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JUNE 2002 IS UP. NOV2002 AND JUNE AND NOVEMBER 2003 IS LEFT. BUT DON'T WORRY , WE'LL COVER IT ALL.

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- Thread starter fatima 007
- Start date

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JUNE 2002 IS UP. NOV2002 AND JUNE AND NOVEMBER 2003 IS LEFT. BUT DON'T WORRY , WE'LL COVER IT ALL.

- Messages
- 73

- Reaction score
- 73

- Points
- 18

WHERE IS JUNE 2003?Ah thankies! I was waiting for someone to post this.

I've solved June 2003, will post it in a few hours. Too tired to type all that atm. :/

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Haha I slept after that.WHERE IS JUNE 2003?

I'll post it now.

- Messages
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please do. We all will be very grateful!Haha I slept after that.

I'll post it now.

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1-B

because mass has magnitude only.

2-B

Work done = Force*distance

kgms^-2 *m = kgm^2 s^-2

3-B

Using cosine rule: c^2 = a^2 + b^2 +2abcosC

c^2 = 10^2 + 10^2 +2(100)cos 120

C = 10 N

4-A

Basic concept that a systematic error can be reduced by checking for zero error.

5-C

because 3% of 330 = 10 m/s

and speed of sound = 330 m/s

6-B

Accuracy: closeness of the measured values to the true value

Precision: closeness of measured values.

7-C

For vertical component: v = u +at

v = usina - gt

and in horizontal component v = u cos a

8-C

In the first 1/3 of the graph, velocity is increasing hence the gradient should be increasing steeply.

In the second 1/3 of the graph, velocity is constant so gradient should be constant, hence a straight line.

In the last 1/3rd of the graph, gradient is decreasing hence line should be least steep.

9-D

because the distance in X is the distance from the ground to the max height after first impact and distance Y is the distance from the max height back to the ground.

10-D

F= ma

Total mass = 4m

hence a = F/4m

According to Newton's 3rd law of motion, Fx= Fy

Fx = ma

3m * F/4m

= 3/4 F

11-B

12-A

m1u1 + m2u2 = (m1 + m2) v

2(8) + 4(2) = (2+4)v

16 + 8 = 6v

24/6 = v

4 m/s =v

13-D

Both the 20 N forces cancel each other out but the 30 N forces are both in clockwise direction hence resultant force is non zero.

The torque is non zero due to both the 20 N forces.

14-B

Vertical component = 10 000- 9000 = 1000 N

15-A

Air resistance increases over time and the resultant force decreases per unit time due to the opposing resistive force acting on it.

16-C

Basic definition.

17-C

P.E. = mgh

= W*q

18-D

2ad = v^2 - u^2

2a (10) = 0 - 10^2

a = -100/20 = -5 m/s^2

2ad = v^2 - u^2

2(-5) d = 0 - 30^2

d = -900/-10 = 90m

19- B

When the wire is stretched beyond its elastic limit, it will get permanently detached, that is, it won't return to original mean position.

The shapes of the graph need to be learnt.

20- A

21-B

Definition: Ultimate tensile stress is the stress at which the material breaks.

22-D

W = k* 3x

k = W/3x

W = ke

W = W/3x *e

3x = e

23-B

Property of transverse waves: They can all be polarised.

24-D

Distance between two successive nodes or two successive anti nodes is half the wavelength.

25-C

All electromagnetic waves travel at the speed of light in vacuum hence speed will be c

and since c = f* lamda

lamda will have to be 2 lamda in order for the value of c to remain constant.

26-B

The time period is 4 boxes on the x axis.

Since 1 box = 2.5 ms, 4 boxes = 4*2.5 = 10 ms

f = 1/T = 1/10*10^-3 =100 Hz

27- C

An interference pattern can be observed in all the conditions except C.

