Question

Water rises in a capillary tube upto a height of $10cm$ whereas mercury depresses in it by $3.42cm$. If the angle of contact and density of mercury are ${135}^{\circ }$ and $13.6gm/cc$ respectively, then the ratio of the surface tension of water and mercury will be nearly:

• A. $5:16$
• B. $16:5$
• C. $2:13$
• D. $13:2$

Answer 1

Answer: (C)

$2:13$

We know that the rise or depression in capillary tube is given by :

$h=\frac{2Tcos\theta }{r\rho g}$

Thus, $h_{water}=\frac{2T_{water}}{{\rho }_{water}gr}$

And $h_{Hg}=\frac{2T_{Hg}cos{45}^{\circ }}{{\rho }_{Hg}gr}$

$⟹\frac{T_{water}}{T_{Hg}}=\frac{h_{water}}{h_{Hg}}\times \frac{{\rho }_{water}}{{\rho }_{Hg}}\times cos{45}^{\circ }$
=$\frac{10}{3.42}\times \frac{1}{13.6}\times \frac{1}{\sqrt{2}}$ = $\frac{2}{13}$