An air bubble of radius 4 5 cm initially at a d... - JAMB Physics 2023 Question

Solution in Own Words:

To find the radius of an air bubble as it rises from a depth of 12 m below the water surface to the surface, we can use Boyle's law, which states that the product of pressure and volume is constant for a given mass of gas at constant temperature. 

Initially, the pressure (\(P_1\)) on the bubble at a depth of 12 m is the sum of atmospheric pressure and the pressure due to the water column above it, which is 22.34 m of water. The initial volume (\(V_1\)) is calculated using the formula for the volume of a sphere. 

As the bubble rises to the surface, the pressure (\(P_2\)) decreases to the atmospheric pressure of 10.34 m of water. The final volume (\(V_2\)) is again calculated using the volume formula. 

Applying Boyle's law (\(P_1V_1 = P_2V_2\)), we can set up an equation and solve for the final radius (\(r_2\)). 

\(r_1\) = 4.5cm, \( P_1\) = total pressure on the bubble at a depth of 12m from the surface.

\(P_1\) = 12 + 10.34 =22.34m

 \(V_1\) = \(\frac{4}{3}\)π× \(r^3_1\)

= \(\frac{4}{3}× π×{4.5^3cm^3}\)

\(P_2\) = 10.34m

\(V_2\)  = \(\frac{4}{3} {π}{r^3_2}\)

From Boyles Law,

\(P_1V_1\)  = \(P_2V_2\) 

22.34× \(\frac{4}{3}× π×{4.5^3}\) = 10.34 × \(\frac{4}{3}×π×{r^3_2}\)

22.34 × \(4.5^3\) = 10.34 × \(r^3_2\)

\(r^3_2\)= \(\^3√196.88\)

\(r^3_2\) = 5.82cm