28-C

d sin theta = n* lamda

d = 1*10^-6

n =1 at the angle 70/2 = 35 degrees

1*10^-6 * sin35 = 1* lamda

5.735 *10^-7 = lamda

~ 574 nm

29-B

Formula E =VQ

30- C

Since the total time period = 8s, we use the mean current i.e. (100+20)/2 = 60 mA

Q = It

= 60 * 8

= 480 mC

31-A

Kirchoff's first law is based on law of conservation of charge.

32-B

V = IR

7.5 = I * 15

7.5/15 = I

0.5 A = I

33- B

Total resistance = T1*T2/ (T1 + T2)

T1 = 10 + 6 = 16

T2 = 6 + 10 = 16

T.R. = 16*16/32 = 8 ohms

34- B

Min V = 0/60 * 9 = O V

Max V = 50/60 * 9 = 7.5 V

Range = 0-7.5 V

35-C

Electric field between two parallel plates is always uniform.

36- B

E = V/D = 700/ 5*10^-3 = 140 000

Direction of field is always towards the plate with a lower p.d. [i.e. away from higher charged plate]

37- B

Again the direction is away from positive charge and towards the negative charge.

38- C

The proton no.s are all the same, hence neutrons differ.

39-B

When a beta particle is emitted, nucleon no. remains same and proton no. increases by 1.

40-D

A proton has a charge of +e

So 2(2/3) + 1(-1/3) = 1 e

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Alright I'll post Nov. 2007 but its going to take some time.please...i would love if someone post Nov 2007 because i knew that there is someone i dont remember his name will post June 2008....and inshallah ill post Nov 2008 later today too so Nov2007 will be the only 1 left

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Thnx very muchAlright I'll post Nov. 2007 but its going to take some time.

- Messages
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thanks a ton!!!!!June 2003

1-B

because mass has magnitude only.

2-B

Work done = Force*distance

kgms^-2 *m = kgm^2 s^-2

3-B

Using cosine rule: c^2 = a^2 + b^2 +2abcosC

c^2 = 10^2 + 10^2 +2(100)cos 120

C = 10 N

4-A

Basic concept that a systematic error can be reduced by checking for zero error.

5-C

because 3% of 330 = 10 m/s

and speed of sound = 330 m/s

6-B

Accuracy: closeness of the measured values to the true value

Precision: closeness of measured values.

7-C

For vertical component: v = u +at

v = usina - gt

and in horizontal component v = u cos a

8-C

In the first 1/3 of the graph, velocity is increasing hence the gradient should be increasing steeply.

In the second 1/3 of the graph, velocity is constant so gradient should be constant, hence a straight line.

In the last 1/3rd of the graph, gradient is decreasing hence line should be least steep.

9-D

because the distance in X is the distance from the ground to the max height after first impact and distance Y is the distance from the max height back to the ground.

10-D

F= ma

Total mass = 4m

hence a = F/4m

According to Newton's 3rd law of motion, Fx= Fy

Fx = ma

3m * F/4m

= 3/4 F

11-B

View attachment 12230

12-A

m1u1 + m2u2 = (m1 + m2) v

2(8) + 4(2) = (2+4)v

16 + 8 = 6v

24/6 = v

4 m/s =v

13-D

Both the 20 N forces cancel each other out but the 30 N forces are both in clockwise direction hence resultant force is non zero.

The torque is non zero due to both the 20 N forces.

14-B

Vertical component = 10 000- 9000 = 1000 N

15-A

Air resistance increases over time and the resultant force decreases per unit time due to the opposing resistive force acting on it.

16-C

Basic definition.

17-C

P.E. = mgh

= W*q

18-D

2ad = v^2 - u^2

2a (10) = 0 - 10^2

a = -100/20 = -5 m/s^2

2ad = v^2 - u^2

2(-5) d = 0 - 30^2

d = -900/-10 = 90m

19- B

When the wire is stretched beyond its elastic limit, it will get permanently detached, that is, it won't return to original mean position.

The shapes of the graph need to be learnt.

20- A

View attachment 12242

21-B

Definition: Ultimate tensile stress is the stress at which the material breaks.

