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Question: A tank \(5m\) high is half-filled with water and then is filled to the top with oil of density \(0.8...

A tank 5m5m high is half-filled with water and then is filled to the top with oil of density 0.85gcm30.85\,g\,c{m^{ - 3}} . The pressure at the bottom of the tank, due to these liquid is:
A. 1.85gdynecm21.85\,g\,dyne\,c{m^{ - 2}}
B. 89.25gdynecm289.25\,g\,dyne\,c{m^{ - 2}}
C. 462.5gdynecm2462.5\,g\,dyne\,c{m^{ - 2}}
D. 500gdynecm2500\,g\,dyne\,c{m^{ - 2}}

Explanation

Solution

Here, we will use the direct formula of pressure at the depth hh of the liquid of density ρ\rho to calculate the pressure at the bottom of the tank that is filled with water and oil. Here, we will add both the pressures due to the water and oil.

Complete step by step answer:
The height of the tank is 5m5\,m . as the tank is half filled with water and half-filled with oil therefore, the height of the water and the oil in the tank will be 2.5m2.5\,m. Therefore, the height of water and the oil is,
h=2.5×100cmh = \,2.5 \times 100\,cm
h=250cm\Rightarrow h = 250\,cm

Now, as given in the question, the density of oil ρoil=0.85gcm3{\rho _{oil}} = 0.85\,g\,c{m^{ - 3}}
Also, the density of water ρwater=1gcm3{\rho _{water}} = 1\,g\,c{m^{ - 3}}
Now, the pressure acting at the bottom of the tank due to water is Pwater{P_{water}} and the pressure acting at the bottom of the tank due to the oil is Poil{P_{oil}} .
Now, using the formula of the pressure PP at the depth hh of the liquid of density ρ\rho , which is given by
P=hρgP = h\rho g
Here, gg is the acceleration due to gravity.

Therefore, in case of both the oil and the water, the formula will become
Pair=hρairg{P_{air}} = h{\rho _{air}}g
Pwater=hρwaterg\Rightarrow {P_{water}} = h{\rho _{water}}g
Now, the total pressure acting at the bottom of the tank is given by
Ptotal=Pair+Pwater{P_{total}} = {P_{air}} + {P_{water}}
Ptotal=hρairg+hρwaterg\Rightarrow \,{P_{total}} = h{\rho _{air}}g + h{\rho _{water}}g
Now, putting the values of height and densities in the above equation, we get
Ptotal=250×0.85×g+250×1×g{P_{total}} = 250 \times 0.85 \times g + 250 \times 1 \times g
Ptotal=212.5g+250g\Rightarrow \,{P_{total}} = 212.5\,g + 250\,g
Ptotal=462.5g\therefore \,{P_{total}} = 462.5\,g

Hence, the total pressure at the bottom of the tank is 462.5gdynecm3462.5\,g\,dyne\,c{m^{ - 3}} .Hence, option C is the correct option.

Note: Here, the pressure on the water and the oil is summed up. This is because the pressure exerted by the oil to the water and this pressure will be transported to the bottom of the tank and it is along the individual pressure of the water at the bottom. As a result, the pressure at the bottom of the tank will be the result of both the pressures.