TY, it can be seen that when increases, both tensions will increase.

Answer: A 9 Upthrust on solid = Weight in air – Weight in liquid

3 3

3

3

25 10 15 N

15

15 900 9.81

1.70 10 m

25

252.55 kg

9.81

2.551500 kg m

1.70 10solid

U

Vg

V

V

mg

m

m

V

Answer: B 10 Work done = Heat produced + gain in gravitational potential energy

600 30 9.81 5.00sin30.0

1340 J

W

Answer: A 11 P = Fv

N 144

30cos2250

250

0.61500

T

T

NF

F

Answer: C 12 Resolving the forces acting on the chair. Vertically, cosT mg --- (1)

Horizontally,

2sinT mr --- (2)

(2)/(1):

4

2

2

1

tan

tan 9.81 tan60

5.0sin60 4.0

1.43 rad s

r

g

g

r

Answer: A

13 Using 2

GMg

R ,

Gravitational field strength at P,

2 2 2

2

22

G MGM GMg

d dd

Answer: B

14 Applying 3 2r T ,

Hence,

2 3

s s

m m

T r

T r

3

2 2ss m

m

rT T

r

3 3

92 2s

s m 7

m

1.43 100.241 29.6 years

5.79 10

rT T

r

Answer: D 15 For an object to escape from the Earth’s surface, loss in K.E. = gain in G.P.E.

21

2

GMmmv

R

1

KER

Q P

P Q

2K R

K R

Q P2K K

Answer: C 16 During melting, rate of heat supplied is

fmlP

t

So, 3

5

f

0.25 10 11 600.5 kg

3.3 10

Ptm

l

Answer: C

5

17 Since the temperature of the water in both beakers is the same, there will not be any net

transfer of heat when the water in beaker B is poured into beaker A. The temperature will remain the same and the water from both beakers will be at thermal equilibrium.

Answer: A 18 In Process 1, as the piston is moved inwards slowly, there will be no change in

temperature of the gas in the cylinder. Hence, this is a constant temperature process,

0T , 0U .

Applying the First Law of Thermodynamics, 0U , Q W . As it is a compression,

work is done on gas, i.e. W is positive. However, as the process takes place slowly, the

gain in energy due to the work done on the gas is lost to the surrounding.

In Process 2, as the piston is moved inwards suddenly, there is insufficient time for heat energy to leave nor be supplied to the gas in the cylinder. Hence, Q = 0.

Applying the First Law of Thermodynamics, U Q W , U W . As it is a

compression, work is done on gas, i.e. W is positive. Hence, U W is positive,

implying increase in internal energy and increase in temperature of gas. By drawing a p-V graph, it can be deduced that the work done on the gas in Process 1 is

lower than that in Process 2.

Answer: C 19 The frequency of the variation of kinetic energy is twice the frequency of oscillation.

If frequency of oscillation is 3

f, frequency of the variation of kinetic energy is

2

3

f.

Answer: C

20 At displacement = 2

A,

22 2 2

4

Au A

At displacement = 0, 2 2 2v A

Therefore 2

3v u

Answer: A

p

V 0

V2 V1

p2

p1

Process 1

Process 2

6

21 80.0400

320

f

v m

280

20

80

20

2

.

.

.

.

2

rad

Answer: C

22 To detect largest amplitude of wave, constructive interference must occur. For

constructive interference to occur, the path difference between S2D and S1D is n, where n is an integer.

When D is moved 1.0 m further away from S1,

Path difference = 868 22 = 2.0 = , where n is 1.

The other options do not satisfy the condition of path difference = n

Answer: C

23 a

Dx

The fringe separation is directly proportion to wavelength. Monochromatic red light has

longer wavelength than monochromatic green light.

Answer: D

24 nsind ,

sind

n

400

101 3d =2.5×106 m

Since sin < 1

544

10550

1052

1

9

6

.n

.n

dn

d

n

The highest order that is observable is 4. Hence, the maximum number of bright fringes

is 9.

Answer: D

7

25 Option A: Alternative units for electric potential are J C−1 and V. Incorrect.

Option B: Unit potential difference exists between any two points, if one joule of work is done in transporting one coulomb of charge between the points. Incorrect.

Option C: Two points in an electric field can be at the same potential, but a unit positive charge placed anywhere on the line joining them (equipotential line) will move in the direction of decreasing electric potential. Incorrect.

Answer: D 26 The point charge is negative as the electric fields point radially inwards. Hence the

electric force acting on X and Y points radially outwards. The further the electrons from the point charge, the weaker the magnitude of the electric force. So, the electric force acting on X is weaker than that on Y.

Answer: C

27 2V W L

2

LR

W

2

2

LR

W L

2L

RV

------------(1)

V is constant.

2

newnew

LR

V

2

5 newLR

V

------------(2)

Solving (1) and (2):

5newL L

Answer: B

28 Energy transferred EQ

E t I

1 5 0 5 4 0 60. . .

180 J

Answer: C 29 The circuit can be redrawn as shown below:

X Y . .

. . .

. Z

8

2

ZY

RR

R R

0.5R

1 1 1

XY ZYR R R R

1 1

0.5R R R

3

5XYR R

Answer: B

30 p.d. across XY 0.5

151.0 0.5

5 V

For balanced condition, E p.d. across XY

5 V

Answer: B

31 Magnetic flux density α I (since o and n are constant).

Reducing the solenoid length to one-third its original length will cause its resistance to correspondingly decrease to one-third its original value. Since the same battery is used,

current I through the solenoid will triple.

Hence magnetic flux density will triple. Answer: C 32 Magnitude of e.m.f. generated is directly proportional to rate of change of magnetic flux

linkage. Since the area of the frame is constant, the e.m.f. generated is directly proportional to the rate of change of magnetic flux density or gradient of the B- t graph. Gradient of the B-t graph is the smallest at Q followed by P and R in ascending order.

Answer: B

33 For sinusoidal p.d.,

P

VR

R

V

P

V

2

2

2V

2

0

2

0

0

rms

9

For rectangular p.d.,

P

P

V

VP

VV

'

rms

10

2

5

5

2

0

2

0

0

Answer D

34 (1) The photon energy is doubled.

' 20.5

hcE

hcE E

(2) The photon momentum is doubled.

' 20.5

hp

hp p

(3) The rate of incidence of photons on the surface is halved.

(0.5 )

' 0.5

N A

t hc

N A N

t hc t

I

I

Answer: D

35 max

hc

For sodium, 34 8

max 19

6.63 10 (3.00 10 )505

2.46(1.60 10 )

nm (green)

For zinc, 34 8

max 19

6.63 10 (3.00 10 )288

4.31(1.60 10 )

nm (UV)

For platinum, 34 8

max 19

6.63 10 (3.00 10 )196

6.35(1.60 10 )

nm (UV)

Therefore only sodium will emit photoelectrons when indigo light is incident.

Answer: B

36 For minimum wavelength, min

AC

hceV

If accelerating potential difference is decreased, min increases.

For characteristic X-rays, characteristic

hcE

10

If target element is replaced by an element of higher atomic number, E increases,

characteristic decreases.

