As physics p1 MCQS YEARLY ONLY.

[Hadi Murtaza]

question number ?

35 ?

 

[Thought blocker]

35 ?

Oh ok, ty
What about the rest ?

 

[Hadi Murtaza]

Oh ok, ty
What about the rest ?

Have u not solved those ?

 

[Thought blocker]

Have u not solved those ?

I did, but m telling of other post...

 

[Thought blocker]

This one posted by her
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_13.pdf

Q15,16,22,36,38
The electricity one is quite uh, hard? I dont really understand tho.

 

[Hadi Murtaza]

I shall try

 

[Thought blocker]

This one posted by her
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_13.pdf

Q15,16,22,36,38
The electricity one is quite uh, hard? I dont really understand tho.

I shall try

I know 16)
the resistive force at speed 0 is the wheel friction, from the graph it is seen that this is 8 kn, since wheel friction is constant at 8 kn, 32(40-8) kn must be the magnitude of wind resistance, thus ratio is 32/8 = 4

 

[Hadi Murtaza]

I know 16)
the resistive force at speed 0 is the wheel friction, from the graph it is seen that this is 8 kn, since wheel friction is constant at 8 kn, 32(40-8) kn must be the magnitude of wind resistance, thus ratio is 32/8 = 4

Only 16 ? wb rest ?

 

[Thought blocker]

Only 16 ? wb rest ?

They need a lot of thinking, I dont think they are simple, I can try question 38 but 15 and 22 and 36 I think I would not know it.

 

[Hadi Murtaza]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_13.pdf

Q15,16,22,36,38
The electricity one is quite uh, hard? I dont really understand tho.

15.
Take moments from da hinge, let F b da reaction force of da spring.
F × 2 = (40 × 9.8) × 5
2F = 1960 N
[ F = 980 N ]

k = 10 kN/m = 10 000 N/m
k = F/x
x = F/k
x = 980/(10 000)
[ x = 0.098 m = 9.8 cm ]
Answer: C

Thought blocker really man ?

 

[Thought blocker]

15.
Take moments from da hinge, let F b da reaction force of da spring.
F × 2 = (40 × 9.8) × 5
2F = 1960 N
[ F = 980 N ]

k = 10 kN/m = 10 000 N/m
k = F/x
x = F/k
x = 980/(10 000)
[ x = 0.098 m = 9.8 cm ]
Answer: C

Thought blocker really man ?

Yes I dk this basic thing, how to calculate... Teach me its concept ._.

 

[Hadi Murtaza]

Yes I dk this basic thing, how to calculate... Teach me its concept ._.

I did dis in M1 -_-

 

[Thought blocker]

I did dis in M1 -_-

You know my M1 SUCKS!

 

[Hadi Murtaza]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_13.pdf

Q15,16,22,36,38
The electricity one is quite uh, hard? I dont really understand tho.

Q.22
Young modulus = Stress/Strain = (F/A) ÷ (L/ΔL) = FL/AΔL
2.1 × 10¹¹ = [ (70 × 9.8) × (20) ] / [ (3.2 × 10⁻⁶ × 2 × 100) × ΔL ]
2.1 × 10¹¹ = 13 720 / (ΔL × 6.4 × 10⁻⁴)
ΔL = 13 720 / (2.1 × 10¹¹ × 6.4 × 10⁻⁴)
ΔL = 13 720 / 1.344 × 10⁸
ΔL = 0.000102 m
[ ΔL = 0.102 mm ]
Answer: C

 

[Thought blocker]

Q.22
Young modulus = Stress/Strain = (F/A) ÷ (L/ΔL) = FL/AΔL
2.1 × 10¹¹ = [ (70 × 9.8) × (20) ] / [ (3.2 × 10⁻⁶ × 2 × 100) × ΔL ]
2.1 × 10¹¹ = 13 720 / (ΔL × 6.4 × 10⁻⁴)
ΔL = 13 720 / (2.1 × 10¹¹ × 6.4 × 10⁻⁴)
ΔL = 13 720 / 1.344 × 10⁸
ΔL = 0.000102 m
[ ΔL = 0.102 mm ]
Answer: C

What about rest ? 36 and 38 ?

 

[Asad Moosvi]

Can someone please help me with question 19? http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_13.pdf

 

[dumb human]

Here is the June 2009.. oh it was a tough one!!

