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 bring the liquid level from H₁ to H₂ will be

A. $$\frac{{2{\text{A}}\sqrt {{{\text{H}}_1}} }}{{{{\text{C}}_{\text{d}}} \times {\text{a}}\sqrt {2{\text{g}}} }}$$

B. $$\frac{{2{\text{A}}\sqrt {{{\text{H}}_2}} }}{{{{\text{C}}_{\text{d}}} \times {\text{a}}\sqrt {2{\text{g}}} }}$$

C. $$\frac{{2{\text{A}}\left( {\sqrt {{{\text{H}}_1}} - \sqrt {{{\text{H}}_2}} } \right)}}{{{{\text{C}}_{\text{d}}} \times {\text{a}}\sqrt {2{\text{g}}} }}$$

D. $$\frac{{2{\text{A}}\left( {{\text{H}}_1^{\frac{3}{2}} - {\text{H}}_2^{\frac{3}{2}}} \right)}}{{{{\text{C}}_{\text{d}}} \times {\text{a}}\sqrt {2{\text{g}}} }}$$

Answer: Option C