Answer: C 37 For production of laser, the photon emitted through stimulated emission is in phase with the incident photon which makes laser coherent. Answer: C 38 Answer: D

39 Binding energy = 2cm

2

2

2

n p nucleus

n p nucleus

nucleus n p

B A Z m Zm m c

BA Z m Zm m

c

Bm A Z m Zm

c

Answer: D

40 ).(..........AA 15000132131

770

2

15

21

210002

1

10002

1

2

1

131

132

51

131

132131

51

131

22412

132

812

131

.A

A

A

AA

:/

)........(..........A

AA

.

.

Answer: D

[Turn over


PHYSICS 9646/02 Higher 2 Paper 2 Structured Questions 17 September 2015 1 hour 45 minutes Candidates answer on the Question Paper. No Additional Materials are required.


Write your name, class and index number on all the work you hand in. Write in dark blue or black pen. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. 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 / 9

2 / 9

3 / 9

4 / 9

5 / 9

6 / 15

7 / 12

Total / 72

2

Data

speed of light in free space, 81000.3 c ms–1

permeability of free space, 70 104 Hm–1

permittivity of free space, 120 1085.8 Fm–1

910361 Fm–1


the Planck constant, 341063.6 h Js




molar gas constant, 31.8R JK–1 mol–1


the Boltzmann constant, 231038.1 k JK–1

gravitational constant, 111067.6 G Nm2 kg–2

acceleration of free fall, 81.9g ms–2

3

[Turn over

Formulae


2

1atuts

asuv 222




Gm



22

0 xx


3

of an ideal gas,




QV

04


transmission coefficient, T α exp 2kd where 2

28

h

EUmk


decay constant,

2

1

693.0

t

4

1 A student takes measurements to determine the acceleration of free fall. He throws a ball

vertically upwards with a velocity of 25 ms−1 from ground level. The variation with time t of the vertical velocity v of the ball is shown in Fig. 1.1.

Fig. 1.1

(a) Suggest why drawing a best fit line helps in reducing random errors.

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

v / ms−1

t / s

5

[Turn over

(b) Using Fig. 1.1, explain

(i) how it can be deduced that air resistance is present in this experiment,

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

(ii) how the magnitude of the acceleration of free fall can be determined.

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

(c) State and explain how the time taken for the upward motion ut compares with the

time taken for the downward motion dt .

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

(d) The air resistance airF acting on a ball of radius r moving with speed v in air is given

by

6airF krv ,

where k is a constant.

The mass of the ball thrown by the student is 10 g with radius 5.0 cm.

(i) Use Fig. 1.1 to calculate the magnitude of the maximum air resistance airF acting

on the ball.

maximum airF = ........................................ N [2]

6

(ii) Hence, determine the value of k, together with its base units.

k = ........................................ [2]

7

[Turn over

2 A rock climber of mass 70 kg is climbing a vertical cliff as shown in Fig. 2.1. He ties

himself to a secure point on the cliff by a rope of unstretched length 15 m. When he has climbed a height of 15 m above the secure point, he slips and falls. The elastic rope

obeys Hooke’s law with a force constant of 200 N m1.

Fig. 2.1

(a) Calculate the speed of the falling climber just before the rope starts to stretch.

speed = ........................................ m s1 [2] (b) Calculate the maximum distance below the secure point that the climber falls.

distance = ........................................ m [3]

15 m

secure point

8

(c) Due to the elasticity of the rope, the climber oscillates up and down until he comes to

a complete stop. Calculate the distance below the secure point when the climber stops moving.

distance = ........................................ m [2]

(d) Climbing ropes must be able to undergo large extensions when stretched. Suggest

why this is so.

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

9

[Turn over

3 (a) (i) Explain what is meant by intensity of a wave.

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

(ii) Fig. 3.1 shows a skyrocket which explodes 100 m above the ground. Three

observers are spaced 100 m apart, at positions A, B and C. The observer at A is directly below the explosion.

Fig. 3.1 Calculate the ratio of the sound intensity heard by the observer at A to the sound

intensity heard by the observer at C.

ratio of the sound intensity = ........................................ [3]

skyrocket

A B C

100 m

100 m 100 m

10

(b) Stationary sound waves can be formed in bottles and pipes.

(i) State the conditions for the establishment of a well-defined stationary wave.

.................................................................................................................................. ........................................................................................................................... [1] (ii) A bottle containing water resonates as air is blown across its top, which is open.

Explain how the fundamental frequency changes as the level of water in the bottle decreases.

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

(iii) A pipe open at both ends resonates at a fundamental frequency f1. When one

end is covered and the pipe is again made to resonate, the fundamental

frequency is f2. Determine the ratio

2

1

f

f.

ratio = ........................................ [2]

11

[Turn over

4 (a) Define magnetic flux density. ....................................................................................................................................... ................................................................................................................................. [1]

(b) A cross-sectional plan view of a solenoid of length 0.300 m is shown in Fig. 4.1.

(i) Based on the indicated direction of current flow in the solenoid, sketch on Fig. 4.1 the magnetic flux pattern within the solenoid. [1]

(ii) The magnetic flux density B (in tesla) inside the solenoid and parallel to its axis is

given by the expression

0B n I

where 0 is the permeability of free space, n is the number of turns per metre

length of the solenoid and I is the current (in amperes) in the solenoid.

Given that there are 900 turns in the solenoid and a current of 4.0 A flows

through the solenoid, calculate the magnetic flux density inside the solenoid.

magnetic flux density = ........................................ T [2]

axis of solenoid

× × × × × × × × × × × × × × × × × × × × × × × × × × × × current into the plane of paper

Fig. 4.1

current out of the plane of paper

0.300 m

. . . . . . . . . . . . . . . . . . . . . . . . . . .

12

(iii) State and explain the effect on the magnetic flux density in (b)(ii) when a ferrous

core is inserted into the solenoid. ..................................................................................................................................

..................................................................................................................................

........................................................................................................................... [2] (c) A current balance with a rectangular frame PQRS pivoted at R and S is shown in Fig.

4.2 (a) and Fig. 4.2 (b). The segments RS and XY are non-conducting. Segment SPQR is a metallic wire and a current flows through it.

The frame PQRS is placed inside the same solenoid in (b) such that it lies within the

magnetic field of the solenoid. When a rider of mass 2.0 g is placed at Y, the frame PQRS is maintained in a horizontal plane and is coaxial with the solenoid.

(i) On Fig. 4.2 (a), draw an arrow to show the direction of current flow in the segment PQ for the plane of the frame to be horizontal. [1]

Fig. 4.2 (a) plan view (not to scale)

rider

Y

X axis of solenoid

P S

× × × × × × × × × × × × × × × × × × × × × × × × × × × ×

current into the plane of paper

R Q


. . . . . . . . . . . . . . . . . . . . . . . . . . .

0.270 m

rider

Q

pivot

R Y

Fig. 4.2 (b) side view (not to scale)

0.030 m

13

[Turn over

(ii) Given that segment PQ has a length of 0.080 m, use your answer in (b)(ii) to

determine the magnitude of current flowing in segment PQ. You may assume that the frame PQRS and segment XY have negligible mass.

current = ........................................ A [2]

14

5 (a) Fig. 5.1 shows some of the energy levels in a sodium atom. An electron moving from

a higher level to a lower level emits electromagnetic radiation.

