Question

An open rectangular tank $1.5\text{m}$ wide, $2\text{m}$ deep and $3\text{m}$ long is half filled with water. It is accelerated horizontally at $3.27{\text{ms}}^{-2}$ in the direction of its length. Determine the depth of water at each end of the tank. $[g=9.81{\text{ms}}^{-2}]$

• A. $0.5\text{m}$ and $1.5\text{m}$
• B. $0.6\text{m}$ and $1.4\text{m}$
• C. $0.7\text{m}$ and $1.3\text{m}$
• D. $1\text{m}$ and $1\text{m}$

Answer 1

The correct option is A $0.5\text{m}$ and $1.5\text{m}$

If a container containing liquid is accelerated with acceleration $a$, then there will be a difference in the height of liquid between horizontal ends, and the angle subtended by the free surface is given by,

$\tan \theta =\frac{a}{g}$

Here, $a=3.27{\text{ms}}^{-2},g=9.81{\text{ms}}^{-2}$

$\Rightarrow \tan \theta =\frac{1}{3}$



$\Rightarrow \frac{x}{\left(\frac{l}{2}\right)}=\frac{1}{3}$

$\Rightarrow x=\frac{l}{6}=0.5\text{m}$

Depth at end $A$

$h_A=1-0.5$

$h_A=0.5\text{m}$

Depth at end $B$

$h_B=1+0.5$

$h_B=1.5\text{m}$

Hence, option $\left(\text{A}\right)$ is correct.