Answer
See answers.
Work Step by Step
a. Apply Bernoulli’s equation to the two ends.
$$P_{sink}+\frac{1}{2}\rho v_{sink}^2+\rho gy_{sink}= P_{pail}+\frac{1}{2}\rho v_{pail}^2+\rho gy_{pail}$$
Assume the pressure at the tube ends is the same.
$$v_{pail}=\sqrt{2g(y_{sink}-y_{pail})}=\sqrt{2(9.80\;m/s^2)(0.40m)}=2.8\;m/s$$
b. The volume flow rate at the bottom end, multiplied by the time to empty the sink, equals the volume of water in the sink.
$$(Av)_{pail}t=V_{sink}$$
$$t=\frac{V_{sink}}{(Av)_{pail}}=\frac{(0.38\;m^2)(0.04\;m)}{\pi(0.0115\;m)^2(2.8\;m/s)}=13\;s$$