AS Physics P1 MCQs Preparation Thread.

[leosco1995]

Any one pls help me out wit dese questions plsss
Q 13 25 26 of N09
Q8 15 20 of J10 variant 12
q5 N10 variant 11

HELP WILL BE APPRECIATED THANKS ALOT

November 2009, Q13)

We know that at the highest point,

Vertical velocity = 0
Horizontal velocity = E cos 45

Resultant = 0² + (E cos 45)²
= 0.5E

Q25) To solve this question, find the maximum order on one side, multiply that by two and then add the central maximum to find the total number of maxima. At the maximum order, sin 90 is used, which is equal to 1.

d sin θ = nλ
10ˉ³/300 = n * 450 * 10ˉ9

n = 7.40

# of orders = 7 * 2 = 14
And central max, so 15.

26) E = F/Q

At the position of charge 'q', the charge will be 'q'

F = E/q

 

[MA1234]

U GUYS ONLY HELP UR FRIENDS... PLZ HELP ME AS WELL

 

[hussamh10]

Moment about Q = force * p.d. from pivot
3.0 = f * 50 mm
f = 60 N

(Q just provides an upward force, so there is no couple and hence we don't use the diameter. Don't be confused about a torque always involving a couple, torque just means turning effect and it could apply to a regular moment also.

You already have the answer with this, but anyway for P:

moment = force * distance from pivot
= 60 * 0.75 mm (force was calculated from Q)
= 4.5 Nm


How does it not make sense? Things which are always conserved in a nuclear reaction are mass-energy, charge/proton # and nucleon number. The number of neutrons aren't always the same.

How did you figure out the distance for finding 4.5 force

 

[xxfarhaxx]

PLZZZZ HELP ME WID MY QUESTION ANYONE PLZZZZ

i jst did xplain ur qustn..c above

 

[xxfarhaxx]

U GUYS ONLY HELP UR FRIENDS... PLZ HELP ME AS WELL

hy i hav answrd nx plnd ur qustn..

 

[leosco1995]

Any one pls help me out wit dese questions plsss
Q 13 25 26 of N09
Q8 15 20 of J10 variant 12
q5 N10 variant 11

HELP WILL BE APPRECIATED THANKS ALOT

June 2010 P12

Q15)

Total work being provided = force * distance
= Fs

Work done in moving the car up the slope mg sin α * s

Efficiency = mgs sin α / Fs
= mg sin α / F

Q20) I can't really help here.. that's the graph of a polymer like rubber. Eventually after you stretch the rubber so much, it starts to get very stiff so there's a tiny change in extension for a large change in force before which it breaks.

 

[hussamh10]

23 The Young modulus of steel is determined using a length of steel wire and is found to have the
value E.
Another experiment is carried out using a wire of the same steel, but of half the length and half
the diameter.
What value is obtained for the Young modulus in the second experiment?
A
2
1
E B E C 2E D 4E

Ans is E because the same material is used this ques is repeate 3 times!!!

 

[leosco1995]

Any one pls help me out wit dese questions plsss
Q 13 25 26 of N09
Q8 15 20 of J10 variant 12
q5 N10 variant 11

HELP WILL BE APPRECIATED THANKS ALOT

Nov 2010 P11 Q5

3000 rev/min

= 50 rev/sec

time period for one rev = 0.02s

For the wave to appear on a 10 cm screen, 0.02/10 = 0.002s.

Then you will divide by 2 for one pulse to appear = 0.001s or 10 ms/cm.

 

[Jaf]

Any one pls help me out wit dese questions plsss
Q 13 25 26 of N09
Q8 15 20 of J10 variant 12
q5 N10 variant 11

HELP WILL BE APPRECIATED THANKS ALOT

ANY ONE PLS RESPOND TO MY QUESTIONS AS WELL PLS :'(

Variant for N 09?

J10)
8 - There is constant acceleration at first. Hence there is supposed to be a curved region on the displacement-time graph. A can be eliminated. Height can not increase with time, D can be eliminated. The steel ball will either reach a terminal velocity or continue with the same acceleration. In C, the speed seems to be decreasing and speed can NOT decrease in a free falling object. Answer is B.