Fig. 5.1 (i) State the equation relating the change in energy E to the wavelength of the

emitted radiation.

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

(ii) The emission spectrum of the sodium atom has two closely-spaced lines at

589.0 nm and at 589.6 nm. They are produced when electrons make transitions from two closely-spaced energy levels at n = 2 to the ground state.

1. On Fig. 5.1, draw an arrow to show the transition that emits radiation of

wavelength 589.6 nm. [1] 2. Calculate the energy of the n = 2 level which emits radiation of wavelength

589.6 nm.

energy = ........................................ eV [2]

−5.138 eV n = 1 (ground state)

n = 2

n = 3

n = 4

15

[Turn over

(b) The scanning tunnelling microscope (STM) applies the phenomenon of quantum

tunnelling in obtaining atomic-scale images of surfaces. Electrons can tunnel through a potential barrier, which is the vacuum gap between the STM probe and the specimen surface.

(i) Fig. 5.2 shows the wave function of an electron on the left side of a potential

barrier. Draw the wave function of the electron within and to the right side of the barrier.

Fig. 5.2 [2]

(ii) One million electrons with kinetic energy 4.50 eV are incident on a barrier of width

0.200 nm and height 10.0 eV. Calculate the number of electrons that can tunnel through the barrier.

number of electrons = ........................................ [3]

potential barrier

16

6 A body which has a temperature above that of its surroundings may lose thermal energy

by conduction, convection and radiation. If the body is placed in air, natural convection currents may be established. To increase the rate of loss of thermal energy, air may be blown past the body. An example is the air-cooled engine of a motorcycle, shown in Fig. 6.1. This type of cooling is known as forced convection.

Fig. 6.1

In order to investigate forced convection, a student obtained a dismantled motorcycle engine and set up the apparatus shown in Fig. 6.2. The engine consists of a metal cylinder fitted with metal fins.

Fig. 6.2

The cylinder of the engine is covered with a lid and initially contained hot oil which is stirred continuously. A constant current of air is directed towards the cylinder. A reading of the temperature θ of the oil is taken every minute as the oil and cylinder is cooled by

the air. The variation with time t of temperature θ is shown in Fig. 6.3.

Room temperature is found to be 25.0 ºC. The student suggests that the rate of decrease in temperature depends on the excess

temperature θE, which is the temperature difference between the engine and its surroundings.

air-cooled engine

thermometer stirrer

oil

metal cylinder

current of air

metal fins

17

[Turn over

Fig. 6.3

(a) Suggest how the metal fins covering the motorcycle engine help in the cooling process. ........................................................................................................................................ ........................................................................................................................................

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

time t / min

temperature θ / ºC

18

(b) Describe, without calculation, the variation with time t of the temperature θ of the oil.

........................................................................................................................................

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

(c) (i) Use Fig. 6.3 to complete Fig. 6.4.

rate of decrease in temperature / K s−1 excess temperature θE / K

0.025 9.0

0.045 16.0

0.068 24.5

............. 37.0

0.142 52.0

0.181 64.0

Fig. 6.4

[3]

19

[Turn over

(ii) Some data from Fig. 6.4 are used to plot the graph of Fig. 6.5.

Fig. 6.5 The relation between the rate of decrease in temperature and excess

temperature θE follows the expression

E

dA

dt

,

where A is a constant. On Fig. 6.5,

1. plot the point corresponding to θE = 37.0 K, [1]

2. draw the line of best fit for all the points. [1]

excess temperature θE / K

rate of decrease in temperature / K s−1

20

(iii) Use the line drawn in (ii) to determine the constant A, with its unit.

A = ........................................ [2]

(iv) Using Fig. 6.5, determine the rate of decrease in temperature when the

temperature of the oil is 55.0 ºC. rate of decrease in temperature = ........................................ K s−1 [2] (v) In theory, the line of best fit in (ii) should pass through the origin of the graph.

Explain why this is so. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ........................................................................................................................... [2]

(d) A motorcycle is fitted with an engine which cools in a similar way to the one used in the experiment. The rider adjusts the engine so that it maintains a constant output power, independent of the speed of the motorcycle. Suggest why the engine is more likely to overheat when travelling uphill than when on level ground. ........................................................................................................................................ ........................................................................................................................................ ........................................................................................................................................

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

21

[Turn over

7 A student decides to investigate the electrical properties of glass. One of the electrical

properties which he chooses to investigate is the resistivity of glass.

You are provided with samples of a particular type of glass in the form of a rectangular block of different length, breadth and thickness as illustrated in Fig. 7.1.

Fig. 7.1

Design an experiment to determine the resistivity of glass. You may use any of the other equipment usually found in a Physics laboratory. You should draw a labelled diagram to show the arrangement of your apparatus. In your account you should pay particular attention to (a) the identification and control of variables,

(b) the equipment you would use,

(c) the procedure to be followed,

(d) how the resistivity of the glass would be measured,

(e) any precautions that would be taken to improve the accuracy and safety of the

experiment.

Diagram

thickness

breadth

length

22

.................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................

....................................................................................................................................................

....................................................................................................................................................

.................................................................................................................................................... ....................................................................................................................................................

.................................................................................................................................................... ....................................................................................................................................................

....................................................................................................................................................

....................................................................................................................................................

.................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................

.................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................

23

[Turn over

.................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... ............................................................................................................................................ [12]

End of paper

1

Answers to 2015 JC2 Preliminary Examination Paper 2 (H2 Physics) Suggested Solutions:

No. Solution Remarks

1(a) Drawing a best fit line minimises the effects of random errors since the best fit line drawn has a weighted mean, where the variation from the actual value may cancel out.

[1] for any reasonable explanation

1(b)(i)

The gradient of the graph represents the acceleration, which is changing in value. This suggested the presence of air resistance, since weight cannot be the only force acting on the ball.

[1]

1(b)(ii) The magnitude of acceleration due to free fall can be determined from the gradient of the graph at 1.75 s. At this time, velocity is zero which means the only force experienced by the ball is its own weight.

[1]

1(c) u dt t .

At a given speed, the net retarding force ( airW F ) when the

body is moving upwards is greater than the net accelerating

force ( airW F ) when the body is moving downwards.

