A reservoir is filled with liquid of density 2000 kgm⁻¹. Calculate the depth at which the pressure in the liquid will be equal to 9100nm⁻² (g = 10ms⁻²).

The problem provides the following information:

• Density of the liquid, $\rho = 2000 \text{ kgm}^{-3}$ (assuming "kgm-1" is a typo and should be $\text{kgm}^{-3}$ for density).
• Pressure, $P = 9100 \text{ Nm}^{-2}$ (assuming "9100nm" is a typo and should be $\text{Nm}^{-2}$ or Pascals for pressure).
• Acceleration due to gravity, $g = 10 \text{ ms}^{-2}$.

The formula for hydrostatic pressure is:

$$P = \rho g h$$

where $P$ is the pressure, $\rho$ is the density of the liquid, $g$ is the acceleration due to gravity, and $h$ is the depth.

To find the depth, we rearrange the formula:

$$h = \frac{P}{\rho g}$$

Substitute the given values into the rearranged formula:

$$h = \frac{9100 \text{ Nm}^{-2}}{(2000 \text{ kgm}^{-3})(10 \text{ ms}^{-2})}$$ $$h = \frac{9100 \text{ Nm}^{-2}}{20000 \text{ kgm}^{-2}\text{s}^{-2}}$$

Since $1 \text{ N} = 1 \text{ kg} \cdot \text{m} \cdot \text{s}^{-2}$, then $1 \text{ Nm}^{-2} = 1 \text{ kg} \cdot \text{m} \cdot \text{s}^{-2} \cdot \text{m}^{-2} = 1 \text{ kg} \cdot \text{m}^{-1} \cdot \text{s}^{-2}$.
So the units cancel out to meters:

$$h = \frac{9100}{20000} \text{ m}$$ $$h = 0.455 \text{ m}$$

The depth at which the pressure in the liquid will be $9100 \text{ Nm}^{-2}$ is $\boxed{\text{0.455 m}}$.