The following bodies are made to roll up (without slipping) the same inclined plane from a horizontal plane: (i) a ring of radius R, (ii) a solid cylinder of radius $$\frac{R}{2}$$ and (iii) a solid sphere of radius $$\frac{R}{4}$$. If, in each case, the speed of the center of mass at the bottom of the incline is same, the ratio of the maximum heights they climb is:

We have three different rigid bodies that roll without slipping and all start the climb with the same linear speed $$v$$ of their centres of mass at the foot of the incline. While climbing, every joule of the initial kinetic energy changes into gravitational potential energy until the bodies momentarily come to rest at their respective highest points. Hence we can write, for any one of them,

$$\text{Initial kinetic energy}= \text{Gain in potential energy}.$$

The total kinetic energy of a rolling body is the sum of its translational part and its rotational part. Stating the formula first,

$$K_{\text{total}}=\frac12\,m\,v^{2}\;+\;\frac12\,I\,\omega^{2},$$

where $$m$$ is the mass, $$I$$ is the moment of inertia about its own axis, $$v$$ is the linear speed of the centre of mass and $$\omega$$ is the angular speed. For rolling without slipping the well-known kinematic relation

$$v=\omega\,r$$

holds, $$r$$ being the radius of the body that is in contact with the plane.

At the highest point the linear speed becomes zero, so the entire kinetic energy converts into gravitational potential energy $$mgh$$. Thus we write

$$mgh=\frac12\,m\,v^{2}\;+\;\frac12\,I\,\omega^{2}.$$

Substituting $$\omega=\dfrac{v}{r}$$ gives

$$mgh=\frac12\,m\,v^{2}+\frac12\,I\left(\frac{v}{r}\right)^{2}=\frac{v^{2}}{2}\left[m+\frac{I}{r^{2}}\right].$$

Dividing both sides by $$mg$$ we arrive at the general expression for the maximum height climbed:

$$h=\frac{v^{2}}{2g}\left[1+\frac{I}{m\,r^{2}}\right].$$

The factor $$\dfrac{I}{m\,r^{2}}$$ is specific to the shape of the body but, importantly, does not depend on the numerical value of the radius itself. We now evaluate this factor for each body.

For the ring (radius $$R$$) we know the standard moment of inertia $$I_{\text{ring}}=m\,r^{2}.$$ Hence

$$\frac{I}{m\,r^{2}}=\frac{m\,r^{2}}{m\,r^{2}}=1,$$

so that

$$h_{\text{ring}}=\frac{v^{2}}{2g}\left(1+1\right)=\frac{v^{2}}{2g}\times2.$$

For the solid cylinder (radius $$\dfrac{R}{2}$$) the standard moment of inertia is $$I_{\text{cyl}}=\dfrac12\,m\,r^{2}.$$ Therefore

$$\frac{I}{m\,r^{2}}=\frac{\dfrac12\,m\,r^{2}}{m\,r^{2}}=\frac12,$$

and

$$h_{\text{cyl}}=\frac{v^{2}}{2g}\left(1+\frac12\right)=\frac{v^{2}}{2g}\times\frac32.$$

For the solid sphere (radius $$\dfrac{R}{4}$$) the standard moment of inertia is $$I_{\text{sphere}}=\dfrac25\,m\,r^{2}.$$ Thus

$$\frac{I}{m\,r^{2}}=\frac{\dfrac25\,m\,r^{2}}{m\,r^{2}}=\frac25,$$

and

$$h_{\text{sphere}}=\frac{v^{2}}{2g}\left(1+\frac25\right)=\frac{v^{2}}{2g}\times\frac75.$$

We now compare the three heights. The common prefactor $$\dfrac{v^{2}}{2g}$$ is the same for every body, so the ratio of maximum heights is governed solely by the parenthetic factors:

$$h_{\text{ring}}:h_{\text{cyl}}:h_{\text{sphere}}=2:\frac32:\frac75.$$

To clear the fractions, multiply each term by $$10$$:

$$2\times10=20,\qquad\frac32\times10=15,\qquad\frac75\times10=14.$$

Therefore

$$h_{\text{ring}}:h_{\text{cyl}}:h_{\text{sphere}}=20:15:14.$$

Hence, the correct answer is Option B.