Three rigid buckets, shown as in the figure (1) (2) and (3), are of identical heights and base areas. further assume that each of these buckets have negligible mass and are full of water. The weights of water in these buckets are denoted as $W_{1}$ , $W_{2}$ , and $W_{3}$ respectively. Also, let the force of water on the base of the bucket be denoted as $F_{1}$ , $F_{2}$ and $F_{3}$ respectively. The option giving an accurate description of the system physics is

W_{2}

W_{1}

W_{3}

and

F_{2}

F_{1}

F_{3}

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W_{2}

W_{1}

W_{3}

and

F_{2}

F_{1}

F_{3}

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W_{2}

W_{1}

W_{3}

and

F_{1}

F_{2}

F_{3}

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W_{2}

W_{1}

W_{3}

and

F_{1}

F_{2}

F_{3}

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Solution

The correct option is D $W_{2}$ > $W_{1}$ > $W_{3}$ and $F_{1}$ = $F_{2}$ = $F_{3}$
Bucket $\to$ identical height

$\to$ identical base area

$\Rightarrow$ $W_{2}$ > $W_{1}$ > $W_{3}$

Force on the base in each case will be equal to = $γ_{w} h A$

Hence $F_{1}$ = $F_{2}$ = $F_{3}$

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