A solid sphere of mass M and radius R is divided into two unequal parts. The first part has a mass of $$\frac{7M}{8}$$ and is converted into uniform disc of radius 2R. The second part is converted into a uniform solid sphere. Let I$$_1$$ be the moment of inertia of the disc about its axis and I$$_2$$ be the moment of inertia of the new sphere about its axis. The ratio I$$_1$$/I$$_2$$ is given by:

Let us denote the density of the original solid sphere by $$\rho$$. Because the sphere is uniform, we have

$$\rho=\dfrac{\text{mass}}{\text{volume}}=\dfrac{M}{\dfrac{4}{3}\pi R^{3}}\;.$$

The sphere is cut into two parts.

First part ⇒ uniform disc

The first part has mass $$\dfrac{7M}{8}$$ and is recast into a uniform disc of radius $$2R$$. For a uniform disc about an axis perpendicular to its plane and passing through its centre, the standard formula is

$$I=\dfrac{1}{2}MR^{2}\;.$$

Here the mass is $$\dfrac{7M}{8}$$ and the radius is $$2R$$, so

$$\begin{aligned} I_{1}&=\dfrac{1}{2}\left(\dfrac{7M}{8}\right)(2R)^{2} \\ &=\dfrac{1}{2}\left(\dfrac{7M}{8}\right)\cdot4R^{2} \\ &=\left(\dfrac{7M}{16}\right)4R^{2} \\ &=\dfrac{7M}{4}\,R^{2}. \end{aligned}$$

Second part ⇒ new solid sphere

The second part has mass $$\dfrac{M}{8}$$ and is melted into another uniform solid sphere of (unknown) radius $$r$$. Using the same density $$\rho$$, we write

$$\dfrac{M}{8}=\rho\left(\dfrac{4}{3}\pi r^{3}\right).$$

Substituting $$\rho=\dfrac{M}{\dfrac{4}{3}\pi R^{3}}$$ from above, we get

$$\dfrac{M}{8}=\left[\dfrac{M}{\dfrac{4}{3}\pi R^{3}}\right]\!\left(\dfrac{4}{3}\pi r^{3}\right) =M\left(\dfrac{r^{3}}{R^{3}}\right).$$

Dividing both sides by $$M$$,

$$\dfrac{1}{8}=\dfrac{r^{3}}{R^{3}} \;\;\Longrightarrow\;\; r^{3}=\dfrac{R^{3}}{8} \;\;\Longrightarrow\;\; r=\dfrac{R}{2}.$$

For a uniform solid sphere about any diameter, the standard formula is

$$I=\dfrac{2}{5}MR^{2}.$$

Replacing $$M$$ by $$\dfrac{M}{8}$$ and $$R$$ by the new radius $$r=\dfrac{R}{2}$$, we have

$$\begin{aligned} I_{2}&=\dfrac{2}{5}\left(\dfrac{M}{8}\right)\left(\dfrac{R}{2}\right)^{2} \\ &=\dfrac{2}{5}\left(\dfrac{M}{8}\right)\dfrac{R^{2}}{4} \\ &=\dfrac{2M R^{2}}{5\cdot8\cdot4} \\ &=\dfrac{2M R^{2}}{160} \\ &=\dfrac{M R^{2}}{80}. \end{aligned}$$

Ratio of the two moments of inertia

$$\begin{aligned} \dfrac{I_{1}}{I_{2}} &=\dfrac{\dfrac{7M}{4}R^{2}}{\dfrac{M R^{2}}{80}} \\ &=\dfrac{7M R^{2}}{4}\times\dfrac{80}{M R^{2}} \\ &=7\times\dfrac{80}{4} \\ &=7\times20 \\ &=140. \end{aligned}$$

Hence, the correct answer is Option A.