would someone please kindly bother to answerhaving problem with mcq no 10 oct/nov 2008.im not getting that howz D the right answer.Need some explanation .
would really appreciate some help .
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As physics p1 MCQS YEARLY ONLY.
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http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf
Kindly tell me the Question 14
for sphere going down
u=0 a=9.81
v=u
s=h
u^2 = 2 x 9.81 x h...ryt?
for sphere going up
u=v a=-9.81
v=0
s=0.5h
0^2 = v^2 + 2 x 0.5h x -9.81...ryt?
now get everything on the right side,simplify,divide the 2 and see what u get!
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can u explain q.21, paper nov 11_12
mark scheme says answer is A but i think the answer is C..
A is wrong coz they say the extension of Q is twice of P...but i dont think thats true...pls just check it out
thanks!
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u first find k the constant so extension for 5N = 14-11=3cm
no find k , F=k x e so k=5/3
no find extension for new force 7N add in k and F =7 in equation and wallah!! u get d answer
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hmm u explained question no.22 but i asked for question no. 21u first find k the constant so extension for 5N = 14-11=3cm
no find k , F=k x e so k=5/3
no find extension for new force 7N add in k and F =7 in equation and wallah!! u get d answer
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Can someone explain the relation with intensity, amplitude and frequency plsssssss
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oopsie i will do it now kay 'sorry
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no problem
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dude da answer is C so u r rite???
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Intensity= square of amplitude and also square of frequency
Used as separate equations.
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yeah C is my answer but mark scheme says its A...i'm a bit
can mark scheme go wrong??!
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dude check again ms says Cyeah C is my answer but mark scheme says its A...i'm a bit
can mark scheme go wrong??!
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AsSalamoAlaikum Wr Wb…
I had been a bit busy with some other things, and managed to do only 15 questions of June 2010 variant 2 [all variants are same, order of the question differs only]. Now looking at the time, I feel bad. Is there anyone who needs it? If anyone wants, I’d continue doing the rest of the paper. If everyone’s gone already, there won’t be any use of me doing it. really sorry….I couldn’t do it.
I had been a bit busy with some other things, and managed to do only 15 questions of June 2010 variant 2 [all variants are same, order of the question differs only]. Now looking at the time, I feel bad. Is there anyone who needs it? If anyone wants, I’d continue doing the rest of the paper. If everyone’s gone already, there won’t be any use of me doing it.
really sorry….I couldn’t do it. please do post it if itz convenient for you right now...it will be very kind of youAsSalamoAlaikum Wr Wb…
I had been a bit busy with some other things, and managed to do only 15 questions of June 2010 variant 2 [all variants are same, order of the question differs only]. Now looking at the time, I feel bad. Is there anyone who needs it? If anyone wants, I’d continue doing the rest of the paper. If everyone’s gone already, there won’t be any use of me doing it.
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June 2010 Second Variant [All variants are the same, only the order of the questions differ.]
1. B
From diagram 1…it is 2.5 on the main scale….and 0.09 on the other one….so the overall reading is 2.59
Diagram 2 shows the zero error which is +0.14.
Hence our actual reading will be 2.59 – 0.14 = 2.45mm
2. D
E = W/Q express W/Q in base units..and simplify..
3. B
Density = mass / volume
Mass = m2 – m1 = 70 – 20 = 50
So error in mass is 1 + 1 = 2g
Add the fractional uncertainties…to find the ‘fractional’ uncertainty in density
2/50 + 0.6/10 = 0.1
So uncertainty in density is 0.1 x 5.0 = 0.5
4. A
Area in the graph is the product quantity on y axis and that on the x axis…[depends on the shape of the graph…may be the product of that multiplied by half, etc.]
In this case, it’ll be power x time, which is energy. Hence answer is A.
5. B
One complete wave is roughly 0.7 cm….that is equal to 7 ms [question states it’s 10ms per cm]. This is equal to 0.007 s .
f=1/t = 1/ 0.007 = 140 Hz [to 2s.f.]
6. C
If you try to imagine, it’s like when we increase the angle, Y will increase and X will decrease.
Mathematically speaking, X is Vcos(angle) and Y is Vsin(angle). As the theta increases, the value if sin theta increases. The opposite happens in the case of cos. Hence….Vcos theta would decrease as the angle will increase…and similarly Vsin theta will increase…so the option is C.
7. A
Pressure x volume is the formula for work done. Hence ans. is A.