[1] for time comparison [1] for explanation

1(d)(i) At 0t s, the ball is moving with the largest speed and

hence experiences the greatest air resistance. gradient of tangent at 0t s,

25

1.2

Since airW F ma ,

25

0.010 9.81 0.0101.2

airF

0.110airF N

[1] for gradient at

0t s

[1] for air resistance

1(d)(ii) 6airF krv

0.110 6 0.050 25k 34.68 10k kgm−1 s−1

[1] for value [1] for units

2(a)

2

1

By the principle of conservation of energy,

1

2

2

2 9.81 30 24.3 m s

mv mgh

v gh

v

[1] for correct application of principle [1] for correct answer

2

2(b)

2

2

2

2

By the principle of conservation of energy,

1

2

170 9.81 30 70 9.81 200

2

100 686.7 20601 0

686.7 686.7 4 100 20601

200

18.2 m

15 18.2 33.2 m

mg h x kx

x x

x x

x

d

[1] for correct application of principle [1] for correct value of x

[1] for correct value of distance

2(c)

Equilibrium of forces is reached when climber comes to a stop

70 9.81 200

3.43 m

15 3.43 18.4 m

mg kx

x

x

d

[1] for correct equation [1] for correct answer

2(d) If the rope can undergo large extensions when stretched, the maximum tension is lower. Hence the force exerted on the climber and acceleration experienced is smaller and less likely to cause injury.

[1] for correct explanation [1] for correct conclusion

3(a)(i) The intensity of the wave is the power carried per unit area perpendicular to the direction of travel of the wave.

[1]

3(a)(ii) Let P be the power from the source

Intensity at A = 21004

P

Distance from source to point C = 22 100200

= 223.607 m

Intensity at C = 26072234 .

P

Ratio = 21004

P÷

26072234 .

P

=

005100

6072232

2

..

[1] Correct expression for intensity at A [1] Correct expression for intensity at C [1] Correct final answer OR [1] Correct relationship between intensity and distance from

source 2

1

rI

[1] Correct substitution into formula [1] Correct final answer

3

3(b)(i) Two progressive waves approach each other in the opposite direction with the same amplitude and frequency (speed).

[1]

3(b)(ii)

As the level of fluid in the bottle decreases, the air column in the bottle increases. The wavelength of the stationary wave increases. Since speed of sound in the air column is constant, the resonant frequency decreases.

[1] Wavelength increases [1] Resonant frequency decreases (with explanation).

3(b)(iii)

When both ends are open,

L

L

2

2

1

1

1

When one end is covered,

L

L

4

4

1

2

2

Since speed of sound, v is constant,

L

vf

21

L

vf

42

22

1 f

f

[1] Correct wavelength expressions [1] Correct final answer

4(a) Magnetic flux density is defined as the magnetic force acting on a conductor of unit length, carrying unit current, placed perpendicularly to the magnetic field.

[1]

4(b)(i)

[1] for correct direction and uniform line spacing within solenoid

4(b)(ii)

0B n I

7 9004 10 4 0

0 300.

.

0 015 T.

[1] for correct substitution [1] for correct answer

× × × × × × × × × × × × × × × × × × × × × × × × × × × × current into the plane of paper


. . . . . . . . . . . . . . . . . . . . . . . . . . .

4

4(b)(iii)

The magnetic field generated by the current in the solenoid partially aligns the domains of magnetic dipoles in the ferrous core, thus creating a much stronger field in the same direction.

[1] for correct reason. [1] for correct effect on magnetic flux density.

4(c)(i)

[1] for current in PQ direction

4(c)(ii)

Take moment about pivot line RS

anti - clockwise moment =clockwise moment

sin90 PQ PQ QR XYB L L mg LI

30 01508 0 080 0 270 2 0 10 9 81 0 030. . . . . . PQI

1 8 A.PQI


5(a)(i) hcE

where h is the Planck constant and c is the speed of light in free space.

[1]

5(a)(ii)1.

[1]

5(a)(ii)2.

34 819

2 9

6.63 10 (3.00 10 )( 5.138) 1.60 10

589.6 10E

2 3.030E eV


−5.138 eV n = 1

n = 2

n = 3

n = 4

Y

X

P S

× × × × × × × × × × × × × × × × × × × × × × × × × × × ×

current into the plane of paper

R Q


. . . . . . . . . . . . . . . . . . . . . . . . . . .

5

5(b)(i)

[1] for correct wave function within the barrier [1] for correct wave function after the barrier

5(b)(ii)

2 31 19

234

8 9.11 10 10.0 4.50 1.6 10

6.63 10k

101.2000 10k m−1

10 92(1.2000 10 )(0.200 10 )T e

38.230 10T

Number of electrons = 3 6(8.230 10 )(1 10 ) 8230

[1] for correct substitution in k

[1] for correct substitution in T and correct answer for T

[1] for correct answer

6(a)

The metal fins covering the outside of the metal cylinder increase the effective surface area facing the incoming current of air. This enables the heat energy from the hot oil to be transferred and carried away by the current of air by forced convection, hence increasing the rate of cylinder cooling.

[1] for effective surface area to help increase rate of cooling by convection

6(b) The temperature of the oil decreases with time at a decreasing rate.


sinusoidal; no change in wavelength

exponentially

decreasing

same height

smaller

amplitude

6

6(c)(i)

Excess temperature = 37.0 K Temperature of oil = 62.0 ºC From the tangent drawn in Fig. 6.3,

Rate of decrease in temperature 188 500.106 K s

6.0 60

[1] for tangent [1] for gradient expression [1] for correct answer (±10 % range)

7

6(c)(ii)1.

[1] for correct plot

6(c)(ii)2. [1] for best fit line

6(c)(iii) From the expression E

dA

dt

, A is the gradient of the

graph in Fig. 6.5.

Hence, 10.168 0.0280.00280 s

60.0 10.0A

[1] for the correct calculation of A

[1] for the correct unit

6(c)(iv) Temperature of oil = 55.0 ºC Excess temperature, θE = 30.0 K From Fig. 6.5, rate of decrease in temperature = 0.084 K s−1

[1] for θE = 30.0 K


6(c)(ii)1.

6(c)(ii)2.

8

6(c)(v) When the excess temperature θE is zero, the temperature θ

of the oil in the metal cylinder and its surrounding is the same. The oil in the metal cylinder is in thermal equilibrium with its surrounding.

According to the relation E

dA

dt

, rate of decrease in

temperature will be zero when θE is zero. Hence, the line of best fit in (ii) should pass through the

origin of the graph.

[1] for correct explanation [1] for correct answer (No marks awarded if no explanation is given)

6(d) When the motorcycle maintains a constant output power travelling uphill, it will be travelling at a lower constant speed compared to when travelling on a level ground. So, the average speed of the cooling air flowing through the air-cooled engine will be lower, resulting in lower rate of engine cooling. Hence, the engine is more likely to overheat when travelling uphill than when on level ground.

[1] for lower constant speed of motorcycle [1] for lower rate of cooling

7 Aim : To determine the resistivity of glass.

Independent variable : Cross-sectional area, A which is the

product of the length and breadth

of the block. Length and breadth

are measured using vernier

callipers or micrometer screw

gauge

Dependent variable : Resistance, R determined by

using a voltmeter and

microammeter.

