One liter of water was poured into a cylindrical vessel with a bottom area s = 100 cm2 and a wooden block weighing

One liter of water was poured into a cylindrical vessel with a bottom area s = 100 cm2 and a wooden block weighing 100 g was allowed to float. Density of water 1000 kg / m3 acceleration of gravity 10 m / s2 Determine the water level in the vessel.

Given:

s = 100 cm ^ 2 = 0.01 m ^ 2 – the area of ​​the bottom of a cylindrical vessel;

V = 1 liter = 0.001 m ^ 3 – the volume of water poured into the vessel;

ro1 = 1000 kg / m ^ 3 – water density;

m = 100 grams = 0.1 kilograms – the mass of a wooden block that was placed in water;

ro2 = 500 kg / m ^ 3 – density of a wooden block;

g = 10 m / s ^ 2 – acceleration of gravity.

It is required to determine the water level in vessel H.

Before the bar was lowered into the water, the water level in the vessel was:

h1 = V / S = 0.001 / 0.01 = 0.1 meter.

A bar immersed in water will displace water in volume:

V1 = m / ro1 = 0.1 / 1000 = 0.0001 m ^ 3.

This volume will increase the water level to a height:

h2 = V1 / s = 0.0001 / 0.01 = 0.01 meter.

Then the water level in the vessel is:

H = h1 + h2 = 0.1 + 0.01 = 0.11 meters = 11 centimeters.

Answer: the water level in the vessel is 11 centimeters.