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MERIDIAN JUNIOR COLLEGE Preliminary Examination Higher 2

_______________________________________________________________________

H2 Physics 9646/1

Paper 1 22 September 2015

1 hour 15 min

_______________________________________________________________________

Class Reg Number

Candidate Name _____________________________

READ THESE INSTRUCTIONS FIRST Do not open this booklet until you are told to do so.

There are forty questions in this section. 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 Optical Mark Sheet (OMS).

Read very carefully the instructions on the OMS.

Write your name and class in the spaces provided on the OMS.

Shade your Index Number column using the following format:

1) first 2 digits is your index number in class (e.g. 5th student is shaded as “05”); 2) ignore the last row of alphabets.

Preliminary Examination Meridian Junior College 22 September 2015 JC2 H2 Physics 2015

2 [Turn over]

Data speed of light in free space c = 3.00 x 10

8 m s

-1

permeability of free space o = 4 x 10-7 H m-1

permittivity of free space ε0 = 8.85 x 10-12 F m-1

= (1/(36)) x 10-9 F m-1

elementary charge e = 1.60 x 10-19 C

the Planck constant h = 6.63 x 10-34 J s

unified atomic mass constant u = 1.66 x 10-27 kg

rest mass of electron me = 9.11 x 10-31 kg

rest mass of proton mp = 1.67 x 10-27 kg

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

the Avogadro constant NA = 6.02 x 1023 mol-1

the Boltzmann constant k = 1.38 x 10-23 J K-1

gravitational constant G = 6.67 x 10-11 N m2 kg-2

acceleration of free fall g = 9.81 m s-2

Formulae uniformly accelerated motion

s = ut + 1

2at2

v2 = u2 + 2as

work done on/by a gas W = pV

hydrostatic pressure p = gh

gravitational potential = Gm

r

displacement of particle in s.h.m. x = xo sin t

velocity of particle in s.h.m. v = vo cos t

= 2 2

o-x x

mean kinetic energy of a molecule of an ideal gas E = 3

2kT

resistors in series R = R1 + R2 + …

resistors in parallel 1/R = 1/R1 + 1/R2 + …

electric potential V = 04

Q

r

alternating current/voltage x = xo sin t

transmission coefficient T exp(-2kd)

where k =

2

2

8 ( )m U E

h

radioactive decay x = xo exp(-t )

decay constant

= 1

2

0.693

t


3 [Turn over]

Answer all 40 questions in this paper and shade your answers on the answer sheet provided. 1 In order to determine the relationship between the weights, W, of the loads hung on a spring

and the extension, x, of the spring, Bobby conducted an experiment by hanging masses of different weights on a spring of a fixed spring constant. The weights of the masses were determined using an electronic balance that has a positive zero error. Without accounting for the zero error, he then plotted a graph of W against x.

What effect does this have on the graph he plotted?

A The gradient of the graph will be smaller than the actual value.

B The gradient of the graph will be larger than the actual value.

C The y-intercept of the graph will occur above the origin.

D The y-intercept of the graph will occur below the origin.

2 A bird has an initial velocity of 20.0 m s-1 in the northerly direction as shown in Fig. 2(a)

below. At a later time, its velocity is 20.0 m s-1 in the westerly direction as shown in Fig. 2(b).

What is the change in velocity that has taken place in this interval, given that directions are indicated by measuring angles anti-clockwise from the direction OX?

A 6.32 m s-1 at an angle of 135° from OX.

B 6.32 m s-1 at an angle of 225° from OX.

C 28.3 m s-1 at an angle of 135° from OX.

D 28.3 m s-1 at an angle of 225° from OX.

3 In a Formula One race, Sebastian Vettel driving the Ferrari race car has to make a pit-stop

to re-fuel. After re-fueling, he starts from rest with a constant acceleration of 11.0 m s-2 and takes 3.5 s to enter the main speedway from the pit area. At the instant Vettel enters the main speedway, another race car on the speedway, driven by Hamilton, traveling at a constant velocity of 69.4 m s-1 passes Vettel’s entering car. What is the total time required for Vettel’s car to catch up with Hamilton from the time he completes re-fueling? Assume that Vettel maintains a constant acceleration throughout this time.

A 5.62 s B 9.12 s C 12.6 s D 16.1 s

20.0 m s-1 20.0 m s-1

O X

Fig. 2(a) Fig. 2(b)


4 [Turn over]

4 An object, initially at rest, moves along a straight line path. The graph below shows the variation of its acceleration, a, with time, t. What is the total displacement of the object until t = 8.0 s?

A 2.0 m B 3.0 m C 6.0 m D 11 m 5 The diagram below shows two spheres undergoing an head-on elastic collision. Sphere A

has mass 2M while sphere B has mass M. Both are moving with speed v towards each

other.

Which of the following statements is incorrect?

A The two spheres cannot come to rest at the same time.

B The total kinetic energy of both spheres is conserved throughout the collision.

C The magnitude of the change in momentum for each sphere is the same after the collision.

D The force exerted by sphere A on B is equal and opposite to the force exerted by B on A during the collision.

6 Some ice cubes are floating in a glass of water. It is noticed that when all the ice cubes

melted, the water level increases slightly. Which of the following could be a possible reason?

A The surrounding air pressure is higher than the normal atmospheric pressure of 1.01 kPa.

B There are some sand particles trapped in the ice cubes.

C The water is mixed with alcohol which lowers its density.

D The ice cubes contains small helium bubbles in them.

v v

2M M

A B

1

-4

t / s

2 4 6 8 0

a / m s-2


5 [Turn over]

7 A man peddles on a bicycle along a horizontal road with a constant speed. Which of the following correctly shows the forces acting on the front and rear wheels due to the road?

A

B

C

D

8 An electron is being projected into a cross field. The cross field consists of an unknown field

X which points downwards, and an unknown field Y which points out of paper.

If the electron passes through the cross field undeflected, which of the following could be the fields X and Y?

Field X Field Y

A magnetic electric

B gravitational electric

C electric magnetic

D electric gravitational

electron

Field X: points downwards Field Y: points out of paper


6 [Turn over]

9 The diagram below shows 2 blocks connected together via a light inextensible string over a smooth pulley. The top block has a mass of 1.5 kg and the bottom block 4.6 kg. The bottom block is pulled with a force of 8.4 N. Assume that all surfaces (except pulley) have a constant frictional force of 1.8 N.

What is the acceleration of the top block?

A 0.49 m s-2 B 0.79 m s-2 C 1.1 m s-2 D 1.4 m s-2 10 A motor was powered by a 10 V cell connected in a circuit shown below.

When the voltmeter and ammeter read 9.5 V and 0.50 A respectively, the motor was able to lift a 0.25 kg toy vertically at a speed of 1.0 m s–1 with an acceleration of 1.0 m s–2. What is the efficiency of the motor at this instant?

A 46 % B 52 % C 57 % D 62 %

11 For an average human, 20 g of blood are pushed into the main arteries in one heartbeat,

increasing the speed from 0.20 m s1 to 0.36 m s1. For a heart pulsing at 80 beats per minute, the average power of the heart pump is

A 0.34 mW B 0.53 mW C 1.2 mW D 1.7 mW

8.4 N

1.5 kg

4.6 kg

smooth

pulley

10 V

A

motor

V


7 [Turn over]

12 The frictional force between the 4.0 kg mass and the plane of the slope is 2.0 N.

If the distance moved by the 10.0 kg mass is 0.15 m, what is the speed gained by the 4.0 kg mass?

A 1.15 m s–1 B 1.28 m s–1 C 1.64 m s–1 D 1.72 m s–1

13 In an amusement park, passengers are seated in a roller coaster that goes round the track

in an anti-clockwise direction. At the instant shown, the roller coaster is slowing down due to resistive forces. Which of the arrows shows the direction of the acceleration?

30°

4.0 kg

10.0 kg

A

B C

D

roller coaster


8 [Turn over]

14 A turntable consists of a rotating horizontal disc moving at a fixed rotational speed. Two small objects P and Q with the same mass, are placed on the turntable as shown, where P is at a distance r1 from the centre O and Q is at a distance r2 from the centre.

Which of the following correctly relates P and Q’s linear speeds and their centripetal forces?

linear speed centripetal force

A same for P and Q greater for Q

B same for P and Q greater for P

C greater for P greater for Q

D greater for P greater for P

15 A double star consists of two stars, one of mass M and the other of mass 2M, revolving

around their common centre of mass. The distance between their centres is R. What is the force acting on the star of mass M and its acceleration?

force acceleration

A

2

2

GM

R

0

B

2

2

GM

R 2

GM

R

C

2

2

2GM

R 2

GM

R

D

2

2

2GM

R 2

2GM

R

16 A stationary object is released from a point at a height of 2R from the surface of the Moon

which has radius R and mass M.

What is the correct expression for the speed of the object on hitting the Moon?

A

4

3

GM

R B

2

3

GM

R C

4GM

R D

2GM

R

Turntable rotating on a horizontal plane


9 [Turn over]

17 The figure below shows two simple pendulum X and Y having the same lengths and same bob sizes and a simple pendulum Z whose length can be varied. The pendulums are made of the materials indicated in the figure below.

The pendulum X and pendulum Y will oscillate after the pendulum Z is set into oscillations in

a plane parallel to the wall.

Which of the graphs below shows the variation of the amplitude, a, for the pendulum X and

pendulum Y with frequency, f, of the pendulum Z when the length of pendulum Z is varied?

