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A hemisphere is half of the sphere cut by the plane passing through its center. The volume of the hemisphere is half of that of the sphere.

A hemisphere with radius r has a volume given as follows:

\[{V_H} = \dfrac{2}{3}\pi {r^3}...........(1)\]

A cone is a right angle triangle that is rotated with any one of its perpendicular sides as the axis in 3-dimensional space. The volume of the cone is one-third of the volume of the cylinder.

The volume of a cone of radius r and height h is given as follows:

\[{V_C} = \dfrac{1}{3}\pi {r^2}h...........(2)\]

The solid in the given question is composed of a hemisphere and a cone. We can calculate its volume by calculating separately the volumes of cones and hemisphere and adding them.

The volume of the hemisphere of radius 1 cm is given by equation (1) as follows:

\[{V_H} = \dfrac{2}{3}\pi {(1)^3}\]

\[{V_H} = \dfrac{2}{3}\pi c{m^3}\]

The volume of the cone of radius 1 cm and height 1 cm is given by equation (2) as follows:

\[{V_C} = \dfrac{1}{3}\pi {(1)^2}(1)\]

\[{V_C} = \dfrac{1}{3}\pi c{m^3}\]

Adding the two volumes, we have:

\[V = {V_H} + {V_C}\]

\[V = \dfrac{2}{3}\pi + \dfrac{1}{3}\pi \]

\[V = \pi c{m^3}\]

Hence, the volume of the solid is \[\pi \] cubic cm.