22-D

W = k* 3x

k = W/3x

W = ke

W = W/3x *e

3x = e

23-B

Property of transverse waves: They can all be polarised.

24-D

Distance between two successive nodes or two successive anti nodes is half the wavelength.

25-C

All electromagnetic waves travel at the speed of light in vacuum hence speed will be c

and since c = f* lamda

lamda will have to be 2 lamda in order for the value of c to remain constant.

26-B

The time period is 4 boxes on the x axis.

Since 1 box = 2.5 ms, 4 boxes = 4*2.5 = 10 ms

f = 1/T = 1/10*10^-3 =100 Hz

27- C

An interference pattern can be observed in all the conditions except C.

28-C

d sin theta = n* lamda

d = 1*10^-6

n =1 at the angle 70/2 = 35 degrees

1*10^-6 * sin35 = 1* lamda

5.735 *10^-7 = lamda

~ 574 nm

29-B

Formula E =VQ

30- C

Since the total time period = 8s, we use the mean current i.e. (100+20)/2 = 60 mA

Q = It

= 60 * 8

= 480 mC

31-A

Kirchoff's first law is based on law of conservation of charge.

32-B

V = IR

7.5 = I * 15

7.5/15 = I

0.5 A = I

33- B

Total resistance = T1*T2/ (T1 + T2)

T1 = 10 + 6 = 16

T2 = 6 + 10 = 16

T.R. = 16*16/32 = 8 ohms

34- B

Min V = 0/60 * 9 = O V

Max V = 50/60 * 9 = 7.5 V

Range = 0-7.5 V

35-C

Electric field between two parallel plates is always uniform.

36- B

E = V/D = 700/ 5*10^-3 = 140 000

Direction of field is always towards the plate with a lower p.d. [i.e. away from higher charged plate]

37- B

Again the direction is away from positive charge and towards the negative charge.

38- C

The proton no.s are all the same, hence neutrons differ.

39-B

When a beta particle is emitted, nucleon no. remains same and proton no. increases by 1.

40-D

A proton has a charge of +e

So 2(2/3) + 1(-1/3) = 1 e

- Messages
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1-C

Use base units and equate both sides.

2-A

Learn the approximations.

3-B

Resultant force is always is opposite direction of its two components in the triangle of forces.

4-D

Precision is the closeness of measured values.

Accuracy is the closeness of measured values to actual value.

5-C

P = fv

f = ma and v = s/t

P = add up all the % uncertainties.

%P = 0.1 + 1 + 1.5 + .5

= 3.1 %

6-A

Due to zero error, the initial value of angle of deflection will be greater than zero at zero current.

7-A

Basic concept that g gravity is the acceleration of free fall.

8-D

The gradient of velocity-time graph is acceleration.

9-B

Area of distance above x axis - area of distance below x axis.

(0.5 * 3 * 30) - (.5 * 2 * 20)

= 25 m

10-D

Driving force - frictional force = ma

12- Fre = 0.6(4)

12- 0.6(4) = Fre

9.6 N = Fre

11- B

12-A

Since the parachutist is falling, height is decreasing over time.

Gradient of distance-time graph is speed.

13-C

Use head to tail method.

14-D

At midpoint, d = 0.8/2 = 0.4m

Moment = F*d

12 = F * 0.4

30 N = F

15-B

Initial K.E. - Final K.E.

(0.5* 1000 *25^2) - (0.5 * 1000 * 5^2)

300 000 J

300 kJ

16 - C

At max d, K.E. = minimum and elastic potential energy = maximum.

17-C

Density = mass/volume

Hence higher density means a higher mass, hence more no. of atoms.

So MpNp > MqNq

18-A

Height at surface =?

Pressure = density * g* h

100 000= 1030 * 9.81 * h

9.896 m = h1

Height at 450 kPa =?

450 000 = 1030 * 9.81 * h

44.535 = h2

Therefore height below surface = 44.535 - 9.896

=34.639 ~ 34.6m

19-B

Basic definition.

