Question

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:

• A. 1:2
• B. 1:3
• C. 1:4
• D. 1:5

Answer 1

Answer: (C)

1:4

When only wooden piece floats, according to law of floatation
weight of the wooden piece = weight of the water displaced
$M_{w}g=\left(\frac{2}{3}V\right)\rho g$ ..........$\left(I\right)$
When man stands on the wooden piece, applying the law of floatation
weight of (wooden piece+man)=weight of the water displaced
$\Rightarrow (M_w+M_m)g=(\frac{2}{3}V+\frac{1}{6}V)\rho g$
$\Rightarrow (M_w+M_m)g=\left(\frac{5}{6}V\right)\rho g$ ..........$\left(II\right)$
Dividing $\left(I\right)$ by $\left(II\right)$, we have
$\Rightarrow \frac{M_w}{M_w+M_m}=\frac{4}{5}\Rightarrow \frac{M_m}{M_w}=\frac{1}{4}$