A body is rolling without slipping on a horizontal surface and its rotational kinetic energy is equal to the translational kinetic energy. The body is
A. Disc
B. Sphere
C. Cylinder
D. Ring

Hint:We will apply the formulas of the translational kinetic energy and the rotational kinetic energy and equate them. Also, since the body is rolling without slipping it follows the relation $v = \omega r$ . We will be able to find the moment of inertia of the moving body which will be compared with the moment of inertia of the bodies given in the options. From the known values of moment of inertia, we will get the answer.

Formula used:
The translational kinetic energy is given by ${K_t} = \dfrac{1}{2}m{v^2}$ whereas the rotational kinetic energy is given by ${K_r} = \dfrac{1}{2}I{\omega ^2}$ where I is the moment of inertia of the body, $\omega $ is the angular velocity and v is the linear velocity.
When the object is rolling without slipping, it follows the relation $v = \omega r$ .

Complete step by step answer:
The translational kinetic energy is given by ${K_t} = \dfrac{1}{2}m{v^2}$ and the rotational kinetic energy is given by ${K_r} = \dfrac{1}{2}I{\omega ^2}$ .
Equating both we get,
$\dfrac{1}{2}I{\omega ^2} = \dfrac{1}{2}m{v^2}$
We also know that the object is rolling without slipping. Hence, it follows the relation $v = \omega r$ .
Substituting in the equation, we get
$\dfrac{1}{2}I{\omega ^2} = \dfrac{1}{2}m{(\omega r)^2}$
Further solving this equation, we get
$I = m{r^2}$

Moment of inertia of the given objects in options are given below


From the table we can clearly see that moment of inertia of the ring matches with the calculated moment of inertia. Thus, the object is a ring.

Hence, D is the correct option.

Note:The moment of inertia given above is along the axis of rotation. For different axes, we have different values of moment of inertia for a single object. In case an extended or a truncated object is given, then we have to apply the basics by calculating the moment of inertia at an infinitesimally small distance and then integrate it over the length of the object.