20- A

Strain energy = 1/2 * F * extension

= 0.5 * 25 * (0.4 - 0.2)

= 2.5 J

21- B

22- B

Learn the wavelengths.

23- B

Max speed = 2pii *a *f

f = ?

s = f* lamda

8 = f (50)

0.16 Hz = f

Max speed = 2pii * 2 * 0.16

=2 m/s

Max K.E. = .5 m* v^2 = 0.5 * 2 * 10^-3 * 2^2

= 4 mJ

24-D

x = lamda * D/ a

Hence increasing the lamda, increases x, that is the fringe separation.

25-B

d sin theta = n* lamda

At n = 3, theta = 45 degrees

d sin45 = 3*lamdaa

0.7 d = 3 lamda

and max angle = 90 , so n =?

d sin90 = n*lamdaa

d = n* lamda

Using the ratio method:

If 0.7d = 3 lamda

1 d =?

cross multiply and you get n as 4.25 hence rounding it off to 4th order.

26- C

Since electric field direction is from higher p.d./charge to lower p.d./charge the electron will be attracted towards the more positively charged plate (or direction) hence to the left.

27-D

Direction of electric field is away from positive charge.

28-B

E = V*Q

E/Q = V

29-D

P = V^2/R

P = 12^2/Rx and P = 6^2/Ry

Rx = 144/P and Ry = 36/P

Rx/Ry = 144/P divided by 36/P

Rx/Ry = 144/36 = 4

30-D

V = IR

6 = I (10 +10)

0.3 A = I

Q = It

0.3 * 60 = 18 C

31-A

32-A

Basic concept of L.D.R.; when light intensity increases, resistance decreases hence voltage decreases too.

33-D

V in = R1/T.R. * V out

4.8 = 10/25 * V out

4.8* 25/10 = V out

12 = V out

34-D

Using the ratio method:

If 1.1 V - 0.7 m

? - 0.9 m

x = 0.9* 1.1/ .7

x = 1.4 V

35-A

The readings won't change because the set up is the same.

36- C

Charge is same since proton no. is same, but mass differs since no. of neutrons differs.

37- C

Basic concept.

38-C

The alpha scattering experiment proves the small size of a gold nucleus.

39-B

In a beta emission, the nucleon no. remains same but the proton no. increases by 1

For 2 beta emissions, proton no will increase by 2 hence 40 +2 = 42

40-A

Momentum = mv, greater m = greater momentum.

hence the particle with the greatest mass from all the choices is A, an alpha particle.

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You're welcome.Thnx very much

November 2007 posted.

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u ppl r really awesom thanks alot.. its really soooo useful...

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thnx again...and ill post nov 2008 soon todayYou're welcome.

November 2007 posted.

Here are the answers OF MAY/JUNE 2001hey ppl can we discuss da M.J.01 papr.. da mark scheme isnt available..

1.B

2.D

3.C

4.C

5.B

6.B

7.B

8.B

9.C

10.C

11.D

12.D

13.A

14.D

15.D

16.D

17.C

18.B

19.C

20.C

21.B

22.C

23.C

24.A

25.A

26.C

27.C

28.D

29.B

30.B

31.B

32.B

33.B

34.C

35.A

36.C

37.C

38.B

39.D

40.B

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here is m.j.01 i dont hv o.n.01^Where can I find the question papers M/J 01 and O/N 01?

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MashAllah this is great. Keep up the awesome work!November 2007

1-C

Use base units and equate both sides.

2-A

Learn the approximations.

3-B

Resultant force is always is opposite direction of its two components in the triangle of forces.

4-D

Precision is the closeness of measured values.

Accuracy is the closeness of measured values to actual value.

5-C

P = fv

f = ma and v = s/t

P = add up all the % uncertainties.

%P = 0.1 + 1 + 1.5 + .5

= 3.1 %

6-A

Due to zero error, the initial value of angle of deflection will be greater than zero at zero current.

7-A

Basic concept that g gravity is the acceleration of free fall.

