Question

A cylindrical vessel is filled with water upto a height of $4\text{m}$. The cross-sectional area of the orifice at the bottom is $\frac{1}{100}$ that of the vessel. Then:

• A. 1.5 minutes is the required time to empty the tank.
• B. Initial velocity of discharge is $8.85\text{m/s}$.
• C. Velocity of discharge will decrease with time.
• D. Same amount of water can be thrown out from the orifice in $45\text{seconds}$ if the level of water is maintained at $4\text{m}$.

Answer 1

Answer: (A), (B), (C), (D)

Same amount of water can be thrown out from the orifice in $45\text{seconds}$ if the level of water is maintained at $4\text{m}$.


Let $t$ be the time taken to empty the tank.
Then, $t=\frac{2A}{a\sqrt{2g}}\sqrt{H}$
$=\frac{2\times 100}{\sqrt{2\times 9.8}}\times \sqrt{4}[\because \frac{A}{a}=100(\text{given}\left)\right]$
$=90.35\text{sec}\simeq 90\text{sec}$
$\approx 1.5\text{minutes}$
Initial velocity of discharge $v_0=\sqrt{2gH}$
$=\sqrt{2\times 9.8\times 4}=8.85\text{m/s}$

As the level of water ($h$) decreases with time, velocity of discharge $v=\sqrt{2gh}$ will also decrease.

If water level is maintained constant, discharge will also remain constant
i.e $Q=a\sqrt{2gH}$
Total volume of water required to be emptied $\left(V\right)=AH$
$\therefore$ Time taken $\left(t_1\right)$ to throw out the same amount of water (volume $V$) at constant rate
$t_1=\frac{V}{Q}=\frac{AH}{a\sqrt{2gH}}=\frac{100\times 4}{\sqrt{2\times 9.8\times 4}}=45.17\text{sec}$
$\therefore t_1\approx 45\text{sec}$
Therefore, options (a), (b), (c) and (d) are correct.