Question

Liquid is kept in a container having square cross section. It is now accelerated from left to right with an acceleration of $5{\text{m/s}}^2$. Due to this the free surface makes some inclination with horizontal. If the length of the square cross section is $10\text{m}$, find the difference in heights of liquid at the left and the right wall. (Take $g=10{\text{m/s}}^2$)

• A. $4\text{m}$
• B. $5\text{m}$
• C. $6\text{m}$
• D. $10\text{m}$

Answer 1

The correct option is B $5\text{m}$

When a fluid kept in a container is accelerated then the free surface makes an angle $\theta$ with the horizontal given by $tan\theta =\frac{a}{g}$.........(1)
Let we call the legth of the cross section as $b$ and the difference in height as $\Delta z$, then we can also express $\theta$ as
$tan\theta =\frac{\Delta z}{b}$........(2)
Comparing (1) and (2)
$\Delta z=b\times \frac{a}{g}$
$\Delta z=10\times \frac{5}{10}$
$\Delta z=5\text{m}$