Question

The work required by a big drop of water of radius R to break up into 8000 tiny drops of equal volume of water is 5.582 KR Joule. Find out the surface tension of water.

Answer 1

∵ Volume of big drop = 8000 x Volume of small drops

Given: ∆W = 5.582πR² 

^{\(\therefore\ T = \frac{\Delta W}{\Delta A} = \frac{5.582\pi R^2}{4\pi \times19 R^2}\)}

 ^{\( = \frac{5.582}{4\times19} = 0.0734\)}

T = 7.3 x 10^-2 N/m