Question

A large open top container of negligible mass and uniform cross sectional area $A$ has a small hole of cross sectional area ${}^{\prime }{a}^{\prime }$ in its side wall near the bottom. The container is kept over a smooth horizontal floor and contains a liquid of density ${}^{\prime }{\rho }^{\prime }$ and mass ${}^{\prime }{m}^{\prime }$. Assuming that the liquid starts flowing through the hole the acceleration of the container will be :

• A. $\frac{2ag}{A}$
• B. $\frac{ag}{A}$
• C. $\frac{2Ag}{a}$
• D. $\frac{Ag}{a}$

Answer 1

The correct option is B $\frac{ag}{A}$

Let at any time the height of water be $x.$

velocity of water flowing through a is $\sqrt{2gx}$

by conservation of momentum , we have.

$m_1{v}_1=m_2{v}_2$

$m=\rho v=\rho (A\times x)$ [$\because$ volume of liquid at

any time is $A\times x$ ]

$\therefore \rho \left(a\right)x\times \sqrt{2gx}=\rho A\times x\times v_2.$

$\Rightarrow v_2=\frac{a\times \sqrt{2gx}}{A}$

$a_1=\frac{d{v}_2}{dt}=\frac{a}{A}\sqrt{2g}\times \frac{1}{2\sqrt{x}}\times \frac{dx}{dt}=\frac{a}{A}\frac{\sqrt{2g}}{2\sqrt{x}}\times \sqrt{2gx}$

$=\frac{ag}{A}$