# Thread: Calculating volume using buoyancy

1. ## Calculating volume using buoyancy

That's the question. Here's the information I have:
Vsphere= pi/6(d^3)
W = mg
Wc + Wo' = 0.650 N
Wc' + Wo' = 0.550 N

I'm unsure of how to do this because it asks to find the volume ONLY BY USING THE two weight measurements. The simpler way to do it would be to measure how much fluid is displaced but that's not allowed apparently so I'm stuck.

I was just wondering if you guys could tell me what to use, maybe some equation or something I could sub in to calculate the volume using only the two weights. I know that this takes a bit of physics knowledge but would greatly appreciate the help.

2. Originally Posted by SportfreundeKeaneKent

That's the question. Here's the information I have:
Vsphere= pi/6(d^3)
W = mg
Wc + Wo' = 0.650 N
Wc' + Wo' = 0.550 N

I'm unsure of how to do this because it asks to find the volume ONLY BY USING THE two weight measurements. The simpler way to do it would be to measure how much fluid is displaced but that's not allowed apparently so I'm stuck.

I was just wondering if you guys could tell me what to use, maybe some equation or something I could sub in to calculate the volume using only the two weights. I know that this takes a bit of physics knowledge but would greatly appreciate the help.
Hej,

1. You only can answer this question if you know the density of the fluid. I assume that you use water as the fluid which has a density of $\displaystyle \rho_{water} = 1\ \frac{g}{cm^3}$

2. The difference of the weights is caused by the buoyant force of the cork. The buoayant force is as great as the weight of the displaced fluid.

3. Obviously the weight of the displaced fluid is 100 cN. Assuming that the fluid is water then the volume of the displaced fluid is 100 cm³ - and that must be the volume of the cork too.

3. Ok, I understand that the volume of the cork is the volume of fluid displaced. But how can I find the amount of fluid displaced in each situation? The volume of the cork will be the volume of water displaced by cork and metal sphere subtracted by the volume of fluid displaced by just the metal sphere in water. I’m not sure what to use to find this though. Where did you get the values 100 cm^3 or 100 N? In this experiment, we’re not given any values of the amount of water in the beaker or how much it increases by after the cork is placed in it.

4. Here's what I've done so far, I know it's wrong because the volume of the cork should be more than the volume of the metal sphere:

In order to find the volume of the cork, the volume of fluid displaced by the metal sphere in water must be found. This amount can be found by using Wc + Wo’.

Wc’ + Wo’ = (0.650 ± 0.010 N)/9.81 m/s² = 0.0662 kg = 66.26 ± 0.10 g
pwater = 1 g/cm³
The volume of the metal sphere will equal the amount of water displaced which is equivalent to mass of water displaced divided by the density of water:
Vcork+sphere = 66.26 g/1g/cm³ = 66.26 ± 0.10 cm³

A similar procedure must be used to find the volume of water displaced by the cork and the metal sphere:

Wc + Wo’ = (0.550 ± 0.010 N)/9.81 m/s² = 0.056 kg = 56.07 ± 0.10 g
pwater = 1 g/cm³
The volume of the metal sphere will equal the amount of water displaced which is equivalent to mass of water displaced divided by the density of water:
Vsphere = 56.07 g/1g/cm³ = 56.07 ± 0.10 cm³

To find the volume of the cork, the value for Vsphere must be subtracted from Vcork+sphere.

Vcork = 66.26 ± 0.10 cm³ - 56.07 ± 0.10 cm³ = 10.19 ± 0.10 cm³

5. Originally Posted by SportfreundeKeaneKent
Here's what I've done so far, I know it's wrong because the volume of the cork should be more than the volume of the metal sphere:

In order to find the volume of the cork, the volume of fluid displaced by the metal sphere in water must be found. This amount can be found by using Wc + Wo’.

Wc’ + Wo’ = (0.650 ± 0.010 N)/9.81 m/s² = 0.0662 kg = 66.26 ± 0.10 g
pwater = 1 g/cm³
The volume of the metal sphere will equal the amount of water displaced which is equivalent to mass of water displaced divided by the density of water:
Vcork+sphere = 66.26 g/1g/cm³ = 66.26 ± 0.10 cm³

A similar procedure must be used to find the volume of water displaced by the cork and the metal sphere:

Wc + Wo’ = (0.550 ± 0.010 N)/9.81 m/s² = 0.056 kg = 56.07 ± 0.10 g
pwater = 1 g/cm³
The volume of the metal sphere will equal the amount of water displaced which is equivalent to mass of water displaced divided by the density of water:
Vsphere = 56.07 g/1g/cm³ = 56.07 ± 0.10 cm³

To find the volume of the cork, the value for Vsphere must be subtracted from Vcork+sphere.

Vcork = 66.26 ± 0.10 cm³ - 56.07 ± 0.10 cm³ = 10.19 ± 0.10 cm³
Hej,

your calculations are correct. I made one ugly typo and all following calculations were wrong: The difference of weights is 0.1 N = 10 cN (cN = centi Newton). The amount of water which has a weight of 10 cN is nearly 10 cm³ (if you use g = 10 (kg*m)/s²)