Answer Scheme Trial SPM Paper 2 2020 part B and C

ANSWER SHEME TRIAL SPM 2020 SECTION B

9. (a) Extension of the spring is directly proportional to force 1
(b) Number of loads in Diagram 9.2 is greater than 9.1 1
The force acting on the spring in Diagram 9.2 is greater than 9.1 1
(c) Extension of the spring in Diagram 9.2 is greater than 9.1 1
The greater the number of loads, the greater the force acting on the
spring 1
The greater the force acting on the spring, the longer the extension of
the spring 1
The forces between atom are attractive force and 1
repulsive force
When the force is applied, the distance between atoms decreases 1
The repulsive force is acted on the atoms 1
When applied force is removed, repulsive force pushes the atoms 1
back to original

NO SCHEME SUB TOTAL
MARK MARK
10(a) To increase or decrease the alternating voltage (Vac)
1 1
(b) Number of turn of secondary in Diagram 10.2 is greater than 10.1
1 1
Output voltage in Diagram 10.2 is greater than Diagram 10.1 1 1
Output current in Diagram 10.1 is greater than Diagram 10.2 1 1
1 1
(i) The higher the number of turn of secondary coil, the higher the
output voltage 11

(ii) The higher the number of turn of secondary coil, the lower the
output current

(c) (i) 1. There is a cutting of flux// change of flux 1
(ii) 2. Produced induced e.m.f // induced current 12

1. Using Fleming’s Right hand rule 1
2. Force is downwards, Magnetic Field from North to South, 12

so the direction of current from Q to P.

(d) Aspect Reasoning

copper wire -resistance is very low 1,1
1, 1
- reduce power loss 1, 1
1, 1 10
thicker wire -resistance is very low
1, 1
- reduce power loss

Use a laminated avoid the Eddy current

core

Use soft iron core - can be magnetized and

demagnetized easily

- avoid hysteresis.

Winding the reduce the leakage of magnetic flux
secondary coil on
top of the primary
coil

TOTAL 20 M

SECTION C

11. (a) (i) Amount of heat required to change 1 kg of liquid into a gas without a change in the 1
(b) (i) temperature.
1
Q = ml 1
Pt = ml 1
200 (2.6 X 60) = 0.09 l (with
l = 31200/0.09 unit)
l = 3.47 x 105 Jkg-1 1
1
(b) (i) (200 W)(2.4 × 60 s) = (0.09) l 1
(with
l = 28800/0.09 unit)
l = 3.2 X 105 J kg-1
1
(ii) Q = mc 1
1
Pt = mc (with
(200 W) [(4.8 – 3.6) × 60 s] = (0.09 kg) c (218 oC – 78 oC) unit)

c = 14400/ 12.6 1,1
c = 1142.86 J kg-1 oC-1
1,1
(c) Explanation
Specification - Cool down faster 1,1
Number of fin blades: - Increases the surface area as so releases heat
greater/more faster 1,1

specific heat capacity: High - Small rise in temperature 1,1
- Absorbed more heat from engine
Size of fan: Big
- Movement of air faster
Rusting rate: Low - more heat can be brought away
- cool down larger area
C is chosen.
. - not easily rust
- long lasting

greater number of fin, high specific heat capacity, big
fan and low rusting rate

NO SCHEME SUB TOTAL
MARK MARK
12(a) The ratio of potential difference to current
(b)(i) Parallel 1 1
1
(ii) 1 = 1 + 1 1 3
R 10 10
R =5Ω 1 3
1
(iii) I = 12 1 1
5 1 (with
unit)
= 2.4 A
1
(c)(i) Increase
1
(ii) Effective / total resistance decreases 1 (with
unit)

1

1

(d) Explanation 1,1
Aspects - High resistance
Tungsten - produce more light 1,1
- high resistivity
Coiled shape - easy to be hot 1,1
- high melting point
1,1
The longer the length, the
higher the resistance 1,1
10
Thin wire - Increase the resistance
- produce more light/heat
- easy to be hot

Low pressure of Nitrogen - Avoid the bulb to breaks
gas - avoid the filament to

U is the most suitable lamp evaporates at high
temperature.
Tungsten filament,
coiled, Thin wire
and low pressure of
nitrogen gas