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A sphere of material of relative density 8 has a concentric spherical cavity and just sinks in water. If the radius of the sphere is $2 \mathrm{~cm}$, then the volume of the cavity is
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$\frac{88}{3} \mathrm{~cm}^3$
Let $r$ be the radius of cavity
$\therefore$ Volume of cavity $=\frac{4}{3} \pi \mathrm{r}^3$
Now, $\frac{4}{3} \pi\left(R^3-r^3\right) d m g=\frac{4}{3} \pi R^3 d_w g$
$\Rightarrow 1-\frac{R^3}{r^3}=\frac{1}{8} \Rightarrow \frac{R^3}{r^3}=1-\frac{1}{8}$
$\Rightarrow \frac{R^3}{r^3}=\frac{7}{8}$
$\Rightarrow r^3=\frac{8}{7} R^3=\frac{7}{8}(2)^3=7$
There Volume of cavity $=\frac{4}{3} \pi \mathrm{r}^3$ $=\frac{4}{3} \times \frac{22}{7} \times 7=\frac{88}{3}$
$\therefore$ Volume of cavity $=\frac{4}{3} \pi \mathrm{r}^3$
Now, $\frac{4}{3} \pi\left(R^3-r^3\right) d m g=\frac{4}{3} \pi R^3 d_w g$
$\Rightarrow 1-\frac{R^3}{r^3}=\frac{1}{8} \Rightarrow \frac{R^3}{r^3}=1-\frac{1}{8}$
$\Rightarrow \frac{R^3}{r^3}=\frac{7}{8}$
$\Rightarrow r^3=\frac{8}{7} R^3=\frac{7}{8}(2)^3=7$
There Volume of cavity $=\frac{4}{3} \pi \mathrm{r}^3$ $=\frac{4}{3} \times \frac{22}{7} \times 7=\frac{88}{3}$
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