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

Step 1: Recall the formula for capillary height
The height of a liquid column in a capillary tube is given by Jurin's law:
$h = \frac{2T \cos \theta}{\rho g r}$
where:
- $h$ is the height of the liquid column
- $T$ is the surface tension of the liquid
- $\theta$ is the contact angle between the liquid and the tube
- $\rho$ is the density of the liquid
- $g$ is the acceleration due to gravity
- $r$ is the radius of the capillary tube

Step 2: Identify constant and variable parameters
For both capillary tubes P and Q, the liquid is water, so $T$, $\rho$, $g$, and $\cos \theta$ (assuming the material of the tubes is the same, leading to the same contact angle) are constant.

Step 3: Establish the relationship between height and radius/diameter
Since $T, \cos \theta, \rho, g$ are constants, we can write:
$h \propto \frac{1}{r}$
Also, diameter $d = 2r$, so $r = \frac{d}{2}$. Substituting this, we get:
$h \propto \frac{1}{d/2} \propto \frac{2}{d} \propto \frac{1}{d}$
Thus, the height of the water level is inversely proportional to the diameter of the capillary tube.
We can write this relationship as $h = \frac{C}{d}$, where $C = \frac{4T \cos \theta}{\rho g}$ is a constant.

Step 4: Apply the relationship to tubes P and Q
For tube P: $h_P = \frac{C}{d_P}$
For tube Q: $h_Q = \frac{C}{d_Q}$

Step 5: Use the given height relationship
We are given that the height of water level in capillary tube P is $\frac{2}{3}$ of the height in capillary tube Q:
$h_P = \frac{2}{3} h_Q$

Step 6: Substitute the expressions for $h_P$ and $h_Q$
Substitute the expressions from Step 4 into the equation from Step 5:
$\frac{C}{d_P} = \frac{2}{3} \left( \frac{C}{d_Q} \right)$

Step 7: Solve for the ratio of diameters
Divide both sides by $C$:
$\frac{1}{d_P} = \frac{2}{3 d_Q}$
Cross-multiply:
$3 d_Q = 2 d_P$
Rearrange to find the ratio $\frac{d_P}{d_Q}$:
$\frac{d_P}{d_Q} = \frac{3}{2}$
So, the ratio of their diameters is $3 : 2$.

Final Answer: The ratio of diameters is $3 : 2$, which matches option (B).

Answer 2

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.