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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 g/c.c. and the angles of contact for mercury and water are 135° and 0° respectively, the ratio of surface tension for water and mercury is

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1 : 0.15

1 : 3

1 : 6.5

1.5 : 1

answer is C.

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Detailed Solution

Height, h = $\frac{2 T \cos θ}{rρg}$

$∴ For water, h_{w} = \frac{2 \times T_{w} \times \cos 0^{0}}{r \times 1 \times g}$ ------(i)

and, for mercury,

$h_{m} = - \frac{2 \times T_{m} \times \cos 135 °}{r \times 13.6 \times g} - - - - - - (ii)$

$∴ \frac{h_{w}}{h_{m}} = \frac{2 \times T_{w} \times 1}{r \times 1 \times g} \times \frac{r \times 13.6 \times g \times \sqrt{2}}{2 \times T_{m} \times 1}$

$[∵ \cos 135 ° = - \frac{1}{\sqrt{2}}]$

$⇒ \frac{10}{3.42} = \frac{T_{w}}{T_{m}} \times 13.6 \times \sqrt{2}$

$∴ \frac{T_{w}}{T_{m}} = \frac{10}{3.42 \times 13.6 \times 1.414} = \frac{1}{6.5}$

Therefore, $T_{w} : T_{m} = 1 : 6.5$

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