Post by: @d!_†oX!© on November 06, 2009, 12:54:41 pm
Work out how much energy is stored in the rope. Take g= 10m/s2 and ignore air resistance.
now i know this is a pretty easy questioon bt i cant get the answer how to do it....the answer is 5250 J....
i did it by subtracting the k.e from the p.e
but why is it so???
guys help plss
Post by: ~~~~shreyapril~~~~ on November 06, 2009, 01:00:56 pm
if not it will have strain energy!!
so if PE then it will be mgh = 60*10*10 = 6000!!
but thatss not the ans so!!
starin energy will be MA = 60*10 = 600!!
but this is also not the ans!!
Post by: @d!_†oX!© on November 06, 2009, 01:01:54 pm
read the ques carefully shrey!!!
Post by: ~~~~shreyapril~~~~ on November 06, 2009, 01:03:35 pm
dude u took the pe wrong cuz it will be 10*20*60]
read the ques carefully shrey!!!
due i have my eyes in place!!!
it says - 10 meters !!
how can you take 20????
Post by: @d!_†oX!© on November 06, 2009, 01:05:08 pm
can u read this????so 10 + 10 = 20 m
now dont come up with some other excuse to cover up plssss....
Post by: ~~~~shreyapril~~~~ on November 06, 2009, 01:07:09 pm
we need to subtract PE by KE beacuse it is starin energy right??
and at the time i guess it is PE and just becoming KE so it needs to be subratctd!!!
i guess
lets wait and watch until astar arives!!
Post by: ~~~~shreyapril~~~~ on November 06, 2009, 01:18:58 pm
60*20*10=12000
0.5*102*60=3000
so 12000-3000 = 9000!!
as you said i have subtracted PE and KE!!
but not the same ans!!
did i do ne mistake!!
i guess i did!!!
Post by: @d!_†oX!© on November 06, 2009, 01:19:52 pm
Post by: ~~~~shreyapril~~~~ on November 06, 2009, 01:20:26 pm
og the mistake!!
it must be 0.5*152*60=6750
so it will be 12000-6750=5250..!!
By the way i also have same wuestion why subtarct???
Post by: Ghost Of Highbury on November 06, 2009, 01:53:43 pm
i guess this is the answer.
Post by: @d!_†oX!© on November 06, 2009, 02:06:45 pm
wt i think is that
elastic energy of the rope+kinetic energy=potential energy
thus, elastic energy=potential -kinetic
i just wanted to confirm it....it'd be gr8 if astar cud say smthng on this....
@shrey dude that's what my question is!!!!!do u like start doing a question sans seeing wat's asked???
Post by: ~~~~shreyapril~~~~ on November 06, 2009, 02:09:32 pm
i did not see and copy!!
By the way your logic has sense aadi!!
Post by: astarmathsandphysics on November 06, 2009, 03:35:22 pm
Post by: Ghost Of Highbury on November 06, 2009, 03:39:18 pm
even i had the same thought in mind...
wt i think is that
elastic energy of the rope+kinetic energy=potential energy
thus, elastic energy=potential -kinetic
i just wanted to confirm it....it'd be gr8 if astar cud say smthng on this....
@shrey dude that's what my question is!!!!!do u like start doing a question sans seeing wat's asked???
y u getting so hyper?
Post by: @d!_†oX!© on November 07, 2009, 06:01:07 am
nd aadi when did i get hyper dude???
Post by: ~~~~shreyapril~~~~ on November 07, 2009, 06:03:13 am
we got the ans but why shud we subtract!!
did astar answered???
Post by: @d!_†oX!© on November 07, 2009, 06:04:34 am
Post by: ~~~~shreyapril~~~~ on November 07, 2009, 06:05:32 am
then i got it!
if it is something else then explain!!
Post by: nid404 on November 07, 2009, 06:07:05 am
k.e lost=p.e gained and vice versa
Post by: @d!_†oX!© on November 07, 2009, 06:07:30 am
our ideas were correct!!
mine was in a scientific form nd his was an explanatory form for the same...
Post by: ~~~~shreyapril~~~~ on November 07, 2009, 06:09:09 am
thanx dudes!!!
Post by: nid404 on November 07, 2009, 06:09:44 am
got it!!
thanx dudes!!!
welcome
Post by: ~~~~shreyapril~~~~ on November 07, 2009, 06:11:00 am
Post by: Ghost Of Highbury on November 07, 2009, 06:20:29 am
yup!!!!
our ideas were correct!!
mine was in a scientific form nd his was an explanatory form for the same...
o_O_o_O
Post by: ~~~~shreyapril~~~~ on November 07, 2009, 06:21:17 am
chil maar!!
Post by: @d!_†oX!© on November 07, 2009, 11:39:52 am
whats the difference between a fixed resistor and a preset resistor??
Post by: nid404 on November 07, 2009, 11:42:06 am
hey guys!!
whats the difference between a fixed resistor and a preset resistor??
fixed resistors r fixed...lol
preset resistor is a type of variable resistor....u needn't know abt it for the igs
Post by: @d!_†oX!© on November 07, 2009, 11:43:15 am
y is it that i always end up asking question that are not needed in my syllabus??!!!!!
anyways cud u still expand a bit more nid??!!!
Post by: nid404 on November 07, 2009, 11:45:34 am
Preset resistors are used in circuits when it is necessary to alter the resistance. Dark/light and temperature sensors usually have these components as the preset resistor allows the circuit to be made more or less sensitive
Post by: @d!_†oX!© on November 07, 2009, 11:48:10 am
thnxx..got it....perhaps it is used with transistors and all....
soo is it a part of the A level curriculum??
