91Ó°ÊÓ

The given data can be listed below as,

• The density of oil is$750kg/m^3$ .
• The mass of each empty barrel is 15.0 kg .
• The volume of oil to fill the barrel is 0.120 $m^3$.
• The density of seawater is$1.03×{10}^{3}kg/m^3$ .
• If the density of oil is $910kg/m^3$and mass of each empty barrel is 32.0 kg .

The buoyant force is readily available whether the object floats or sinks.

The total density of the oil fill barrel is equal to the sum of the density of the barrel and density of the oil. It is express as,

$\begin{array}{r}{\mathrm{ÒÏ}}_{\text{Total}}={\mathrm{ÒÏ}}_{\text{bamel}}+{\mathrm{ÒÏ}}_{\text{oil}}\\ {\mathrm{ÒÏ}}_{\text{Total}}=\frac{m_{\text{bamel}}}{V}+{\mathrm{ÒÏ}}_{\text{oil}}\end{array}$

Substitute all the values in the above equation.

$\begin{array}{r}{\mathrm{ÒÏ}}_{\text{Total}}=\frac{15.0\mathrm{kg}}{0.120{\mathrm{m}}^3}+750\mathrm{kg}/{\mathrm{m}}^3\\ =875\mathrm{kg}/{\mathrm{m}}^3\end{array}$

The total density is less than the seawater density i.e. (${\mathrm{ÒÏ}}_{Total}<{\mathrm{ÒÏ}}_{sea}$), hence the oil filled barrel floats.

The oil fill barrel is floating, and then the buoyant force exerted by sea water is equal to the weight of barrel i.e.

$\begin{array}{r}{F}_B=W\\ {\mathrm{ÒÏ}}_{\text{sea}}×V_{\text{submerged}}×g=m_{\text{Total}}×g\\ {\mathrm{ÒÏ}}_{\text{sea}}×V_{\text{submerged}}=m_{\text{Total}}\end{array}$

Here is the sum of mass of the barrel and oil inside.

$\begin{array}{r}{\mathrm{ÒÏ}}_{\text{sea}}×V_{\text{submerged}}={\mathrm{ÒÏ}}_{\text{Total}}×V\\ \frac{V_{\text{submerged}}}{V}=\frac{{\mathrm{ÒÏ}}_{\text{Total}}}{{\mathrm{ÒÏ}}_{\text{sea}}}\end{array}$

The fraction of the barrel volume that will be more than the water surface is then express as,

$1−\frac{V_{\text{submerged}}}{V}=1−\frac{{\mathrm{ÒÏ}}_{\text{Total}}}{{\mathrm{ÒÏ}}_{\text{sea}}}$

Substitute all the values in the above equation.

$\begin{array}{r}1−\frac{V_{\text{submerged}}}{V}=1−\frac{875\mathrm{kg}/{\mathrm{m}}^3}{1.03×{10}^3\mathrm{kg}/{\mathrm{m}}^3}\\ =0.15\end{array}$

Hence the fraction of the barrels volume that will be over the water surface is 0.15.

The total density of the oil fill barrel is equal to the sum of the density of the barrel and density of the oil.If the density of oil is$910kg/m^3$ andmass of each empty barrel is 32.0 kg .

$\begin{array}{r}{\mathrm{ÒÏ}}_{\text{Total}}={\mathrm{ÒÏ}}_{\text{barrel}}+{\mathrm{ÒÏ}}_{\text{oil}}\\ {\mathrm{ÒÏ}}_{\text{Total}}=\frac{m_{\text{barrel}}}{V}+{\mathrm{ÒÏ}}_{\text{oil}}\end{array}$

Substitute all the values in the above equation.

$\begin{array}{r}\text{Total}=\frac{32.0\mathrm{kg}}{0.120{\mathrm{m}}^3}+910\mathrm{kg}/{\mathrm{m}}^3\\ =1176.67\mathrm{kg}/{\mathrm{m}}^3\end{array}$

The total density is greater than the seawater density i.e. (${\mathrm{ÒÏ}}_{Total}<{\mathrm{ÒÏ}}_{sea}$), hence the oil filled barrel sinks.

To stability the forces, an upward force need to be exerted at the barrel such that the network force at the barrel is zero. This force is the tension T inside the rope. So,

$T+F_B=W$

Substitute the value of$F_B$ and w in the above equation.

$\begin{array}{r}T+{\mathrm{ÒÏ}}_{\text{sea}}×V×g=m_{\text{Total}}×g\\ T+{\mathrm{ÒÏ}}_{\text{sea}}×V×g={\mathrm{ÒÏ}}_{\text{Total}}×V×g\\ T=\left({\mathrm{ÒÏ}}_{\text{Total}}−{\mathrm{ÒÏ}}_{\text{sea}}\right)Vg\end{array}$

Substitute all the values in the above equation.

$\begin{array}{r}T=\left(1176.67\mathrm{kg}/{\mathrm{m}}^3−1.03×{10}^3\mathrm{kg}/{\mathrm{m}}^3\right)×0.120{\mathrm{m}}^3×9.81\mathrm{N}/\mathrm{kg}\\ =172.65\mathrm{N}\end{array}$

Hence the barrel sinks 172.65 N .