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 disc is moved away from the opening at a short distance. 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

The correct option is A $\frac{1}{2}$

$F_1=\left(\rho gh\right)A$

Now,

When disc is displaced

$P_0+\rho gh+\frac{1}{2}{\rho V}^2=\frac{F}{A}+P_0$

$F_2=A(\rho gh+\frac{1}{2}\rho {\left(\sqrt{2g}h\right)}^2)$

$F_2=A\left(2\rho gh\right)$

So, $\frac{F_1}{F_2}=\frac{1}{2}$