The frequency `(n)` of vibration of a string is given as ` n = (1)/( 2 l) sqrt((T)/(m))` , where `T` is tension and `l` is the length of vibrating string , then the dimensional formula for m is

Options

A. `[M^(0) L^(1) T^(1)]`
B. `[M^(0) L^(0) T^(0)]`
C. `[M^(1) L^(-1) T^(0)]`
D. `[ML^(0) T^(0)]`

Explanation

To find the dimensional formula for mass \( m \) in the equation for the frequency of vibration of a string, we start with the given formula:

\[ n = \frac{1}{2l} \sqrt{\frac{T}{m}} \]

### Step 1: Rearranging the Equation
First, we square both sides to eliminate the square root:

\[ n^2 = \left(\frac{1}{2l}\right)^2 \cdot \frac{T}{m} \]

This simplifies to:

\[ n^2 = \frac{T}{4l^2 m} \]

### Step 2: Isolating \( m \)
Next, we rearrange the equation to solve for \( m \):

\[ m = \frac{T}{4l^2 n^2} \]

### Step 3: Finding Dimensions
Now, we need to find the dimensions of each component in the equation:

1. **Tension \( T \)**: The dimension of tension is the same as that of force. The dimensional formula for force is:

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

2. **Length \( l \)**: The dimensional formula for length is:

\[ [l] = L \]

3. **Frequency \( n \)**: The dimensional formula for frequency is the reciprocal of time. Since frequency is defined as \( n = \frac{1}{T} \) (where \( T \) is the time period), we have:

\[ [n] = T^{-1} \]

### Step 4: Substituting Dimensions
Now, substituting the dimensions back into the equation for \( m \):

\[ m = \frac{T}{4l^2 n^2} \]

Substituting the dimensions we found:

\[ [m] = \frac{M L T^{-2}}{4 \cdot L^2 \cdot (T^{-1})^2} \]

### Step 5: Simplifying the Expression
Now, simplifying the right side:

\[ [m] = \frac{M L T^{-2}}{4 \cdot L^2 \cdot T^{-2}} = \frac{M L T^{-2}}{4L^2 T^{-2}} = \frac{M}{4L} \]

Thus, the dimensional formula for mass \( m \) simplifies to:

\[ [m] = M L^{-1} \]

### Final Result
Therefore, the dimensional formula for \( m \) is:

\[ \boxed{M L^{-1}} \]