Two identical hollow spheres of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) are joined together and the combination is rotated about an axis tangential to one sphere and perpendicular to the line connecting their centres. The moment of inertia of the combination is \(\\{\mathrm{A}\\} 10 \mathrm{MR}^{2}\) \(\\{\mathrm{B}\\}(4 / 3) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(32 / 3) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\}(34 / 3) \mathrm{MR}^{2}\)

The moment of inertia \(I\) for a single hollow sphere about an axis through its center can be given as \(2/3 MR^2\).

Since the axis passes through the edge of one of the spheres and is perpendicular to the line joining the centers, for the sphere with the axis passing through it, the moment of inertia stays as \(2/3 MR^2\). For the other sphere, we have to apply the parallel axis theorem. It states that the moment of inertia about any axis parallel to and a distance d away from the axis through the centre of mass of an object is given by \(I+Md^2\), where \(I\) is the moment of inertia about the axis through the center of mass, \(M\) is the object's mass, and \(d\) is the distance from the center of mass to axis. In our case, \(d\) is equal to the diameter of the sphere which is \(2R\), thus for the other sphere, it is \(2/3 MR^2 + M(2R)^2\).

To obtain the total moment of inertia for the entire system, we must sum the moment of inertia of both spheres. So, the total moment of inertia is \(2/3 MR^2 + 2/3 MR^2 + 4MR^2 = 10 MR^2\).

Comparing the result with the given options, we find that the total moment of inertia is equal to option (A) \(10 MR^2\), which is the correct answer.