8. B
Gradient of the graph gives velocity. v = u + at ; u = 0 and a is constant and negative (g), so v increases with t, and is negative…
The only graph whose gradient shows this is B. The tangents drawn, will be more steeper….and the gradient is negative in B. A has constant gradient, so that’s wrong. C has negative gradient BUT it decreases with time. D has increasing gradient, but it’s positive.
9. C
Draw a tangent at t=3 and find the gradient.
10. D
The collision is elastic. This is the usual case that happens.
11. A
Apply the momentum conservation principle..
momentum before collision = momentum after collision
60m + (-40)m = (m + m) v
^simplify…v = 20/2 = 10
12. A
3 forces, if form a closed triangle, resultant will be zero.
13. C
Upthrust, is always cuz of pressure difference.
14. C
Find the resultant force, 10 -4 = 6
6=ma
m = 2
a = 6/2 = 3
15. D
D is the right option..Efficiency = useful power/total power…
Power = force x velocity
For constant velocity, we shud have the force mgsina, useful power will be this force..
Total power is the total force x velocity…
1. B
From diagram 1…it is 2.5 on the main scale….and 0.09 on the other one….so the overall reading is 2.59
Diagram 2 shows the zero error which is +0.14.
Hence our actual reading will be 2.59 – 0.14 = 2.45mm
2. D
E = W/Q express W/Q in base units..and simplify..
3. B
Density = mass / volume
Mass = m2 – m1 = 70 – 20 = 50
So error in mass is 1 + 1 = 2g
Add the fractional uncertainties…to find the ‘fractional’ uncertainty in density
2/50 + 0.6/10 = 0.1
So uncertainty in density is 0.1 x 5.0 = 0.5
4. A
Area in the graph is the product quantity on y axis and that on the x axis…[depends on the shape of the graph…may be the product of that multiplied by half, etc.]
In this case, it’ll be power x time, which is energy. Hence answer is A.
5. B
One complete wave is roughly 0.7 cm….that is equal to 7 ms [question states it’s 10ms per cm]. This is equal to 0.007 s .
f=1/t = 1/ 0.007 = 140 Hz [to 2s.f.]
6. C
If you try to imagine, it’s like when we increase the angle, Y will increase and X will decrease.
Mathematically speaking, X is Vcos(angle) and Y is Vsin(angle). As the theta increases, the value if sin theta increases. The opposite happens in the case of cos. Hence….Vcos theta would decrease as the angle will increase…and similarly Vsin theta will increase…so the option is C.
7. A
Pressure x volume is the formula for work done. Hence ans. is A.
8. B
Gradient of the graph gives velocity. v = u + at ; u = 0 and a is constant and negative (g), so v increases with t, and is negative…
The only graph whose gradient shows this is B. The tangents drawn, will be more steeper….and the gradient is negative in B. A has constant gradient, so that’s wrong. C has negative gradient BUT it decreases with time. D has increasing gradient, but it’s positive.
9. C
Draw a tangent at t=3 and find the gradient.
10. D
The collision is elastic. This is the usual case that happens.
11. A
Apply the momentum conservation principle..
momentum before collision = momentum after collision
60m + (-40)m = (m + m) v
^simplify…v = 20/2 = 10
12. A
3 forces, if form a closed triangle, resultant will be zero.
13. C
Upthrust, is always cuz of pressure difference.
14. C
Find the resultant force, 10 -4 = 6
6=ma
m = 2
a = 6/2 = 3
15. D
D is the right option..Efficiency = useful power/total power…
Power = force x velocity
For constant velocity, we shud have the force mgsina, useful power will be this force..
Total power is the total force x velocity…
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If it takes like another 2 hours or so, will it be ok for you? I'll be glad if it helps you..please do post it if itz convenient for you right now...it will be very kind of you
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any one who can post may june 2011 ?
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Can you tell mw, why you have to half the wavelength of the open ended tube compare with the first close ended tube?No frequency found lower than the 1 of the resonance means, that this is the fundemental harmonic, for tube X, needs a minimum of a quarter wavelength for a resonance, so conider both tubes have a length of 12m for example, so tube X will be 12m, so the stationary wave inside, which is just a quarter wavelength is 12m, so the wavelength will be 12*4 =48m , conisder v (velocity) = 300, so v=f*lamda , 300=f*48, so f=6.25 ...... for tube Y u gotta do the same thing, the minimum wavelength for resonance to occur is half wavelength,, taking the length of the tube as 12, so half wavelength will be 12m, so the wavelength will be 12*2= 24m,, v=f*lamda, 300=f*24, f=12.5,, so the answer will be 2F