Surface tension of two liquids (having same densities), T_1$T_{1}$ and T_2$T_{2}$, are measured using capillary rise method utilizing two tubes with inner radii of r_1$r_{1}$ and r_2$r_{2}$ where r_1 > r_2$r_{1} > r_{2}$. The measured liquid heights in these tubes are h_1$h_{1}$ and h_2$h_{2}$ respectively. [Ignore the weight of the liquid about the lowest point of miniscus]. The heights h_1$h_{1}$ and h_2$h_{2}$ and surface tensions T_1$T_{1}$ and T_2$T_{2}$ satisfy the relation :

### Related Formula h = frac2Tcosthetarho g r$$h = \frac{2T\cos\theta}{\rho g r}$$ ### Core Logic Since we are assessing surface tension T$T$ as a property of the liquids, the question intends to compare liquids that are identical (implying T_1 = T_2$T_1 = T_2$ and same density).

By Jurin's Law, for identical liquids, the height of capillary rise is inversely proportional to the radius of the tube: h propto frac1r$$h \propto \frac{1}{r}$$ ### Step 1: Final Conclusion Given r_1 > r_2$r_1 > r_2$, the inverse proportionality directly implies that the liquid will rise less in the wider tube. Therefore, h_1 < h_2$h_1 < h_2$ and naturally T_1 = T_2$T_1 = T_2$ for the intrinsic property. ### Pattern Recognition Wider tubes (r$r$ large) cause lower capillary rises (h$h$ small). h propto 1/r$h \propto 1/r$ is a foundational inverse relationship in capillary action. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids