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anyone here knows how to solve mcq 35 of o/n 2004??
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provision of links of qs and ms wil help alot.....anyone here knows how to solve mcq 35 of o/n 2004??
well plenty of thm thn lets start from thisupload any particular question..???
can you please solve question 26? URGENTLY. Answer is BMysteRyGiRl and Jaf for qn 25 use formula: d x sintheta = n x lambda ; where theta = 90 ; sin90 = 1
so we r first going to count the number of orders which is n...... so n= d/lambda
d = 1 / N , N in metres would be 3 x 10 ^5 , therefore d= 1/(3 x 10^5)
so d/lambda = (1/3x10^5) / (450 x 10 ^-9)
ans is 7 , THEN u multiply that by 2, bcoz this was just on one side of the original beam, u get 14, add 1 to this because u r including the undeviated beam, u get 15 and hence D is the answer
well plenty of thm thn lets start from this
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf
Q.10
well plenty of thm thn lets start from this
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf
Q.10
can you please solve question 26? URGENTLY. Answer is B
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_12.pdf
Thanks alotwhen they started they are same
then
look at time 18 second there is phase diffeernce of 180.....
means the phase difference has increased gradually and linearly.... frm o to 180...
now a cycle compromise of 2pi means 360
now difference of 1/8 of cycle wil wil be 360/8 = 45 degrees...
if phase diff. of 180 is produced in 18sec. then 45 wil be produced in 4.5 sec..
180..................18sec
45..................... xsec
what is the ans?????please assist urgentlyView attachment 12992
Awhat is the ans?????
please assist urgentlyView attachment 12992
that's what i thought so too bt the ms says 0.5 is the correct answerAlthough I am not sure, I believe the answer is 2/3. The way I did it was:
Density (P) = Pressure/g3x
Density (Q) = Pressure/g2x
Density of P/Density of Q = (Pressure/g3x) / (Pressure/g2x)
= Pressure/g3x * g2x/Pressure
The Pressure for both liquids is the same (they are in equilibrium) and so it g. Thus, Pressure, g and x can be cancelled out, leaving 2/3.
I am not, however, sure whether that is the right answer. If you could give me the references for this, I could check the marking scheme and confirm...
I think if, rather than taking the base of the tube, you take x as your starting point, you will get the right answer because density of P would Pressure/g2x and density of Q would be Pressure/gx, which gives 0.5.
thanx a lotSoldier313 Jaf
take the second dotted line from the bottom (first x) as the zero level. since the pressure acting on that level in both the columns is the same, and the liquid does not level off in each of the bars in the u tube. so 1x of Q exerts the same pressure as 2x of P.
so ρ = P / gh
ρ is directly proportional to 1/h
for P, h = 2x >> ρ /2
for Q, h = x >> ρ
so P / Q = 0.5 / 1 = 1/2
thank youokay so this is simple
for question 10, the rate of change of momentum is the resultant force ( newtons second law) so u find the gradient of the graph
so u apply the formula (p1-p2)/(t2-t1) which gives u B.
now question 34. The light intensity is increasing. Hence the resistance of the LDR falls. Pd is directly proportional to the resistance, so if the resistance falls, so does the pd across it. Now the new pd can be anything lower than 4V and for that there is only one option that is A.
Question 36..u find the current in the whole circuit 3=I(4+2), I=0.5
Terminal potential difference is the pd the cell provides after taking in account the pd being lost in the internal resistance..that is V=0.5x2=1. 1 volt is being lost, hence terminal pd is 3-1=2
to find power output just use the formula V2/R= 2^2/4=1
hence the answer is C. Hope this helped and GOOD LUCK for tomorrow
buh cu u tell me the method /??provision of links of qs and ms wil help alot.....
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