91Ó°ÊÓ

The upward force experienced by an object immersed in a fluid is equal to the weight of the fluid displaced by the immersed portion of the body. This force is called as buoyant or up thrust force.

The depth to which the cylinder is submerged is equal to the length of the stretching of the spring.

In the case of static equilibrium, the net force acting on the cylinder is equal to zero.

$\mathrm{Î\pounds }{F}_y=0$

$F_{\mathrm{B}}+F_{\mathrm{s}}−mg=0$

${\mathrm{ÒÏ}}_{\mathrm{w}}Ayg+ky=mg$

$y=\frac{mg}{{\mathrm{ÒÏ}}_{\mathrm{w}}Ag+k}$

Here, $F_B$is buoyant force due to liquid of density ${\mathrm{ÒÏ}}_{\mathrm{w}}$, $F_S$is spring force, $m$is mass, $g$is acceleration due to gravity,$A$is cross-sectional area of the cylinder, $y$is length of submerging of cylinder, and$K$ is spring constant of the spring.

Calculate the length of the submerging of cylinder.

Substitute $1.0\mathrm{kg}\text{for}m,9.8\mathrm{m}/{\mathrm{s}}^2,1000\mathrm{kg}/{\mathrm{m}}^3\text{for}{\mathrm{ÒÏ}}_{\mathrm{w}},\mathrm{Ï€}(0.025\mathrm{m}{)}^2\text{for}A\text{, and}35\mathrm{N}/\mathrm{m}\mathrm{for}\mathrm{k}$

in the equation

$y=\frac{\left(1.0\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}{\left(1000\mathrm{kg}/{\mathrm{m}}^3\right)\mathrm{Ï€}(0.025\mathrm{m}{)}^2\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)+35\mathrm{N}/\mathrm{m}}$

$=0.181\mathrm{m}$

$=0.181\mathrm{m}\left(\frac{100\mathrm{cm}}{1\mathrm{m}}\right)$

$=18.1\mathrm{cm}$