The frequency of vibration of string is given by `v=p/(2l)[F/m]^(1//2)`. Here `p` is number of segments in the string and `l` is the length. The dimensional formula for `m` will be

Options

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

Explanation

To find the dimensional formula for \( m \) in the equation \( v = \frac{p}{2l} \left( \frac{F}{m} \right)^{1/2} \), we can follow these steps:

### Step 1: Rearranging the equation
We start with the given equation:
\[ v = \frac{p}{2l} \left( \frac{F}{m} \right)^{1/2} \]
To isolate \( m \), we can rearrange the equation:
\[ \left( \frac{F}{m} \right)^{1/2} = \frac{2lv}{p} \]
Now, squaring both sides gives:
\[ \frac{F}{m} = \left( \frac{2lv}{p} \right)^2 \]

### Step 2: Expressing \( m \)
Now, we can express \( m \) in terms of \( F \), \( l \), and \( v \):
\[ m = \frac{F}{\left( \frac{2lv}{p} \right)^2} = \frac{F \cdot p^2}{4l^2v^2} \]

### Step 3: Finding the dimensional formula for \( F \), \( l \), and \( v \)
Next, we need to find the dimensional formulas for the quantities involved:
- The dimensional formula for force \( F \) is:
\[ [F] = [M][L][T^{-2}] \]
- The dimensional formula for length \( l \) is:
\[ [l] = [L] \]
- The dimensional formula for velocity \( v \) is:
\[ [v] = \frac{[L]}{[T]} = [L][T^{-1}] \]

### Step 4: Substituting the dimensional formulas
Now substituting these into the expression for \( m \):
\[ m = \frac{F \cdot p^2}{4l^2v^2} \]
Ignoring the constants (as they are dimensionless), we focus on the dimensional part:
\[ [m] = \frac{[F]}{[l^2][v^2]} = \frac{[M][L][T^{-2}]}{[L^2][L^2][T^{-2}]} = \frac{[M][L][T^{-2}]}{[L^4][T^{-2}]} = [M][L^{-3}] \]

### Final Answer
Thus, the dimensional formula for \( m \) is:
\[ [m] = [M][L^{-3}] \]