Question

If the container filled with liquid gets accelerated horizontally, find the value of acceleration $a$ for which point $x$ is exposed to air as shown in the figure.

• A. $\frac{3g}{4}$
• B. $\frac{4g}{3}$
• C. $\frac{2g}{3}$
• D. $\frac{3g}{2}$

Answer 1

The correct option is B $\frac{4g}{3}$



We have,
$tan\theta =\frac{a}{g}$
Also, from the diagram,
$tan\theta =\frac{L}{\frac{3L}{4}}=\frac{4}{3}$
$\Rightarrow \frac{a}{g}=\frac{4}{3}$
$\therefore a=\frac{4g}{3}$.