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

Options

A. `1:2`
B. `1:3`
C. `1:4`
D. `1:5`

Answer

C

Explanation

Let `M` and `m` be the masses of the wooden piece and the man `V` the volume of the cylinder and `rho` the density of water. Then by law of floatation
`Mg=(2Vrho)/3gimpliesM=(2Vrho)/3`……..i
and `(M+m)=(2/3+1/6)Vrhog=(5Vrho)/6g`
`implies(M+m)=(5Vrho)/6`
From eqn i and ii we get `m/M=1/4`