A large water storage container is filled to a depth of $3\mathrm{.}0$ m $.$ The volume above the water is filled with air. In order to pump the water from the container, the air pressure in the container is increased by $3\mathrm{.}0{10}^5\frac{N}{m^2}.$ What happens to the pressure in the water at a depth of $1\mathrm{.}0$ m when the air pressure is increased?

It increases by $2\mathrm{.}0{10}^5\frac{N}{m^2},$ since it is at $\frac{2}{3}$ the depth from the bottom of the container.

It increases by $3\mathrm{.}0{10}^5\frac{N}{m^2}.$

It increases by $3\mathrm{.}0{10}^5\frac{N}{m^2}+9\mathrm{.}8{10}^3\frac{N}{m^2}$ since it is $1\mathrm{.}0$ m deep in the water.

Nothing, it is still at a depth of $1\mathrm{.}0$ m and the pressure in the water depends only on the depth.

It increases by $1\mathrm{.}0{10}^5\frac{N}{m^2},$ since it is at $\frac{1}{3}$ the depth of the water.