91Ó°ÊÓ

Buoyancy is an upward force equal to the weight of the water displaced by the body. Here the buoyant force on the ice is equal to the weight of the ice. It is expressed in Newtons.

The specific gravity of iceis \(\left( {{\rm{S}}{{\rm{G}}_{{\rm{ice}}}}} \right) = 0.917\).

The specific gravity of seawater is \(\left( {{\rm{S}}{{\rm{G}}_{{\rm{seawater}}}}} \right) = 1.025\).

As we have discussed above,

\(\begin{aligned}{F_{{\rm{buoyant}}}} &= {W_{{\rm{ice}}}}\\{m_{\scriptstyle{\rm{seawater}}\atop\scriptstyle{\rm{submerged}}}}g &= {m_{{\rm{ice}}}}g\\{m_{\scriptstyle{\rm{seawater}}\atop\scriptstyle{\rm{submerged}}}} &= {m_{{\rm{ice}}}}\end{aligned}\)

Here \({m_{\scriptstyle{\rm{seawater}}\atop\scriptstyle{\rm{submerged}}}}\)is the mass of the seawater displaced by the submerged iceberg and\({m_{{\rm{ice}}}}\) is the mass of the ice.

We know that the ratio of mass and volume is equal to density. Therefore,

\({\rho _{{\rm{seawater}}}}{V_{{\rm{seawater}}}} = {\rho _{{\rm{ice}}}}{V_{{\rm{ice}}}}\)

But in the question, specific gravity is given, which is equal to the ratio of the density of the fluid to the density of the standard fluid (example, water), i.e.,

\({\left( {SG} \right)_{{\rm{seawater}}}} = \frac{{{\rho _{{\rm{seawater}}}}}}{{{\rho _{{\rm{water}}}}}};\;{\left( {SG} \right)_{{\rm{ice}}}} = \frac{{{\rho _{{\rm{ice}}}}}}{{{\rho _{{\rm{water}}}}}}\)

So, we can write,

\(\begin{aligned}{\left( {SG} \right)_{{\rm{seawater}}}}{\rho _{{\rm{water}}}}{V_{\scriptstyle{\rm{submerged}}\atop\scriptstyle{\rm{ice}}}} = {\left( {SG} \right)_{{\rm{ice}}}}{\rho _{{\rm{water}}}}{V_{{\rm{ice}}}}\\{\left( {SG} \right)_{{\rm{seawater}}}}{V_{\scriptstyle{\rm{submerged}}\atop\scriptstyle{\rm{ice}}}} &= {\left( {SG} \right)_{{\rm{ice}}}}{V_{{\rm{ice}}}}\\{V_{\scriptstyle{\rm{submerged}}\atop\scriptstyle{\rm{ice}}}} &= \frac{{{{\left( {SG} \right)}_{{\rm{ice}}}}}}{{{{\left( {SG} \right)}_{{\rm{seawater}}}}}}{V_{{\rm{ice}}}}\end{aligned}\)

Substitute the values in above formula.

\(\begin{aligned}{V_{\scriptstyle{\rm{submerged}}\atop\scriptstyle{\rm{ice}}}} &= \frac{{0.917}}{{1.025}}{V_{{\rm{ice}}}}\\{V_{\scriptstyle{\rm{submerged}}\atop\scriptstyle{\rm{ice}}}} &= 0.895\;{V_{{\rm{ice}}}}\end{aligned}\)

The fraction of an iceberg above the surface of the water is

\(\begin{aligned}{V_{{\rm{above}}}} &= {V_{{\rm{ice}}}} - {V_{{\rm{submerged}}}}\\{V_{{\rm{above}}}} &= 0.105\;{V_{{\rm{ice}}}}\\{V_{{\rm{above}}}} &= 10.5\% \end{aligned}\)

Hence, the percent of an iceberg above the surface of the water is \(10.5\% \).