So there are 10 fishes in a tank..

Depends on whether you remove the fish or not

LOL. You got me LOOOOL.


but... what if the other fish killed that fish, what then?

Scimi

decrease ?
dead fish moves up over the surface. Part of the dead fish body will be out of the water, reducing the volume of water in the tank (volume of water = water + volume of fishes inside the water).
I need to make a real test to be sure this answer is correct (any idea how to kill a fish ?)

format_quote Originally Posted by Hulk

Haha good one! But actually the water level will rise either way because 9 fishes will cry!

Lool ! good one.

format_quote Originally Posted by Hulk

Haha good one! But actually the water level will rise either way because 9 fishes will cry!

lol but it would make no difference to the water level

darn, a-levels physics, buyoancy and archimedes principles. darn I knew how to solve this!

........................

ok guys here's the answer.

when the fish is alive, it is more dense than water, hence it sinks and stays beneath water surface. the volume of water it displaces would be equal to its volume.

now when the fish dies, it becomes less dense due to build up of gasses but its weight (assuming) will remain the same even though volume has increased (density = mass / volume). The volume of fluid displaced by this fish will always exert a greater force than the weight of the fish, and hence that is why it will float!!

now since the fish volume has increased, it will float. but now not all of the fish is floating. some of its body will be submerged, while the rest will be above the surface of water. The submerged volume of fish is what we are concerned with. This submerged volume of fish will displace an equal volume of water that is enough to exert upward force to keep the fish floating. Since this force is greater than the force exerted when the fish was fully submerged, we have to think which variable in the force changed. Force is given by Force = fluid density x gravity x volume of object

now since all the factors are same whether the fish is submerged or floating except volume of object or fish, in order for force to be greater for keeping the fish floating, the volume displaced by fish has to increase. Hence the water level will rise when the fish is floating, compared to when it was alive and deep down into the water.

feels good to answer this.

format_quote Originally Posted by CosmicPathos

ok guys here's the answer.

when the fish is alive, it is more dense than water, hence it sinks and stays beneath water surface. the volume of water it displaces would be equal to its volume.

now when the fish dies, it becomes less dense due to build up of gasses but its weight (assuming) will remain the same even though volume has increased (density = mass / volume). The volume of fluid displaced by this fish will always exert a greater force than the weight of the fish, and hence that is why it will float!!

now since the fish volume has increased, it will float. but now not all of the fish is floating. some of its body will be submerged, while the rest will be above the surface of water. The submerged volume of fish is what we are concerned with. This submerged volume of fish will displace an equal volume of water that is enough to exert upward force to keep the fish floating. Since this force is greater than the force exerted when the fish was fully submerged, we have to think which variable in the force changed. Force is given by Force = fluid density x gravity x volume of object

now since all the factors are same whether the fish is submerged or floating except volume of object or fish, in order for force to be greater for keeping the fish floating, the volume displaced by fish has to increase. Hence the water level will rise when the fish is floating, compared to when it was alive and deep down into the water.

feels good to answer this.

GEEEEK !!!

Scimi