A solid sphere and a solid cylinder of same mass and radius are rolling on a horizontal surface without slipping. The ratio of their radius of gyrations respectively $\left(k_{\text {sph }}: k_{\text {cyl }}\right)$ is $2: \sqrt{x}$. The value of $x$ is ____________ .

Let the mass of both the solid sphere and the solid cylinder be $M$, and let their common radius be $R$. The moment of inertia $I$ of a solid sphere and a solid cylinder are given by: <br/><br/> $I_{sph} = \frac{2}{5}MR^2$ <br/><br/> $I_{cyl} = \frac{1}{2}MR^2$ <br/><br/> The radius of gyration $k$ is related to the moment of inertia $I$ by the formula $I = Mk^2$. Therefore, we can find the radius of gyration for both the solid sphere and the solid cylinder using their respective moments of inertia: <br/><br/> $k_{sph}^2 = \frac{I_{sph}}{M} = \frac{2}{5}R^2$ <br/><br/> $k_{cyl}^2 = \frac{I_{cyl}}{M} = \frac{1}{2}R^2$ <br/><br/> Now, let's find the ratio of their radius of gyrations: <br/><br/> $\frac{k_{sph}}{k_{cyl}} = \frac{2}{\sqrt{x}}$ <br/><br/> Squaring both sides: <br/><br/> $\frac{k_{sph}^2}{k_{cyl}^2} = \frac{4}{x}$ <br/><br/> Substituting the expressions for $k_{sph}^2$ and $k_{cyl}^2$: <br/><br/> $\frac{\frac{2}{5}R^2}{\frac{1}{2}R^2} = \frac{4}{x}$ <br/><br/> Simplifying and solving for $x$: <br/><br/> $\frac{2}{5} \cdot \frac{2}{1} = \frac{4}{x}$ <br/><br/> $\frac{4}{5} = \frac{4}{x}$ <br/><br/> Thus, $x = 5$.