Question

There is a hole of area $A$ at the bottom of a cylindrical vessel. Water is filled up to a height $h$ and water flows out in $t$ seconds. If water is filled to a height $4 \, h$ , then it will flow out in time

• A. $2 \, t$
• B. $4 \, t$
• C. $16 \, t$
• D. None of these

Answer 1

Answer: (A)

The volume of water in the vessel of the base area $A$ and height $h$ is $V=Ah$ . The average velocity of outflowing water when the height of water changes from $h$ to 0 is
$v=\frac{\sqrt{2 \, g h} + 0}{2}=\frac{\sqrt{2 g h}}{2}$
$\therefore V=A \, v \, t$ ...(i)
When the vessel is filled to height 4 $h$ , then the volume in the vessel
$=4V=4A \, vt=4A\frac{\sqrt{2 g h}}{2}\times t \, $
If $t$ is the time taken for the outflowing liquid and $v_{1}$ is the average velocity of outflowing liquid then
$4V=A \, v_{1}t_{1}$
or $t_{1}=\frac{4 \, V}{A \, v_{1}}=\frac{4 \, A \sqrt{2 g h} \times t \times 2}{2 \times A \times \sqrt{2 g \times 4 h}}=2t$