[edu.joshuatly.com] kedah stpm trial 2010 physics [w ans] [0453a76a]

8/12/2019 [Edu.joshuatly.com] Kedah STPM Trial 2010 Physics [w Ans] [0453A76A]

http://slidepdf.com/reader/full/edujoshuatlycom-kedah-stpm-trial-2010-physics-w-ans-0453a76a 1/45

htt // d h tl /


http://slidepdf.com/reader/full/edujoshuatlycom-kedah-stpm-trial-2010-physics-w-ans-0453a76a 2/45http://edu.joshuatly.com/http://www joshuatly com


http://slidepdf.com/reader/full/edujoshuatlycom-kedah-stpm-trial-2010-physics-w-ans-0453a76a 3/45http://edu.joshuatly.com/htt // j h tl


http://slidepdf.com/reader/full/edujoshuatlycom-kedah-stpm-trial-2010-physics-w-ans-0453a76a 5/45http://edu.joshuatly.com/http // j h tl


http://slidepdf.com/reader/full/edujoshuatlycom-kedah-stpm-trial-2010-physics-w-ans-0453a76a 7/45http://edu.joshuatly.com/http://www.joshuatly.com


http://slidepdf.com/reader/full/edujoshuatlycom-kedah-stpm-trial-2010-physics-w-ans-0453a76a 10/45http://edu.joshuatly.com/http // j h tl



http://edu.joshuatly.com/http://www.joshuatly.com


http://slidepdf.com/reader/full/edujoshuatlycom-kedah-stpm-trial-2010-physics-w-ans-0453a76a 16/45http://edu.joshuatly.com/htt // j h tl



http://edu.joshuatly.com/



MARKING

SCHEME

KEDAH

PHYSICS TRIAL

STPM 2010




ANSWER SCHEME – PHYSICS TRIAL 2010 PAPER 1

1 D 11 A 21 A 31 A 41 B

2 D 12 D 22 A 32 D 42 D

3 A 13 D 23 D 33 C 43 D

4 B 14 D 24 A 34 B 44 A

5 B 15 C 25 D 35 C 45 D

6 B 16 B 26 C 36 C 46 B7 B 17 A 27 D 37 C 47 A

8 D 18 C 28 B 38 B 48 D

9 B 19 C 29 A 39 D 49 B

10 B 20 A 30 C 40 D 50 B




MARKING SCHEME

KEDAH

PHYSICS TRIALSTPM 2010




2

PAPER 2

No Solution Marks

PART A

1 (a)

(b)

0.2100

75

kW 5.1

Fv P

11.38.950

1500

ms F

P v

1

1

1

1,1

2

2

324

GT

r M

23

s

E

E

s

E

s

T

T

r

r

M

M

2

7

63

8

11

102.3

104.2

108.3

105.1

E

s

M

M

5105.3 E

s

M

M

1

1

1

1

3 (a)

3 (b)

g

l T 2

8.9

12 T

sT 01.2

ol l h 10cos

)10cos(2 02 l l g v

)10cos(2 ol l g v

)10cos1)(8.9(2 ov

155.0 msv

1

1

1

1

1

1

4 (a)

4 (b)

4 (c)

Young’s modulus =

l e A

F

= A

l

e

F

=

=

Strain energy = Fe2

1

= )100.2(1002

1 3

= 0.10 J

Assumption: Extension is within the limit of proportionality.

1

1

11

1http://edu.joshuatly.com/http://www.joshuatly.com



3

5 (a)

(b)

CR

t

oeQQ

(1 – 0.63) Qo = Qo 6108.1

75.0

C e

6108.1

75.0

C = ln 0.37

C = 4.19 x 10-7 F

U =C

Q2

2

1 ………………………… (1)

U’ =C

Q 2)37.0(

2

1 ……………………(2)

(2) (1)U

U ' = 0.372

U’ = 0.137U Therefore, % of energy remains = 13.7%

1

1

1

1

1

6 (a)

(b)

(c)

Gain =

k

k

1

10+ 1 = 11

V in =k k

k

1.08.1

8.1

x 2

= 29.1

8.1 = 1.89 V

V o = 1.89 x 11 = 20.8 VSaturation occur, voltmeter reading = 9 V

V o = 112508.1

8.1

k k

k

= 0.76 V

1

1

11

1

1

7 (a)

(b)

(c)

min

hc = eV

V =e

hc

min

=

V 1910

834

106.11054.1

100.31063.6

= 1.81 x 104 V

For smallest angle of diffraction, n = 12d sin 160 = 1(1.54x10-10)

= 2.79 x 10-10 m

1

1

1




4

Each corner of a cube contains8

1 of an ion. Thus, only one ion occupies a

volume of d 3.