Controlled variable:

- Thickness of glass block is kept constant

- Keep temperature of glass constant

Diagram :

[1] for stating A

[1] for stating R

[1] for

controlled variable

[1] for high voltage

supply, variable

resistor, micro-

ammeter and

voltmeter

[1] for metal plates

in contact with the

glass block

t

metal plate metal plate

V

A

glass

High voltage supply

9

Procedure :

a) Set up the apparatus as shown above.

b) Measure the thickness t of the glass using a

micrometer screw gauge or a pair of vernier

callipers.

c) Measure the length L and breadth B of the glass

block using a micrometer screw gauge or a pair of

vernier callipers and hence determine the cross-

sectional area A which is the product of the length

and breadth.

d) Use metal plates to make contact with the largest

cross-sectional area of the glass block so that a

larger current will flow across the glass.

e) Determine the resistance R, using the relation R =I

V

where V is the potential difference across the

thickness of the glass block as read by the voltmeter

and I is the current through the thickness of the glass

block as read by the microammeter,

f) Repeat steps (c) to (e) to obtain 6 sets of readings

for R, A(=LB) and A

1, with t kept constant.

g) Using the relation R = A

t, plot a graph of R against

A

1 where gradient of the graph = t and

resistivity = t

gradient.

OR

Plot Analysis

R against A

t

= gradient

R

1 against A =

tgradient

1

R

t against A =

gradient

1

Further details :

Precautions to improve accuracy :

- Take more than one reading of cross-sectional area and

find the average of the readings.

- Ensure good contact between circuit and glass by having

[1]

[1]

[1]

[1]

[1]

[1] for one

appropriate

precaution to

improve accuracy

10

metal plates cover the whole cross-sectional area of the

glass block.

- Ensure the metal plates are not too large as compared to

the cross-sectional area as to cause short-circuiting.

- Readings on the voltmeter and microammeter should be

taken quickly so as to keep the temperature of glass

constant.

- Keep the glass dry before performing the experiment as

water conducts electric current.

Safety precautions :

- Switch off the high voltage supply before changing the

sample of glass block in the circuit to prevent

electrocution..

- Use rubber gloves / rubber boots when switching off the

high voltage supply.

- Wear thick gloves in handling glass to prevent cuts by

glass.

[1] for one

appropriate safety

precaution



PHYSICS 9646/03 Higher 2 Paper 3 Longer Structured Questions 23 September 2015 2 hours Candidates answer on the Question Paper. No Additional Materials are required.


Write your name, class and index number on all the work you hand in. Write in dark blue or black pen. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Section A Answer all questions.

Section B Answer any two questions.

You are advised to spend about one hour on each section. 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 / 8

3 / 8

4 / 8

5 / 8

6 / 20

7 / 20

8 / 20

Total / 80

2


Data

speed of light in free space, 81000.3 c ms–1

permeability of free space, 70 104 Hm–1

permittivity of free space, 120 1085.8 Fm–1

910361 Fm–1


the Planck constant, 341063.6 h Js




molar gas constant, 31.8R JK–1 mol–1


the Boltzmann constant, 231038.1 k JK–1

gravitational constant, 111067.6 G Nm2 kg–2

acceleration of free fall, 81.9g ms–2

3


Formulae


2

1atuts

asuv 222




Gm



22

0 xx


3

of an ideal gas,




QV

04


transmission coefficient, exp 2T kd where 2

28

h

EUmk


decay constant,

2

1

693.0

t

4


Section A

Answer all questions in this section. 1 (a) Explain what is meant by gravitational field strength.

........................................................................................................................................ ........................................................................................................................................

................................................................................................................................. [1] (b) The variation with distance from the surface of Pluto of the gravitational potential is

shown in Fig. 1.1.

Fig. 1.1

distance from surface of Pluto / 106 m

gravitational potential

/ 105 J kg−1

5


(i) Use Fig. 1.1 to determine the magnitude of the gravitational field strength at the surface of Pluto.

gravitational field strength = ................................ N kg−1 [2]

(ii) A meteorite hits the surface of Pluto and ejects a lump of ice of mass 112 kg vertically upwards with velocity 680 m s−1.

Use Fig. 1.1, or otherwise, determine the maximum height reached by the ice above the surface of Pluto, assuming no dissipative forces.

maximum height = ................................... m [3] (c) In 2015, the New Horizons space probe passes between Pluto and its moon, Charon,

as shown in Fig. 1.2. Pluto has a mass seven times that of Charon and their separation is 1.96 × 107 m. Pluto and Charon may be considered to be point masses with their masses concentrated at their centres.

Fig. 1.2 (not to scale)

1.96 × 107 m

d

Charon Pluto space probe

6


Calculate the distance d from the centre of Pluto where the resultant gravitational

force acting on the probe is zero. d = .................................... m [2]

7


2 Fig. 2.1 shows a 0.400 kg block attached to a spring with spring constant 18.5 N m−1 at

its equilibrium position. By pulling the block to the lowest point and then releasing it, the block undergoes free oscillations.

Fig. 2.1 (a) Explain what is meant by free oscillations.

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

(b) The period T of the oscillations is given by

2m

Tk

,

where m is the mass of the block and k is the spring constant of the spring. (i) Given that the block has a speed of 1.40 m s−1 when passing the equilibrium

position, determine the amplitude of the oscillations.

amplitude = ........................................ m [2]

block

highest point

lowest point

equilibrium position

spring

8


(ii) Calculate the maximum tension in the spring during the oscillations.

maximum tension = ........................................ N [3] (c) (i) On Fig. 2.2, sketch a graph to show the variation with displacement x of the

velocity v of the block. Label the graph with appropriate numerical values.

Fig. 2.2

[1]

(ii) The block is now replaced with one of mass larger than 0.400 kg and pulled down by the same amplitude calculated in (b)(i).

On Fig. 2.2, sketch a second graph to show the variation with displacement x of the velocity v of the block with the larger mass.

Label the second graph Z. [1]

v / m s−1

x / m 0

9


3 Circuits M and N are each set up using a battery of four identical cells as shown in

Fig. 3.1. The voltmeter readings for the respective circuits are as shown. The voltmeters in both circuits are identical and can be assumed to be ideal. YZ is a uniform resistance

wire of length 1.0 m and resistance 3.5 . The resistance of resistor R is 3.0 .

(a) Explain why the voltmeter in circuit M records a larger reading than the voltmeter in

circuit N. ........................................................................................................................................ ........................................................................................................................................ ........................................................................................................................................

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

(b) Calculate the internal resistance r of a single cell.

r = ........................................ [2]

Fig. 3.1

.

. 3.0

circuit M

4.0 V V

Y

Z

R

circuit N

3.0 V V

.

.

10


(c) One of the cells in the battery is reversed in both circuits M and N as shown in

Fig. 3.2. The voltmeter readings are now different.

For circuit N, determine for the new arrangement,

(i) the current drawn from the battery,

current = ........................................ A [2]

(ii) the efficiency of power transfer of the battery.

efficiency = ........................................ % [2]

.

. 3.0

V

circuit M

Y

Z

R

.

. circuit N

V

Fig. 3.2

11


4 (a) Fig. 4.1 shows a long vertical wire carrying a constant current. A small coil of wire

with its horizontal axis perpendicular to the vertical wire, is placed near the vertical wire.

Fig. 4.1 A sensitive voltmeter is connected to the small coil of wire. State (i) if the voltmeter registers a reading and justify your answer, ........................................................................................................................................ ........................................................................................................................................

........................................................................................................................................ ................................................................................................................................. [3]

(ii) two different ways in which an e.m.f. may be induced in the coil.

