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Reynold's Number

Three identic...

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 $F_{C}$ are the total forces acting on the base of vessels A, B and C respectively, then

A

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

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B

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

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C

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

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D

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

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Solution

The correct option is A
$F_{A} = F_{B} = F_{C}$

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

$F_{A} = W + f_{B} = W + m g$

$F_{B} = W + f_{B} = W + m g$

$F_{C} = W + f_{B} + N = W + m g$

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

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