Therefore, no. of ions in 1 m3 of NaCl =3

1

d =

3101079.2

1

= 4.6 x 1028

1

1

8 (a)

8 (b)

(i) random: At any instant, the chance of a radioactive atom decaying is the

same for all radioactive atoms./ Cannot be predicted

(ii) spontaneous: The process cannot be controlled by physical conditions or

external factors

(i) Number of proton decay in 5 year = 510

103

2

1

34

33

= 1 proton

(ii) mc =

h

m =c

h

m/me = 1.0

1

1

1

1

1

PART B

9 (a) Newton’s second law of motion :

Net or resultant force acting on an object is directly proportional to the rate of

change of momentum.

During collision suppose F1 and F2 are acting on two objects m1 and m2.

According to Newton’s second law :

F1 =t

umvm 1111

F2 =t

umvm 2222

According to Newton’s third law :

F1 = -F2

1

1




5

t

umvm 1111 = -t

umvm 2222

m1v1 – m1u1 = - m2v2 + m2u2

m1u1 + m2u2 = m1v1 + m2v2

1

1

9 (b) (i) F

A 2T

2T T

B

3.0g

3.0g

(ii) For A :

F - 3T - 3.0g = 3(7.0)

For B :

2T - 3.0g = 3(10.5)

Solving the 2 equations :

F = 141.9 N

T = 30.5 N

1

1

1

1

1

1

9 (c) (i) For motion from A to B

fd = ½ mvo2 - ½ mvB

2

mgd = ½ mvo2 - ½ mvB

2 [ f = mg ]

gd = ½ vo2 - ½ vB

2

0.3(9.81)(2.0) = ½ (4)2 - ½ vB2

vB = 2.06 m s-1

(ii) For motion from B to C

½ mvC2 - ½ mvB

2 = mgh

vC2 - vB

2 = 2gh

vC = 4.89 m s-1

1

1

1

1

1

10 (a) (i) Number of independent ways in which a molecule can possess energy.

(ii) Boltzmann constant

(iii) Number of molecules in 1 mol of gas = N A where N A is the Avogrado

number.

Total kinetic energy of 1 mol of ideal gas = bT f

2( N A)

= RT f

2

, R is the molar gas constant

Internal energy of a gas

= total kinetic energy of molecules + total potential energy of the

molecules

For 1 mol of an ideal gas,

1

1

1




6

Internal energy = total kinetic energy of molecules, as potential energy = 0

Therefore for 1 mol of an ideal gas,

Internal energy, U = RT f

2

Therefore

U = 2

f

R T

According to the first law of thermodynamics, Q = U + W

When a gas absorbs energy at constant volume, Q = U , as W = 0

Therefore Q = U =2

f R T

Molar heat capacity at constant volume, C V =T

Q

=2

f R

1

1

10 (b)(i) Using

2

2

1

1

T

V

T

V to find the final temperature

2

'2

273

'

T

V V

T 2 = 546 K

Q = C p T

= ( 2

f R + R) T

=2

7(8.31) (546 – 273) ( substitution)

= 7940 J (with unit)

(ii) Q = W as U = 0 for isothermal process.

W = RT ln ('

'2

V

V )

= (8.31) (273) (ln 2) (substitution)

= 1570 JTherefore Q = 1570 J (with unit)

1

1

1

1

1

1

10 (c) (i) Let the temperature of the junction be

(0.05) (25 - ) = 0.13 ( - 0) (substitution)

= 6.9 oC (with unit)

(ii) rate of heat flow through rubber

= k A temperature gradient

= 0.13 100 10-4

3102

09.6

(substitution)

= 4.5 J s-1 (with unit)

1

1

1




7

5.0 mm

B

+ + + + + + + + +

11 (a) Mutual induction is the phenomenon where an e.m.f. is induced in a conductor

when the current in a neighbouring conductor is changing.

Primary coil (solenoid): N2 turns, length l , radius r .

Secondary coil (narrow): N1 turns, radius r .

