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A sphere of relative density $\sigma$ and diameter $D$ has concentric cavity of diameter $d$. The ratio of $\frac{D}{d}$, if it just floats on water in a tank is :
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The correct answer is:
$\left(\frac{\sigma}{\sigma-1}\right)^{1 / 3}$
weight $(w)=\frac{4}{3} \pi\left(\frac{\mathrm{D}^3-\mathrm{d}^3}{8}\right) \sigma \mathrm{g}$
Buoyant force $\left(\mathrm{F}_{\mathrm{b}}\right)=1 \times \frac{4}{3} \pi\left(\frac{\mathrm{D}^3}{8}\right) \cdot \mathrm{g}$
For Just Float $\Rightarrow \mathrm{w}=\mathrm{F}_{\mathrm{b}}$
$\begin{aligned} & \Rightarrow\left(D^3-d^3\right) \sigma=D^3 \\ & \Rightarrow 1-\frac{d^3}{D^3}=\frac{1}{\sigma} \\ & \Rightarrow 1-\frac{1}{\sigma}=\left(\frac{d}{D}\right)^3 \\ & \Rightarrow\left(\frac{\sigma}{\sigma-1}\right)^{\frac{1}{3}}=\left(\frac{D}{d}\right)\end{aligned}$
Buoyant force $\left(\mathrm{F}_{\mathrm{b}}\right)=1 \times \frac{4}{3} \pi\left(\frac{\mathrm{D}^3}{8}\right) \cdot \mathrm{g}$
For Just Float $\Rightarrow \mathrm{w}=\mathrm{F}_{\mathrm{b}}$
$\begin{aligned} & \Rightarrow\left(D^3-d^3\right) \sigma=D^3 \\ & \Rightarrow 1-\frac{d^3}{D^3}=\frac{1}{\sigma} \\ & \Rightarrow 1-\frac{1}{\sigma}=\left(\frac{d}{D}\right)^3 \\ & \Rightarrow\left(\frac{\sigma}{\sigma-1}\right)^{\frac{1}{3}}=\left(\frac{D}{d}\right)\end{aligned}$
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