Post by: ~~~~shreyapril~~~~ on November 07, 2009, 11:54:12 am
then no need to know!!
but extra info is not at all dangerous!!! :P :P
Post by: nid404 on November 07, 2009, 11:56:03 am
There r two kinds of variable resistors i know of- Control and preset..(extra info) :P
Post by: ~~~~shreyapril~~~~ on November 07, 2009, 11:57:47 am
Post by: @d!_†oX!© on November 07, 2009, 12:02:38 pm
if yes, then what exactly???its such a vast topic.....
Post by: nid404 on November 07, 2009, 12:03:51 pm
do we need to know fusion and fission???
if yes, then what exactly???its such a vast topic.....
You don't really need to know anything in detai. Just the chain reaction in the nuclear reactor
Post by: ~~~~shreyapril~~~~ on November 07, 2009, 12:04:56 pm
true fussion is splitting and fission takes place in sun nd is abt joiing!!!
Post by: @d!_†oX!© on November 07, 2009, 12:10:07 pm
Post by: ~~~~shreyapril~~~~ on November 07, 2009, 12:22:25 pm
ne time!!
Post by: nid404 on November 07, 2009, 12:22:57 pm
ohhkkk....Thanks guys...
abbe mensun not
Post by: ~~~~shreyapril~~~~ on November 07, 2009, 02:15:10 pm
same here!!
Post by: @d!_†oX!© on November 07, 2009, 02:16:17 pm
A mercury thread of a thermometer is 12mm long when placed in melting ice, and 82mm long when placed in boiling water.
Find the:
a)length of the thread at 50oC
b)temperature when the length is 61mm long
plss guys, i need a full-fledged explanation for this cuz whenever i see this type of question after say 2-3 days, i tend to again do it wrong....plsss tell me a way to remember it forever....
Thanks guys!! :) :)
Post by: ~~~~shreyapril~~~~ on November 07, 2009, 02:17:49 pm
solve it!!
and then similarly do
12:0::61:x
solve for x and get the ans!!
Post by: @d!_†oX!© on November 07, 2009, 02:20:25 pm
12:0
which is 12/0
which is infinte.....apply ur brain a little b4 replying man!!! :P
Post by: ~~~~shreyapril~~~~ on November 07, 2009, 02:21:29 pm
82 naa
take 82:100
Post by: @d!_†oX!© on November 07, 2009, 02:23:11 pm
Post by: cooldude on November 07, 2009, 02:30:08 pm
okay guys...another one...
A mercury thread of a thermometer is 12mm long when placed in melting ice, and 82mm long when placed in boiling water.
Find the:
a)length of the thread at 50oC
b)temperature when the length is 61mm long
plss guys, i need a full-fledged explanation for this cuz whenever i see this type of question after say 2-3 days, i tend to again do it wrong....plsss tell me a way to remember it forever....
Thanks guys!! :) :)
look adi we knw that b/w 12mm and 82mm lies 100.c. cuz melting point of ice=0 and boiling point of water=100. therefore in this scale 1mm=(82-12)/100
=>7/10=0.7
=>1.c=0.7mm
therfore at 50.c length=12+(0.7*50)=47mm
and then the next part=>
1.c=0.7mm
therefore the length for which there is a temperature change=>61-12=49mm
therfore the temperature at this=>49/0.7=70.c
just do it by simple unitary method
got it?
Post by: @d!_†oX!© on November 07, 2009, 02:39:10 pm
0.7C - 1 mm
so 1C = 1/0.7 mm
so wt i didnt understand is y u did (50*0.7)
cuz 50 is the temp and 0.7 is also the temp at 1mm....so shudn't it be 50*0.7
bt if i do that den the answer is wrong....plsss dude tell me some way to get this thing rite into my head!!!
Post by: cooldude on November 07, 2009, 02:44:34 pm
see, acc. to u
0.7C - 1 mm
so 1C = 1/0.7 mm
so wt i didnt understand is y u did (50*0.7)
cuz 50 is the temp and 0.7 is also the temp at 1mm....so shudn't it be 50*0.7
bt if i do that den the answer is wrong....plsss dude tell me some way to get this thing rite into my head!!!
dude 0.7 is not the temperature at 1 mm it is the value of the temperature of that changes for each increase in 1mm. theyve just given u 12mm-82mm to confuse u. this is the same as 1mm-70mm
cuz theyve started from 12 u hve to add 12 to (50*0.7), theres no hard nd fast way to answer it, uve gotta understand wat there askin and then answer accordingly
Post by: holychalice on November 07, 2009, 02:46:22 pm
check again
Post by: @d!_†oX!© on November 07, 2009, 02:46:28 pm
nd cooldude...i'm kinda beginning to understand it now...
Post by: nid404 on November 07, 2009, 02:48:29 pm
I will try my best to explain this
Given
at 0oC=12mm
& and 100oC=82mm
difference is 70. So if the reading was.....if for 0 and 70 mm, then length would have been 70/2=35mm for 50 degrees....exactly half
so when the startin point is 12oC...u will add 12 to the difference/2=47mm. I hope ur able to figure out the logic.
