Check the correctness of the relation `h =(2 sigma cos theta)/(r^2 dg),` where h is height , `sigam` is surface tension, `theta` is angle of contact, r is radius, d is density and g in acceleration due to gravity.

Explanation

To check the correctness of the relation \( h = \frac{2 \sigma \cos \theta}{r^2 d g} \), we will analyze the dimensions of both sides of the equation. If the dimensions match, the relation is dimensionally correct; if they do not, the relation is incorrect.

### Step-by-Step Solution

1. **Identify the dimensions of each variable:**
- **Height (h)**: The dimension of height is length, denoted as \( [h] = L^1 \).
- **Surface Tension (\( \sigma \))**: Surface tension is defined as force per unit length. The dimension of force is \( [F] = M^1 L^1 T^{-2} \). Therefore, the dimension of surface tension is:
\[ [\sigma] = \frac{[F]}{[L]} = \frac{M^1 L^1 T^{-2}}{L^1} = M^1 L^0 T^{-2} \]
- **Angle of Contact (\( \theta \))**: The angle is dimensionless, so:
\[ [\theta] = M^0 L^0 T^0 \]
- **Radius (r)**: The radius is also a length, so:
\[ [r] = L^1 \]
- **Density (d)**: Density is mass per unit volume. The dimension of volume is \( L^3 \), so:
\[ [d] = \frac{M^1}{L^3} = M^1 L^{-3} \]
- **Acceleration due to Gravity (g)**: The dimension of acceleration is:
\[ [g] = L^1 T^{-2} \]

2. **Calculate the dimensions of the right-hand side (RHS)**:
The RHS of the equation is:
\[ \frac{2 \sigma \cos \theta}{r^2 d g} \]
Since \( 2 \) is a constant, it does not affect dimensions. The dimensions can be calculated as follows:
- **Cosine of angle**: \( [\cos \theta] = M^0 L^0 T^0 \)
- **Dimensions of \( r^2 \)**:
\[ [r^2] = (L^1)^2 = L^2 \]
- **Dimensions of \( d \)**:
\[ [d] = M^1 L^{-3} \]
- **Dimensions of \( g \)**:
\[ [g] = L^1 T^{-2} \]

Now substituting these into the RHS:
\[ [\text{RHS}] = \frac{[M^1 L^0 T^{-2}] \cdot [M^0 L^0 T^0]}{[L^2] \cdot [M^1 L^{-3}] \cdot [L^1 T^{-2}]} \]
Simplifying this:
\[ = \frac{M^1 L^0 T^{-2}}{L^2 \cdot M^1 L^{-3} \cdot L^1 T^{-2}} = \frac{M^1 L^0 T^{-2}}{M^1 L^{2 - 3 + 1} T^{-2}} = \frac{M^1 L^0 T^{-2}}{M^1 L^0 T^{-2}} = M^0 L^0 T^0 \]
Thus, the dimensions of the RHS are dimensionless:
\[ [\text{RHS}] = M^0 L^0 T^0 \]

3. **Compare with the left-hand side (LHS)**:
The LHS is simply \( h \), which has the dimension:
\[ [\text{LHS}] = [h] = M^0 L^1 T^0 \]

4. **Conclusion**:
Since the dimensions of the LHS \( (M^0 L^1 T^0) \) and the RHS \( (M^0 L^0 T^0) \) do not match, we conclude that the relation is not dimensionally correct.

### Final Answer:
The relation \( h = \frac{2 \sigma \cos \theta}{r^2 d g} \) is dimensionally incorrect.