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
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Correct Answer: Option D
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
There is an explanation video available below.
\(r_1\) = 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
There is an explanation video available below.