Question

A machine is blowing spherical soap bubbles of different radii filled with helium gas. It is found that, if the bubbles have a radius smaller than $1 \,cm$, then they sink to the floor in still air. Larger bubbles float in the air. Assume that the thickness of the soap film in all bubbles is uniform and equal. Assume that the density of soap solution is same as that of water $(= 1000 \,kg\, m ^{-3})$. The density of helium inside the bubbles and air are $0.18 \,kg\, m^{-3}$ and $1.23 \,kg \,m^{-3}$, respectively. Then, the thickness of the soap film of the bubbles is (Note $1\, \mu m = 10^{-6}\, m)$

• A. $0.50\, \mu m$
• B. $1.50\, \mu m$
• C. $7.00\, \mu m$
• D. $3.50 \,\mu m$

Answer 1

Answer: (D)

For a soap bubble floating in air, Gravitational force = Buoyant force
$\Rightarrow g$ (mass of helium + mass of soap film)
= Weight of air displaced by bubble ... (i)

Let $r =$ inner radius of soap bubble and
$t =$ thickness of film.
Then, from Eq. (i), we have
$\Rightarrow \frac{4}{3} \pi r^3 \times \rho_{He} + 4\pi r^2 \times t \times \rho_{\text{soap}}$
$ = \frac{4}{3} \pi r^3 \times \rho_{air}$
Substituting values in above equation, we get
$\Rightarrow \frac{4}{3} \times \pi \times (10^{-2})^3 \times 0.18 + 4\pi \times (10^{-2})^2$
$ \times t \times 1000$
$ = \frac{4}{3} \times \pi \times (10^{-2})^3 \times 1.23$
Rearranging, we get
$\Rightarrow 4\pi (10^{-2}) \cdot t \cdot 1000 = \frac{4}{3} \pi (10^{-6})(1.80)$
$\Rightarrow (10^5)t = 0.35$
or $ t = 3.5 \times 10^{-6} m$
$= 3.50 \,\mu m$