Question

Water is filled in a conical flask up to height $20cm$. If the radius of base is $10cm$ and the atmospheric pressure is $1.01\times {10}^{5}Pa$. Find the force exerted on bottom of the flask?

Answer 1

Step 1:Given data:

Height of the water,$h=20cm$, Base radius of conical flask, $r=10cm$, atmospheric pressure, $P_0=1.01\times {10}^{5}Pa$

Take acceleration due to gravity,$g=10m/s^2$ and the density of water, $\rho =1000Kg/m^3$

Step 2:Fnding Force

The pressure at the surface of the water is equal to the atmospheric pressure $P_0$,

The pressure at the bottom is, $P=P_0+h\rho g$
$=1.01\times {10}^5+(0.20\times 1000\times 10)$
$=1.01\times {10}^5+.02\times {10}^5$

$=(1.01+.02)\times {10}^5$

Therefore $P=1.03\times {10}^{5}Pa$
The area of the bottom =$\pi r^2$$=3.14\times {(0.1)}^2$ $=0.0314m^2$
Therefore the force on the bottom of the conical flask, $F=P\times A=P\times \pi r^2$
$F=1.03\times {10}^5\times 0.0314$

$F=3230N$
Hence, the force exerted on bottom of the flask $3230N$.