Topic: Fluid Mechanics
A tank of uniform cross section, sectional cross area (A) containing liquid upto height ($H_1$) has an orifice of cross sectional area (a) at its bottom. The time required to empty the tank completely will be
- A. $\frac{2A\sqrt{H_{1}}}{C_{d}a\sqrt{2g}}$
- B. $\frac{2AH_{1}}{C_{d}a\sqrt{2g}}$
- C. $\frac{2AH_{1}^{3/2}}{C_{d}a\sqrt{2g}}$
- D. $\frac{2AH_{1}^{2}}{C_{d}a\sqrt{2g}}$
Answer: A. $\frac{2A\sqrt{H_{1}}}{C_{d}a\sqrt{2g}}$
Explanation: Let the height of water at any instant be $h$, and in time $dt$, the height changes by $dh$. The volume flow rate equates to $-A dh = C_{d}\sqrt{2gh} \cdot a \cdot dt$. Integrating both sides from the initial height $H_1$ to final height $H_2 = 0$ gives $t = \frac{2A\sqrt{H_1}}{aC_{d}\sqrt{2g}}$.