Answer
A total height of $~~2.80~m~~$ of water in the long arm will put the seal on the verge of popping.
Work Step by Step
We can find the required pressure on the seal:
$P = \frac{F}{A}$
$P = \frac{9.80~N}{5.00\times 10^{-4}~m^2}$
$P = 1.96\times 10^4~N/m^2$
We can find the required additional height of the water:
$P = \rho~g~h$
$h = \frac{P}{\rho~g}$
$h = \frac{1.96\times 10^4~N/m^2}{(1000~kg/m^3)(9.80~m/s^2)}$
$h = 2.00~m$
Since $d = 0.800~m$, a total height of $~~2.80~m~~$ of water in the long arm will put the seal on the verge of popping.
