Gold, which has a density of 19.32 g/cm^{3}, is the most ductile

a) Using the provided information about the density of gold, the sample size, thickness, and the following
equations and conversion factors, find the area of the gold leaf:
${\displaystyle V=ell\cdot w\cdot h=A\cdot h}$
${\displaystyle m=p\cdot V}$
${\displaystyle 1\mu ={10}^{-6}m}$
${\displaystyle Gol{d}_p=19.32\frac{g}{c}{m}^3}$
First, find the volume of the sample and then find the area of the sample.
$$V=\frac{m}{p}=\frac{27.6g}{19.32g/c{m}^3}\cdot (\begin{array}{c}\frac{0.01m}{1cm}\end{array}{)}^3$$
${\displaystyle \approx 1.429\times {10}^{-6}{m}^3}$
${\displaystyle V=A\cdot h\Rightarrow A=\frac{V}{h}=\frac{1.429\times {10}^{-6}{m}^3}{1x{10}^{-6}m}\approx 1.429m^2}$
b) Using the provided information from part a), the radius of the cylinder, and the following equation for the
volume of a cylinder, find the length of the fiber
${\displaystyle V=\pi r^{2}h\Rightarrow h=\frac{V}{\pi r^2}}$
${\displaystyle h=\frac{1.429\times {10}^{-6}{m}^3}{\pi \cdot {(2.5\times {10}^{-6}m)}^2}\approx 72778m}$

From the equation(1 - 8), the density (p) of the gold is defined as the mass of the gold (m) per its volume (V):
${\displaystyle p=\frac{m}{V}=19.32\frac{g}{c}{m}^3\left(1\right)}$
(a) The volume of the thin leaf is found to be;
$V_{\ell eaf}=(area)(thickness)=A_{\ell eaf}\times t_{\ell eaf}(2)$
From equation (1), the volume of the thin leaf will be;
$V_{\ell eaf}=\frac{m_{\ell eaf}}{p_{\ell eaf}}=\frac{27.63g}{19.32g/c{m}^3}=1.43c{m}^3$
Now, by converting the volume to SI units, we get
$$V_{Leaf}=(1.43c{m}^3)(\begin{array}{c}\frac{1m}{100cm}\end{array}{)}^3=1.43\times {10}^{-6}{m}^3$$
The thickness of the thin leaf: ${\displaystyle t_{elleaf}=1.00\mu m=1\times {10}^{-6}m}$ Therefore, from the equation (2), we can find the area of the thin leaf as following;
${\displaystyle A_{elleaf}=\frac{V_{Leaf}}{t_{Leaf}}=\frac{1.43\times {10}^{{}_6}{m}^3}{1\times {10}^{-6}m}1.43m^2}$
(b) The volume of cylindrical fiber is equal to the volume of the thin leaf (same gold)
${\displaystyle V_{Fiber}=V_{Leaf}}$
As known, the volume of the cylindrical fiber is;
$V_{Fiber}=(cross-sectionalarea)(length)=A\cdot {\ell }_{Fiber}(3)$
This cross-section area is circular and given by: ${\displaystyle A=\pi r^2,wherer=2.5\mu m=2.5\times {10}^{-6}m}$ Thus from equation (3), we have
${\ell }_{Fiber}=\frac{V_{Fiber}}{\pi r^2}=\frac{1.43\times {10}^{-6}{m}^3}{(3.14)(2.5\times {10}^{-6}m{)}^2}=7.286\times {10}^{4}m$
${\displaystyle \therefore el{l}_{Fiber}=7.286\times {10}^{4}m\text{or}72.866km}$