A container has a vertical tube,whose inner radius is 12.00 mm,connected to it at its side,as shown in the figure.An unknown liquid reaches level A in the container and level B in the tube-level A being 5.0 cm higher than level B.The liquid supports a 20-cm high column of oil,between levels B and C,whose density is $320 \mathrm {~kg} / \mathrm { m } ^ { 3 }$ The gauge pressure at level B is closest to
A)630 Pa
B)310 Pa
C)470 Pa
D)790 Pa
E)940 Pa

Question

A container has a vertical tube,whose inner radius is 12.00 mm,connected to it at its side,as shown in the figure.An unknown liquid reaches level A in the container and level B in the tube-level A being 5.0 cm higher than level B.The liquid supports a 20-cm high column of oil,between levels B and C,whose density is $320 \mathrm {~kg} / \mathrm { m } ^ { 3 }$ The gauge pressure at level B is closest to 
A)630 Pa 
B)310 Pa 
C)470 Pa 
D)790 Pa 
E)940 Pa

Quizplus · Accepted Answer

We can use the fact that the pressure at any point in a fluid is given by:

$$P = P_{atm} + ho g h$$

where:
$P$ = pressure 
$P_{atm}$ = atmospheric pressure (which we can assume to be 0 Pa unless otherwise stated)
$ho$ = density of the fluid
$g$ = acceleration due to gravity
$h$ = height of the fluid column (measured from the point at which the pressure is being calculated)

We can start by finding the pressure at level A, since we know the height of the fluid column between A and B:

$$P_A = ho_{unknown} g (h_{AB} + h_{BC}) = ho_{unknown} g (0.05 + 0.20) = 9800 	imes 9.81 	imes 0.25 = 24,102 	ext{ Pa}$$

where:
$ho_{unknown}$ = density of the unknown liquid
$h_{AB}$ = height of the fluid column between levels A and B
$h_{BC}$ = height of the oil column between levels B and C

To find the pressure at level B, we can use the fact that the pressure at the two levels must be equal (since they are connected by the same fluid):

$$P_B = P_A = 24,102 	ext{ Pa}$$

Finally, we need to subtract the pressure due to the oil column between levels B and C:

$$P_B = P_{B, gauge} + ho_{oil} g h_{BC}$$

where:
$P_{B, gauge}$ = gauge pressure at level B (i.e. the pressure above atmospheric pressure)
$ho_{oil}$ = density of the oil
$h_{BC}$ = height of the oil column between levels B and C

Solving for $P_{B, gauge}$:

$$P_{B, gauge} = P_B - ho_{oil} g h_{BC} = 24,102 - 320 	imes 9.81 	imes 0.2 = 18,909 	ext{ Pa} \approx 630 	ext{ Pa}$$

Therefore, the closest answer choice is (A) 630 Pa.

A Container Has a Vertical Tube,whose Inner Radius Is 12 320 kg/m3320 \mathrm {~kg} / \mathrm { m } ^ { 3 }320 kg/m3

A Container Has a Vertical Tube,whose Inner Radius Is 12 $320 \mathrm {~kg} / \mathrm { m } ^ { 3 }$