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**Hint**We will use the concept of analogy of translatory motion and rotatory motion. We will find the equivalent relations for both the motions. Finally, we will find their ratio.

**Formulae Used**\[{K_t} = \frac{1}{2}m{v^2}\] And \[{K_r} = \frac{1}{5}m{v^2}\]

**Step By Step Solution**

Let the mass of the sphere be $m$ , its velocity be $v$ .

Now,

The translational kinetic energy , \[{K_t} = \frac{1}{2}m{v^2}\]

Then,

For the rotatory motion,

Moment of inertia $I$ is analogical to mass in translational motion.

Thus,

For sphere, \[I = \frac{2}{5}m{r^2}\]

Similarly,

Angular velocity $\omega $ is analogical to velocity in translational motion.

Thus,

For Sphere, $\omega = \frac{v}{r}$

Here,

$r$ is the radius of the sphere.

Now,

Rotational kinetic energy, \[{K_r} = \frac{1}{2}I{\omega ^2}\]

Thus,

Substituting the values, we get

\[{K_r} = \frac{1}{2} \times \frac{2}{5}m{r^2}\frac{{{v^2}}}{{{r^2}}}\]

Thus, we get

\[{K_r} = \frac{1}{5}m{v^2}\]

Now,

\[\left( {{K_t} + {K_r}} \right) = \frac{1}{2}m{v^2} + \frac{1}{5}m{v^2} = \frac{7}{{10}}m{v^2}\]

Thus,

\[{K_t}:\left( {{K_t} + {K_r}} \right) = \frac{1}{2}:\frac{7}{{10}} = 5:7\]

Hence,

**The answer is (2).**

**Additional Information**The moment of inertia we discussed is a parameter which comes from the observation that a rotating body acts as if all its mass is concentrated at a single point. Also the radius through which it rotates deviates from the original position of the actual one.

The translational motion and the rotatory motion are analogous at every aspect of parameters starting from radius to centripetal force.

**Note:**For calculating the rotatory kinetic energy, we assumed that the sphere was rotating about a fixed axis perpendicular to its plane and passing through its center. We can also take it to be random. But the calculations then become very clumsy. Though the answer will be the same.

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