Question

Width of the base of the accelerated container shown in the figure is: [Assume $g=10{\text{m/s}}^2$]

• A. $10\text{m}$
• B. $5\text{m}$
• C. $6\text{m}$
• D. $8\text{m}$

Answer 1

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

Let the width of base of container be $x$.
The free surface of the liquid will get tilted and become perpendicular to the net force (of pseudo force $\&$ weight) on the fluid mass.


From the triangle shown in figure $\Delta ABC$,
$\text{tan}\theta =\frac{5}{x}...\left(1\right)$
For horizontally accelerated fluid,
$\text{tan}\theta =\frac{a}{g}$
$\Rightarrow \text{tan}\theta =\frac{10}{10}=1...\left(2\right)$
From Eq.$\left(1\right)$ and $\left(2\right)$,
$\frac{5}{x}=1$
$\therefore x=5\text{m}$