QJEEM24Ph275b: JEE Main 2024 Physics - Class 11 JEEM24Ph275b Solution: The excess pressure inside a soap bubble is thrice the excess…

To find the ratio between the volumes of the first and the second soap bubble, we need to understand the relation between the excess pressure inside a soap bubble and its volume.

The excess pressure ( $P$ ) inside a soap bubble is given by the formula:

$P = \frac{4T}{r}$

where $T$ is the surface tension of the soap solution, and $r$ is the radius of the soap bubble. For a soap bubble, the factor of 4 comes from having two surfaces (inner and outer), each contributing $2T/r$ to the pressure.

Given that the excess pressure inside the first soap bubble ( $P_1$ ) is thrice the excess pressure inside the second soap bubble ( $P_2$ ), we can write:

$P_1 = 3P_2$

Substituting the formula for excess pressure, we get:

$\frac{4T}{r_1} = 3 \times \frac{4T}{r_2} \Rightarrow \frac{1}{r_1} = 3 \times \frac{1}{r_2} \Rightarrow \frac{r_2}{r_1} = 3$

Next, we calculate the ratio between their volumes. The volume ( $V$ ) of a sphere (or a bubble) is given by:

$V = \frac{4}{3}\pi r^3$

Thus, the volume ratio of the first bubble ( $V_1$ ) to the second bubble ( $V_2$ ) is:

$\frac{V_1}{V_2} = \frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3} = \left(\frac{r_1}{r_2}\right)^3$

Since we found that $\frac{r_2}{r_1} = 3$ , it then follows that:

$\frac{V_1}{V_2} = \left(\frac{1}{3}\right)^3 = \frac{1}{27}$

Therefore, the correct option is:

Option B: $1: 27$