Two vessels A and B have the same base area a | vedclass

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(C) The pressure $P$ at the base of a vessel containing a liquid of density $\rho$ at a height $h$ is given by $P = \rho gh$.The thrust $T$ (force) ex

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$1$

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(C) The pressure $P$ at the base of a vessel containing a liquid of density $ho$ at a height $h$ is given by $P = ho gh$.The thrust $T$ (force) exerted on the base of the vessel is the product of the pressure $P$ and the base area $A_{base}$.Thus,$T = P 	imes A_{base} = ho gh 	imes A_{base}$.In the given problem,both vessels $A$ and $B$ have the same base area $(A_{base})$ and contain water to the same height $(h)$. Since the liquid is the same (water),the density $ho$ is also the same for both.Therefore,the thrust at the base of vessel $A$ is $T_A = ho gh A_{base}$ and the thrust at the base of vessel $B$ is $T_B = ho gh A_{base}$.The ratio of the thrusts is $\frac{T_A}{T_B} = \frac{ho gh A_{base}}{ho gh A_{base}} = 1:1$.

Two vessels $A$ and $B$ have the same base area and contain water to the same height,but the mass of water in $A$ is four times that in $B$. The ratio of the liquid thrust at the base of $A$ to that at the base of $B$ is (in $:1$)

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