Chapter 9, Problem 68QAP

To determine

(a)

The increase volume of diamond once it is removed from the chamber.

Answer to Problem 68QAP

The increase volume of diamond once it is removed from the chamber is $2.1\times {10}^{-8}{m}^3$.

Explanation of Solution

Given:

Initial Pressure will act like volume stress $P=\frac{F}{A}=58\times {10}^{9}N/m^2$

Bulk modulus of diamond $B=4.43\times {10}^{11}N/m^2$

Let $\Delta V$ be the increase volume

Density of diamond $\rho =3.52g/c{m}^3$

One carat mass $m=0.200g$

Volume of 2.8-carat diamond $V=\frac{2.8\times m}{\rho }=\frac{2.8\times 0.200g}{3.52g/c{m}^3}=0.16c{m}^3=0.16\times {10}^{-6}{m}^3$

Formula used:

Bulk Modulus $B=\frac{volumestress}{volumestrain}=\frac{F/A}{\Delta V/V}$

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in the formula,

  $\begin{array}{l}B=\frac{F/A}{\Delta V/V}\\ or,\Delta V=\frac{F/A}{B/V}\\ or,\Delta V=\frac{58\times {10}^{9}N/m^2}{\left(4.43\times {10}^{11}N/m^2\right)}\times \left(0.16\times {10}^{-6}{m}^3\right)\\ or,\Delta V=2.1\times {10}^{-8}{m}^3\end{array}$

Hence, the increase volume of diamond once it is removed from the chamber is $2.1\times {10}^{-8}{m}^3$.

Conclusion:

Thus, the increase volume of diamond once it is removed from the chamber is $2.1\times {10}^{-8}{m}^3$.

To determine

(b)

The increase diamond radius.

Answer to Problem 68QAP

The increase diamond radius is $1.71mm$.

Explanation of Solution

Given:

The increase volume of diamond once it is removed from the chamber is $2.1\times {10}^{-8}{m}^3$.

Let $\Delta r$be the increase radius of spherical diamond ball.

Formula used:

  $Increasevolumeofsphericalball\left(\Delta V\right)=\frac{4}{3}\pi {\left(\Delta r\right)}^3$

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in the formula,

  $\begin{array}{l}\left(\Delta V\right)=\frac{4}{3}\pi {\left(\Delta r\right)}^3\\ or,\Delta r=\sqrt[3]{\frac{3\left(\Delta V\right)}{4\times \pi }}\end{array}$

  $\begin{array}{l}or,\Delta r=\sqrt[3]{\frac{3\times \left(2.1\times {10}^{-8}{m}^3\right)}{4\times \pi }}\\ or,\Delta r=1.71\times {10}^{-3}m=1.71mm\end{array}$

Hence, the increase diamond radius is $1.71mm$.

Conclusion:

Thus, the increase diamond radius is $1.71mm$.