question_answer

A ring is cut form a platinum tube 8.5 cm internal and 8.7 cm external diameter. It is supported horizontally from a pan of balance so that it comes in contact with the water in a glass vessel. What is the surface tension of water if an extra 3.97 g weight is required to pull it away from water? \[\left( g=980cm/{{s}^{2}} \right)\]

A) 71.23 dyne/cm  

B) 72.13 dyne/cm

C) 31.27 dyne/cm  

D) 21.37 dyne/cm

Correct Answer: B

Solution :

The ring is in contact with water along its inner and outer circumference so, when pulled out the total force on it due to surface tension will be

\[F=T\left( 2p{{r}_{1}}+2p{{r}_{2}} \right)\]

\[\Rightarrow T=\frac{mg}{2\pi \left( {{r}_{1}}+{{r}_{2}} \right)}\left[ \because F=mg \right]\]

i.e. \[T=\frac{3.97\times 980}{3.14\times \left( 8.5+8.7 \right)}\]

\[=72.13dyne/cm\]