The length for 50 degrees would be right in the middle of 12 and 82
Post by: @d!_†oX!© on November 07, 2009, 02:50:42 pm
i really thing cooldude's explanation is gud...bt i really dont wanna forget it again...so pls guys try nd give me ur working ways to..so dt i have some more ways to do it nd i do not forget...
Post by: Ghost Of Highbury on November 07, 2009, 02:52:06 pm
theyre asking 61mm but thats frm their scale, from our scale it wud be...61-12 = 49
so
70mm - 100
49mm - 70
Post by: cooldude on November 07, 2009, 02:52:34 pm
Post by: @d!_†oX!© on November 07, 2009, 02:53:35 pm
@aadi dts wat i wanna know dude...how did ya get 70 for 49 mm??????
Post by: cooldude on November 07, 2009, 02:55:30 pm
yaaa dts rite cooldude...
@aadi dts wat i wanna know dude...how did ya get 70 for 49 mm??????
dude!!!!!!!!!!! i also explained the second part...look!!!!!!!!!!
ill paste it for u--
1.c=0.7mm
1mm=1/0.7.c
therefore the length for which there is a temperature change=>61-12=49mm
therfore the temperature at this=>49*(1/0.7)=70.c
just do it by simple unitary method
and also since the temperature starts from 0, 70 is ur final ans but cuz the length starts from 12 so u hve to consider 12 in every calculation u make wich evry1 is trying to tell u
Post by: Ghost Of Highbury on November 07, 2009, 02:56:36 pm
ok look..
in the range of 12 to 82
they are asking 61
61 is 49 above from 12
so for our scale which 0 to 70, we have to go 49 above from 0 = 0+49 = 49
got it?
Post by: @d!_†oX!© on November 07, 2009, 02:59:47 pm
@aadi u till didnt get my point man..i know till here...now how did u get 70 fro 49?? is my question.....
Post by: nid404 on November 07, 2009, 03:00:34 pm
so
assuming
100o=70mm(82-12)
then xo= 61mm-12mm
x= 4900/70=70o
Post by: cooldude on November 07, 2009, 03:02:00 pm
@ cooldude chill man...i've seen urs..nd i got it....i said m just looking for alternate ways...!!!stay calm man...take a deep breath in.........and out......cool??!!
@aadi u till didnt get my point man..i know till here...now how did u get 70 fro 49?? is my question.....
k dude, By the way just look at my last post ive told u a few things, let other ppl give their views nw
Post by: Ghost Of Highbury on November 07, 2009, 03:03:11 pm
@ cooldude chill man...i've seen urs..nd i got it....i said m just looking for alternate ways...!!!stay calm man...take a deep breath in.........and out......cool??!!
@aadi u till didnt get my point man..i know till here...now how did u get 70 fro 49?? is my question.....
dude, thats the easier part.
its simple ratio proportion
70mm - 100
49mm - 100/70 * 49 = 70
got it?
Post by: @d!_†oX!© on November 07, 2009, 03:06:25 pm
Thanks cooldude, nid, aadi, nd shrey(for trying just without even seeing the question)
Post by: Ghost Of Highbury on November 07, 2009, 03:07:29 pm
ook guys!!!ya i got it now...will be back when i forget it again!!
Thanks cooldude, nid, aadi, nd shrey(for trying just without even seeing the question)
ur welcome
and chill man!..wat if shrey cudnt get it? its k..happens sometimes yaar!. credit nahi dena hai to mat de yaar! :)
Post by: @d!_†oX!© on November 07, 2009, 03:08:19 pm
no personal grudges dude!! ;D :D
Post by: nid404 on November 07, 2009, 03:08:50 pm
Post by: Ghost Of Highbury on November 07, 2009, 03:09:14 pm
wasnt this a MCQ question?
Post by: cooldude on November 07, 2009, 03:11:17 pm
lol..its k.. :P
wasnt this a MCQ question?
nope, its from the igcse study guide
Post by: @d!_†oX!© on November 07, 2009, 03:11:36 pm
Post by: @d!_†oX!© on November 07, 2009, 03:27:04 pm
okk..so here i go again..
The definition of sound energy is "longitudinal pressure waves that travel through a compressible material"
now my question is that sounds travel through solids...but solids are incompressible...
another addition....sounds waves are a series of compressions and rarefactions....in air, they compress the air molecules...but how do they do that i solid???
Post by: nid404 on November 07, 2009, 04:13:20 pm
longitudinal waves is not in much detail in the ig syllabus....yeh link mein jaa ke chck the motion of particles....ull understand better
Post by: Ghost Of Highbury on November 07, 2009, 04:16:47 pm
Post by: nid404 on November 07, 2009, 04:18:17 pm
Post by: @d!_†oX!© on November 07, 2009, 04:24:13 pm
Post by: moon on November 07, 2009, 05:05:29 pm
Post by: cooldude on November 07, 2009, 05:27:20 pm
plz, can u help in jun.08 Q4(c), and Jun.07 Q10 (b) need explanation for these questions. thanks.
