Question 52

[msamadia]

I am lost in the explanation... can someone please explain this...

i don't understand how the 1/400 was derived using the eqn.

thanks!!!

[mcat_premed3832]

The question is asking for the change in pressure. The equation that you are responsible for equates change in pressure with the product of density, gravity and change in height. Since we know density (the density of water should be memorized in both units of g/ml and kg/m^3) and gravity, the problem is to determine the change in height.

So the question comes down to this: if a volume of 0.5 m^3 was displaced then by what height "h" did the water level rise?

Keeping in mind that the cross section of the tank must be a circle, then the volume must be the area of a circle times the height. The area of a circle is given by pi r squared. Because the passage gives the radius of the tank is 8 m, then you can see how the calculation was made.

Notice that if you divide volume (m^3) by area (m^2), you end up just with height in meters (= Dimensional Analysis).

If there is a passage on fluids in a real MCAT, all of the concepts above would be commonly tested.

[jvuofm1709]

Did the question want the change in pressure at the bottom of the big section of the tower or the small section?

The way I calculated at first, was I looked at it like a hydraulics problem. If .5 m^3 of water was displaced, then the force of the wooden lump must be= (p*V*g)= .5*1000*10=5000 N. So, this force changes the pressure differently for both the wide pipe of the tower and the thin pipe. Using F/A=pressure, for the wide pipe deltaP= 5000/8^2*pi=25 pa. However, at the bottom of the thin pipe deltaP=5000/.1^2*pi=159154.9 pa.

Using the method explained in the answer, it would seem the pressure change should be the same at all levels of the pipe, why is their a discrepancy between the two methods?

[jellywing_2058]

Hydraulically, the weight that the thin pipe would have to hold on to is going to be lesser; therefore, the pressure going to be less than 159154.0 pa at the base of it. In fact, it is going to be the same as that at the base of the wide pipe, that is = 25 Pa.