Question

A capilary tube of radius $0.5mm$ is dipped in a liquid vertically in a liquid of surface tension $0.04N/mm$ and relative density $0.8g/cc.$ Calculate the height of capillary rise$,$ if the angle of contact is ${10}^0$ $\left(giveng=9.8m/s^2\right)$

• A. $2.01\times {10}^{-2}m$
• B. $4.01\times {10}^{-2}m$
• C. $6.41\times {10}^{-7}m$
• D. $5\times {10}^{-5}m$

Answer 1

The correct option is A $2.01\times {10}^{-2}m$

$r=0.5mm=0.5\times {10}^3$

$=5\times {10}^{-4}m$

$T=0.04N/m$

$\rho =0.8gm/cc$

$=\frac{0.8\times {10}^{-3}}{{10}^{-6}}$

$\theta =10⁰$

$g=9.8m/s²$

$T=rh\rho g/2cos\theta$

$\therefore h=2Tcos\theta /rg\rho$

$\therefore h=2\times 0.04\cos \frac{10}{5}\times 800\times 9.8$

$h=2\times 0.04\times \frac{0.9848}{5}\times 800\times 9.8\times {10}^{-4}$ m

$h=2.01\times {10}^{-2}m$

$\therefore$ The height of the capillary rise is $2.01\times {10}^{-2}m$

Hence,

option $\left(A\right)$ is correct answer.