A B

C D

Wall

String

Polystyrene Lead Copper

X Y Z

Wall

Y

X

a

f

Y

X

a

f

Y

X

a

f

X

a

f

Y


10 [Turn over]

18 Two containers of volumes 4.0 m3 and 6.0 m3 contain an ideal gas at pressures of 100 Pa and 50 Pa respectively. Their temperatures are equal and they are joined by a tube of negligible volume. The gas flows from one container to the other with no change in temperature. What is the final pressure?

A 70 Pa B 75 Pa C 80 Pa D 150 Pa

19 A system of ideal gas undergoes a change of state from (p1,V1) to (p2,V2) by either process

(i) or process (ii).

Which of the following statements is correct?

A The work done by the gas is greater in (i) than in (ii).

B The heat supplied to the gas is greater in (i) than in (ii).

C The change in internal energy of the system in both processes is the same.

D If (i) is such that pV is kept constant, it is an adiabatic process.

20 The variation of the acceleration a with the position x of air molecules along a sound wave is

shown below. Taking rightwards as positive, at which position is a rarefaction at the instant shown?

21 The distance between 2 points of a progressive transverse wave having a phase difference

of 3

is 40 cm. If the frequency of the wave is 200 Hz, what is the speed of the wave?

A 240 m s-1 B 480 m s-1 C 24000 m s-1 D 48000 m s-1

(p1,V1)

(p2,V2)

V

p

(i)

(ii)

A

B

C

D

a

x


11 [Turn over]

22 When a two-slit arrangement was set up to produce a superposition pattern on a screen using a monochromatic source of green light, the fringes were found to be too close together for accurate observation. It would be possible to increase the separation of the fringes by

A replacing the light source with a monochromatic source of red light.

B increasing the distance between the source and the slits.

C decreasing the distance between the slits and the screen.

D increasing the distance between the two slits.

23 A standing wave is set up on a stretched string XY as shown in the diagram below.

At which point(s) will the oscillation be out of phase with that at P?

A 2 only

B 1 and 2 only

C 1 and 3 only

D 1, 2 and 3

24 A dipole is a pair of one negative charge and one positive charge of equal magnitude. The

electric field of a dipole is shown below. In which direction does the force act on an electron when at point X?

A D

B C

X

P 1 2 3 X Y


12 [Turn over]

25 The radius of a lithium 7

3Li nucleus is 2.3 x 10-15 m, and the radius of a proton is

1.2 x 10-15 m.

What is the magnitude of the force acting on the proton when it is just in contact with the lithium nucleus?

A 480 N B 130 N C 56 N D 19 N

26 The material of a wire has resistivity 1.3 x 10-8 Ω m. The wire has diameter 0.50 mm and its

length is just enough to enable it to be wound tightly round an insulating rod 30 times. The rod has diameter 1.5 cm.

What is the resistance of the wire?

A 1.1 x 101 Ω B 1.1 x 10-5 Ω C 7.0 x 10-4 Ω D 9.4 x 10-2 Ω

28 What is the value of the potential at point X in the circuit below?

A − 7.0 V B − 3.9 V C 3.1 V D 4.9 V

27 When a 4.0 Ω resistor is connected between the terminals of a certain cell, a 2.0 A current flows. When the 4.0 Ω resistor is replaced by one of 2.0 Ω, the current is 3.0 A. The e.m.f. and internal resistance of the cell are respectively

A 15 V, 4.0 Ω B 12 V, 2.0 Ω C 10 V, 1.0 Ω D 8.0 V, zero

wire wound round rod 30 times insulating rod

diameter 1.5 cm

8.0 V

12.0 Ω

8.0 Ω

2.0 Ω

8.0 Ω

6.0 Ω

X


13 [Turn over]

29 The diagram below shows a circuit with a uniform wire XY of length 1.0 m and resistance

1.0 Ω connected in series with a cell of e.m.f. 3.0 V and internal resistance 0.50 Ω. The circuit is used to measure the e.m.f. generated by a thermocouple.

When the e.m.f generated by the thermocouple is 0.20 V and the galvanometer gives a zero reading, the position of the jockey will be

A 10 cm away from X.

B 20 cm away from X.

C 80 cm away from X.

D 90 cm away from X.

30 A positively charged particle enters a region of magnetic field at an angle θ as shown.

How will the time taken for the particle to move from point X to Y change when the following

changes are made independent of one another?

magnetic flux density increased angle θ increased

A unchanged increase

B decrease unchanged

C increase decrease

D unchanged decrease

0.50 Ω 3.0 V

0.20 V

thermocouple

X Y

+q

v

B field

X Y

θ


14 [Turn over]

31 A wire of length 3.5 cm is placed on a vertical plane in a region of magnetic field. Uniform

magnetic field of flux density 0.080 T is directed out of the plane as shown. The wire makes

an angle θ with the horizontal and carries a current I of 4.0 A in the direction as shown.

Which of the following is true about the magnitude of the force which the field exerts on the wire at different angles of θ from 0° to 360°? (assuming that the rod remains in the magnetic field throughout)

A The force is constant at 0 N.

B The force varies between 0 N and 0.0112 N.

C The force is constant at 0.0112 N.

D The force varies between 0.0112 N and 0.0224 N.

32 A magnetic field of flux density 1.5 x 10-4 T passes through a coil of wire placed horizontally.

The field makes an angle of 30° with the horizontal, as shown in the diagram below. The coil has 200 turns and an area of 2.0 x 10-4 m2.

What is the change in magnetic flux linkage if the coil is rotated 180° through axis XY?

A 0 Wb B 1.0 x 10-5 Wb C 3.0 x 10-6 Wb D 6.0 x 10-6 Wb

1.5 x 10-4 T

X Y 30°

θ

I


15 [Turn over]

33 A bar magnet is suspended above a coil. The magnet is given an initial displacement and

then released at time t = 0 s, so that it swings as a pendulum, as shown in the diagram

below.

Which of the following graphs represents the variation with time t of the e.m.f. E induced in

the coil for one complete oscillation?

A B

C D

N

V

coil

E

t

E

t

E

t

E

t


16 [Turn over]

34 The diagram shows an a.c. supply connected to 4 diodes and a resistor R.

Which of the following graphs best represents the variation with time t of the potential difference V across the resistor?

A B

C D

35 The ratio of the primary turns to the secondary turns in an iron-cored transformer is 1:20.

The secondary coil is connected to a 5.0 Ω resistor, and the r.m.s. current in the primary coil is 10 A.

What is the value of the maximum power dissipated by the resistor?

A 1.25 W B 2.50 W C 2.0 x 105 W D 4.0 x 105 W

V

t

V

t

V

t

V

t

R


17 [Turn over]

36 Electrons are emitted from a metal surface when a beam of monochromatic light is incident on it. The intensity of the beam of light is then doubled. Which of the following statements is true?

A Photons of other frequencies are emitted.

B The number of electrons emitted remains the same.

C The maximum kinetic energy of the emitted electrons doubles.

D The change in momentum of each photon absorbed remains the same.

37 The work function of a metal is 2.7 eV. Electromagnetic radiation of frequencies ranging

from 1.0 x 1014 Hz to 9.0 x 1014 Hz is incident on the surface of the metal.

What is the maximum kinetic energy of the electrons ejected from the surface of the metal?

A 1.0 eV B 2.3 eV C 3.1 eV D 6.4 eV

38 Below are energy level schemes of possible laser materials. The “pump” is where excitation

takes place. The “laser transition” indicates the transition where lasing should occur. The transition “fast decay” is fast compared to the pumping or excitation process as well as the lasing transition.

Which of these level schemes is best for facilitating population inversion?

A B

C D

pump laser

transition

fast decay

fast decay

pump

fast decay

laser transition

pump laser

transition pump

laser transition

fast decay


18 [Turn over]

39 The number of disintegrations per second in a sample of radioactive nuclide depends on A number of atoms per mole of the nuclide.

B chemical state of the nuclide.

C temperature of the nuclide.

D the decay constant.

40 A detector is used for monitoring an α-source and a reading of 150 counts per minute is

observed. After a time equal to the half-life of the α-source, the reading has fallen to 80 counts per minute. If a 10 mm thick lead sheet is inserted between the α-source and the detector, the

reading would probably be

A 0 B 10 C 20 D 25

End of Paper

2015 MJC JC2 H2 Physics Prelim P1

1

Paper 1 Suggested Solutions:

1 C 11 C 21 B 31 C

2 D 12 B 22 A 32 D

3 B 13 C 23 B 33 B

4 C 14 D 24 A 34 C

5 B 15 D 25 C 35 B

6 D 16 A 26 D 36 D

7 B 17 A 27 B 37 A

8 A 18 A 28 D 38 A

9 A 19 C 29 A 39 D

10 C 20 B 30 A 40 B

MCQ 1: C

W (actual reading) – zero error = kx W = kx + zero error Hence the y intercept of the graph will occur above the origin.

MCQ 2: D

2 2 -1

f iΔv = v +(-v )= (20)+(20)= 28.3 m s

at an angle of 225o anti-clockwise from OX.

MCQ 3: B Vettel’s speed when Hamilton passes him, v = (11)(3.5) = 38.5 m s-1

When they are next to each other,

2 250 100038.5 (11/ 2) ( )

60 60t t t

t = 5.62 s Total time taken = 3.5 +5.62 = 9.12 s

O x

-

where vi = 20 m s-1

Final velocity vf = 20.0 m s-1

45o


2

MCQ 4: C

One can plot the corresponding velocity-time graph and calculate its area to obtain the displacement of the object.

