If a solid sphere of mass 5 kg and a disc of mass 4 kg have the same radius, then the ratio of moment of inertia of the disc about a tangent in its plane to the moment of inertia of the sphere about its tangent will be $$\frac{x}{7}$$. The value of $$x$$ is _____.

Since the solid sphere has mass $$M_s = 5\text{ kg}$$ and the disc has mass $$M_d = 4\text{ kg}$$, both with radius $$R$$, we first determine their moments of inertia about tangents.

For the disc, the moment of inertia about a diameter is $$\frac{1}{4}M_dR^2$$. Now using the parallel axis theorem to shift the axis to a tangent in its plane (a distance $$R$$ from the center) gives

$$I_{\text{disc}} = \frac{1}{4}M_dR^2 + M_dR^2 = \frac{5}{4}M_dR^2 = \frac{5}{4}\,(4)\,R^2 = 5R^2.$$

Similarly, the moment of inertia of the sphere about a diameter is $$\frac{2}{5}M_sR^2$$. Substituting into the parallel axis theorem at distance $$R$$ yields

$$I_{\text{sphere}} = \frac{2}{5}M_sR^2 + M_sR^2 = \frac{7}{5}M_sR^2 = \frac{7}{5}\,(5)\,R^2 = 7R^2.$$

Therefore, the ratio of the disc’s moment to the sphere’s moment is

$$\frac{I_{\text{disc}}}{I_{\text{sphere}}} = \frac{5R^2}{7R^2} = \frac{5}{7} = \frac{x}{7},$$

which gives $$x = 5\,. $$