Physics 31, If 15 more minutes were provided

[Wolfgangs]

Guys, about the Paper 31. There was a question that gave us the atmospheric pressure and told us to calculate the pressure (I don't remember the question exactly, but it was the second question in the section with the manometer of 250mm). Does anyone know the question and how it is solved?

 

[Sojiro]

then the pressure 120000 pa +the atmospheric pressure the second question

 

[Alex10]

The paper was great - finished in like 45 minutes.

 

[Sojiro]

no sorry the pressure isnt 120000 the density was 13600 so the pressure is 34000 pa and the second part 34000 + 102000=136000pa

 

[Alex10]

no sorry the pressure isnt 120000 the density was 13600 so the pressure is 34000 pa and the second part 34000 + 102000=136000pa

Yeah that is correct

 

[Sojiro]

Yeah that is correct

 

[lisaann]

Actually you'll subtract both pressure cause at the same level
P(atmospheric)=P(mercury)+P(gas)

 

[Sarah Jawrdawi]

how many max u would loose???

I'll get around 60/80 but I'm not worried cause I did very well in paper 6 and hopefully I'll do good in paper 1

 

[silentkiller97]

Actually you'll subtract both pressure cause at the same level
P(atmospheric)=P(mercury)+P(gas)

yes you are right I did the same

 

[Wolfgangs]

Actual Pressure = Atmospheric pressure + excess pressure (this is the equation according to my text book and used this in many past test papers done in our school). The pressure in (a) is the excess since it was the pressure of the 250mm. While 2.10x10^5 Pa is the atmospheric pressure. So the question wanted the total pressure which is the actual pressure so you must add them to get it since two pressures were acting. Also, does anyone remember what was the (b) question exactly since I forget the question. Hope someone will reply soon.

 

[Wolfgangs]

Also, you can notice that in Paper 31 May/June 2011. There is a similar question i.e. question no. 3 (b) (iii). They got the excess pressure and then added it to the atmospheric pressure. There is also in Paper 33 October/November 2012 (I think Q3. too).

 

[Mohab88]

Guys, about the Paper 31. There was a question that gave us the atmospheric pressure and told us to calculate the pressure (I don't remember the question exactly, but it was the second question in the section with the manometer of 250mm). Does anyone know the question and how it is solved?

Atmospheric pressure + pressure in part A

 

[Shada Mohammed]

Actually you'll subtract both pressure cause at the same level
P(atmospheric)=P(mercury)+P(gas)

yeah that's right

 

[kevinjohn]

hey guys... any comment on the pressure question?like the final question which asks whst the total pressure acting on the flask....
do we add or subtract the pressure value obtained and the atmospheric pressure?

 

[kevinjohn]

yeah that's right

cos many ppl have different asnwers.. some say subtract the values

 

[ahmed soliman]

no sorry the pressure isnt 120000 the density was 13600 so the pressure is 34000 pa and the second part 34000 + 102000=136000pa

correct but there was a conversion because height is in mm and denseity was in m cubed

 

[ahmed soliman]

no sorry the pressure isnt 120000 the density was 13600 so the pressure is 34000 pa and the second part 34000 + 102000=136000pa

correct but there was a conversion because height is in mm and denseity was in m cubed

 

[ahmed soliman]

was the force on left side (AB) in induction question anticlockwise ???

 

[Sarah Jawrdawi]

Yes

was the force on left side (AB) in induction question anticlockwise ???

 

[Sojiro]

correct but there was a conversion because height is in mm and denseity was in m cubed

yes i know it will be 250mm divide 1000 =0.25 m then used in the equation