Within the first 3 s, the object gains a speed of

m s-1

For the next 3 s, the object maintains the same speed.

Finally, between 6 s and 7 s,

it experiences deceleration.

m s-1

Total area under graph

= 6 m

MCQ 5: B

During the collision, some of the kinetic energy is converted to elastic potential energy. Total kinetic energy is only conserved before and after the collision, but not during the collision. MCQ 6: D

Due to the helium bubbles, the volume of water displaced will be smaller than without the bubbles since the upthrust needed is less than the weight of the ice itself. Thus when the ice melts, the melted water will occupy a bigger volume than the originally displaced volume. Hence the water level increases. In the other options, the weight of water displaced is greater than the weight of the ice itself. MCQ 7: B

The force acting on each wheel is a resultant of the normal contact force and the frictional force, hence the forces should be diagonal. On rear wheel, the frictional force points forward. On front wheel, the frictional force points backward. MCQ 8: A Due to field X, the magnetic force points out of paper. Due to field Y, the electrostatic force points into paper. Hence both forces can cancel out. Options B and D are incorrect, as the forces due to both fields will point in perpendicular directions, hence they cannot cancel out. Option C is incorrect as both electrostatic and magnetic forces point upwards.

210

v

atuv

2v

142

v

atuv

2v

1223222

1

1

-4

a/m s-2

t/s 2 4 6 8 0

2

v/m s-1

t/s

2 4 6 0

-2 8


3

MCQ 9: A Since the string is light and inextensible, both blocks have the same magnitude of acceleration. Considering horizontal forces acting on the blocks, Top block:

1.8 1.5T a

Bottom block:

8 4 1 8 1 8 4 6. . . T . a

Solving, -20.49 m s

2.5 N

a

T

MCQ 10: C

0 25 9 81 1 0 2 7025

2 7025 1 2 7025

4 75

56 9

out

in

out

in

F mg ma

F . ( . . ) .

P Fv . .

P IV .

PEff . %

P

MCQ 11: C

2 20 5 0 020 0 36 0 21 2 mW

60 80

E . . ( . . )P .

t /

MCQ 12: B

2

-1

10 9 81 0 15 0 5 14 4 9 81 0 15 30 2 0 15

1 28 m s

. . . v . . sin .

v .

MCQ 13: C. Friction opposes motion therefore it acts towards the right. Centripetal acceleration acts towards the centre. Hence the net acceleration is shown by C. MCQ 14: D

Since P and Q have the same angular velocity, the linear speed (v r ) of P is greater

than Q as r1>r2. Also, centripetal force 2

F mr is greater for P than Q as r1>r2.

1.5 kg

4.6 kg

1.8 N

1.8 N

1.8 N

8.4 N

T

T


4

MCQ 15: D

2

2 2

on mass M 2

2 2

2

GM( M ) GMF

R R

F GMa

M R

MCQ 16: A

Conservation of energy, E Final = E Initial

210

2 3

4

3

GMm GMmmv

R R

GMv

R

MCQ 17: A

Z is the driver causing oscillations of X and Y. By varying the length of Z, the driver frequency is being changed. Since X is made of polystyrene, it is subjected to air resistance, so its motion is more damped, thus the curve is less sharp. MCQ 18: A

1 1 1

1 1 2 21 2

1 2

1 2

(100)(4.0) (50)(6.0),

400 300 700

( )

70070Pa

10

f f

f

PV n RT

PV P Vn n

RT RT RT RT

n nRT RT RT

PV n n RT

P

MCQ 19: C

(A)is wrong as the area under the graph for (ii) is larger than (i), hence more work is done by the gas in (ii). (B) is wrong as from (A), heat supplied must be larger in (ii) than (i) as the change in internal energy of the gas is the same. (C) is correct since both processes have similar initial states and final states, the change in internal energy for both processes are the same. (D) is wrong as when PV is kept constant, it is an isothermal process. MCQ 20: B Using SHM, acceleration is proportional to displacement but in the opposite direction. Hence, displacement position graph is a –sin graph and B is a rarefaction. MCQ 21: B

Since phase difference of 3

= 40cm, the wavelength = 40 x 6 = 2.40 m

Thus, speed = = 200 x 2.40 = 480 ms-1


5

MCQ 22: A

Dx

a

x

MCQ 23: B Points between 2 nodes are in phase and π out of phase with points between the next 2

nodes. MCQ 24: A Direction of FE on an electron is opposite the direction of E-field at that point. MCQ 25: C

F = 2

1 2

04

QQ

r

=

2

-19 -19

-12 -15 -15

3 1.6 10 1.6 10

4 8.85 10 1.2 10 2.3 10

= 56 N MCQ 26: D

R = L/A = (N2D/2)/(d2/4) = 4ND/d2 = (4)(1.3 x 10-8)(30)(1.5x10-2)/(0.5x10-3)2

= 9.4 x 10-2 MCQ 27: B

E = I x (R+r) E = 2 x ( 4 +r) --- Eqn (1)

E = 3 x ( 2+r) --- Eqn (2)

8 + 2r = 6 + 3r r = 2 E = 2 x (4 + 2) = 12 V


6

MCQ 28: D

The potential at point A is 8 V (since the other end of the battery is earthed). Therefore to create a potential difference of 3.09 V across the first pair of resistors, the potential at X must be 8.00 – 3.09 = 4.90 V

MCQ 29: A

XY

1V 3 2 V

1 0.5

0.2 balance length

2 100

Balance length = 10 cm

MCQ 30: A

Analyzing the horizontal component, The time is dependent on the horizontal displacement and the horizontal speed When B field is increased, since horizontal displacement and horizontal speed does not change, hence time should not change When angle is increased, horizontal speed decreases, and since horizontal displacement remains unchanged, hence time increases. MCQ 31: C The force is constant at 0.0112 N as the current flowing in the conductor and the B field are always at right angles to one another.

MCQ 32: D

Φf = NBA cos 60° Φi = -NBA cos 60° ΔΦ = NBA cos 60°-(-NBA cos 60°) = 2 NBA cos 60° = 2 (200) (1.5 x 10-4) (2.0 x 10-4) cos 60° = 6.0 x 10-6 Wb


7

MCQ 33: B

Consider the motion of the magnet from left to right (half a complete cycle): As the magnet moves into the region of the coil, magnetic flux linkage through the coil increases. As the magnet is leaving the coil, magnetic flux linkage through the coil decreases. Hence, by lenz’s law, the induced e.m.f. will be of opposite polarity. Note that as the magnetic flux density is always pointing downwards, there is no change in the trace between the first and second half of the cycle. Option D represents the variation in magnetic flux linkage with time. MCQ 34: C Full wave rectification MCQ 35: B

, ,

110 0.5

20 A

p

s rms p rms

s

NI I

N

20.5

max

5.0 = 1.25 W

P = 2<P> = 2 x 1.25 = 2.50 W

P

MCQ 36: D

By de Broglie’s equation, p=h/λ Since intensity has no bearing on the initial momentum of each photon, and the final momentum of each absorbed photon is still zero, the change in momentum remains the same. MCQ 37: A

Most energetic photon = (6.63 x 10-34)(9.0 x 1014) = 5.967 x 10-19 J = 3.73 eV hf = Φ + EKmax

EKmax = 3.73 – 2.7 = 1.0 eV MCQ 38: A

The fast decay transitions in scheme A facilitates population inversion best because it causes the upper lasing level to be populated and the lower lasing level to be emptied quickly. MCQ 39: D

A N , where is the decay constant.

MCQ 40: B Lead plate will remove the count rate due to α-source. After one half-life, count rate decreased by 70 counts. This implies the α-source provided an

original count of 140. Hence background count is 10.

This question paper consists of 23 printed pages (including 1 blank page)

MERIDIAN JUNIOR COLLEGE

Preliminary Examination Higher 2

_______________________________________________________________________

H2 Physics 9646/2


1 hour 45 min

_______________________________________________________________________

Class Reg Number

Candidate Name _____________________________

READ THESE INSTRUCTIONS FIRST This booklet contains 8 questions.

Do not open this booklet until you are told to do so. Answer all questions.

Write your answers on this question booklet in the blanks provided.

Write your name and class on page 19 of the booklet. Pages 19 to 23 will be submitted separately.

INFORMATION FOR CANDIDATES

The number of marks is given in brackets [ ] at the end of each question or part question. Marks will be deducted if units are not stated where necessary or if answers are not quoted to the appropriate number of significant figures.

All working for numerical answers must be shown. You are reminded of the need for good English and clear presentation of your answers.

Examiner’s Use

Section A

Q1 /7

Q2 /8

Q3 /8

Q4 /7

Q5 /8

Q6 /8

Q7 /14

Q8 /12

Deductions

Total /72


2


8 m s

-1



= (1/(36)) x 10-9 F m-1












s = ut + 1

2at2

v2 = u2 + 2as




r



= 2 2

o-x x


2kT




Q

r



where k =

2

2

8 ( )m U E

h


decay constant

=

1

2

0.693

t


3

Answer all the questions in the spaces provided.

