Question

A thin square plate of side $4m$ is placed horizontally $2m$ below the surface of water. Calculate the thrust on one of the surface of the plate.
(Given: Atmospheric pressure,$P_{atm}=1.013\times {10}^{5}N/m^2$
Density of water, ${\rho }_w$ = $1000kg/m^3$,
Acceleration due to gravity, $\left(g\right)=9.8m{s}^{-1}$)

• A. $19.344\times {10}^{5}N$
• B. $19.344\times {10}^{3}N$
• C. $17.344\times {10}^{5}N$
• D. $21.344\times {10}^{5}N$

Answer 1

The correct option is A

$19.344\times {10}^{5}N$

Given :
Depth of plate in the water from the surface, $h=2m$
Density of water, ${\rho }_w=1000kg/m^3$,
Acceleration due to gravity, $g=9.8m{s}^{-1}$
Atmospheric pressure, $P_{atm}=1.013\times {10}^{5}N/m^2$

Area of plate,
$A=4m\times 4m$
$A=16m^2$

Total pressure at a depth $h$ below the surface of water is given by:
$P=P_{atm}+h{\rho }_{w}g$
$P=(1.013\times {10}^5)+(2\times {10}^3\times 9.8)$
$P=1.209\times {10}^{5}Pa$

Thrust on surface of plate due to liquid above it is:
$F=PA$
$F=1.209\times {10}^{5}Pa\times 16m^2$
$F=19.344\times {10}^{5}N$