Question

An open rectangular tank 5m × 4m × 3m
high containing water upto a height of 2m is accelerated horizontally along the larger side. If the acceleration is increased by 20%, the percenatage of water spilt over after increasing the acceleration is

Answer 1

For a fluid in a container moving with acceleration $a_0$



$\tan {\theta }_0=\frac{a_0}{g}$
Just before the water spills, as shown in the figure

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


$\tan \theta =\frac{3-y}{5}\Rightarrow y=3-5\tan \theta =3-5\times 0.48=0.6mVolumeofliquidbeforespilling=\frac{1}{2}\times 2\times 5\times 4+1\times 5\times 4=40m^{3}Volumeofliquidafterspilling=\frac{1}{2}\times 2.4\times 5\times 4+0.6\times 5\times 4=36m^3\therefore percentageofliquidspilled=\frac{40-36}{40}\times 100=10\%$