Calculating Volume of Steel for 230-L Steel Drum with Gasoline

[Nemo's]

Homework Statement

A 230-l steel drum is completely full of gasoline. What total volume of steel material can be used in making the drum if the gasoline filled drum is to float in fresh water with the whole volume immersed?

Homework Equations

ρwater=1000 kgm^-3
ρgasoline=680 kgm^-3
ρsteel = 7800 kgm^-3

The Attempt at a Solution

Fb=mdrum*g
ρwater*Vdrum=msteel+mgasoline
ρwater*(Vsteel+Vgasoline)=(ρV)steel+(ρV)gasoline
Then taking Vsteel subject and substituting with ρgasoline = 230*10^-3
Is this method correct ? I don't know whether the 230-l is the total volume of the drum or is it just the volume of gasoline?

 

[SteamKing]

"A 230-l steel drum is completely full of gasoline." What does this sentence mean to you?

 

[Nemo's]

that the steel drum can take up to 230-l of gasoline and that it's completely full. So the volume of gasoline must equal 230-l right ?
But my tutor used 230-l as both the volume of gasoline and the total volume of the drum. I just can't see why he did that.

 

[Chestermiller]

that the steel drum can take up to 230-l of gasoline and that it's completely full. So the volume of gasoline must equal 230-l right ?
But my tutor used 230-l as both the volume of gasoline and the total volume of the drum. I just can't see why he did that.

With all due respect to your tutor, the volume of gasoline is 230 liters. What is the weight of the gasoline? Let V represent the volume of steel. In terms of V, what is the weight of the steel. The total volume submerged is 230 + V. In terms of V, what is the mass of water displaced? What is the weight of the water displaced? If the whole drum volume is just immersed, how does the weight of water displaced compare with the weight of the gasoline plus steel?

 

[Nemo's]

O.K so I guess I'm doing it the right way. Thanks a lot