Physics: Post your doubts here!

[killer22]

Y is da answer not c but d???
View attachment 28829

Long explanation. Just memorise the answer.

 

[1597.2217]

can someone please help me with Q 12,19,27,30,36 from Winter 2012 ...... http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_12.pdf

Q12-
Remember the equation: F=Change in momentum/time
Change in momentum:
0.5x12 = o.5x-8
Change is 10
10/0.1 = 100
So (D)

Q19-
P = F/A
F = P x A
F = 4000 x 500/100x100 [Convert cm^2 to m^2]

Wd = F x s
200 x 0.3 = 60

Ans is (A)

Q27-
Y = FxL/AxX
L and X are constant for both.
(Y) of X = 1/2 (Y) of Y
So F/x = F/2x [Don't include L and A because they are same for both.]
You see that the ratio is 1:2 so (B)

Q30-
Sorry no idea, don't have Sin O here.
Otherwise 1/300x10^-3 x Sin O = n x 690x10^-9

Q36-
P = I^2 x R
Max = 400
Min = 100
There are 5 max and 4 min, so
5 x 400 + 4 x 100 / 9
= 266.6

Comes close to (C) of 250


Good luck tomorrow. Please remember me in yours prayers.

 

[1597.2217]

Y is da answer not c but d???
View attachment 28829

Because (t2-t1)^2 is the square of the time difference.
However, t2^2 - t1^2 will you you difference of squares of both answers, which not the same value as above.

Lets take T1 as 3 and T2 as 5.
(5-3)^2 = 4.000
5^2 - 3^2 = 16.000

 

[Lyfroker]

Q4-

LHS = RHS
The units must be cancelled out so that nothing remains. First, you cancel out all the constants:
L x 1/P = Q x T^2 [We know L is m and T is s]
m x 1/P = Q x s^2 [To cancel the s^2, Q must have s^-2]
Now substitute the units given in options:
(B) m x m^-2 = m x s^-2 x s
Giving m = m

Q5-

Basically what they are asking is to first find the PERCENTAGE UNCERTAINTY. Eg, for A- 0.002/2.043 x 100% gives 0.097%.
Find out for other options. You will see that B- has 3.7% and D- has 3.85%. But it doesn't end here. (B) is d, and in the equation the d is multiplied by d again. So multiply the uncertainty giving you value greater than (D).

Q15-

Total AC moment = Total C moment.
The rings are of same mass.
7xm + Yxm = 5x2m [Subs m as 1 for simplicity]
7 + y = 10
y = 3
[8 from pivot minus 3 = 5.]
Ans is hence (C)


Good luck tomorrow. Remember me in yours prayers please.

Thnx a lot n same to u

 

[Lyfroker]

Q#9, 11, 18, 24 & 37
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_11.pdf

 

[6Astarstudent]

Q#9, 11, 18, 24 & 37
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_11.pdf

#9: V^2=U^2 + 2 as
so original equation 0 = U^2 + 2ax
increase in initial velocity by 20% so 0 = (1.2U)^2 + 2ax = 1.44U^2 + 2ax
X is proportional to U so answer is C, 1.44x

#18 solved by Gain in potential energy of m1 minus loss in potential energy in m2. the change in energy is the work done by the motor.
so Ep=mgh and change in Ep = m1gh - m2gh = (m1-m2)gh.
question is asking for power (rate of change of work) so divide the work done by time. (m1-m2)gh/t = (m1-m2)gv ansewr is D

#24 since it is a progression wave, picture another wave less than or equal to π/2 infront of it (direction of wave). notice only B and D is moving up. and then the velocity is the gradient of the wave so greater gradient at B so answer it B.

#37
The Voltmeter is parallel to both resistors, not just one. so there is no change in pd so answer is A

 

[Zaibyraja]

Could someone please help me with may/June/12/2012 pleasee.
Mcq no 13,18 and 24.
Here's a link to the paper: http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf

 

[6Astarstudent]

Could someone please help me with may/June/12/2012 pleasee.
Mcq no 13,18 and 24.
Here's a link to the paper: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_12.pdf

#13
F(net force) = ma
F(net) = 2x9.81(the weight) - 6(friction) = 13.62
total mass = 8+2 = 10
F=ma
A=F/M = 13.62/10 = 1.4 so answer is A

#18
efficiency = output power/input power
in this case its a generator so output power is P=VI = 230 x 32=7360
input power = E(potential) / time = Mgh/t = 200 x 8 x 9.81 = 15696
efficency = 7360/15696 x 100%=47% answer is D

#24
simply find the area inclosed by the graph and x = 0 to x= 7mm

 

[Zaibyraja]

#13
F(net force) = ma
F(net) = 2x9.81(the weight) - 6(friction) = 13.62
total mass = 8+2 = 10
F=ma
A=F/M = 13.62/10 = 1.4 so answer is A

