Question

An open cubical tank was initially fully filled with water. When the tank was accelerated on a horizontal plane along one of its side it was found that one third of volume of water spilled out. The acceleration was

• A. $g/3$
• B. $2g/3$
• C. $3g/2$
• D. None

Answer 1

The correct option is B $2g/3$

Let each side of cube be of length '$H$'.

Given,Volume of water spilled out = $\frac{1}{3}\times H^3$ --- (1)

If the surface dip by height '$h$'

Then,the volume of spilled liquid = area of triangle$\times$side of cube

= $\frac{1}{2}\times h\times H^2$ --- (2)

From (1),(2) $\frac{1}{3}\times H^3$ = $\frac{1}{2}\times h\times H^2$

$h=\frac{2}{3}\times H$

Angle made by the surface with horizontal be $\theta$,

$\tan \theta$ = $\frac{a}{g}$ (See concept)

From figure, $\tan \theta$ = $\frac{h}{H}$

So, $a=g\times \frac{h}{H}$

$a=\frac{2g}{3}$