15 - Actual work done by the car in moving up the slope is Fs.
Theoretical work that needs to be done to move the car is mg(sinα)s.
So [mg(sinα)s]/Fs is the efficiency (the s's cancel).

20 - The easiest way to do this is to think of what happens when you stretch something made of rubber. At one point, stretching very hard will produce little extension. This is true only in A.

N10)
5 - 3000 revolutions per minute = 3000/60 = 50 revolutions per second. This is the frequency of revolution.
Time period is 1/frequency. Time period = 1/50 = 0.02s = 20ms

A is out of question. 1 cm = 1 s = 1000ms. 1 cm will show 50 cycles.
B seems appropriate as one cycle will be seen every 2cm.
C 1 cm = 100μs = 0.1 ms. 200cm will be needed to show on cycle.
D This is even smaller than C so there's no point in checking this one.

 

[leosco1995]

How did you figure out the distance for finding 4.5 force

The diameter was given in the question..

 

[freakybandi]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w06_qp_1.pdf
qs 27??? plz

 

[leosco1995]

Hey there. Thank you so much. Mashallah, you're like a physics genius. Honestly, we had a very bad teacher for our AS level, so yeah I pretty much suck at physics. Thanks again!

Thanks. And you're welcome. I love Physics MCQs (even the O-level ones too).

 

[xxfarhaxx]

time for ma doubts nw ... pls do solve them.....plsssssssssss!!!!!!!!!!
Q- 12, 25, 28, 31, 33, 34

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s06_qp_1.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s06_ms_1.pdf

pls clear ma doubts...postin fr da 2nd tym!!!!!

 

[kinglynx]

 

[kinglynx]

 

[kinglynx]

 

[kinglynx]

can someone please help with those three pictures? theyre the right way around and everything

 

[geek101]

kinglynx
here I is directly proportional to 1/d^2 and the graph which shows this is D. and it can not be C cuz when ^2 is zero there is gonna be zero intensity as well, but in C when d^2 is zero...intensity is max, thats wrong.

second pic: http://www.animations.physics.unsw.edu.au/jw/SHM.htm see this...

3rd: ok the strain energy = work done
take the first part as a triangle and the second as a trapezium...
so the area = 0.5 x 500 x 10 x 10^-3 = 2.5
and 0.5 (550 + 500) x 2 x 10^-3 = 1.05
so the wd = 2.5 + 1.05 = 3.55

 

[xxfarhaxx]

pls clear ma doubts...postin fr da 2nd tym!!!!!

guyssssssssss ....hllllllllppppppp!!!!!!!!!

 

[xyz!]

pls clear ma doubts...postin fr da 2nd tym!!!!!

Q12 :-
p(b4 collision)=3mv-2mv=mv
total mass aftr collisn=m+3m=4m
let the speed after collision be say x
and v noe dat p(b4 collisn)=p(after collisn)
thus, mv=4mx
so x=mv/4m=v/4 ==> A


Q.28:-
first considr whts given
so, wavelnth=7*10^-7m (red light)
x=2mm= 2*10^-3m (we divided by 2 bcos 4 is the sepratn btwn 2 bright fringes and nt btwen 2 adjacnt bright and dark fringes)
a= 1mm=1*10^-3 (sumthng v usually consider ven not specified)
hence by using the formula lambda=ax/D, we get D=2.86m

then, for the next conditn,
a=0.5*10^-3m
D=2*2.86=5.714 (m considerng entire calculatr values here)
lambda=7*10^-3 m
again by using the formula v get, x=8*10^-3m= 8mm
bt v hav to calculate the separation btwn 2 bright fringes again so multiply the answer by 2
thus the ans is 16mm==> D


Q.31:-
I=4.8
Q=elementary charge= 1.6*10^-19
v noe dat Q=It
so t=Q/I
bt v hav to find t^-1 (see the units given,its s^-1)
so we will take it to be I/Q= 4.8/(charge)=3*10^19 ==>C


Q.34:-
E=I(R+r)
12=I*(3+1)
so I=3
Power=(I^2)*R
=(3^2)*3
=27 ==> A