Fluid Mechanics Viscosity Question 320
Question: A water film is formed between two straight parallel wires of 10 cm length 0.5 cm apart. If the distance between wires is increased by 1 mm. What will be the work done? (surface tension of water $ =72dyne/cm $ )
Options:
A) 36 erg
B) 288 erg
C) 144 erg
D) 72 erg
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Work done = Surface tension $ \times $ increase in area of the film $ W=S\times \Delta A $ $ Increaseinarea=Finalarea-initialarea $ $ =10\times (0.5+0.1)-10\times 0.5=1cm^{2} $ $ \therefore ,,,W=72\times 2\times 1=144erg $ [$ \therefore $ There are 2 free surfaces; $ \therefore ,,\Delta A=2\times 1 $ ].
BETA