1 (a) A projectile was fired from a stationary train at a speed of 50 m s-1 and at an angle of

20° above the horizontal. Given that the projectile originates from a height of 2.1 m above the ground, calculate

(i) the time taken by the projectile to reach the ground, and

t = ………………………………. s [2]

(ii) the speed of the projectile 2.0 s after it was fired.

speed = ……………………………. m s-1 [3]

(b) While the train was accelerating forward, a bored passenger decided to carry out a

physics experiment. Facing the front of the train, he marked the spot where he was standing at. He then leaped vertically upwards into the air. State and explain if he landed in front of the mark, on the mark or behind the mark.

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

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

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

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

[2]


4

2 (a) (i) State Newton’s first law of motion.

…………………………………………………………………………………….………

…………………………………………………………………………………….………

……………………………………………………………………………………. [2] (ii) Define inertia.

…………………………………………………………………………………….………

…………………………………………………………………………………….………

……………………………………………………………………………………. [1] (iii) Discuss the relationship between inertia and Newton’s first law of motion.

…………………………………………………………………………………….………

…………………………………………………………………………………….………

……………………………………………………………………………………. [1]


5

(b) Fig. 2.1 below shows a tower crane which is in equilibrium. The crane is made up of the tower (section AB), jib (section CD) and a counter-weight. The tower and jib are connected by a fulcrum along their central axes. The end of the jib (D) is also connected to the top of the tower (A) by a metal cable such that angle DAB is a right

angle. The jib is tilted at an angle of 18o to the horizontal. The jib has a mass of 1200 kg and its centre of mass is 5.8 m to right of the fulcrum along the length of the jib. The length of the jib from the fulcrum to D is 18.5 m, and

the metal cable is taut with a tension of 3400 N. A counter-weight of 3800 kg is placed with its centre of mass 7.9 m to the left of the fulcrum.

Fig. 2.1

(i) The crane is used to lift a load of 2450 kg at a constant velocity. Determine the

distance d of this load from the fulcrum.

d = ………………………………. m [2]

fulcrum

counter-weight

A

B

C

load

d

D

7.9 m 5.8 m

centre of

mass of jib

metal cable

18.5 m

18o


6

(ii) The rope holding the load is slowly lengthened to lower the load into a big tank of water without touching the base of the tank. Discuss if the jib can still maintain its equilibrium when the load is submerged.

…………………………………………………………………………………………….

…………………………………………………………………………………………….

……………………………………………………………………………………….……

……………………………………………………………………………………….……

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


7

3 (a) Fig. 3.1 shows the equipotential lines for Earth, where point A is at a potential of - 4.0 x 10 7 Jkg−1 and points B and C are at a potential of - 5.0 x 10 7 Jkg−1.

Fig. 3.1

(i) On Fig. 3.1, draw the equipotential line for the gravitational potential of - 4.5 x 10 7 Jkg−1. [1]

(ii) Calculate the work done by the gravitational field in bringing a body of mass

2500 kg from A to B.

work done = ........................................ J [2]

- 5.0 x 10 7 Jkg−1

- 4.0 x 10 7 Jkg−1

Earth

A

B

C


8

(b) (i) Information related to the Earth and the Sun is given below.

5103.3Earthofmass

Sunofmass

110Earthofradius

Sunofradius

Given that the escape speed from the Earth is 4101.1 m s−1, show that the

escape speed from the Sun is 6.0 × 105 m s−1. [2]

(ii) The surface temperature of the Sun is about 6000 K and hydrogen is the most

abundant element in the Sun’s atmosphere. Using suitable calculations, explain why the hydrogen atom can be found in the Sun’s atmosphere, assuming that the hydrogen atom behaves like an ideal gas and has a mass of 1.0 u.

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

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

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

[3]


9

4 The eardrum is a very thin membrane that separates the external ear from the middle ear as shown in Fig. 4.1. When sound waves fall on the eardrum, the drum vibrates.

Fig. 4.1

Fig. 4.2 shows the variation of displacement y with time t of a sound wave incident on a person's eardrum. It can be assumed that the eardrum vibrates with the same frequency and amplitude as the incident sound wave.

(a) Calculate the frequency of the oscillating eardrum.

frequency = ………………………………. Hz

[1]

Fig. 4.2

y / 10-12

m

t / ms 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5

10

-5

5

-10

0


10

(b) Given that the mass of the ear drum is approximately 1.0 x 10-4 kg, determine the maximum kinetic energy of the vibrating eardrum.

maximum kinetic energy = …………………………… J [2] (c) Given that the eardrum has an area of approximately 6.5 x 10-5 m2, determine the

maximum pressure difference between the outer and middle ear.

maximum pressure difference = ………………………………. Pa [3] (d) Hence suggest why intense loud sound can damage your eardrum.

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

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


11

5 Fig. 5.1 illustrates two parallel charged plates, X and Y, placed a distance 40 mm apart in a vacuum, with plate X at a potential of 50 V and plate Y at a potential of -50 V. A proton is projected from position A with an initial speed of 1.3 x 105 m s-1, at an angle above the horizontal. The proton passes through point B before it hits plate Y. Points A and B are

10 mm away vertically from plates X and Y respectively.

Fig. 5.1 (not drawn to scale) (a) State and explain the value of the potential along the centre line between the plates, in

absence of any other charged bodies.

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

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

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

……………………………………………………………………………………….……….. [2] (b) Determine the magnitude of the electric field strength between the plates.

electric field strength = ………………………………. V m-1 [2]

- 50 V

+ 50 V

40 mm

A

B

plate Y

plate X

centre line

10 mm

10 mm


12

(c) Calculate the speed of the proton as it passes point B.

speed = ………………………………. m s-1 [3] (d) On Fig 5.1, sketch a possible path traced by the proton as it travels from

point A to plate Y. [1]


13

6 A pair of concentric coils is shown in Fig. 6.1.

Fig. 6.1

The outer coil A is connected to a variable power supply by the terminals XY. The variation with

time t of the current I in coil A is shown in Fig. 6.2.

Fig. 6.2

(a) Using Faraday’s law, explain why an e.m.f. is induced in coil B.

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

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

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

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

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

…………………………………………………………………………………........………… [3] (b) State an instant where the induced e.m.f. in coil B is a maximum value.

…………………………........ ms [1]

X

Y

coil A

coil B to external circuit

10 0 20 30

t / ms

I / A


14

(c) Explain your choice of answer for (b).

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

…………………………………………………………………………………........………… [1] (d) Explain whether the direction of induced current in coil B flows in the same or opposite

direction as the current in coil A from t = 10 ms to t = 15 ms.

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

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

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

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

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

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


15

7 Light Dependent Resistors (LDRs) are commonly used in light-sensor circuits to turn on or off relay systems. The resistance of an LDR varies when illuminated with light of different intensities, measured with the unit lux. An experiment is carried out to investigate the variation of resistance R of an LDR with

intensity I. Fig. 7.1 shows the readings obtained.

I / lux R / Ω

10 360000

50 49000

1000 1100

5000 150

35000 17

100000 4

Fig. 7.1

Fig. 7.2 is a graph of some of the data of Fig. 7.1.

(a) On Fig. 7.2, (i) plot the points corresponding to I = 1000 lux and I = 100000 lux.

(ii) draw the line of best fit for all the points. [2]

1

10

100

1000

10000

100000

1000000

1 10 100 1000 10000 1000001 10 100 1000 10000 100000

I / lux

1000000

100000

10000

1000

100

10

1

R / Ω

Fig. 7.2


16

(b) Explain whether the graph shows a linear relationship between R and I.

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

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

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

(c) The LDR is connected in a circuit with a 12 V e.m.f. source and a variable resistor as shown

in Fig. 7.3. A lighting system can be connected to this circuit (either across AB or BC) and

then installed in an open field, such that the lighting system can be turned on automatically when the intensity of sunlight falls below a trigger level.

Fig. 7.3 (i) Explain whether the lighting system should be connected across AB or BC.

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

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

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

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

……………………………………………………………………………………………………… [3] (ii) The variation in intensity of sunlight in the field with time is shown in Fig. 7.4.

0

5000

10000

15000

20000

25000

30000

35000

40000

0 200 400 600 800 1000 1200 1400 1600 1800 2000 2200 2400

12 V

A B C

I / lux

40000

30000

20000

10000

0

time of day

12:00 AM 06:00 AM 12:00 PM 6:00 PM 12:00 AM

Fig. 7.4


17

Given that the lighting system turns on at 6:00 pm,

1. using Fig. 7.4, state the time at which the lights will be turned off the next day.

time = ………..................... [1]

2. state the resistance of the LDR at 6:00 pm.

R = ………..................... Ω [1]

3. determine the resistance of the variable resistor, given that the lighting system is

turned on when the potential difference across the lighting system reaches 3.0 V.

resistance ………..................... Ω [2]

4. calculate the current through the variable resistor.

current ………..................... A [1] (iii) A technician wrongly connects the circuit as shown in Fig. 7.5.

Fig. 7.5 State and explain the effect on the lights in the field.

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

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

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

Vout to

lighting system

12 V


18

Blank Page


19

Class Reg Number

Candidate Name _____________________________

This section contains Q8 (Planning) of Paper 2.

You are advised to spend about 30 minutes on this question. You are required to detach this section and submit it separately. Please turn over for Q8.

Examiner’s Use

Q8 /12


20

It is recommended that you spend about 30 minutes on this question.

