A hollow spherical shell at outer radius R floats just submerged under the water surface. The inner radius of the shell is r. If the specific gravity of the shell material is 27 \over 8 w.r.t water, the value of r is :

Properties of Matter PYQ — JEE Main 2020 · 5 Sep · Shift 1 · Q49 with Solution

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Correct answer: D

If the hollow spherical shell floats just submerged under water, then its entire outer volume is submerged.

So, by buoyancy:

$\text{Weight of shell} = \text{Weight of displaced water}$

Outer radius $= R$ , inner radius $= r$ .

Hence volume of shell material is

$V_{\text{shell}} = \frac{4}{3}\pi (R^3-r^3)$

Let density of water be $\rho_w$ .

Given specific gravity of shell material with respect to water is

$\frac{\rho_{\text{material}}}{\rho_w} = \frac{27}{8}$

So,

$\rho_{\text{material}} = \frac{27}{8}\rho_w$

Therefore weight of shell is proportional to

= \frac{27}{8}\rho_w \cdot \frac{4}{3}\pi (R^3-r^3) g$$ 4. **Buoyant force when just submerged** Since the outer sphere is fully submerged, displaced water volume is $$V_{\text{disp}} = \frac{4}{3}\pi R^3$$ Hence buoyant force is proportional to $$\rho_w \cdot \frac{4}{3}\pi R^3 g$$ 5. **Apply floating condition** Equating weight and buoyant force: $$\frac{27}{8}\rho_w \cdot \frac{4}{3}\pi (R^3-r^3) g = \rho_w \cdot \frac{4}{3}\pi R^3 g$$ Cancel common factors $\rho_w, \frac{4}{3}\pi, g$: $$\frac{27}{8}(R^3-r^3)=R^3$$ So, $$R^3-r^3 = \frac{8}{27}R^3$$ $$r^3 = R^3\left(1-\frac{8}{27}\right) = R^3\cdot \frac{19}{27}$$ Thus, $$r = R\left(\frac{19}{27}\right)^{1/3}$$ 6. **Compare with options** Now, $$\left(\frac{19}{27}\right)^{1/3} \approx 0.889$$ and $$\frac{8}{9}R \approx 0.889R$$ Indeed, $$\left(\frac{8}{9}\right)^3 = \frac{512}{729} \approx 0.702$$ while $$\frac{19}{27} = \frac{513}{729}$$ So the exact value is $$r = R\left(\frac{19}{27}\right)^{1/3}$$ which is **extremely close** to $\frac{8}{9}R$, and clearly corresponds to option **D** in the intended MCQ. 7. **Final answer** $$\boxed{\frac{8}{9}R}$$ (Strictly, the exact expression is $R\left(\frac{19}{27}\right)^{1/3}$, but among the given options, $\frac{8}{9}R$ is the matching/intended answer.)