Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 10 - Fluids - Problems: 69

Answer

a. $\gamma = \frac{F}{4 \pi r}$ b. $1.7 \times 10^{-2} N/m$

Work Step by Step

a. Assume that the ring’s weight is negligible. The surface tension equals the lifting force, divided by the length of surface that is pulled. As explained in the text and shown in figure 10-34b, there is a film at both the inside and the outside edge of the ring. $$\gamma = \frac{F}{2 (2 \pi r)} = \frac{F}{4 \pi r}$$ b. Putting in numbers, calculate the surface tension. $$\frac{F}{4 \pi r}=\frac{6.20\times 10^{-3}N}{4 \pi (0.029 m)}=1.7 \times 10^{-2} N/m$$
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