A ring is cut from a platinum tube having 10 cm internal and 11 cm external diameter. It is supported horizontally from a pan of balance. so that it comes in contact with water in a glass vessel. What is the surface tension of water if an extra, 4.752gram weight is required to pull it away from water ? $\left(\mathrm{g}=10 \mathrm{m}/{\mathrm{s}}^2\right) :$

Given,

$\mathrm{Internal} \mathrm{diameter} = 10\mathrm{cm} , \mathrm{External} \mathrm{diameter} = 11\mathrm{cm}$

$\mathrm{extra} \mathrm{mass} = 4.752\mathrm{gmwt}$

We  know that the internal and external circumferences of a platinum tube's surface tension force equal its weight.

$\mathrm{T\pi }(\mathrm{internal} \mathrm{diameter}+\mathrm{External} \mathrm{diameter})=\mathrm{mg}$

$\Rightarrow \mathrm{T\pi }(10+11)=4.752\times 1000\mathrm{cm}/{\mathrm{s}}^2$

$\Rightarrow \mathrm{T}=\frac{4752\times {10}^3}{3.14\times 21}=72.06\mathrm{dyne}/cm=72\mathrm{N}/\mathrm{m}$

Hence the correct answer is 72N/m.