Question

There is small hole in the bottom of a fixed container containing a liquid upto ${}^{\prime }{h}^{\prime }$. The top of the liquid as well as the hole at the bottom are exposed to atmosphere. As the liquid comes out of the hole. Then,
(Area of the hole is ${}^{\prime }{a}^{\prime }$ and that of the top surface is ${}^{\prime }{A}^{\prime }$) :

• A. the top surface of the liquid accelerates with acceleration $=g$
• B. the top surface of the liquid accelerates with acceleration $=g\frac{a^2}{A^2}$
• C. the top surface of the liquid retards with retardation $=g\frac{a}{A}$
• D. the top surface of the liquid retards with retardation $=\frac{g{a}^2}{A^2}$

Answer 1

The correct option is D the top surface of the liquid retards with retardation $=\frac{g{a}^2}{A^2}$

The velocity of fluid at the hole is
$V_2=\sqrt{\frac{2gh}{1+\left(a^2/A^2\right)}}$
Using continuity equation, at the two cross-sections $A$ and $B$ :


$V_{1}A=V_{2}a$
$\Rightarrow V_1=\frac{a}{A}{V}_2$
$\Rightarrow$ Acceleration of top surface
$a_1=-V_1\frac{d{V}_1}{dh}=-\left(\frac{a}{A}\right)V_2\frac{d}{dh}\left(\frac{a}{A}{V}_2\right)$
$a_1=-\frac{a^2}{A^2}{V}_2\frac{d{V}_2}{dh}$
$a_1=-\frac{a^2}{A^2}\sqrt{2gh}\sqrt{2}g.\frac{1}{2\sqrt{h}}\Rightarrow a_1=-\frac{g{a}^2}{A^2}$