8 When a body is rotating about some axis, it has a tendency to keep rotating at same angular speed. This tendency is called the rotational inertia and is measured by a physical quantity

called the moment of inertia, I, of the body. The moment of inertia Inet of two bodies rotating

about an axis is the sum of their individual moment of inertia I1 and I2 about the same axis (i.e.

Inet = I1 + I2).

Fig. 8.1 shows a typical setup used in the determination of moment of inertia of a rotating object. The object to be rotated is attached to a free-rotating axle as illustrated. A string, with one end fixed to the axle, is wound round the axle and has a mass m attached to the other

end. As the mass falls from rest, the string unwinds and rotates the setup. From the law of conservation of energy, the moment of inertia of the rotating setup is given by the following.

gt

hmr

2

21

2 …… Eqn. 1

where r is the radius of the axle, g is the acceleration of free fall and t is the time taken for the mass m to move through a height h.

By first manipulating Eqn. 1 to a suitable form, design a laboratory experiment to determine the moment of inertia of a bicycle wheel. The following equipment is available:

- bicycle wheel - standard masses

- rotatable axle with support (as shown in Fig. 8.1) - long non-extensible string

and any other equipment available in a school laboratory.

You should draw a labelled diagram showing the arrangement of apparatus you would use. In

your account, you should pay particular attention to

(a) the equipment you would use,

(b) the procedure to be followed,

(c) the control of variables,

(d) how you would obtain various sets of data by varying the mass in order to determine the moment of inertia,

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

[Total marks for this question: 12]

rotatable axle

support with

bearings support with bearings

rotating object

mass m

Fig. 8.1

string


21

Diagram:

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End of Paper

MJC Prelims 2015 H2 Physics

1

H2 Physics Paper 2 Suggested Solutions:

1 (a) (i) Taking the vertical component and upwards as positive

2

2

1s ut at

2

12.1 50 sin 20 t 9.81 t

2

t 3.61s or 0.119 s

(rejected)

M1 A1

(ii) Horizontal velocity after 2 s = 050cos20 = 46.98 ms-1

Using y yv u at

050sin 20 9.81(2)yv

= -2.52 m s-1 or 2.52 m s-1 downwards.

speed at that point = 2 246.98 ( 2.52)

= 47.0 m s-1

M1

M1

A1

(b) When the passenger jumps into the air, he stops accelerating horizontally while the train continues its horizontal acceleration. Hence, the train will travel a longer horizontal distance than the man during his jump. Therefore, he will land behind the mark.

M1

A1

2 (a) (i) Newton's first law of motion states that a body will continue in its state of

rest or uniform motion in a straight line unless an external resultant force acts on it. [B2]

(ii) Inertia is the tendency of things to resist changes in motion. [B1] OR Inertia is the property of an object to stay in a state of rest of uniform motion.

(iii) Inertia is basically an alternative statement to Newton’s first law of motion, however, Newton’s first law of motion made reference to how motion of a body can be changed with a resultant force. [B1]

(b) (i) Taking moments about the fulcrum,

o o

o o

1200 9.81 (5.8)cos18 2450 9.81 cos18

3400 18.5 sin18 3800 9.81 7.9 cos18 [M1]

10.3 m [A1]

d

d

(iii) When immersed in water, upthrust acts on the load hence the tension in the rope holding the load will decrease, decreasing the clockwise moment. [B1] Hence there will be a net anti-clockwise moment and the jib may not be in equilibrium. [B1] OR


2

When immersed in water, upthrust acts on the load hence the tension in the rope holding the load will decrease, decreasing the clockwise moment. [B1] However, the tension in the metal cable will also decrease, hence decreasing the anti-clockwise moment and the jib can still be in equilibrium. [B1]

3 (a)(i)

[1] for line drawn nearer to the 7100.5 J kg−1 equipotential line

(a)(ii) Work done by external force

B Am

72500 5 0 4 0 10. .

102 5 10. J

Work done by gravitational field 102 5 10. J

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

(b)(i) Since

E

E

Earthfromescape R

GMv

2 , we have

ES

SE

Sunfromescape

Earthfromescape

RM

RM

v

v

5

4

103.3

110101.1

Sunfromescapev

56 0 10

escape from Sunv . m s–1

[C1] [M1] for correct manipulation and substitution [A0] for correct answer

J kg−1

J kg−1

Earth

A

B

C


3

(b)(ii) Kinetic energy of an atom of mass m is given as

kTmv smr2

3

2

1 2

...

60001038.131066.11 232

...27

smrv

4... 102.1 smrv m s–1

Since the r.m.s. speed of the hydrogen atom is less than the escape speed, hydrogen atoms can be found in the Sun’s atmosphere.

[M1] [M1] [A1] for correct deduction

4 (a)

3

3

1 11.0 10 = 1000

1 10f

T

s-1 [A1]

(b) Maximum kinetic energy = ½ m(

0x )2

= ½ 4 12 2(1.0 10 )(2 1000 11 10 ) [M1]

= 192.4 10 J [A1]

(c) Magnitude of max acceleration =

2 12(2 1000) 11 10 [M1]

Max pressure difference = ma / area

=

4 2 12

-5

1.0 10 (2 1000) 11 10

6.5 x 10

[M1]

= 6.68 x 10-4 Pa [A1]

(d) Since the ear drum is very thin, loud or intense sound will have a larger amplitude which cause greater pressure/force to be exerted on eardrum, damaging it. [A1]

5 (a) The electric field strength between 2 parallel plates is uniform. Hence, in absence of any other charged bodies, the electric potential changes linearly between the 2 plates (or the potential gradient is constant). [B1] Midway between +50 V and -50 V would then be 0V. [B1]

(b) E =

V

d

=

3

50 50

40 10 [M1]

= 2500 V m-1 [A1]

(c) By Conservation of Energy,

Loss in EPE =q V = 19 201.6 10 100

40

[C1]

OR Loss in EPE q E d = 19 201.6 10 2500

1000

Gain in KE = Loss in EPE


4

2 2

227 2 27 5 19

1 1

2 2

1 1 201.67 10 1.67 10 1.3 10 1.6 10 100

2 2 40[M1]

f i

f

mv mv q V

v

vf = 1.63 x 105 m s-1 [A1]

(d) [B1 for correct shape (parabolic, acceleration upwards)]

Note that the start point should NOT have a horizontal gradient and end point should NOT be vertical.

6 (a) The changing current in coil A produces a changing magnetic flux density (B) since

magnetic flux density is directly proportional to the current. [B1] Hence, the alternating current flowing coil A produces a constantly changing magnetic flux linkage in coil B. [B1] Since Faraday’s law states that the induced e.m.f. is directly proportional to the rate of change of magnetic flux linkage, an e.m.f. is induced in coil B. [B1]

[3] (b) t = 5 ms, 15 ms or 25 ms

[1]

(c) Since the gradient of the tangent of the graph at this point is the largest, it corresponds to greatest rate of change of magnetic flux linkage. Hence induced e.m.f. is a maximum.

[1]

(d) From t = 10 ms to 15 ms, current in coil A is decreasing. The magnetic flux linkage through coil B is decreasing. [M1] According to Lenz’s law, the induced emf will be directed such that the current it causes to flow in the coil B to oppose this change in coil B. [M1] Hence, the current in coil B flows in the same direction as coil A. [A1]

[3]

- 50 V

+ 50 V

A

B

plate Y

plate X


5

7

(a) (i) Correct points plotted [B1]

(ii) Correct BFL with even scatter of points on both sides of the line [B1] (b) Since lgR-lgI gives a straight line graph, lg R is linearly related with lg I [M1]

R and I do not vary linearly. [A1]

Alt: Since the scales are non-linear, a straight line graph shows that R and I do not vary

linearly.

(c) The resistance of the LDR increases as intensity decreases. [M1] By potential-divider rule, the higher the resistance of the LDR, the larger the potential difference across BC. [M1] Hence, the system should be connected across BC. [A1]

(ii) 1. 7:30 am [B1]

2. Intensity = 20 000 lux.

Reading off Fig. 8.2, R = 30 Ω [B1]

3. Using potential-divider rule,

12

303.0 12 M1

30

90 Ω A1

LDRout

LDR

RV

R r

r

r

4. I = V / R

= 9.0 / 90

= 0.10 A [M1]

(ii) Since the system is connected in parallel to the e.m.f. source, the potential difference across the system is always 12 V. [M1] Hence, the lights will always be turned on. [A1]

1

10

100

1000

10000

100000

1000000

1 10 100 1000 10000 100000


6

Proposed solutions to Q8 Planning

Item Description

Basic

Procedure

(BP) [2]

Vary standard mass/height AND measure time taken to obtain moment of inertia of rotating objects.

[BP1]

Determine from gradient of graph to obtain I. [BP2]

Diagram

(D) [1]

Diagram of workable arrangement that has correct labels of equipment and variables. [D1]

Method of

Varying and

Measuring/

Determining

the IV.

Measuring/

determining

the DV

Analyzing

the data

(M) [3]

Obtain moment of inertia for apparatus without bicycle wheel [M1]

Repeat experiment for 6 sets of readings

Plot a graph of t2 against 1/m

Determine the moment of inertia Iapp from gradient = h

d g

2

8

Obtain moment of inertia for apparatus with bicycle wheel [M2]

Repeat experiment for 6 sets of readings

Plot a graph of t2 against 1/m

Determine the moment of inertia Icombine from gradient = h

d g

2

8

Calculate moment of inertia for bicycle wheel [M3]

Determine moment of inertia of the bicycle wheel from

Iwheel = Icombine – Iapp.

