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The fraction of a floating object of volume $V_0$ and density $d_0$ above the surface of a liquid of density $d$ will be
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The correct answer is:
$\frac{d-d_0}{d}$
For the floatation $V_0 d_0 g=V_{i n} d g \Rightarrow V_{i n}=V_0 \frac{d_0}{d}$
$\therefore \quad V_{a w t}=V_0-V_{m n}=V_0-V_0 \frac{d_0}{d}=V_0\left[\frac{d-d_0}{d}\right]$
$\Rightarrow \frac{V_{\text {out }}}{V_0}=\frac{d-d_0}{d}$.
$\therefore \quad V_{a w t}=V_0-V_{m n}=V_0-V_0 \frac{d_0}{d}=V_0\left[\frac{d-d_0}{d}\right]$
$\Rightarrow \frac{V_{\text {out }}}{V_0}=\frac{d-d_0}{d}$.
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