Question

In air, a charged soap bubble of radius $'R'$ breaks into $27$ small soap bubbles of equal radius $'r'$ Then the ratio of mechanical force acting per unit area of big soap bubble to that of a small soap bubble is

• A. $\frac{1}{81}$
• B. $\frac{3}{1}$
• C. $\frac{1}{3}$
• D. $\frac{9}{1}$

Answer 1

Answer: (C)

Key Idea The force per unit area is pressure Pressure inside a soap bubble of radius $R$. is given by
$P=\frac{4 T}{R}$
where, $T=$ surface tension
and $R=$ radius of the drop
Pressure inside a soap bubble, $P=\frac{4 T}{R}$.
If a bubble is break into 27 small soap bubbles then the volume of single bubble of radius $R$ and the combined volume of 27 bubbles of radius $r$ would be constant.
$27 \times$ volume of small bubbles $=$ volume of larger bubble
$\Rightarrow 27\left(\frac{4}{3} \pi r^{3}\right)=\frac{4}{3} \pi R^{3}$
$\Rightarrow 27 r^{3}=R^{3}$
$\Rightarrow r=\frac{R}{3}$...(i)
Now, the pressure inside smaller soap bubble,
$P_{\text {small }}=\frac{4 T}{r}=\frac{12 T}{R}$ (using the relation)
and similarly $P_{\text {large }}=\frac{4 t}{R}$
$\therefore $ Ratio of pressure of the smaller and larger soap bubble is given as,
$\frac{P_{larger}}{P_{small}} = \frac{4T}{R} \times\frac{R}{12T} = \frac{1}{3}$
$P_{\text{larger}} :P_{\text{small}} = 1:3$
Hence, the ratio of mechanical force acting per unit area of big soap bubble to that of a small bubble is 1: 3