If the excess pressure inside a soap bubble is balanced by an oil column of height of \(2~\text{mm},\) then the surface tension of the soap solution will be:
(the radius of the soap bubble, \(r=1~\text{cm}\) and density of oil, \(d=0.8~\text{gm/cm}^3\) )
1. \(3.9~\text {N/m}\) 
2. $\mathrm{}$\(3.9\times 10^{-2}~\text{N/m}\)$\mathrm{}$
3. $\mathrm{}$\(3.9\times 10^{-3}~\text{N/m}\)$\mathrm{}$
4. \(3.9​​​​~\text{dyne/m}\) 

The energy needed to break a drop of radius \(R\) into \(n\) drops of radii \(r\) is given by:
1. \(4 πT ( nr ^2 - R ^2 )\)
2. \(\frac{4}{3} \pi \left(r^{3} n - R^{2}\right)\)
3. \(4 πT \left(R^{2} -nr^{2}\right)\)
4. \(4 πT \left(nr^{2}+R^{2} \right)\)$$