First slide

Surface tension and surface energy

Question

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

Moderate

A

56.25 dyn/cm
B

70.80 dyn/cm
C

63.35 dyn/cm
D

60 dyn/cm

Solution

$\begin{array}{l} Total force on the ring due to surface tension is given by F = σ (2 π r_{1} + 2 π r_{2}) \\ a n d excess weight is compensated by surface tension force \\ ∴ σ (2 π r_{1} + 2 π r_{2}) = m g \\ \Rightarrow [2 π \times \frac{8.7}{2} + 2 π \times \frac{8.5}{2}] σ = 3.103 \times 980 \\ ⇒ σ = \frac{3.103 \times 980 \times 7}{22 \times 17.2} dyn / cm = 56.25 dyn / cm \end{array}$

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