Find the extra pressure 2.5 km down in the sea, taking the density of water as 1.0 × 10³ kg/m³…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Find the extra pressure $2.5\ \text{km}$ down in the sea, taking the density of water as $1.0 \times 10^{3}\ \text{kg/m}^3$ . By what percentage is the water squeezed there, and what is its new density? (Bulk modulus of water $2.2 \times 10^{9}\ \text{Pa}$ .)

Need a reminder? The key ideaShow

Bulk modulus: $B = -\dfrac{p}{\Delta V/V}$ ; compressibility $k = \dfrac{1}{B}$ . Water: $B = 2.2 \times 10^{9}\ \text{Pa}$ . A gas at a steady temperature has $B = p$ . For a cube, each edge shrinks by a third of the fractional change of volume.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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