Control of other

Variables

(CV) [2]

Ensure the same height from the ground is used throughout the experiment by measuring height and

adjusting the number of times the string is wound. [CV1]

Use the same rotating axle throughout the experiment AND diameter of axle measured using a

vernier caliper / micrometer screw gauge. [CV2]

Other details

(R+S = max 4)

(Reliability)

(R)

Steps taken to reduce random error/uncertainty

Take average diameter of the axle at different points where the string will be wound over. [R1]


7

Item Description

[Max 3] Take average time taken for object to fall from rest to the ground. [R2]

Use of photogates attached to a digital timer and placed in correct positions to reduce human errors.

[R3]

Experiment with different standard masses to start the apparatus rotating such that it will give a time

of fall between 4 and 10 seconds. This is to reduce the percentage uncertainty due to errors in the

recording of time of fall. [R4]

Ensure that the bicycle wheel is rotating upright throughout the experiment by making sure that the

wheel is fitted properly on the axle. [R5]

Safety

Precautions

(S)

[Max 2]

Place a thin rubber mat below the mass to prevent damage to the ground/mass. [S1]

Do not place foot below the mass as fast falling mass can hit the foot and cause injury. [S2]


8

Diagram

Procedure

1. Setup the apparatus as shown the diagram without the bicycle wheel.

2. Tie a long string to a hole in the axle and wind the remaining length around the axle. The

string should be longer than the height of mass from the ground.

3. Measure the diameter of the axle at three different points where the string will be wound

over with a vernier caliper. Take the average [R1] and record this as d. [CV2]

4. Photogate A should be positioned such that it starts the digital timer just when mass

starts to fall and photogate B should be positioned such that it stops the digital timer

when the mass hits the ground. [R3] (alternative: use stop watch)

5. Measure the height of the mass from the rubber mat on the ground using a metre rule.

Record this as h. Keep this height constant throughout the experiment by adjust the

number of times the string is wound. [CV1]

6. Experiment with different standard masses to start the apparatus rotating such that it will

give a time of fall between 4 and 10 seconds. This is to reduce the percentage

uncertainty due to errors in the recording of time of fall. [R4] Record the standard mass

used as m. [M1]

7. Drop the mass and record the time taken as shown on digital timer for the mass to fall

from rest to the ground. Repeat this step to obtain another time. Calculate the average

time [R2] and record it as t. [M1]

8. Repeat steps 5 to 7 to obtain a total of six sets of data using different standard masses

m.

9. Plot a graph of t2 against 1/m.

10. Since h h

td g m g

2

2

8 1 2, the moment of inertia of the apparatus Iapp can be obtained

from the gradient of the graph using gradient = h

d g

2

8 [M1]

11. Attach the bicycle wheel to the same rotating axle [CV2] as shown in the diagram.

Ensure wheel is properly fitted such that it rotates upright throughout the experiment. [R5]

rotatable axle support with bearings

support with

bearings

mass

Fig. A

rubber mat

h

bicycle wheel

photogate A

photogate B

digital timer


9

12. Repeat steps 5 to 10 for to obtain the combined moment of inertia of the setup Icombine.

[M2]

13. The moment of inertia of the bicycle wheel Iwheel can be obtained from Iwheel = Icombine – Iapp.

[M3]

Safety Considerations

Place a thin rubber mat below the mass to prevent damage to the ground/mass. [S1]

Do not place foot below the mass as fast falling mass can hit the foot and cause injury. [S2]

Derivation (for info):

By conservation of energy

f f

f f

f f

mgh mv

v vmv

d d

2 2

2

2

2

1 1

2 2

4 21 1

2 2

f i

average

f

v v hv

t

hv

t

md gt

h

2 2

2

2

14 2

This question paper consists of 21 printed pages

MERIDIAN JUNIOR COLLEGE Preliminary Examination Higher 2

_______________________________________________________________________

H2 Physics 9646/3


2 hours

_______________________________________________________________________

Class Reg Number

Candidate Name _____________________________

READ THESE INSTRUCTIONS FIRST This booklet contains Sections A and B of the Preliminary Examination Paper 3. Do not open this booklet until you are told to do so. Section A Answer all questions.

Section B Answer any two questions.

You are advised to spend about one hour on each section. Write your answers on this question booklet in the blanks provided. INFORMATION FOR CANDIDATES

The number of marks is given in brackets [ ] at the end of each question or part question. Marks will be deducted if units are not stated where necessary or if answers are not quoted to the appropriate number of significant figures.

All working for numerical answers must be shown. You are reminded of the need for good English and clear presentation of your answers.

Examiner’s Use

Section A

Q1 / 10

Q2 / 15

Q3 / 10

Q4 / 5

Section B

Circle the questions you

have attempted

Q5 /20

Q6 /20

Q7 /20

Deductions

Total /80


2


8 m s

-1



= (1/(36)) x 10-9 F m-1












s = ut + 1

2at2

v2 = u2 + 2as




r



= 2 2

o-x x


2kT




Q

r



where k =

2

2

8 ( )m U E

h


decay constant

=

1

2

0.693

t


3

Section A Answer all the questions in this section.

(c) A monatomic ideal gas is put through the cycle of operations as shown in the PV diagram

below, AB is an isothermal process and BC is an adiabatic process. The temperature of

the gas at A is 300 K.

(i) The first law of thermodynamics may be expressed in terms of the equation ∆U = Q + W

Identify each of the terms in the equation.

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………….

[1]

1 (a) State what is meant by specific latent heat of fusion.

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

…………………………………………………………………………………………………. [1] (b) Using the kinetic model for matter, explain why the value of the specific latent heat of

vaporisation is larger than the value of the specific latent heat of fusion for any particular substance.

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

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

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

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

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


4

(ii) Determine the change in temperature, ΔT , of the gas from process B to C.

ΔT =…………………………K [3] (iii) Complete the table with either the value ‘0’, or the symbol ‘+’ or ‘–‘ to indicate the

signs of the thermodynamic quantities for the cycle ABCDA. [3]

Process ΔU Q W ΔT

A to B

B to C

C to D

D to A

2 (a) State two conditions necessary for the superposition of two waves to give rise to a well-

defined interference pattern.

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

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

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

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


5

(c) The frequency of the signal generator, is gradually decreased until a minimum intensity

of sound is observed at P.

(i) Calculate the frequency at which this is observed.

frequency = ……………………Hz [2]

(b) Fig. 2.1 shows two point sources of sound S1 and S2 separated by a distance of 1.7 m.

They are connected to a signal generator such that they produce tones of the same

frequency, phase and amplitude. The frequency of the signal generator can be varied

between 100 Hz to 800 Hz.

Fig. 2.1

Between S1 and S2 is a point P such that P is 0.50 m from S1.

Determine the phase difference between the waves from the sources arriving at P with both S1 and S2 switched on when the frequency of the generator was set to 800 Hz. (speed of sound = 330 m s-1)

phase difference = ……………………rad [3]

S1 S2 P

1.7 m

0.50 m


6

(ii) Determine the total number of points between S1 and S2 at which sound of minimum intensity is observed.

number of points = ……………… [3] (iii) Each of the point sources of sound has an output power of 600 W. Calculate the

resultant intensity of the sound at point P.

resultant intensity = ……………………W m-2 [3] (iv) State and explain how the intensity of the sound at the minima at P will be affected

if the power of S2 is slightly increased.

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

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

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

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

[2]


7

3 A cylindrical rod of mass 50 g is resting freely on two horizontal smooth conducting rails Y and Z of negligible width. Rails Y and Z are placed 0.15 m apart and connected to a 24 V battery of

negligible internal resistance. A uniform magnetic field of flux density 0.55 T is applied perpendicularly out of the paper as shown in Fig. 3.1. The rod has a diameter of 0.010 m and length 0.20 m. The resistivity of the rod is 2.3 x 10-3 Ω m.

(a) Explain what is meant by a magnetic flux density of 0.55T.

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

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

……..…………………………………………………………………………………………… [1] (b) (i) State and explain the direction in which the cylindrical rod moves when the switch

is closed.

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

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

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

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

…………………………………………………………………………………............ [2] (ii) Show that the current through the rod is 5.5 A. [2]

Fig. 3.1

0.15 m

Y Z

rod

24 V


8

(iii) Hence or otherwise, determine the initial acceleration experienced by the rod.

acceleration = …………………………………. m s-2 [2] (iv) State and explain the change if any to the acceleration of the rod as it moves on the

rail.

…………………………………………………………………………………………………

…………………………………………………………………………………………………

…………………………………………………………………………………………………

…………………………………………………………………………………………………

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


9

4 (a) Draw a labelled energy band diagram for a semiconductor that has been doped with Group V element. [2]

(b) Explain, using band theory, how the electrical conductivity of the doped semiconductor change when temperature is gradually increased from 0 K to room temperature.

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

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

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

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

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

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

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


10

Section B Answer two questions in this section.

5 (a) State Newton’s second law of motion.

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

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

…………………………………………………………………………………………………. [2] (b) A man of weight 700 N stands on a weighing scale in an elevator. As the elevator moves,

the reading on the weighing scale varies. In the following scenarios, state how the reading on the weighing scale varies as compared to the man’s weight (i.e. more than / less than / same as his weight of 700 N).

Scenario State more / less / same.

