MCQ
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:
  • $\left(\frac{\sigma}{\sigma-1}\right)^{\frac{1}{3}}$
  • B
    $\left(\frac{\sigma+1}{\sigma-1}\right)^{\frac{1}{3}}$
  • C
    $\left(\frac{\sigma-1}{\sigma}\right)^{\frac{1}{3}}$
  • D
    $\left(\frac{\sigma-2}{\sigma+2}\right)^{\frac{1}{3}}$

Answer

Correct option: A.
$\left(\frac{\sigma}{\sigma-1}\right)^{\frac{1}{3}}$
a
$\text { weight (w) }=\frac{4}{3} \pi\left(\frac{D^3-d^3}{8}\right) \sigma g$

Buoyant force $\left(F_b\right)=1 \times \frac{4}{3} \pi\left(\frac{D^3}{8}\right) \cdot g$

For Just Float $\Rightarrow \mathrm{W}=\mathrm{F}_{\mathrm{b}}$

$\Rightarrow\left(\mathrm{D}^3-\mathrm{d}^3\right) \sigma=\mathrm{D}^3$

$\Rightarrow 1-\frac{\mathrm{d}^3}{\mathrm{D}^3}=\frac{1}{\sigma}$

$\Rightarrow 1-\frac{1}{\sigma}=\left(\frac{\mathrm{d}}{\mathrm{D}}\right)^3$

$\Rightarrow\left(\frac{\sigma}{\sigma-1}\right)^{\frac{1}{3}}=\left(\frac{\mathrm{D}}{\mathrm{d}}\right)$

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