Question

A cylindrical wooden floats whose base area $S$ and the height $H$ drifts on the water surface. Density of wood $d$ and density of water is $p$. What minimum work must be performed to take the float out of the water?

• A. $\frac{3Sg{d}^2{H}^2}{2p}$
• B. $\frac{Sg{d}^2{H}^2}{2p}$
• C. $\frac{Sg{d}^2{H}^2}{p}$
• D. $\frac{2Sg{d}^2{H}^2}{3p}$

Answer 1

The correct option is B $\frac{Sg{d}^2{H}^2}{2p}$

Apply the condition of equilibrium.

Applying Newton's law in vertical direction

$\Rightarrow mg=F_B\Rightarrow dSH\times g=pSh\times g$

$\Rightarrow h=\frac{dH}{p}$

Find the minimum work to take the float out of the water.
Now when force $F$ is applied, for minimum work $a=0$ (for $a=0$, $F$ is minimum)

$F-mg+pSxg=0$

$F=mg-pSxg$

$W=\int Fdx=\int (mg-pSxg)dx$

$=mg{\int }_0^{h}dx-pSg{\int }_0^{h}xdx$

$W=mgh-\frac{pSg{h}^2}{2}=\left(dsH\right)gh-\frac{pSg{h}^2}{2}$

Putting the value of $h$,

$W=\frac{Sg{d}^2{H}^2}{2p}$