When a stress (which is pressure) is applied to a solid, it will deform slightly. The resulting deformation is called a strain. There are several ways of stressing a solid. In general, stresses and strains are related through an elastic modulus E, as long as we remain within the elastic limit of the solid. The general formula is:
$$
E=\frac{\text { stress }}{\text { strain }}
$$
The stress is always an applied pressure, and the strain is always a fractional change in some dimension, such as length or volume.
If pressure P is applied equally to all sides of a solid, it changes its volume by amount $\Delta \mathrm{V}$. The pressure and resulting change in volume are related through the equation:
$$
\mathrm{B}=-\frac{\mathrm{P}}{\Delta \mathrm{~V} / \mathrm{V}}
$$
where V is the original volume of the solid, and B is a constant called the bulk modulus that depends on the material making up the solid.
As a second example of stress and strain, consider a cylindrical wire of length L and cross-sectional area A , attached to a ceiling with one end dangling. If a force F pulls the wire downward, perpendicular to A , it will stretch the wire by an amount $\Delta \mathrm{L}$. The amount of stretch is related to the amount of applied force through:
$$
\mathrm{Y}=\frac{\mathrm{F} / \mathrm{A}}{\Delta \mathrm{~L} / \mathrm{L}},
$$
where Y is called Young's modulus and is a constant that depends on material. Another important elastic modulus is called the shear modulus, S . We will not define S here; but if the solid is isotropic, $\mathrm{Y}, \mathrm{B}$, and S are related through:
$$
S=\frac{Y}{2(1+\sigma)} \quad B=\frac{Y}{3(1-2 \sigma)}
$$
where $\sigma$ is a dimensionless constant that depends on the material.
Useful information:
$$
\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2, \rho_{\text {water }}=1000 \mathrm{~kg} / \mathrm{m}^3, \mathrm{~B}_{\mathrm{Al}}=7.5 \times 10^{10} \mathrm{~Pa}
$$
When a force $F$ is applied to an aluminum wire of length L and cross-sectional area A , it stretches by $\Delta \mathrm{L}$. If the same force is applied to an aluminum wire of length L and cross-sectional area 2 A , it stretches by:
A. $\frac{\Delta \mathrm{L}}{4}$
B. $\frac{\Delta \mathrm{L}}{2}$
C. $2 \Delta \mathrm{~L}$
D. $4 \Delta \mathrm{~L}$