When a long glass capillary tube of radius $0.015 \mathrm{~cm}$ is dipped in a liquid, the liquid rises to a height of $15 \mathrm{~cm}$ within it. If the contact angle between the liquid and glass to close to $0^{\circ}$, the surface tension of the liquid, in millinewton $\mathrm{m}^{-1}$, is $\left[\rho_{\text {(liquid) }}=900 \mathrm{kgm}^{-3}\right.$, $g=10 \mathrm{~ms}^{-2}$ ] (Give answer in closest integer)________
(101)
Given : Radius of capillary tube,
$r=0.015 \mathrm{~cm}=15 \times 10^{-5} \mathrm{~mm}$
$h=15 \mathrm{~cm}=15 \times 10^{-2} \mathrm{~mm}$
Using, $h=\frac{2 T \cos \theta}{\rho g r}$ $\left[\cos \theta=\cos 0^{\circ}=1\right]$
Surface tension,
$T=\frac{r h \rho g}{2}=\frac{15 \times 10^{-5} \times 15 \times 10^{-2} \times 900 \times 10}{2}=101 \mathrm{milli}$
newton $\mathrm{m}^{-1}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.