JAMB - Physics (2023 - No. 19)

An air bubble of radius 4.5 cm initially at a depth of 12 m below the water surface rises to the surface. If the atmospheric pressure is equal to 10.34 m of water, the radius of the bubble just before it reaches the water surface is
6.43 cm
8.24 cm
4.26 cm
5.82 cm

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 × \(r^3_2\)

 \(r^3_2 = \sqrt[3]{196.88}\)

⇒ \(r_2\) = 5.82cm

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