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 $\frac{27}{8}$ w.r.t water, the value of $r$ is:

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$\frac{8}{9} R$

$\frac{1}{3} R$

$\frac{4}{9} R$

$\frac{2}{3} R$

answer is B.

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Detailed Solution

$\begin{array}{l} F o r f l o a t i n g, \\ F_{B} = m g \\ \Rightarrow V_{i m m e r s e d} ρ_{w} g {=V}_{m a t e r i a l} ρ_{m a t e r i a l} g \\ \Rightarrow \frac{4}{3} {πR}^{3} ρ_{w} g = \frac{4}{3} π (R^{3} - r^{3}) \frac{27}{8} ρ_{w} g \\ \Rightarrow R^{3} = (R^{3} - r^{3}) (\frac{27}{8}) \\ \Rightarrow {8R}^{3} = 27 R^{3} - 27 r^{3} \\ \Rightarrow r = \frac{19^{\frac{1}{3}}}{3} R = \frac{3 (19^{\frac{1}{3}})}{9} R \approx \frac{8}{9} R \end{array}$