Three identical vessels `A, B` and `C` contain same quantity of liquid. In each vessel balls of different densities but same masses are placed. In vessel `A`, the ball is partly submerged, in vessel `B`, the ball is completely submerged but floating and in vessel `C`, the ball has sunk to the base. If `F_(A), F_(B)` and `F_(C)` are the total forces acting on the base of vessels `A, B` and `C`, respectively, then

Options

A. `F_(A)=F_(B)=F_(C)`
B. `F_(A)ltF_(B)ltF_(C)`
C. `F_(A)=F_(B)ltF_(C)`
D. `F_(A)ltF_(B)ltF_(C)`

Answer

A

Explanation

`W=` weight of liquid
`f_(B)=` buoyant force on the ball
`mg=` weight of the ball
`N=` normal reaction between the ball and the surface
The free body diagrams of the balloon each vessel are as follows

At base reaction force of buoyant force will act in downward direction.
The force acting at the base of each tank are
`F_(A)=W+f_(B)=W+mg`
`F_(B)=W+f_(B)=W+mg`
`F_(C)=W+F_(B)+N=W+mg`
Thus `F_(A)=F_(B)=F_(C)`