Physics: Post your doubts here!

[hmlahori]

November 2005, Paper 1, Q.5;
Q.5) let rogue be p
p = m/v
therefore p = m/(l * b * h)
delta p/p = delta m/m + delta l/l + delta b/b + delta h/h
therefore delta p/p = 0.1/25 + 0.01/2 + 0.01/2 + 0.01/1
therefore delta p/p = 0.004 + 0.002 + 0.005 + 0.01 Calculation part is a little bit tricky so I suggest you redo this question too.
The value of density is given. And the question is asking for the uncertainty in the result of the calculated density. And what you've calculated now is the fractional uncertainty.
therefore delta p/2.5 = 0.021
therefore delta p = 0.05 gcm^-3 Ans C

Thank you so much..... all thats been a great help .... specially that barrel question explanation..... thanks.

 

[hmlahori]

s05 q. 14) see the resultant and all other forces initiate from the same point. the resultant is coming outwards from the paper u gotta imagine it in 3D. in 2d most likely the right direction is D​

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w05 qp 5) uncertainty in density / density = uncertainity over mass + Uncer. / length + Uncer / breath + Uncert./ height. equate the values and you will get 0.05 g/ cm^3​

7) s= ut + 1/2 gt^2 . ut is zero as initial velocity is xero. g = 2s / t^2 right so now we need to subtract both times in order to get one expression giving acceleration : 2s / t2^2- t1^2 D option.​

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16 ) calculate the weight component along the slope which will be 1 into 10^3 into sin 30. multiply the force now with distance i.e 5 m. u will get B 2.5 into 10^3 Nm​

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oct/ nov o9 ) q.9 i dnt knw q.14 dnt knw, q. 28 ) the polarity has to be negative as the weight is acting downwards the electric force must be upwards in order to balance it so it stays at equilibrium position. E = F / q , q = F/E i.e B​

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s 10) q.22 its c , Infra-red radiation. u need to remember the wavelengths and frequencies of visible spectrum..plus the size of slit should always be very close to the wavelength in order to see diffraction.​

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26) the force acts on every charge present in the available field region, the direction of force is parallel to the field. fact.​

27) zero A because work done = force into distance travelled in the direction of force. the direction of force is in the middle due to centripetal force but the charge is moving elsewhere. so work done zer0​

w10) q 18) Power = Fv , upwards resultant force > F= mg , so (m1-m2) gv​

q 24) ​

you need to identify water waves as transverse waves. For transverse waves, point particles move up and down and NOT left and right. ​

So each of the points A, B , C and D will move up and down ONLY, so do not get tricked that they will move to the right.​

First of All, B and D are wrong because they are the maximum displacement , so obviously, the acceleration is max and speed is zero.​

To get the right answer you have to image the wave moving. If you just shift this wave to the right abit, you will notice that A will be at the bottom of the shifted wave and C will be at the top. Thus, C is the point with the maximum speed going upwards.​

q. 37) u can see that the voltmeter is place across the power supply, the voltmeter will give us the e.m.f. not voltage against any of the two resistors.-----------------------------------------------------------------------------------------------------------------------------------------------------------------------​

s 11 paper 11) q. 15 ) loss in g.pe = gain in kinetic energy plus work done against friction​

2 . 9.8 . 3 = 1/2 . 2 . v^2 + 5 into 7​

58.86 - 35 = v^2 v = 4.9 m/s​

34 I don't know sorry.​

Thanks you've been a great help just one more question
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_w05_qp_1.pdf
in question 7 of this paper why is the answer not C cuz that also makes sense.

 

[Scafalon40]

HAHAHA Its very important. one decimal can change ur life.
man i also don't get what i said. i m happy to loose marks in this question. i dnt knw what it is. sorry

Lets try and discuss it.
Al right so we know that change in momentum is =m(change in velocity)
So that means that the change in velocity must be (4+0.8) for the answer to be correct
But change in velocity being (-0.8-4) makes more sense, since we have been taught that change in velocity=(v-u)
So the overall problem is with the sign
The illogical thing would be that the change is velocity is actually -(-0.8-4), hence the change the change in momentum is +58.76
LOL that made no sense
GOD WHAT AM I TO DO?

