Question

The opening near the bottom of the vessel shown in the figure has an area A. A disc is held against the opening to keep the liquid from running out. Let $F_1$ be the net force on the disc applied by liquid and air in this case. Now the liquid comes out and strikes the disc inelastically. Let $F_2$ be the force exerted by the liquid in this condition. Then $F_1/F_2$ is

• A. $\frac{1}{2}$
• B. $1$
• C. $\frac{2}{1}$
• D. $\frac{1}{4}$

Answer 1

Answer: (A)

$\frac{1}{2}$

Let $\rho$ be the density of the liquid. Then
$F_1=\left(\Delta P\right)A=\rho ghA$
In the second case, $F_2=$ rate of change of momentum
$=\rho A{u}^2=\rho A(\sqrt{2gh}{)}^2=2\rho ghA$
$\therefore \frac{F_1}{F_2}=\frac{1}{2}$