Answer:
Using \[p=\frac{2T}{r}\]
(excess pressure in a drop)
\[r=\frac{2T}{p}=\,\frac{2\times
\,7.28\,\times \,{{10}^{-2}}}{2.33\,\times \,{{10}^{3}}}\]
\[=6.25\times
{{10}^{-5}}\,m\]
You need to login to perform this action.
You will be redirected in
3 sec