• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

AS Physics P1 MCQs Preparation Thread.

Messages
102
Reaction score
38
Points
38
t
B. Consider the momentum of the sand+car as a whole. Also, consider Newton's first law with regard to the car. There's no accelerating force acting on the car at Y so it can not gain speed.
thnx mate
 

Jaf

Messages
321
Reaction score
232
Points
53
And finally, 20. Now this has to be the toughest question in the whole paper; I still haven't actually figured it out. But from what I understand; the boy applies a pressure which is a tenth of the atmospheric pressure atop the straw.

Which means that p0/10 is the pressure applied.
You're misreading the question.
The question says:
"[...]the child is capable of lowering the pressure at the top of the straw by 10%. [...]"
This means that the pressure on top is 0.9p₀.

What's the argument now? :p
 
Messages
224
Reaction score
183
Points
53
You're misreading the question.
The question says:
"[...]the child is capable of lowering the pressure at the top of the straw by 10%. [...]"
This means that the pressure on top is 0.9p₀.

What's the argument now? :p

My answer is correct by the Mark Scheme. And your methodology is wrong by the Examiner Report. And my methodology is correct by my mind. ;)

The pressure we need to put into the equation is 0.1p₀. Please understand. Pressure at the top is 0.9p₀; but that doesn't cause the water to move up. If the total pressure at the top caused the water to move up, water would be sprouting out from the straw even without the child applying any pressure on it. And with greater intensity! - since without the child sucking on the straw the pressure would be even greater at 1.0p₀! You would see McDonald's flooded with drinks the moment any kids put straws in their drinking glasses!

... Now wouldn't that be a sight to behold? :cautious:

We need the effective pressure at the top; the one that will cause the water in the straw to move up. That is not 0.9p₀, its 0.1p₀, since that's the pressure change caused by the kid that causes water to move up the straw. The pressure change is by 10%, and 10% of p₀ is 0.1p₀.

I hope that clears things up insha' Allah!
 
Messages
398
Reaction score
233
Points
53
Saad (سعد) Jaf this question...argh! the question says the child is capable of reducing the pressure at the top by 10% so the pressure at the top is now 0.9Po and it gives you the right answer when you substitute it into P=pgh! but why does the examiner report say so! is it cuz there asking about the maximum length, so it means that the lowest possible pressure is to be used, becuz the lowest pressure will give you the max lenght....which is 0.9 Po cuz the child cant reduce it more, and hence the answer is B. :confused:
 
Messages
224
Reaction score
183
Points
53
Saad (سعد) Jaf this question...argh! the question says the child is capable of reducing the pressure at the top by 10% so the pressure at the top is now 0.9Po and it gives you the right answer when you substitute it into P=pgh! but why does the examiner report say so! is it cuz there asking about the maximum length, so it means that the lowest possible pressure is to be used, which is 0.9 Po cuz the child cant reduce it more, and hence the answer is B. :confused:

Read my reply above. I'm pretty sure that's how it works... and Allah knows best.
 
Messages
398
Reaction score
233
Points
53
Read my reply above. I'm pretty sure that's how it works... and Allah knows best.

the pressure can be 0.9Po on the top, without the water spurting out...it just means that the straw is really long!
so you mean to say the answer is supposed to be A...
 
Messages
971
Reaction score
532
Points
103
3. Intensity = energy/(area * time) (also given in the question)
= kgm²s¯² / s¯1 * m2
= kgs¯3
Hence the correct answer is D.

8. You should know how to “convert” displacement/acceleration/velocity graphs to one another. Anyway, here is how this should be:

At the first stage, the speed is constant. This means that the object’s displacement is increasing at a uniform rate. At the second stage, the speed is decreasing, this logically means that the displacement is increasing, but at a decreasing rate. This means that eventually, at S there will be a straight horizontal line. The only graph that shows these 2 stages correctly is C.

28.

y = λd / a = 6.0 * 10-9 / a (this is for the original wavelength of 600nm)
ay = 6.0 * 10-9

Now we want to find the new value of d,
Y = (4.0 * 10-9 * d) / a

Substitute a from the old equation, a and y will cancel out and the value of d turns out to be 1.5m.

31. The best way to solve this is to see each option, and find out which one is correct.
A is wrong because the word force shouldn't be there, it should be replaced with energy.
B is wrong because the word energy should be replaced with power.

