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Byju's Answer

Standard XII

Physics

Pascal's Law

The vessel sh...

Question

The vessel shown in the figure has two sections of area of cross section $A_{1}$ and $A_{2}$ . A liquid of density $ρ$ fills both the section upto a height $h$ in each. Consider atmospheric pressure $P_{0}$ also. Find
(a) the pressure at the base of vessel
(b) the force exerted by the liquid on the base of the vessel
(c) the downward force exerted by the walls of the vessel at the level $B$ .

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Solution

(a) $P_{B a s e} = P_{0} + ρ g 2 h$
(b) Force applied by the liquid on the base $= (P_{0} + 2 ρ g h) A_{2}$
(c) Downward force by the walls of the vessel at level $B (N_{B}) =$ Pressure at $B \times$ Area at the level $B$ .
$= (P_{0} + h ρ g) (A_{2} - A_{1})$
Alternatively
$P_{0} A_{1} + N_{B} + (A_{1} h + A_{2} h) ρ g = (P_{0} + 2 ρ g h) A_{2}$
$N_{B} = P_{0} A_{2} + 2 ρ g h A_{2} - P_{0} A_{1} - A_{1} h_{1} ρ g - A_{2} h ρ g$
$= (P_{0} + h ρ g) (A_{2} - A_{1})$

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(a) PBase=P0+ρg2h(b) Force applied by the liquid on the base =(P0+2ρgh)A2(c) Downward force by the walls of the vessel at level B(NB)= Pressure at B× Area at the level B.=(P0+hρg)(A2−A1)AlternativelyP0A1+NB+(A1h+A2h)ρg=(P0+2ρgh)A2NB=P0A2+2ρghA2−P0A1−A1h1ρg−A2hρg=(P0+hρg)(A2−A1)