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Awesome physics with tcer Alina@Rajabio 13 Ogos 2019 1

Scheme Answer

PAPER 2 (STRUCTURE) NO. ANSWER MARK / NOTE

1. (a) Vernier calliper 1 (b) Negative zero error // systematic error 1 (c) = 2.20 + 0.04 = 2.24 cm

Actual reading = 2.24 – (-0.08) = 2.32 cm 1

1 (awu) TOTAL 4 MARKS

NO. ANSWER MARK / NOTE

2. (a) The force that opposes the motion 1 (b) F = ma

F = 1 000 (2) = 2 000 N

1

1 (awu) (c) Air resistance 1 (d) Move at constant velocity 1

TOTAL 5 MARKS NO. ANSWER MARK / NOTE

3. (a) (i) Pascal’s Principle 1 (ii) Transmission of pressure is slower // less effective // the force

is used to compress the bubble 1

(b) (i)

1

1 (awu)

(ii)

1 1 (awu)

TOTAL 6 MARKS

-4

4

F 15P = = A 5.2 x 10

P = 2.88 x 10 Pa4 -4F =PA = 2.88 x 10 (6.2 x 10 )

F = 17.856 N



4. (a) The lowest temperature in theory in which the pressure and the kinetic energy of gas molecules are zero

1

(b) • Temperature increases, Kinetic energy will increases • Collisions between the molecule increase, thus the volume

will increases

1 1

(c) (i) Charles’ law 1 (ii)

1

(d)

1


1 2

1 2

2

32

V V=T T

V35 =40+273 70 + 273V = 38.35 m



5. (a) Angle of incidence in denser medium when refracted angle is 90o

1

(b) (i)

2

(ii) Refraction 1 (c) (i)

1 1

(ii) Distance from the base of the beaker decrease Apparent depth increase

1 1


6. (a) One color // One wavelength // lambda // frequency 1 (b) (i) a1 is shorter than a2 // a1 < a2 1 (ii) x1 is longer (than x2) // x1> x2 1 (iii) When a increase, x decrease 1 (iv) Wavelength λ // distance between double slit to screen, D 1 (c) Interference 1 (d) Constructive interference correspond to bright fringes

Destructive interference correspond to dark fringes 1 1

TOTAL 8 MARKS

H 20η = =h 12

η =1.67



7. (a) Ratio of potential difference to current 1 (b) (i)

1 2

(ii) Parallel (Diagram 7.2) Effective resistance is low, more current flow

1 1

(c) (i) More number of dry cells More current flow

1 1

(ii) Series Voltage lost (voltage drop) will decrease

1 1


8. (a) A temporary magnet when current flow through a conductor 1 (b) (i) Number of turns of coil wire // number of dry cells 1 (ii)

1

(c) (i) Bigger size of plate Can give more space for trapping iron scraps

1 1

(ii) Soft iron core plate Can be magnetized and demagnetized easily

1 1

(iii) Bigger diameter of wire for coil Reduces the resistance so produce bigger current

1 1

(iv) Higher number of turns Produce stronger magnetic fields

1 1

(v) Crane B 1 TOTAL 12 MARKS

SeriesR = 2+2+2 = 6 Ω

Parallel1 1 1 1 = + +R 2 2 2R = 0.667 Ω


CHAPTER: FORCE & MOTION NO. ANSWER MARK / NOTE

1. (a) Rate of change of displacement 1 (b) (i) Constant velocity // a = 0 ms-2 1 (ii)

1

(c) Displacement (area under the graph) 1 TOTAL 4 MARKS


2. (a) The product of the applied force and the displacement in the direction of the applied force

1

(b) (i) W = Fs = 110 (0.5) = 55 J // Nm

1

1 (awu) (ii) F = ma

110 - 100 = 10a a = 1 ms-2

1



3. (a) P = lift force Q = weight

1 1

(b) FL = W OR Lift force = Weight // P = Q // P + Q = 0 // P – Q = 0

1

(c) (i) FR = 150 kN 1 (ii)

1

1 (awu)

TOTAL 6 MARKS

F = 200 kN - 150 kNnett = 50 kNF = manett

50 kN -2 a = = 10 ms5000


CHAPTER: LIGHT NO. ANSWER MARK / NOTE

1. (a) The bending of light ray at the boundary as it travels from one medium to another of different optical densities

1

(b) (i) Angle refraction of the glass blocks in Diagram 4.1 > in Diagram 4.2

1

(ii) Density of the glass blocks in Diagram 4.2 > in Diagram 4.1 1 (iii) Density increase, angle of refractin decrease 1 (c)

1

1 (awu) (d) Decrease 1


2. (a) Concave mirror 1 (b) Reflection 1 (c) Virtual, Upright, Magnified 1 (d) Same Size 1


3. (a) Convex mirror 1 (b) Wider field of view // wider vision 1 (c)

2

TOTAL 4 MARKS

Sin iη = Sin rSin30 = Sin19

= 1.536


CHAPTER: ELECTRONIC NO. ANSWER MARK / NOTE

1. (a) Transistor n-p-n 1 (b) (i) IB + IC = IE 1 (ii)

1

1 (awu)

(c) (i) The bulb will not light up Resistance R2 is small // V2 is small

1 1

(ii) Current amplifier 1 TOTAL 7 MARKS


2 (a) (i) Diagram 2.1 – anode of the diode is connected to positive terminal of dry cell.

