Question

(a) Explain what is meant by the term 'pressure'. With the aid of a labelled diagram, describe a simple experiment using a tall, cylindrical tank to demonstrate that pressure varies with depth in a fluid. (b) An aquarium has a rectangular base that measures 60 cm by 30 cm. Its height is 45 cm and sea water of density 1150 kg/m^3 is used to fill it. (i) Calculate the weight of water in the aquarium and the pressure on a fish at the bottom of the aquarium due to the water alone. (ii) Calculate the pressure on a fish at the bottom of the aquarium if atmospheric pressure of 100 kPa acts on the surface of the water (Acceleration due to gravity g = 10 m s^-2)

Answer 1

a)

Force on unit area normal to it is Pressure.

Attach three pipes with up turned nozzles to the cylinder carrying water as shown in above figure.

The water jet coming from the bottom pipe raises almost to the top. The water jet coming from the middle pipe also raises to the top , there will be no water jet coming from the top pipe. The speed with which water comes out of these pipes depends upon pressure in the water at the depth where the pipe is located. For the top pipe pressure is almost equal to the atmospheric pressure. Hence no water jet. As the depth of water increases , pressure increases there by water jet at bottom reaches to greater height.

b)

i) Volume of aquarium $V=60*30*45=81000\,cm^3=0.081\,m^3$

Density of sea water $\rho=1150\,kg/m^3$

Mass of water in the aquarium is $m=\rho *V=1150*0.081=93.15\,kg$

Weight of water in the aquarium is $mg=\rho *V*g=1150*0.081*10=931.5\,N$

Presurre on fish at bottom due to water alone is $P=\rho g h = 1150*10*0.45=5175\,Pa$

ii)

Presurre on fish at bottom of aquarium $P=P_0+\rho g h =100000+5175=105.2\,kPa$