What is the specific gravity of the disk?

[sona1177]

Homework Statement

A cylindrical disk has volume 8.97 x 10^-3 m^3 and mass 8.16 kg. The disk is floating on the surface on some water with its flat surfaces horizontal. The area of each flat surface is .640 m^2. What is the specific gravity of the disk? How far below the water level is its bottom surface? How far above the water level is its top surface?

I know that the S.G. is .9097, and the density is 909.70 kg/m^3. So Vsubmerged = (Density of disk x Volume of disk)/density of fluid.

I got a value of .00816 m^3.

Then from Vsubmerged=Area x depth under water =.00816/.640=.01275 m

The last question is the one I am unsure about.

I subtracted the value of the object from V submerged to get 8.1 x 10^-4 for the volume not submerged (above water) and got 8.1 x 10^-4. So then V =Ad and V/A= 8.1 10^-4/.640=.01266 m.

I am not sure that I did the last part correctly, can you please tell me if I took the right approach?

Homework Equations

The Attempt at a Solution

 

[Sleve123]

I agree with your answers until the last part, where you say 0.00816/.64 = 0.01266 m

I think you've just missed a zero, so it would be 0.001266 m

Check it by adding the two depths and multiplying by the base area to give the original volume.

 

[sona1177]

I agree with your answers until the last part, where you say 0.00816/.64 = 0.01266 m

Thanks!