How to prepare for physics paper 1(cie)??????

[Joseph_COOL]

Q22 Nov 09 plz

[Joseph_COOL]

anyone plz Q22 and Q26 November O9

[halosh92]

anyone plz Q22 and Q26 November O9

Q26) n lamda= d sin thita
its fallin at 90 degrees (normal) as they say so thita =90
sin 90= 1 ( check with ure calculator)
we have to find "n"
 n * (450*10^-9) = 1/(300* 10^3)
(300 mm as i mentioned previously it would be 1/300 but because its mm so u have to change m to mm so 300 *10^3...so it would be 1/(300*10^3)
we dont consider sin 90 in the equation because its =1
anyways
when u solve the equation n=7.4
so basically u have 7 maximas..BUT..they asked for the total number of maximas
u have 7 maxima up and 7 maxima down thats 14...and dont forget the central main maxima where u find the white light on the centre of the screen thats a maxima
so 14 + 1= 15 maxima in toal
got it?

[Joseph_COOL]

thanks very much halosh u got an easy way of explanation

[halosh92]

thanks very much halosh u got an easy way of explanation

glad to help dude 

[matabd]

Uhhh  . .
Lol i dont think i would know how to explain it easier. Thats what i think Sorry

[halosh92]

anyone plz Q22 and Q26 November O9

Q22)Y.M=STRESS/STRAIN
stress=F/A
stress=F/pie (d/2)^2
stress=20/ ( pie *(([5*10^-4])/2)^2)
strain= stress/Y.M
strain = stress(the value we calculated above substitute here)/ (2*10^11)
starin= answer A
basically the question is asking for estension/original length which is nothing but strain

[matabd]

Question 22:

First use the formula E=Fl/ax (E= Young Modulus, F=Tension l = original length A=Cross sectional area and x=extension)

First find A
A=Pi/4 D^2
A=Pi/4 (5x10^-4)^2
A= 1.9635x10^-7

thus if u rearrange the formula you get x/l = F/AE
Yes ? just rearrange the above formula i gave you. Then u put in the values for F, A and E
so x/l = 20/(1.9635x10^-7)(2x10^11)

thus u get the fractional change in length, to get Percentage you want to multiply the value by 100.
So u get Answer B (x/l is just the fraction how it changes, we want to find the percentage of original length)

[aldehyde1612]

Firstly for the Pressure One:

What i think is that Since the RHS new pressure is Patm.  - pgh and the LHS pressure is Patm+pgh. Thus the Increase is equal to RHS-LHS so you get 2pgh.

im a bit late.. but i dont get that... why did u take Patm on the LHS???
i just thought since the difference in the levels of liquid in the arms is taken as 'h' in hpg... when one side goes down by h the other side increases by h... so the difference from the level on the LHS to the level on the RHS side is 2h.

[matabd]

im a bit late.. but i dont get that... why did u take Patm on the LHS???
i just thought since the difference in the levels of liquid in the arms is taken as 'h' in hpg... when one side goes down by h the other side increases by h... so the difference from the level on the LHS to the level on the RHS side is 2h.

Correct. Thats what many of us, thought initially. However maybe u can say that since only the LHS increases by h, the RHS should also increase by h, thus the increase in P should just be Rho-gh. But maybe the formula actually is Delta P (change in P) = rho-g-Delta h (change in h)

[Joseph_COOL]

thanks halosh and matabd
i missed the negative in 10^-4 LOL

[matabd]

No Probs. Good Luck for tomorrow. Lol im supposed to do the Exam also.


Good Luck to everyone

[Joseph_COOL]

good luck to u to matabd 

can anyone explain Q36 in June 07??

[Joseph_COOL]

PLZ someone explain Q36 June 07 and Q25 Nov 07

[Joseph_COOL]

PLZ someone explain Q36 June 07 and Q25 Nov 07
Really need the explanation PLZZZ