A solid sphere and solid cylinder of identical radii approach an incline with the same linear velocity. Both roll without slipping all throughout. The two climbs maximum heights \[{{h}_{sph}}\] and \[{{h}_{cyl}}\] on the incline. The ratio \[\dfrac{{{h}_{sph}}}{{{h}_{cyl}}}\] is given by:-

\[A.\,\dfrac{14}{15}\]
\[B.\,\dfrac{4}{5}\]
\[C.\,1\]
\[D.\,\dfrac{2}{\sqrt{5}}\]

Hint: This question is based on the combination of the concepts of the moment of inertia, the kinetic energy and the potential energy. We will use a direct formula that relates all the above terms for solid cylinder and sphere. Then, the ratio of the heights is calculated using the same.
Formula used:
\[K{{E}_{total}}=\dfrac{1}{2}\left( {{I}_{center}}+M_{R}^{2} \right)\dfrac{{{V}^{2}}}{{{R}^{2}}}\]

Complete answer:
The formulae used are:
The moment of inertia of the solid cylinder about the central axis is,
\[I=\dfrac{1}{2}M{{R}^{2}}\]
Where M is the mass of the cylinder and R is the radius of the cylinder.
The moment of inertia of the solid sphere about the central axis is,
\[I=\dfrac{2}{5}M{{R}^{2}}\]
Where M is the mass of the cylinder and R is the radius of the cylinder.
The potential energy is given by the formula,
\[PE=mgh\]
Where m is the mass, g is the gravitational constant and h is the height.
The kinetic energy is given by the formula,
\[KE=\dfrac{1}{2}m{{v}^{2}}\]
Where m is the mass and v is the velocity.

There are two methods to solve this problem.
Method I: Direct method.
\[\begin{align}
  & \dfrac{1}{2}m{{v}^{2}}(1+\eta )=mgh \\
 & \Rightarrow h\propto 1+\eta \\
\end{align}\]
Where \[\eta \]is the distance from the central axis.
The ratio of the heights of the solid sphere and solid cylinder are given to be as follows,
\[\begin{align}
  & \dfrac{{{h}_{sph}}}{{{h}_{cyl}}}=\dfrac{1+{{\eta }_{sph}}}{1+{{\eta }_{cyl}}} \\
 & \Rightarrow \dfrac{{{h}_{sph}}}{{{h}_{cyl}}}=\dfrac{1+\dfrac{2}{5}}{1+\dfrac{1}{2}} \\
\end{align}\]
Continue further calculation.
\[\begin{align}
  & \dfrac{{{h}_{sph}}}{{{h}_{cyl}}}=\dfrac{\dfrac{7}{5}}{\dfrac{3}{2}} \\
 & \Rightarrow \dfrac{{{h}_{sph}}}{{{h}_{cyl}}}=\dfrac{14}{15} \\
\end{align}\]
Method II: Indirect method.
This indirect method is the elaborated form of the direct method.
In the case of a solid sphere,
\[\dfrac{1}{2}m{{v}^{2}}+\dfrac{1}{2}\times \dfrac{2}{5}m{{R}^{2}}\times \dfrac{{{v}^{2}}}{{{R}^{2}}}=mgh{}_{sph}\] …… (1)
In the case of a solid cylinder,
\[\dfrac{1}{2}m{{v}^{2}}+\dfrac{1}{2}\times \dfrac{1}{2}m{{R}^{2}}\times \dfrac{{{v}^{2}}}{{{R}^{2}}}=mgh{}_{cyl}\]…… (2)
Take the ratio of the equations (1) and (2).
\[\dfrac{{{h}_{sph}}}{{{h}_{cyl}}}=\dfrac{1+\dfrac{2}{5}}{1+\dfrac{1}{2}}\]
Continue further calculation.
\[\begin{align}
  & \dfrac{{{h}_{sph}}}{{{h}_{cyl}}}=\dfrac{\dfrac{7}{5}}{\dfrac{3}{2}} \\
 & \Rightarrow \dfrac{{{h}_{sph}}}{{{h}_{cyl}}}=\dfrac{14}{15} \\
\end{align}\]
Therefore, the ratio\[\dfrac{{{h}_{sph}}}{{{h}_{cyl}}}\]is given by, \[\dfrac{14}{15}\].
As, the value of the ratio\[\dfrac{{{h}_{sph}}}{{{h}_{cyl}}}\]is obtained to be equal to\[\dfrac{14}{15}\], thus, the option (A) is correct.

So, the correct answer is “Option A”.

Note:
Both the methods mentioned above are the same, but one method has a little elaboration involved. If we are perfect with the formulae, then, we can use the final formula. There is no need to take care of the units, as we are supposed to find the ratio.