If a body has equal amount of rotational kinetic energy and translational kinetic energy while rolling without slipping on a horizontal surface.

Then body is a

Solution:

Let $K_R$ be the rotational kinetic energy and $K_T$ be the translational kinetic energy.

Given that $K_R = K_T$.

We know that:

Rotational kinetic energy $K_R = \frac{1}{2}I\omega^2$

Translational kinetic energy $K_T = \frac{1}{2}Mv^2$

Where $I$ is the moment of inertia, $\omega$ is the angular velocity, $M$ is the mass, and $v$ is the linear velocity of the center of mass.

For rolling without slipping, the relation between linear and angular velocity is $v = R\omega$, or $\omega = \frac{v}{R}$, where $R$ is the radius of the body.

Setting $K_R = K_T$:

$\frac{1}{2}I\omega^2 = \frac{1}{2}Mv^2$

$I\omega^2 = Mv^2$

Substitute $\omega = \frac{v}{R}$ into the equation:

$I\left(\frac{v}{R}\right)^2 = Mv^2$

$I\frac{v^2}{R^2} = Mv^2$

Since $v \neq 0$, we can cancel $v^2$ from both sides:

$\frac{I}{R^2} = M$

$I = MR^2$

Now, let's check the moment of inertia for each given option about an axis passing through its center of mass and perpendicular to the plane of rotation (for ring/disc/cylinder) or through its center (for sphere):

1. Ring: For a ring, the moment of inertia $I_{ring} = MR^2$.

2. Disc: For a disc, the moment of inertia $I_{disc} = \frac{1}{2}MR^2$.

3. Sphere (Solid): For a solid sphere, the moment of inertia $I_{sphere} = \frac{2}{5}MR^2$.

4. Uniform cylinder: A uniform cylinder is essentially a disc if rolling about its central axis. So, $I_{cylinder} = \frac{1}{2}MR^2$.

Comparing the derived condition $I = MR^2$ with the moments of inertia of the given bodies, only the Ring satisfies this condition.

Thus, the body is a Ring.

The final answer is $\boxed{\text{Ring}}$