Question

If $r=5\times {10}^{-4}m\rho ={10}^{3}Kg{m}^{-3},g=10m/s^2,T=0.11N{m}^{-1}$ the radius of the drop when it detached from the dropper is approximately:

• A. $1.4\times {10}^{-3}m$
• B. $3.3\times {10}^{-3}m$
• C. $2.0\times {10}^{-3}m$
• D. $4.1\times {10}^{-3}m$

Answer 1

The correct option is A $1.4\times {10}^{-3}m$

$mg={\int }_o^{2\pi r}T\sin \theta ds$

$m=$ Mass of bubble that is getting detached.

$T=$ Surface tension.

$\therefore \rho \times \frac{4}{3}\pi R^2\times g=T\times 2\pi r\times \frac{r}{R}$

$\frac{2}{3}\times \rho R^{3}g=T\frac{r^2}{R}$

$\Rightarrow R^4=\frac{3T.r^2}{2\rho g}$

$=\frac{3\times 0.11\times (5\times {10}^{-4}{)}^2}{2\times {10}^3\times 10}$

$=\frac{3\times 0.11\times 25\times {10}^{-8}}{2\times {10}^4}$

$R=1.4\times {10}^{-3}m$