A rectangular tank is filled to a depth of 10m with freshwater and open to air at atmospheric pressure. It is fitted with two drain plugs on the bottom of the tank, one at the left side and one at the right side.
Both plugs have chains attached and are removed by lifting the chains.
The left-side plug is a flat circular disk of 50cm diameter and 1cm height.
The right-side plug is a hemisphere and is also 50cm in diameter.
Q1. What is the upwards force required to lift the left- side circular disk plug?
Q2. What is the upwards force required to lift the right-side hemisphere plug?

1. 39068.07 N

2. 19534.036 N

Explanation:

depth of water h = 10 m

atmospheric pressure Patm = 101325 Pa

density of water p = 1000 kg/m^3

acceleration due to gravity g = 9.81 m/s^2

pressure due to depth of water = pgh

P = 1000 x 9.81 x 10 = 98100 Pa

total pressure on the bottom of the tank is Patm + p = 101325 + 98100 = 199425 Pa

Left plug has diameter = 50 cm = 0.5 m

radius = 0.5/2 = 0.25 m

height = 1 cm = 0.01 m

height below tank surface = 10 - 0.01 = 9.99

pressure at this depth =  1000 x 9.81 x 9.99 = 98001.9 Pa

total pressure = Patm + P = 101325 + 98001.9 = 199326.9 Pa

surface area of plug = π = 3.142 x = 0.196

force required to lift left plug = pressure x area

F =  199326.9 x 0.196 = 39068.07 N

The right side is a hemisphere with the same diameter, therefore surface area is half of the left plug

A = 0.196/2 = 0.098

force F required to lift right plug =  199326.9  x 0.098 = 19534.036 N

i don't know that i agree with that but what's the question ?