(a) First of all, let's convert the gold's density into appropriate units:
[tex]d=19.32 g/cm^3 = 19320 kg/m^3[/tex]
and the mass as well:
[tex]m=7.583 g=7.583 \cdot 10^{-3} kg[/tex]
From density and mass, we can find the volume of the leaf, V:
[tex]V= \frac{m}{d}= \frac{7.583 \cdot 10^{-3} kg}{19320 kg/m^3} =3.9\cdot 10^{-7}m^3[/tex]
We know that the thickness is [tex]d=3.061 \mu m=3.061 \cdot 10^{-6} m[/tex], and the volume is the product between the thickness and the area: [tex]V=A d[/tex], so we can find the area:
[tex]A= \frac{V}{d}= \frac{3.9\cdot 10^{-7}m^3}{3.061 \cdot 10^{-6} m} =0.127 m^2[/tex]
(b) The radius of the cylinder is [tex]r=2.5 \mu m=2.5 \cdot 10^{-6} m[/tex], therefore its area is
[tex]A=\pi r^2 = 1.96\cdot 10^{-11} m^2[/tex]
For a cylinder, the volume is the product between the length L and the area A: V=AL, therefore we can find the length L (the volume is the one calculated at the previous step):
[tex]L= \frac{V}{A} = \frac{3.9\cdot 10^{-7}m^3}{1.96\cdot 10^{-11} m^2} =1.99\cdot 10^4 m[/tex]