okay so first june 07 q10 b--
see if we want to change the temperature at which the lamp switches on, we have to change the value of resistance R1. if we increase the value of the resistance then more voltage will flow through the transistor and if we decrease the voltage then less voltage will flow through the transistor. therefore if we assume that the amount of heat falling on the thermistor is constant then when we increase or decrease the value of R1 so will the value of the voltage passing through the tranistor will change. so if we want to make the lamp switch on at a different temperature we change the value of R1 because if the heat is falling on the thermistor and thus its resistance decreasing then if the base voltage is enough to switch on the transistor it will light however decreasing the value of the resistor R1 will cause the voltage flowing through the transistor to be too less to switch it on (this is if we want to make the temperature at which the lamp lights more; just explain the opposite thing if u want the temperature at which the transistor lights lower than it is and also u dnt need to expalin all this, ive done this to help u understand it and if u want a shortened version ill be happy to help). therefore in short the temperature at which the lamp lights can be changed by changing the value of R1. (the lamp lights when the base voltage becomes enough for the lamp to switch on)
hope it helps
p.s. i think ive explained a bit too much though
okay now the second q--
see if the temperature increases so will the kinetic energy of the particles. therefore they will exert more pressure in the tube and therefore if the pressure increases the switch will switch on. this is because when we pressed the rubber cover the same thing was achieved but when there is a higher temperature it comes to the same thing
Note--when the transistor switches on it will in turn switch on the lamp
Post by: nid404 on November 07, 2009, 05:38:47 pm
okk....but this sounds like a compromise sorts....i mean it shud be in the syllabus dude...
dude u really wanna know stuff that's not in the ig syllabus.....maybe ur inquisitive and wanna know more but this is not the right time......ur too late
Post by: zara on November 07, 2009, 05:39:29 pm
Post by: @d!_†oX!© on November 07, 2009, 05:44:40 pm
dude u really wanna know stuff that's not in the ig syllabus.....maybe ur inquisitive and wanna know more but this is not the right time......ur too latei am inquisitive dude.....but u know what i am actually getting scared now...what about u???
Post by: nid404 on November 07, 2009, 05:45:26 pm
i am inquisitive dude.....but u know what i am actually getting scared now...what about u???
I'm done wid my igz :D.....my AS in June :-X
Post by: Ghost Of Highbury on November 07, 2009, 05:45:33 pm
Post by: zara on November 07, 2009, 05:46:37 pm
i am inquisitive dude.....but u know what i am actually getting scared now...what about u???one piece of advice...
dont get tooo inquistive or ull mess up!
just study whats there in the syllabus.
Post by: cooldude on November 07, 2009, 05:47:19 pm
hey cooldude, that was an amazing explanation. thanks a lot! because that cleared my concepts of transistor too! +rep to u!
Thanks!!!!!!!!! i was just wondering whether my explanantion was a bit too detailed and whether ppl wud understand it properly..Thanks for the rep ;) happy to help
Post by: cooldude on November 07, 2009, 05:47:40 pm
wow cooldude!
Thanks
Post by: nid404 on November 07, 2009, 05:48:12 pm
Post by: Ghost Of Highbury on November 07, 2009, 05:48:52 pm
Post by: cooldude on November 07, 2009, 05:51:51 pm
oh yes and cooldude...that explanation ws really good :)
Thanks
Post by: @d!_†oX!© on November 07, 2009, 05:53:30 pm
Post by: cooldude on November 07, 2009, 06:01:29 pm
awesome explanation dude....
Thanks
Post by: moon on November 07, 2009, 06:08:20 pm
Post by: cooldude on November 07, 2009, 06:22:32 pm
Thanx 4 ur awesome explanation, but i still need J08 Q4 c 2nd varient and 4 that question y did u choose R1 and not R2??? ???
okay for y i chose r1 and not r2----
this is cuz r1 is just in place to prevent too much current flowing throught the transistor and damaging it. r1 and the thermistor r acting as a potential divider; when we change the resitance of one of the 2 the voltage across the thermistor or r1 changes depending upon which components resistance we have changed. as u knw the voltage flowing across a component is governed by the resistance of the component. since we have to change the value at which the temperature at which the bulb lamps we have to change the resistance of r1. r1 is a part of the potential divider, r2 is not
ill just answer ur other q with my next post
Post by: @d!_†oX!© on November 07, 2009, 06:30:34 pm
see, just remember that to alter the temperature at which the light switches on, u gotta change the resistances as cooldude said...
so there are two resistors in the circuit, nd u can't change the resistance of the thermistor cuz it depends on the temperature only...so to change the voltage at which the transistor switches on, we have to change the resistance of R1....
hope u get it!!!
Post by: cooldude on November 07, 2009, 06:37:18 pm
we knw that (p1*v1)=(p2*v2);
therefore 12000*75=300000*v2
therefore v2=30mm
the distance of the piston from the closed end was=75mm
the distance of the piston from the closed end now is=30mm
therefore the distance it has been moved=75-30-45mm
Post by: @d!_†oX!© on November 07, 2009, 06:37:50 pm
see
P1*V1=P2*V2
P1=1.2*10^5 Pa
V1=Area*75
P2=3*10^5 Pa
V2=length*Area
because u know the area is same
thus put the values in the formula and find out the length
length=30mm
now, because the piston was 75mm from the closed end and is now 30mm from the closed end, thus the distance moved=75mm - 30mm =40mm
Post by: @d!_†oX!© on November 07, 2009, 06:38:30 pm
Post by: cooldude on November 07, 2009, 06:43:15 pm
sorry cooldude...didn't see that u already posted.... ::)
k np
Post by: moon on November 07, 2009, 07:02:25 pm
Post by: cooldude on November 07, 2009, 07:29:58 pm
Thanx a lot for both of u as u helped me to understand these questions... :) :)
np, ur welcum feel free to ask some more qs
Post by: @d!_†oX!© on November 08, 2009, 04:09:15 am
i didnt knw which varaint u were askin for, neway ur ans---By the way cooldude...ur method is wrong cuz 75mm is not the volume but the length....u gotta show it in the volume format as i did in my post....
we knw that (p1*v1)=(p2*v2);
therefore 12000*75=300000*v2
therefore v2=30mm
the distance of the piston from the closed end was=75mm
the distance of the piston from the closed end now is=30mm
therefore the distance it has been moved=75-30-45mm
just look back cuz if u show this working in ur p3, u will lose some of ur marks there....
