Question

A straight piece of wire $5\text{cm}$ long is placed horizontally on the surface of water and is gently pulled up. It is found that a force of $728\text{dyne}$ in addition to the weight of the wire is required for the purpose. What is the surface tension of the water?

• A. $0.72\text{dyne/cm}$
• B. $72.8\text{dyne/cm}$
• C. $7.28\text{N/m}$
• D. $75\text{dyne/cm}$
  

Answer 1

The correct option is B $72.8\text{dyne/cm}$


Wire will be attracted from both sides along its length by the water molecules.
Length of wire, $l=5\text{cm}$
$\therefore$ Length of wire on which surface tension is acting, $L=2\times l=2\times 5\text{cm}=10\text{cm}$
The extra force $728\text{dyne}$ is required to counter the force due to surface tension, on the wire.
$\Rightarrow$Force $\left(F\right)$ experienced due to surface tension $\left(T\right)$ is expressed as,
$F=T\times \left(2l\right)$
$728\text{dyne}=T\text{dyne/cm}\times 10\text{cm}$
$\therefore T=72.8\text{dyne/cm}$