A cooking pot has a spherical bottom, while the upper part is a truncated cone. Its vertical cross-section is shown in the figure.If the volume of food increases by 15% during cooking, what is the maximum initial volume of food that can be cooked without spoiling ?
It is clear that we have to evaluate the volume of the cooking pot, first. One important observation that I had made was that that whole cooking pot is a part of the spherical sector(of the sphere of which the bottom is a part). So if we can evaluate the volume of the spherical sector and subtract the volume of the truncated cone problem will be solved. However I am unable to evaluate the volume of the spherical sector.
Any other method of solving the sum will also be accepted.
As a twelfth standard student, a well-explained solution (preferably with diagram) will be necessary for my clear understanding.
So first consider the bottom spherical part. We know that the distance in the y (vertical) direction from the top of the spherical part to the bottom is 20cm and that the distance across is 40. It should be easy to see that this implies that the radius is 20cm and that we are dealing with a half-sphere. Thus, we can use the formula for the the volume of a sphere $\frac{4}{3}*\pi*r^3$ and divide by 2 where r = 20cm. Next we need the conical part. We can think of this as two cones, one is missing however, which is the top part. So, we can calculate the big cone and subtract the small cone to get the almost-cone's volume. We can use the standard formulae here too. $V=\pi*r^2*h$ So we have a height of 20 (explained later) and r =20. Finally we need the small cone, which has a radius of merely 10. But we don't have the height? However, we do have that the cone is the same cone as before right? So the slope of this missing cone must be the same. So it must have half the height (of the total) because it has half the radius. So we have $V=\pi*r^2*h$ with r = 10 and h =10. As a result we have $\frac{4}{6}*\pi*20^3 + \pi*20^2*20 - \pi*10^2*10$