91Ó°ÊÓ

The expression of the pressure is

$P=\mathrm{ÒÏ}gh$

Here, $\mathrm{ÒÏ}$ is the density, $g$is the acceleration due to gravity, and $h$is the height.

The expression of pressure is

$P=\mathrm{ÒÏ}gh$

Since the cylinder being pushed down a distance is$x$

Substitute $x$for $h$in above equation

$P=\mathrm{ÒÏ}gx$

The expression for the force is

$F=PA$

Here, $P$is the pressure and $A$is the area

Substitute$\mathrm{ÒÏ}gx$for$P$

$F=\mathrm{ÒÏ}gxA$

Hence the required force is$\mathrm{ÒÏ}gxA$

At the depth $h$the force that needs to be overcome to sink the cylinder is

The expression for the force is

$F=PA$

Substitute $\mathrm{ÒÏ}gh$for $P$

$F=\mathrm{ÒÏ}ghA$

The infinitesimal work is

$dW=Fdh$

Substitute $\mathrm{ÒÏ}ghA$for $F$

$dW=\left(\mathrm{ÒÏ}ghA\right)dh$

Therefore, the work to sink the cylinder until the depth $x$is,

$W={∫}_0^x \left(\mathrm{ÒÏ}ghA\right)dh$

$=A\mathrm{ÒÏ}g{\left(\frac{h^2}{2}\right)}_0^x$

$=A\mathrm{ÒÏ}g\left(\frac{x^2}{2}\right)$

$=\frac{A\mathrm{ÒÏ}g{x}^2}{2}$

The area is $A=\mathrm{Ï€}{r}^2$

Here r is the radius

Substitute $\frac{d}{2}\text{for}r$

$A=\mathrm{Ï€}{\left(\frac{d}{2}\right)}^2$

Here d is the diameter

Substitute $\mathrm{Ï€}{\left(\frac{d}{2}\right)}^2$for A in the equation $W=\frac{A\mathrm{ÒÏ}g{x}^2}{2}$

$W=\frac{\left(\mathrm{Ï€}{\left(\frac{d}{2}\right)}^2\right)\mathrm{ÒÏ}g{x}^2}{2}$

$=\frac{\mathrm{ÒÏ}\mathrm{Ï€}g{d}^2{x}^2}{8}$

Substitute $1000\mathrm{kg}/{\mathrm{m}}^3\text{for}\mathrm{ÒÏ},9.8\mathrm{m}/{\mathrm{s}}^2\text{for}g,4\mathrm{cm}\text{for}d\text{, and}10\mathrm{cm}\text{for}x\text{.}$

$\begin{array}{r}W=\frac{\left(1000\mathrm{kg}/{\mathrm{m}}^3\right)\mathrm{Ï€}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right){\left(\left(4\mathrm{cm}\right)\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)\right)}^2{\left(\left(10\mathrm{cm}\right)\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)\right)}^2}{8}\\ \end{array}$

$=0.062J$