Question

An open rectangular tank ($5m\times 4m\times 3m$) containing water upto a height of $2m$ as shown, is accelerated $\left({\text{a}}_0\right)$ horizontally along the larger side, and the water is just about to spill



If the acceleration is increased by 20%, then the percenatage of water spill is (Answer upto two decimal places)

Answer 1

For a fluid in a container moving with acceleration ${\text{a}}_0$
$\tan {\theta }_0=\frac{{\text{a}}_0}{g}$
Just before the water spills, as shown in the figure

Volume of water inside the tank remains constant
$\therefore \frac{3+y_0}{2}\times 5\times 4=5\times 2\times 4or{y}_0=1m\tan {\theta }_0=\frac{3-y_0}{5}=\frac{2}{5}=0.4\text{Since}\tan {\theta }_0=\frac{a_0}{g},a_0=0.4g=4m{s}^{-2}$
When acceleration is increased by 20%,
a = 1.2$a_0$ = 0.48g
$\tan \theta =\frac{a}{g}=0.48$
After spilling,


$\tan \theta =\frac{3-y}{5}\Rightarrow y=3-5\tan \theta =3-5\times 0.48=0.6m\text{Volume of liquid before spilling}=2\times 5\times 4=40m^3\text{Volume of liquid after spilling}=\frac{1}{2}\times 2.4\times 5\times 4+0.6\times 5\times 4=36m^3\therefore \text{percentage of liquid spilled}=\frac{40-36}{40}\times 100=10\%$