A ring is cut from a platinum tube 8.5 cm internal and 8.7 cm external diameter. It is supported horizontally from the pan of a balance, so that it comes in contact with the water in a glass vessel. If an extra 3.103 gf is required to pull it away from water, the surface tension of water is
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
56.25 dyn/cm
b
70.80 dyn/cm
c
63.35 dyn/cm
d
60 dyn/cm
answer is A.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Total force on the ring due to surface tension is given by F=σ2πr1+2πr2and excess weight is compensated by surface tension force ∴σ2πr1+2πr2=mg⇒2π×8.72+2π×8.52σ=3.103×980⇒σ=3.103×980×722×17.2dyn/cm=56.25 dyn/cm