Question

Angle of inclination of the water surface from horizontal for the given accelerated container shown in the figure is [Assume $g=10{\text{m/s}}^2$]

• A. ${45}^{\circ }$
• B. ${30}^{\circ }$
• C. ${60}^{\circ }$
• D. ${70}^{\circ }$

Answer 1

The correct option is A ${45}^{\circ }$

For a horizontally accelerated fluid, the inclination of the free surface of fluid is given by,
$\text{tan}\theta =\frac{a}{g}...\left(i\right)$
The above result comes after applying equilbrium condition on the liquid mass with respect to container, due to the reason that free surface of a liquid is always oriented perpendicular to the net force on fluid mass.


$\Rightarrow \text{tan}\theta =\frac{10}{10}$
$\Rightarrow \text{tan}\theta =1$
$\therefore \theta ={\tan }^{-1}1={45}^{\circ }$
Angle of inclination of the free surface of water with the horizontal is ${45}^{\circ }$.