A glass rod of diameter `d_(1)=1.5 mm` in inserted symmetricaly into a glas capillary with inside diameter `d_(2)=2.0` mm. Then the whole arrangement is vertically oriented and bruoght in contact with the surface of water. To what height will the liquid rise in the capillary? Surface tension of water `=73xx10^(-3)N//m`

A glass rod of diameter `d_(1)=1.5 mm` in inserted symmetricaly into a

1.	A glass rod of diameter `d_(1)=1.5 mm` in inserted symmetricaly into a glas capillary with inside diameter `d_(2)=2.0` mm. Then the whole arrangement is vertically oriented and bruoght in contact with the surface of water. To what height will the liquid rise in the capillary? Surface tension of water `=73xx10^(-3)N//m`
Answer» If `R` is radius of meniscs, then `(2T)/R=hrhog` Here `R=(r_(2)-r_(1))/(costheta)` `theta` being angle of contact `r_(1)=` radius of glass rod `r_(2)=`radius of capillary, `(2Tcostheta)/(r_(2)-r_(1))=hrhog` or `h=(2Tcostheta)/((r_(1)-r_(2))rhog)` Here `r_(1)=(d_(1))/2, r_(2)=(d_(2))/2` `:. h=(4Tcostheta)/((d_(2)-d_(1))rhog)` Substituting given values and `theta~=0^@` for water glass interface we have `h=(4xx73xx10^(-3)cos0^@)/((2.0-1.5)xx10^(-3)xx10^(3)xx9.8)=60xx10^(-3)m=6cm`

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