 

[confused123]

Thanks you've been a great help just one more question
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_w05_qp_1.pdf
in question 7 of this paper why is the answer not C cuz that also makes sense.

Your welcome. nice mcq's helping me as well, yeah dats a reasonable argument but i think (a+b)^2 = a square plus 2ab + b^2 , this will not be the case in D option.

 

[hmlahori]

Your welcome. nice mcq's helping me as well, yeah dats a reasonable argument but i think (a+b)^2 = a square plus 2ab + b^2 , this will not be the case in D option.

Right yes that does make sense
well thanks again.

 

[confused123]

Lets try and discuss it.
Al right so we know that change in momentum is =m(change in velocity)
So that means that the change in velocity must be (4+0.8) for the answer to be correct
But change in velocity being (-0.8-4) makes more sense, since we have been taught that change in velocity=(v-u)
So the overall problem is with the sign
The illogical thing would be that the change is velocity is actually -(-0.8-4), hence the change the change in momentum is +58.76
LOL that made no sense
GOD WHAT AM I TO DO?

lolz lets not try and move on & utilize our brilliant brains on some other mcq's. plus i think dakrnessinme is right dat ms and examiner reports can be wrong sometimes. our formula is correct.

 

[hmlahori]

Fellows need explanations to these questions...they're really confusing
Thanks..

http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s11_qp_11.pdf Q34

http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_w09_qp_11.pdf Q9,14

 

[ABDSyed]

http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_w10_qp_51.pdf
PLz someone know how to do this Uncertaintiy Q2b
Does Any Know ?
?
?
?
?
????????????????????????????????????????????????PLz

 

[Scafalon40]

lolz lets not try and move on & utilize our brilliant brains on some other mcq's. plus i think dakrnessinme is right dat ms and examiner reports can be wrong sometimes. our formula is correct.

So basically you are questioning the the examining body whose papers you have given in your O'levels, and whose papers you are about to give in your A'levels; the very examining body which gives you your grades after years of hard work; the examining body which has world wide recognition, and the body which is affiliated with one of the most prestigious, distinguished, respected and notable Universities of the World.​

It makes mistakes, eh?
.
.
.
.
.
.
It just may!

 

[confused123]

Fellows need explanations to these questions...they're really confusing
Thanks..

http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s11_qp_11.pdf Q34

http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_w09_qp_11.pdf Q9,14

14 I don't know

 

[confused123]

So basically you are questioning the the examining body whose papers you have given in your O'levels, and whose papers you are about to give in your A'levels; the very examining body which gives you your grades after years of hard work; the examining body which has world wide recognition, and the body which is affiliated with one of the most prestigious, distinguished, respected and notable Universities of the World.​

It makes mistakes, eh?
.
.
.
.
.
.
It just may!

hahaha Bombastic! Loved it i hope u don't get a warning by mods caz of having fun here

yes man the matter of fact is dat this dignified, honourable, disgusting, money making machines, world wide status uni, disrespected, renowned and so on examining body can make MISTAKES!!

 

[hmlahori]

14 I don't knowView attachment 6417View attachment 6418

This makes things so much clearer.
But u no for the resistance of the cube question i was taking the area to be 6(v^1/3)^2 because thats how you calculate the surface area of a cube right.
or else my method was fine. well thanks anyways....

 

[confused123]

This makes things so much clearer.
But u no for the resistance of the cube question i was taking the area to be 6(v^1/3)^2 because thats how you calculate the surface area of a cube right.
or else my method was fine. well thanks anyways....

yeah it was hard to figure out at the first glance. near to impossible to do it right if u get this for the 1st time in exams...ur welcome. stay blessed

 

[confused123]

This makes things so much clearer.
But u no for the resistance of the cube question i was taking the area to be 6(v^1/3)^2 because thats how you calculate the surface area of a cube right.
or else my method was fine. well thanks anyways....

14 )

 

[rideronthestorm]

why do we make tangents from the origin again ?? ... can we make it from any other points when finding gradient like in practical papers :/ ?

 

[confused123]

why do we make tangents from the origin again ?? ... can we make it from any other points when finding gradient like in practical papers :/ ?

yeah i also wonder why? :/ u try i give up.

 

[rideronthestorm]

plz explain all of these tooo

 

[rideronthestorm]

 

[rideronthestorm]

 

[rideronthestorm]