Question

In the given situation what should be the minimum horizontal acceleration given to the vessel so that liquid does not come out of the hole when it is open?

• A. $g$
• B. $g\left(H/l\right)$
• C. $g\left(2h/l\right)$
• D. $g(H-h)l$

Answer 1

The correct option is C $g\left(2h/l\right)$

Area of profile above the hole must be constant

$\tan \theta =\frac{a}{g}hl=\frac{1}{2}l\times x\Rightarrow x=2h\tan \theta =\frac{2h}{l}=\frac{a}{g}a=g\left(\frac{2h}{l}\right)$