answer:

First, let us calculate the buoyant force. We know that the force is equal to the weight of water in the volume of the body. The volume of block is

V = \dfrac{P}{\rho g} = \dfrac{1300\,\mathrm{N}}{2400\,\mathrm{kg/m}^3\cdot 9.81\,\mathrm{N/kg}} = 0.055\,\mathrm{m}^3.V=

ρg

P

=

2400kg/m

3

⋅9.81N/kg

1300N

=0.055m

3

.

The buoyant force will be \rho_{\text{w}}gV = 541.7\,\mathrm{N}ρ

w

gV=541.7N .

The system floats, so there should another force that is directed upwards and is equal to 1300-541.7 = 758.3\,\mathrm{N}.1300−541.7=758.3N. This force is equal to the buoyant force that acts on the volume of air under the water line. Therefore, this volume is

V_1 = \dfrac{758.3\,\mathrm{N}}{1000\,\mathrm{kg/m}^3\cdot9.81\,\mathrm{N/kg}} = 0.077\,\mathrm{m}^3.V

1

=

1000kg/m

3

⋅9.81N/kg

758.3N

=0.077m

3

. The height of such a layer is h_1 = \dfrac{V_1}{\pi D^2/4} = 0.8h

1

=

πD

2

/4

V

1

=0.8 m.

The pressure of such a layer of water should be balanced by the pressure in air in the tank:

p_1 = \rho_{\text{w}} g h_1 = 7882\,\mathrm{Pa}.p

1

=ρ

w

gh

1

=7882Pa. We should take the atmospheric pressure p_0p

0

into account to get the total pressure 10882 10882 Pa.

According to the ideal gas law, p_0V_0 = p_1V_1, \;\; p_0\cdot\pi \dfrac{D^2}{4} h = p_1\cdot\pi \dfrac{D^2}{4} h_2p

0

V

0

=p

1

V

1

,p

0

⋅π

4

D

2

h=p

1

⋅π

4

D

2

h

2

, where h_2h

2

is the height of air column. h_2 = h \cdot{p_0}{p_1} = 1.67\,\mathrm{m}.h

2

=h⋅p

0

p

1

=1.67m. Therefore, the depth of water will be only 1.8-1.67=0.131.8−1.67=0.13 m

Explanation:

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