1: D...basic stuff
2: C...opp/adj = tan ( theta) so opp=adj tan(theta) = 20tan( 30) =11.5
3: B...the readings are -6.5 and 14...so 14- (-6.05) so 20.5
4: B...P=VI= 1.2 x 0.48
5: D..he siad terminal velocity means velocity is constant and so d must have a constant increase..all other 3 graphs have d starts to decrease so D is correct
6: C...obvious from the graph
7: A...basic stuff...
8: C...air resistance is neglected so constant horizontal velocity and constant vertical acceleration
9: B...rate of changeof momentum= mv-mu, so (0.1 x -30) - (0.1 x 20)= -5....we used - 30 because it was travelling in the opposite direction
10: C...the momentum of body m must equal the momentum of body 2m..this means that if body m has half the mass so it must have double the velocity so ratio of kinetic energy of m /2m = ( 0.5 m x 2v^2)/ (0.5 x 2m x v^2) so it will be 2/1
11: A..the ball will not fall diagonaly..so S must equal to Q...according to the question..he didnt say if the ball will flought orwill sink so i thought that the upthrust is the difference of the pressure betwwn the water and the weight of the object so R > than P
12: D...he said it is falling with uniform velocity so no acceleration so no resultant force..and he said it is still falling horizontally so no resultant torque..so the conditions for equillibrium are found
13: ooh ill blow from this one and i may need some 1 to explain it to me..thnx in advance
14: C...K= F/v^2 means F/v x v and F = P/v so K = P/v/ (v x v) so K=(P/v) x v^-2 and then K = P x v^-1 x v^-2 so P x v^-3 means P/v^
15: B...container X lost half of the water means it lost half the height and half THE MASS so m/2 g h/2 so 2 x 2=4 so ( mgh )/4
16 and 17 are basic and simple stuff
18: D...at first the pressure was pgh..then the height was raised by another h so pgh + h = pg2h
19: A...B and C are wrong because they are brittle so they will break..steel is a mixture of 2 substances so it have the melting point of 2 substances which are iron and carbon...so aluminium is a single metal susbtance and so it will melt first
20: D...young modulus is ( Fl)/ (Ae) he said they are the same material so same young modulus..and same extension..then tension is the F so F= A/l ( no need for the young modulus and the extensoin so ...( l/A) /( 2l /0.5A) continue and u will get 4 / 1
21: A...if u draw a straight line from the origin to the same point the drawn graph reaches and took 0.5 x 17 x 0.30 it will equal 2.7..and the graph shows that thre is much spce between it and the line we drew so it must be < 2.7..so C and are wrong..B is wrong because it is about or approximatley the same as 2.7 so it alo must be less than 2.6 so A
22 and 23 are also basic and simple stuff
24: B...the distance between 2 adjacent nodes orantinodes is half a wavelength..so the distance between a node and an anti node is the quarter
25: A..N lines per metre means 1/N...then substituting into the equation it will be (1/N) sin (theta)= n ( lambda)
26: B...as shown 1.5 lambda = 2.1m so one wavelength is 1.4..then v=f lambda = 80 x 1.4 = 112
27: A...both are attracted to the +ve plate so both are -ve
28: B...E is proportional to V so doubling V doubles E......E is inverseley proportional to d so halfing d doubles E so B
29: D...v^2 = 2as..a=F/m so v^2 = 2 x (F/m) s and F= EQ means Ee as in the question..so v^2/2 = (Ee/m) s continue and u will get D
30: A...I=Q/t..simple question
31: D...W=QV so Q= 7.2 x 10^4 / 12 = 6000
32: A...basic EMF definition
33: B....I = E1 - E2 / R so ( 3 - 1.2) / 9 = 0.2
34: D..simple question..total resistorsin series = R1+R2+R3..in parrallel...1/Rt= 1/R1 + 1/R2 + 1/R3
35: D...light intensity and temperature increases so resistance in the LDR and thermistor decrease so the other resistor resistance increases
36 and 37 are simple questoins
38: C...just subtract the proton number from the nucleon number of each
39: C...simple question
40: D...he said a neutron so proton number must be zero and nucleon number is one

can u explain A? I dont get it .... =(

 

[dumb human]

Here is the June 2009.. oh it was a tough one!!

