Question

A rectangular vessel is filled with water and oil in equal proportion (by volume) , the oil being twice lighter than water. Show that the force on each side wall of the vessel will be reduced by one fifth if the vessel is filled only with oil. (Assume atm. pressure to be negligible)

Answer 1

@iambatman @hartnn @ganeshie8 Any help will be greatly appreciated.

Answer 2

As a start, I've done the following:

For rect. vessel containing water & oil both :

Pressure (Total) = \(\rho_{w} g x + \rho_{o} g x \)

And we are given with specific gravity of oil = \(\cfrac{\rho_o}{\rho_w} = \cfrac{1}{2} = 0.5 \) . Thus, \(2 \rho_o = \rho_w \)

Therefore, P (Total) = \(2 \rho_o gx + \rho_o gx = 3 \rho_o gx\)

For the vessel containing only oil :

P (Total) = \(\rho_o g (2x) = 2 \rho_o gx\)

Where : 2x being the height of the vessel.