An open rectangular tank $5m\times 4m\times 3$ m high containing water upto a height of 2m is accelerated horizontally along the longer side.  $(g=10m/s^2,{\rho }_{water}=1000kg/m^3)$

a) Volume of water inside the tank remains constant 
$\left(\frac{3+y_0}{5}\right)5\times 4=5\times 2\times 4$  or  $y_0=1m$
$\therefore \tan {\theta }_0=\frac{3-1}{5}=0.4$ 
 
Since, $\tan {\theta }_0=\frac{a_0}{g},$  therefore $a_0=0.4g=4m/s^2$ 
b) When acceleration is increased by 20%
                                $a=1.2a_0=0.48g$
                                $\tan \theta =\frac{a}{g}=0.48$
 
Now,  $y=3-5\tan \theta =3-5\left(0.48\right)=0.6m$
Fraction of water spilt over 
                   $=\frac{4\times 2\times 5-\frac{(3+0.6)}{2}\times 5\times 4}{2\times 5\times 4}=0.1$
Percentage of water spilt over = 10%