Given as : `h=(2 S cos theta)/(r rhog)` where S is the surface tension of liquid, r is the radius of capillary tube. `rho` is the density and g is acceleration due to gravity then dimensional formula for S is:

Options

A. `[ML^(0)T^(-2)]`
B. `[M^(0)LT^(-2)]`
C. `[ML^(2)T^(-2)]`
D. `[M^(0)L^(0)T^(-3)]`

Explanation

To find the dimensional formula for surface tension \( S \) given the equation:

\[ h = \frac{2 S \cos \theta}{r \rho g} \]

we can follow these steps:

### Step 1: Rearranging the Equation
From the equation, we can isolate \( S \):

\[ S = \frac{h r \rho g}{2 \cos \theta} \]

Since \( \cos \theta \) and the constant 2 are dimensionless, we can ignore them for dimensional analysis:

\[ S = h \cdot r \cdot \rho \cdot g \]

### Step 2: Identifying Dimensions
Now, we need to identify the dimensions of each variable:

- **Dimension of \( h \)**: Height is a length, so its dimension is \( [h] = L \).
- **Dimension of \( r \)**: The radius is also a length, so its dimension is \( [r] = L \).
- **Dimension of \( \rho \)**: Density is mass per unit volume, so its dimension is \( [\rho] = M L^{-3} \).
- **Dimension of \( g \)**: Acceleration due to gravity is acceleration, which has dimensions \( [g] = L T^{-2} \).

### Step 3: Combining Dimensions
Now, we can substitute these dimensions back into the equation for \( S \):

\[ [S] = [h] \cdot [r] \cdot [\rho] \cdot [g] \]

Substituting the dimensions we found:

\[ [S] = (L) \cdot (L) \cdot (M L^{-3}) \cdot (L T^{-2}) \]

### Step 4: Simplifying the Expression
Now, we simplify the expression:

\[ [S] = L^2 \cdot M L^{-3} \cdot L T^{-2} \]

Combining the powers of \( L \):

\[ [S] = M \cdot L^{2 - 3 + 1} \cdot T^{-2} = M L^0 T^{-2} \]

Thus, the dimensional formula for surface tension \( S \) is:

\[ [S] = M L^0 T^{-2} \]

### Final Answer
The dimensional formula for surface tension \( S \) is \( M L^0 T^{-2} \).

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