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

long vertical

wire

small coil of wire direction of current

axis of coil

12


(b) Fig. 4.2 shows a simple iron-cored transformer.

Fig. 4.2

Explain why the e.m.f. induced in the secondary coil is not in phase with the current in the primary coil.

........................................................................................................................................ ........................................................................................................................................

........................................................................................................................................ ................................................................................................................................. [3]

load

a.c. supply

primary

coil secondary

coil

~

13


5 (a) (i) Explain what is meant by doping and state its effect on the conductivity of a

semiconductor.

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

(ii) Use band theory to distinguish between p-type and n-type doping.

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

(b) Fig. 5.1 shows p-type and n-type semiconductor materials placed together to form a

junction.

Fig. 5.1

(i) On Fig. 5.1,

1. use the symbols + and − to label the charges in the depletion region for the

p-type and n-type sides of the junction. [1] 2. draw the symbol for a battery, connected so as to decrease the width of the

depletion region. [1]

p-type material n-type material

14


(c) Fig. 5.2 shows the variation with potential difference V of current I for a

semiconductor diode.

Fig. 5.2

(i) State how the resistance of the diode changes with temperature.

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

(ii) Explain why I remains zero for very small values of V.

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

I

V

0

0

15


Section B

Answer two questions from this section. 6 (a) State the relation between force and momentum.

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

(b) Fig. 6.1 shows a stream of water projected from a hose of radius 0.800 cm, and

travelling at speed v towards a stationary 2.50 kg container resting on a smooth floor. The stream of water strikes the container at an angle of 60.0° to the vertical and falls vertically down without splashing.

Fig. 6.1 (i) Use Newton’s laws of motion to explain why the water exerts a horizontal force

on the container.

..................................................................................................................................

..................................................................................................................................

..................................................................................................................................

.................................................................................................................................. ........................................................................................................................... [3]

(ii) Calculate the speed of the water v if the container’s average acceleration is

1.10 m s−2. The density of water is 1000 kg m−3.

v = ........................................ m s−1 [3]

container

stream of water

60.0° v

hose

16


(iii) As the container accelerates to the left, a toy that is hanging inside the container

by a string makes at an angle α to the vertical, as shown in Fig. 6.2.

Fig. 6.2

Calculate angle α.

α = ........................................ ° [2] (iv) The container collides with a barrier and stops suddenly. The string breaks and

the toy falls onto the floor of the container. Fig. 6.3 shows the velocities of the toy before and after impact with the floor. Fig. 6.4 shows the variation with time of the resultant force on the toy during the impact.

Fig. 6.3

floor of container

25.0° 20.9°

3.18 m s−1 3.31 m s−1 toy

α

17


Fig. 6.4

1. Calculate the magnitude of the average resultant force on the toy.

average resultant force = ........................................ N [1] 2. Calculate the magnitude of the change in velocity of the toy.

change in velocity = ........................................ m s-1 [2]

0.0

resultant

force / N

2.0

4.0

6.0

8.0

10.0

12.0

40

10 20 30 0 50

t / ms

18


3. Calculate the mass of the toy.

mass = ........................................ kg [2] (c) Fig. 6.5 shows a 3.50 g bullet fired horizontally at two blocks resting on a smooth

surface. The bullet takes 42.13 10 s to pass through the first block of mass 1.20 kg.

It then embeds itself in the second block of mass 1.80 kg. The blocks move off with speeds of 0.630 m s−1 and 1.40 m s−1 respectively as shown in Fig. 6.5.

Fig. 6.5

(i) Neglecting the mass removed from the first block by the bullet, calculate the bullet’s original speed u and its speed v immediately after it emerges from the first

block.

u = ........................................ m s−1

v = ........................................ m s−1 [4]

0.630 m s−1 1.40 m s−1

1.20 kg 1.80 kg

bullet

19


(ii) Calculate the magnitude of the average force on the bullet when it is passing

through the first block.

average force = ........................................ N [2]

20


7 (a) Explain what is meant by an uniform electric field.

........................................................................................................................................

................................................................................................................................. [1] (b) Fig. 7.1 shows two parallel conducting plates P and Q each of length 7.5 mm. The

separation of the plates is 2.5 mm. The electric potentials of plates P and Q are −V and 0 V respectively. An electron enters the region between the plates at an angle of 45º with a speed of 6.0 × 106 m s−1 at point A and exits at point B as shown.

Fig. 7.1

(i) Calculate the time for which the electron is between the plates.

time = ................................... s [1]

(ii) Show that the potential difference between the plates is 68.3 V. [3]

2.5 mm

45º

P

Q 0 V

A B

7.5 mm

6.0 × 106 m s−1

y

−V

displacement d

along plates

centre line

between plates

21


(iii) Hence, or otherwise, determine the electron’s distance of closest approach y to

plate P. y = ................................... m [3]

(iv) On Fig. 7.2, sketch, with appropriate values, the variation with the displacement d

along the plates of the

1. electric potential energy,

2. kinetic energy, of the electron.

Fig. 7.2 [3]

(v) Explain why, in the absence of any other charged bodies, the potential will be

2

V along the centre line between the plates.

.................................................................................................................................. ..................................................................................................................................

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

energy / J

d / mm 0

0

22


(c) Four point charges, Q1, Q2, Q3 and Q4, each of charge −0.050 μC, are held in a square array, as shown in Fig. 7.3. A small conducting sphere with a charge of −0.060 μC is placed at X, 30.0 mm vertically above the centre C of the array where it is held stationary by electric forces. The sphere is 50.0 mm from each of the four charges as shown.

Fig. 7.3

Calculate

(i) the magnitude of the force acting on the sphere due to charge Q1, force = ................................... N [2]

(ii) the mass of the sphere, mass = ................................... kg [3]

(iii) the electric potential at X.

electric potential = .................................. V [2]

conducting sphere

Q1

Q2 Q3

Q4

30.0 mm 50.0 mm

C

X

23


8 (a) Helium-3 is an isotope that is commonly used in the detection of neutrons. It captures

a neutron and produces a hydrogen nuclide and a tritium (hydrogen-3) nuclide which may be represented by the equation

3 1 1 3

2 0 1 1He n H H

The following data is given:

binding energy per nucleon of tritium nuclide = 2.8273 MeV binding energy per nucleon of helium-3 nuclide = 2.5727 MeV mass of tritium nuclide = 3.016049 u mass of helium-3 nuclide = 3.016029 u mass of hydrogen nuclide = 1.007827 u

(i) Define binding energy per nucleon.

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

(ii) Using your answer in (i) and appropriate data values, explain whether energy is

supplied or released in this reaction.

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

(iii) Calculate the energy supplied or released in this reaction.

energy = ........................................ J [2]

24


(v) Hence, calculate the mass of a neutron in unified atomic mass units.

mass of neutron = ........................................ u [3]

(b) 131

52Te decays by emission to 131

53I . 131

53I is not stable, and decays by emission to the

stable isotope Xe, but the half-life for this decay is very much longer than that for the

decay of 131

52Te . A sample of pure 131

52Te is prepared at time 0t , and Fig. 8.1 shows

the variation with time t of lg A, where A is the activity of the whole sample.

