AS Physics P1 MCQs Preparation Thread.

[Abu mota]

#32
Power = I^2 R
so power in Y = (I/2)^2 x 2R = I^2/2R
power in X = I^2 R
Y/X = 1/2 so answer is B

ty, could u plz help me with the other questions?

 

[LeQuavina]

Energy is always conserved so
Initial energy = final energy
so Ek = new Ek + change in Ep
(mu^2)/2 = (mv^2)/2 + mgh
cancel out all the m ( masses)
(u^2)/2 = (v^2)/2 + gh
28^2 /2 = v^2 / 2+ 9.81x22
392 = v^2 /2 + 215.82
v^2 /2 = 176.18
v^2 = 352.36
v = 18.8
answer is B

Thanks a lot.

 

[6Astarstudent]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_11.pdf Q.9,Q.27,Q.29,Q.33
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_11.pdf Q.18,Q.34 why arent the wires in the cable considered to be parallel i.e shouldnt we divide the R by 2 not multiply it by 2.

#34
wire is in a loop ie. supply -> wire -> relay -> wire -> supply -> wire...... etc. so its in series

#18
Energy is always conserved so
Initial energy = final energy
so Ek = new Ek + change in Ep
(mu^2)/2 = (mv^2)/2 + mgh
cancel out all the m ( masses)
(u^2)/2 = (v^2)/2 + gh
28^2 /2 = v^2 / 2+ 9.81x22
392 = v^2 /2 + 215.82
v^2 /2 = 176.18
v^2 = 352.36
v = 18.8
answer is B

for the s10
#9
answer is B because the ball is falling down with constant acceleration so graph will be an reversed parabola
#27
distance between centre and Q, centre and P is both the radius which is the same. so there is no net workdone. answer is A
#29
field lines represent movement of positive charge, since electron is negative so it will mvoe in opposite direction of field line towards observer so answer is D

#33
there are lots of ways for this question
I find V the easiest to use in an explanation
PQR all have same resistance
Q,R have same V while P has 2x of their V so 2V
Power = (V^2)/R
so Power ratio of P Q R =((2v)^2)/R: (V^2)/R : (V^R)/R or 4:1:1
total power is 12, divided to 6 parts so Power of R is 2W

 

[Obsidian Fl1ght]

Hey anyone in progressive waves, anti nodal lines... does constructive or destructive interference take place on them?

 

[6Astarstudent]

Hey anyone in progressive waves, anti nodal lines... does constructive or destructive interference take place on them?

constructive interference

 

[Obsidian Fl1ght]

constructive interference

Thanx
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[LeQuavina]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_13.pdf
question 24

 

[6Astarstudent]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_13.pdf
question 24

"trolley is held at rest" so there is no net force
so force from P must equal force from Q
F=kx so force in P = 60x0.4=24
Force in Q

 

[Suno Yaar]

HELP HELP QUESTION 7: http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w05_qp_1.pdf QUICK PLEASE

 

[6Astarstudent]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_13.pdf
question 24

"trolley is held at rest" so there is no net force
so force from P must equal force from Q
F=kx so force in P = 60x0.4=24
Force in Q must equal 24 too
x= F/k = 24/120 = 0.2 answer is B

 

[Suno Yaar]

N

can anyone provide me marking schemes of ON 2002 physics p1 ???

no

 

[Suno Yaar]

"trolley is held at rest" so there is no net force
so force from P must equal force from Q
F=kx so force in P = 60x0.4=24
Force in Q must equal 24 too
x= F/k = 24/120 = 0.2 answer is B

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w05_qp_1.pdf help question 7

 

[6Astarstudent]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w05_qp_1.pdf help question 7

kinematics equation
s = ut + (at^2)/2
since initial speed is 0 so
s = (at^2)/2
substitute the values in
x = (a(t1)^2)/2
x + h = (a(t2)^2)/2
solve the simultaneous equations so
h = (a(t2)^2)/2 - (a(t1)^2)/2 = ((t2)^2 - (t1)^2) x a / 2

a = 2h/((t2)^2 - (t1)^2) so answer is D

 

[Obsidian Fl1ght]

What r the no. pf wavelengths of visible light in one metre?

 

[Suno Yaar]

i

kinematics equation
s = ut + (at^2)/2
since initial speed is 0 so
s = (at^2)/2
substitute the values in
x = (a(t1)^2)/2
x + h = (a(t2)^2)/2
solve the simultaneous equations so
h = (a(t2)^2)/2 - (a(t1)^2)/2 = ((t2)^2 - (t1)^2) x a / 2

a = 2h/((t2)^2 - (t1)^2) so answer is C

did the same!!!!!!!!!!!!! but the answer is D :/

 

[6Astarstudent]

i
did the same!!!!!!!!!!!!! but the answer is D :/

I mean the answer is D sorry typo

 

[Suno Yaar]

oh

kinematics equation
s = ut + (at^2)/2
since initial speed is 0 so
s = (at^2)/2
substitute the values in
x = (a(t1)^2)/2
x + h = (a(t2)^2)/2
solve the simultaneous equations so
h = (a(t2)^2)/2 - (a(t1)^2)/2 = ((t2)^2 - (t1)^2) x a / 2

a = 2h/((t2)^2 - (t1)^2) so answer is C

oh lol. your working is right. i just saw your answer/ your working is right and your answer si right. the answer you got is in part D.
you accidentally wrote C lol.

 

[Obsidian Fl1ght]

Dude someone answer this question kindly!
What's the order of no. of wavelengths of visible light in one metre?

 

[6Astarstudent]

What r the no. pf wavelengths of visible light in one metre?

number of wavelengths of visible light in one metre?
if so, wavelength of visible light is 400nm->700nm
1/400nm and 1/700nm so roughly 1.4 to 2.5 x 10^6

so around 10 to the order of 6

 

[Suno Yaar]

i

Dude someone answer this question kindly!
What's the order of no. of wavelengths of visible light in one metre?

idk