Q22P - Fundamentals Of Physics

Question

Gold, which has a density of $\mathbf{19}\mathbf{.}\mathbf{32}\mathbf{ }\mathbf{g}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}$, is the most ductile metal and can be pressed into a thin leaf or drawn out into a long fiber. (a) If a sample of gold, with a mass of 27.63 g, is pressed into a leaf of 1.000 µm thickness, what is the area of the leaf? (b) If, instead, the gold is drawn out into a cylindrical fiber of radius 2.500 µm, what is the length of the fiber?

Step-by-Step Solution

Answer

(a). The area of the leaf is $1.430 {\mathrm{m}}^2$

(b). The length of the fiber is $7.284 \mathrm{x} {10}^4$ 

Step 1: Given data

The density of gold, $19.32 \mathrm{g}/{\mathrm{cm}}^3$

The mass of gold, $\mathrm{m}=27.63 \mathrm{g}$

The thickness of a leaf, $\mathrm{t}=1.000\mathrm{\mu m}$

The radius of fiber,  $\mathrm{r}=2.500 \mathrm{\mu m}$ 

Step 2: Understanding the density of a material

The density of a material (in this case gold) is defined as the mass per unit volume. In this problem, the volume of gold is equal to the volume of gold pressed into a leaf and long fiber. 

The expression for density is given as: 

$\mathrm{p}=\frac{\mathrm{m}}{\mathrm{v}}$                                                                                         … (i)

Here, $\mathrm{p}$ is the density, $\mathrm{m}$ is the mass and $\mathrm{v}$  is the volume.

Step 3: Determination of volume of gold

Using equation (i), the volume of gold is calculated as: 

$$\begin{gathered} V=\frac{m}{p} \\ =\frac{27.63 \mathrm{g}}{19.32\mathrm{g}/{\mathrm{cm}}^3} \\ =1.430 {\mathrm{cm}}^3 \end{gathered}$$

 

Now, convert the volume ^{$1.430 {\mathrm{cm}}^3$}into ${\mathrm{m}}^3$.

 

$1 {\mathrm{cm}}^3 =1\times {10}^{-6} {\mathrm{m}}^3$

 

Therefore,

$$\begin{gathered} V=1.430 {\mathrm{cm}}^3 \times \frac{1\times {10}^{-6} {\mathrm{m}}^3}{1 {\mathrm{cm}}^3} \\ =1.430\times {10}^{-6} {\mathrm{m}}^3 \end{gathered}$$

Step 4: (a) Determination of the area of a leaf

Convert the thickness $1.000 \mathrm{\mu m}$into m.

 

$1.000 \mathrm{\mu m} =1\times {10}^{-6} \mathrm{m}$

 

The expression for the area of the leaf is, 

$A=\frac{V}{t}$

 

Substitute the values in the above expression. 

 

$A=\frac{1.430\times {10}^{-6} {\mathrm{m}}^3}{1\times {10}^{-6} \mathrm{m}}$

 

Thus, the area of the leaf is $1.430 {\mathrm{m}}^2$.

Step 5: (b) Determination of the length of the fiber

The volume of the cylinder is given as: 

$V=A\times L$                                                                                       … (ii)

 

Here, A is the cross section area and L is the length.

 

The cross-section area of cylinder is given as: 

$A=\pi r^2$                                                                                        … (iii)

 

Here, r is the radius of the cylinder.

 

From equation (ii) and (iii), 

$L=\frac{V}{\pi r^2}$                                                                                       … (Iv)

 

Substitute the values of r and V in equation (iv).

$$\begin{gathered} L=\frac{1.430\times {10}^{-6} {\mathrm{m}}^3}{3.142\times {(2.500\times {10}^{-6} \mathrm{m})}^2} \\ =7.284\times {10}^4 \mathrm{m} \end{gathered}$$