A thin rectangular orifice of width b and depth d connects two reservoirs with a difference in liquid levels H. Neglecting losses other than coefficient of discharge Cd, the theoretical discharge expression is:

Introduction / Context:
Flow through thin-plate orifices is a staple topic in hydraulics. When the head difference across the plate is H and the orifice has width b and depth d (rectangular opening), the discharge is estimated from the product of effective area and theoretical jet velocity, corrected by the discharge coefficient Cd.

Given Data / Assumptions:

• Thin, sharp-edged rectangular orifice of width b and depth d.
• Head difference across the plate is H.
• Coefficient of discharge = Cd accounts for contraction and velocity effects.

Concept / Approach:
The theoretical velocity of efflux under head H is v_th = sqrt(2 * g * H). The geometrical area of the opening is A = b * d. Actual discharge equals Q = Cd * A * v_th = Cd * b * d * sqrt(2 * g * H).

Step-by-Step Solution:
Compute area: A = b * d.Compute theoretical jet speed: v_th = sqrt(2 * g * H).Multiply and apply Cd: Q = Cd * b * d * sqrt(2 * g * H).

Verification / Alternative check:
Dimensional analysis confirms units: [Q] = L^3/T from (L * L) * sqrt(L/T^2) = L^2 * L^0.5 / T = L^2.5/T (wait), with Cd making no unit change. But since sqrt(2 g H) has units L/T and A has L^2, Q has L^3/T as required.

Why Other Options Are Wrong:
Cd * b * sqrt(2 g H): missing the depth d (area incomplete).Cd * b * d * (2 g H): missing the square root; would massively overpredict Q.Cd * b * d^2 * sqrt(2 g H): area wrongly taken as b d^2.

Common Pitfalls:

• Using head to the center of orifice when the head varies significantly across depth; for small d relative to H, taking uniform H is acceptable, otherwise integrate over depth.
• Confusing Cd with Cc or Cv individually.

Final Answer:
Cd * b * d * sqrt(2 * g * H)