Tanks $A$ and $B$ open at the top contain two different liquids upto certain height in them. A hole is made…
Tanks $A$ and $B$ open at the top contain two different liquids upto certain height in them. A hole is made to the wall of each tank at a depth $h$ from the surface of the liquid. The area of the hole in $B$ is twice that of in $A$. If the liquid mass flux through each hole is equal, then the ratio of the densities of the liquids respectively, is
$1$
$\frac{3}{2}$
$\frac{2}{3}$
$\frac{1}{2}$
Solution
Let the velocity of liquid from hole in $A$ is $v_1$ and velocity of liquid from hole in $B$ is $v_2$, then from equation of continuity
$\begin{aligned}
A_1 v_1 & =A_2 v_2 \\
A v_1 & =2 A v_2 \\
v_2 & =\frac{v_1}{2}
\end{aligned}$
Volume of liquid coming out per second from hole in tank $A=A_1 v_1=A v_1$
$\therefore \quad \text { Mass } m_1=A v_1 \rho_1$
Similarly,
$\begin{aligned}
m_2 & =A_2 v_2 \rho_2 \\
& =2 A \cdot \frac{v_1}{2} \rho_2 \\
& =A v_1 \rho_2
\end{aligned}$
But
$\begin{aligned}
& m_1=m_2 \\
& A v_1 \rho_1=A v_1 \rho_2 \\
& \frac{\rho_1}{\rho_2}=1 \\
&
\end{aligned}$
^