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Physics MCQs thread.

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sniper7137 said:
Hi. Can anyone explain Q12 and Q26 of October November 2009 variant 1? And Q15 and Q23 of May June 2009?

Q 26, Nov. 2009 V1

Total maxima = 2xhighest maxima +1

highest maxima = line spacing/wavelength

line spacing = 1/(300x10^-3)

highest maxima = 1/(300x10^-3)/(450x10^-9)

=7.4 x 10^6

total maxima = 2(7.4 x 10^6) +1 = 14.8 x 10^7 = 15 [when rounded off]
 
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nerd007 said:
mrpaudel said:
@nerd07: for 34....u use wire of length 800m..therefore has resistance of 800*0.005=4ohm...so minimun e.m.f=V+Ir=16+(0.6*4)=18.4V thus C.!! i dont know rest of the questions!!

Thanks for taking out the time to solve my questions. Anyone know how to do 5? looks like a tough one.

i didnot understand the question ...haha..!! and hit thanks button if it has helped u dude..!!
 
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sniper7137 said:
Hi. Can anyone explain Q12 and Q26 of October November 2009 variant 1? And Q15 and Q23 of May June 2009?


Q15 May June 2009

Since both mass and height are decreasing to half when water goes to vessel Y.

Therefore potential energy loss = 0.5m x g x 0.55 = mgh/4
 
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Two equal masses travel towards each other on a frictionless air track at speeds of 60 cm s–1 and
40 cm s–1. They stick together on impact.
60cms–1 40cms–1
What is the speed of the masses after impact?
A 10 cm s–1 B 20 cm s–1 C 40 cm s–1 D 50 cm s–1
how to find speed of mases ....??
 
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mrpaudel said:
nerd007 said:
mrpaudel said:
@nerd07: for 34....u use wire of length 800m..therefore has resistance of 800*0.005=4ohm...so minimun e.m.f=V+Ir=16+(0.6*4)=18.4V thus C.!! i dont know rest of the questions!!

Thanks for taking out the time to solve my questions. Anyone know how to do 5? looks like a tough one.

i didnot understand the question ...haha..!! and hit thanks button if it has helped u dude..!!


lol.. I did for one answer.. think of it as for all your help =P
 
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Mobeen said:
http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_w10_qp_12.pdf
qs 13,14.26 help.... plzzzzzz

13D...gravitation is always equal!!
14D...calculate force in that direction in which person is pushing..which is 400N...so work done=400*1.5+150=750
 
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mrpaudel said:
nerd007 said:
mrpaudel said:
@nerd07: for 34....u use wire of length 800m..therefore has resistance of 800*0.005=4ohm...so minimun e.m.f=V+Ir=16+(0.6*4)=18.4V thus C.!! i dont know rest of the questions!!

Thanks for taking out the time to solve my questions. Anyone know how to do 5? looks like a tough one.

i didnot understand the question ...haha..!! and hit thanks button if it has helped u dude..!!

yeah..1 is enough..!!
 
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Mobeen said:
Two equal masses travel towards each other on a frictionless air track at speeds of 60 cm s–1 and
40 cm s–1. They stick together on impact.
60cms–1 40cms–1
What is the speed of the masses after impact?
A 10 cm s–1 B 20 cm s–1 C 40 cm s–1 D 50 cm s–1
how to find speed of mases ....??

m1u1+m2u2=mv( Now m taking 60m/s as positive..so 40m/s will b negative...these are jus direction...so while calculating u need to care bout sign..if u do assume a direction as positive..other direction will b negative..so)

60m-40m=2mv
or 20=2v'
or v=10m/s!!
 
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Mobeen said:
http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_w10_qp_12.pdf
qs 13,14.26 help.... plzzzzzz


Q 13

The gravitational pull on the cube will not change even if it immersed in water, that is, the weight will not change hence it is D.


Q 26

since distance between two nodes or anti nodes = 1/2 lamda

therefore lamda = 2 x 15x10^-3

= 30 x 10^-3

speed = 3 x 10^8

use the formula s = f x lamda

f = 3 x 10^8/(3x10^-3)

= 1x 10^10

so answer is C
 
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can u xplain me of q.18 of this year??
i m doing 2 give ur ans.....
plz if u know 18 then...
 
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