Question

A glass rod of radius $1mm$ is inserted symmetrically into a glass capillary tube with inside radius $2mm$. Then the whole arrangement is brought in contact of the surface of water. Surface tension of water is $7\times {10}^{-2}N/m$. To what height will the water rise in the capillary?$(\theta =0^{\circ })$

Answer 1

Since Force = $T(2\pi r_1+2\pi r_2)$

which is equal to $\varrho gh\times \pi (r_2^2-r_1^2)$

$T2\pi (r_1+r_2)=\varrho gh\pi (r_2^2-r_1^2)$

$7\times {10}^{-2}\times 2\left(3mm\right)=1000\frac{kg}{m^3}\times 9.8\frac{m}{s^2}\times h\left(3mm\right)$

$h=\frac{14\times {10}^{-2}}{9.8\times {10}^3\times {10}^{-3}}m=1.42cm$