A sphere of radius R is cut from a larger solid sphere of radius 2R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is:   

                                                                        

The problem asks for the ratio

$I\_\{\backslash text\{small\}\}\; /\; I\_\{\backslash text\{rest\}\}$

where the small sphere of radius R is removed from a solid sphere of radius 2R. Both moments of inertia are taken about the vertical Y‑axis that passes through the centre of the large sphere.

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**1. Masses of the two spheres**

The density ρ is uniform, so the masses are proportional to the volumes.

$M\_\{\backslash text\{big\}\}\; =\; \backslash rho\; \backslash frac\{4\}\{3\}\backslash pi\; (2R)^\{3\}=\; \backslash rho\; \backslash frac\{32\}\{3\}\backslash pi\; R^\{3\}$

$M\_\{\backslash text\{small\}\}\; =\; \backslash rho\; \backslash frac\{4\}\{3\}\backslash pi\; R^\{3\}=\; \backslash rho\; \backslash frac\{4\}\{3\}\backslash pi\; R^\{3\}$

Thus

$M\_\{\backslash text\{small\}\}\; =\; \backslash frac\{1\}\{8\}\backslash ,M\_\{\backslash text\{big\}\}$

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**2. Geometry of the cut**

From the diagram the small sphere touches the outer surface of the large one. Therefore the distance between the two centres is

$d\; =\; 2R\; -\; R\; =\; R$

Hence the centre of the small sphere is a distance R from the Y‑axis (which goes through the centre of the big sphere).

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**3. Moment of inertia of the small sphere about the Y‑axis**

For a solid sphere about any diameter:

$I\_\{\backslash text\{cm\}\}\; =\; \backslash frac\{2\}\{5\}\; M\; R^\{2\}$

Using the parallel‑axis theorem for the offset d = R:

$I\_\{\backslash text\{small\}\}\; =\; I\_\{\backslash text\{cm\}\}\; +\; M\_\{\backslash text\{small\}\}\; d^\{2\}\; =\; \backslash frac\{2\}\{5\}M\_\{\backslash text\{small\}\}R^\{2\}\; +\; M\_\{\backslash text\{small\}\}R^\{2\}\; =\; \backslash left(\backslash frac\{2\}\{5\}+1\backslash right)M\_\{\backslash text\{small\}\}R^\{2\}\; =\; \backslash frac\{7\}\{5\}M\_\{\backslash text\{small\}\}R^\{2\}$

Substituting \(M_{\text{small}} = M_{\text{big}}/8\):

$I\_\{\backslash text\{small\}\}\; =\; \backslash frac\{7\}\{5\}\backslash cdot\backslash frac\{M\_\{\backslash text\{big\}\}\}\{8\}R^\{2\}\; =\; \backslash frac\{7\}\{40\}\backslash ,M\_\{\backslash text\{big\}\}R^\{2\}$

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**4. Moment of inertia of the original (big) sphere about the Y‑axis**

$I\_\{\backslash text\{big\}\}\; =\; \backslash frac\{2\}\{5\}M\_\{\backslash text\{big\}\}(2R)^\{2\}\; =\; \backslash frac\{2\}\{5\}M\_\{\backslash text\{big\}\}\; \backslash cdot\; 4R^\{2\}\; =\; \backslash frac\{8\}\{5\}\backslash ,M\_\{\backslash text\{big\}\}R^\{2\}\; =\; \backslash frac\{64\}\{40\}\backslash ,M\_\{\backslash text\{big\}\}R^\{2\}$

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**5. Moment of inertia of the remaining part**

The rest of the solid is the original sphere minus the removed piece:

$I\_\{\backslash text\{rest\}\}\; =\; I\_\{\backslash text\{big\}\}\; -\; I\_\{\backslash text\{small\}\}\; =\; \backslash frac\{64\}\{40\}M\_\{\backslash text\{big\}\}R^\{2\}\; -\; \backslash frac\{7\}\{40\}M\_\{\backslash text\{big\}\}R^\{2\}\; =\; \backslash frac\{57\}\{40\}M\_\{\backslash text\{big\}\}R^\{2\}$

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**6. Required ratio**

$\backslash frac\{I\_\{\backslash text\{small\}\}\}\{I\_\{\backslash text\{rest\}\}\}\; =\; \backslash frac\{\backslash frac\{7\}\{40\}M\_\{\backslash text\{big\}\}R^\{2\}\}\{\backslash frac\{57\}\{40\}M\_\{\backslash text\{big\}\}R^\{2\}\}\; =\; \backslash frac\{7\}\{57\}$

Thus the ratio of the moment of inertia of the cut‑out smaller sphere to that of the remaining part about the Y‑axis is **\(\displaystyle \frac{7}{57}\)**, which matches the provided correct option.