A charge q is situated at the centre of the open end of a...

According to Gauss's law, the total electric flux through a closed surface enclosing a charge $q$ is $\dfrac{q}{\varepsilon_0}$.

The given cylindrical vessel is open at one end, and the charge $q$ is placed at the centre of this open end.

To apply Gauss's law, we can construct a symmetrical closed surface by imagining an identical inverted cylindrical vessel placed over the open end.

The charge $q$ is now enclosed at the centre of this combined closed cylindrical surface.

The total flux through this closed surface is $\Phi_{\text{total}} = \dfrac{q}{\varepsilon_0}$.

Due to symmetry, the flux through the original cylindrical vessel is exactly half of the total flux.

Therefore, the electric flux through the surface of the vessel is $\Phi = \dfrac{1}{2} \Phi_{\text{total}} = \dfrac{q}{2\varepsilon_0}$.