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(B) Let $r_{1} = 3 \, cm$ and $r_{2} = 4 \, cm$ be the radii of the two soap bubbles. Let $r$ be the radius of the new bubble formed after coalescence

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$5$

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(B) Let $r_{1} = 3 \, cm$ and $r_{2} = 4 \, cm$ be the radii of the two soap bubbles. Let $r$ be the radius of the new bubble formed after coalescence.Since the process is isothermal,the number of moles of air remains constant. The pressure inside a soap bubble of radius $R$ is given by $P = P_{atm} + \frac{4T}{R}$. In a vacuum,$P_{atm} = 0$,so $P = \frac{4T}{R}$.Using the ideal gas law $PV = nRT$,for a constant temperature,$PV$ is constant.For the two bubbles: $P_{1}V_{1} + P_{2}V_{2} = PV$.Substituting $P = \frac{4T}{R}$ and $V = \frac{4}{3}\pi R^{3}$:$(\frac{4T}{r_{1}})(\frac{4}{3}\pi r_{1}^{3}) + (\frac{4T}{r_{2}})(\frac{4}{3}\pi r_{2}^{3}) = (\frac{4T}{r})(\frac{4}{3}\pi r^{3})$.This simplifies to $r_{1}^{2} + r_{2}^{2} = r^{2}$.Substituting the values: $3^{2} + 4^{2} = r^{2} \Rightarrow 9 + 16 = r^{2} \Rightarrow r^{2} = 25$.Thus,$r = 5 \, cm$.

$A$ soap bubble in vacuum has a radius $3 \, cm$ and another soap bubble in vacuum has radius $4 \, cm$. If two bubbles coalesce under isothermal conditions,then the radius of the new bubble will be .............. $cm$.

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