JEE Physics 2026 Question with Solution

Given: Two liquids have the same density. Capillary tubes have inner radii $r_1$ and $r_2$ with $r_1 > r_2$. The corresponding rise heights are $h_1$ and $h_2$.

Find: The correct relation among $h_1, h_2, T_1, T_2$.

Using Jurin's law, the capillary rise is

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

So for the same liquid conditions, or when comparing under the same contact conditions and same density,

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

Hence,

$$h_1 r_1 = h_2 r_2$$

Given $r_1 > r_2$, the height must satisfy

$$h_1 < h_2$$

The solution concludes that the surface tensions are equal, so

$$T_1 = T_2$$

Therefore, the correct option is C.

Given: Capillary rise is measured in two tubes of radii $r_1$ and $r_2$ where $r_1 > r_2$.

Find: Whether $h_1$ is greater or smaller than $h_2$, and the relation between $T_1$ and $T_2$.

From Jurin's law,

$$h \propto \frac{1}{r}$$

Therefore, a smaller radius gives a larger rise.

Since $r_2 < r_1$, tube $2$ is narrower, so the liquid rises more in tube $2$. Thus,

$$h_2 > h_1$$

or equivalently,

$$h_1 < h_2$$

The extracted solution states that the comparison corresponds to equal surface tensions, hence

$$T_1 = T_2$$

So the required relation is $h_1 < h_2$ and $T_1 = T_2$, which is option C.