(III) Estimate the diameter of a steel needle that can just barely remain on top of water due to surface tension.

For an object to remain floating on top of the water, the weight of the object acting downwards is balanced by the forces acting on the object in the upward direction.

Assuming a needle having mass \(m\), diameter \(d\) and length \(l\) is lying on top of the water. The surface tension force \(F\) is acting at an angle of \(\theta \) from the vertical axis.

The free-body diagram of the needle on top of water is given below,

The surface tension force acting on the needle is given by,

\(F = \gamma l\)

Here, \(\gamma \) is the surface tension acting on the water surface.

Balancing all the forces acting on the needle in the vertical direction,

\(\begin{array}{l}mg = 2F\cos \theta \\mg = 2\gamma l\cos \theta \end{array}\)

Since the angle \(\theta \) is very small, so \(\theta = 0^\circ \) and \(\cos \theta = \cos 0^\circ \).

The above equation becomes,

\(\begin{array}{l}mg = 2\gamma l\cos 0^\circ \\mg = 2\gamma l \times 1\\mg = 2\gamma l\end{array}\)…… (i)

Also, the expression for the mass of the needle is given by,

\(\begin{array}{l}m = \rho V\\m = \rho \times \frac{\pi }{4}{d^2}l\end{array}\)

Substitute the value of mass in the equation (i),

\(\begin{array}{c}\rho \times \frac{\pi }{4}{d^2}lg = 2\gamma l\\{d^2} = \frac{{2\gamma }}{{\rho \times \frac{\pi }{4}g}}\\d = \sqrt {\frac{{8\gamma }}{{\pi \rho g}}} \end{array}\)

Thus, the diameter of the needle to remain at the top of the water is \(\sqrt {\frac{{8\gamma }}{{\pi \rho g}}} \).