Density of the liquid, \(\rho = 1150 \texttt{ kg m}^{-3}\)

Depth, \(d = 100 \texttt{ m}\)

Formula:

Pressure, \(P = \rho g d\)

Where, \(g\) is the acceleration due to gravity, which is \(9.8 \texttt{ m s}^{-2}\).

Calculation:

Substituting the values, we get:

$$P = (1150 \texttt{ kg m}^{-3})(9.8 \texttt{ m s}^{-2})(100 \texttt{ m})$$

$$P = 1.13 \times 10^6 \texttt{ Pa}$$

Answer:

The pressure 100 m below the surface is \(1.13 \times 10^6 \texttt{ Pa}\).