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(B) Let $P_1$ and $V_1$ be the pressure and volume of the bubble at the bottom,and $P_2$ and $V_2$ be the pressure and volume at the top surface.Since

Vedclass · Accepted Answer

$5: 6$

Vedclass · Answer

(B) Let $P_1$ and $V_1$ be the pressure and volume of the bubble at the bottom,and $P_2$ and $V_2$ be the pressure and volume at the top surface.Since the temperature is constant,we use Boyle's Law: $P_1 V_1 = P_2 V_2$.The pressure at the bottom is $P_1 = P_{atm} + h \rho g = 10 \ m + 7.28 \ m = 17.28 \ m$ of water.The pressure at the top is $P_2 = P_{atm} = 10 \ m$ of water.The volume of a spherical bubble is $V = \frac{4}{3} \pi r^3$,so $V \propto r^3$.Substituting this into Boyle's Law: $P_1 r_1^3 = P_2 r_2^3$.Thus,$\frac{r_1^3}{r_2^3} = \frac{P_2}{P_1} = \frac{10}{17.28} = \frac{1000}{1728}$.Taking the cube root of both sides: $\frac{r_1}{r_2} = \sqrt[3]{\frac{1000}{1728}} = \frac{10}{12} = \frac{5}{6}$.Therefore,the ratio of the radii is $5: 6$.

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