First slide

Surface tension

Question

Water rises to a height of 10 cm in a capillary tube, and mercury falls to a depth of 3 . 42 cm in the same capillary tube. If the density of mercury is 13 .6 and the angle of contact is 135^o, the ratio of surface tension for water and mercury is

Moderate

A

1:0.5
B

1:3
C

1:6.5
D

1.5:1

Solution

$T = \frac{ρrgh}{2 \cos θ} for water T_{1} = \frac{rg \times 10}{2 \cos θ} = 5 rg for mercury T_{2} = \frac{rg \times 3 \cdot 42 \times 13 \cdot 6}{2 \cos 135^{\circ}} = 6 \cdot 5 \times 5 rg \frac{T_{1}}{T_{2}} = \frac{1}{6 \cdot 5} or T_{1} : T_{2} = 1 : 6 \cdot 5$

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