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Physics Pp:3-Variant 2- Discussing Answers!!

Did you finish the whole paper in time??

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kboss

kboss

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mamoon123 said:
Uh, were we supposed to write "CD"?
I only wrote D. :c
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duh...:p
it was supposed to be a wavefront so it has to be a line...and for that 2 points are needed at a minimum :p
but i think the examiner may give u marks if the rest of the reasoning is correct!!! :D
 
mamoon123

mamoon123

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And this is what happens when you panic after seeing a bulb question ._.
Silly me.
 
Aay-Jay Bird <3

Aay-Jay Bird <3

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Chucky said:
must mention the type clockwise or anti clockwise.

it is anti clockwise by the way.
when we were all young we played with toys like these, funny we should sit for an exam with a question regarding one too.
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Isn't it supposed to be a balancing toy, though? :confused:
 
N

n0n4m3

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Question 2: Now that I've seen the drawing for the bird question, it will rotate anti-clockwise. Explanation: anti-clcockwise-moment is greater than clockwise-moment, so there's a nett anti-clockwise moment and the bird will rotate back to equilibrium position.

Question 9: "Fail-short" current is given to be 0.9A and resistance of each lamp at this current has increased to 5% its original value. Original value: 8 ohm so the new value would be 8.4 ohm. After this it's just simple math.
V=RI
240=(8.4 x X)(0.9) is approximately 31.7. 40-31.7=8.3 bulbs. But since you cannot have an 8.3 bulb. the fuse will blow after the 9th one blows.

Question 11: Subtract the background radiation. 52-14=38. So for it to be half of 38, it's 38/2 =19. Add back 14, you get 33. And from there just read the graph, it gives you around 1.6 days.
 
J

jsdeo

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n0n4m3 said:
Question 9: "Fail-short" current is given to be 0.9A and resistance of each lamp at this current has increased to 5% its original value. Original value: 8 ohm so the new value would be 8.4 ohm. After this it's just simple math.
V=RI
240=(8.4 x X)(0.9) is approximately 31.7. 40-31.7=8.3 bulbs. But since you cannot have an 8.3 bulb. the fuse will blow after the 9th one blows.
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I think the question asked for the maximum number of bulbs that can blow before the fuse blows, though? I'm not completely sure... if this is the case then it would be 8. Otherwise, you are correct.
 
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n0n4m3

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jsdeo said:
I think the question asked for the maximum number of bulbs that can blow before the fuse blows, though? I'm not completely sure... if this is the case then it would be 8. Otherwise, you are correct.
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Yes that's what the question asked for. But with 8 bulbs, the current has not exceeded the fail-short current rating. Let's work backwards. 32 bulbs still working. So resistance is 8.4 x 32 = 268.8. So current in circuit is I = V/R which is 0.89A. Almost exceeded but not yet. When 9th bulb blows, it would definitely exceed.
 
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Sherry Berry

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kareem said:
so
so if i lose 15-25 marks on paper 3 , lets say 10 on paper 1 and like 8 on paper 6 does that mean its an A ?
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yea seems like but i don't think B .
Well you might even knock on an A* ;p
Cuz the gt's gotta b low this time ..hopeful :)
 
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jsdeo

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n0n4m3 said:
Yes that's what the question asked for. But with 8 bulbs, the current has not exceeded the fail-short current rating. Let's work backwards. 32 bulbs still working. So resistance is 8.4 x 32 = 268.8. So current in circuit is I = V/R which is 0.89A. Almost exceeded but not yet. When 9th bulb blows, it would definitely exceed.
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Oh, I meant to say, I thought the question asked for the maximum number of bulbs that can blow WITHOUT the fuse blowing... sorry about that.
 
A

Ashay

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Chucky said:
the answer is 8
let the total number of bulbs be x.
5% of original resistance is 8.4 ohms
voltage is same (source same) and the max current before the fuse melts is 0.9A
therefore current= voltage/total resistance
0.9 = 240/(x * 8.4)
the answer comes in 31.77 , round that off its 32. so the total number of bulbs is 32. difference 40-32= 8 bulbs


It's upwards, as the current induced in the coil , produces a magnetic field that OPPOSES the motion on the bar magnet.
therefore when it is entering the solenoid, it will produce north to repel
when it is leaving the solenoid, it will produce south to attract, overall all the forces exerted being upwards.



it's way more. you have to deduct background radiation which was around 14 counts per minute.
you find that from the graph (it is the point after which the counts do not decrease anymore)
52-14= 38
38/2= 19 counts per minute
find out how long it took to come to 19 counts per minute.
should be more than 3 days.


when it is low tide, meaning the sea water is at a lowe level, water moves FROM RIGHT to LEFT



I'm not sure about this, but I heard the red ray is supposed to refract more, so you draw one line refracting and a faint line for a partial reflection as well.
if you have drawn the reflection, you get one mark.



OR THEN NOT, remember to get that sequence correct in your drawing.
Y was 0, Z was 1.

hope that helped.
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It can't be more than 3 days because you have to add background radiation(14) to 19 so from the graph you have to see the time when the count rate is 33, not 19
 
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