In Borda mouthpiece, the coefficient of contraction is 

Explanation:

Borda's mouthpiece is a short cylindrical tube projecting inward. The edge of the tube must be relatively thin and sharp to ensure perfect contraction, and its length, about ½ diameter, such that the jet will not touch the sides of the tube.

Pressure force at the entrance of the mouth piece

F = ρ × g × H × a , where a is the area of the mouthpiece.

H = distance of center of gravity of the mouthpiece from the free surface.

from Newton's second law of motion, the rate of change of momentum = net force 

ρ × a × vc × (vc - 0) = ρ × g × H × a 

ρ × ac × vc2 = ρ × g × H × a ..........(A) 

applying Bernoulli's equation

\(\frac{{{P_1}}}{{\rho g}} + \frac{{V_1^2}}{{2g}} + {z_1}+ {h_f} = \frac{{{P}}}{{\rho g}} + \frac{{V^2}}{{2g}} + {z} \)

\( \frac{{v_c^2}}{{2g}} = H \) ⇒ \(v_c = \sqrt {2gH}\) 

P1 = P2 = Patm ​= 0, hf  = 0 (since the fluid is frictionless)

from eqn. (A) 

ρ × ac × 2g × H = ρ × g × H × a

ac = \(\frac{a}{2}\) ⇒ \(C_c =\frac{a_c}{a} = 0.5\)