UNIT 4
Density and Pressure
Example 1
Example 2
Example 3
Example 4
Example 5
Example 6
Example 7

 

Density and Pressure in Fluids
Example 1

example1

Fish can maintain their depth in water by adjusting the air content of their air sacs to make its average density the same as that of the water. Suppose that with its air sacs collapsed, a fish has a density of 1200 kgm-3 and mass 4kg.
(a) How much should the fish expanded its body volume to float in water of density 1008 kgm-3 at a depth of 10.25m? Density of air at atmospheric pressure = 1.02 kgm-3, g = 9.81ms-2, Atmospheric pressure = 101 KPa.
(b) Does the weight of the air make any significant difference in the volume change?

Let Volume of fish with its air sacs expanded = Va
Volume of fish with its air sacs collapsed = V
Mass of fish (ma) = 4kg
Mass of air in air sac (ma) = makg
Depth = 10.25m
Density of water = 1008 kgm-3
Density of air = ρa
Density = mass/volume

(a) Pressure at 10.25m = ρwgh + Atmospheric Pressure
Pressure at 10.25m = 202 KPa
Density of air at 202KPa = 1.02 kgm-3 X [202KPa / 101Kpa]
Density of air at 202KPa = 2.05 kgm-3
Mass of air in air sac = [(Va - V) X Density of air]
Mass of air in air sac = [(2.05 kgm-3) X (Va - V)]
Mass of air in air sac = [(2.05Va kg) - (2.05V kg)]
Density of fish with its air sacs collapsed = 4 kg /V
1200kgm-3 = 4kg /V
V = 3.333 X 10-3 m3
Density of fish with its air sacs expanded = [(4kg + makg) / (Va) ]
Density of fish with its air sacs expanded = {[(4kg + (2.05Va kg) - (2.05V kg)] / (Vam3) ]}
Density of fish with its air sacs expanded = {[(4kg + (2.05Va kg) - (2.05 X 3.33kg X 10-3m3)] / (Vam3)]}
Density of fish with its air sacs expanded = (2.05kg) + {[(4kg - (2.05 X 3.33kg X 10-3m3)] / (Vam3)]}
But Density of fish with its air sacs expanded = Density of the water
So 1008kgm-3 = (2.05kgm-3) + {[4kg - (2.05 X 3.33kg X 10-3m3)] / (Va)]}
(1008kgm-3 - 2.05kgm-3) = {[4kg - (2.05 X 3.33kg X 10-3m3)] / (Va)]}
(1008kgm-3 - 2.05kgm-3) = {[4kg - (2.05 X 3.33kg X 10-3m3)] / (Va)]}
[(1008kgm-3 - 2.05kgm-3) X (Va)] = {[4kg - (2.05 X 3.33kg X 10-3m3)]}
(Va) = {[4kg - (2.05 X 3.33kg X 10-3m3)] / [(1008kgm-3 - 2.05kgm-3)]

(Va) = 3.970 X 10-3m3
% increase = [(Va) - (V)] / (V)
% increase = 19.1%

(b) If the weight of the air is considered negligible:
Density of fish with its air sacs expanded = Density of the water
So 1008kgm-3 = [(4kg ) / (Va)]
(Va) = [(4kg ) / (1008kgm-3)]
(Va) = 3.968 X 10-3m3
(Va) would 0.1% smaller.

   
<
>
Concept by Kishore Lal. Programmed by Kishore Lal... Copyright © 2015 Kishore Lal. All rights reserved.