A large wooden piece in the form of a cylinder floats on water with two-thirds of its length immersed. When a man stands on its upper surface, a further one-sixth of its length is immersed. The ratio between the masses of the man and the wooden piece is

Let M and m be the masses of the wooden piece and the man, V the volume of the cylinder and p the density of water. Then by law of floatation,

$\begin{array}{l} \mathrm{Mg}=\frac{2\mathrm{V\rho }}{3}\mathrm{g}\Rightarrow \mathrm{M}=\frac{2\mathrm{V\rho }}{3} .......\left(\mathrm{i}\right)\\ \text{ and }(\mathrm{M}+\mathrm{m})\mathrm{g}=\left(\frac{2}{3}+\frac{1}{6}\right)\mathrm{V\rho g}=\frac{5\mathrm{V\rho }}{6}\mathrm{g}\\ \Rightarrow (\mathrm{M}+\mathrm{m})=\frac{5\mathrm{V\rho }}{6} ........\left(\mathrm{ii}\right)\end{array}$

From Eqs. (i) and (ii), we get $\frac{\mathrm{m}}{\mathrm{M}}=\frac{1}{4}$.