1: D...basic stuff
2: C...opp/adj = tan ( theta) so opp=adj tan(theta) = 20tan( 30) =11.5
3: B...the readings are -6.5 and 14...so 14- (-6.05) so 20.5
4: B...P=VI= 1.2 x 0.48
5: D..he siad terminal velocity means velocity is constant and so d must have a constant increase..all other 3 graphs have d starts to decrease so D is correct
6: C...obvious from the graph
7: A...basic stuff...
8: C...air resistance is neglected so constant horizontal velocity and constant vertical acceleration
9: B...rate of changeof momentum= mv-mu, so (0.1 x -30) - (0.1 x 20)= -5....we used - 30 because it was travelling in the opposite direction
10: C...the momentum of body m must equal the momentum of body 2m..this means that if body m has half the mass so it must have double the velocity so ratio of kinetic energy of m /2m = ( 0.5 m x 2v^2)/ (0.5 x 2m x v^2) so it will be 2/1
11: A..the ball will not fall diagonaly..so S must equal to Q...according to the question..he didnt say if the ball will flought orwill sink so i thought that the upthrust is the difference of the pressure betwwn the water and the weight of the object so R > than P
12: D...he said it is falling with uniform velocity so no acceleration so no resultant force..and he said it is still falling horizontally so no resultant torque..so the conditions for equillibrium are found
13: ooh ill blow from this one and i may need some 1 to explain it to me..thnx in advance
14: C...K= F/v^2 means F/v x v and F = P/v so K = P/v/ (v x v) so K=(P/v) x v^-2 and then K = P x v^-1 x v^-2 so P x v^-3 means P/v^
15: B...container X lost half of the water means it lost half the height and half THE MASS so m/2 g h/2 so 2 x 2=4 so ( mgh )/4
16 and 17 are basic and simple stuff
18: D...at first the pressure was pgh..then the height was raised by another h so pgh + h = pg2h
19: A...B and C are wrong because they are brittle so they will break..steel is a mixture of 2 substances so it have the melting point of 2 substances which are iron and carbon...so aluminium is a single metal susbtance and so it will melt first
20: D...young modulus is ( Fl)/ (Ae) he said they are the same material so same young modulus..and same extension..then tension is the F so F= A/l ( no need for the young modulus and the extensoin so ...( l/A) /( 2l /0.5A) continue and u will get 4 / 1
21: A...if u draw a straight line from the origin to the same point the drawn graph reaches and took 0.5 x 17 x 0.30 it will equal 2.7..and the graph shows that thre is much spce between it and the line we drew so it must be < 2.7..so C and are wrong..B is wrong because it is about or approximatley the same as 2.7 so it alo must be less than 2.6 so A
22 and 23 are also basic and simple stuff
24: B...the distance between 2 adjacent nodes orantinodes is half a wavelength..so the distance between a node and an anti node is the quarter
25: A..N lines per metre means 1/N...then substituting into the equation it will be (1/N) sin (theta)= n ( lambda)
26: B...as shown 1.5 lambda = 2.1m so one wavelength is 1.4..then v=f lambda = 80 x 1.4 = 112
27: A...both are attracted to the +ve plate so both are -ve
28: B...E is proportional to V so doubling V doubles E......E is inverseley proportional to d so halfing d doubles E so B
29: D...v^2 = 2as..a=F/m so v^2 = 2 x (F/m) s and F= EQ means Ee as in the question..so v^2/2 = (Ee/m) s continue and u will get D
30: A...I=Q/t..simple question
31: D...W=QV so Q= 7.2 x 10^4 / 12 = 6000
32: A...basic EMF definition
33: B....I = E1 - E2 / R so ( 3 - 1.2) / 9 = 0.2
34: D..simple question..total resistorsin series = R1+R2+R3..in parrallel...1/Rt= 1/R1 + 1/R2 + 1/R3
35: D...light intensity and temperature increases so resistance in the LDR and thermistor decrease so the other resistor resistance increases
36 and 37 are simple questoins
38: C...just subtract the proton number from the nucleon number of each
39: C...simple question
40: D...he said a neutron so proton number must be zero and nucleon number is one

Thought blocker help bro =(

 

[Thought blocker]

can u explain A? I dont get it .... =(

Thought blocker help bro =(

Question to batao

 

[dumb human]

Question to batao

1st questiom per stuck!!! lol