Post by: Ghost Of Highbury on November 08, 2009, 05:40:59 am
Post by: @d!_†oX!© on November 08, 2009, 05:52:14 am
By the way aadi, do u know the formula for the focal length...???
i know for convex lens, just wanna know if it is the same for concave as well...
Post by: Ghost Of Highbury on November 08, 2009, 05:59:12 am
diverging lenses are assigned negative focal lengths.
Post by: @d!_†oX!© on November 08, 2009, 06:02:31 am
Post by: Ghost Of Highbury on November 08, 2009, 06:11:04 am
dint i write negative?
Post by: @d!_†oX!© on November 08, 2009, 06:12:41 am
Post by: HalpPlease on November 08, 2009, 06:18:08 am
Post by: @d!_†oX!© on November 08, 2009, 06:21:13 am
f- focal length
So-distance of object from the center
Si-distance of image from the center
Post by: cooldude on November 08, 2009, 06:48:05 am
its very simple halp...see i the exam u wud not be asked to use this formula,....u'd be asked to make a sketch diagram..bt this is for ur own reference nd checking....simply put the values in the formula nd take out the required thing
f- focal length
So-distance of object from the center
Si-distance of image from the center
u shud specify from the centre of the lens, oh yeh and about that q, i cancelled the area like aadi says. but if u want to knw the area of cross-section, its the same, if u really want to explain it u shud also say that the area of cross-section of the piston stays constant cuz the examiner can assume that the area of cross section of the cylinder remains constant (just to make it look as u knw what ur talkin about ;) )
Post by: @d!_†oX!© on November 08, 2009, 06:52:36 am
Post by: Ghost Of Highbury on November 08, 2009, 07:05:31 am
help
Post by: @d!_†oX!© on November 08, 2009, 07:08:54 am
Post by: Ghost Of Highbury on November 08, 2009, 07:09:51 am
Post by: HalpPlease on November 08, 2009, 07:12:04 am
Post by: @d!_†oX!© on November 08, 2009, 07:13:45 am
@aadi....dude y u getting hot?i was just telling ya...dts just a habit to use many punctuations...lol..
see i used it for halp as well unintentionally...
Post by: @d!_†oX!© on November 08, 2009, 09:52:27 am
Resistance= (constant*length)/area of cross-section
???
just saw it in the study guide....i know its simple to remember and use both...m just asking for curiosity....
Post by: nid404 on November 08, 2009, 09:53:19 am
Post by: @d!_†oX!© on November 08, 2009, 09:54:33 am
bt never seen any question on this in any past paper b4.... :-[
Post by: nid404 on November 08, 2009, 09:55:50 am
ohhh...ohkkay....
bt never seen any question on this in any past paper b4.... :-[
yes there r questions in past papers....mostly MCQs but yes some in theory too
Post by: @d!_†oX!© on November 08, 2009, 09:59:40 am
Post by: nid404 on November 08, 2009, 10:04:12 am
challo theek hai..thanks a lot~~~
ur welks
Post by: Nk on November 08, 2009, 11:06:24 am
Post by: @d!_†oX!© on November 08, 2009, 11:07:57 am
Post by: astarmathsandphysics on November 08, 2009, 11:08:10 am
http://www.astarmathsandphysics.com/igcse_physics_notes_menu.html
Post by: xXFaKeXx on November 08, 2009, 11:12:07 am
Post by: @d!_†oX!© on November 08, 2009, 11:13:53 am
will be mentioned in the question if required in the calculation
Post by: astarmathsandphysics on November 08, 2009, 11:14:18 am
http://freeexampapers.com/Dndex.php?d=SUdDU0UvUGh5c2ljcy9DSUUvUmVzb3VyY2Vz&catagory=
Post by: xXFaKeXx on November 08, 2009, 11:15:35 am
Post by: @d!_†oX!© on November 08, 2009, 11:22:39 am
as in the book says its used when a time delay is required....bt wt exactly is a time delay...as in my question is why is the time delay required...any examples of circuits where a time delay is required??
Post by: astarmathsandphysics on November 08, 2009, 11:35:00 am
Post by: cooldude on November 08, 2009, 11:35:45 am
hey guys temme....wats the use of a capacitor in a circuit...
as in the book says its used when a time delay is required....bt wt exactly is a time delay...as in my question is why is the time delay required...any examples of circuits where a time delay is required??
it is in an industrial oven. if we opened the door of the oven then a alarm would sound but if the door was opened and closed within the time it took for the capacitor to charge or discharge no alarm would sound. an alarm would be needed if there was a malfunction and if the oven door opened accidentally
Post by: @d!_†oX!© on November 08, 2009, 11:36:25 am
nd u too cooldude... :)
Post by: cooldude on November 08, 2009, 11:37:27 am
ohhhhh.....yeahh....okkk...Thanks sir!! ;D
nd u too cooldude... :)
By the way its all in the guide..if uve read it thoroughly :P
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 12:07:55 pm
By the way adi i had read the question!!
and tahnk u for mentioninng me in thank list!! :P
Post by: cooldude on November 08, 2009, 12:14:46 pm
are you all refering to physics guide which is red in colour ???
