Physics - Fluid Question with Solution | TestHub

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

$v=\frac{\sqrt{2 \mathrm{g}h}+0}{2}=\frac{\sqrt{2\mathrm{g}h}}{2}$

$\therefore V=A v t$...(i)

When the vessel is filled to height 4$h$, then the volume in the vessel

$=4V=4A vt=4A\frac{\sqrt{2\mathrm{g}h}}{2}\times t$

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$t_1=\frac{4 V}{A{ v}_1}=\frac{4 A\sqrt{2\mathrm{g}h}\times t\times 2}{2\times A\times \sqrt{2\mathrm{g}\times 4h}}=2t$