When current I flows in primary coil, magnetic flux density in the

solenoid is

B = µonI ( n = N2/l )

magnetic flux linkage in the secondary coil

Φ = N1BA

= N1(l

I N 20 )( 2r )

Mutual inductance, M = Φ/I

M = l

r N N 2

210

2

1

1

1

11 (b)(i) V = IR = I x ρ

A

l

5 = I x 5 x 10-2 x 10 x 10-3 x6105.05

1

v

5 = 200I

I = 0.025 A

= 25 mA

(ii) I = Aven

22196 104106.1105.05

025.0

Aen

I v

= 25/16

= 1.563

= 1.6 m s-1

(iii) F = Bqv

= 0.050 x 1.6 x 10-19 x 1.56

= 1.248 x 10-20

= 1.2 x 10-20 N

(iv) VH = Bvd

= 0.05 x 1.56 x 5 x 10-3

= 0.39 mV

OR VH =ntq

BI

=19322

106.1105.0104

025.0050.0

= 0.00039

= 0.039 mV

1

1

1

1

1

1

1

1




8

0.50 mm

10 mm

I

- - - - - - - - - - - - --

correct faces ----------- (1)

correct signs ----------- (1)

12 (a) (i)

Applying Kirchhoff’s 2nd rule to upper loop

I1 5 + (I1 + I2) (2 + 1) = 12 + 6

Applying Kirchhoff’s 2nd rule to lower loop

I2 9 + (I1 + I2) (2 + 1) = 8 + 6

Ammeter reading = I1 = 2.0 A

(ii) I2 = 0.67 A

Voltmeter reading = - (2 + 0.67) x 2 + 8

= 2.66 V

( or any logical method )

1

1

1

1

1

1

12 (b)(i) χ c =

rms

rms

I

V

=2

oV .rms I

1

=2

6.

6.1

1

= 2.7 Ω

(ii) χ c = fC 2

1

C =7.22

01.0

= 5.9 x 10-4 F

1

1

1

1

1

1

I1

I2

I1 + I2




9

(iii)

Show cosine graph

Show value 2.3 and – 2.3

Show value 0.01

(0 mark if axes not label and no unit shown)

1

11

13 (a) (i) Threshold frequency is the minimum frequency required to emit an electron

from the surface of a metal.

(ii) Work function is the minimum work required to emit an electron from the

surface of a metal

1

1

(b) - Photoelectrons can only be emitted if the frequency of the radiation is greaterthan the threshold frequency of the metal

- Photoelectric emission is an instantaneous process

- The kinetic energy of the photoelectron emitted is independent of the intensity

of the radiation

** any 2 answers 2

(c) (i) maximum kinetic energy, ½ mvmax = hc/ λ - W

= (6.63 x 10-34) )1013.4

100.3(

7

8

- 20(1.6 x 10-19)

= 1.62 x 10-19 J

(ii) ½ mvmax2 = 1.62 x 10-19 J

mvmax = 3119 1011.91062.12

de Broglie wavelength, λ =max

mv

h =

3119

34

1011.91062.12

1063.6

= 1.22 x 10-9 m

(iii) If n is the number of electrons per second falling on the caesium surface,

n

hc = 5.0 x 25 x 10-4

n =

834

47

100.31063.6

10250.51013.4

1

1

1

1

1

0.01 0.02

I/A

t/s

2.3

-2.3




10

the number of electrons emitted =100

60 x

834

47

100.31063.6

10250.51013.4

= 1.56 x 1016

1

1

(d)

(i) electric potential energy = r

e

0

2

4 =

1112

219

1031.51085.84

1060.1

= -4.34 x 10-18 J

(ii) from2

0

22

4 r

e

r

mv

kinetic energy, ½ mv2 =r

e

0

2

8

=

1112

219

1031.51085.88

1060.1

= 2.17 x 10-18 J

1

1

1

1

14 (a)

14 (b)

14 (c)

(i) the difference between the total mass of the constituent nucleons and the

mass of the nucleus .

(ii) the energy required to separate completely all the nucleons in the nucleus

E = Δm c2

E = binding energy

Δm = mass defect

c = speed of light in vacuum

( i ) N 14

7 + He4

2 O17

8 + H 1

1

(ii) Δm = [ (2.82282 + 0.16735) x 10-26 ] – [ ( 2.32500 + 0.66466) x 10-26 ]

= 5.1 x 10-30 kg

Minimum kinetic energy of the alpha particle = Δm c2

= 5.1 x 10-30 x (3 x 108)2

= 4.59 x 10-13 J

(i) Usable energy per fission = 200 x 106 x 1.6 x 10-19 x100

20

= 6.4 x 10-12 J

1

1

1

1

2

1

1

1

1




11

Number of fission required per second =12

8

104.6

107

x

x

= 1.094 x 1020

Number of uranium atom used in one day = 1.094 x 1020 x 24 x 60 x 60

= 9.45 x 1024

(ii) Mass of uranium used in each day =23

24

1002.6

1045.9

x

x (0.235)

= 3.68 kg

1

1

1

1

[edu.joshuatly.com] kedah stpm trial 2010 physics [w ans] [0453a76a]

Documents