A uniform ring and uniform solid sphere rolls down same inclined plane by same distance. If ratio of their translational kinetic energies is 7/x then x is (Given mass and radius of ring and sphere are equal)

The correct answer is 10.

Step-by-Step Derivation:

Let the mass of both the uniform ring and the uniform solid sphere be $m$, and their radius be $R$.
When an object rolls down an inclined plane of height $h$ without slipping, it conservation of mechanical energy dictates that the loss in gravitational potential energy is converted entirely into total kinetic energy ($K$).

For both objects starting from rest, the total kinetic energy acquired after rolling down the same vertical height $h$ is:
$K=mgh$

The total kinetic energy of a rolling body is the sum of its translational kinetic energy ($K_T$) and rotational kinetic energy ($K_R$):
$K=K_T+K_R$
Using the formulas for kinetic energies, we have:
$K_T=\frac{1}{2}m{v}^2$
and
$K_R=\frac{1}{2}I{\omega }^2$
where $I$ is the moment of inertia, $v$ is the linear velocity of the center of mass, and $\omega$ is the angular velocity.

For rolling without slipping, $v=\omega R$, or $\omega =\frac{v}{R}$. Substituting this into the rotational kinetic energy expression gives:
$K_R=\frac{1}{2}I{\left(\frac{v}{R}\right)}^2=\frac{1}{2}\left(\frac{I}{R^2}\right)v^2$

Expressing the total kinetic energy in terms of translational kinetic energy:
$K=\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{I}{R^2}\right)v^2=\frac{1}{2}m{v}^2\left(1+\frac{I}{m{R}^2}\right)=K_T\left(1+\frac{I}{m{R}^2}\right)$
Since $K=mgh$, the translational kinetic energy is:
$K_T=\frac{mgh}{1+\frac{I}{m{R}^2}}$

1. Translational kinetic energy of the uniform ring:
The moment of inertia of a uniform ring is $I_{\text{ring}}=m{R}^2$.
Substituting this value:
$K_{T,\text{ring}}=\frac{mgh}{1+\frac{m{R}^2}{m{R}^2}}=\frac{mgh}{1+1}=\frac{mgh}{2}$

2. Translational kinetic energy of the uniform solid sphere:
The moment of inertia of a uniform solid sphere is $I_{\text{sphere}}=\frac{2}{5}m{R}^2$.
Substituting this value:
$K_{T,\text{sphere}}=\frac{mgh}{1+\frac{\frac{2}{5}m{R}^2}{m{R}^2}}=\frac{mgh}{1+\frac{2}{5}}=\frac{mgh}{\frac{7}{5}}$

3. Ratio of their translational kinetic energies:
Now, we calculate the ratio of $K_{T,\text{ring}}$ to $K_{T,\text{sphere}}$:
$\frac{K_{T,\text{ring}}}{K_{T,\text{sphere}}}=\frac{\frac{1}{2}mgh}{\frac{5}{7}mgh}=\frac{1}{2}\times \frac{7}{5}=\frac{7}{10}$

Given in the problem, this ratio is equal to $\frac{7}{x}$:
$\frac{7}{10}=\frac{7}{x}\Rightarrow x=10$