Elevator moving with constant speed

Elevator moving up and speeding up

Elevator moving up and slowing down

Elevator moving down and speeding up

Elevator moving down and slowing down [2]

(c) State and explain whether the person described in the following scenarios is

experiencing apparent weightlessness. (i) A parachutist who just opened her parachute in mid-air.

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

……………………………………………………………………………………………….

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


11

(ii) A boy being kept “afloat” above a big fan blowing air upwards with a high velocity (see Fig. 5.1).

Fig. 5.1

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

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

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

(d) State the principle of conservation of momentum.

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

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

……………………………………………………………………………………………….... [2] (e) Fig. 5.2 below shows 2 rubber balls colliding.

Ball A has a mass of 0.53 kg and moving to the right with a speed of 7.2 m s-1. Ball B has a mass of 0.34 kg and moving to the left with a speed of 9.6 m s-1. The collision is head-on and elastic. After the collision, Ball A and Ball B moves with velocities vA and vB respectively. Take rightwards as positive for this entire question.

Fig. 5.2

Big fan blowing air upwards

7.2 m s-1 9.6 m s-1

0.34 kg 0.53 kg

A B

vA vB A B

Before

collision

After

collision


12

(i) Show that vA = –5.9 m s-1 and vB = 11 m s-1. [3]

(ii) During the collision, it can be assumed that the forces acting on each of the balls

vary linearly. Given that t1 is the time which the balls first touch and t2 is the time which the balls

first separate, sketch on Fig. 5.3,

1. a solid line to show the force acting on Ball A;

2. a dotted line to show the force acting on Ball B. [2]

Fig. 5.3

F / N

t / s t1 t2 0


13

(iii) Hence, or otherwise, determine the maximum force experienced by each of the balls given that the duration of collision, (t2 – t1) = 0.42 s.

maximum force on Ball A = ………………….. N maximum force on Ball B = ………………….. N [3] (iv) On Fig. 5.4, sketch a labelled graph to show the variation with time of the velocity

of

1. Ball A using a solid line;

2. Ball B using a dotted line. [2]

Fig. 5.4

v / m s-1

t / s t1 t2 0


14

6 (a) Two cool gases, A and B, placed in a specially designed chamber are separated by a thick transparent glass as shown in Fig. 6.1. To excite the atoms of gas A, electrons with kinetic energy of 5.00 × 10-19 J each is directed as shown.

Fig. 6.1

Fig. 6.2 shows the energy states within the atoms of gas A, with level A1 being the lowest energy state.

Fig. 6.2

(i) State the highest energy level in which the atoms of gas A can be excited to in this setup.

level number = ………………… [1]

(ii) Determine the shortest wavelength of the photons emitted from gas A. State the region of the electromagnetic spectrum in which the radiation occurs.

wavelength = ………………… m

region: …………………………………… [3]

thick transparent

glass

beam of

electrons

Gas A Gas B

P

Energy/ 10-19

J Level number

0.31 0.00

0.78

1.36

2.42

5.45 A1

A2

A3

A4 A5 A6


15

(iii) The transparent glass in Fig. 6.1 is designed such that it allows only photons but not electrons to pass through it. Fig. 6.3 shows the energy states within the atoms of gas B, with level B1 being the lowest energy state. Electromagnetic radiation of various wavelengths is detected at point P.

Fig. 6.3

1. State all the transitions that resulted in the electromagnetic radiation detected at P.

transitions: ……………………………………………………

…………………………………………………… [2]

2. Electrons are also detected at point P. Determine the kinetic energy of these electrons.

kinetic energy = ………………… J [2]

3. Hence calculate the wavelength of the electrons.

wavelength = ………………… m [2]

Energy/ 10-19

J Level number

0.40

0.00

4.20 B1

2.00 B2

1.17 B3

B4 B5


16

(b) Fig. 6.4 shows how X-rays are produced inside an X-ray tube. The electrons, emitted

at the filament, are accelerated from rest using an accelerating voltage V to hit a target

of heavy metal at the anode and as a result X-rays are produced.

Fig. 6.4

Fig. 6.5 is the X-ray spectrum which shows the variations of the relative intensity of the

emitted radiation with its wavelength.

Fig. 6.5

(i) State and explain what does the series of sharp radiation peaks in Fig. 6.5 suggests about the energy levels of the electrons in the target.

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

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

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

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

accelerating

voltage, V

filament

anode

target

electrons

cooling

liquid

X-rays

Relative

Intensity

Wavelength / 10-11 m 2.0 4.0 6.0 0


17

(ii) Determine the maximum energy of X-ray photons emitted.

maximum energy = ………………… J [2]

(iii) Calculate the accelerating voltage V.

V = ………………… V [2]

(iv) On Fig. 6.5, sketch a graph to show how intensity of the X-rays emitted varies with wavelength if the current in the filament is reduced. [2]

(v) The target is replaced with another metal of higher atomic number. State and explain the effect on the maximum energy of the X-ray photons emitted.

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

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

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

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


18

7 (a) Nuclear reactions that occur spontaneously are able to produce significant amounts of energy. Discuss why this is not a violation of conservation of energy.

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

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

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

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

[2]

(b) Explain clearly two differences between fusion and fission.

(i) ………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

(ii)

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………..

[2]


19

(c) Fig. 7.1 shows a graph of mass per nucleon vs nucleon number.

Fig 7.1

Discuss an advantage and a disadvantage of using fusion reactions as opposed to fission reactions in a nuclear power plant in terms of energy considerations. You should use Fig 7.1, where possible, to support your answer.

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

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

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

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

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

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

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

(d) Besides in power plants, nuclear reactors are also used in ice-breaker ships deployed in

the Arctic Ocean for expeditions. The largest ever built, the Artika, uses two nuclear reactors that provide the propulsion engine with a total of 52 MW of propulsion.

(i) When breaking ice, the Artika can travel at a speed of 3.0 km h -1. Determine the force generated by the propulsion engine.

Force generated = …………………………. N [2]

Ma

ss p

er

nu

cle

on

Nucleon number


20

(ii) The main source of nuclear fuel used in the Artika is uranium-235. A uranium-235 nuclide bombarded with a slow-moving neutron will undergo fission to produce barium-141 and krypton-92 as the daughter nuclides with 3 other neutrons. Determine the energy released from the fission of one uranium-235 nucleus, using the following information: B.E. per nucleon of U-235: 7.5909 MeV B.E. per nucleon of Ba-141: 8.3260 MeV B.E. per nucleon of Kr-92: 8.5128 MeV

Energy per fission of U-235 = …………………. J [3] (iii) Given that the mass of a U-235 nucleus is 235.0439 u, determine the mass of fuel

used by the ship in one day of ice breaking.

Mass of U-235 used per day ………………… kg [3]


21

(iv) One of the daughter nuclides, Kr-92, is highly unstable and decays readily to form Rb-92. Suggest the type of decay that occurs.

Type of decay: ……………………………………………………………….. [1]

(v) Kr-92 has an atomic number of 36. Hence, write out the nuclear reaction equation

that describes the above decay.

………………………………………………………………………………….. [2] (vi) Given that the half-life of Kr-92 is 1.84 s, determine the decay constant of Kr-92.

Decay constant ………………… s-1 [2] End of Paper

1 JC2 H2 Physics Prelim Exam 2015 Solutions

MJC JC2 H2 Physics Prelim Examinations 2015 Paper 3 Suggested Solutions:

1a) The specific latent heat of fusion fL is defined as the quantity of heat required to

change a unit mass of a substance from the solid phase to the liquid phase without any change in temperature.

*Do not accept 1K and do not accept 1kg

[B1]

1b) The specific latent heat of vaporisation is larger than the specific latent heat of fusion as: Inter-molecular bonds In the melting process, molecules need only to break down the structure into a less-ordered arrangement of molecules (The molecular bonds are still present between the molecules, they are just weakened). In the vaporisation process, however, the molecular bonds are completely broken and this process requires a greater amount of energy than melting requires. Work done against atmosphere As a solid melts into a liquid, its volume does not increase substantially and very little energy is required to do work against the atmosphere as a result of this small volume increase. As a liquid vaporises into a gas, the increase in volume is very much larger and a lot more energy is required to do work against the atmosphere due to the large expansion.

[B1]

[B1]

1ci)

∆U = Q + W

∆U = increase in internal energy of the gas Q = heat gained by the gas W = work done on the gas

[B1]

ii)

1 0 3 0 5 0 1 0

300

300

500

500 300 200

( . )( . ) ( . )( . )

C CA A

A C

C

B A

c

PV nRT

P VP V

T T

T

T T K

T K

T K

[C1]

[A1]

[B1]

iii)

Process ΔU Q W ΔT

AB 0 - + 0

BC + 0 + +

CD - - 0 -

DA + + - +

Any 1 row wrong deduct 1 m.

[3]


Q2 (a) 1. Sources must be coherent, where the 2 waves have a constant phase difference.

2. The 2 waves must be unpolarised or polarised in the same plane

3. The amplitude of both waves must be similar.

[1 mark for each point.]

(b)

2x

1 2 0 52

330 800

. .

/

[M1] [C1 for x ]

10 7 rad. [ [A1]

(c) (i) For destructive interference,

1 5x

.

0 467. [M1]

707 Hzf [A1]

(ii) Internodal distance = 0.467 / 2 = 0.2335 m [C1]

Number on right of P = 1.2/0.2335=5 [M1 for either]

Number on left of P = 0.5/0.2335=2

Total number of nodes = 2 + 5 + 1 = 8 [A1]

(iii)

24

PI

r

1 2

600190 99

4 0 5I .

