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(B) The excess pressure $\Delta P$ in a soap bubble of radius $R$ is given by $\Delta P = \frac{4S}{R}$,where $S$ is the surface tension.Let the exces

Vedclass · Accepted Answer

$1: 8$

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(B) The excess pressure $\Delta P$ in a soap bubble of radius $R$ is given by $\Delta P = \frac{4S}{R}$,where $S$ is the surface tension.Let the excess pressure in the first bubble be $P_1 = P$ and its radius be $R_1 = R$.Then,$P = \frac{4S}{R}$.Let the excess pressure in the second bubble be $P_2 = 2P$ and its radius be $R_2 = x$.Then,$2P = \frac{4S}{x}$.Substituting the value of $P$ from the first equation into the second:$2 \left( \frac{4S}{R} \right) = \frac{4S}{x}$$\Rightarrow \frac{2}{R} = \frac{1}{x}$$\Rightarrow x = \frac{R}{2}$.The volume $V$ of a spherical bubble is given by $V = \frac{4}{3} \pi R^3$.Therefore,the ratio of their volumes is:$\frac{V_1}{V_2} = \frac{\frac{4}{3} \pi R_1^3}{\frac{4}{3} \pi R_2^3} = \left( \frac{R_1}{R_2} \right)^3 = \left( \frac{R}{R/2} \right)^3 = (2)^3 = 8$.Wait,the question asks for the ratio of the first to the second. If $P_1 = P$ and $P_2 = 2P$,then $R_1 = R$ and $R_2 = R/2$. The ratio of volumes $V_1 : V_2 = R_1^3 : R_2^3 = R^3 : (R/2)^3 = 1 : 1/8 = 8 : 1$. However,if the question implies the ratio of the smaller to the larger,it is $1:8$. Given the options,the ratio is $1:8$.

The excess pressure in a soap bubble is double that in another one. The ratio of their volumes is .............

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