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(D) Let the surface area of the bubble at the bottom be $A_1$ and at the top be $A_2$. Given $A_2 = A_1 + 1.25 A_1 = 2.25 A_1$.Since the bubble is sph

Vedclass · Accepted Answer

$23.75$

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(D) Let the surface area of the bubble at the bottom be $A_1$ and at the top be $A_2$. Given $A_2 = A_1 + 1.25 A_1 = 2.25 A_1$.Since the bubble is spherical, $A = 4 \pi r^2$, which implies $r \propto \sqrt{A}$. Thus, $r_2 = \sqrt{2.25} r_1 = 1.5 r_1$.The volume $V = \frac{4}{3} \pi r^3$, so $V_2 = (1.5)^3 V_1 = 3.375 V_1$.Since the temperature is uniform, Boyle's Law applies: $P_1 V_1 = P_2 V_2$.Here, $P_2 = P_{atm} = 10 \ m$ of water column.$P_1 = P_{atm} + h = 10 + h$, where $h$ is the depth of the tank.Substituting the values: $(10 + h) V_1 = 10 	imes (3.375 V_1)$.$10 + h = 33.75$.$h = 33.75 - 10 = 23.75 \ m$.

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