Q. A ring is cut from a platinum tube 8.5 cm internal diameter and 8.7 cm external diameter. It is supported horizontally from a pan of a balance so, that it comes in contact with the water is in glass vessel. If an extra 3.47 g-wt is required to pull it away from water, surface tension of water is
A) $72.07\ dyne\ cm^{- 1}$
B) $70.80\ dyne\ cm^{- 1}$
C) $65.35\ dyne\ cm^{- 1}$
D) $60.00\ dyne\ cm^{- 1}$
Explanation
(a) Force on the ring due to surface tension of water
$= \left( \pi D_{1} + \pi D_{2} \right)S = mg$
So $S = \frac{mg}{\pi(D_{1} + D_{2})} = \frac{3.47 \times 980}{(22/7) \times (8.5 + 8.7)}$
$= 72.07\ dyne\ cm^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.