Question

Two vessels A and B have the same base area and contain water to the same height, but the mass of water in A is four times that in B. The ratio of the liquid thrust at the base of A to that at the base of B is:

A. $4:1$
B. $2:1$
C. $1:1$
D. $16:1$

Answer 1

Hint: The pressure exerted by a liquid at the bottom of a container is $P = h\rho g$. Where, $h$ and $\rho $ are the height and density of the liquid in the container. $g$ is the acceleration due to gravity and it is $9.8m.{s^{ - 2}}$.
The thrust on the base of a container due to any liquid is $F = P \cdot A$. Where, $A$ is the area of the base of the container.

Complete step by step answer:
It is given that the base area of the two vessels A and B are equal. Let $A$ is the base area.
Also, the two vessels contain water to the same height $(H)$.
The pressure exerted by water at the bottom of the vessel A is ${P_A} = H\rho g$.
The pressure exerted by water at the bottom of the vessel B is ${P_B} = H\rho g$.
We know that the thrust due to any liquid at the bottom of a container is equal to the product of pressure exerted by the liquid and the base area of the container.
Since both the vessels have the same pressure at the bottom and the same base area, the ratio of the liquid thrust at the base of vessels A and B will be $1:1$.
i.e., $\dfrac{{{F_A}}}{{{F_B}}} = \dfrac{{{P_A} \cdot A}}{{{P_B} \cdot A}}$
But ${P_A} = {P_B}$
$ \Rightarrow \dfrac{{{F_A}}}{{{F_B}}} = 1:1$
Hence, the correct option is (C) $1:1$.

Note:The density for particular liquid is constant for a given temperature and pressure. At room temperature the density of water is $998.2kg/{m^3}$ .
The pressure is the ratio of the force applied on a surface of a body and the surface area of the body over which the force is applied.