Calculate the density of the material of which the sphere is made

AI Thread Summary
A hollow sphere with an inner radius of 8.0 cm and an outer radius of 9.0 cm floats half submerged in a liquid with a specific gravity of 0.80. The density of the liquid is calculated to be 800 kg/m^3, and the volume of the sphere's shell material is derived from the difference in volumes of the outer and inner spheres. The buoyant force, equal to the weight of the displaced liquid, is crucial for determining the sphere's weight and subsequently its density. Corrections were made regarding the calculations of the sphere's weight and the volume of the displaced liquid, leading to a reevaluation of the density of the material of the sphere. Properly calculating the volume and weight is essential for accurate density determination.
anyone1979
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A hollow sphere of inner radius 8.0 cm and outer radius 9.0 cm floats half submerged in a liquid of specific gravity 0.80. Calculate the density of the material of which the sphere is made.

Density of water = 1 * 10^3 kg/m^3
inner radius = 8.0 cm = 8 * 10 ^-2 m
outer radius = 9.0 cm = 9 * 10 ^-2 m
specific gravity = Density of liquid / density of water
Density of liquid = specific gravity * density of water = .80*1000 = 800 kg/m^3
V = (4 pi / 3) * (outer radius^3 * inner radius^3) = 9.09 *10 ^-4 m^3

weight of liquid = density of liquid * V * gravity = 7.127 N (buoyant force)

weight of sphere = weight of liquid + buoyant force = 7.127 + 7.127 = 14.254 N

mass of sphere = weight of sphere / gravity = 14.254/9.8 = 1.454 kg

density of sphere = mass of sphere / V = (1.454/(9.09 * 10^-4)) = 1.6 * 10^3 kg/m^3



-----All the work seems right, but I think the answer is not right. especially the calculation of the weight of the sphere. Can anybody help?





The Attempt at a Solution

 
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I suppose the sphere is filled with air? In that case the volume of the displaced liquid is just (4 pi/3)*outer radius^3 but only half of the sphere is submerged so the displaced volume is (4 pi/3)*outer radius^3/2.

The weight of the displaced volume is equal to to total weight of the sphere.
the mass of the sphere is then easy.

to find the density you will then have to use the volume of only the shell (as you did)
 
anyone1979 said:
A hollow sphere of inner radius 8.0 cm and outer radius 9.0 cm floats half submerged in a liquid of specific gravity 0.80. Calculate the density of the material of which the sphere is made.

Density of water = 1 * 10^3 kg/m^3
inner radius = 8.0 cm = 8 * 10 ^-2 m
outer radius = 9.0 cm = 9 * 10 ^-2 m
specific gravity = Density of liquid / density of water
Density of liquid = specific gravity * density of water = .80*1000 = 800 kg/m^3
OK.
V = (4 pi / 3) * (outer radius^3 * inner radius^3) = 9.09 *10 ^-4 m^3
This is the volume of the shell material (not the sphere). I assume you meant to write R_o^3 - R_i^3. You'll need the volume of the sphere as well.

weight of liquid = density of liquid * V * gravity = 7.127 N (buoyant force)

weight of sphere = weight of liquid + buoyant force = 7.127 + 7.127 = 14.254 N
Not sure what you're doing here: The buoyant force equals the weight of the displaced fluid. What's the volume of fluid displaced?

And the buoyant force also equals the weight of the sphere, since its floating.

mass of sphere = weight of sphere / gravity = 14.254/9.8 = 1.454 kg

density of sphere = mass of sphere / V = (1.454/(9.09 * 10^-4)) = 1.6 * 10^3 kg/m^3
Once you properly calculate the weight of the sphere, this method will give you the density of the material.
 
kamerling said:
I suppose the sphere is filled with air? In that case the volume of the displaced liquid is just (4 pi/3)*outer radius^3 but only half of the sphere is submerged so the displaced volume is (4 pi/3)*outer radius^3/2.

The weight of the displaced volume is equal to to total weight of the sphere.
the mass of the sphere is then easy.

to find the density you will then have to use the volume of only the shell (as you did)

Thanks, I just realized it...
I calculated all wrong
 
Last edited:
Doc Al said:
OK.

This is the volume of the shell material (not the sphere). I assume you meant to write R_o^3 - R_i^3. You'll need the volume of the sphere as well.


Not sure what you're doing here: The buoyant force equals the weight of the displaced fluid. What's the volume of fluid displaced?

And the buoyant force also equals the weight of the sphere, since its floating.


Once you properly calculate the weight of the sphere, this method will give you the density of the material.


Thanks...I realized my error now.
I made a mistake on the calculation
 
Last edited:
Sorry guys, I had a brain fart.
I just realized my mistake...You guys are both right.
Thanks for all your help
 
anyone1979 said:
This is the volume of the sphere material:
V = (4 pi / 3) * (outer radius^3 * inner radius^3) = 9.09 *10 ^-4 m^3
You mean:
V = \frac{4 \pi}{3} (R_{outer}^3 - R_{inner}^3)

This is the volume of the sphere:
Vs = (4 pi / 3) * (outer radius^3) = 3.1 * 10 ^-3 m^3 = volume of displaced liquid
That's the volume of the sphere. Is the entire sphere submerged?

weight of liquid = density of liquid * V * gravity = (buoyant force)
Wl = 800 * (3.1 * 10 ^-3) *9.8 = 24.3 N
Correct this.

So what about the volume of the sphere material since I am trying to find the density of the material of the sphere...I was thinking I needed to use that volume.
You'll need the volume of shell material when calculating the density. First find the mass.
 
Thanks for all your help...
 
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