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AS Physics P1 MCQs Preparation Thread.

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Please, someone answer this. I'd be reallyy grateful.
U.PNG
the answers are A and D respectively.
 
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go to page 79 of this thread and ull find me explaining the whole NOv 2011 P1 variant 1 and 2..and ull find that i explained this question with details that will help u alot
please solve this and explain how do we solve wheatstone bridge circuits for effective resistance and current on each component?
 

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m/j/09/01
mcq no. 2 ,10,11 n 14
plzzzzzzzzzzzzzzzz anyone !:cry:

Q 2
you have horizontal component 20 and angle 30 so use tan
tan 30 = x/20 x =11.5

Q10
well first right what we will now
so X: 0.5mv^2=E
and Y: has 0.5*2m*l^2 (l is the speed because it can't be same as X has)
since it is an explosion they both have same KE so will try to find l for Y
so l= sqrt(E/m)
v= sqrt(2E/m) now divide l by v so we get l is sqrt(0.5)*v

no put l back in equation for Y and we get 0.25mv^2
and X will be 0.5mv^2

now divide X by Y to get 2/1

Q11
answer is A because P has to be less than R for upthrust to take place and S+ Q play no role in upthrust so are equal

Q14
answer is C
first find units of K from F/V^2 which is kg/m

then compare it with answer to get C
 
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Can someone please explain this?
ans is C

View attachment 12951

answer is C because
T is the largest becasue if you look closely you will see it it near to the pivot and due to it being near requires more force to lift the door as well as it is at an angle so requires more even force

H is second larget because it supporting the door (reaction force) and is also at angle

which leaves W to be the smallest
 
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go to page 79 of this thread and ull find me explaining the whole NOv 2011 P1 variant 1 and 2..and ull find that i explained this question with details that will help u alot
For Q.11:why did you use the second mass equal to 20 kg whereas in the question it says that one mass of 20 kg....collides with a mass of 12 kg??plzz help:(
 
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Please, someone answer this. I'd be reallyy grateful.
View attachment 12955
the answers are A and D respectively.

i remember doing question with my teacher and he said there is something missing in the information given so we weren't able to do it

Q25
use equation f=v/4l for closed tube and f=v/2l for open tube where v is a constant speed of air in this case

so f ' =v/2l
and f=v/4l

divide f ' by f to get 2 which is D
 
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btw u missed part of the question abt PE of Q which is less than P by 50kj

ok so use the equation PE = KE + WD ( i am omitting the kilo unit so it faster to calculate)

at Q: P-50 = KE +10

but we don't know P so for P: PE= KE +WD and work done from P to Qis 10
so PE is 5 so P=15

15-50=KE+10

so KE is 45kj

next time post paper and ms as well as diagram
 
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variant 2,ques 29
The formula for use here is X = lambda x distance between screen and slits / distance between slits. Here, X stands for the distance between 2 CONSECUTIVE dark OR bright fringes. So for example if you take the distance between 2 dark fringes from the second diagram, it will be around 3mm ( 3x10^-3 m ). Substituting all these values into the formula you'll get:

3x10^-3 = lambda x 5 / 0.9x10^-3

Solve this and you'll get answer as B.
 
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FOR Q8)
Calculate individual velocities .
between X & Y (40/12)=3.33 ms^-1 AND from Y to Z (40/6)=6.67 ms^-1
Next calculate the average time taken (12+6)/2=9 sec... find acc. It comes like (6.67-3.33)/9 = 0.37 ms^-2
But its uniform acceleration, and the formula V=d/t, isn't this for costt. velocity?
 
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