AS Physics P1 MCQs Preparation Thread.

[Capricedcapri]

anyone here knows how to solve mcq 35 of o/n 2004??

 

[Soldier313]

MysteRyGiRl and Jaf for qn 25 use formula: d x sintheta = n x lambda ; where theta = 90 ; sin90 = 1
so we r first going to count the number of orders which is n...... so n= d/lambda
d = 1 / N , N in metres would be 3 x 10 ^5 , therefore d= 1/(3 x 10^5)
so d/lambda = (1/3x10^5) / (450 x 10 ^-9)
ans is 7 , THEN u multiply that by 2, bcoz this was just on one side of the original beam, u get 14, add 1 to this because u r including the undeviated beam, u get 15 and hence D is the answer

 

[leadingguy]

anyone here knows how to solve mcq 35 of o/n 2004??

provision of links of qs and ms wil help alot.....

 

[perkypearl]

upload any particular question..???

well plenty of thm thn lets start from this
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s11_qp_11.pdf
Q.10

 

[oldfashionedgirl]

MysteRyGiRl and Jaf for qn 25 use formula: d x sintheta = n x lambda ; where theta = 90 ; sin90 = 1
so we r first going to count the number of orders which is n...... so n= d/lambda
d = 1 / N , N in metres would be 3 x 10 ^5 , therefore d= 1/(3 x 10^5)
so d/lambda = (1/3x10^5) / (450 x 10 ^-9)
ans is 7 , THEN u multiply that by 2, bcoz this was just on one side of the original beam, u get 14, add 1 to this because u r including the undeviated beam, u get 15 and hence D is the answer

can you please solve question 26? URGENTLY. Answer is B

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_12.pdf

 

[leadingguy]

well plenty of thm thn lets start from this
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf
Q.10

a body moving with velocity u take that as positive direction momentum wil be mu (initial momentum)
now when it strikes and returns with same speed, velocity wil be different as it depends on direction so velocity wil be -u momentum becoms -mu (final momentum)

now change is always determined by final - initial .... so do the same
final - initial = ans
-mu - (mu) it wil be - 2mu ans is D .... have not seen ms yet but hopefully am rite

 

[Jaf]

well plenty of thm thn lets start from this
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf
Q.10

D -2mu

This is because if it rebounded and is going in the opposite direction, and for example, the change was just mu, the ball would stop because mu - mu = 0. Since it is also going in the opposite direction at the same speed as before, it is -2mu

 

[leadingguy]

can you please solve question 26? URGENTLY. Answer is B

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_12.pdf

when they started they are same
then
look at time 18 second there is phase diffeernce of 180.....

means the phase difference has increased gradually and linearly.... frm o to 180...
now a cycle compromise of 2pi means 360

now difference of 1/8 of cycle wil wil be 360/8 = 45 degrees...

if phase diff. of 180 is produced in 18sec. then 45 wil be produced in 4.5 sec..

180..................18sec
45..................... xsec

 

[oldfashionedgirl]

when they started they are same
then
look at time 18 second there is phase diffeernce of 180.....

means the phase difference has increased gradually and linearly.... frm o to 180...
now a cycle compromise of 2pi means 360

now difference of 1/8 of cycle wil wil be 360/8 = 45 degrees...

if phase diff. of 180 is produced in 18sec. then 45 wil be produced in 4.5 sec..

180..................18sec
45..................... xsec

Thanks alot

 

[Soldier313]

please assist urgently

 

[leadingguy]

please assist urgentlyView attachment 12992

what is the ans?????

 

[Soldier313]

what is the ans?????

A

 

[Jaf]

please assist urgentlyView attachment 12992

Although I am not sure, I believe the answer is 2/3. The way I did it was:

Density (P) = Pressure/g3x
Density (Q) = Pressure/g2x

Density of P/Density of Q = (Pressure/g3x) / (Pressure/g2x)
= Pressure/g3x * g2x/Pressure

The Pressure for both liquids is the same (they are in equilibrium) and so it g. Thus, Pressure, g and x can be cancelled out, leaving 2/3.

I am not, however, sure whether that is the right answer. If you could give me the references for this, I could check the marking scheme and confirm...

 

[Soldier313]

Although I am not sure, I believe the answer is 2/3. The way I did it was:

Density (P) = Pressure/g3x
Density (Q) = Pressure/g2x

Density of P/Density of Q = (Pressure/g3x) / (Pressure/g2x)
= Pressure/g3x * g2x/Pressure

The Pressure for both liquids is the same (they are in equilibrium) and so it g. Thus, Pressure, g and x can be cancelled out, leaving 2/3.

I am not, however, sure whether that is the right answer. If you could give me the references for this, I could check the marking scheme and confirm...

that's what i thought so too bt the ms says 0.5 is the correct answer

ref : qn 20..... on/04

 

[Jaf]

I think if, rather than taking the base of the tube, you take x as your starting point, you will get the right answer because density of P would Pressure/g2x and density of Q would be Pressure/gx, which gives 0.5.

 

[geek101]

Soldier313 Jaf
take the second dotted line from the bottom (first x) as the zero level. since the pressure acting on that level in both the columns is the same, and the liquid does not level off in each of the bars in the u tube. so 1x of Q exerts the same pressure as 2x of P.
so ρ = P / gh
ρ is directly proportional to 1/h
for P, h = 2x >> ρ /2
for Q, h = x >> ρ
so P / Q = 0.5 / 1 = 1/2

 

[Soldier313]

I think if, rather than taking the base of the tube, you take x as your starting point, you will get the right answer because density of P would Pressure/g2x and density of Q would be Pressure/gx, which gives 0.5.

Soldier313 Jaf
take the second dotted line from the bottom (first x) as the zero level. since the pressure acting on that level in both the columns is the same, and the liquid does not level off in each of the bars in the u tube. so 1x of Q exerts the same pressure as 2x of P.
so ρ = P / gh
ρ is directly proportional to 1/h
for P, h = 2x >> ρ /2
for Q, h = x >> ρ
so P / Q = 0.5 / 1 = 1/2

thanx a lot

 

[Oliveme]

okay so this is simple
for question 10, the rate of change of momentum is the resultant force ( newtons second law) so u find the gradient of the graph
so u apply the formula (p1-p2)/(t2-t1) which gives u B.
now question 34. The light intensity is increasing. Hence the resistance of the LDR falls. Pd is directly proportional to the resistance, so if the resistance falls, so does the pd across it. Now the new pd can be anything lower than 4V and for that there is only one option that is A.
Question 36..u find the current in the whole circuit 3=I(4+2), I=0.5
Terminal potential difference is the pd the cell provides after taking in account the pd being lost in the internal resistance..that is V=0.5x2=1. 1 volt is being lost, hence terminal pd is 3-1=2
to find power output just use the formula V2/R= 2^2/4=1
hence the answer is C. Hope this helped and GOOD LUCK for tomorrow

thank you

 

[KurayamiKimmi]

please explain question no 17 and 24 please
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_12.pdf
ans 17-C
24-B

 

[Soldier313]

copper and glass.....which of these two is crystalline?

 

[Capricedcapri]

provision of links of qs and ms wil help alot.....

buh cu u tell me the method /??