Question

A dry clean needle of diameter 'd' and density $\rho$ when carefully placed on the surface of water remains floating. If T is the surface tension of water, then maximum value for the diameter 'd' of the needle for enabling it to float will be:

• A. $d=\sqrt{\frac{8\rho \pi }{Tg}}$
• B. $d=\sqrt{\frac{4\rho \pi }{Tg}}$
• C. $d=\sqrt{\frac{8T}{\rho \pi g}}$
• D. data incomplete

Answer 1

Answer: (C)

$d=\sqrt{\frac{8T}{\rho \pi g}}$

Let the length of pin is L

Now upward force due to surface tension =downward force due to mg

$T\times 2L=\frac{\pi d^{2}L\rho g}{4}$

So $d=\sqrt{\frac{8T}{\rho \pi g}}$