Question

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:

• A. $4/9R$
• B. $8/9R$
• C. $1/3R$
• D. $2/3R$

Answer 1

The correct option is B

$8/9R$

Step 1: Given data in the form of a diagram:

Specific gravity = $\frac{{\rho }_s}{{\rho }_w}=\frac{27}{8}$ [${\rho }_s,{\rho }_w$ is the density of shell and water respectively]

Step 2: Finding the value of $r$:

The volume of immersed liquid, $V_i=\frac{4}{3}{\mathrm{\pi R}}^3$

The volume of the shell, $V_s=\frac{4}{3}\mathrm{\pi }\left({\mathrm{R}}^3-{\mathrm{r}}^3\right)$

In equilibrium,

$\frac{4}{3}{\mathrm{\pi R}}^3{\mathrm{\rho }}_{\mathrm{w}}\mathrm{g}=\frac{4}{3}\mathrm{\pi }\left({\mathrm{R}}^3-{\mathrm{r}}^3\right){\mathrm{\rho }}_{\mathrm{s}}\mathrm{g}$ [Where, $R$ outer radius of the shell, $r$ is inner radius,$g$ is acceleration due to gravity]

$R^3=\left(R^3-r^3\right)\frac{27}{8}$

$8R^3=27R^3-27r^3$

$27r^3=19R^3$

$\frac{r^3}{R^3}=\frac{19}{27}$

$r={\left(\frac{19}{27}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

$r\approx \frac{8}{9}R$

Hence, option $\left(B\right)$ is correct.