$12\mathrm{.}2\mathrm{.}1($a$)$ A ship lock has vertical sides. Water enters or leaves it through a conduit

area of cross$-$sectional area A$.$ The flow through this conduit is given by Q$=$

C$\_($d$)$A$(2$gz$)^((1)/(2)),$ where z is the variable difference in level between the water

surface in the lock and that outside. Prove that the time required for the water

level in the lock to drop from z$\_(1)$ to z$\_(2)$ is

t$=(2$A$\_($s$))/($C$\_($d$)$A$\backslash$sqrt$(2$g$))($z$\_(1)^((1)/(2))-$z$\_(2)^((1)/(2)))$

$($Note: If the lock is being filled, the signs must be reversed.$)($b$)$ Suppose the

lock is $400$ ft long by $100$ ft wide, and the discharge coefficient for the conduit is

$0\mathrm{.}50.$ If the water surface in the lock is initially $40$ ft below the level of the

surface of the water upstream, how large must the conduit be if the lock is to be

filled in $10$min?