An air bubble of radius 4.5 cm initially at a depth of 12 m below

Explanation

\(r_1\) = 4.5cm , \( P_1\) =is the 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