A film of water is formed between two straight parallel wires of length 10 cm each separated by 0.5 cm. If their separation is increased by 1 mm while still maintaining their parallelism, how much work will have to be done (Surface tension of water = 7 .2 x 10-2 N/m)
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
7.22×10-6 Joule
b
1.44×10-5 Joule
c
2.88×10-5 Joule
d
5.76×10-5 Joule
answer is B.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Increment in area of soap film = A2-A12×[(10×0.6)-(10×0.5)]×10-4 =2×10-4 m2Work done = T×∆A= 7.2×10-2×2×10-4 = 1.44×10-5 J