Hence, Increase in pressure on the liquid in the wider tube is
$P = \dfrac{F}{A} = \dfrac{mg}{A}$
Let, $H$ is the level of water to which piston will move, thus
$P = H\rho g$
$H = \dfrac{P}{\rho g}$
$H = \dfrac{mg}{A\rho g} = \dfrac{m}{A\rho}$  
$H = \dfrac{10000}{1000}$.......($\because \rho = 1$)
$\therefore H = 10 cm$ from surface of tube.

Since, $(H-h) = 50cm, H = 60cm = 0.6 m$
The velocity of effluent from the hole is
$v = \sqrt {h\rho g}$
$v = (\sqrt {2gh}) $ 
$v = \sqrt {(2)(9.8)(0.6) } = 3.4 ms^{-1}$