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
A
6.43 cm
B
8.24 cm
C
4.26 cm
D
5.82 cm
Verified Correct Answer: D.
Physical Principles & Numerical Solution
\(r_1\) = 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
\(V_1\) = \(\frac{4}{3}\)\pi\times \(r^3_1\)
= \(\frac{4}{3}\times \pi\times{4.5^3cm^3}\)
\(P_2\) = 10.34m
\(V_2\) = \(\frac{4}{3} {\pi}{r^3_2}\)
from boyles law:
\(P_1V_1\) = \(P_2V_2\)
⇒ 22.34\times \(\frac{4}{3}\times \pi\times{4.5^3}\) = 10.34 \times \(\frac{4}{3}\times\pi\times{r^3_2}\)
⇒ 22.34 \times \(4.5^3\) = 10.34 \times \(r^3_2\)
⇒ \(r^3_2 = \sqrt[3]{196.88}\)
⇒ \(r_2\) = 5.82cm
Source: WAEC / JAMB / NECO Classroom Archive (Record #67831)