Background radiation can be ignored.

Fig. 8.1

lg(A / Bq)

t / hours

25


(i) Define half-life.

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

(ii) Write down the mass number and atomic number of the isotope Xe.

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

(iii) Explain why the amount of 131

53I in the sample first increases and then decreases.

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

(iii) Using Fig 8.1, determine the half-life of 131

52Te .

half-life = ........................................ hours [2]

(iv) Explain the significance of the flat part of the graph (after 6 hourst ).

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

(v) Hence, calculate the half-life of 131

53I .

half-life = ........................................ hours [3]

End of paper

1

Answers to 2015 JC2 Preliminary Examination Paper 3 (H2 Physics)

Suggested Solutions:


1(a)

The gravitational field strength at a point in space is the gravitational force per unit mass experienced by a test mass placed at that point.


1(b)(i)

Applying d

gdr

,

From the tangent drawn in Fig. 1.1, gravitational field strength at the surface of Pluto,

5

1

6

0 7.0 100.538 N kg

1.3 0 10

dg

dr

[1] for tangent drawn [1] for gradient expression and correct answer

1(b)(ii) By loss in KE = gain in GPE,

21

2mv m

2

f i

1

2v

[1] for correct expression

-8.0

-7.0

-6.0

-5.0

-4.0

-3.0

-2.0

-1.0

0.0

0.0 1.0 2.0 3.0 4.0 5.0 6.0

2


22 5 5 1

f i

1 1680 7.0 10 4.69 10 J kg

2 2v

From Fig. 1.1, maximum height from surface of Pluto = 6.0 × 105 m

[1] for correct numerical

substitution and f


1(c) Let D be the distance between Pluto and Charon. Zero resultant gravitational force => zero resultant gravitational field strength

Using CP

2 2( )

GMGM

d D d

,

2

C

2

P

( ) MD d

Md

C

P

MD d

d M

11

7

D

d

771.96 10

1.42 10 m1 1

1 17 7

Dd

[1] for correct expression [1] for numerical substitution and answer

2(a) Oscillations in the absence of an external force.

[1]

2(b)(i) Angular frequency 12 18.5

6.80 rad s0.400

k

T m

Amplitude max 1.40

6.80

v

= 0.206 m

[1] for correct substitution for angular frequency [1] for correct substitution for amplitude

2(b)(ii) Maximum acceleration = 2(6.80) (0.206)

At lowest point, T mg ma

2(0.400)(9.81) (0.400)(6.80) (0.206)T

7.73 NT

[1] for correct substitution for maximum acceleration [1] for correct substitution [1] for correct answer

3


2(c)(i)

[1] for correct shape and values Allow e.c.f.

2(c)(ii)

[1] for same amplitude and smaller maximum speed

3(a) In circuit M, there is no current since the voltmeter has infinite resistance. Hence the potential difference across the voltmeter is the e.m.f of the 4 cells. In circuit N, the effective resistance is not infinite and hence there is a current through the circuit. As such there is a voltage drop across the internal resistance of the cells. Hence the potential difference the voltmeter read is lower than the e.m.f.

[1] [1]

v / m s−1

x / m 0 +0.206 −0.206

+1.40

−1.40

Z

v / m s−1

x / m 0 +0.206 −0.206

+1.40

−1.40

4


3(b) Using V rI

3 0 3 0

4 0 3 0 43 0 3 5

. .. . r

. .

0 13 r .


3(c)(i) e.m.f of battery 2.0 V

eff

3 0 3 50 1346 4

3 0 3 5

. .R .

. .

3 0 3 5

0 1346 43 0 3 5

2 1538

. ..

. .

.

Current drawn 2 0

2 1538

.

.

0 93 A.

[1] for correct effective resistance [1] for correct current

3(c)(ii) ext

e.m.f.

efficiency 100P

%P

2ext extP I R

2 3 0 3 5

0 92863 0 3 5

1 3929 W

. ..

. .

.

ext

e.m.f.

efficiency = 100P

%P

1 3929100

1 3929100

0 9286 2

75

.%

I

.%

.

%

[1] for Pext


4 (a)(i) The constant current in vertical wire produces a constant magnetic field, hence the magnetic flux linkage of the coil does not change. According to Faraday’s law, the e.m.f. induced is proportional to the rate of change of magnetic linkage. Since there is no change in the magnetic flux linkage, the voltmeter does not register a reading.

[1] [1] [1]

4 (a)(ii) Vary the current in the vertical wire by switching the current on and off or use an alternating current. Move the coil towards or away from the vertical wire or move the vertical wire towards or away from the coil. Rotate the coil so that it cuts the magnetic flux.

[1] each for any 2 ways

4 (b) The magnetic flux in the iron core is in phase with the current in the primary coil.

[1]

5


The magnetic flux in the core is not in phase with the rate of cutting of magnetic flux in the secondary coil (or the rate of change of magnetic flux linkage in the secondary coil). The rate of cutting of magnetic flux in the secondary coil (or the rate of change of magnetic flux linkage in the secondary coil) is proportional to the induced e.m.f. in the secondary coil. Hence the e.m.f. induced in the secondary coil is not in phase with the current in the primary coil.

[1] [1]

5(a)(i) Doping is a process where impurities are added into an intrinsic semiconductor, so as to increase the conductivity (or decrease the resistivity) of the material.

[2] for underlined key phrases

5(a)(ii) p-type doping creates an acceptor level right above the valence band, where electrons in the valence band are able to excite to this level, leaving behind holes in the valence band. n-type doping creates a donor level right below the conduction band, where electrons from this level are able to excite into the conduction band, creating more conduction electrons in conduction band.

[1] [1] [Note: Award one mark if only 1 type of doping is described.]

5(b)(i)1 [1]

5(b)(i)2

[1]

5(c)(i) The resistance of the diode decreases with increasing temperature.

[1]

5(c)(ii) At very small values of V, the electric field provided by the

battery is insufficient to overcome the internal electric field set up in the depletion region. Hence, current is zero.

[1]

6(a) The rate of change of momentum of a body is directly proportional to the resultant force acting on it, and takes place in the direction of the resultant force.

[1]


+ +

−−


6

6(b)(i) There is a change in horizontal momentum of the stream of water. Therefore the trolley is exerting a horizontal force on the water, according to Newton’s Second law of motion. By Newton’s Third law of motion, the water exerts an equal and opposite force on the trolley.

[1] [1] [1]

6(b)(ii)

2

2 2

2 2

Force on water

0 sin60

sin60

Force on container

sin60

w

c

dpF

dt

v r v

v r

F ma

v r ma

2 2 2sin60.0 (1000) (0.800 10 ) 2.50(1.10)v

3.97v m s−1

[1] for correct Fw [1] for Fc = ma


6(b)(iii) sinma T

cosmg T

1.10tan

9.81

a

g

6.40


6(b)(iv)1.

Magnitude of average resultant force

3

3

1(11.4)(48 10 )

2

48 10F

5.7 NF


6(b)(iv)2.

