Gujarati
Hindi
9-1.Fluid Mechanics
hard

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

A

$\frac{{{{\left( {28} \right)}^{1/3}}}}{3}$

B

$\frac{{{{\left( {30} \right)}^{1/3}}}}{3}$

C

$\frac{{{{\left( {26} \right)}^{1/3}}}}{3}$

D

$\frac{{{{\left( {24} \right)}^{1/3}}}}{3}$

Solution

From law of flotation

$\frac{4}{3} \pi \mathrm{R}^{3} \times 1 \times \mathrm{g}=\frac{4}{3} \pi\left(\mathrm{R}^{3}-\mathrm{r}^{3}\right) \rho \mathrm{g}$

$\mathrm{R}^{3}=\left(\mathrm{R}^{3}-\mathrm{r}^{3}\right) \times 27$

$\frac{r^{3}}{R^{3}}=\frac{26}{27} \Rightarrow \frac{r}{R}=\frac{(26)^{1 / 3}}{3}$

Standard 11
Physics

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