Let Volume of fish with its air sacs expanded = V_{a}

Volume of fish with its air sacs collapsed = V

Mass of fish (m_{a}) = 4kg

Mass of air in air sac (m_{a}) = m_{a}kg

Depth = 10.25m

Density of water = 1008 kgm^{-3}

Density of air = ρ_{a}

Density = ^{mass}/_{volume}

(a) Pressure at 10.25m = ρ_{w}gh + 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 = [(V_{a} - V) X Density of air]

Mass of air in air sac = [(2.05 kgm^{-3}) X (V_{a} - V)]

Mass of air in air sac = [(2.05V_{a} 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} m^{3}

Density of fish with its air sacs expanded = [(4kg + m_{a}kg) / (V_{a}) ]

Density of fish with its air sacs expanded = {[(4kg + (2.05V_{a} kg) - (2.05V kg)] / (V_{a}m^{3}) ]}

Density of fish with its air sacs expanded = {[(4kg + (2.05V_{a} kg) - (2.05 X 3.33kg X 10^{-3}m^{3})] / (V_{a}m^{3})]}

Density of fish with its air sacs expanded = (2.05kg) + {[(4kg - (2.05 X 3.33kg X 10^{-3}m^{3})] / (V_{a}m^{3})]}

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^{-3}m^{3})] / (V_{a})]}

(1008kgm^{-3} - 2.05kgm^{-3}) = {[4kg - (2.05 X 3.33kg X 10^{-3}m^{3})] / (V_{a})]}

(1008kgm^{-3} - 2.05kgm^{-3}) = {[4kg - (2.05 X 3.33kg X 10^{-3}m^{3})] / (V_{a})]}

[(1008kgm^{-3} - 2.05kgm^{-3}) X (V_{a})] = {[4kg - (2.05 X 3.33kg X 10^{-3}m^{3})]}

(V_{a}) = {[4kg - (2.05 X 3.33kg X 10^{-3}m^{3})] / [(1008kgm^{-3} - 2.05kgm^{-3})]

(V_{a}) = 3.970 X 10^{-3}m^{3}

% increase = [(V_{a}) - (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 ) / (V_{a})]

(V_{a}) = [(4kg ) / (1008kgm^{-3})]

(V_{a}) = 3.968 X 10^{-3}m^{3}

(V_{a}) would 0.1% smaller.