AS Physics P1 MCQs Preparation Thread.

[salvatore]

4. Angular deflection decreases. That means points get closer together (so not much deflection is needed).
8. Since it's a velocity/time graph (and not speed/time), the area will be displacement an displacement is a vector. The upper part will be positive displacement and the lower part will be negative displacement.
13. When taking moments about a pivot, make sure you take the PERPENDICULAR distance from the pivot.
31. Find the rate of flow using Q=It. The direction will be from negative to positive; they didn't ask for the conventional current direction, that's why.

Thank you for your reply..

I haven't understood qn no. 13, could you please show me the working?

For qn no. 31, I found the rate of flow as t = Q/I = (1.6 x 10^-19) / 4.8 = 3.3 x 10^-20. That is not correct right?

 

[Sohail A. Razzak]

I have doubts in M j 2012 paper 11 question 12,18,23
And doubts in M j 2012 paper 12 question 6,13,23,24,26,28,29,30.
Would be grateful if sumbudy could explain the questions.!

 

[Sohail A. Razzak]

....can sumbody explain!please help

 

[Sohail A. Razzak]

mcq 29: power are same, so use P=V^2/R
cancel powers nd then (12^2)/6^2 = Rx/Ry u will solve to get 4...
mcq 17: electrical to light is only 7J hence...7/(93+7) * 100= 7 %

Could u please answer my doubts??would be grateful!

 

[Aima]

View attachment 28613 ....can sumbody explain!please help

Driving Force = ma
The driving force is caused due to the 1 kg mass moving downwards hence 1x9.81 = 9.81 and the mass is the mass of the entire system = 3 kg. From this, we get the acceleration 3.27. Substitute in the equation v^2= u^2 + 2as and you'll get the speed as 1.8 (taking initial speed as 0 and the distance as 0.5 m) thus A.

 

[Sohail A. Razzak]

Driving Force = ma
The driving force is caused due to the 1 kg mass moving downwards hence 1x9.81 = 9.81 and the mass is the mass of the entire system = 3 kg. From this, we get the acceleration 3.27. Substitute in the equation v^2= u^2 + 2as and you'll get the speed as 1.8 (taking initial speed as 0 and the distance as 0.5 m) thus A.

JazakALLAH hu khair.thanx.

 

[Sohail A. Razzak]

This one is pretty simple:
You will use the given weight to find the mass... (20/10 =2kg)
Then use the resultant force and equate it to ma (10-4=2a , solve for a (6/2 = 3 ms-2)

I have doubts in M j 2012 paper 11 question 12,18,23
And doubts in M j 2012 paper 12 question 6,13,23,24,26,28,29,30.
Would be grateful if sumbudy could explain the questions.!

 

[Lyfroker]

q# 13, 18, 28 & 33
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w04_qp_1.pdf

some1 plz.....

 

[Lyfroker]

q# 9 & 18
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s04_qp_1.pdf

 

[Tkp]

some1 plz.....

q# 13, 18, 28 & 33
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w04_qp_1.pdf

13: B..W is in the middle of the rod..so ( W x d=F x d) so W = (300 x 2)/1.25 = 480
18: A..( 9000 x 40) - ( 20000 x 12)= 120Kj
28: D ... the distance between the slits and the screen is indirectly proportional to the wavelength..so if the wavelength is decreased so the distance must increase and so D is the answer cause it is the only one greater than 1.00m
33: C... R= pl/A = pl/(pie) (0.5d)^2..then p x 2l / (pie) x (2d/2)^2.....from 0.5d squared to 2d/2 squared is multiplication by 4 so R will be 0.5 then I=V/R so 2V/0.5R then 4I so V

 

[Tkp]

q# 9 & 18
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s04_qp_1.pdf

Q9: D...first Vertically..get the time using s=ut + 0.5 at^2...so 1.25= 0.5 x 9.81 x t^2...t=0.5, then v=d/t...10/0.5= 20 ms^-1
Q 18: C... work done = Force in direction of motion x distance moved so 50kg x 9.80 ms^-2 x 1.6 = 784 about 780J

 

[itallion stallion]

no probz...okay its like the current thru ammeter is not changed b/c their no changes r done there...nd resistance ov variable resistor increases so p.d across it shud increase hence on the other hand the p.d shud decrease hence C...

when the resistance of variable resistor will increase less current will flow through that loop,so wont the reading of ammeter increase as that left current will take the low resistance loop and pass through the loop in which ammeter is connected.thanks!

