First slide

Pressure in a fluid-mechanical properties of fluids

Question

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 arc the total forces acting on the base of vessels A, B and C, respectively, then

Moderate

A

$F_{A} < F_{B} < F_{C}$
B

$F_{A} = F_{B} = F_{C}$
C

$F_{A} = F_{B} < F_{C}$
D

$F_{A} > F_{B} > F_{C}$

Solution

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 balls in each vessel are as follows.
At base, reaction force of buoyant force will act in downward direction.
The forces acting at the base of each tank are
$\begin{matrix} F_{A} = W + f_{B} = W + m g \\ F_{B} = W + f_{B} = W + m g \\ F_{C} = W + f_{B} + N = W + m g \\ Thus, F_{A} = F_{B} = F_{C} \end{matrix}$

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