( . ) [either: C1]

2 2

60033 16

4 1 2I .

( . )

2

190 99 33 16netI . . [M1]

-265 2 W m. [A1]

(iv) The intensity at the minima will be lower [A1] because with an increase in

amplitude from S2, the waves can better cancel each other out when they meet exactly out of phase. [M1]


3 (a) A magnetic flux density of 0.55 T means that a force of 0.55 N will act on a wire of

length 1 m with current 1 A passing through it when wire is placed perpendicular to the magnetic field [B1]

(b) (i) The rod moves downwards.

When the switch is closed, current flows in the rod. As the rod is placed in a magnetic field, it experiences a magnetic force. And by Fleming’s LHR, the rod the direction of the magnetic force is downwards.

[B1] [B1]

(ii)

3

23

0 152 3 10 4 39

10 10

2

.( . )( ) .

lR

A

[M1]

24 4 39

5 46

( . )

. A

V IR

I

I

[B1]

(iii) 0 55 5 46 0 15 0 450

0 4509 0

0 050

-2

. . . . N

.. ms

.

m

m

F BIL

Fa

m

[M1]

[A1] (iv) Acceleration will decrease [B1]

When the rod moves, there is rate of change of magnetic flux linkage, hence there will be an induced emf in the rod that opposes the emf of the battery. [B1] This reduces the current, which then causes the force and acceleration to decrease. [B1]

4 (a)

Conduction Band

Valence Band


[B1] Award mark only if there are more electrons in the conduction band as compared

to holes in valence band.

[B1] Award mark only if donor level is drawn closer to conduction band labelled.

(b) At zero kelvin, the valence band is full and the conduction band is empty, thus the semiconductor has zero conductivity. [B1]

When the temperature is increased slightly, the electrons from the donor levels just below the conduction band are thermally excited to the conduction band, increasing conductivity. [B1] As the temperature is further increased, some electrons from the valence band are thermally excited into the conduction band, thus creating holes in the valence band. This increase in charge carriers also contributes towards the increase in conductivity. [B1

Section B

5 (a) Newton’s second Law of Motion states the rate of change of momentum of a body

is directly proportional to the resultant force acting on it and the change takes place in the direction of the force. [B2]

(b)

Scenario State more / less / same.

Elevator moving with constant speed Same

Elevator moving up and speeding up More

Elevator moving up and slowing down Less

Elevator moving down and speeding up Less

Elevator moving down and slowing down More [2]

[B2] for all correct. [B1] for any 3 or more correct

(c) (i) No. [A1] When the parachute just opened, the drag force upwards due to air on the parachute is much larger than the weight of the parachutist. Hence the parachutist experiences a net force upwards. [M1] She is accelerating upwards and hence cannot be experiencing weightlessness since she is not free-falling at acceleration due to gravity.

(ii) No. [A1] The drag force upwards due to air on the boy is equal and opposite to the weight of the boy. Hence the net force on him is zero. [M1] He is not experiencing weightlessness since he is not free-falling at acceleration due to gravity.

(d) The principle of conservation of momentum states that the total momentum of a system of objects remains constant provided no resultant external force acts on the system.


(e) (i) By conservation of momentum,

0.53 0.34 0.53 7.2 0.34 9.6

0.53 0.34 0.552 [B1]

A B

A B

v v

v v

Relative speed of approach = relative speed of separation

9.6 7.2

16.8 [B1]

or 16.8

A B

A B

B A

v v

v v

v v

Substituting,

-1

-1

0.53 0.34 16.8 0.552

5.931 5.9 m s

5.931 16.8 10.87 11 m s

A A

A

B

v v

v

v

(ii)

Fig. 5.3

(iii) Considering Ball A,

on A

0.53 5.9 7.2Average [M1]

0.42

16.57 N

ApF

t

Hence, max on A 16.57 2 33.1 N [A1]F

By Newton’s third law, Fon B = 33.1 N [A1]

OR Considering Ball A,

on A, max

on A, max

Area under graph for ball A

1 0.53 5.9 7.2 [M1]

2

33.1 N [A1]

Ap

F t

F

By Newton’s third law, Fon B = 33.1 N [A1]

[B1]

F / N

t / s t1 t2 0

[B2] – correct shape and symmetrical vertically and horizontally OR [B1] – correct shape but not symmetrical Labels not marked.


(iv)

Fig. 5.4

6 (a) (i) (−5.45 + 5.00) x 10

-19 = −0.45 x 10

-19 J

Hence the highest energy level which the atom can be excited to is A4 (i.e. −0.78 x 10

-19 J)

[B1]

(ii) Highest energy photon emitted from A4 to A1 transition E = ((−0.78) –(−5.45)) x 10

-19

4 1

hc A A

6.63 3.0 . . [M1]

. m [M1]

34 8

19

7

10 100 78 5 45 10

4 26 10

Region: visible light (accept blue or violet light)

[A1]

(iii) 1. De-excitation in Gas A will result in photons of the following energy

A4 to A1 photon energy : 4.67 x 10-19 J



Only photon from transition A2 to A1 (photons of energy 3.03 x 10

-19) matches the energy difference in Gas B (between B1 to B3). This

will result in excitation of atoms in gas B to B3. The de-excitation of atoms from B3 to the lower states results in the emission of photons detected at P. Transitions: B3 to B1, B3 to B2, B2 to B1

[B1 – only 2 transitions stated correctly] OR [B2 – 3 transitions stated correctly]

v / m s-1

t / s t1 t2 0

[B1] – correct shape (nonlinear line during collision) [B1] – correct labels and

intersection is positive

7.2

11

–5.9

–9.6


2. Highest energy photon bombarding B E = ((−0.78) –(−5.45)) x 10

-19

= 4.67 x 10-19

J Ionisation energy of gas B = 4.20 x 10

-19 J

Kinetic energy of electron = (4.67 – 4.20) x 10

-19

= 4.7 x 10-20

J

[M1] [A1]

3.

K

K

h p

pE

m

h

m

h

E m

6.63 [M1]

. 9.11

. m [A1]

2

2

34

20 31

9

2

2

2

10

2 4 7 10 10

2 3 10

Note: No ecf for EK > 4.1 x 10

-14 J as it suggests that the electron is travelling

faster than the speed of light.

(b) (i) The sharp peaks occur at discrete (or specific) wavelengths of radiation which show that the energies of the emitted X-ray photons are quantised. The emission of the quantised photons suggests that energy transitions in the target atom must be quantised (or specific). Thus the energy levels of the atoms in the metal must be discrete.

[B1] [B1]

(ii) From graph, smallest wavelength λ0 = 1.2 x 10-11

m

K max

hcE

6.63 3.0 [M1]

1.2

.

. J [A1]

0

34 8

11

14

14

10 10

10

1 658 10

1 7 10

(iii) Assuming that the initial KE of electrons at the filament is negligible as compared to the final KE at the target. Loss in EPE = Gain in KE

- -. V . [M1]

V .

. V [A1]

19 14

5

5

1 6 10 1 658 10

1 036 10

1 0 10


(iv)

[B1 – Overall relative intensity is lower with same characteristic peaks OR minimum wavelength] OR [B2 – Overall relative intensity is lower with same characteristic peaks AND minimum wavelength]

(v) The maximum energy of the photons emitted will remain the same. Its value depends on the accelerating voltage which is constant. The maximum energy of the emitted photons is independent of the metal target.

[B1] [B1]

7 (a) During a spontaneous nuclear reaction, there is mass defect as the total

mass of the product is lesser than that of the original nuclides.

The excess mass is converted into energy, according to 2E mc ,

which accounts for the energy released during the reaction.

[B1] [B1]

(b) Any two of the following:

Fusion is the combining of two light nuclei to produce a heavier nucleus while fission is the disintegration of a heavy nucleus into two lighter nuclei of approximately equal masses. Fusion cannot occur spontaneously but fission can. Fission produces radioactive materials while fusion does not.

[B1] each

(c) From Fig. 7.1., fusion reactions are more efficient as the energy

produced per nucleon is greater than that of fission. This can be seen from the steep gradient of the left portion of the graph, which represents fusible nuclides in comparison to the gentle gradient of the right portion of the graph, which represents the fissile nuclides. However, it does not occur spontaneously as it requires large amounts of energy to overcome coulomb repulsion.

[B1] [B1] [B1]


(d) (i)

.

6

7

300052 10

3600

6 2 10 N

P Fv

F

F

[M1] [A1]

(ii)

. . .

.

6 6 6

11

(C1 mark for finding B.E. of the nuclides correctly)

Energy released,

B.E. of products B.E. of reactants

141 8 3260 10 92 8 5128 10 235 7 5909 10

173 MeV

2 77 10 J

E

[C1] [M1] [A1]

(iii) Total number of reactions needed

.

.

6

11

23

52 10 24 3600

2 77 10

1 622 10

Total mass

. . .

.

23 27

2

1 622 10 235 0439 1 66 10

6 33 10 kg

[M1] [M1] [A1]

(iv) β-decay [A1] (v) 92 92 0

36 37 1Kr Rb e

1 mark for correct identification of beta particle 1 mark for atomic number of Rb

(vi) ln ln

.

.

1

2

-1

2 2

1 84

0 377 s

t

[M1] [A1]

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