Question

The cubit is an ancient unit of length based on the distance between that elbow and the tip of the middle finger of the measurer. Assume that the distance ranged from 43 to 53 cm, and suppose that ancient drawings indicate that a cylindrical pillar was to have a length of 9 cubits and a diameter of 2 cubits. For the state range, what are the lower value and the upper value, respectively for the cylinder's volume in cubic meters?

Answer 1

Hint: We can do the question either by converting the given parameters first and then finding the volume, or by initially finding the volume and then performing the conversion. Here, we are given the length and the diameter. Therefore, we follow the latter approach.

Complete step by step solution:
This problem gives us the maximum and minimum values for the length of a cubit. We just need to convert the units from centimeters to meters, write our equation, and plug in the max and min.
We’re given that h is 9 cubits. Hence we multiply it with the minimum and maximum length of a cubit i.e. ${{h}_{\min }}=9(43cm)(\dfrac{{{10}^{-2}}cm}{1cm})=3.87m$
Similarly, ${{h}_{\max }}=9(53cm)(\dfrac{{{10}^{-2}}cm}{1cm})=4.77m$
The column is between 3.9m and 4.8 m
We know that the formula to find the volume of a cylinder is
$V=\pi {{r}^{2}}h$, where $h$ is the length or the height of the cylinder and $r$ is the radius of the cylinder.
We're given the diameter, but we can use $\dfrac{d}{2} = r = 1.0$ cubit.
Substituting, we get:
$
\Rightarrow {{V}_{\min }}=\pi {{(0.43)}^{2}}(3.87)=2.248{{m}^{2}} \\
\Rightarrow {{V}_{\max }}=\pi {{(0.53)}^{2}}(4.77)=4.209{{m}^{2}} \\
$
Hence the value of ${{V}_{\min }}$ is $2.248{{m}^{2}}$ and that of ${{V}_{\max }}$ is $4.209{{m}^{2}}$.

Note: To convert between absolute uncertainty and percent uncertainty, we’ll use this formula (m = measurement, $\Delta u$ = absolute uncertainty):
$m\pm \Delta u=m\pm (\dfrac{\Delta u}{m}\times 100%)$)