(II) How much pressure is needed to compress the volume of an iron block by 0.10%? Express your answer in \({\bf{N/}}{{\bf{m}}^{\bf{2}}}\), and compare it to atmospheric pressure \(\left( {{\bf{1}}{\bf{.0 \times 1}}{{\bf{0}}^{\bf{5}}}\;{\bf{N/}}{{\bf{m}}^{\bf{2}}}} \right)\).

When pressure acts on an object from all sides, its volume decreases by an amount directly proportional to the original volume (\({V_0}\)) and the change in the pressure\(\left( {\Delta P} \right)\), i.e., \(\Delta V = - \frac{1}{B}{V_0}\Delta P\).

Here, B is the constant of proportionality, termed the bulk modulus, and the minus sign indicates that the volume decreases on increasing pressure.

In this problem, the percentage change in volume of the iron block is directly proportional to the pressure needed to compress it. The value of bulk modulus of the iron is\(B{\bf{ = 90 \times 1}}{{\bf{0}}^{\bf{9}}}\;{\bf{N/}}{{\bf{m}}^{\bf{2}}}\).

If\({V_0}\)is the original volume of the iron block, and\(\Delta V\)is the volume change, the percentage chain in the volume of the block is as follows:

\(\begin{array}{c}\frac{{\Delta V}}{{{V_0}}} \times 100 = 0.10\\\frac{{\Delta V}}{{{V_0}}} = 0.10 \times {10^{ - 2}}\end{array}\)

The value of atmospheric pressure is\({P_{\rm{a}}} = 1.0 \times {10^5}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}\).

Let\(\Delta P\)be the pressure needed to change the volume of the iron block by\(\Delta V\).

The change in volume of the iron block when pressure\(\Delta P\)is applied on it is calculated as follows:

\(\begin{array}{c}\Delta V = - \frac{1}{B}{V_0}\Delta P\\\Delta P = - B\frac{{\Delta V}}{{{V_0}}}\\ = - \left( {90 \times {{10}^9}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}} \right) \times \left( {0.10 \times {{10}^{ - 2}}} \right)\\ = - 9.0 \times {10^7}\;{\rm{N/}}{{\rm{m}}^2}\end{array}\)

Thus, the pressure of \(9.0 \times {10^7}\;{\rm{N/}}{{\rm{m}}^2}\) is needed to compress the volume of the iron block by 0.10%. Here, the negative sign indicates the volume is compressed.

The ratio of pressureneeded to compress the volume of the iron block by 0.10% and the atmospheric pressure is as follows:

\(\begin{array}{c}\frac{{\Delta P}}{{{P_{\rm{a}}}}} = \frac{{9.0 \times {{10}^7}\;{\rm{N/}}{{\rm{m}}^2}}}{{1.0 \times {{10}^5}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}}}\\ = 9.0 \times {10^2}\\ = 900\end{array}\)

Thus, you can say that the pressure needed to compress the volume of an iron block by 0.10% is 900 times the atmospheric pressure.