A student writes the equation for the capillary rise of a liquid in a tube as `h=(rhog)/(2Tcos theta)`, where r is the radius of the capillary tube, `rho` is the density of liquid, T is the surface tension and `theta` is the angle of contact. Check the correctness of the equation using dimensional analysis?

Explanation

To check the correctness of the equation for capillary rise \( h = \frac{\rho g r}{2T \cos \theta} \) using dimensional analysis, we will analyze the dimensions of both sides of the equation.

### Step 1: Identify the Left-Hand Side (LHS)
The left-hand side of the equation is \( h \), which represents the capillary height. The dimension of height is:
\[ [L] \quad \text{(length)} \]

### Step 2: Identify the Right-Hand Side (RHS)
The right-hand side of the equation is \( \frac{\rho g r}{2T \cos \theta} \). We will break this down into its components.

1. **Density (\( \rho \))**: The dimension of density is mass per unit volume:
\[ [\rho] = [M L^{-3}] \]

2. **Acceleration due to gravity (\( g \))**: The dimension of acceleration is:
\[ [g] = [L T^{-2}] \]

3. **Radius (\( r \))**: The dimension of radius is:
\[ [r] = [L] \]

4. **Surface tension (\( T \))**: Surface tension is defined as force per unit length. The dimension of force is:
\[ [F] = [M L T^{-2}] \]
Therefore, the dimension of surface tension is:
\[ [T] = \frac{[F]}{[L]} = \frac{[M L T^{-2}]}{[L]} = [M T^{-2}] \]

5. **Cosine of angle (\( \cos \theta \))**: The cosine of an angle is dimensionless:
\[ [\cos \theta] = [1] \]

### Step 3: Combine the Dimensions on the RHS
Now, we can combine the dimensions on the right-hand side:
\[ \text{RHS} = \frac{\rho g r}{2T \cos \theta} \]
Substituting the dimensions we found:
\[ = \frac{[M L^{-3}] [L T^{-2}] [L]}{[M T^{-2}] [1]} \]

### Step 4: Simplify the RHS
Now, let's simplify the expression:
\[ = \frac{[M L^{-3} L^2 T^{-2}]}{[M T^{-2}]} \]
This simplifies to:
\[ = \frac{[M L^{-1} T^{-2}]}{[M T^{-2}]} = [L^{-1}] \]

### Step 5: Compare LHS and RHS
Now we compare the dimensions of LHS and RHS:
- LHS: \( [L] \)
- RHS: \( [L^{-1}] \)

### Conclusion
Since the dimensions of LHS and RHS are not equal, we conclude that the equation \( h = \frac{\rho g r}{2T \cos \theta} \) is **not dimensionally correct**.