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A liquid of density 830 kg. m^-3 rises t...

A liquid of density `830 kg. m^-3` rises through `0.0893 m` in a capillary tube of diameter `1.68xx10^-4` m. Determine the surface tension of the liquid. Take the angle of contact as `0^@`.

Text Solution

Verified by Experts

We know that, `T=(rh rho g)/( 2 cos theta)`
Here, `r=(1.68xx10^-4)/(2)=8.4xx10^-5 m`,
`h= 0.0893 m, rho=830 kg. m^-3 and theta = 0^@`.
`therefore T=(8.4xx10^-5xx0.0893xx830xx9.8)/(2cos0^@)=0.0305 N. m^-1`.
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