Answer:
Assume that [tex]\rm g= 9.81\; N\cdot kg^{-1}[/tex]; [tex]\rho(\text{Water}) = \rm 1000\;kg\cdot m^{-3}[/tex].
Density of the disk: approximately [tex]\rm 2.19\times 10^{3}\; kg\cdot m^{-3}[/tex].
Weight of the disk: approximately [tex]\rm 245\;N[/tex].
Buoyant force on the disk if it is submerged under water: approximately [tex]\rm 112\; N[/tex].
The disk will sink when placed in water.
Explanation:
Convert the dimensions of this disk to SI units:
- Diameter: [tex]d = \rm 25\; inches = (25\times 0.3048)\; m = 0.762\;m[/tex].
- Thickness [tex]h = \rm 2.5\; cm = (2.5\times 0.01)\; m = 0.025\;m[/tex].
The radius of a circle is 1/2 its diameter:
[tex]\displaystyle r = \rm \frac{1}{2}\times 0.762\;m = 0.381\; m[/tex].
Volume of this disk:
[tex]V(\text{disk}) = \pi\cdot r^{2}\cdot h = \pi\times 0.381^{2}\times 0.025 \approx 0.0114009\; m^{3}[/tex].
Density of this disk:
[tex]\displaystyle \rho(\text{disk}) = \frac{m}{V} = \rm \frac{25\; kg}{0.0114009\; m^{3}} = 2.19\times 10^{3}\;kg\cdot m^{-3}[/tex].
[tex]\rho(\text{disk}) >\rho(\text{water})[/tex] indicates that the disk will sink when placed in water.
Weight of the object:
[tex]W(\text{disk}) = m\cdot g = \rm 25\times 9.81 = 245.25\; N[/tex].
The buoyant force on an object in water is equal to the weight of water that this object displaces. When this disk is submerged under water, it will displace approximately [tex]\rm 0.0114009\; m^{3}[/tex] of water. The buoyant force on the disk will be:
[tex]\begin{aligned}F(\text{buoyant force}) &= W(\text{Water Displaced}) \\& = \rho\cdot V(\text{Water Displaced})\cdot g\\ & = \rm 1\times 10^{3}\; kg\cdot m^{-3}\times 0.0114009\; m^{3}\times 9.81\; N\cdot kg^{-1}\\ &\approx \rm 112\; N\end{aligned}[/tex].
The size of this disk's weight is greater than the size of the buoyant force on it when submerged under water. As a result, the disk will sink when placed in water.
