What fraction of an iron anchor’s weight will be supported by buoyant force when submerged in saltwater?

Fluid statics, often known as hydrostatics, is a branch of fluid mechanics that investigates the state of balance of a floating and submerged body, as well as the pressure in a fluid, or imposed by a fluid, on an immersed body.

The ratio of the object's density to the density of the fluid in which it is immersed determines the fraction submerged in the fluid.

The density of water${\rho }_w$is${\text{1025 kg/m}}^{\text{3}}$.

The density of iron$\rho$is${\text{7800 kg/m}}^{\text{3}}$.

The fraction of an iron anchor’s weight will be ratio of the buoyant force and the weight of the iron anchor.

$R=\frac{F_B}{W}$

Here, the buoyant force is the weight of the fluid displaced.

The Buoyant force air acting on him is equal to the weight of the fluid which is displaced by that object

$$\begin{gathered} F_B=Weight of fluid displaced \\ =M_{w}g \\ \frac{F_B}{W}=\frac{M_{w}g}{Mg} \\ \frac{F_B}{W}=\frac{M_w}{M} \\ \frac{F_B}{W}=\frac{{\rho }_w{V}_w}{\rho V}\left(Now, V_w=V\right) \end{gathered}$$

The volume of the liquid displaced will be equal to the volume of the anchor .

$$\begin{gathered} \frac{F_B}{W}=\frac{{{\rho }_w}_w}{\rho } \\ \frac{F_B}{W}=\frac{{\rho }_w}{\rho } \\ \frac{F_B}{W}=\frac{1025}{7800} \\ \frac{F_B}{W}=0.13 \end{gathered}$$

Therefore, fraction of an iron anchor’s weight will be ratio of the buoyant force and the weight of the iron anchor is $0.13$ or $13\%$.