By the way adi i had read the question!!
and tahnk u for mentioninng me in thank list!! :P
yep the mike folland one
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 12:16:20 pm
yep the mike folland onethats a coool book!!
Post by: moon on November 08, 2009, 12:41:40 pm
Post by: nid404 on November 08, 2009, 12:46:47 pm
plz can u help in Jun.01 Q.8 a(ii) and how to measure theamplitude in compression and rarefaction?
don't hav the paper...could u upload it
Post by: @d!_†oX!© on November 08, 2009, 12:53:01 pm
Post by: moon on November 08, 2009, 12:56:12 pm
Post by: Ghost Of Highbury on November 08, 2009, 12:58:10 pm
wat is da minimum voltage required to switch on the transistor???????
is it 0.6V
if no, what is 0.6V for?
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 01:00:57 pm
it is 0.6W
Post by: cooldude on November 08, 2009, 01:01:21 pm
is it 0.6V
if no, what is 0.6V for?
aadi its given in the guide for that specific model, it varies from model to model, for the one which u say ( i think ur talking about the guides transistor) then for that it is. they shud give u the voltage in the exam
Post by: Ghost Of Highbury on November 08, 2009, 01:08:33 pm
R = V/I => 4/0.4 = 10 ohms
@cooldude - ohh..okk..thanks
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 01:10:39 pm
charge flowing is 0.6*20=1.2C
and i am just solving the current ka problem!!
Post by: @d!_†oX!© on November 08, 2009, 01:12:00 pm
Post by: Ghost Of Highbury on November 08, 2009, 01:12:22 pm
adithe ans is easy!
charge flowing is 0.6*20=1.2C
and i am just solving the current ka problem!!
abbe oye, i have solved the question, and that 0.6V ka is minimum volts for transistors..not this q ka..check the quoted post dude!!
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 01:13:07 pm
got it!!
By the way wats the current then!!
0.4 rite?
Post by: Ghost Of Highbury on November 08, 2009, 01:14:05 pm
Post by: holychalice on November 08, 2009, 01:16:04 pm
nd yea .. aadi has don't d rest
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 01:16:19 pm
i did not read the resistance ka part!!
yu solved first two i did the last one!!
its k!!
easy question i guess!!
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 01:17:26 pm
d charge ans is 8C ,,, 0.4*20 = 8charge = power * tmie!!!
nd yea .. aadi has don't d rest
that is 0.6*20!!!
this time i read the question!!!
Post by: holychalice on November 08, 2009, 01:20:10 pm
charge = 0.4 * 20 = 8C
???
nd power = Itv = Charge * Voltage
Post by: Ghost Of Highbury on November 08, 2009, 01:20:36 pm
charge = power * tmie!!!
that is 0.6*20!!!
this time i read the question!!!
abbe oye, Q = I *t
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 01:21:02 pm
will ticker tape question come in the paer!!!
i have seen one such question in 2008 winter !!
dude pls hel[p!!
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 01:21:37 pm
abbe oye, Q = I *ttoh fir wat is power * time!!!??
Post by: moon on November 08, 2009, 01:23:26 pm
R = V/I => 4/0.4 = 10 ohms
Why did we find the resistance by 4/0.4 instead of 1.5/0.4 ??? ??????
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 01:25:01 pm
didnt notice it!!
Post by: xXFaKeXx on November 08, 2009, 01:28:19 pm
Post by: Ghost Of Highbury on November 08, 2009, 01:29:49 pm
sin r / sin i if from optically dense medium to less dense medium
Post by: xXFaKeXx on November 08, 2009, 01:30:49 pm
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 01:32:35 pm
they had asked resistance for lamp!!
u did for the whole thing!!
Post by: Ghost Of Highbury on November 08, 2009, 01:34:20 pm
i too think it shud be 1.5/.4
astar may help?
Post by: holychalice on November 08, 2009, 01:34:43 pm
4V is supplied to d circuit so 4 volts will be taken nd not 1.5
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 01:36:03 pm
they are telling to find resistance for lamp not the whole thing!!!
read the q carefully!!!
Post by: Ghost Of Highbury on November 08, 2009, 01:37:08 pm
but the voltage is different in series circuit, so it shudhere its only asked for the lamp, so it shud be 1.5/0.4
Post by: @d!_†oX!© on November 08, 2009, 01:38:47 pm
Can som1 plz tell me wen 2 use sin r/sin i instead of sin i /sin r to calculate the refractive index?u know what it wuold be even better if u use the formula
n1sin1=n2sin2
instead of the above....cuz u will never go wrong wid this...
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 01:39:26 pm
i am being too ,lazy to go back and see it!!
we can check the ms if possible!!
Post by: Ghost Of Highbury on November 08, 2009, 01:39:52 pm
u know what it wuold be even better if u use the formulaadi, dont ignore the q, wat do u think it shud be 1.5/0.4 rite?
n1sin1=n2sin2
instead of the above....cuz u will never go wrong wid this...
Post by: holychalice on November 08, 2009, 01:40:05 pm
but i guess dat cud be d only possible reason ...
read d first part .. d wire nd lamp is connected to 4V supply ...
d lamp is marked with ... 1.5V nd 0.6W ...
nd ms says 10ohms 4/0.4 = 10
Post by: @d!_†oX!© on November 08, 2009, 01:40:43 pm
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 01:41:09 pm
it is for the whole thing!!
and also in series the voltage changes!!
so i guess it must be 1.5 only!!
