Question

The fraction of a floating body of volume $V_o$ and density $d_o$ above the surface of a liquid of density $d$ will be -

• A. $\frac{d}{d_o}$
• B. $\frac{d_o}{d}$
• C. $\frac{d-d_o}{d}$
• D. $\frac{d}{d-d_o}$

Answer 1

The correct option is C $\frac{d-d_o}{d}$

Let the floating body be a cube of volume $V_o$, density $d_o$ and immersed volume inside the liquid be $V_{in}$.
So, the volume of cube outside the liquid will be $V_{out}=V_o-V_{in}$.

In equilibrium, $F_b=W$
$\Rightarrow V_{in}dg=V_o{d}_{o}g$
$\Rightarrow V_{in}=\frac{V_o{d}_o}{d}$

Also, $V_{out}=V_o-V_{in}=V_o-\frac{V_o{d}_o}{d}$
$\Rightarrow V_{out}=\frac{V_{o}d-V_o{d}_o}{d}$
$\Rightarrow \frac{V_{out}}{V_o}=\frac{d-d_o}{d}$ $[$Required fraction$]$