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 $4h$ , then it will flow out in time

• A. $2t$
• B. $4t$
• C. $16t$
• 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
$v=\frac{\sqrt{2gh}+0}{2}=\frac{\sqrt{2gh}}{2}$
$\therefore V=Avt$ ...(i)
When the vessel is filled to height 4 $h$ , then the volume in the vessel
$=4V=4Avt=4A\frac{\sqrt{2gh}}{2}\times t$
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 $t_1=\frac{4V}{A{v}_1}=\frac{4A\sqrt{2gh}\times t\times 2}{2\times A\times \sqrt{2g\times 4h}}=2t$