The relation from buoyant forceis given by,
\(\begin{array}{c}{W_{\rm{w}}} = {W_{\rm{P}}} + {W_{\rm{L}}}\\12\left( {{\rho _{\rm{w}}}Vg} \right) = Nmg + 12\left( {{\rho _{\rm{L}}}Vg} \right)\\N = \frac{{12V\left( {{\rho _{\rm{w}}} - {\rho _{\rm{L}}}} \right)}}{m}\\N = \frac{{12\left( {\frac{{\pi {d^2}}}{4}L} \right)\left( {{\rho _{\rm{w}}} - SG \times {\rho _{\rm{w}}}} \right)}}{m}\end{array}\)
Here, \({W_{\rm{w}}}\) , \({W_{\rm{P}}}\) and \({W_{\rm{L}}}\) is the weight of the water displaced, passengers and logs respectively, \(V\) is the volume of logs, \(m\) is the mass of the passengers, \(N\) is the number of passengers, \({\rho _{\rm{w}}}\) is the density of water, \({\rho _{\rm{L}}}\) is the density of logs and \(g\) is the gravitational acceleration.
On plugging the values in the above relation,
\(\begin{array}{l}N = \frac{{12\left( {\frac{{\pi {{\left( {45\;{\rm{cm}} \times \frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)}^2}}}{4}\left( {6.5\;{\rm{m}}} \right)} \right)\left( {1000\;{\rm{kg/}}{{\rm{m}}^3} - 0.60 \times 1000\;{\rm{kg/}}{{\rm{m}}^3}} \right)}}{{\left( {68\;{\rm{kg}}} \right)}}\\N \approx 73\end{array}\)
Thus, the required number of people are \( \approx 73\).
