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(D) For a soap bubble to float in air,the gravitational force must be equal to the buoyant force.Gravitational force = Buoyant force$g 	imes (	ext{m

Vedclass · Accepted Answer

$3.50$

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(D) For a soap bubble to float in air,the gravitational force must be equal to the buoyant force.Gravitational force = Buoyant force$g 	imes (	ext{mass of helium} + 	ext{mass of soap film}) = 	ext{Weight of air displaced by bubble}$Let $r$ be the inner radius of the soap bubble and $t$ be the thickness of the film.The volume of helium is $V_{He} = \frac{4}{3} \pi r^3$.The volume of the soap film is $V_{film} \approx 4 \pi r^2 t$ (since $t \ll r$).The equation of equilibrium is:$\frac{4}{3} \pi r^3 ho_{He} g + (4 \pi r^2 t) ho_{soap} g = \frac{4}{3} \pi r^3 ho_{air} g$Dividing by $g$ and rearranging:$4 \pi r^2 t ho_{soap} = \frac{4}{3} \pi r^3 (ho_{air} - ho_{He})$$t = \frac{r}{3} \frac{(ho_{air} - ho_{He})}{ho_{soap}}$Given $r = 1 \, cm = 10^{-2} \, m$,$ho_{air} = 1.23 \, kg \, m^{-3}$,$ho_{He} = 0.18 \, kg \, m^{-3}$,and $ho_{soap} = 1000 \, kg \, m^{-3}$.$t = \frac{10^{-2}}{3} 	imes \frac{(1.23 - 0.18)}{1000}$$t = \frac{10^{-2}}{3} 	imes \frac{1.05}{1000} = \frac{1.05 	imes 10^{-5}}{3} = 0.35 	imes 10^{-5} \, m = 3.5 	imes 10^{-6} \, m$.Since $1 \, \mu m = 10^{-6} \, m$,the thickness $t = 3.50 \, \mu m$.

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