Question

The excess pressure inside a soap bubble is twice the excess pressure inside a second soap bubble. The volume of the first bubble is n times the volume of the second where n is
(a) 4
(b) 2
(c) 1
(d) 0.125

Answer 1

The correct Answer is:-

(d) 0.125

Answer 2

The excess pressure inside a soap bubble is twice the excess pressure inside a second soap bubble. The volume of the first bubble is n times the volume of the second where n is 4

Explanation:

Step 1:

Suppose excess pressure within the soap bubble = P

Excess pressure inside soap bubble = 2P

Let First Bubble Radius = x

Let Second Bubble Radius = R

So, in first bubble, excess pressure = $2 P=\frac{4 S}{x}$

In second bubble, excess pressure = $P=\frac{4 S}{r}$

Thus,

$2\left(\frac{4 S}{r}\right)=\frac{4 S}{x}$

$x=\frac{r}{2}$

Step 2:

Level of initial soap bubble $(\mathrm{V} 1)=\frac{4}{3 \pi \mathrm{x}^{3}}$

Volume of last soap bubble $(\mathrm{V} 2)=\frac{4}{3 \pi \mathrm{R}^{3}}$

Step 3:

Volume Ratio

$\frac{V 1}{V 2}=\frac{\frac{4}{3 \pi x^{3}}}{\frac{4}{3 \pi R^{3}}}$

$x^{3}=n R^{3}$

$=\left(\frac{R}{2}\right)^{3}=n R^{3}$

$n=\frac{1}{8}=0.125$

The volume of the first bubble is therefore n times that of the second where n is $0.125$