C

Step 1: Consider a small body of mass \( m \) sliding down from the top of a hemispherical surface of radius \( r \). At any point of the body sliding down, we can analyze the forces acting on it. The gravitational force acting on the body is \( mg \) (downward) and provides the centripetal force necessary for it to travel in a circular path.

Step 2: The height \( h \) above the ground at which the block loses contact with the hemisphere can be expressed in terms of the angle \( \theta \) that the radius makes with the vertical line. We can express the height as: \( h = r - r \cos(\theta) \).

Step 3: At the point of losing contact, the normal force becomes zero. The centripetal force needed to keep the block in circular motion, \( \frac{mv^2}{r} \), will then equal the component of gravitational force acting towards the center, which is \( mg \cos(\theta) \).

Step 4: Therefore, we can equate these forces: \( \frac{mv^2}{r} = mg \cos(\theta) \).

Step 5: The potential energy at the top converts into kinetic energy and potential energy at height \( h \): \( mgh = \frac{1}{2} mv^2 \) gives us \( gh = \frac{v^2}{2} \).

Step 6: Substituting the value of v from Step 4, we get \( g(r - h) = \frac{g (r \cos(\theta))}{2} \).

Step 7: Solving for \( h \) gives us \( h = \frac{r}{2} \).

Therefore, the height at which the body loses contact with the surface of the sphere is \( h = \frac{r}{2} \). Thus, the answer is option C.