D

To determine the smallest horizontal velocity \( v \) required for the disc to leave the hemisphere without sliding down, we can use the concept of centripetal acceleration and forces acting on the disc.

1. **Understanding the Forces:** When the disc is at the top of the hemisphere, the forces acting on it are the gravitational force \( mg \) acting downwards and the normal force \( N \) from the hemisphere acting perpendicular to the surface. For the disc to leave the surface, the normal force must become zero.

2. **Centripetal Force Requirement:** At the point just before the disc loses contact, the only force providing the required centripetal force is the weight component acting towards the center of the hemisphere. For a small angle \( \theta \) at the top, the radial (centripetal) acceleration \( a_c \) is given by \( \frac{v^2}{R} \), where \( R \) is the radius of the hemisphere.

3. **Setting Up the Equation:** Therefore, at the top: \( mg = \frac{mv^2}{R} \). As the mass cancels out, we get:
\[ g = \frac{v^2}{R} \]
or, rearranging gives:
\[ v^2 = gR \]
and taking the square root:
\[ v = \sqrt{gR} \]

4. **Final Calculation:** The disc must have a velocity that provides a centripetal acceleration equal to the gravitational force acting downwards when it is at the top of the hemisphere. Hence, the required minimum horizontal velocity is given by the equation derived above.

Therefore, Option D, which is \( v = \sqrt{gR} \), is the correct answer.