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

Two liquids with the same densities and surface tensions $$T_1$$ and $$T_2$$ rise to heights $$h_1$$ and $$h_2$$, respectively, in capillary tubes of radii $$r_1 > r_2$$.

We first recall the capillary rise formula:

$$h = \frac{2T\cos\theta}{\rho g r}$$

Assuming that both liquids have the same contact angle (so that $$\cos\theta = 1$$) and the same density, we can rearrange to find

$$T = \frac{h\rho g r}{2}$$, which shows that $$T \propto hr$$.

Because the surface tensions are equal ($$T_1 = T_2$$), it follows that

$$h_1 r_1 = h_2 r_2$$, and since $$r_1 > r_2$$, we have $$h_1 < h_2$$.

This conclusion matches Option C, which states $$h_1 < h_2$$ and $$T_1 = T_2$$.

The correct answer is Option C.