Question

A liquid is poured into a vessel at rest with the hole in a wall closed by a valve. It is filled to a height $H$. The distance of hole from the top surface is $h$. With what horizontal acceleration should the vessel be moved so that the liquid does not come out when valve is opened ? (Given : $l$ = length of base)

• A. $2gh$
• B. $g$
• C. $\frac{2gh}{l}$
• D. $none$

Answer 1

The correct option is C $\frac{2gh}{l}$

Given, That the hole is present at a distance $h$ from the top.

Let the horizontal acceleration applied to prevent the liquid from coming out of vessel through hole be $a,$

Due to the acceleration, the liquid surface in the vessel will tilt to angle $\theta$ with horizontal axis,

For liquid not to come out, the liquid should go $2h$ height above the hole in the opposite end.

So, we get $tan\theta =\frac{2h}{l}=\frac{a}{g}$

$\Rightarrow a=\frac{2gh}{l}$

Therefore option $C$ is correct.