Work Done in Blowing a Liquid Drop or Soap Bubble
Category : JEE Main & Advanced
(1) If the initial radius of liquid drop is \[{{r}_{1}}\] and final radius of liquid drop is \[{{r}_{2}}\] then
\[W=T\times \] Increment in surface area
\[W=T\times 4\pi [r_{2}^{2}-r_{1}^{2}]\] [drop has only one free surface]
(2) In case of soap bubble
\[W=T\times 8\pi [r_{2}^{2}-r_{1}^{2}]\] [Bubble has two free surfaces]
You need to login to perform this action.
You will be redirected in
3 sec