Post by: nid404 on November 08, 2009, 01:43:31 pm
Post by: @d!_†oX!© on November 08, 2009, 01:44:15 pm
P=IV
I=P/V=0.4 A
b)
R=V/I=1.5/0.4=3.75 ohms
c)
Q=It=0.4*20=8C
all's well...what's the problem you guys are having in the question??
Post by: Ghost Of Highbury on November 08, 2009, 01:45:38 pm
Post by: @d!_†oX!© on November 08, 2009, 01:46:18 pm
its not on fep..plss check nd let me know too
Post by: moon on November 08, 2009, 01:46:45 pm
Post by: Ghost Of Highbury on November 08, 2009, 01:46:56 pm
dude, if it was that easy, we wont be arguing here, lol :P check the ms, its 4/0.4
check d quoted post
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 01:47:56 pm
didnt get it>???
Post by: @d!_†oX!© on November 08, 2009, 01:48:50 pm
the markscheme's answer is 10 ohms y??? ???how do u say that it is 10 ohms moon???????
Post by: wassup on November 08, 2009, 01:49:28 pm
Post by: nid404 on November 08, 2009, 01:49:53 pm
Post by: @d!_†oX!© on November 08, 2009, 01:51:06 pm
now all i think is that there can be an error in the ms too...
cuz what we are doing is the right method!!
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 01:52:08 pm
which paper??june 2001 Q8>>>
Post by: wassup on November 08, 2009, 01:52:51 pm
Post by: nid404 on November 08, 2009, 01:53:13 pm
june 2001 Q8>>>
thank u....m back home.....ill check..
Post by: @d!_†oX!© on November 08, 2009, 01:53:40 pm
see it's taking V as 4 volts whereas the lamp does not get the complete 4V....it cannot be like this...our answer is right!!!!!
Post by: wassup on November 08, 2009, 01:54:14 pm
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 01:54:20 pm
i stil did not get why 4/0.4!!????
wat if ms is wrong!! ;)
Post by: astarmathsandphysics on November 08, 2009, 01:55:10 pm
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 01:55:54 pm
di toxic attach it here agin for sir!!
Post by: @d!_†oX!© on November 08, 2009, 01:58:08 pm
Post by: wassup on November 08, 2009, 01:59:18 pm
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 02:00:03 pm
chil maro yaar!!
do baar nahin ek baar thik tha!!
Post by: astarmathsandphysics on November 08, 2009, 02:00:10 pm
b)Use V=IR for the lamp 1.5=0.4*R so R=1.5/0.4=3.75ohms did everyone get this wrong?
c)Q=It=0.4*20=8C
Post by: @d!_†oX!© on November 08, 2009, 02:01:02 pm
Post by: nid404 on November 08, 2009, 02:01:20 pm
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 02:02:31 pm
ms is wrong again!!!
Post by: @d!_†oX!© on November 08, 2009, 02:03:13 pm
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 02:03:53 pm
but is it 1.5 or 4!??
i still feel it must be 1.5!!!
Post by: nid404 on November 08, 2009, 02:04:31 pm
even i'd go for 1.5V/0.4
Post by: @d!_†oX!© on November 08, 2009, 02:04:48 pm
Post by: cooldude on November 08, 2009, 02:05:32 pm
now we come to the point!!
ms is wrong again!!!
By the way does evry1 knw that this ms is not written by cie, the qs r solved by the person who made this document. so it is very likely that the person can be wrong. i did the maths papers of 1993-2003 and in one q that guy got an ans wrong so dnt go exactly by what this ms says
Post by: nid404 on November 08, 2009, 02:06:40 pm
which part are u toking about rite now???
the same damn resistance thingy
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 02:07:06 pm
Post by: @d!_†oX!© on November 08, 2009, 02:07:26 pm
Post by: wassup on November 08, 2009, 02:12:43 pm
mj03 q4b, q9
mj02 q9a
thanks in adv.
Post by: @d!_†oX!© on November 08, 2009, 02:13:37 pm
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 02:15:08 pm
like 2*105:80::x:25!!
solve for x you'll get the ans!!
Post by: astarmathsandphysics on November 08, 2009, 02:17:38 pm
Post by: astarmathsandphysics on November 08, 2009, 02:18:51 pm
Post by: wassup on November 08, 2009, 02:19:08 pm
4b is like you hav to do it with ratio propotion!!!
like 2*105:85::x:25!!
solve for x you'll get the ans!!
ya i know this but the ans is 64 in d ms n by this method d ans is coming out 2 b 62500 Pa.???
Post by: wassup on November 08, 2009, 02:21:18 pm
it'd be gr8 if u cud post the questionshere plss....
u can see d q's here:
mj02 http://www.freeexampapers.us/IGCSE/Physics/CIE/2002%20Jun/0625_s02_qp_3.pdf
mj03 http://www.freeexampapers.us/IGCSE/Physics/CIE/2003%20Jun/0625_s03_qp_3.pdf
Post by: nid404 on November 08, 2009, 02:21:30 pm
I just spotted I passed 6000 posts.
congrats ;D
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 02:22:36 pm
ii) power1=power2 so 48=0.4x!!
solve for x x=120 volts!!
b)i) the transformer being 100% efficient means that power in primary is eaul to power in secondary!! it converts all the elcerticity withouht ne or less wastage!!
ms:no/very little energy/power lost or energy/power in =
energy/power out
ii)ms:any mention of magnetic field
changing magnetic field
field passes through core or secondary coil
induces voltage in secondary coil
number of turns on secondary determines voltage
output
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 02:25:15 pm
mj02 q9aconnect the 1 volt supply to Y's of CRO then measure the readings!!
the maximum value is 1!!
acc. mark the sacle!!