Now from C and D, use the equation P = I2R.
P = IV
V = P/I

This is the ratio of power dissipated to current, so C is correct.

34. C. You have to learn these graphics (diode, filament lamp, ohmic conductor and thermistor/semi-conductor type material). Google them if you don’t already know them.

37. The easiest way to solve these questions is by using some fake values of “I” entering the parallel junction. You know that the current is divided into the inverse proportional of the resistances in parallel, so that means if say, you have 6 A entering the junction, 4 A would go the 2 Ohm resistor and 2 A to the 6 Ohm resistor in the 1st parallel combination (lets call this “block 1”).

In “block 2”, 6 A is also entering the junction, and 3 A is shared equally between the resistors. I1 is 4 A and I2 is 2 A, and I1 > I2. Since you also have the currents entering the resistors, you can easily find out the values of V too and compare them.
 
Messages
4,609
Reaction score
3,903
Points
323
37. The easiest way to solve these questions is by using some fake values of “I” entering the parallel junction. You know that the current is divided into the inverse proportional of the resistances in parallel, so that means if say, you have 6 A entering the junction, 4 A would go the 2 Ohm resistor and 2 A to the 6 Ohm resistor in the 1st parallel combination (lets call this “block 1”).

In “block 2”, 6 A is also entering the junction, and 3 A is shared equally between the resistors. I1 is 4 A and I2 is 2 A, and I1 > I2. Since you also have the currents entering the resistors, you can easily find out the values of V too and compare them.

Thanks alot :) :D
explanation needed for this one :)
 
Messages
255
Reaction score
165
Points
43
hey guys when r we going to start the preparation...there is less than 2 weeks left
 
Messages
971
Reaction score
532
Points
103
Thanks alot :) :D
explanation needed for this one :)
OK, see this diagram:

ei17hv.png


I have used a fake value of 6A (you can use anything) entering each junction (they have to be the same at each junction of course). In parallel, the currents are divided into the inverse proportionals of the resistors. I have named each resistor "A", "B", "C" and "D". In block A, the total resistance is 6+3 = 9 Ohms. A will get (6/9) * 6 = 4A for example. I did the calculations for the rest of the resistors, too. To find the voltage, of the resistors, just use the formula V = IR (I being the current entering each resistor).

I hope this helped.
 
Messages
4,609
Reaction score
3,903
Points
323
OK, see this diagram:

ei17hv.png


I have used a fake value of 6A (you can use anything) entering each junction (they have to be the same at each junction of course). In parallel, the currents are divided into the inverse proportionals of the resistors. I have named each resistor "A", "B", "C" and "D". In block A, the total resistance is 6+3 = 9 Ohms. A will get (6/9) * 6 = 4A for example. I did the calculations for the rest of the resistors, too. To find the voltage, of the resistors, just use the formula V = IR (I being the current entering each resistor).

I hope this helped.


YES !!! :D got it thanks alot :) it did help thanks again :)
 
Messages
47
Reaction score
9
Points
18
9702_s04_qp_1.pdf
question 25. I don't know if this question has been discussed before coz the thread is too long to search. Can you kindly help. Any help will be greatly appreciated. Thanks in advance:)
 
Messages
267
Reaction score
184
Points
43
9702_s04_qp_1.pdf
question 25. I don't know if this question has been discussed before coz the thread is too long to search. Can you kindly help. Any help will be greatly appreciated. Thanks in advance:)
the answer is B coz their is a trough before point P so as the wave propagates point P will move downward..
for Q as it is at its maximum point it will be stationary.......
 
Messages
47
Reaction score
9
Points
18
thanks for the quick reply........your explanation was correct but the corresponding answer choice was A not B.
 
Messages
267
Reaction score
184
Points
43
the answer is B coz their is a trough before point P so as the wave propagates point P will move downward..
for Q as it is at its maximum point it will be stationary.......
my bad...........had to be A.........(y)
 
Messages
224
Reaction score
183
Points
53
the pressure can be 0.9Po on the top, without the water spurting out...it just means that the straw is really long!
so you mean to say the answer is supposed to be A...

... I dunno. >.>

I guess I'll note down this particular question and discuss it with my Sir tomorrow. We'll see what becomes of it.
 
Messages
267
Reaction score
184
Points
43
Question 3 and 18 plz with description............(y)
 

Attachments

  • 9702_s02_qp_1.pdf
    163.4 KB · Views: 15
Top