What is the density of the liquid if a 10.2 N force acts on a 10 cm2 plug closing the hole

What is the density of the liquid if a 10.2 N force acts on a 10 cm2 plug closing the hole in the bottom of the vessel at a depth of 1.3 m?

S = 10 cm2 = 0.001 m2.

h = 1.3 m.

F = 10.2 N.

g = 10 m / s2.

ρ -?

To close the hole in the bottom of the vessel, it is necessary that the pressure on different sides of the hole is the same.

Water presses on the plug with hydrostatic pressure, the value of which is expressed by the formula: P = ρ * g * h. Where ρ is the density of the liquid, g is the acceleration of gravity, h is the height of the liquid, in our case the depth.

From the side of the ship, the same pressure must be applied to the plug, the value of which is expressed by the formula: P = F / S, where F is the force, S is the area of ​​the plug that closes the hole.

ρ * g * h = F / S.

ρ = F / S * g * h.

ρ = 10.2 N / 0.001 m2 * 10 m / s2 * 1.3 m = 784.6 kg / m3.

Answer: the ship is in a liquid with a density of ρ = 784.6 kg / m3.