A solid sphere of specific gravity $27$ has a concentric spherical cavity and it just sinks in water. The ratio of cavity radius to that of outer radius of sphere is
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From law of flotation
$\frac{4}{3} \pi \mathrm{R}^{3} \times 1 \times \mathrm{g}=\frac{4}{3} \pi\left(\mathrm{R}^{3}-\mathrm{r}^{3}\right) \rho \mathrm{g}$
$\mathrm{R}^{3}=\left(\mathrm{R}^{3}-\mathrm{r}^{3}\right) \times 27$
$\frac{r^{3}}{R^{3}}=\frac{26}{27} \Rightarrow \frac{r}{R}=\frac{(26)^{1 / 3}}{3}$
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