Answer
(a) $\sqrt{2gd}$
(b) same
Work Step by Step
(a) According to the Bernoulli's equation
$P_1+\frac{1}{2}\rho v_1^2+\rho gy_1=P_2+\frac{1}{2}\rho v_2^2+\rho gy_2$
$\implies P_1+\frac{1}{2}(0)^2+\rho gd=P_2+\frac{1}{2}\rho v_2^2+\rho g(0)$
$\implies P_1+\rho gd=P_2+\frac{1}{2}\rho v_2^2$
This simplifies to:
$v_2=\sqrt{2gd}$
This is the required speed of water leaving the siphon.
(b) We know that the speed of a fluid changes when the cross sectional area of the siphon through which it flows changes. If the cross sectional area of the siphon is constant then the speed of the fluid remains constant. Thus, we conclude that the speed of the water is the same.
