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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


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Comments ( 1 )

  1. Madhavan Palmani
    Madhavan Palmani :
    4 years ago

    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

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