Question

A wide tank of cross section area $A$ containing a liquid to a height $H$has an orifice as it base of area $\frac{A}{3}$.The initial acceleration of top surface of liquid is

• A. $\frac{g}{2}$
• B. $\frac{g}{4}$
• C. $\frac{g}{8}$
• D. $\frac{g}{9}$

Answer 1

The correct option is C $\frac{g}{8}$

Let v be the velocity at the surface of liquid. Then from continuity principle, velocity at the orifice will be $3v$

From Bernouli's principle, we have,

$P+\frac{1}{2}p{v}^2=pgH=p+\frac{1}{2}p(3v{)}^2+0$

where P = atmospheric pressure, is the density of the liquid.

Solving above equation for v, we have

$V^2=gh/4$

differentiating both sides with respect to time, we get

$2v\frac{dv}{dt}=g/4\frac{dh}{dt}$

$==>2v\frac{dv}{dt}=-gv/4\text{as}\frac{dh}{dt}=-v$

$==>a=-g/8$

Magnitude of acceleration is $\frac{g}{8}$