Question

An alligator waits for prey by floating with only the top of its head exposed, so that the prey cannot easily see it. One way it can adjust the extent of sinking is by controlling the size of its lungs. Another way may be by swallowing stones (gastrolithes) that then reside in the stomach. Figure shows a highly simplified model (a “rhombohedron gater”) of mass$\mathbf{130}\mathbf{}\mathit{k}\mathit{g}$that roams with its head partially exposed. The top head surface has area$\mathbf{0}\mathbf{.20}\mathbf{}{\mathit{m}}^{\mathbf{2}}$. If the alligator were to swallow stones with a total mass of$\mathbf{1}\mathbf{.0}\mathit{\%}$of its body mass (a typical amount), how far would it sink?

Short answer

The depth by which the alligator sinks is$6.5\times {10}^{-3}\text{m}$$6.5\times {10}^{-3}\text{m}$

Step-by-step solution

 Step 1: The given data

I)i) Mass of alligator,$m=130kg$

ii Top head surface area of alligator,$A=0.20m^2$

iii) Swallow stones has massof $1.0\text{\%}$mass of alligator

Understanding the concept of Archimedes Principle

Using Archimedes' principle, we can say that the weight of the body that sinks or is submerged in a liquid is equal to the weight of that liquid, which is displaced. Also, the mass and volume are directly proportional to each other.

Formula:

Buoyant force exerted by fluid on a body, ${\text{F}}_{\text{b}}={\text{ρ}}_{\text{w}}\times \text{g}\times \text{v}$ (i)

Density of a material,$\text{ρ}=\frac{\text{m}}{\text{v}}$ (ii)

Calculation of depth at which alligator swims

Using equation (i) and the given values, we get

$\begin{array}{c}{m}_A\times g={\rho }_w\times g\times v\text{(}\because {\text{F}}_b{\text{=m}}_A\times \text{g)}\\ m_A={\rho }_w\times v\\ 130=1000v\\ v=0.13m^3\end{array}$

Now,themass is increased by$1.0\text{\%}$. Then the volume will increase by the same.

So, new increased volume is

$\begin{array}{c}\text{V}=0.01\text{v+v}\\ =\left(0.13\right)\left(1.01\right)\\ =1.3\times {10}^{-3}{m}^3\end{array}$

Now, fromthetop head surface area and above volume, we can find the distance/depth by whichthealligator sinks:

$\begin{array}{c}Volume=Area\times Depth\\ 1.3\times {10}^{-3}=0.20\times Depth\\ Depth=6.5\times {10}^{-3}m\end{array}$

Hence, the alligator sinks at a depth of$6.5\times {10}^{-3}m$$6.5\times {10}^{-3}m$