Q151 / 200 · TS EAMCET 2002 · Physics

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

  • Correct answer
    A
    $1$
  • B
    $\frac{3}{2}$
  • C
    $\frac{2}{3}$
  • D
    $\frac{1}{2}$

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 $$ \begin{aligned} & m_1=m_2 \\ & A v_1 \rho_1=A v_1 \rho_2 \\ & \frac{\rho_1}{\rho_2}=1 \\ & \end{aligned} $$

Hence, the correct answer is option: 'A'.

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