91Ó°ÊÓ

Given data can be listed below,

• Mass of cylindrical buoy w = 950kg.
• Diameter of the buoy D=0.900m.
• Mass of man standing at top m = 80kg.

When the object is in a line parallel to the liquid's centre of gravity, they become buoyant. However, the centre of gravity of the superseded object is frequently not the same. The force is just the same as the weight of the liquid being replenished by the object.

ByArchimedes’ principle, the additional force will be equal to the extra weight on the buoy and its volume can be given by,

$V\text{=}\frac{m}{\mathrm{ÒÏ}}$

Here, m is mass of the man, and ÒÏ is the density of water which is 1080kg/m³.

Substitute 80 kg for m, and 1080 kg/m³ for ÒÏ the values in above equation,

$\begin{array}{c}V\text{=}\frac{80\text{ k²µ}}{1030\text{ k²µ}/{\text{m}}^{\text{3}}}\\ \text{=}0.07767{\text{″¾}}^{\text{3}}\end{array}$

The volume can also be expressed as,

$\begin{array}{l}V\text{=}dA\\ d\text{=}\frac{V}{A}\\ d\text{=}\frac{V}{\frac{\mathrm{Ï€}}{4}{D}^2}\end{array}$

Here, dis the additional distance and A is the area of buoy.

Substitute 0.07767 m³ for V, and 0.9 m for D in above equation,

$\begin{array}{c}d\text{=}\frac{0.07767{\text{″¾}}^{\text{3}}}{\frac{\mathrm{Ï€}}{4}{\left(0.900\text{″¾}\right)}^2}\\ d\text{=}0.122\text{″¾}\end{array}$

Thus, the additional distance buoy will sink is 0.122m.