An air bubble of radius 4.5 cm initially at a depth of 12...
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
Correct answer is D
r1 = 4.5cm , P1 =is the total pressure on the bubble at a depth of 12m from the surface.
P1 = 12 + 10.34 =22.34m
V1 = 43π× r31
= \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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