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UNIT TWO — PAPER ONE

MARK SCHEME

Question 1 COMMENT 1 KC I UK

(a) The force of attraction or repulsion between two point charges is directly proportional to the product

1

of the charges and inversely proportional to the square of the distance between the charges.

(b)

ite (Gauge) — – ye wire

(1)

— The strong E-field around the wire charges the

dust and smoke particles —ve (1)

— The negatively charged particles are collected on the +ve plate (1)

— Plate is shaken — debris falls to bottom of

chimney where it is collected (1)

(c) +41AC —511C • < • >

F6 F5 (1)

F = F5 F6 = 9x109 x4X10-6 >C3x104 9X109 x4x10-6 X6X104

(5X10-2 )2 (10x10-2 )2

—72 + 21.6 = —50.4N (1)

To the left (1)

4

2

3

TOTAL 1515

Syllabus Objectives: M1:1.2, 5.1

002474/SPEC/MS/2001

CARIBBEAN EXAMINATIONS COUNCIL

ADVANCED PROFICIENCY EXAMINATION

PHYSICS

SPECIMEN PAPER

MARK SCHEME

UNIT 2

PAPER 01

Q = CV

Q= CV = C IV + qv + c3v

= C 1 + C2 + C3

(c) (1) V = 10 V

Q=CV=4x 10 =4011C

(iii) Ce = 121AF

Qe = 4011C

V = —Q

= 40

= 3.33V C 12

3

1

1

(1)

(1)

(1) 3

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UNIT TWO – PAPER ONE

MARK SCHEME

COMMENT I KC I UK Question 2

magnitude of charge on one of the conductors (a) C=

(1) p.d. across the conductors

Written statement (1) (b)

Vi Ici L 1C3 T TC 2 T

NU_ l c T

Q = CV

2

TOTAL IS IS

Syllabus Objectives: MI: 6.2, 6.4

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UNIrr TWO - PAPER ONE

MARK SCHEME

COMMENT KC UK Question 3

(a) Magnetic flux per unit area

(1) =BA

(1) Accept equation 3

Tesla (T)

(1)

(b) Shape 1 Direction 1 2

(c) (i) B = go/a + ALA

27cra 27crb

= 2x4n-x10-7 x30 271x0.5x10-2

= 2.4 x 10-3T

B = rb ra

4nx10-7 x30( 1 1 2n L1x10-2 2x10-2 )

= 3.0 x 10-4 T

3

(1)

(1)

2

TOTAL

5 1 5

Syllabus Objectives:M2: 7.1, 7.2, 7.3

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UNIT TWO — PAPER ONE

MARK SCHEME

Question 4

(a) Peak – maximum value rms – dc equivalent

Inns =

(b) (i) P =12R= /02 RSinwt

P = 10 2 R

COMMENT I KC I UK

3

2

(c) (i) f = 20 .11 0-3 = 50Hz

V =6"0V P

6 V =-

42-= 4.2V

(1)

3

thns

All +ve Shape 1 Period 1 2

TOTAL

5 I 5

Syllabus objectives: M2: 2.1, 2.2, 2.3

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UNIT TWO — PAPER ONE

MARK SCHEME

Question 5

(a) Voltage at the power station is stepped up by a transformer (1) --> Reduction in power loss (1)

(b) X — soft-iron core (1) Y — secondary coil (1)

COMMENT I KC I UK

3

2

(c)

(1)

(1)

Voltage is stepped down to consumer

(i) v = v pp ss

V / 24 x3 =0.6A s s

P V 120

(ii) N—

N Vs

(1) (1)

N p = 120 = 5

Ns 24

(1)

2

(1)

3

TOTAL 1 51 5

Syllabus Objectives: M2: 2.4, 2.5, 2.6

HA — 1 FA- 1 FA - 1 Connection 1

2

4

(1)

(c) (i)

BA A3 B2 A2 B 1 A1

Co (1)

A

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UNIT TWO - PAPER ONE

MARK SCHEME

COMMENT I KC I UK

Question 6

(a) HA — Adds 1 S column only (two inputs)

2

FA— Adds the other columns (3 inputs)

(b) (i) A B Q

0 0 0 0 1 1

1 1 0 1 1 1 0

(ii) A3 A2 A1 B3 B2 B1 (1) 1

1 1 0 1 0 1

TOTAL 5 5

Syllabus Objectives: M2: 5.2, 5.6, 5.7

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UNIT TWO - PAPER ONE

MARK SCHE I

COMMENT I KC I UK

Question 7

(a) (i) Heated filament emits electrons (1)

(ii) High Z number, high m.p. (1) 3

(iii) 99% of electrons energy converted to heat .-. cooling (1)

(iv) Ern- radiation is produced by accelerating or decelerating electrons (1)

:. when the electrons are stopped by the metal 2 X-rays are produced (1)

hc (v) eV = A — (1)

hc . 6.63x10 -34 x3x10 8

2.07 x10-11m ( 1) 2 eV 1.6x10-19 x60x10-'

(b) I = Ioe-gx (1)

= 0.5 =

in 0.5 x = (1) 3

-0.27

= 2.6 cm (1)

1 TOTAL 1 5 1 5

Syllabus Objectives: M3:1.2, 1.10, 1.12

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UNIT TWO-PAPER ONE

MARK SCHEME

COMMENT I KC I UK Question 8

(a) - Pass electrons through a crystal (1) - Crystal acts as a diffraction grating (1) - Diffraction pattern is observed (1) 3

(b) E —1

mv2 =

1 m2 v2(1) k -

2m

p my

k - 2m 2

(c) (i) = (1)

h 6.63x10 34 3 P= = (1) 1X10-9

= 6.63 x 10-25 kg ms-1 (1)

l

(6.63 x10-"g

) 2

(11) Ek- (1) 2 X 9.1X10-31

= 2.42 x 10-19J (1) I I 2

TOTAL 15 IS

Syllabus Objectives: M3: 1.19, 1.20, 1.21

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UNIT TWO — PAPER ONE

MARK SCHEME

Question 9 COMMENT KC UK

1 (a) Energy to separate nucleous

(b) (i) A Z 50 (1) 1

(ii) When large nuclei split B E increases (1)

— Increase BE —> Energy released (1) 3

— When small nuclei fuse B E increases (1)

(c) Mass of protons = 8 x 1.00728 = 8.05824 (1) Mass of neutrons = 8 x 1.00866 = 8.06928 (1) Mass of electrons = 8 x 0.00055 = 0.00440 (1)

m = 16.13192 (1)

BE = (1E13192 —15_99492) 931 =0.137x931 5 = 128 MeV (1)

TOTAL 1515

Syllabus objective: M3:3.1, 3.2, 3.5, 3.6