OR Diagram 2.2 – anode of diode is connected to the negative terminal of dry cell.

1

(ii) Bulb in diagram 2.1 lights up, bulb in Diagram 2.2 does not lights up.

1

(iii) The bulb does not light up when the positive diode is connected to negative battery // reversed biased

OR The bulb light up when the positive diode is connected to positive battery // forward biased

1

(b) (i)

3

(ii)

1

(iii) Capacitor 1 TOTAL 8 MARKS

2 2

2

2

V R=6 R +1500R =750Ω


ESSAY SECTION B

SPM 2013

NO. ANSWER MARK / NOTE 1 (a) Current 1

(b) (i) • Time taken to charge the van de Graaff generator in Diagram 10.1 < Diagram 10.2

• Quantity of charge produced on the dome in Diagram 10.1 < Diagram 10.2

• Angle of deflection of the micrometer pointer in Diagram 10.1 < Diagram 10.2

1 1 1

(ii) • The longer the time taken to charge the Van de Graaff generator, the greater the quantity of charge produced on the dome.

• The greater the quantity of charge produced on the dome, the greater the current produced.

1 1

(c) • coil shape, length increases • resistance increases • heat produced increases • heat energy convert to light energy

1 1 1 1

(d) Modification

Characteristics Explanation Small cross-sectional area of wire

Higher resistance

Longer length of wire Resistance increase More number of turns of wire

Potential difference decrease along the wire

Higher conductivity Current can flow between the slider and wire

High resistance Small current flow

10

TOTAL 20 MARKS


SPM 2006

NO. ANSWER MARK / NOTE 2 (a) A magnetic field is a region in which a conductor will

experience a magnetic force 1

(b)

• the number of turn of the coils in Diagram 10.2 > Diagram

10.1 • the pattern of the iron filings in Diagram 10.2 is denser

than Diagram 10.1 • the angle of deflection of the ammeter indicator in

Diagram 10.2 > Diagram 10.1

1 1 1

(c) (i) • The closer the pattern of the iron filings, the greater the strength of the magnetic field

1

(ii) • The number of turns of the coil increase, the greater the strength of the magnetic field

1

(d) • current flow, magnetic field is form around copper strip • due to the different direction of current flow,

S à R and P à Q • the direction of the magnetic field between the two strips

are in the same direction • this will produce a stronger magnetic region between the

strips, hence the strips repel each other

1 1 1 1

(e) (i) • when switched on, the flow of current in the coil forms magnetic field

• the magnetic field of coil intersect the magnetic field of permanent magnet

• catapult force formed at each side of the coil • a pair of equal but opposite forces is formed which rotates

the armature

1 1 1 1

(ii) Modification

Characteristics Explanation

use a diode and capacitor / set a rectification circuit

to change a.c to d.c

replace slip rings to commutator

to reverse the direction of the current flow in the armature

increase the number of turns of the coil

to increase the strength of the turning force

6

TOTAL 20 MARKS


ESSAY SECTION C

TRIAL KEDAH 2011

NO. ANSWER MARK / NOTE 1 (a) Radioisotopes are unstable isotopes which decay and give out

radioactive emission 1

(b) When water level is too high • water will block the radioactive emission • detector receive less radioactive emission

Outlet valve open because

• detector detect low reading • outlet valve controller activated and send signal to open

outlet valve

1 1

1 1

(c)

Decision making

Characteristics Explanation High initial activity So that it will higher than

background radiation Type of radioactive emission is beta

It have moderate penetrating power and less danger

Long half-life of the radioisotope

Can last longer /

Change from solid to liquid at high temperature

Always in solid state and easier to handle

Iron-60 High initial activity radioactive emission is beta Long half-life Change from solid to liquid at high temperature

10

(d) (i) The number of neutrons in an atom of radioisotope Br = 83 – 35 = 48

1

(ii) Krypton 1 (iii) decay equation for radioisotope

𝐵𝑟 → 𝐾𝑟 + 𝑒()*

+,-+

+.-+

2

(iv) activity of radioisotope T after 9.6 hours, 9.6 / 2.4 = 4 half life 384 à 192 à 96 à 48 à 24 24 counts per minute

1

1 TOTAL 20 MARKS


TRIAL KEDAH 2013

NO. ANSWER MARK / NOTE 2 (a) (i) 45° 1 (ii) n = 1 / sin 42o

= 1.49 1 1

(iii) Correct path for light ray A Correct path for light ray B

EXTRA INFO:

1 1

(iv) Total internal reflection 1 (b) • The layer of air near the ground is hotter

• The layer of air near the ground is less dense • Light ray from the sky passing the layers of air is refracted

away from the normal • Light ray undergoes total internal reflection when the

incident angle is greater than the critical angle

1 1 1

1


(c)

Decision making

Characteristics Explanation Long To produce a higher magnification Big To receive more light // to produce a

brighter image Convex As a magnifying glass // To magnify

the image produced by the objective lens

High To produce bigger / magnified image

R Objective lens with long focal length and big diameter and convex lens for eyepiece with high power

10

TOTAL 20 MARKS

scheme answer-awesome physics@rajabio 13 ogos …...awesome physics with tcer alina@rajabio 13 ogos...

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