91Ó°ÊÓ

When a force (F) is applied to an object, the length of the object gets changed by an amount proportional to the original length (\({l_0}\)) and inversely proportional to the cross-sectional area (A), i.e., \(\Delta l = \frac{1}{E}\frac{F}{A}{l_0}\).

Here, E is the constant of proportionality, termed the elastic modulus or Young’s modulus; the ratio of force per unit area is termed stress; the ratio of change in length to the original length is called strain.

In this problem, the ratio of change in the height of the marble column to the original height of the marble column is strain.

The original height of the marble column is\({h_0} = 8.6\;{\rm{m}}\).

The strain in the marble column is\(3.5 \times {10^{ - 6}}\).

Let \(\Delta h\) be the change in the height of the marble column.

The strain in the marble column is calculated as follows:

\(\begin{array}{c}Strain = \frac{{\Delta h}}{{{h_0}}}\\\Delta h = Strain \times {h_0}\\ = \left( {3.5 \times {{10}^{ - 6}}} \right) \times \left( {8.6\;{\rm{m}}} \right)\\ = 3.0 \times {10^{ - 6}}\;{\rm{m}}\end{array}\)

Thus, the height of the marble column is changed by \(3.0 \times {10^{ - 6}}\;{\rm{m}}\).