Question

A fixed container having area of cross-section 1 $m^2$ contains water .The initial height of water in container is 1m .The container has an orifice of small cross section area near its bottom. Find the ratio of $R_1/R_2$ where $R_1$ and $R_2$ are the rates of change of height of the liquid in the container when the heights of liquid in container are 1 m and 0.25 m respectively. Area of office is $0.001m^2$.

Answer 1

The rate of fall of water in the cylinder is given by,

$d{h}_1/dt=A_{tube}/A_{cylinder}{V}_1$ = $R_1$

Since, the ratio of areas is constant, the ratio will just be a ratio of the velocities.

The velocity of flow is given by, $v=\sqrt{2gh}$

Therefore,

$R_1/R_2=\sqrt{h_1/h_2}$

$R_1/R_2=\sqrt{1/0.25}$ = 2