Question

The capillary rise of water in a tube of diameter $0.70mm$ is $4cm.$ If $g=10m{s}^{-2}$, the surface tension of water is $.........N{m}^{-1}$.

• A. $70\times {10}^{-3}$
• B. $70$
• C. $0.35\times {10}^{-2}$
• D. $7\times {10}^{-3}$

Answer 1

Answer: (A)

$70\times {10}^{-3}$

In the capillary rise of water, the rise of water in the capillary is given as below

$h=\frac{2T\cos \theta }{\rho gr}(taking\cos \theta =1)$

where, h is rise of water in capillary

$T$ is surface tension of water

$\rho$ is density of water

$g$ is acceleration due to gravity

$r$ is radius of capillary

So, putting values in above equation

$r=\frac{0.70}{2\times 1000}=3.5\times {10}^{-4}m$

$\frac{4}{100}=\frac{2T}{1000\times 10\times 3.5\times {10}^{-4}}$

$T=\frac{4\times 3.5\times {10}^{-4}\times 100}{2}=70\times {10}^{-3}N{m}^{-1}$

therefore $T=70X{10}^{-3}N/m$

Option A is correct.