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 mass130kgthat roams with its head partially exposed. The top head surface has area0.20m2. If the alligator were to swallow stones with a total mass of1.0%of its body mass (a typical amount), how far would it sink?

Short Answer

Expert verified

The depth by which the alligator sinks is6.5×103m6.5×103m

Step by step solution

01

 Step 1: The given data

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

ii Top head surface area of alligator,A=0.20m2

iii) Swallow stones has massof 1.0%mass of alligator

02

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, Fb=ρw×g×v (i)

Density of a material,ρ=mv (ii)

03

Calculation of depth at which alligator swims

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

mA×g=ρw×g×v(Fb=mA×g)mA=ρw×v130=1000vv=0.13m3

Now,themass is increased by1.0%. Then the volume will increase by the same.

So, new increased volume is

V=0.01v+v=(0.13)(1.01)=1.3×103m3

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

Volume=Area×Depth1.3×103=0.20×DepthDepth=6.5×103m

Hence, the alligator sinks at a depth of6.5×103m6.5×103m

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