#18
efficiency = output power/input power
in this case its a generator so output power is P=VI = 230 x 32=7360
input power = E(potential) / time = Mgh/t = 200 x 8 x 9.81 = 15696
efficency = 7360/15696 x 100%=47% answer is D

#24
simply find the area inclosed by the graph and x = 0 to x= 7mm

Thankyou so much. On the last mcq however the answer I get is A whereas the mark scheme says B :/ and if someone could do 17 as well I'd really apprciate that. The answer to that is B

 

[6Astarstudent]

Thankyou so much. On the last mcq however the answer I get is A whereas the mark scheme says B :/

wierd, I just did calculation again and it is B.
You can always use the last resort by counting the squares. I counted 243 squares and each area of square is 5x10^-4 so I get 0.1215 J which is B

 

[hope4thebest]

Ans is A. Can someone explain?

 

[Oceanic]

View attachment 28827

Both questions please!
Answers are
17: A
18: B

17 is a because only that body cannot stay in that position ie it will move down the slope.
No clue about 18 though
But which app is this ?

 

[Tkp]

17 is a because only that body cannot stay in that position ie it will move down the slope.
No clue about 18 though
But which app is this ?

for no. 18
v2=u2-2gs
v2=28^2-2*9.81*22=19
so ans is B

 

[Kyusam]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf
Ok so I'm done with every single thing except this question which is literally haunting me :'( plzzz plzzzz plzzzz if some1 knows it answers it pleaseeeeee it's Q:30

 

[6Astarstudent]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_12.pdf
Ok so I'm done with every single thing except this question which is literally haunting me :'( plzzz plzzzz plzzzz if some1 knows it answers it pleaseeeeee it's Q:30

ok first you find the wavelength λ which is 2x the distance between 2 antinodes X and Y. so
λ= 2x 33cm = 66cm = 0.66m
and then v=f λ so f= v/ λ=330/0.66=500
we know that period (T) = 1/f = 1/500 = 0.2ms for 1 oscillation
so answer is B

 

[Zarif009]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_11.pdf

Question 6
This is the only problem i have out of the entire Question Paper someone please solve this thanks

 

[6Astarstudent]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_11.pdf

Question 6
This is the only problem i have out of the entire Question Paper someone please solve this thanks

The question states that "X is measured with a percentage uncertainty of ±1 % of its value at all temperatures." meaning that the percentage uncertainty is unchanged, so A and B is eliminated.
then you need to apply some logic here, will the actual uncertainty get bigger as you approach 100 or 0?.
you can quickly do some math here, 100x1% = ±1 uncertainty while 0x1% = 0
so answer is C, least actual uncertainty when temperature close to 0

 

[Zarif009]

The question states that "X is measured with a percentage uncertainty of ±1 % of its value at all temperatures." meaning that the percentage uncertainty is unchanged, so A and B is eliminated.
then you need to apply some logic here, will the actual uncertainty get bigger as you approach 100 or 0?.
you can quickly do some math here, 100x1% = ±1 uncertainty while 0x1% = 0
so answer is C, least actual uncertainty when temperature close to 0

The question states that "X is measured with a percentage uncertainty of ±1 % of its value at all temperatures." meaning that the percentage uncertainty is unchanged, so A and B is eliminated.
then you need to apply some logic here, will the actual uncertainty get bigger as you approach 100 or 0?.
you can quickly do some math here, 100x1% = ±1 uncertainty while 0x1% = 0
so answer is C, least actual uncertainty when temperature close to 0

Wow thank you so much i should've read the question more carefully. Didn't see that percentage uncertainty is the same!

 

[TaffsAsLevel]

Two springs P and Q both obey Hooke’s law. They have spring constants 2k and k respectively.
The springs are stretched, separately, by a force that is gradually increased from zero up to a
certain maximum value, the same for each spring. The work done in stretching spring P is WP,
and the work done in stretching spring Q is WQ.
How is WP related to WQ?
A WP =
 0.25WQ B WP = 0.5
 WQ C WP = 2WQ D WP = 4WQ

How can it B???? Common sense it should be C but why markscheme said B?@?@?

OH MY GODDD NEVERMIND

first we need to find out the extension for both the springs
force is the same for both the springs that is equal to 'F'
force for P F=2k*x
x=F/2k
force for Q F=kx
x=F/k
Wp=1/2*k*x^2=1/2*2k*(F/2k)^2
Wq=1/2*k*x^2=1/2*k*(F/k)^2
Wp/Wq=1/2*2k*(F/2k)^2÷1/2*k*(F/k)^2
Wp=1/2Wq Ans

 

[1597.2217]

Good Luck everyone. There isn't anytime to clarify any more doubts, so just memorize the answer of what you find hard. Good luck again, remember me in yours prayers. Will really need them for P1s.

@Lyfroker
@IGCSEstudent2012

Good luck In sha Allah. Please remember me in yours prayers.