Question

Two capillary tubes P and Q are dipped vertically in water. The height of water level in capillary tube P is \(\frac{2}{3}^{rd}\) of the height in capillary tube Q. The ratio of their diameter is _______.

• A. 2 : 3
• B. 3 : 2
• C. 3 : 4
• D. 4 : 3

Answer 1

The Correct Option is B

The height of the water level in a capillary tube is given by $h = \frac{2T\cos\theta}{r\rho g}$, where:

• $T$ is the surface tension,
• $\theta$ is the contact angle,
• $r$ is the radius of the capillary tube,
• $\rho$ is the density of the liquid,
• $g$ is the acceleration due to gravity.

Let $h_P$ and $h_Q$ be the heights of water in capillary tubes P and Q, respectively, and $r_P$ and $r_Q$ be their radii. We are given that $h_P = \frac{2}{3}h_Q$.

Since $T$, $\theta$, $\rho$, and $g$ are the same for both tubes, we have:

$h_P \propto \frac{1}{r_P}$ and $h_Q \propto \frac{1}{r_Q}$

$\frac{h_P}{h_Q} = \frac{r_Q}{r_P}$

$\frac{2}{3} = \frac{r_Q}{r_P}$

Since the diameter $d = 2r$, the ratio of their diameters is the same as the ratio of their radii:

$\frac{d_P}{d_Q} = \frac{r_P}{r_Q} = \frac{3}{2}$

The correct answer is (B) 3 : 2.