May/June 2013 physics p31...Doneee...Hurrayyyy...!!!

[kevinjohn]

it was the 4 ohm resistor

the question of the power of the lamps, what was the answer ?

 

[yamaner]

4 ohms has higher power

i made it wrong i said 12 ohms because higher resistance higher voltage and higher voltage higher power as P=VI as v increase p increase, how many marks i will lose all of it?

 

[kevinjohn]

i made it wrong i said 12 ohms because higher resistance higher voltage and higher voltage higher power as P=VI as v increase p increase, how many marks i will lose all of it?

not really sure dude. cos u gave the ans as 12+ ur explaination is completely wrong.... srry man

 

[yamaner]

not really sure dude. cos u gave the ans as 12+ ur explaination is completely wrong.... srry man

y mean i will lose all of the mark?

 

[Wolfgangs]

I mention the equation P=I^2 X R. But the answer was wrong. Do you think I will get marks for it?

 

[vpatelvpatel]

Yes , i think that's exactly what i did, you have to add atmospheric pressure and the pressure you got not subtract them.

follow the link and read the manometer part. You will see why we have to subtract AND NOT ADD.
http://www.kentchemistry.com/links/GasLaws/manBar.htm
http://www.upscale.utoronto.ca/PVB/Harrison/Manometer/Manometer.html

 

[kevinjohn]

if u have mentioned the formula P=I^2 xR THEN U MAY GET A MARK

y mean i will lose all of the mark?

 

[Wolfgangs]

The first link mentions in Case 2 that was our case "Thus, in this case the pressure of the gas that is trapped in the closed end of the tube is greater than atmospheric pressure by the amount of pressure exerted by the column of liquid of height h." So the pressure is greater than the atmospheric pressure which means addition must occur not subtraction. In the second link, you'll notice that they mentioned p(gas) = p(atm) + p(h2) which has the exact same diagram as ours. This diagram with open tube is the same diagram as the one in Paper 31. Also, the height which was 250mm was high too. So, overall addition must occur and I'm sure of it and you proved it with those links.

 

[h4z3rd]

The first link mentions in Case 2 that was our case "Thus, in this case the pressure of the gas that is trapped in the closed end of the tube is greater than atmospheric pressure by the amount of pressure exerted by the column of liquid of height h." So the pressure is greater than the atmospheric pressure which means addition must occur not subtraction. In the second link, you'll notice that they mentioned p(gas) = p(atm) + p(h2) which has the exact same diagram as ours. This diagram with open tube is the same diagram as the one in Paper 31. Also, the height which was 250mm was high too. So, overall addition must occur and I'm sure of it and you proved it with those links.

i added them too!
do you get a 4.4 x 10^5 result?

 

[Ahmad Mubaid]

The first link mentions in Case 2 that was our case "Thus, in this case the pressure of the gas that is trapped in the closed end of the tube is greater than atmospheric pressure by the amount of pressure exerted by the column of liquid of height h." So the pressure is greater than the atmospheric pressure which means addition must occur not subtraction. In the second link, you'll notice that they mentioned p(gas) = p(atm) + p(h2) which has the exact same diagram as ours. This diagram with open tube is the same diagram as the one in Paper 31. Also, the height which was 250mm was high too. So, overall addition must occur and I'm sure of it and you proved it with those links.

man the atmospheric pressure was higher than the gas iam sure

 

[Wolfgangs]

man the atmospheric pressure was higher than the gas iam sure

Dude, you can notice that in the diagram there was excess 250mm was given to you. That's the reason you calculate it's pressure first. With that you p(h2) and the p(atm) was given. So, you simply add them.

 

[vpatelvpatel]

Im not sure if you're correct, because the diagram was case 3. not case 2.

The first link mentions in Case 2 that was our case "Thus, in this case the pressure of the gas that is trapped in the closed end of the tube is greater than atmospheric pressure by the amount of pressure exerted by the column of liquid of height h." So the pressure is greater than the atmospheric pressure which means addition must occur not subtraction. In the second link, you'll notice that they mentioned p(gas) = p(atm) + p(h2) which has the exact same diagram as ours. This diagram with open tube is the same diagram as the one in Paper 31. Also, the height which was 250mm was high too. So, overall addition must occur and I'm sure of it and you proved it with those links.

 

[vpatelvpatel]

I am completely sure that the mercury was the other way round. The gas pressure was less than atmospheric pressure because it pushed the mercury less, thus making the level go the other way round. So we have to subtract 100000Pa- 34000Pa so we get 66000Pa, making it lower than atmospheric pressure.

The first link mentions in Case 2 that was our case "Thus, in this case the pressure of the gas that is trapped in the closed end of the tube is greater than atmospheric pressure by the amount of pressure exerted by the column of liquid of height h." So the pressure is greater than the atmospheric pressure which means addition must occur not subtraction. In the second link, you'll notice that they mentioned p(gas) = p(atm) + p(h2) which has the exact same diagram as ours. This diagram with open tube is the same diagram as the one in Paper 31. Also, the height which was 250mm was high too. So, overall addition must occur and I'm sure of it and you proved it with those links.

 

[Wolfgangs]

I am completely sure that the mercury was the other way round. The gas pressure was less than atmospheric pressure because it pushed the mercury less, thus making the level go the other way round. So we have to subtract 100000Pa- 34000Pa so we get 66000Pa, making it lower than atmospheric pressure.

Until we get the PDF of this paper. No one is sure until we get one. I hope someone will get the sample soon for us.

 

[Wolfgangs]

Plus, I remember that it was case 2 since there an excess pressure involved. But still no one is sure and we all have weak memories XP. Anyway, does anyone have the PDF that shows the Threshold for the October/November 2012 Physics.

 

[ahmed soliman]

stop talking about the atm question (addition or subtraction) !!!!!!!!!!
it carries 1 mark only
instead of arguing about this 1 mark
discuss about different questions which carry 2-3 marks
anyway i added them

 

[Wolfgangs]

stop talking about the atm question (addition or subtraction) !!!!!!!!!!
it carries 1 mark only
instead of arguing about this 1 mark
discuss about different questions which carry 2-3 marks
anyway i added them

Really? I thought that it carried three marks, which made me worried a lot. Thanks for informing me XP

 

[shazmina]

Yeah... it was so easy....

vry easy..........

 

[Ahmad Mubaid]

guys the alpha and beta where do they curve in the last question

 

[Mohamed Alhendi]

such an easy paper evn a baby cn do it....

32 was hard :')