In FIGURE CP14.74, a cone of density p₀ and total height l floats in a liquid of density pբ . The height of the cone above the liquid is h. What is the ratio h/l of the exposed height to the total height?

Video transcript

Hi everyone. In this practice problem, we're being asked to determine the ratio of height of the expo surface of an object to the entire height of that particular object represented by H divided by capital H. A cylindrical object made of a material with a density of row is partially submerged in a liquid with a density of sigma. The object has a height of capital H and the portion of the cylinder that remains above the liquid surface has a height of H as shown in the figure we are being asked to determine the ratio of the height of the expo surface H to the entire height of the object. Capital H. And the options given are a H divided by capital H to be equals to one minus row squared, divided by sigma squared B H divided by capital H equals one plus row squared, divided by sigma squared C H divided by capital H equals one minus row divided by sigma and lastly DH divided by capital H equals one plus row divided by stigma. Since the cylindrical object is floating in the static equilibrium in the liquid, the buoyant force F B has to equal the gravitational force F G. In this case, F G, let's look into F G first F G will equals to mass multiplied by gravity. So mass is going to equal to row multiplied by volume multiplied by gravitational acceleration. For F G row is the density of the cylindrical object. V is the volume of the cylindrical object and G is the gravitational acceleration. The volume of the cylindrical object is given by V equals to a multiplied by the height of the overall object which is capital H A is the area of the base and H is the overall height of the cylinder. So A is represented by pi R squared. So pi R squared or multiplied by capital H will be the volume of the cylindrical object. So substituting the volume equation to our F G equation, we will then get F G to be equals to row multiplied by pi R squared capital H multiplied by G. So this is going to be the equation for F G and now we will look into our buan force. So the buoyant force is the gravitational force of the displaced fluid. The volume of the displaced fluid is going to be the volume of the cylindrical object of height capital H minus the volume of part of the cylindrical object above the fluid, namely the volume of the cylindrical object with a height of small H or just H. So the volume of the cylindrical object of height small H is just going to be V equals to pi or fee apostrophe is going to be equals to pi multiplied by R multiplied by small H and be displaced is then going to equals to V, the normal V with the capital H minus V apostrophe. So that will be pi R squared, capital H minus pi R squared small H. So in this case, P displaced by taking the pi R squared out of the equation or in front, we will then get P displaces to be equals to pi R squared multiplied by an open parenthesis capital H minus small H, the buoyant force is given by F B. So F B, you wanna recall that boy force is equals to the um density which in this case, the density of the liquid, which is going to be sigma multiplied by V displaces multiplied by the gravitational acceleration. And in this case, we want to substitute defeat this place equation. So we'll have sigma multiplied by pi R squared multiplied by open parenthesis capital H minus small H closed parentheses multiplied by G. This will be our equation for F B. So now we wanna uh put F B and F G both into the F B equals F G condition for our neutral equilibrium condition. So substituting F G and F B, we will then get F B equals to F G F B is sigma multiplied by pi R squared multiplied by open parentheses, capital H minus H D multiplied by G equal that way, a row multiplied by pi R squared multiplied by capital H multiplied by G. And now I am going to cross out the pi R squared from both sides and also the gravitational acceleration G from both sides as well. And then from there, our equation will then come out to just be row multiplied by open parentheses. Capital H minus H equals to row multiplied by capital H. So in this case, we will want to next remove or put the Sigma onto the other side to make it simple for us. So we will then get capital H minus small H equals to row divided by Sigma multiplied by capital H. We wanna next divide everything in with capital H or put the capital H onto the either side. So then we will have first capital H divided by capital H, we will then get one and then we will have minus H or small H divided by capital H. So that will be a small H divided by capital H to then be equals to row divided by sigma. We wanna keep the H divided by capital H on one side. And I wanna put that in as the positive. So we will then have H divided by capital H to then equals to one minus row divided by Sigma. So that will be the final equation that we will have, which is H divided by capital H to be equals to one minus row divided by Sigma which will correspond to option C in our answer choices. So option C will be the answer to this particular practice problem. And I'll be it for this video. If you guys still have any sort of confusion on this one, please make sure to check out our other lesson videos on similar topics. But other than that, that will be it for this video. Thank you.