Question

A large container with open top and uniform cross-sectional area $\left(A\right)$ has a small hole of cross-sectional area $a$ in its side wall near the bottom. The container is kept on a smooth horizontal platform and contains a liquid of density $\rho$ and mass $m$. If the liquid starts flowing out of the hole at time $t=0$, the initial acceleration $\left(\alpha \right)$ of the container is

• A. $\frac{ga}{A}$
• B. $\frac{gA}{a}$
• C. $\frac{2ga}{A}$
• D. $\frac{gA}{2a}$
  

Answer 1

The correct option is C $\frac{2ga}{A}$

Let $h$ be the initial height of the liquid of density $\rho$ in the container of cross-sectional area $A$.



Velocity of the liquid flowing out of the hole is
$v=\sqrt{2gh}$
Rate of discharge of volume of liquid is,
$\Rightarrow \frac{dV}{dt}=Q=av$
Rate of discharge of mass of liquid is given by,
$\frac{dm}{dt}=\rho \frac{dV}{dt}=\rho av$
Let $a=$ cross-sectional area of the hole
Rate of change of momentum of liquid is,
$\frac{dP}{dt}=\frac{d\left(mv\right)}{dt}=v\frac{dm}{dt}$
$\Rightarrow \frac{dP}{dt}=v\left(\rho av\right)=\rho a{v}^2$
According to Newton's second law,
$F=\frac{dP}{dt}$
From Newton's $3^{\text{rd}}$ law, the force on tank will be in a direction opposite to that of velocity of efflux.
$\Rightarrow$The force exerted on tank by the hole is,
$F=\rho a{v}^2$
Mass of liquid in container is
$\Rightarrow \text{mass}=\rho \times \text{volume}=\rho Ah$
Initial acceleration( Let $\alpha$) of tank will be,
$\alpha =\frac{F_{\text{net}}}{\text{mass}}=\frac{\rho a{v}^2}{Ah\rho }=\frac{a{v}^2}{Ah}$
$\Rightarrow \alpha =\frac{a\times 2gh}{Ah}$
$(v=\sqrt{2gh})$
$\therefore \alpha =\frac{2ga}{A}$
Hence $\alpha =\frac{2ga}{A}$ is the required initial acceleration of tank