Chapter 9, Problem 44P

To determine

Pressure needed to compress the iron block by 0.10% volume. Compare it with atmospheric pressure.

Answer to Problem 44P

Solution: $\text{9 }\times \text{ 1}{0}^{\text{7}}\text{N}/{\text{m}}^{\text{2}}$ of pressure is required to compress the volume iron block by $0.10\%$ .

The pressure required for compress the iron block is 900 times greater than atmospheric pressure.

Explanation of Solution

Given: Volume change is 0.10% which is $0.001$

The Bulk Modulus value for steel is $K=90\times {10}^9{\text{N/m}}^{\text{2}}$ .

Formula Used:

The Bulk modulus of elasticity is defined as ratio bulk stress to bulk strain, which is given by

  $K=\frac{dP}{dV}V$

Where K is bulk modulus, V is the volume $\text{dV}$ is the change in volume, $\text{dP}$ is change in pressure.

Calculations: Substitute all the values in above equation and solve for change in pressure

  $\begin{array}{l}K=\frac{dP}{dV}V\\ dP=\frac{K}{\frac{V}{dV}}\\ dP=\frac{90\times 10^9}{100/0.1}\\ dP=9\times 10^{7}N/m^2\end{array}$

  $\text{9 }\times \text{1}{0}^{\text{7}}\text{N}/{\text{m}}^{\text{2}}$ of pressure is required to compress the volume iron block by 0.10 %.

Comparing this pressure with atmospheric pressure

  $\frac{Pressure}{Amosphericpressure}=\frac{9\times 10^7}{1\times 10^5}=900$

The pressure required for compress the iron block is 900 times greater than atmospheric pressure.

Conclusion: $\text{9 }\times \text{1}{0}^{\text{7}}\text{N}/{\text{m}}^{\text{2}}$ of pressure is required to compress the volume iron block by 0.10 %

The pressure required for compress the iron block is 900 times greater than atmospheric pressure.