The container shown in the figure(Figure 1) is filled with oil. It is open to the atmosphere on the left.

**Answer 1**

The ol type is not specified. Density of oil varies from less than 800 kg / m^3 to more than 920 kg/m^3

We can only assume a ‘mean’ density of 850 kg / m^3

Part A:

Point A is 50cm below liquid surface.

P = h. d. g = 0.5 x 850 x 9.81 = 4169 Pa

Part B: A and B are 50 cms apart.

So, pressure difference is again 4169 Pa

Part C : A and C are 50 cms apart.

So, pressure difference is once again 4169 Pa

**Answer 2**

Assuming you have the same book as me, the density of oil is 900 kg/m^3

Relevant equations:

P=Po+hdg

Where P is the pressure you are trying to find, Po is any known pressure, h is the difference in height in meters, d is the density, and g is the acceleration of gravity.

Known:

g = 9.8m/s^2

d = 900kg/m^3

Po = 101300 Pa (atmospheric pressure)

Since the left side of the tank is open to the atmosphere, the pressure at the top left of the tank is going to be equal to the pressure of the atmosphere. This is why we use the atmospheric pressure as our Po.

Calculations:

Pa = Po + hdg

= 101300 + (1 – .5)(900)(9.8)

= 105710 Pa

= 106 kPa (using 3 significant figures)

Since B and C are the same height, the answers will be the same

Pb = Pc = Po + hdg

= 101300 + (1 – 0)(900)(9.8)

= 110120 Pa

Pb – Pa = 110120 – 105710

= 4410 Pa

= 4.4 kPa (to two significant figures)

Pc – Pa = Pb – Pa

= 4.4 kPa

Source(s): Doing the same homework. These answers were all marked right on mine.

**Answer 3**

Po + (Density)gd

= 1.013*10^5+900*9.8*.5

=1.057 then subtract the gauge pressure of 101.3kPa

this should roughly give you

4400 Pa

that is EXACTLY the right Pa

Source(s): In Course right now (PHYS 1)