Question

Some liquid is filled in a cylindrical vessel of radius $R$. Let $F_1$ be the force applied by the liquid on the bottom of the cylinder. Now the same liquid is poured into a vessel of uniform square cross-section of side $R$. Let $F_2$ be the force applied by the liquid on the bottom of this new vessel. Then:

• A. $F_1=\pi F_2$
• B. $F_1=\frac{F_2}{\pi }$
• C. $F_1=\sqrt{\pi }{F}_2$
• D. $F_1=F_2$

Answer 1

The correct option is D $F_1=F_2$

Given,
$\text{Volume of liquid in cylindrical vessel = Volume of liquid in cubical vessel}$
$\pi R^2\times H_{\text{cylinder}}=R^2\times H_{\text{cube}}$
$\Rightarrow H_{\text{cube}}=\pi H_{\text{cylinder}}...\left(i\right)$

$F_1$ = Force applied by the liquid on the bottom of the cylindrical vessel.
$F_1=\left(P_{\text{bottom}}\right)\times$ Base area
or $F_1=\left(\rho g{H}_{\text{cyclinder}}\right)\times \pi R^2$

$F_2$ = Force applied by the liquid on the bottom of the cubical vessel.
$F_2=\left(P_{\text{bottom}}\right)\times$ Area
or $F_2=\left(\rho g{H}_{\text{cube}}\right)\times R^2$
Therefore,
$\frac{F_1}{F_2}=\frac{\rho g{H}_{\text{cylinder}}\times \pi R^2}{\rho g{H}_{\text{cube}}\times R^2}$
Using eqn (i)
$\frac{F_1}{F_2}=\frac{\rho g{H}_{\text{cylinder}}\times \pi R^2}{\rho g\left(\pi H_{\text{cylinder}}\right)\times R^2}$
$\therefore F_1=F_2$