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# How can you weigh a living fish?

## Question

How can you weigh a living fish( which is more than 500 gm),found in an Aquarium?

You are not allowed to:

1- take the fish out of water or out of the aquarium.

2- move the aquarium or weigh it.

3- replace the water with new one.

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Can we make the fish stand still?

tie a

weight the fish and let it rest on a scale at the bottom of the
aquarium, take note of the number you see on the scale, go to another
aquarium under the same conditions and put a dummy fish of your own in
it that you can control it's weight and try to make it give the same
result as the previous experiment...

OR

Put

a scale inside the aquarium and take the water out (you still didn't
technically replace the water) and weigh the fish

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if there are no restrictions other than those imposed,

measure the height of the water, then remove the water and put it into another aquarium, allowing you to calculate it's volume.

you can now derive the fish's weight based on the height (depth), length, and width of the initial aquarium.

the fish'll probably die, though, so you'll have to be quick with your calculations to detemine its weight while it's still alive.

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The problem says that we can't take the fish out of the aquarium, or put in new water. However, it doesn't say we can't take the water out of the aquarium. So, remove all the water, quickly weigh the fish, and then put back the water. Simple!

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Let it inside a plastic bag, tie a string on the opening (let no air),

fish in the bag of water is heavier than the bag of water without the fish.

Tie the end of string to a weigh hook with the fish bag fully submerged.

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Let it inside a plastic bag, tie a string on the opening (let no air),

fish in the bag of water is heavier than the bag of water without the fish.

Tie the end of string to a weigh hook with the fish bag fully submerged.

Yes...thats was my idea...thanks

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Capture the fish inside an open, buoyant container (something like a large pot with some inherent buoyancy), so that the fish is contained within the pot which is now floating in the aquarium. Measure the water level both inside and outside the pot. Release the fish, and fill the pot with aquarium water until it sits as deeply in the water as before (the "outside" water level is the same). Note the water level within the pot again, and how this differs from the interior water level as measured with the fish. You can determine the weight of the additional water by calculating its volume (current interior water depth minus interior water depth with fish included, multiplied by the cross-sectional area of the pot...) and multiplying by the known density of water. This is the weight of the fish.

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You would also need to account for the water displaced by the fish to get an accurate calculation. So, a measure of the fish's volume would be required. I believe this is also the case for the "fish in a plastic bag" solution -- it is important to know how much water is displaced by the fish in order to determine the weight.

To modify my approach: capture fish in open, buoyant pot. Measure interior and exterior water levels. Release fish, fill pot with water until exterior water level is the same as before (pot sits as deeply in the water of the aquarium as before). Measure the new interior water level, and calculate the amount of "extra" water added to the pot to reach the same weight as when the pot held the fish (including the volume of water the fish originally displaced). Then, using the known (or easily measurable) density of the water, you can determine the weight of the "extra" water, and this is the weight of the fish.

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Capture the fish inside an open, buoyant container (something like a large pot with some inherent buoyancy), so that the fish is contained within the pot which is now floating in the aquarium. Measure the water level both inside and outside the pot. Release the fish, and fill the pot with aquarium water until it sits as deeply in the water as before (the "outside" water level is the same). Note the water level within the pot again, and how this differs from the interior water level as measured with the fish. You can determine the weight of the additional water by calculating its volume (current interior water depth minus interior water depth with fish included, multiplied by the cross-sectional area of the pot...) and multiplying by the known density of water. This is the weight of the fish.

A very good idea...

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Let it inside a plastic bag, tie a string on the opening (let no air),

fish in the bag of water is heavier than the bag of water without the fish.

Tie the end of string to a weigh hook with the fish bag fully submerged.

I've been pondering this one... and I am not sure if it would work without lifting the bag out of the aquarium (at least if I am picturing it correctly). If the bag (with or without the fish) is submerged in the aquarium, then a weight hook would be useful for measuring its

buoyancy, but not its weight. A bag full of water, submerged in water, would have virtually zero apparent weight (the plastic of the bag and the material of the string would contribute a little, technically); if the fish had his air bladder balanced right, the bag with the fish inside would also have virtually zero apparent weight, at least while underwater. (In fact, if the fish had an overinflated air bladder, the bag with fish inside might appear to have negative weight while submerged, as the fish would tend to float and push the bag to the surface.)
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Let it inside a plastic bag, tie a string on the opening (let no air),

fish in the bag of water is heavier than the bag of water without the fish.

Tie the end of string to a weigh hook with the fish bag fully submerged.

I've been pondering this one... and I am not sure if it would work without lifting the bag out of the aquarium (at least if I am picturing it correctly). If the bag (with or without the fish) is submerged in the aquarium, then a weight hook would be useful for measuring its

buoyancy, but not its weight. A bag full of water, submerged in water, would have virtually zero apparent weight (the plastic of the bag and the material of the string would contribute a little, technically); if the fish had his air bladder balanced right, the bag with the fish inside would also have virtually zero apparent weight, at least while underwater. (In fact, if the fish had an overinflated air bladder, the bag with fish inside might appear to have negative weight while submerged, as the fish would tend to float and push the bag to the surface.)

Yes, that is right.. the bag of water with fish inside will sink (give submerge weight ) but this is not the "out of the water weight" . Also i agree, a bloat fish makes the bag less dense than water ,good physics thanks.

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