The average density of the body of a fish is 1080 kg/m³ . To keep...

Question

The average density of the body of a fish is 1080 kg/m³ . To keep from sinking, a fish increases its volume by inflating an internal air bladder, known as a swim bladder, with air. By what percent must the fish increase its volume to be neutrally buoyant in fresh water? The density of air at 20°C is 119 kg/m³.

Accepted Answer

Hi, everyone in this practice problem, we are being asked to determine a percentage of increase of a toy's volume to achieve neutral buoyancy in water. We will have a specially designed underwater toy with an average density of 1210 kg per meter cube which causes the toy to sink in the water to prevent it from sinking. The toy has a built-in inflatable chamber that fills with air and when inflated, the chamber will contain air at a density of 1.18 kg per meter cube at 25 degrees Celsius. We're being asked to determine the percentage of increase of the toys volume to achieve neutral buoyancy in the water. The options given are a 50% b 14% C 21% and lastly D 38%. So to achieve neutral buoyancy in the water, the force due to the gravity for from the toy or F G must be equal to the buoyant force or F B. So F G must be equals to F B for neutral buoyancy. So the force due to gravity of the toy or F G is given by row of the toy multiplied by the volume of the toy to get the mass of the toy multiplied by the gravitational acceleration. And since we have an inflatable chamber, we want to multiply the, uh we wanna plus that with the row of the air multiply by the volume of the inflatable chamber or V I multiply that as well with the gravitational acceleration. So row T is the density of the toy V T is the volume of the toy row is the density of the air V I is the volume of the inflated chamber chamber of the toy. And G is the gravitational acceleration. On the other hand, the buoyant force which is the force of gravity on the volume of water displaced by the toy is given by F B which is equals to row of the water multiplied by the volume of the toy submerge multiplied by G plus row of the water multiplied by the volume of the inflatable multiplied by she, we can simplify the F B by um pulling row W and G out. So we will get the equation for F B to equals to row W multiplied by G multiplied by open parentheses V D plus V I. So then we wanna use the neutral buoyancy um condition and substitute the equation for F B and F G into the condition in order for us to uh finally derive an equation for V I which is the uh inflatable chamber volume. So what we wanna do is we wanna use the, the condition of F G equals to F B and we have our F G two B row T multiplied by V T multiplied by G plus row, multiplied by V I multiplied by G equals to row W multiplied by G multiplied by open parentheses. V T plus V I close parentheses. So I am going to cross out all the GS because we have GS in all terms here. So we can neglect the gravitational acceleration here. So next, the equation that we have is then going to be row T multiplied by V T plus row, multiplied by V I equals row W, I'm gonna distribute this multiplied by V T plus row. W multiplied by V I, I'm going to lump up the terms with V I together and V T. So we will have row T multiplied by V T minus row, W multiplied by V T equals to row W multiplied by V I minus row multiplied by V I. I am going to pull both V T and V I out. So what we will have is then row T minus row, W in parenthesis multiplied by V T will equals to open parenthesis row W minus row, close parentheses multiplied by V I. As I said, we are interested to find the equation for V I. So then V I will equals to open parenthesis row T minus row, W multiplied by V T all of that divided by open parenthesis row W minus row er just like so, so we can actually then substitute all of our information that is given in the problem statement into this equation right here to then solve for V I in terms of BT. So let's go and do exactly that. We will have V I to then equals two row T is the row of the toy, which is given to be 1000 210 kg per meter cube minus the row of water, which is we know to be 1000 kg per meter cube multiply that by V T divide that we open parenthesis row of water, which is going to be 1000 kg divided by a meter cube minus row of air which is 1. kg per meter cube just like. So, so now um calculating of this, we can get V I to then equals two uh 210 divided by 998.82 multiplied by V T which will come out to a value to 0.21 multiplied by V T. So in this case, we know that V I has to be 0.21 times V T in order for it to achieve neutral buoyancy. So the increase in the toys volume has to be to then solve for V I in terms of V T. So let's go and do exactly that we will have V I to then equals to row T is the row of the toy which is given to be 1000 210 kg per meter cube minus the row of water, which is we know to be 1000 kg per meter cube multiply that by V T divide that with open parenthesis row of water, which is going to be 1000 kg divided by a meter cube minus row of air which is 1. kg per meter cube just like. So, so now um calculating of this, we can get V I to then equals two uh 210 divided by 998.82 multiplied by V T which will come out to a value to 0.21 multiplied by V T. So in this case, we know that V I has to be 0.21 times V T in order for it to achieve neutral buoyancy. So the increase in the toys volume has to be V I equals.

Key Concepts

Buoyancy

Density

Volume Increase for Neutral Buoyancy

Watch next