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Physics MCQs thread.

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kakarocks said:
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w09_qp_11.pdf here is the link , Please help with question 23 and 28 guys :D
for Q.23 i got this link from another thread, just scroll down to the end of the page, and try to remember the order:
http://en.wikibooks.org/wiki/A-level_Ph ... _a_wave%3F

for Q.28
force on the charged liquid is balanced by it's weight
F=mg
(F=q*E), so,
qE=mg
as per the question, they asked for the ratio of charge/mass (q/m),
so arrange the equation and you'll get (q/m)=(g/E).
But I also have doubt on finding the polarity.(This is Milikan's oil drop experiment).
 
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ivorydale said:
kakarocks said:
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w09_qp_11.pdf here is the link , Please help with question 23 and 28 guys :D
for Q.23 i got this link from another thread, just scroll down to the end of the page, and try to remember the order:
http://en.wikibooks.org/wiki/A-level_Ph ... _a_wave%3F

for Q.28
force on the charged liquid is balanced by it's weight
F=mg
(F=q*E), so,
qE=mg
as per the question, they asked for the ratio of charge/mass (q/m),
so arrange the equation and you'll get (q/m)=(g/E).
But I also have doubt on finding the polarity.(This is Milikan's oil drop experiment).

polarity is simple, it has to be negative because there is a +ive charge on top and since its at equilibrium the +ive charge is attracting it upwards [which only happens if its -ive polarity] where as the weight is pulling it downwards, and the forces are balanced.
 
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ivorydale said:
kakarocks said:
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w09_qp_11.pdf here is the link , Please help with question 23 and 28 guys :D
for Q.23 i got this link from another thread, just scroll down to the end of the page, and try to remember the order:
http://en.wikibooks.org/wiki/A-level_Ph ... _a_wave%3F

for Q.28
force on the charged liquid is balanced by it's weight
F=mg
(F=q*E), so,
qE=mg
as per the question, they asked for the ratio of charge/mass (q/m),
so arrange the equation and you'll get (q/m)=(g/E).
But I also have doubt on finding the polarity.(This is Milikan's oil drop experiment).

The Polarity is easy , G is down so the F due to E must be up hence it's a negative charge. Thanks though :D
 
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bloooooo said:
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_11.pdf
PLZ!!!! Question no.37 of this paper!!!!
I think answer B is corrent but according to marking scheme answer A is correct but this does not make any sense to me. And i did it in some other paper and i got b answer. :fool: :fool: :fool: :fool:


PLZ................URGENTLY!!!!!!!!

It's b . Because at Position X the volt meter give P.s across both resistors which is 4 . at position y it give it across 1 resistor which is 2 so the answer is B
 
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Mssamgirl15 said:
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w08_qp_1.pdf

OCTOBER/NOVEMBER 2008 QUESTION 1
Can anyone help me with question number 1. PLEASE????


HE is asking to calculate the frequency since Frequency is "cycles per second" i.e distance travelled in terms of wavelength so dovode 3x 10^8 by 600 x 10^-9
 
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Mssamgirl15 said:
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w08_qp_1.pdf

OCTOBER/NOVEMBER 2008 QUESTION 1
Can anyone help me with question number 1. PLEASE????
you have to find the frequency here as it is number of wavelengths travelled in 1s. v=f*wavelength. u hev got wavelength 600*10^-9 and v is speed of light C= 3*10^8 So u get frequency= 5*10^14.
 
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A moving body undergoes uniform acceleration while travelling in a straight line between points
X, Y and Z. The distances XY and YZ are both 40 m. The time to travel from X to Y is 12 s and
from Y to Z is 6.0 s.
What is the acceleration of the body?
A 0.37 B) 0.49 C) 0.56 D) 1.1
plzz dis question 2
 
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intel1993 said:
A moving body undergoes uniform acceleration while travelling in a straight line between points
X, Y and Z. The distances XY and YZ are both 40 m. The time to travel from X to Y is 12 s and
from Y to Z is 6.0 s.
What is the acceleration of the body?
A 0.37 B) 0.49 C) 0.56 D) 1.1
plzz dis question 2

That's the one i'm asking as well, it's quite confusing . And my brain is dead since it's 12 45 already and tommorow is the exam!
 
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kakarocks said:
intel1993 said:
A moving body undergoes uniform acceleration while travelling in a straight line between points
X, Y and Z. The distances XY and YZ are both 40 m. The time to travel from X to Y is 12 s and
from Y to Z is 6.0 s.
What is the acceleration of the body?
A 0.37 B) 0.49 C) 0.56 D) 1.1
plzz dis question 2

That's the one i'm asking as well, it's quite confusing . And my brain is dead since it's 12 45 already and tommorow is the exam!

someone please help with this one, I know it has something to do with ut+w/2 at^2 ...but i forgot how to do it..
 
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its b because the road applies a force upward to counter the weight as well as fiction force which is in the forward direction as the wheels are spinning clockwise. the resultant of these two forces [ which look just like A and D] will be B.
 
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