## College Physics (4th Edition)

The fish must inflate the swim bladder to a volume of $1.7\times 10^{-7}~m^3$
We can find the volume $V$ of the fish with a deflated bladder: $V = \frac{m}{\rho} = \frac{0.010~kg}{1080~kg/m^3} = 9.26\times 10^{-6}~m^3$ We can find the volume $V'$ which will give the fish a density of $1060~kg/m^3$: $V' = \frac{m}{\rho} = \frac{0.010~kg}{1060~kg/m^3} = 9.43\times 10^{-6}~m^3$ The volume of the inflated swim bladder is equal to the difference of $V'$ and $V$: $V'-V = (9.43\times 10^{-6}~m^3)-(9.26\times 10^{-6}~m^3)$ $V'-V = 1.7\times 10^{-7}~m^3$ The fish must inflate the swim bladder to a volume of $1.7\times 10^{-7}~m^3$.