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 $$= 1.4 \times 10^{11} Nm^{-2}$$)

We need to find the volume contraction of a solid copper cube under hydraulic pressure.

The bulk modulus $$K$$ is defined as the ratio of volumetric stress to volumetric strain: $$K = -\frac{P}{\Delta V / V} = \frac{P \cdot V}{|\Delta V|}$$ where $$P$$ is the applied pressure, $$V$$ is the original volume, and $$\Delta V$$ is the change in volume.

The negative sign indicates that an increase in pressure causes a decrease in volume (contraction).

Rearranging gives $$|\Delta V| = \frac{P \cdot V}{K}$$.

The cube has edge length 10 cm = 0.1 m, so its original volume is $$V = (0.1)^3 = 10^{-3} \text{ m}^3$$.

With an applied pressure of $$P = 7 \times 10^6$$ Pa and bulk modulus $$K = 1.4 \times 10^{11}$$ N/m$$^2$$, substitution gives $$|\Delta V| = \frac{7 \times 10^6 \times 10^{-3}}{1.4 \times 10^{11}} = \frac{7 \times 10^3}{1.4 \times 10^{11}} = 5 \times 10^{-8} \text{ m}^3$$.

Since $$1 \text{ m} = 10^3 \text{ mm}$$, then $$1 \text{ m}^3 = (10^3)^3 = 10^9 \text{ mm}^3$$ and thus $$|\Delta V| = 5 \times 10^{-8} \times 10^9 = 50 \text{ mm}^3$$.

The negative sign indicates contraction, so the magnitude of volume contraction is 50 mm$$^3$$.

The answer is 50 mm$$^3$$.