Post by: wassup on November 08, 2009, 02:27:46 pm
9)ai) power=VI=24*2=48W!!
ii) power1=power2 so 48=0.4x!!
solve for x x=120 volts!!
b)i) the transformer being 100% efficient means that power in primary is eaul to power in secondary!! it converts all the elcerticity withouht ne or less wastage!!
ms:no/very little energy/power lost or energy/power in =
energy/power out
ii)ms:any mention of magnetic field
changing magnetic field
field passes through core or secondary coil
induces voltage in secondary coil
number of turns on secondary determines voltage
output
sorry fr this but i wnted 2 know d ans fr 6th question of mj03.
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 02:28:32 pm
Post by: moon on November 08, 2009, 02:34:11 pm
june 2001 Q8>>>
At last found the correct answer!!!!! Thanx 4 answering. :) :) :) :)
Post by: wassup on November 08, 2009, 02:35:01 pm
np i'll do it and tell ya within short time!!
thanks but can u explain d mj02 q9 properly plzzzzzzzz???
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 02:37:39 pm
i have tried to to it!!
hope you get it!!
Post by: wassup on November 08, 2009, 02:44:24 pm
Post by: astarmathsandphysics on November 08, 2009, 02:49:54 pm
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 02:50:18 pm
i guess it is better if you do this for 4b
volume*pressure=k!!
so 2*105*80=k
so now new volume =k/25
= 2*105*80/25
=6.4*105
hope ya get it!!
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 02:51:41 pm
thanks but can u explain d mj02 q9 properly plzzzzzzzz???just put the connections in y input!!
dc gives you the max value that is 1V
mark the max value as 1V and then half as 0.5!!
and so on!!
Post by: wassup on November 08, 2009, 02:52:34 pm
y wats it??
i guess it is better if you do this for 4b
volume*pressure=k!!
so 2*105*80=k
so now new volume =k/25
= 2*105*80/25
=6.4*105
hope ya get it!!
i did this only but the ans in d ms is 64 ??? ??? ???
i know how 2 do it. just wanted 2 know how is it 64 or is d ms wrong again?
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 02:56:21 pm
no editing:
(b) any attempt to use p x v = constant or correct
proportion
fraction 2 x 80/25 seen
p = 6.4 x 10 (Pa)
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 02:59:57 pm
direct form ms!!i guess they have missed 6.4 *105
no editing:
(b) any attempt to use p x v = constant or correct
proportion
fraction 2 x 80/25 seen
p = 6.4 x 10 (Pa)
this is the true ans!!
i guess the ms is wrong again!!!
Post by: @d!_†oX!© on November 08, 2009, 03:00:34 pm
the ms has misprinted nd forgotten to add 5 there
Post by: wassup on November 08, 2009, 03:00:52 pm
Post by: @d!_†oX!© on November 08, 2009, 03:01:58 pm
Question 4
(a) The confusion between particles and molecules was less common than with previous questions on
this topic. However there were considerable numbers of candidates who thought that the dust
particles were “lighter than the heavy air molecules”. Many answers were spoilt by extending this
wrong concept to “fast moving dust particles colliding with air molecules”. However, there were
many excellent answers seen.
(b) Large numbers of fully correct answers were seen, generally well set out and with all the working
shown. The most common error was the use of 2 x 25/80 instead of 2 x 80/25.
Answer: (b) 6.4 x 105 Pa.
Post by: ~~~~shreyapril~~~~ on November 08, 2009, 03:02:30 pm
Post by: wassup on November 08, 2009, 03:08:17 pm
hey wassup check this out, direct from the er:
Question 4
(a) The confusion between particles and molecules was less common than with previous questions on
this topic. However there were considerable numbers of candidates who thought that the dust
particles were “lighter than the heavy air molecules”. Many answers were spoilt by extending this
wrong concept to “fast moving dust particles colliding with air molecules”. However, there were
many excellent answers seen.
(b) Large numbers of fully correct answers were seen, generally well set out and with all the working
shown. The most common error was the use of 2 x 25/80 instead of 2 x 80/25.
Answer: (b) 6.4 x 105 Pa.
thanks adi
Post by: @d!_†oX!© on November 08, 2009, 03:09:19 pm
Post by: Mari mc on November 08, 2009, 03:21:42 pm
Post by: Ghost Of Highbury on November 08, 2009, 03:22:27 pm
Post by: @d!_†oX!© on November 08, 2009, 03:22:35 pm
Post by: neelcullen on November 08, 2009, 03:30:14 pm
Post by: nid404 on November 08, 2009, 03:39:06 pm
could anyone explain wat's charles law?? :S
charles law is when pressure is constant for an ideal gas the volume is directly proportional to the absolute temperature
Post by: ~~~~shreyapril~~~~ on November 09, 2009, 05:56:49 am
u dont take tension just remmeber that:
volume is directly propotional to temp
and presure is inversley to volume!!
Post by: xXFaKeXx on November 09, 2009, 06:01:16 am