Question

1. A 1 kg fish lives at a depth of 300 m in seawater. The specific density of the fish, excluding its swim bladder, is 1.07 A To be neutrally buoyant at this depth, what fraction of the fish's volume must its swim bladder fill? B What is the gas pressure inside the swim bladder at this depth? C Suppose that the fish suddenly descends to 400 m. The gas in its swim bladder will contract under pressure, and the fish will no longer be neutrally buoyant, but will tend to want to sink. What is the net vertical force on the fish at this point? (Assume that the amount of gas in the swim bladder doesn't change.) D The swim attempts to maintain its depth at 400 m by swimming in the horizontal direction. It angles its body up at a 10° angle relative to horizontal, and then applies enough thrust force to just cancel the vertical force. How much thrust must it supply? E While supplying this amount of thrust at this angle, the fish will move horizontally. Assume that the fish has a frontal area of 30 cm2 and a drag coefficient of CD 0.3. Calculate the forward velocity of the fish if it is swimming stably at this depth with its body angle maintained at 10°.

Answer 1

A the specific density of the fish is 1.07 that means the weight of the fish is 7% more than water so if the gas bladder of the fish is 7% the volume of the fish main body then the buoyancy force will be the same as the weight of the fish, so 7%.

B the pressure of gas in the bladder will be same as the surroundings for the walls of bladder to stay in equllibrium, so       p_atm+$\rho$gh = 10⁵ + 10³ *9.81* 300 = 30.43*10⁵ N/m² or 30.43 atm

C pressure of water at 400 meter deep is 10⁵ + 10³ *9.81* 400 =40.24*10⁵ N/m² or 40.24 atm then we know from ideal gas equation pv = nrt and there is no significant change in temperature so p₁ v₁ = p₂ v₂ then final volume =initial volume*30.43/40.24 = 0.756* initial volume ie, 5.3% volume of fish

weight of fish is 1 kg then 1/1.07 is the volume of the fish without bladder : total volume at 400 m depth is (1/1.07)*1.053 since the specific density of water is 1 this is the net buoyancy on the fish ie (1/1.07)*1.053 = 0.984Kg.

net weight is 1 kg so net vertical force on the fish is 1-0.984 = 0.016Kg = 0.1568N

D so for the fish to stay in the same depth the vertical component of the thrust it is applying is same as in C ,if the thrust force is assume to be F then F sin10 = 0.1568N

F = 0.1568/sin10 = 0.9N

E frontal area is 30 and cd is 0.3 then aerodynamic area is 9 cm² ,horizontal component of the thrust is 0.9 cos10 = 0.886N for drag force we have substituting known values v² = $\frac{2Fd}{A Cd}$ = 0.197

v= 0.443m/s