Water rises to a height of 10cm 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 13.6gm/cc and the angle of contact is 135°. The ratio of surface tensions of mercury and water is x $\times$ 10^-1

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Detailed Solution

$T= \frac{rdgh}{2 \cos θ}; Tα \frac{dh}{\cos θ} \frac{T_{1}}{T_{2}} = \frac{d_{1}}{d_{2}} \cdot \frac{h_{1}}{h_{2}} \frac{\cos 135^{\circ}}{\cos 0^{\circ}}$

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