[Physics] Diameter of a steel needle that can just float on water due to surface tension

forceshomework-and-exercisessurface-tension

γ water = 0.072 p steel = 7.8 x $10^3$
I solve the problem like this :

mg = surface tension force

p.V.g= γ.2πrL

p.π$r^2.L.g$ =γ.2πrL

r= (2γ)/(pg)

the result is different from my book. Where did i go wrong here?

Best Answer

The right hand side of your equation $\rho.V.g= γ.2πrL$ is incorrect.
Ignoring the ends of the needle you need the total length of the needle in contact with the water.
On the left hand side express the volume of the needle $V$ in terms of appropriate variables and you will find an equation which will enable you to find the radius $r$ of the needle, the length $L$ of the needle having cancelling out.
Hence you can evaluate the diameter of the needle.

Related Question