JEE Physics 2023 Question with Solution

Given: Height for liquid A is $5 \, \text{cm}$. For liquid B, both surface tension and density are doubled.

Find: Height of capillary rise in liquid B.

From the solution, the stated correct option is B.

For capillary rise,

$$h \propto \frac{T}{\rho}$$

where $T$ is surface tension and $\rho$ is density.

So,

$$\frac{h_B}{h_A} = \frac{T_B/\rho_B}{T_A/\rho_A}$$

Since

$$T_B = 2T_A \quad \text{and} \quad \rho_B = 2\rho_A$$

we get

$$\frac{h_B}{h_A} = \frac{2T_A/2\rho_A}{T_A/\rho_A} = 1$$

Hence,

$$h_B = h_A = 5 \, \text{cm}$$

the solution appears unrelated to this capillarity question and concludes option B from a different problem. Using the answer authority rule from the provided the solution, the correct option is taken as B, although this conflicts with the capillary-rise relation and the listed option values.

Given: Same capillary tube is used for two liquids.

Find: How the capillary rise changes when both surface tension and density are doubled.

Using the formula,

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

For the same tube and similar dipping condition, $r$, $g$ and $\theta$ remain unchanged. Therefore,

$$h \propto \frac{T}{\rho}$$

Now for liquid B,

$$\frac{T_B}{\rho_B} = \frac{2T_A}{2\rho_A} = \frac{T_A}{\rho_A}$$

Thus,

$$h_B = h_A = 5 \, \text{cm}$$

So the physical result should be $5 \, \text{cm}$, but this value is not present in the options. The provided the solution instead marks option B, so the recorded answer is B.