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Physics help needed For AS level

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Hi there

Can any one help me with the following MCQ's regarding deformaion of solids for AS level physics. I have a test.

Question No 21 OCT/NOV 2009 Paper 11
Question No 18,20,21 MAY/JUNE 2009 paper 1
Question No 24 MAY/JUNE 2008 Paper 1
Question No 23 OCT/NOV 2004 paper1



Thank you in advance and God bless you.

please reply fast any one!!!!!! :?:
 
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Question No 21 OCT/NOV 2009 Paper 11

well if u see the line is a bit curved...so to find the strain energy, area under the curve (its not a triangle so u hav to add the small area too! )
so its 0.5*(2/1000)*100=0.1J plus the small area above the triangle
so u estimate that if the energy of the triangle is 0.1J so the energy of the small area must be really small like 0.01

so u add both energies, to get the area under curve...which is 0.11J




Question No 24 MAY/JUNE 2008 Paper 1

YOUNG MODULUS IS ALWAYS THE SAME FOR THE SAME MATERIAL!!!!
SO THE ANSWER IS (C) CUZ THE SAME STEEL IS USED!


Question No 23 OCT/NOV 2004 paper1

same as before.
u make a line where the graph is starting and ending...this makes a triangle.
fined the area under this triangle........1/2 * 5 * 40= 100 J

the actual area is less than 100 J so the answer is (A)
 
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Thank you very much!!!!

thanks alot!!!

actually i dont consider the subject difficult. it is all about technique rather than content. if u know the correct technique in answering the questions you will succeed. any how thanks a lot.

What about the MAY/June 2009 paper 1 question 18,20? :D :)

what is your qualification?

again thanks
 
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JUNE 09

18...I DONNO :shock:
20....P has l lenght and A cross section....so its extension is E
Q has 2l...which means 2 times more extension and A/2 so 2 times again more extention...so in total 4E

the ratio of EXTENTION IN P : EXTENSION IN Q is 1:4

bt the force(tension) needed to extend P will be more, cuz it has less extention so the ratio of TENSION IN P : TENSION IN Q is 4:1...which is (D)

21......again the same :D

divide the grph into 3 areas, area of a triangle, rectangle and trapezium

area of triangle..u make a straight line, some are is extra, (in excess :p lol) = 0.275 J
area of rectangle=0.6J
and area of trapizum =1.15J...u make a line to make a full trapezium
the area that is in excess in triangle is apprx. equal to the area which trapizium is lacking
so u just add these 3 areas to get 2 J :)

ummm, im an As student, i'll give my AS exams this may InshaAllah
 
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q 18
on ryt handside the liquid rises by h meaning that it falls by h on left handside so the height difference is 2h and ANS is therefore D
Q 20
ITS JUST proportionality...E=Fl/Ae e(extension;f force;a.. area)
rearrange this to give F=EeA/L ......WHERE e AND E are constant so lets call them k giving us the final reLAtion
F=kA/L......NOW FROM THIS equation l is inversley proprtion to force and A directly proportional to F
 
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Well, I need some help with P1 Nov 2005 no. 34, please. Would be appreciated, thanks.
 
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MW24595 said:
Well, I need some help with P1 Nov 2005 no. 34, please. Would be appreciated, thanks.
In the diagram, actually the gradient (or slope) of the curve represents the resistance. Gradient = I / V , so it equals the reciprocal (hope you know this word) of the resistance. That is, the steeper the slope, the smaller the resistance, and vice-versa. So in the diagram the smallest resistance is corresponding to the steepest slope - the answer is B.
 
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Yes, I thought so, as well, and that seems to be the correct answer as far as I (or you for that matter, too) can tell, B corresponds to the greatest reciprocal of R and so represents the lowest resistance. However, it seems that according to the Mark Scheme, the answer is C. And I have no clue as to how that could be right.
 
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caution:never write the reciprocal of I-V curve gradient idea for resistance in exam....its nt acceptable...........it cn help u sumtimes bt neva write this in any explanation
 
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