Answer
(a) $11N$
(b) same
Work Step by Step
(a) We know that
$F=\rho g(h_1A_{base}+h_2A_{ring})$
We plug in the known values to obtain:
$F=(1000Kg/m^3)(9.81m/s^2)(0.18m)(24\times 10^{-4}m^2)-(0.09m)(72\times 10^{-4}m^2)=11N$
(b) We know that the total weight of water in container 2 is given as
$W=mg$
$W=\rho Vg$
$\implies W=\rho g(h_1A_{base}+h_2A_{ring})$
We plug in the known values to obtain:
$W=(1000Kg/m^3)(9.81m/s^2)(0.18m)(24\times 10^{-4}m^2)-(0.09m)(72\times 10^{-4}m^2)$
$W=11N$
Thus, the result from part (a) is equal to the weight of the water in container 2.
