The excess pressure inside the first soap bub | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

Excess pressure in first soap bubble,

$\mathrm{p}_{1}=\frac{4 \mathrm{T}}{\mathrm{r}_{1}}$

excess pressure inside second bubble,

$\mathrm{p}_

Vedclass · Accepted Answer

$1 : 27$

Vedclass · Answer

Excess pressure in first soap bubble,

$\mathrm{p}_{1}=\frac{4 \mathrm{T}}{\mathrm{r}_{1}}$

excess pressure inside second bubble,

$\mathrm{p}_{2}=\frac{4 \mathrm{T}}{\mathrm{r}_{2}}$

On dividing these, we get

$\frac{p_{1}}{p_{2}}=\frac{r_{2}}{r_{1}}$

but $\mathrm{p}_{1}=3 \mathrm{p}_{2} \Rightarrow \frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\frac{1}{3}$

$\Rightarrow\left(\frac{r_{1}}{r_{2}}\right)^{3}=\frac{1}{27}$

So, ratio of their volumes is,

$\frac{{\frac{4}{3}\pi {\rm{r}}_1^3}}{{\frac{4}{3}\pi {\rm{r}}_2^3}} = \frac{{{{\rm{v}}_1}}}{{{{\rm{v}}_2}}} \Rightarrow \frac{{{{\rm{v}}_1}}}{{{{\rm{v}}_2}}} = \frac{1}{{27}}$

The excess pressure inside the first soap bubble is three times that inside the second bubble then, the ratio of volume of the first to the second bubble will be

Similar Questions

A boat carrying a number of large stones is floating in a water tank. What would happen to the water level if a few stones are unloaded into water

A sphere of mass $M$ and radius $R$ is falling in a viscous fluid. The terminal velocity attained by the falling object will be proportional to

The height of water in a tank is $H$. The range of the liquid emerging out form a hole in the wall of the tank at a depth $\frac{{3H}}{4}$ from the upper surface of water, will be