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)
$\frac{F_1}{F_2}=\frac{\rho g{H}_{\text{cylinder}}\times \pi R^2}{\rho g\left(\pi H_{\text{cylinder}}\right)\times R^2}$
$\therefore F_1=F_2$

Alternate Solution: For those shapes whose cross sectional area is perpendicular to base area, force exerted by the liquid of same amount on the base area of different containers will be same. This is because of the reason that entire weight of liquid is supported by base of container for such shapes whose cross sectional area is perpendicular to base area