A cube of metal is subjected to a hydrostatic pressure of $4 \,GPa$. The percentage change in the length of the side of the cube is close to (Given bulk modulus of metal, $\left. B =8 \times 10^{10} Pa \right)$

Question

The volume contraction of a solid copper cube of edge length 10 cm, when subjected to a hydraulic pressure of $ 7 \times 10^6 $ Pa, would be ________________________ mm$^3$. (Given bulk modulus of copper = $ 1.4 \times 10^{11} $ N m$^{-2}$)

Answer 1

To solve the problem of determining the percentage change in the length of the side of a cube when subjected to a hydrostatic pressure, we use the concept of the bulk modulus.

The bulk modulus $ B $ is defined as the ratio of the change in pressure $ \Delta P $ to the relative change in volume ($\frac{\Delta V}{V}$):

B = -\frac{\Delta P}{\frac{\Delta V}{V}}

Given:

Bulk modulus, B = 8 \times 10^{10} \, \text{Pa}.
Pressure applied, \Delta P = 4 \, \text{GPa} = 4 \times 10^9 \, \text{Pa}.

We need to find the percentage change in the side length of the cube. For small deformations, the volume change relation for a cube is:

\frac{\Delta V}{V} \approx 3 \frac{\Delta L}{L}

Where:

\Delta L\) is the change in the side length.
L\) is the original side length.

Rearrange the formula for bulk modulus to find the relative change in volume:

\frac{\Delta V}{V} = -\frac{\Delta P}{B}

Substituting the given values:

\frac{\Delta V}{V} = -\frac{4 \times 10^9}{8 \times 10^{10}} = -\frac{1}{20}

This implies:

3 \frac{\Delta L}{L} = -\frac{1}{20}

Therefore, the change in side length is:

\frac{\Delta L}{L} = -\frac{1}{60}

The percentage change in length, i.e., the absolute value of the relative change multiplied by 100, is:

\left|\frac{\Delta L}{L} \right| \times 100 = \frac{1}{60} \times 100 \approx 1.67\%

Thus, the percentage change in the length of the cube's side is approximately 1.67%. Therefore, the correct answer is 1.67.

Answer 2