Question

The radius of the base of a circular cylindrical holding tank is 8 times the radius of the base of a circular cylindrical pipe that empties into it. If they have the same length, say L feet, the volume of the tank is how many times the volume of the pipe?

Answer 1

We will determine the times the volume of the cylinder is bigger as follows:

*First: We determine the volume of the pipe:

`V_p=\pi r^2h`

*Second: We determine the volume of the holding tank:

`V_t=\pi(8r)^2h`

*Third: Since the length(h) will be the same for both cylinders, we will assing an arbitrary value for it [When we assign this value no matter what number it might be, except of course 0, the result will be the same]. In our case, to make it "easier" or more intuitive we will work with h = 1, so, we would have:

`V_(p1)=\pi r^2`

&

`V_(t1)=64\pi r^2`

*Fourth: We determine the scale factor:

`\pi r^2x=64\pi r^2\Rightarrow x=(64\pi r^2)/(\pi r^2)\Rightarrow x=64`

So, the volume of the holding tank is 64 times greater than the volume of the tube. [Option 4]