8-D

The gradient of velocity-time graph is acceleration.

9-B

Area of distance above x axis - area of distance below x axis.

(0.5 * 3 * 30) - (.5 * 2 * 20)

= 25 m

10-D

Driving force - frictional force = ma

12- Fre = 0.6(4)

12- 0.6(4) = Fre

9.6 N = Fre

11- B

View attachment 12245

12-A

Since the parachutist is falling, height is decreasing over time.

Gradient of distance-time graph is speed.

13-C

Use head to tail method.

14-D

At midpoint, d = 0.8/2 = 0.4m

Moment = F*d

12 = F * 0.4

30 N = F

15-B

Initial K.E. - Final K.E.

(0.5* 1000 *25^2) - (0.5 * 1000 * 5^2)

300 000 J

300 kJ

16 - C

At max d, K.E. = minimum and elastic potential energy = maximum.

17-C

Density = mass/volume

Hence higher density means a higher mass, hence more no. of atoms.

So MpNp > MqNq

18-A

Height at surface =?

Pressure = density * g* h

100 000= 1030 * 9.81 * h

9.896 m = h1

Height at 450 kPa =?

450 000 = 1030 * 9.81 * h

44.535 = h2

Therefore height below surface = 44.535 - 9.896

=34.639 ~ 34.6m

19-B

Basic definition.

20- A

Strain energy = 1/2 * F * extension

= 0.5 * 25 * (0.4 - 0.2)

= 2.5 J

21- B

View attachment 12247

22- B

Learn the wavelengths.

23- B

Max speed = 2pii *a *f

f = ?

s = f* lamda

8 = f (50)

0.16 Hz = f

Max speed = 2pii * 2 * 0.16

=2 m/s

Max K.E. = .5 m* v^2 = 0.5 * 2 * 10^-3 * 2^2

= 4 mJ

24-D

x = lamda * D/ a

Hence increasing the lamda, increases x, that is the fringe separation.

25-B

d sin theta = n* lamda

At n = 3, theta = 45 degrees

d sin45 = 3*lamdaa

0.7 d = 3 lamda

and max angle = 90 , so n =?

d sin90 = n*lamdaa

d = n* lamda

Using the ratio method:

If 0.7d = 3 lamda

1 d =?

cross multiply and you get n as 4.25 hence rounding it off to 4th order.

26- C

Since electric field direction is from higher p.d./charge to lower p.d./charge the electron will be attracted towards the more positively charged plate (or direction) hence to the left.

27-D

Direction of electric field is away from positive charge.

28-B

E = V*Q

E/Q = V

29-D

P = V^2/R

P = 12^2/Rx and P = 6^2/Ry

Rx = 144/P and Ry = 36/P

Rx/Ry = 144/P divided by 36/P

Rx/Ry = 144/36 = 4

30-D

V = IR

6 = I (10 +10)

0.3 A = I

Q = It

0.3 * 60 = 18 C

31-A

View attachment 12248

32-A

Basic concept of L.D.R.; when light intensity increases, resistance decreases hence voltage decreases too.

33-D

V in = R1/T.R. * V out

4.8 = 10/25 * V out

4.8* 25/10 = V out

12 = V out

34-D

Using the ratio method:

If 1.1 V - 0.7 m

? - 0.9 m

x = 0.9* 1.1/ .7

x = 1.4 V

35-A

The readings won't change because the set up is the same.

36- C

Charge is same since proton no. is same, but mass differs since no. of neutrons differs.

37- C

Basic concept.

38-C

The alpha scattering experiment proves the small size of a gold nucleus.

39-B

In a beta emission, the nucleon no. remains same but the proton no. increases by 1

For 2 beta emissions, proton no will increase by 2 hence 40 +2 = 42

40-A

Momentum = mv, greater m = greater momentum.

hence the particle with the greatest mass from all the choices is A, an alpha particle.

(I was about to start with this after about an hour or two, but it's good you did and posted it first. )

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