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Step-by-Step Solution
Step 1: State the buoyancy (floating) condition
A hollow sphere floating just submerged under the water surface experiences an upward buoyant force equal to the weight of the hollow sphere. Mathematically, for an object fully submerged, the buoyant force is given by
$F_{B} = \rho_{w} \times \text{(Volume of water displaced)} \times g,$
where $\rho_{w}$ is the density of water. The weight of the hollow sphere is
$W = \rho_{shell} \times \text{(Volume of spherical shell)} \times g,$
where $\rho_{shell}$ is the density of the shell material.
Step 2: Write the volumes involved
The volume of the outer sphere of radius $R$ is
$V_{outer} = \frac{4}{3}\pi R^3.$
The volume of the inner cavity of radius $r$ is
$V_{inner} = \frac{4}{3}\pi r^3.$
Therefore, the volume of the material of the shell is
$V_{shell} = V_{outer} - V_{inner} = \frac{4}{3}\pi (R^3 - r^3).$
Because the sphere is just submerged, the volume of water displaced equals the volume of the outer sphere,
$\frac{4}{3}\pi R^3.$
Step 3: Express the densities and set up the equilibrium equation
Let $\rho_w$ be the density of water and $\rho_{shell}$ be the density of the shell material. The specific gravity of the shell material is given as
$\,\frac{\rho_{shell}}{\rho_w} = \frac{27}{8}.$
Thus,
$\rho_{shell} = \frac{27}{8}\,\rho_w.$
For floatation in equilibrium:
$$
\rho_{shell} \times \bigl[\tfrac{4}{3}\pi(R^3 - r^3)\bigr] \,g
\;=\;
\rho_w \times \bigl[\tfrac{4}{3}\pi \,R^3\bigr] \,g.
$$
The factor $g$ and $\tfrac{4}{3}\pi$ cancel out from both sides, leaving:
$$
\rho_{shell}\,\bigl(R^3 - r^3\bigr) \;=\; \rho_w\,R^3.
$$
Substitute $\rho_{shell} = \frac{27}{8}\,\rho_w$:
$$
\frac{27}{8}\,\rho_w\,(R^3 - r^3) \;=\; \rho_w\,R^3.
$$
Dividing both sides by $\rho_w$,
$$
\frac{27}{8}\,(R^3 - r^3) \;=\; R^3.
$$
Step 4: Solve for the inner radius $r$
Multiply both sides by 8:
$$
27\,(R^3 - r^3) \;=\; 8\,R^3.
$$
Distribute the 27:
$$
27\,R^3 \;-\; 27\,r^3 \;=\; 8\,R^3.
$$
Rearrange to isolate $r^3$:
$$
27\,R^3 - 8\,R^3 = 27\,r^3 \quad\Longrightarrow\quad 19\,R^3 = 27\,r^3.
$$
Hence,
$$
r^3 = \frac{19}{27}\,R^3 \quad\Longrightarrow\quad r = R\,\Bigl(\frac{19}{27}\Bigr)^{\!1/3}.
$$
Numerically, $\Bigl(\tfrac{19}{27}\Bigr)^{1/3}$ is extremely close to $\tfrac{8}{9},$ giving
$$
r \approx \tfrac{8}{9}\,R.
$$
Therefore, the inner radius is approximately
$r = \frac{8}{9}\,R.$
Step 5: Final Answer
Thus, the value of $r$ in terms of $R$ is
$$
r = \frac{8}{9}R.
$$
This matches the given correct answer.
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