$m$ and radius $r$) rolling down an incline without slipping and a point mass $m$ sliding down the same incline is $\frac{5}{7}$.

Let $a$ be the acceleration of the center of mass of the sphere and $g$ be the acceleration due to gravity. Then, using the condition of no slipping, we can write:

$a = \frac{5}{7} g$

The force causing the acceleration of the center of mass of the sphere is the component of the gravitational force along the incline, which is:

$F = mg\sin\theta$

where $\theta$ is the angle of inclination. By Newton's second law, we have:

$F = ma$

For a point mass sliding down the same incline, the force causing the acceleration is simply the component of the gravitational force along the incline, which is also $mg\sin\theta$. Therefore, the acceleration of the point mass is:

$a_p = \frac{F}{m} = g\sin\theta$

The ratio of the accelerations is:

$\frac{a}{a_p} = \frac{5}{7}\cdot\frac{1}{\sin\theta}$

To find the angle $\theta$, we need to use the condition of no slipping, which is:

$a = \frac{5}{7} g = r\alpha$

where $\alpha$ is the angular acceleration of the sphere. The torque causing the angular acceleration is the component of the gravitational torque along the axis of rotation, which is:

$\tau = mgr\sin\theta$

where $m$ is the mass of the sphere. By the rotational analog of Newton's second law, we have:

$\tau = I\alpha$

where $I$ is the moment of inertia of the sphere. For a solid sphere rotating about its diameter, we have:

$I = \frac{2}{5} mr^2$

Substituting for $\tau$ and $I$, we get:

$mgr\sin\theta = \frac{2}{5} mr^2\alpha$

Simplifying, we get:

$\alpha = \frac{5}{2}\frac{g}{r}\sin\theta$

Substituting for $\alpha$ in the condition of no slipping, we get:

$a = \frac{5}{7} g = r\alpha = \frac{5}{2}g\sin\theta$

Solving for $\sin\theta$, we get:

$\sin\theta = \frac{5}{14}$

Substituting in the ratio of accelerations, we get:

$\frac{a}{a_p} = \frac{5}{7}\cdot\frac{1}{\sin\theta} = \frac{5}{7}\cdot\frac{14}{5} = \frac{70}{35} = 2$

Therefore, the ratio of the accelerations is 2.