 

[sweetjinnah]

when the resistance of variable resistor will increase less current will flow through that loop,so wont the reading of ammeter increase as that left current will take the low resistance loop and pass through the loop in which ammeter is connected.thanks!

nope....

 

[sweetjinnah]

I have doubts in M j 2012 paper 11 question 12,18,23
And doubts in M j 2012 paper 12 question 6,13,23,24,26,28,29,30.
Would be grateful if sumbudy could explain the questions.!

may june 2012 p11 :
mcq 12) first calculate force( weight) of man nd barrel..., then net force = 1200-800= 400N
by using F=ma calculate a for both barrel nd man.u vl get a= 3.333 nd 5 respectively...then change in acceleration 5- 3.33=1.67 approx.2m/s^2
using knowledge of pe= ke , v= squareroot 2gh u vl get ...v= squareroot 2*2*9 half of height is used cz the balance made ...the ans u get is v= 6m/s...
18) efficiency= power output / power input .....use p=e/t so u vl get B....
23) young modulus = Fl/eA
so F/e * l/A HENCE F/e = spring constant so the ans isC....hope u get these....

 

[sweetjinnah]

I have doubts in M j 2012 paper 11 question 12,18,23
And doubts in M j 2012 paper 12 question 6,13,23,24,26,28,29,30.
Would be grateful if sumbudy could explain the questions.!

may june 2012 p12...:
mcq 6) the grid shows pulse lasting for 2cm , 1cm= 1microsecond ....so 2cm = 2 microsec..ans A
13) F net= W- friction
ma= mg- f
(8+2)* a = (2*9.8) + (-6) the ans is A
23)pressure of gas= the level line nd using rho g h+ atm pressure = pressure of gas (the height u take is the vertical height)
24)area under the graph nd using an estimate to get the correct answer...
26)time= distance/speed
time= total distance is 150+150/(3*10^8) = 1 microsec hence D
28)360 degrees = one lamda
x degrees = 3/4 lamda cross multiplication to get 270 degrees hence D...
29)v= f lamda so f= c/3 lamda
T= 1/f put value of f u vl get 3 lamda / c hence C...
30) okay in this X nd Y are two maxima nd the distance b/w two adjacent antinodes( where their is maxima ) is 1/2 lamda...= 33 cm
calculate lamda from the eq nd then put it into v= f lamda to get f= 5oo Hz...then calculate Time period from T= 1/f = 2ms....so according to this only B is correct...hope u got these...

 

[Sohail A. Razzak]

May ALLAH bless u with lots of succes...waz worried since yesterday...Sincere duas for u...JazakALLAH khair.

 

[Jasmine 12]

I am having problem in november 09/12 can anyone help ,e in these questions q4,8,9,12,13,14,15

 

[TaffsAsLevel]

Ah, my apologies. I'll try again: The question asks for the point on the I/V graph at which resistance is smallest. We know that the gradient of an I/V graph is 1/R. When R is smallest, 1/R (the gradient) is biggest. We need to check and see which point has a tangent to it with the biggest gradient:

The line at C is the steepest of all, and so will have the biggest gradient (1/R) and the smallest R.

I thought it's tangent to the curve? Can you tell me why you start them all from the origin? I thought answer was B for that reason

 

[Jasmine 12]

I am having problem in november 09/12 can anyone help ,e in these questions q8,9,12,13,14,15

Anyone.??? /:

 

[studen12345]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_13.pdf
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