Question

A tank of uniform cross-section 'A', open at the top contains two immiscible, incompressible, non-viscous ideal liquids P and Q of densities D and 2D respectively to height H each as shown. Atmospheric pressure is $P_0$. A pit exists near the base of the tank with its vertical walls indicated. AT time t = 0, a hole of area a (<<A) is punched in the vertical wall at a height h (<H) so the efflux falls at B. (a) Calculate the height h (b) Find the total mass of the liquid that falls on the pit. (c) Calculate the time for which the liquid Q falls on the pit (d) the time for which the liquid P falls on the pit.

Answer 1

(a) The height $h$ is $\frac{H}{2}$. (b) The total mass of the liquid that falls on the pit is $ADH \left(\frac{4-\sqrt{5}}{2}\right)$. (c) The time for which the liquid Q falls on the pit is $\frac{A}{a} \sqrt{\frac{H}{2g}} \left(2 - \sqrt{\frac{1+\sqrt{5}}{2}}\right)$. (d) The time for which the liquid P falls on the pit is $\frac{A}{a} \sqrt{\frac{H}{g}} (\sqrt{2}-1)$.

Answer 2

(a) Torricelli's law and projectile motion equations are used to find the efflux velocity and range. Equating the given range $\sqrt{2}H$ to the calculated range leads to a quadratic equation for $h$, yielding $h=H/2$.

(b) The condition for the liquid to fall on the pit is analyzed by comparing the range with the pit's location. The mass is calculated by integrating the density over the volume of liquid that falls within the pit's horizontal limits, considering both liquid Q and P.

(c) The time for liquid Q to fall on the pit is found by integrating the rate of volume flow, considering the velocity change as the liquid level drops from $H$ to $\frac{1+\sqrt{5}}{4}H$.

(d) Similarly, the time for liquid P to fall on the pit is calculated by integrating the flow rate as its level drops from $2H$ to $3H/2$.