2 2 23.18 3.31 2(3.18)(3.31)cos 25.0 20.9v

2.53v m s−1


6(b)(iv)3.

vF m

t

3

2.535.7

48 10m

0.108 kgm


6(c)(i) PCM for interaction between bullet and first block: 3 33.50 10 0 3.50 10 (1.20)(0.630)u v

PCM for interaction between bullet and second block:

3 33.50 10 0 (3.50 10 1.80)(1.40)v

[1] for correct substitution [1] for correct substitution

7

1937 m su

1721 m sv

[1] for correct answer [1] for correct answer

6(c)(ii) Magnitude of the impulse on bullet = magnitude of impulse on first block

= (1.20)(0.630)

Average force

= 6

(1.20)(0.630)

213 10

= 3550 N


7(a) An electric field of force is a region in space around a charged object in which a charge would experience a constant electrostatic force.


7(b)(i) Time for which the electron is between the plates,

x

x

3

6

9

9

7.5 10

6.0 10 cos 45

1.768 10 s

1.77 10 s

st

u


7(b)(ii) In a uniform E-field,

VE

x

uniform acceleration

Applying Newton’s 2nd Law, eF m a

E

e e e

F eE e Va

m m m x

Using 2

y y y

1

2s u t a t ,

From A to B, y 0s ,

2

y y

10

2u t a t

y y

10

2u a t

y

e

10

2

e Vu t

m x

e

6 31 3

19 9

2

2 6.0 10 sin45 9.11 10 2.5 10

1.60 10 1.768 10

68.32 V

68.3 V

yu m xV

et

[1] for uniform a

[1] for correct expression and numerical substitution [1] for correct answer

8

7(b)(iii)

Consider vertical motion, Loss in KE = Gain in EPE

2

e y E

E y

1work done by on electron

2

m u F

F s

2

e y

y

E

2

e y

E

231 6 3

19

3

2

2

9.11 10 6.0 10 sin45 2.5 10

2 1.60 10 68.32

1.875 10 m

1.875 mm

m us

F

m u x e VF

e V x

Distance of closest approach,

32.5 1.875 0.625 mm 6.25 10 my

[1] for correct expression and numerical substitution [1] for sy [1] for correct y

7(b)(iv)1. & 2.

Initial KE of electron

2

e

231 6

17

1

2

19.11 10 6.0 10

2

1.64 10 J

m u

Max EPE = KE lost at highest point

17 2

e y

217 31 6

18

11.64 10

2

11.64 10 9.11 10 6.0 10 sin45

2

8.20 10 J

m u

[1] for correct EPE [1] for correct KE [1] for correct labelling of values on axis

energy / J

d / mm

0

0 7.5

1.64 × 10−17

3.75

8.20 × 10−18

(iv)2.

KE

(iv)1.

EPE

9

7(b)(v) The electric field strength between parallel plates is uniform. Since the electric field strength is the negative of the potential

gradient, i.e. dV

Edr

, the potential decreases uniformly

from plate Q (0 V) to plate P (−V). Hence, the centre

equipotential line, which is the mid-way between the 0 V and

–V plates, must be at2

V .

Alteratively: The electric potential is defined as the energy per unit charge for a charged particle placed in an electric field. When an electron is moved from plate P (−V) to the centre line, work done is positive with magnitude e∆V, its electric potential decreases by ∆V. When the electron is moved from plate Q

(0 V) to the centre line, work done is negative with magnitude e∆V, its electric potential increase by ∆V. Hence, the centre

equipotential line, which is the mid-way between the 0 V and

–V plates, must be at 2

V .

[1] for uniform electric field between parallel plates [1] for uniform decrease in potential [1] for definition of potential [1] for work associated when moving the electron to the centre position

7(c)(i) Using Coulomb’s Law,

1 XE 2

o

6 6

212 3

2

2

4

0.050 10 0.060 10

4 8.85 10 50.0 10

1.079 10 N

1.08 10 N

Q QF

x


7(c)(ii) Vertical component of FE due to Q1,

2

3

3

cos

3 1.079 10

5

6.474 10 N

6.47 10 N

y EF F

As sphere is in equilibrium,

y4mg F

3y

3

4 4 6.474 10

9.81

2.64 10 kg

Fm

g

[1] for correct Fy


7(c)(iii) Electric potential at X due to the 4 charges, [1] for correct expression and numerical substitution

10

X

o

6

12 3

4

4

44

0.050 10 4

4 8.85 10 50.0 10

3.597 10

3.60 10 V

QV

x

[1] for correct answer with the correct sign

8(a)(i) Binding energy per nucleon is the minimum energy per nucleon required to split up a nucleus into its constituent nucleons.

[1]

8(a)(ii) The BE per nucleon of tritium is higher than that of helium-3. This means that the products of the reaction is more stable than the reactants hence energy will be released.

[1] for correct comparison of BE/nucleon [1] for stating energy released

8(a)(iii)

13

Energy released Difference in total BE

3 2.8273 3 2.5727

0.764 MeV

1.22 10 J

[1] for taking difference of total BE [1] for correct answer

8(a)(iv)

13

27 2

Energy released is equivalent to mass difference between

reactants and products

1.22 103.016029 1.007827 3.016049

1.66 10

1.008664

n

n

u m u uc

m u

[1] for converting energy released to mass difference [1] for correct equation [1] for correct answer

8(b)(i) Half-life of a radioactive isotope is the average time taken for a sample of atoms to decay to half their initial number.

[1] for correct definition

8(b)(ii) Mass number is 131 and atomic number is 54.

[2] for correct answers

8(b)(iii) Te-131 decays quickly to I-131 and the amount increases due to the short half-life as compared to the long half-life of decay of I-131. When almost all the Te-131 has decayed, the amount of I-131 starts to decrease as it decays slowly to Xe.

[1] fast decay of Te-131 [1] slow decay of I-131 after all the Te has decayed

8(b)(iii) 0

12

0

11

12

12

lg 12

1.0 10 Bq

5.0 10 Bq

lg 11.7

From the graph, half-life is 0.4 hours

A

A

A

A

[1] for calculating activity at half-life [1] for determining half-life

8(b)(iv) The activity remains relatively constant after 6 hours. Almost all the Te must have decayed into I. The half-life of I is long, hence the activity remains constant for a long period of time as compared to Te.

[1] All Te decayed into I [1] The activity remains constant due to long half life

11

8(b)(v)

9

12

0

15

0

0

9 15

6 1

5

1/2 6

lg 9.32

2.09 10 Bq

ln21.0 10

0.4 60 60

2.08 10

Since all Te decayed to I,

2.09 10 2.08 10

1.00 10 s

ln26.90 10 s

1.00 10

192 hours

A

A

N

N

A N

T

[1] for calculating activity of I [1] for calculating number of undecayed atoms of I [1] for correct half-life

Documents

1 hour 15 minutes - Weeblyscore-in-chemistry.weebly.com/uploads/4/8/7/1/48719755/pjc_2015_prelim.pdf1.66u 10 27 kg rest mass of electron, 9.11u10 31 m e kg rest mass of proton, 1.67