A tank of uniform cross-sectional area (A) containing liquid upto height (H1) has an orifice of cross-sectional area (a) at its bottom. The time required to empty the tank completely will be
A. $$\frac{{2{\text{A}}\sqrt {{{\text{H}}_1}} }}{{{{\text{C}}_{\text{d}}} \times {\text{a}}\sqrt {2{\text{g}}} }}$$
B. $$\frac{{2{\text{A}}{{\text{H}}_1}}}{{{{\text{C}}_{\text{d}}} \times {\text{a}}\sqrt {2{\text{g}}} }}$$
C. $$\frac{{2{\text{AH}}_1^{\frac{3}{2}}}}{{{{\text{C}}_{\text{d}}} \times {\text{a}}\sqrt {2{\text{g}}} }}$$
D. $$\frac{{2{\text{AH}}_1^2}}{{{{\text{C}}_{\text{d}}} \times {\text{a}}\sqrt {2{\text{g}}} }}$$
Answer: Option A
This Question Belongs to Mechanical Engineering >> Hydraulics And Fluid Mechanics In ME
should be.
Volume/(Discharge).
= AH/(Cd*a*(2gH)^.5).
= (A*(H)^.5)/(Cd*a*(2g)^.5).
Why is there an extra 2 in the numerator