Answer
(a) $\rho_{liq} gAx$
(b) $0.062J$
Work Step by Step
(a) We know that the pressure applied at a distance $x$ is given as
$p_{x}=\rho gx$
Now the applied force is given as
$F=p_{x}A$
$\implies F=\rho_{liq} gAx$
(b) We know that
$W=\frac{Ag\rho }{2}x^2$
$\implies W=\frac{\pi D^2 g\rho}{8}x^2$
We plug in the known values to obtain:
$W=\frac{\pi(0.04)^2(9.8)(1000)}{8}(0.1)^2$
$\implies W=0.062J$
